symmetry math (sm) by jack kuykendall breaking the broken-symmetry (bs) math’s dash (-), cross (+)...
TRANSCRIPT
SYMMETRY MATH (SM)By Jack Kuykendall
Breaking the Broken-Symmetry (BS) Mathrsquos
Dash (-) Cross (+) Codes
Symmetry Math and Logic
Why is Symmetry Math NeededBecause the BS Rule-of-Signs a (-)(-) = (+) is illogical and produces many incorrect answers
The BS number line is a Broken-Symmetry number line
From ldquoWhat Is Mathematicsrdquo by Courant and Robbins ndash page 55The Rule of Signs (-1) (-1) = +1
Which we set up to govern the multiplication of negative integers is a consequence of our desire to preserve the distribution law
a(b+c) = ab + ac
It took a long time for mathematicians to realize that the ldquoRule-of-signsrdquo together with all the other definitions governing negative integers and fractions
They were created by us in order to attain the freedom of operation while preserving the fundamental laws of arithmetic
Cannot be ldquoPROVEDrdquo
According to Professor Philip Kanarev there has not been a major discovery in physics or chemistry by ldquopeer-reviewing-academiardquo due to THEORY since the early 1900rsquos
There have been virtually no science-changing discoveries since the theories of quantum mechanics and relativity were introduced
One of the reasons is that the math being used is BS math and it operates on Broken-Symmetry and can only provide usable answers
BS math cannot be used to describe reality in space or time Space is symmetrical Space cannot depend on the direction one orientates a coordinate system
BS mathematicians should have discovered the problem Instead they invented symbols and definitions and bypassed the real problem ie [(imaginary numbers i2 = -1) and (absolutely values |-X|=+X)]
The Mathematical Principle of Error Joseph A Rybczyk 1996
Once an error enters a calculation all calculations after that point become an extension of the error
Studying the dash-cross codes of BS math has allowed me to break the codes and understand the reasons for the errors
Dash sign (-) CODES In Math used as
22
1 15 004255
bull (-) the number of zeros to the right of a decimal point 10-5=000001
bull (-) a subtraction operator
bull (-) a direction in space labeled negative (whatever that means)
bull (-) an exponent to mean divide
bull (-) describe the negative half of a sin or cos graph
bull (-) describe the negative half of an ldquoerdquo and ldquo1erdquo graph
bull (-) describe the negative side of all numbers raised to a power x
In Physics
bull (-) An Electron has been labeled as negative
bull (-) Anti-Particles have been labeled as negative
In Chemistry
bull (-) amp (+) Thermochemistry
Cross sign (+) CODES
(+) a direction in space labeled positive (whatever that means)
(+) an addition operator
(+) used to show the number of zeros to the left of a decimal point 10+5 = 100000
In Physics and Chemistry
bull (+) Protons have been Labeled as positive
bull (+) numerous other particles have been Labeled as positive
bull (+) amp (-) Thermochemistry
In Math used
The same dash symbol (-) is used for the numerous different math operations
The definitions established for the use of the dash symbol (-) do not distinguish between their different operations
In many math operations the dash symbol is changed to mean one of the other meanings
It is amazing that math has proceeded to its current level of use with this illogical use of a symbol
The same cross symbol (+) is used for numerous different math operations and produces the same illogical answers
A Summary of the BS Broken-Symmetry math
1 1 cosh tanhln
x xe x xe
bull Originated with broken-symmetry in the XYZ plane
bull Originated with mirror-broken-symmetry in the XZ plane
bull Created a Rule-of-Signs that cannot be proved (-)(-)=(+) (+)(+)=(+) (-)(+)=(-)
bull Created imaginary numbers to compensate for broken-symmetry
bull Created absolute values to change illogical negative number answers into positive number answers
bull Produces broken symmetry graphs for many functions
It took four years of study to discover why a (-)(-)=(+) only works in an imaginary world where space in the dash (-) direction is different from space in the cross (+) direction
Started working on the problem in August of 2001
Data points created using BS math produce graphs that are not symmetrical dash side different from cross side
However if a real problems data points follow a non symmetrical graph the graph can provide usable answers
This is why no one discovered the problem
Solved in January of 2005
Symmetry-Math(SM) vs Broken-Symmetry(BS) Math
In SM if we specify that the dash sign (-) means only subtraction and the cross sign (+) means only addition and an appropriate symbol is used for a direction in space space becomes symmetrical and math becomes logical
Math operators and directions in space are not the same and the same symbol should not be used to represent them
Letrsquos start with the (dash)multiplied by a (dash)=(cross) in the BS system
Usable answers may be obtained but it will be for illogical reasons
multiplied by a ( ) ( ) ( )subtraction Subtraction addition
rightleft left
dash dash cross
positivenegative negative
Symmetry-Math The subtraction of a direction is equal to the opposite direction The answers are correct using correct logic
opposite direction direction
SM
(negative)(negative) = (positive)
( ) ( ) ( )subtraction
BS
There is NO multiplication of a subtraction operator by a direction in space
There is just the subtraction of a direction in space
If instead of labeling the left side of a coordinate system as a negative (-) the same as a subtraction operator we label it with an arrow ( ) to represent the direction then a subtraction from that arrow direction will be in the opposite direction ( )
( ) mutiplied by a ( ) ( )
negative Positive negative
subtraction addition subtraction
dash cross dash
left right left
A subtraction operator multiplied by an addition operator is equal to a subtraction operator This is illogical
BS Math
A negative multiplied by a positive is equal to a negative This is illogical
A direction to the left multiplied by a direction to the right is a direction to the left This is illogical and Einsteins math error in special relativity
A dash times a cross is equal to a dash This is illogical
A vector moving to the left multiplied by a vector moving to the right is equal to ONLY to a vector moving to the left that is eight orders of magnitude faster thanThe speed of light
opposite direction subtraction direction
SM ( ) ( )
BS (negative)(positive) = (negative)
Symmetry Math The subtraction of a direction is equal to the opposite direction
There is no multiplication of a subtractionoperator times a direction in space
( ) multiplied by a ( ) ( )
positive positive positiveaddition addition additioncross cross cross
direction direction direction
What is the meaning of multiplying addition operators Addition operators are not multiplication operators
BS Math
() multiplied by a direction ( )=
number
() multiplied by a direction ( )=
number
A number times a direction maintains the same direction
( ) ()
dirdiradd
( ) ()
dirdiradd
The addition of a direction is in the same direction There is no multiplication by addition operators
Symmetry Math
Future math books need to eliminate the use of the cross sign (+) to represent something labeled a positive direction in space and the dash sign (-) to represent something labeled a negative direction in space Space does not have positive and negative directions
Symbols that are logical and have no illogical representation should be adopted SM uses either arrows or symbols
Symmetry-Math Number Line
Symmetry-Math Rules
The dash sign (-) will have only one use subtraction
The cross sign (+) will have only one use addition
Directions have an arrow and a number
All observers see the same direction and magnitude
bull There is no multiplication of arrows
amp ampamp ampampamp
( 3 ) ( 5 ) ( 6 ) ( 65 ) (17256 ) ( 32993 )
dir dirdir dirdirdir
Negative and Positive Directions in Space
If I asked you to point to a negative direction in space which way would you point
Hopefully you will realize that there is no such thing as a negative direction in space
If I asked you to point to a positive direction in space which way would you point
Again hopefully you will realize that there is no such thing as a positive direction in space
Objects on the right side of the x-axis are positive (+)(+) = (+)
Objects on the left side of the x-axis are negative (-)(-) = (+) negative math is different from positive math This is illogical
Objects on the top of the y-axis are positive
Objects on the bottom of the y-axis are negative
Objects in the front of the z-axis are positive
Objects in the back of the z-axis are negative
Again BS math of the positive direction (x right y up and z front) is different from the BS math of the negative direction (x left y down and z back)
BS math is Broken-Symmetry
Mirror image is broken symmetry in the x and z axis If the left amp right-axis are
reversed symmetry is broken Obs-1 math is different from obs-2
If the front amp back-axis is reversed symmetry is broken Obs-1 math is different from obs-2
If the top amp bottom-axis are
reversed symmetry is not broken Obs-1 math is the same as obs-2
Math answers cannot depend on which side of a number line an observer sit
Absolute values for Displacement
Even for a simple displacement of an object BS math invented a definition and absolute values to obtain something labeled a positive number answer
I----------I----------I----------I----------I----------I----------I----------I----------I
-4 -3 -2 -1 0 +1 +2 +3 +4 a gtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtb cltltltltltltltltltltltltltltltltltltltltltltltltltltltb
If an object starts at ( 3)ia X and moves to ( X 3) and the to ( 1)b fb c X
BS Math provides a usable answer without a definition or absolute values
However If an object starts at
( 1) and moves to ( X 3) and the to ( 3)i b fa X b c X
Then absolute values and a definition must be used
---------I----------I----------I----------I----------I----------I----------I---------- -4 -3 -2 -1 0 +1 +2 +3 +4
a gtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtb cltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltb
BS Math makes up a definition that all displacements are positive Since this example provides an answer with something labeled a negative the positive definition must be applied and (-2) is changed to (+2) A definition is needed to arrive at a useable answer This is Illogical
In SM absolute values and a meaningless definition of ldquopositiverdquo are removed In SM objects and directions in space are defined by the direction and magnitude of the resultant of arrows
With SM you get total distance traveled by the object and the final direction of the displacement
22222 2 bababbaababababa If w e su b stitu te p o s itiv e n u m b e rs in to th e eq u a tio n th e an sw er w ill b e co rrec t If (a = 5 ) an d (b = 3 )
2
2
8
5 3 5 3 5 3 5 5 5 3 3 5 3 3 25 15 15 9 64
8 8 8 8 8 8 8 8 8 8 64
W h en tw o p o sitiv e n u m b ers a re a d d ed an d sq u a re d th e B S m a th d is tr ib u tiv e law p ro v id es a co rre c t an sw er 1
Illogical and Incorrect BS math for the distributive law
222 2 bababbabbaaabababa B y d e fin itio n in B S m a th n o n u m b er sq u a red c a n ev e r b e n eg a tiv e T h e re fo re n o sq u a red n u m b er c an ev e r b e a v e c to r m o v in g in th e d a sh ( -) o r n eg a tiv e d ire c tio n if it is sq u a red In B S m a th an d ad d itio n o p e ra to r m u ltip lied b y a su b trac tio n o p e ra to r is A L W A Y S eq u a l to a su b tra c tio n o p e ra to r If th is h ad n o t b e en u sed fo r th e la s t fo u r h u n d red y e a rs y o u w o u ld d ie lau g h in g a t th e u tte r ab su rd ity o f th e la s t s ta tem e n t T h is is illo g ica l an d v io la te s sy m m etry
If a = 5 an d b = 3 42222352
22
4930253)3)(5(253)5)(3()3)(5(5353535 22222 A s lo n g a s ldquo ardquo is g re a te r th an ldquob rdquo th e B S m a th d is tr ib u tiv e law p ro v id es co rrec t an sw ers 1
B S m a th b re a k s d o w n w h e n ldquo b rdquo is g rea ter th a n ldquo a rdquo
If a = 3 an d b = 5
2 2
2
2
3 5 2 4
2 ( 2) ( 2) 4
2 ( )( 2) ( )( 2) 4
A n y co m b in a tio n ex cep t th e m id d le o n e is illo g ica l an d in co rre c t B y d e fin itio n th e an sw er can n o t b e -4 in B S m a th B S m a th is in co rrec t
4253095)5)(3(235)3)(5()5)(3(3535353 22222 T h is is w h ere th e d is tr ib u tiv e la w p ro v id es in co rrec t an sw ers I t p ro d u ces an an sw er o f + 4 b e cau se b y d e fin iti o n n o n u m b er sq u a red in B S m a th c an b e a n eg a tiv e n u m b er U sin g in co rre c t B S m a th th e d is tr ib u tiv e law w ill g iv e an in co rrec t an sw ers o f + 4 T h is is in co rrec t an d illo g ica l 1
Since there can be no negative direction from a squared term in BS math BS math should be modified or abandoned
The BS distributive law should be modified or abandoned
The numbers +1+2+3+4 and ndash1-2-3-4hellip should be abandoned
BS math gives incorrect answers because their Rule-of-Signs allows multiplying an arrow going in one direction by an arrow going in another direction
Clearly this is not logical
Arrows should be used that are specific for directions
W h at d o es it m e an to m u ltip ly ( 4) b y (4 ) A s w e k n o w 1
m u ltip lica tio n is ju s t ad d itio n W h a t w o u ld ( 4) tim e (4 ) b e 2
In BS math the answer is by definition -16 unless you square -4 and then it is +16
In SM you cannot multiply arrows going in opposite directions
The middle two terms are not logical You cannot multiply opposite directions
This is a MAJOR error in BS math They multiply a dash [(-) a direction to the left] by a cross [(+) a direction to the right] And then ldquoby definitionrdquo label the answer dash (-) This is illogical and produces incorrect answers
2
2 2
This is Einsteinrsquos math error
in Special relativity
Magnitude 4 Direction 1 Magnitude 3 Direction 2 Magnitude 5 Direction 3 Relative to an observer at 0 4
Magnitude 4 Direction 1 Magnitude 3 Direction 2 Magnitude 5 Direction 3 Relative to an observer at 0 4
Magnitude and Direction (MD) An arrow has both a magnitude andA direction relative to a 2nd objectobserver
bull Magnitude and velocity (MV) relative to a 2nd objectobserver
bull Magnitude And acceleration (MA) relative to a 2nd objectobserver2
v da
t t
Introducing a new symbol the ldquoamprdquo sign The ldquoamprdquo sign means resultant ADD all the Arrows for each direction and then SUBTRACT to find the final resultant
1 1amp 1 2 2amp 2 4 If =
1 amp1 2 2 amp2 4 If =
amp 5 amp 3 2 If gt
amp 3 amp 5= 2 If gt
amp 0 3 amp 3=0 if =
only the resultant equals zero
the vectors do not - they are still there
Does not exist There is no multiplication of arrows
There are two arrows One has a magnitude of 3 One has a magnitude of 5 The maximum these two arrows can be is 8
Only the resultant of arrows exist
An SM arrow has a direction and a magnitude
Arrows and magnitudes are still there even if the resultant is zero
There are no negative SM Arrows
There are no positive SM Arrows
There are no ZERO SM Arrows A zero would negate the definition of an SM Arrow something with no magnitude and no direction cannot be the definition of an arrow that is defined as having magnitude and direction
bull A resultant is the addition and subtraction of the differences in magnitude and direction arrows
A B and C are the arrows (a) A amp B exist(b) A amp B amp C exist(c) A-B Do Not Exist in Sm there are no negatives or positives(d) A+B-C Do Not Exist in SM there are no negatives or positives
There are only resultants when there are multiple arrows
SM Graphs versus BS Graphs
Numbers Raised to Powers
When numbers are raised to powers the negative sides of all BS math graphs are ILLOGICAL and INCORRECT
Incorrect BS Math for 2x
2x
4 16 correct3 8 correct2 4 correct1 2 correct0 1 correct-1 005 incorrect-2 025 incorrect-3 0125 incorrect-4 00625 incorrect
negative side of graph is incorrectpositive side of graph is correct
16
8
42105025012500625
-4 -3 -2 -1 0 1 2 3 4
po
sit
ive
negative positive
2x BS Math
correctincorrect
In BS math the dash sign (-) used in this example represents three different operations They arebull (-) Subtractionbull (-) A direction to the leftbull (-) As an exponent to mean divide by
All positive numbers produce correct results and graphs
All negative numbers produce incorrect results and graphs
A direction of three units to the left (-3) is placed in the equation as an exponent
A direction in space (3 units toward the left) changes a whole number to a fraction by using the dash (-) as an exponent rather than a direction in space
This is illogical and incorrect BS math should have performed this operation as
Unfortunately the answers to problems can come out correct because data points of a real problem fit the curve
However the answers will not be due to logical reasoning
Because the equation produced a curve that real problem data points follow it has been falsely believed that the equations and graphs are correct
1
2
x
Incorrect BK graph for (12)x
(12)x
4 00625 correct3 0125 correct2 025 correct1 1 correct0 1 correct-1 2 incorrect-2 4 incorrect-3 8 incorrect-4 16 incorrect
negative side of graph is incorrectpositive side of graph is correct
00625012502505124
8
16
-5 -4 -3 -2 -1 0 1 2 3 4 5P
osit
ive
negative positive
BS Math for (12)x
correctincorrect
Incorrect BS math and graph for
All graphs are symmetrical with SM
Two BS Math Equations that Produce Incorrect Answers
2 0x x starting equation 2
2
x x x x
x x
Add lsquoxrsquo to both sides
2(1) 1 let lsquoxrsquo = 1
2 1 This is an incorrect answer 2(0)-(0)=0 The only answer that BK math can give is x=0
1
a b starting equation ( )( ) ( )( )a a a b multiply both sides by ( )a
( )( ) ( )( ) ( )( ) ( )( )a a b b a b b b subtract ( )( )b b from both sides
( )( ) ( )( )a b a b a b b factor ( )a b from both sides
( )a b b divide both sides by ( )a b
( )b b b replace ( )a with ( )b 2b b simplify 2(1) (1) Let ( )b = 1
2 1 Again BS math yields and incorrect answer 1
This incorrect result is a problem with the distributive law SM explains how this incorrect answer is obtained and the problem with the distributive law that allows the error
Page 148 ndash149 Barry Mazur ndash Imagining Numbers
3
2
31
Do the computation carefully on paper using the rules we agreed to Then ponder your answer which should be something of a surprise to you donrsquot stop there Think of what your answer might possibly mean or might imply
Solve this equation using the distributive law
JK definitely a surprise the answer wasincorrect and illogical
Gerhard Gade University Professor doing math at Harvard University
Here is the BS math answer that Barry Mazur calculated
2
31
2
31
2
31
1 3 1 3 1 3 3 3 2 2 3 1 3
2 2 2 2 4 4 4 4 4 4 2 2
dash
1 3 1 3 1 31
2 2 2 2 4 4
3 3
4 4
When the distributive law is used with the BS Rule-of-Signs and one of the signs is a (-) dash sign the middle two terms always cancel each other This leads to the incorrect answer of (-1) in the above equation
Barry Mazur did not go back and check his answer with the original equation If he had he would have known that the answer was incorrect
3
1 31
2
check answer by setting equal
3
331 3
12
1 31
2
Take the Cube root of both sides
1 3 2 multiply both sides by 2
3 3 subtract 1 from both sides
2 2
3 3 square both sides
3 9 This is an incorrect answer A simple equation check could have shown that the distributive law using BS math provides an incorrect answer
1
BS Illogical and Incorrect Numbers Raised to Powers
nC D n DC
642 DDD
In BS math only a positive number can have an answer if ldquonrdquo is an even number (246---)
All negative numbers have no solutions
642 DDD
With SM just add a direction arrow and any number gives the correct answer
or n nC D C D
3 3( 3 8 8 or 8 2 or 2 )n D C
( 2 16 C= 16 or 16 4 or 4 )n D
Since a negative number has no square root answer in BS math BS mathematicians invented imaginary numbers
i2=-1
24414 iiC n
These equations are easily solved without the use of imaginary numbers with SM math
If 4 then 4 2x
If 4 then 4 2x
The 360-degree circle is a 4400 years old extremely outdated system based on a year of 12 months of 30 days each that is 360 days
SM will use a more logical system that will have a circle of 6283185307o
d
( )
(deg )
d r radians
d r rees
the arc distance around
the circumference of a circle
d
radiusr
2 2
2 2
d r
dv rd t t
v dda rd t t t
1
d
r
( )av tv
The average angular velocity of a body is the rate of its angular displacement
( )0
liminstt
vt
The instantaneous angular velocity is the limit of the ratio
t
as t approaches zero
( )d
av
d
tv
The average circumference velocity of the body is the rate of its distance traveled d divided by the time t
dvv
r
1 dd vdrv
t t t r r
The equation of the relationship between the angular velocity and the velocity of a point on that body
( )o
av
v v v
t ta
The average angular acceleration is the change in the angular velocity
( )0
liminstt
v
ta
The instantaneous angular acceleration is the
limit of the ration of v
t
as t approaches
zero
daa
r
1
1
d
d s d
vva
t r tv a
r t r
The equation of the relationship between the angular acceleration and the acceleration of a point on the body
21
2v a t
If a body with an initial angular velocity ov has a constant
angular acceleration a it will turn through an angle in a time t
1
Straight Line Symbol Units Rotational Symbol Units Distance d m Angular
Distance radians or degrees
Velocity dt
v m
s Angular Velocity
V
t
radians or degrees
s
Acceleration vat
2ms
Angular Acceleration
VA
t
2
radians or degrees
s
Mass M Kg Mass distribution
dM 2Kgm
Force actionA Ma
2
Kgm
sNewton
Torque
dT M A 2
2
Kgm
s
Momentum Mv Kgm
s
Angular Momentum
dM V 2Kgm
s
Work (Force)(distance)
Ad 2
Kgmm
sNm Joule
Work T
Using AS arrow math to solve and direction in space problems using the equation 1 2
2 2 2 2 2 2( ) amp and ( ) amp 3
4
2 2 2
2 2
( ) amp
(5 ) 4 3
(5 ) 16 9
(5 ) 25
(5 ) (5 )
2 2(5 ) 4 3
(5 ) 16 9
(5 ) 25
(5 ) (5 )
In SM math Pythagorasrsquo theorem provides both magnitude and direction
In BS math Pythagorasrsquo theorem provides magnitude only
time set equal to zero
23
( ) ( 0 )10 1
x t xx
3( ) ( 0)
10 1W t f
In BS math this is an equation for a traveling wave
must use a square term in order to make the curve symmetric
NO SQUARE TERM IS NECESSARY
In SM math when t is set to ldquo0rdquo
This is a major fault in BS math
x f(x)
-4 0019
-3 0033
-2 0073
-1 0273
0 3000
1 0273
2 0073
3 0033
4 0019 -4 -3 -2 -1 0 1 2 3 400
05
10
15
20
25
30
35
BS wave (must use square term)
snap shot of wave at x = 0
Am
plit
ude
lt to gt lt to gt f(ltgt)
lt -4 0073
lt -3 0097
lt -2 0143
lt -1 0273
origin 0 3000
gt 1 0273
gt 2 0143
gt 3 0097
gt 4 0073lt0073 lt0097 lt0143 lt0273
3000
0273gt 0143gt 0097gt 0073gt
Wave moving
snap shot of wave
Am
plit
ude
lt4 lt3 lt2 lt1 0 1gt 2gt 3gt 4gt
3
2
1
In SM NO squared terms are needed
SPRING
An Action is needed to stretch or compress a spring to a starting point where it can execute cycle motion
tsA (ts to start) is the work required to move a mass attached to a spring a distance (d1) from
its initial rest position RAK 0 the spring constant is the stiffness of the spring
o A is attraction o R is repulsion o
ARRA
00 is for the back and forth motion through the zero initial rest position A
stretched spring goes from ldquoArdquo to ldquo0rdquo to ldquoRrdquo to ldquoArdquo to ldquo0rdquo to ldquoRrdquo d1 is the distance a mass is moved from its initial rest position
0
0 1ts A RA K d
Once ( )mad
is performed to move the mass to position d1 (against the restraint of the spring) a of the ( )mad
is transferred into the spring as potential attractionrepulsion (PAR) There is always
more ( )mad
performed than is stored as PAR When the spring is released from the restraining Ats the PAR starts the oscillation Conservation of ( )mad
requires that the PAR and the motion ( )mad
( )( )madM be a constant
( )amp ( ) constantmadPAR M 02
0 0
1
2A R A RPAR K d
02
( ) 0 0
1( )
2mad A R A RM m v
0 0 02 2 2
0 0 0 1
2 2 21
2 2 21
1 1 1amp
2 2 2
1 1 1amp
2 2 2
amp
A R A R A RK d m v K d
Kd mv Kd
Kd mv Kd
0
1d is frac12 the height from the base to the peak of a cycle
An experiment of moving paper under a pendulum through a small angle filled with sand that can escape through a small hole in the bottom shows that a pendulum cycling back and forth follows a curve that is mathematically analogous to a lsquosinrsquo or lsquocosrsquo
The projection of the curved area onto the straight line exactly duplicate the back and forth motion of CM This allows for the math of lsquosinrsquo or lsquocosrsquo to be used
SM will use lsquosinrdquo math It could use lsquocosrsquo or a combination of both lsquosinrsquo and lsquocosrsquo Using just the lsquosinrsquo math makes the understanding and math easier
0 sin 0 0SM 0 sin157 1SM The diagram at the left shows a spring being stretched from its (zero rest) position-2 and the PAR being released at position-1 There are 4 sections (1 to 2 2 to 3 3 to 4 and 4 to5) that alternate between maximum acceleration and zero velocity to max velocity and zero acceleration
At position-1 there is maximum acceleration and zero velocity o The lsquosinrsquo values from 0o to 157o will be used for velocity (values from 0 to 1) o The lsquosinrsquo values from 157o to 0o will be used for acceleration (values from 1 to 0)
At position-2 there is maximum velocity and zero acceleration Acceleration changes to deceleration
o The lsquosinrsquo values from 157o to 0o will be used for deceleration (values from 1 to 0) o The lsquosinrsquo values from 0o to 157o will be used for acceleration (values from 0 to 1)
At position-3 there is maximum acceleration and zero velocity o The lsquosinrsquo values from 0o to 157o will be used for velocity (values from 0 to 1) o The lsquosinrsquo values from 157o to 0o will be used for acceleration (values from 1 to 0)
At position-4 there is maximum velocity and zero acceleration Acceleration changes to deceleration
o The lsquosinrsquo values from 157o to 0o will be used for deceleration (values from 1 to 0) o The lsquosinrsquo values from 0o to 157o will be used for acceleration (values from 0 to 1)
At position-5 the cycle starts again
2
)539(
cT
MK
spring constant (K) when mass (M) and cycle-time (T) are known
)(
))(89)((
111 md
uKgM
d
MG
d
FK sts
spring constant (K) when mass (M) and amplitude ( d1) are known
2 21( )
Kv d d
m
velocity (v) at any point during the cycle
velocity
dforM
Kdv
max
012 velocity (v2) maximum (when d = 0 at position-2 and 4)
acceleration (a1) maximum when spring constant (K) mass (M) and amplitude (d1) are known
acceleration (a1) maximum when amplitude (d1) and cycle-time (Tc) are known
acceleration (a) at any point during the cycle-time (T)
1
151
286
v
dTc
cycle-time (Tc) when amplitude (d1) and velocity (v1) are known
K
MTc 28651
cycle-time (Tc) when mass (M) and spring constant (K) are known
fTc
1
cycle-time (Tc) when the frequency is known
cTf
1
frequency (f) when the cycle-time (Tc) is known
of cycles
elapsed timef
frequency (f) when the number of cycles in an elapsed time is known
d1 amplitude (distance from rest to start position)
Also frac12 the distance from the base to the peak
3131 oror dM
Ka
251
3131 539
c
oror T
da
2539
cT
da
In December of 2006 I discovered the mathematical error in Einsteinrsquos Special Relativity Einstein used BS math and that math only works in an imaginary world
It cannot describe real symmetrical space
Einsteinrsquos Special Relativity should be abandoned and a theory of real space using SM should be used to replace it
Numerous articles on the Internet show Einsteinrsquos theory to be incorrect but no one in academia is willing to publish the information
Figure 1-10 A moving particle observed from two frames of reference in uniform translation
Einsteinrsquos Math Error in Special Relativity Transformation of Velocity (From Eisenberg- Fundamentals of Modern Physics pages 27-29) Consider the particle shown in figure (1-10) moving with velocity v as
seen in a frame of reference O
dx dy dz
Vx Vy Vzdt dt dt
dx dy dz
V x V y V zdy dt dt
We would like to evaluate the velocity v of the particle as seen in the frame of reference 0 which is itself moving relative to 0 with velocity v Measured in the 0 frame the velocity vector of the particle has com ponents
The same velocity vector as measured in the 0 frame has components
2
2
2
2
2
1
1
x vtx
v
cy y
z z
vxt
ctv
c
Now from equations (1-13) we know that the relation between the primed and the unprimed coordinates and times is
Take the differential of these equations remembering that v is a constant This gives
2
2
2
2
2
1
1
dx vdtdx
v
cdy dy
dz dz
vxdt
cdtv
c
2
2
2 2 2 2 2
2
2
2
11 ( )
1( ) 1
1
1
1
dx vdt
v dxdx vdt v
dx dx vdtc dt dtvdx vdx vdx v dxdt dt dt dt
dt dtc c c c
VxV x
Vx
VxV x
vv
vv
c
x
v
c
cV
( )
coordinate of particle relative to moving primed system
coordinate of particle relative to stationary system
velocity of primed coordina
V x
Vx
te system relative to stationaryv sys
tem
Page 28 Fundamentals of Modern Physics
(1-16)
According to Einstein this equation tells you how to transform the observed velocity from one frame of reference to another frame of reference First we note that as Vc and vc approachzero the equation approaches those which would be derived from the Galilean transformation
2 2
( ) 1 01 1 ( )
Vx v Vx vV
VxV x
Vx v
v cVxx
v
c
v
cc
Another property of this equation is that it is impossible to choose V and v such that Vrsquo the magnitude of the velocity which is seen in the new frame is greater than c
1
2
09 ( 09 ) 18009876543
( 09 ) 1811 (09 )
c c cV c c
cc
c
Example As seen by O particle 1 has velocity 09c in the direction of positive x and particle 2 has velocity 09c in the negative x direction
Einstein equation however predicts To evaluate the velocity of particle 1 with respect to particle 2 we transform from the O frame to the Orsquo frame moving in the negative x direction with velocity v = -09c using equation 1-16 We obtain
JK LOGIC will observe the particles separating at 18c
A confusing INCORRECT and ILLOGICAL answer
left is a dir rightsub of a
1
2
1
09 ( 09 ) 18009876543
181( 09 )1 ( 09 )
The multiplication of dir left time a dir right is ONLY a dir left
09
dir
direction leftdirection rightsubtraction
c c cV c c
cc
c
V
dir
22
22
09 18 18009876543
181( 081)1(09)(09)
1
left rightsub
sub
c c c cc c
ccc
c
Analyzing Einsteinrsquos BS math Equation
Using SM math a logical answer can be obtained
Cannot multiply a subtraction operator times a direction in space
Can only subtract and change its direction
Cannot multiply a direction to the left times a direction to the right
Cannot multiply opposite directions in space
Can only solve for the resultant
rightleft
subtraction
1
subtraction
2
sub of a dir to the left is a dir to the right
1 resultant 0 sub 2
2sub 2
2
09 09
09 amp 091
09 09 18 1
(0)1
(09) amp (09)1
c cV
c cc
c c cV
cc
cc
8
181
cc
SM math predicts a logical answer using logic
21st century scientists state in many articles that Einsteins General Relativity and Quantum Mechanics are incompatible
They are not only incompatible they are both mathematically incorrect
They both use illogical BS math
Any equations in math that use imaginary numbers should be abandoned
Imaginary numbers are only necessary in the current BS math system
This means that Schroumldingers equations should be abandoned
Any equation in quantum mechanics that uses imaginary number should be abandoned
Prof Kanarev has developed logical physics and chemistrytheories without the use of Quantum mechanics or relativity
A partial list of his theories coverbull Mechanics ndash replaces Newtonbull Photon - bull Electrons ndash Protons- Neutronsbull Spectroscopybull Nucleusbull Atoms amp Moleculesbull Electrodynamicsbull Heatbull Curvature of Space ndash replaces Schwarzcchildrsquos equation
Kanarevrsquos Hypothesis
bull Electron align linearly with proton They do NOT spin or form clouds around the nucleus
bull Neutrons are in the center of atom Proton are outside of Neutrons
bull The wavelength of all particles is the experimental radius
bull Double slit experiment explained correctly
bull Heisenbergrsquos inequality explained correctly
bull is the radius of approach of the magnetic lines around the torus of the electron
( 1 137)
bull Why water expands when it freezes
bull Electron clusters form lightning
rbull The meaning of Plancks constant is explained
22 2 kg m
h m f mr f consts
2210254 42m r kgm bull If the mr-constant describes a ring the equations
describe experiments bull If the mr-constant describes a wave the equations do
NOT describe experiments
Kanarevrsquos 5 Equations of Mechanics
Kanarevrsquos 1st Equation Acceleration ampmv maMma R R
Kanarevrsquos 2nd Equation Uniform motion amp mv maMma R R
maM mvR R
Kanarevrsquos 3rd Equation
Kanarevrsquos 4th Equation Equal and opposite reaction
1 21 21 2
12
a m
m a m ama
Kanarevrsquos 5th Equation 1 2 3 Mi M M M Mna R R R R R
The sum of the Actions working on a moving body is never equal to zero
Kanarevrsquos Photon model is a rolling ring
His equations accurately describe known experiments with photonsbull Electromagnetic radiation should be called photon radiationbull Diffraction and interference are the same phenomenabull Correctly explains the double slit experiment without quantum equationsbull Correctly explains the meaning of Heisenbergrsquos inequality
e
2 (2)(3142)(2817 15 ) 10007297352568
R 2426 12 137mag
xp
r m
m
Charge
Magnetic action
Proton
The intensity of the magnetic action near the geometrical center of the proton ring is so great that we theorize that it is the strong nuclear force
(9383 8)(1602 19)8476 14
4 (4)(3142)(1411 26)p
pp
EB T
M
(9395 8)(1602 19)1035 15
4 (4)(3142)(0966 26)N
NN
EB T
M
Neutron
1a Hydrogenavi
Why is Symmetry Math NeededBecause the BS Rule-of-Signs a (-)(-) = (+) is illogical and produces many incorrect answers
The BS number line is a Broken-Symmetry number line
From ldquoWhat Is Mathematicsrdquo by Courant and Robbins ndash page 55The Rule of Signs (-1) (-1) = +1
Which we set up to govern the multiplication of negative integers is a consequence of our desire to preserve the distribution law
a(b+c) = ab + ac
It took a long time for mathematicians to realize that the ldquoRule-of-signsrdquo together with all the other definitions governing negative integers and fractions
They were created by us in order to attain the freedom of operation while preserving the fundamental laws of arithmetic
Cannot be ldquoPROVEDrdquo
According to Professor Philip Kanarev there has not been a major discovery in physics or chemistry by ldquopeer-reviewing-academiardquo due to THEORY since the early 1900rsquos
There have been virtually no science-changing discoveries since the theories of quantum mechanics and relativity were introduced
One of the reasons is that the math being used is BS math and it operates on Broken-Symmetry and can only provide usable answers
BS math cannot be used to describe reality in space or time Space is symmetrical Space cannot depend on the direction one orientates a coordinate system
BS mathematicians should have discovered the problem Instead they invented symbols and definitions and bypassed the real problem ie [(imaginary numbers i2 = -1) and (absolutely values |-X|=+X)]
The Mathematical Principle of Error Joseph A Rybczyk 1996
Once an error enters a calculation all calculations after that point become an extension of the error
Studying the dash-cross codes of BS math has allowed me to break the codes and understand the reasons for the errors
Dash sign (-) CODES In Math used as
22
1 15 004255
bull (-) the number of zeros to the right of a decimal point 10-5=000001
bull (-) a subtraction operator
bull (-) a direction in space labeled negative (whatever that means)
bull (-) an exponent to mean divide
bull (-) describe the negative half of a sin or cos graph
bull (-) describe the negative half of an ldquoerdquo and ldquo1erdquo graph
bull (-) describe the negative side of all numbers raised to a power x
In Physics
bull (-) An Electron has been labeled as negative
bull (-) Anti-Particles have been labeled as negative
In Chemistry
bull (-) amp (+) Thermochemistry
Cross sign (+) CODES
(+) a direction in space labeled positive (whatever that means)
(+) an addition operator
(+) used to show the number of zeros to the left of a decimal point 10+5 = 100000
In Physics and Chemistry
bull (+) Protons have been Labeled as positive
bull (+) numerous other particles have been Labeled as positive
bull (+) amp (-) Thermochemistry
In Math used
The same dash symbol (-) is used for the numerous different math operations
The definitions established for the use of the dash symbol (-) do not distinguish between their different operations
In many math operations the dash symbol is changed to mean one of the other meanings
It is amazing that math has proceeded to its current level of use with this illogical use of a symbol
The same cross symbol (+) is used for numerous different math operations and produces the same illogical answers
A Summary of the BS Broken-Symmetry math
1 1 cosh tanhln
x xe x xe
bull Originated with broken-symmetry in the XYZ plane
bull Originated with mirror-broken-symmetry in the XZ plane
bull Created a Rule-of-Signs that cannot be proved (-)(-)=(+) (+)(+)=(+) (-)(+)=(-)
bull Created imaginary numbers to compensate for broken-symmetry
bull Created absolute values to change illogical negative number answers into positive number answers
bull Produces broken symmetry graphs for many functions
It took four years of study to discover why a (-)(-)=(+) only works in an imaginary world where space in the dash (-) direction is different from space in the cross (+) direction
Started working on the problem in August of 2001
Data points created using BS math produce graphs that are not symmetrical dash side different from cross side
However if a real problems data points follow a non symmetrical graph the graph can provide usable answers
This is why no one discovered the problem
Solved in January of 2005
Symmetry-Math(SM) vs Broken-Symmetry(BS) Math
In SM if we specify that the dash sign (-) means only subtraction and the cross sign (+) means only addition and an appropriate symbol is used for a direction in space space becomes symmetrical and math becomes logical
Math operators and directions in space are not the same and the same symbol should not be used to represent them
Letrsquos start with the (dash)multiplied by a (dash)=(cross) in the BS system
Usable answers may be obtained but it will be for illogical reasons
multiplied by a ( ) ( ) ( )subtraction Subtraction addition
rightleft left
dash dash cross
positivenegative negative
Symmetry-Math The subtraction of a direction is equal to the opposite direction The answers are correct using correct logic
opposite direction direction
SM
(negative)(negative) = (positive)
( ) ( ) ( )subtraction
BS
There is NO multiplication of a subtraction operator by a direction in space
There is just the subtraction of a direction in space
If instead of labeling the left side of a coordinate system as a negative (-) the same as a subtraction operator we label it with an arrow ( ) to represent the direction then a subtraction from that arrow direction will be in the opposite direction ( )
( ) mutiplied by a ( ) ( )
negative Positive negative
subtraction addition subtraction
dash cross dash
left right left
A subtraction operator multiplied by an addition operator is equal to a subtraction operator This is illogical
BS Math
A negative multiplied by a positive is equal to a negative This is illogical
A direction to the left multiplied by a direction to the right is a direction to the left This is illogical and Einsteins math error in special relativity
A dash times a cross is equal to a dash This is illogical
A vector moving to the left multiplied by a vector moving to the right is equal to ONLY to a vector moving to the left that is eight orders of magnitude faster thanThe speed of light
opposite direction subtraction direction
SM ( ) ( )
BS (negative)(positive) = (negative)
Symmetry Math The subtraction of a direction is equal to the opposite direction
There is no multiplication of a subtractionoperator times a direction in space
( ) multiplied by a ( ) ( )
positive positive positiveaddition addition additioncross cross cross
direction direction direction
What is the meaning of multiplying addition operators Addition operators are not multiplication operators
BS Math
() multiplied by a direction ( )=
number
() multiplied by a direction ( )=
number
A number times a direction maintains the same direction
( ) ()
dirdiradd
( ) ()
dirdiradd
The addition of a direction is in the same direction There is no multiplication by addition operators
Symmetry Math
Future math books need to eliminate the use of the cross sign (+) to represent something labeled a positive direction in space and the dash sign (-) to represent something labeled a negative direction in space Space does not have positive and negative directions
Symbols that are logical and have no illogical representation should be adopted SM uses either arrows or symbols
Symmetry-Math Number Line
Symmetry-Math Rules
The dash sign (-) will have only one use subtraction
The cross sign (+) will have only one use addition
Directions have an arrow and a number
All observers see the same direction and magnitude
bull There is no multiplication of arrows
amp ampamp ampampamp
( 3 ) ( 5 ) ( 6 ) ( 65 ) (17256 ) ( 32993 )
dir dirdir dirdirdir
Negative and Positive Directions in Space
If I asked you to point to a negative direction in space which way would you point
Hopefully you will realize that there is no such thing as a negative direction in space
If I asked you to point to a positive direction in space which way would you point
Again hopefully you will realize that there is no such thing as a positive direction in space
Objects on the right side of the x-axis are positive (+)(+) = (+)
Objects on the left side of the x-axis are negative (-)(-) = (+) negative math is different from positive math This is illogical
Objects on the top of the y-axis are positive
Objects on the bottom of the y-axis are negative
Objects in the front of the z-axis are positive
Objects in the back of the z-axis are negative
Again BS math of the positive direction (x right y up and z front) is different from the BS math of the negative direction (x left y down and z back)
BS math is Broken-Symmetry
Mirror image is broken symmetry in the x and z axis If the left amp right-axis are
reversed symmetry is broken Obs-1 math is different from obs-2
If the front amp back-axis is reversed symmetry is broken Obs-1 math is different from obs-2
If the top amp bottom-axis are
reversed symmetry is not broken Obs-1 math is the same as obs-2
Math answers cannot depend on which side of a number line an observer sit
Absolute values for Displacement
Even for a simple displacement of an object BS math invented a definition and absolute values to obtain something labeled a positive number answer
I----------I----------I----------I----------I----------I----------I----------I----------I
-4 -3 -2 -1 0 +1 +2 +3 +4 a gtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtb cltltltltltltltltltltltltltltltltltltltltltltltltltltltb
If an object starts at ( 3)ia X and moves to ( X 3) and the to ( 1)b fb c X
BS Math provides a usable answer without a definition or absolute values
However If an object starts at
( 1) and moves to ( X 3) and the to ( 3)i b fa X b c X
Then absolute values and a definition must be used
---------I----------I----------I----------I----------I----------I----------I---------- -4 -3 -2 -1 0 +1 +2 +3 +4
a gtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtb cltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltb
BS Math makes up a definition that all displacements are positive Since this example provides an answer with something labeled a negative the positive definition must be applied and (-2) is changed to (+2) A definition is needed to arrive at a useable answer This is Illogical
In SM absolute values and a meaningless definition of ldquopositiverdquo are removed In SM objects and directions in space are defined by the direction and magnitude of the resultant of arrows
With SM you get total distance traveled by the object and the final direction of the displacement
22222 2 bababbaababababa If w e su b stitu te p o s itiv e n u m b e rs in to th e eq u a tio n th e an sw er w ill b e co rrec t If (a = 5 ) an d (b = 3 )
2
2
8
5 3 5 3 5 3 5 5 5 3 3 5 3 3 25 15 15 9 64
8 8 8 8 8 8 8 8 8 8 64
W h en tw o p o sitiv e n u m b ers a re a d d ed an d sq u a re d th e B S m a th d is tr ib u tiv e law p ro v id es a co rre c t an sw er 1
Illogical and Incorrect BS math for the distributive law
222 2 bababbabbaaabababa B y d e fin itio n in B S m a th n o n u m b er sq u a red c a n ev e r b e n eg a tiv e T h e re fo re n o sq u a red n u m b er c an ev e r b e a v e c to r m o v in g in th e d a sh ( -) o r n eg a tiv e d ire c tio n if it is sq u a red In B S m a th an d ad d itio n o p e ra to r m u ltip lied b y a su b trac tio n o p e ra to r is A L W A Y S eq u a l to a su b tra c tio n o p e ra to r If th is h ad n o t b e en u sed fo r th e la s t fo u r h u n d red y e a rs y o u w o u ld d ie lau g h in g a t th e u tte r ab su rd ity o f th e la s t s ta tem e n t T h is is illo g ica l an d v io la te s sy m m etry
If a = 5 an d b = 3 42222352
22
4930253)3)(5(253)5)(3()3)(5(5353535 22222 A s lo n g a s ldquo ardquo is g re a te r th an ldquob rdquo th e B S m a th d is tr ib u tiv e law p ro v id es co rrec t an sw ers 1
B S m a th b re a k s d o w n w h e n ldquo b rdquo is g rea ter th a n ldquo a rdquo
If a = 3 an d b = 5
2 2
2
2
3 5 2 4
2 ( 2) ( 2) 4
2 ( )( 2) ( )( 2) 4
A n y co m b in a tio n ex cep t th e m id d le o n e is illo g ica l an d in co rre c t B y d e fin itio n th e an sw er can n o t b e -4 in B S m a th B S m a th is in co rrec t
4253095)5)(3(235)3)(5()5)(3(3535353 22222 T h is is w h ere th e d is tr ib u tiv e la w p ro v id es in co rrec t an sw ers I t p ro d u ces an an sw er o f + 4 b e cau se b y d e fin iti o n n o n u m b er sq u a red in B S m a th c an b e a n eg a tiv e n u m b er U sin g in co rre c t B S m a th th e d is tr ib u tiv e law w ill g iv e an in co rrec t an sw ers o f + 4 T h is is in co rrec t an d illo g ica l 1
Since there can be no negative direction from a squared term in BS math BS math should be modified or abandoned
The BS distributive law should be modified or abandoned
The numbers +1+2+3+4 and ndash1-2-3-4hellip should be abandoned
BS math gives incorrect answers because their Rule-of-Signs allows multiplying an arrow going in one direction by an arrow going in another direction
Clearly this is not logical
Arrows should be used that are specific for directions
W h at d o es it m e an to m u ltip ly ( 4) b y (4 ) A s w e k n o w 1
m u ltip lica tio n is ju s t ad d itio n W h a t w o u ld ( 4) tim e (4 ) b e 2
In BS math the answer is by definition -16 unless you square -4 and then it is +16
In SM you cannot multiply arrows going in opposite directions
The middle two terms are not logical You cannot multiply opposite directions
This is a MAJOR error in BS math They multiply a dash [(-) a direction to the left] by a cross [(+) a direction to the right] And then ldquoby definitionrdquo label the answer dash (-) This is illogical and produces incorrect answers
2
2 2
This is Einsteinrsquos math error
in Special relativity
Magnitude 4 Direction 1 Magnitude 3 Direction 2 Magnitude 5 Direction 3 Relative to an observer at 0 4
Magnitude 4 Direction 1 Magnitude 3 Direction 2 Magnitude 5 Direction 3 Relative to an observer at 0 4
Magnitude and Direction (MD) An arrow has both a magnitude andA direction relative to a 2nd objectobserver
bull Magnitude and velocity (MV) relative to a 2nd objectobserver
bull Magnitude And acceleration (MA) relative to a 2nd objectobserver2
v da
t t
Introducing a new symbol the ldquoamprdquo sign The ldquoamprdquo sign means resultant ADD all the Arrows for each direction and then SUBTRACT to find the final resultant
1 1amp 1 2 2amp 2 4 If =
1 amp1 2 2 amp2 4 If =
amp 5 amp 3 2 If gt
amp 3 amp 5= 2 If gt
amp 0 3 amp 3=0 if =
only the resultant equals zero
the vectors do not - they are still there
Does not exist There is no multiplication of arrows
There are two arrows One has a magnitude of 3 One has a magnitude of 5 The maximum these two arrows can be is 8
Only the resultant of arrows exist
An SM arrow has a direction and a magnitude
Arrows and magnitudes are still there even if the resultant is zero
There are no negative SM Arrows
There are no positive SM Arrows
There are no ZERO SM Arrows A zero would negate the definition of an SM Arrow something with no magnitude and no direction cannot be the definition of an arrow that is defined as having magnitude and direction
bull A resultant is the addition and subtraction of the differences in magnitude and direction arrows
A B and C are the arrows (a) A amp B exist(b) A amp B amp C exist(c) A-B Do Not Exist in Sm there are no negatives or positives(d) A+B-C Do Not Exist in SM there are no negatives or positives
There are only resultants when there are multiple arrows
SM Graphs versus BS Graphs
Numbers Raised to Powers
When numbers are raised to powers the negative sides of all BS math graphs are ILLOGICAL and INCORRECT
Incorrect BS Math for 2x
2x
4 16 correct3 8 correct2 4 correct1 2 correct0 1 correct-1 005 incorrect-2 025 incorrect-3 0125 incorrect-4 00625 incorrect
negative side of graph is incorrectpositive side of graph is correct
16
8
42105025012500625
-4 -3 -2 -1 0 1 2 3 4
po
sit
ive
negative positive
2x BS Math
correctincorrect
In BS math the dash sign (-) used in this example represents three different operations They arebull (-) Subtractionbull (-) A direction to the leftbull (-) As an exponent to mean divide by
All positive numbers produce correct results and graphs
All negative numbers produce incorrect results and graphs
A direction of three units to the left (-3) is placed in the equation as an exponent
A direction in space (3 units toward the left) changes a whole number to a fraction by using the dash (-) as an exponent rather than a direction in space
This is illogical and incorrect BS math should have performed this operation as
Unfortunately the answers to problems can come out correct because data points of a real problem fit the curve
However the answers will not be due to logical reasoning
Because the equation produced a curve that real problem data points follow it has been falsely believed that the equations and graphs are correct
1
2
x
Incorrect BK graph for (12)x
(12)x
4 00625 correct3 0125 correct2 025 correct1 1 correct0 1 correct-1 2 incorrect-2 4 incorrect-3 8 incorrect-4 16 incorrect
negative side of graph is incorrectpositive side of graph is correct
00625012502505124
8
16
-5 -4 -3 -2 -1 0 1 2 3 4 5P
osit
ive
negative positive
BS Math for (12)x
correctincorrect
Incorrect BS math and graph for
All graphs are symmetrical with SM
Two BS Math Equations that Produce Incorrect Answers
2 0x x starting equation 2
2
x x x x
x x
Add lsquoxrsquo to both sides
2(1) 1 let lsquoxrsquo = 1
2 1 This is an incorrect answer 2(0)-(0)=0 The only answer that BK math can give is x=0
1
a b starting equation ( )( ) ( )( )a a a b multiply both sides by ( )a
( )( ) ( )( ) ( )( ) ( )( )a a b b a b b b subtract ( )( )b b from both sides
( )( ) ( )( )a b a b a b b factor ( )a b from both sides
( )a b b divide both sides by ( )a b
( )b b b replace ( )a with ( )b 2b b simplify 2(1) (1) Let ( )b = 1
2 1 Again BS math yields and incorrect answer 1
This incorrect result is a problem with the distributive law SM explains how this incorrect answer is obtained and the problem with the distributive law that allows the error
Page 148 ndash149 Barry Mazur ndash Imagining Numbers
3
2
31
Do the computation carefully on paper using the rules we agreed to Then ponder your answer which should be something of a surprise to you donrsquot stop there Think of what your answer might possibly mean or might imply
Solve this equation using the distributive law
JK definitely a surprise the answer wasincorrect and illogical
Gerhard Gade University Professor doing math at Harvard University
Here is the BS math answer that Barry Mazur calculated
2
31
2
31
2
31
1 3 1 3 1 3 3 3 2 2 3 1 3
2 2 2 2 4 4 4 4 4 4 2 2
dash
1 3 1 3 1 31
2 2 2 2 4 4
3 3
4 4
When the distributive law is used with the BS Rule-of-Signs and one of the signs is a (-) dash sign the middle two terms always cancel each other This leads to the incorrect answer of (-1) in the above equation
Barry Mazur did not go back and check his answer with the original equation If he had he would have known that the answer was incorrect
3
1 31
2
check answer by setting equal
3
331 3
12
1 31
2
Take the Cube root of both sides
1 3 2 multiply both sides by 2
3 3 subtract 1 from both sides
2 2
3 3 square both sides
3 9 This is an incorrect answer A simple equation check could have shown that the distributive law using BS math provides an incorrect answer
1
BS Illogical and Incorrect Numbers Raised to Powers
nC D n DC
642 DDD
In BS math only a positive number can have an answer if ldquonrdquo is an even number (246---)
All negative numbers have no solutions
642 DDD
With SM just add a direction arrow and any number gives the correct answer
or n nC D C D
3 3( 3 8 8 or 8 2 or 2 )n D C
( 2 16 C= 16 or 16 4 or 4 )n D
Since a negative number has no square root answer in BS math BS mathematicians invented imaginary numbers
i2=-1
24414 iiC n
These equations are easily solved without the use of imaginary numbers with SM math
If 4 then 4 2x
If 4 then 4 2x
The 360-degree circle is a 4400 years old extremely outdated system based on a year of 12 months of 30 days each that is 360 days
SM will use a more logical system that will have a circle of 6283185307o
d
( )
(deg )
d r radians
d r rees
the arc distance around
the circumference of a circle
d
radiusr
2 2
2 2
d r
dv rd t t
v dda rd t t t
1
d
r
( )av tv
The average angular velocity of a body is the rate of its angular displacement
( )0
liminstt
vt
The instantaneous angular velocity is the limit of the ratio
t
as t approaches zero
( )d
av
d
tv
The average circumference velocity of the body is the rate of its distance traveled d divided by the time t
dvv
r
1 dd vdrv
t t t r r
The equation of the relationship between the angular velocity and the velocity of a point on that body
( )o
av
v v v
t ta
The average angular acceleration is the change in the angular velocity
( )0
liminstt
v
ta
The instantaneous angular acceleration is the
limit of the ration of v
t
as t approaches
zero
daa
r
1
1
d
d s d
vva
t r tv a
r t r
The equation of the relationship between the angular acceleration and the acceleration of a point on the body
21
2v a t
If a body with an initial angular velocity ov has a constant
angular acceleration a it will turn through an angle in a time t
1
Straight Line Symbol Units Rotational Symbol Units Distance d m Angular
Distance radians or degrees
Velocity dt
v m
s Angular Velocity
V
t
radians or degrees
s
Acceleration vat
2ms
Angular Acceleration
VA
t
2
radians or degrees
s
Mass M Kg Mass distribution
dM 2Kgm
Force actionA Ma
2
Kgm
sNewton
Torque
dT M A 2
2
Kgm
s
Momentum Mv Kgm
s
Angular Momentum
dM V 2Kgm
s
Work (Force)(distance)
Ad 2
Kgmm
sNm Joule
Work T
Using AS arrow math to solve and direction in space problems using the equation 1 2
2 2 2 2 2 2( ) amp and ( ) amp 3
4
2 2 2
2 2
( ) amp
(5 ) 4 3
(5 ) 16 9
(5 ) 25
(5 ) (5 )
2 2(5 ) 4 3
(5 ) 16 9
(5 ) 25
(5 ) (5 )
In SM math Pythagorasrsquo theorem provides both magnitude and direction
In BS math Pythagorasrsquo theorem provides magnitude only
time set equal to zero
23
( ) ( 0 )10 1
x t xx
3( ) ( 0)
10 1W t f
In BS math this is an equation for a traveling wave
must use a square term in order to make the curve symmetric
NO SQUARE TERM IS NECESSARY
In SM math when t is set to ldquo0rdquo
This is a major fault in BS math
x f(x)
-4 0019
-3 0033
-2 0073
-1 0273
0 3000
1 0273
2 0073
3 0033
4 0019 -4 -3 -2 -1 0 1 2 3 400
05
10
15
20
25
30
35
BS wave (must use square term)
snap shot of wave at x = 0
Am
plit
ude
lt to gt lt to gt f(ltgt)
lt -4 0073
lt -3 0097
lt -2 0143
lt -1 0273
origin 0 3000
gt 1 0273
gt 2 0143
gt 3 0097
gt 4 0073lt0073 lt0097 lt0143 lt0273
3000
0273gt 0143gt 0097gt 0073gt
Wave moving
snap shot of wave
Am
plit
ude
lt4 lt3 lt2 lt1 0 1gt 2gt 3gt 4gt
3
2
1
In SM NO squared terms are needed
SPRING
An Action is needed to stretch or compress a spring to a starting point where it can execute cycle motion
tsA (ts to start) is the work required to move a mass attached to a spring a distance (d1) from
its initial rest position RAK 0 the spring constant is the stiffness of the spring
o A is attraction o R is repulsion o
ARRA
00 is for the back and forth motion through the zero initial rest position A
stretched spring goes from ldquoArdquo to ldquo0rdquo to ldquoRrdquo to ldquoArdquo to ldquo0rdquo to ldquoRrdquo d1 is the distance a mass is moved from its initial rest position
0
0 1ts A RA K d
Once ( )mad
is performed to move the mass to position d1 (against the restraint of the spring) a of the ( )mad
is transferred into the spring as potential attractionrepulsion (PAR) There is always
more ( )mad
performed than is stored as PAR When the spring is released from the restraining Ats the PAR starts the oscillation Conservation of ( )mad
requires that the PAR and the motion ( )mad
( )( )madM be a constant
( )amp ( ) constantmadPAR M 02
0 0
1
2A R A RPAR K d
02
( ) 0 0
1( )
2mad A R A RM m v
0 0 02 2 2
0 0 0 1
2 2 21
2 2 21
1 1 1amp
2 2 2
1 1 1amp
2 2 2
amp
A R A R A RK d m v K d
Kd mv Kd
Kd mv Kd
0
1d is frac12 the height from the base to the peak of a cycle
An experiment of moving paper under a pendulum through a small angle filled with sand that can escape through a small hole in the bottom shows that a pendulum cycling back and forth follows a curve that is mathematically analogous to a lsquosinrsquo or lsquocosrsquo
The projection of the curved area onto the straight line exactly duplicate the back and forth motion of CM This allows for the math of lsquosinrsquo or lsquocosrsquo to be used
SM will use lsquosinrdquo math It could use lsquocosrsquo or a combination of both lsquosinrsquo and lsquocosrsquo Using just the lsquosinrsquo math makes the understanding and math easier
0 sin 0 0SM 0 sin157 1SM The diagram at the left shows a spring being stretched from its (zero rest) position-2 and the PAR being released at position-1 There are 4 sections (1 to 2 2 to 3 3 to 4 and 4 to5) that alternate between maximum acceleration and zero velocity to max velocity and zero acceleration
At position-1 there is maximum acceleration and zero velocity o The lsquosinrsquo values from 0o to 157o will be used for velocity (values from 0 to 1) o The lsquosinrsquo values from 157o to 0o will be used for acceleration (values from 1 to 0)
At position-2 there is maximum velocity and zero acceleration Acceleration changes to deceleration
o The lsquosinrsquo values from 157o to 0o will be used for deceleration (values from 1 to 0) o The lsquosinrsquo values from 0o to 157o will be used for acceleration (values from 0 to 1)
At position-3 there is maximum acceleration and zero velocity o The lsquosinrsquo values from 0o to 157o will be used for velocity (values from 0 to 1) o The lsquosinrsquo values from 157o to 0o will be used for acceleration (values from 1 to 0)
At position-4 there is maximum velocity and zero acceleration Acceleration changes to deceleration
o The lsquosinrsquo values from 157o to 0o will be used for deceleration (values from 1 to 0) o The lsquosinrsquo values from 0o to 157o will be used for acceleration (values from 0 to 1)
At position-5 the cycle starts again
2
)539(
cT
MK
spring constant (K) when mass (M) and cycle-time (T) are known
)(
))(89)((
111 md
uKgM
d
MG
d
FK sts
spring constant (K) when mass (M) and amplitude ( d1) are known
2 21( )
Kv d d
m
velocity (v) at any point during the cycle
velocity
dforM
Kdv
max
012 velocity (v2) maximum (when d = 0 at position-2 and 4)
acceleration (a1) maximum when spring constant (K) mass (M) and amplitude (d1) are known
acceleration (a1) maximum when amplitude (d1) and cycle-time (Tc) are known
acceleration (a) at any point during the cycle-time (T)
1
151
286
v
dTc
cycle-time (Tc) when amplitude (d1) and velocity (v1) are known
K
MTc 28651
cycle-time (Tc) when mass (M) and spring constant (K) are known
fTc
1
cycle-time (Tc) when the frequency is known
cTf
1
frequency (f) when the cycle-time (Tc) is known
of cycles
elapsed timef
frequency (f) when the number of cycles in an elapsed time is known
d1 amplitude (distance from rest to start position)
Also frac12 the distance from the base to the peak
3131 oror dM
Ka
251
3131 539
c
oror T
da
2539
cT
da
In December of 2006 I discovered the mathematical error in Einsteinrsquos Special Relativity Einstein used BS math and that math only works in an imaginary world
It cannot describe real symmetrical space
Einsteinrsquos Special Relativity should be abandoned and a theory of real space using SM should be used to replace it
Numerous articles on the Internet show Einsteinrsquos theory to be incorrect but no one in academia is willing to publish the information
Figure 1-10 A moving particle observed from two frames of reference in uniform translation
Einsteinrsquos Math Error in Special Relativity Transformation of Velocity (From Eisenberg- Fundamentals of Modern Physics pages 27-29) Consider the particle shown in figure (1-10) moving with velocity v as
seen in a frame of reference O
dx dy dz
Vx Vy Vzdt dt dt
dx dy dz
V x V y V zdy dt dt
We would like to evaluate the velocity v of the particle as seen in the frame of reference 0 which is itself moving relative to 0 with velocity v Measured in the 0 frame the velocity vector of the particle has com ponents
The same velocity vector as measured in the 0 frame has components
2
2
2
2
2
1
1
x vtx
v
cy y
z z
vxt
ctv
c
Now from equations (1-13) we know that the relation between the primed and the unprimed coordinates and times is
Take the differential of these equations remembering that v is a constant This gives
2
2
2
2
2
1
1
dx vdtdx
v
cdy dy
dz dz
vxdt
cdtv
c
2
2
2 2 2 2 2
2
2
2
11 ( )
1( ) 1
1
1
1
dx vdt
v dxdx vdt v
dx dx vdtc dt dtvdx vdx vdx v dxdt dt dt dt
dt dtc c c c
VxV x
Vx
VxV x
vv
vv
c
x
v
c
cV
( )
coordinate of particle relative to moving primed system
coordinate of particle relative to stationary system
velocity of primed coordina
V x
Vx
te system relative to stationaryv sys
tem
Page 28 Fundamentals of Modern Physics
(1-16)
According to Einstein this equation tells you how to transform the observed velocity from one frame of reference to another frame of reference First we note that as Vc and vc approachzero the equation approaches those which would be derived from the Galilean transformation
2 2
( ) 1 01 1 ( )
Vx v Vx vV
VxV x
Vx v
v cVxx
v
c
v
cc
Another property of this equation is that it is impossible to choose V and v such that Vrsquo the magnitude of the velocity which is seen in the new frame is greater than c
1
2
09 ( 09 ) 18009876543
( 09 ) 1811 (09 )
c c cV c c
cc
c
Example As seen by O particle 1 has velocity 09c in the direction of positive x and particle 2 has velocity 09c in the negative x direction
Einstein equation however predicts To evaluate the velocity of particle 1 with respect to particle 2 we transform from the O frame to the Orsquo frame moving in the negative x direction with velocity v = -09c using equation 1-16 We obtain
JK LOGIC will observe the particles separating at 18c
A confusing INCORRECT and ILLOGICAL answer
left is a dir rightsub of a
1
2
1
09 ( 09 ) 18009876543
181( 09 )1 ( 09 )
The multiplication of dir left time a dir right is ONLY a dir left
09
dir
direction leftdirection rightsubtraction
c c cV c c
cc
c
V
dir
22
22
09 18 18009876543
181( 081)1(09)(09)
1
left rightsub
sub
c c c cc c
ccc
c
Analyzing Einsteinrsquos BS math Equation
Using SM math a logical answer can be obtained
Cannot multiply a subtraction operator times a direction in space
Can only subtract and change its direction
Cannot multiply a direction to the left times a direction to the right
Cannot multiply opposite directions in space
Can only solve for the resultant
rightleft
subtraction
1
subtraction
2
sub of a dir to the left is a dir to the right
1 resultant 0 sub 2
2sub 2
2
09 09
09 amp 091
09 09 18 1
(0)1
(09) amp (09)1
c cV
c cc
c c cV
cc
cc
8
181
cc
SM math predicts a logical answer using logic
21st century scientists state in many articles that Einsteins General Relativity and Quantum Mechanics are incompatible
They are not only incompatible they are both mathematically incorrect
They both use illogical BS math
Any equations in math that use imaginary numbers should be abandoned
Imaginary numbers are only necessary in the current BS math system
This means that Schroumldingers equations should be abandoned
Any equation in quantum mechanics that uses imaginary number should be abandoned
Prof Kanarev has developed logical physics and chemistrytheories without the use of Quantum mechanics or relativity
A partial list of his theories coverbull Mechanics ndash replaces Newtonbull Photon - bull Electrons ndash Protons- Neutronsbull Spectroscopybull Nucleusbull Atoms amp Moleculesbull Electrodynamicsbull Heatbull Curvature of Space ndash replaces Schwarzcchildrsquos equation
Kanarevrsquos Hypothesis
bull Electron align linearly with proton They do NOT spin or form clouds around the nucleus
bull Neutrons are in the center of atom Proton are outside of Neutrons
bull The wavelength of all particles is the experimental radius
bull Double slit experiment explained correctly
bull Heisenbergrsquos inequality explained correctly
bull is the radius of approach of the magnetic lines around the torus of the electron
( 1 137)
bull Why water expands when it freezes
bull Electron clusters form lightning
rbull The meaning of Plancks constant is explained
22 2 kg m
h m f mr f consts
2210254 42m r kgm bull If the mr-constant describes a ring the equations
describe experiments bull If the mr-constant describes a wave the equations do
NOT describe experiments
Kanarevrsquos 5 Equations of Mechanics
Kanarevrsquos 1st Equation Acceleration ampmv maMma R R
Kanarevrsquos 2nd Equation Uniform motion amp mv maMma R R
maM mvR R
Kanarevrsquos 3rd Equation
Kanarevrsquos 4th Equation Equal and opposite reaction
1 21 21 2
12
a m
m a m ama
Kanarevrsquos 5th Equation 1 2 3 Mi M M M Mna R R R R R
The sum of the Actions working on a moving body is never equal to zero
Kanarevrsquos Photon model is a rolling ring
His equations accurately describe known experiments with photonsbull Electromagnetic radiation should be called photon radiationbull Diffraction and interference are the same phenomenabull Correctly explains the double slit experiment without quantum equationsbull Correctly explains the meaning of Heisenbergrsquos inequality
e
2 (2)(3142)(2817 15 ) 10007297352568
R 2426 12 137mag
xp
r m
m
Charge
Magnetic action
Proton
The intensity of the magnetic action near the geometrical center of the proton ring is so great that we theorize that it is the strong nuclear force
(9383 8)(1602 19)8476 14
4 (4)(3142)(1411 26)p
pp
EB T
M
(9395 8)(1602 19)1035 15
4 (4)(3142)(0966 26)N
NN
EB T
M
Neutron
1a Hydrogenavi
From ldquoWhat Is Mathematicsrdquo by Courant and Robbins ndash page 55The Rule of Signs (-1) (-1) = +1
Which we set up to govern the multiplication of negative integers is a consequence of our desire to preserve the distribution law
a(b+c) = ab + ac
It took a long time for mathematicians to realize that the ldquoRule-of-signsrdquo together with all the other definitions governing negative integers and fractions
They were created by us in order to attain the freedom of operation while preserving the fundamental laws of arithmetic
Cannot be ldquoPROVEDrdquo
According to Professor Philip Kanarev there has not been a major discovery in physics or chemistry by ldquopeer-reviewing-academiardquo due to THEORY since the early 1900rsquos
There have been virtually no science-changing discoveries since the theories of quantum mechanics and relativity were introduced
One of the reasons is that the math being used is BS math and it operates on Broken-Symmetry and can only provide usable answers
BS math cannot be used to describe reality in space or time Space is symmetrical Space cannot depend on the direction one orientates a coordinate system
BS mathematicians should have discovered the problem Instead they invented symbols and definitions and bypassed the real problem ie [(imaginary numbers i2 = -1) and (absolutely values |-X|=+X)]
The Mathematical Principle of Error Joseph A Rybczyk 1996
Once an error enters a calculation all calculations after that point become an extension of the error
Studying the dash-cross codes of BS math has allowed me to break the codes and understand the reasons for the errors
Dash sign (-) CODES In Math used as
22
1 15 004255
bull (-) the number of zeros to the right of a decimal point 10-5=000001
bull (-) a subtraction operator
bull (-) a direction in space labeled negative (whatever that means)
bull (-) an exponent to mean divide
bull (-) describe the negative half of a sin or cos graph
bull (-) describe the negative half of an ldquoerdquo and ldquo1erdquo graph
bull (-) describe the negative side of all numbers raised to a power x
In Physics
bull (-) An Electron has been labeled as negative
bull (-) Anti-Particles have been labeled as negative
In Chemistry
bull (-) amp (+) Thermochemistry
Cross sign (+) CODES
(+) a direction in space labeled positive (whatever that means)
(+) an addition operator
(+) used to show the number of zeros to the left of a decimal point 10+5 = 100000
In Physics and Chemistry
bull (+) Protons have been Labeled as positive
bull (+) numerous other particles have been Labeled as positive
bull (+) amp (-) Thermochemistry
In Math used
The same dash symbol (-) is used for the numerous different math operations
The definitions established for the use of the dash symbol (-) do not distinguish between their different operations
In many math operations the dash symbol is changed to mean one of the other meanings
It is amazing that math has proceeded to its current level of use with this illogical use of a symbol
The same cross symbol (+) is used for numerous different math operations and produces the same illogical answers
A Summary of the BS Broken-Symmetry math
1 1 cosh tanhln
x xe x xe
bull Originated with broken-symmetry in the XYZ plane
bull Originated with mirror-broken-symmetry in the XZ plane
bull Created a Rule-of-Signs that cannot be proved (-)(-)=(+) (+)(+)=(+) (-)(+)=(-)
bull Created imaginary numbers to compensate for broken-symmetry
bull Created absolute values to change illogical negative number answers into positive number answers
bull Produces broken symmetry graphs for many functions
It took four years of study to discover why a (-)(-)=(+) only works in an imaginary world where space in the dash (-) direction is different from space in the cross (+) direction
Started working on the problem in August of 2001
Data points created using BS math produce graphs that are not symmetrical dash side different from cross side
However if a real problems data points follow a non symmetrical graph the graph can provide usable answers
This is why no one discovered the problem
Solved in January of 2005
Symmetry-Math(SM) vs Broken-Symmetry(BS) Math
In SM if we specify that the dash sign (-) means only subtraction and the cross sign (+) means only addition and an appropriate symbol is used for a direction in space space becomes symmetrical and math becomes logical
Math operators and directions in space are not the same and the same symbol should not be used to represent them
Letrsquos start with the (dash)multiplied by a (dash)=(cross) in the BS system
Usable answers may be obtained but it will be for illogical reasons
multiplied by a ( ) ( ) ( )subtraction Subtraction addition
rightleft left
dash dash cross
positivenegative negative
Symmetry-Math The subtraction of a direction is equal to the opposite direction The answers are correct using correct logic
opposite direction direction
SM
(negative)(negative) = (positive)
( ) ( ) ( )subtraction
BS
There is NO multiplication of a subtraction operator by a direction in space
There is just the subtraction of a direction in space
If instead of labeling the left side of a coordinate system as a negative (-) the same as a subtraction operator we label it with an arrow ( ) to represent the direction then a subtraction from that arrow direction will be in the opposite direction ( )
( ) mutiplied by a ( ) ( )
negative Positive negative
subtraction addition subtraction
dash cross dash
left right left
A subtraction operator multiplied by an addition operator is equal to a subtraction operator This is illogical
BS Math
A negative multiplied by a positive is equal to a negative This is illogical
A direction to the left multiplied by a direction to the right is a direction to the left This is illogical and Einsteins math error in special relativity
A dash times a cross is equal to a dash This is illogical
A vector moving to the left multiplied by a vector moving to the right is equal to ONLY to a vector moving to the left that is eight orders of magnitude faster thanThe speed of light
opposite direction subtraction direction
SM ( ) ( )
BS (negative)(positive) = (negative)
Symmetry Math The subtraction of a direction is equal to the opposite direction
There is no multiplication of a subtractionoperator times a direction in space
( ) multiplied by a ( ) ( )
positive positive positiveaddition addition additioncross cross cross
direction direction direction
What is the meaning of multiplying addition operators Addition operators are not multiplication operators
BS Math
() multiplied by a direction ( )=
number
() multiplied by a direction ( )=
number
A number times a direction maintains the same direction
( ) ()
dirdiradd
( ) ()
dirdiradd
The addition of a direction is in the same direction There is no multiplication by addition operators
Symmetry Math
Future math books need to eliminate the use of the cross sign (+) to represent something labeled a positive direction in space and the dash sign (-) to represent something labeled a negative direction in space Space does not have positive and negative directions
Symbols that are logical and have no illogical representation should be adopted SM uses either arrows or symbols
Symmetry-Math Number Line
Symmetry-Math Rules
The dash sign (-) will have only one use subtraction
The cross sign (+) will have only one use addition
Directions have an arrow and a number
All observers see the same direction and magnitude
bull There is no multiplication of arrows
amp ampamp ampampamp
( 3 ) ( 5 ) ( 6 ) ( 65 ) (17256 ) ( 32993 )
dir dirdir dirdirdir
Negative and Positive Directions in Space
If I asked you to point to a negative direction in space which way would you point
Hopefully you will realize that there is no such thing as a negative direction in space
If I asked you to point to a positive direction in space which way would you point
Again hopefully you will realize that there is no such thing as a positive direction in space
Objects on the right side of the x-axis are positive (+)(+) = (+)
Objects on the left side of the x-axis are negative (-)(-) = (+) negative math is different from positive math This is illogical
Objects on the top of the y-axis are positive
Objects on the bottom of the y-axis are negative
Objects in the front of the z-axis are positive
Objects in the back of the z-axis are negative
Again BS math of the positive direction (x right y up and z front) is different from the BS math of the negative direction (x left y down and z back)
BS math is Broken-Symmetry
Mirror image is broken symmetry in the x and z axis If the left amp right-axis are
reversed symmetry is broken Obs-1 math is different from obs-2
If the front amp back-axis is reversed symmetry is broken Obs-1 math is different from obs-2
If the top amp bottom-axis are
reversed symmetry is not broken Obs-1 math is the same as obs-2
Math answers cannot depend on which side of a number line an observer sit
Absolute values for Displacement
Even for a simple displacement of an object BS math invented a definition and absolute values to obtain something labeled a positive number answer
I----------I----------I----------I----------I----------I----------I----------I----------I
-4 -3 -2 -1 0 +1 +2 +3 +4 a gtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtb cltltltltltltltltltltltltltltltltltltltltltltltltltltltb
If an object starts at ( 3)ia X and moves to ( X 3) and the to ( 1)b fb c X
BS Math provides a usable answer without a definition or absolute values
However If an object starts at
( 1) and moves to ( X 3) and the to ( 3)i b fa X b c X
Then absolute values and a definition must be used
---------I----------I----------I----------I----------I----------I----------I---------- -4 -3 -2 -1 0 +1 +2 +3 +4
a gtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtb cltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltb
BS Math makes up a definition that all displacements are positive Since this example provides an answer with something labeled a negative the positive definition must be applied and (-2) is changed to (+2) A definition is needed to arrive at a useable answer This is Illogical
In SM absolute values and a meaningless definition of ldquopositiverdquo are removed In SM objects and directions in space are defined by the direction and magnitude of the resultant of arrows
With SM you get total distance traveled by the object and the final direction of the displacement
22222 2 bababbaababababa If w e su b stitu te p o s itiv e n u m b e rs in to th e eq u a tio n th e an sw er w ill b e co rrec t If (a = 5 ) an d (b = 3 )
2
2
8
5 3 5 3 5 3 5 5 5 3 3 5 3 3 25 15 15 9 64
8 8 8 8 8 8 8 8 8 8 64
W h en tw o p o sitiv e n u m b ers a re a d d ed an d sq u a re d th e B S m a th d is tr ib u tiv e law p ro v id es a co rre c t an sw er 1
Illogical and Incorrect BS math for the distributive law
222 2 bababbabbaaabababa B y d e fin itio n in B S m a th n o n u m b er sq u a red c a n ev e r b e n eg a tiv e T h e re fo re n o sq u a red n u m b er c an ev e r b e a v e c to r m o v in g in th e d a sh ( -) o r n eg a tiv e d ire c tio n if it is sq u a red In B S m a th an d ad d itio n o p e ra to r m u ltip lied b y a su b trac tio n o p e ra to r is A L W A Y S eq u a l to a su b tra c tio n o p e ra to r If th is h ad n o t b e en u sed fo r th e la s t fo u r h u n d red y e a rs y o u w o u ld d ie lau g h in g a t th e u tte r ab su rd ity o f th e la s t s ta tem e n t T h is is illo g ica l an d v io la te s sy m m etry
If a = 5 an d b = 3 42222352
22
4930253)3)(5(253)5)(3()3)(5(5353535 22222 A s lo n g a s ldquo ardquo is g re a te r th an ldquob rdquo th e B S m a th d is tr ib u tiv e law p ro v id es co rrec t an sw ers 1
B S m a th b re a k s d o w n w h e n ldquo b rdquo is g rea ter th a n ldquo a rdquo
If a = 3 an d b = 5
2 2
2
2
3 5 2 4
2 ( 2) ( 2) 4
2 ( )( 2) ( )( 2) 4
A n y co m b in a tio n ex cep t th e m id d le o n e is illo g ica l an d in co rre c t B y d e fin itio n th e an sw er can n o t b e -4 in B S m a th B S m a th is in co rrec t
4253095)5)(3(235)3)(5()5)(3(3535353 22222 T h is is w h ere th e d is tr ib u tiv e la w p ro v id es in co rrec t an sw ers I t p ro d u ces an an sw er o f + 4 b e cau se b y d e fin iti o n n o n u m b er sq u a red in B S m a th c an b e a n eg a tiv e n u m b er U sin g in co rre c t B S m a th th e d is tr ib u tiv e law w ill g iv e an in co rrec t an sw ers o f + 4 T h is is in co rrec t an d illo g ica l 1
Since there can be no negative direction from a squared term in BS math BS math should be modified or abandoned
The BS distributive law should be modified or abandoned
The numbers +1+2+3+4 and ndash1-2-3-4hellip should be abandoned
BS math gives incorrect answers because their Rule-of-Signs allows multiplying an arrow going in one direction by an arrow going in another direction
Clearly this is not logical
Arrows should be used that are specific for directions
W h at d o es it m e an to m u ltip ly ( 4) b y (4 ) A s w e k n o w 1
m u ltip lica tio n is ju s t ad d itio n W h a t w o u ld ( 4) tim e (4 ) b e 2
In BS math the answer is by definition -16 unless you square -4 and then it is +16
In SM you cannot multiply arrows going in opposite directions
The middle two terms are not logical You cannot multiply opposite directions
This is a MAJOR error in BS math They multiply a dash [(-) a direction to the left] by a cross [(+) a direction to the right] And then ldquoby definitionrdquo label the answer dash (-) This is illogical and produces incorrect answers
2
2 2
This is Einsteinrsquos math error
in Special relativity
Magnitude 4 Direction 1 Magnitude 3 Direction 2 Magnitude 5 Direction 3 Relative to an observer at 0 4
Magnitude 4 Direction 1 Magnitude 3 Direction 2 Magnitude 5 Direction 3 Relative to an observer at 0 4
Magnitude and Direction (MD) An arrow has both a magnitude andA direction relative to a 2nd objectobserver
bull Magnitude and velocity (MV) relative to a 2nd objectobserver
bull Magnitude And acceleration (MA) relative to a 2nd objectobserver2
v da
t t
Introducing a new symbol the ldquoamprdquo sign The ldquoamprdquo sign means resultant ADD all the Arrows for each direction and then SUBTRACT to find the final resultant
1 1amp 1 2 2amp 2 4 If =
1 amp1 2 2 amp2 4 If =
amp 5 amp 3 2 If gt
amp 3 amp 5= 2 If gt
amp 0 3 amp 3=0 if =
only the resultant equals zero
the vectors do not - they are still there
Does not exist There is no multiplication of arrows
There are two arrows One has a magnitude of 3 One has a magnitude of 5 The maximum these two arrows can be is 8
Only the resultant of arrows exist
An SM arrow has a direction and a magnitude
Arrows and magnitudes are still there even if the resultant is zero
There are no negative SM Arrows
There are no positive SM Arrows
There are no ZERO SM Arrows A zero would negate the definition of an SM Arrow something with no magnitude and no direction cannot be the definition of an arrow that is defined as having magnitude and direction
bull A resultant is the addition and subtraction of the differences in magnitude and direction arrows
A B and C are the arrows (a) A amp B exist(b) A amp B amp C exist(c) A-B Do Not Exist in Sm there are no negatives or positives(d) A+B-C Do Not Exist in SM there are no negatives or positives
There are only resultants when there are multiple arrows
SM Graphs versus BS Graphs
Numbers Raised to Powers
When numbers are raised to powers the negative sides of all BS math graphs are ILLOGICAL and INCORRECT
Incorrect BS Math for 2x
2x
4 16 correct3 8 correct2 4 correct1 2 correct0 1 correct-1 005 incorrect-2 025 incorrect-3 0125 incorrect-4 00625 incorrect
negative side of graph is incorrectpositive side of graph is correct
16
8
42105025012500625
-4 -3 -2 -1 0 1 2 3 4
po
sit
ive
negative positive
2x BS Math
correctincorrect
In BS math the dash sign (-) used in this example represents three different operations They arebull (-) Subtractionbull (-) A direction to the leftbull (-) As an exponent to mean divide by
All positive numbers produce correct results and graphs
All negative numbers produce incorrect results and graphs
A direction of three units to the left (-3) is placed in the equation as an exponent
A direction in space (3 units toward the left) changes a whole number to a fraction by using the dash (-) as an exponent rather than a direction in space
This is illogical and incorrect BS math should have performed this operation as
Unfortunately the answers to problems can come out correct because data points of a real problem fit the curve
However the answers will not be due to logical reasoning
Because the equation produced a curve that real problem data points follow it has been falsely believed that the equations and graphs are correct
1
2
x
Incorrect BK graph for (12)x
(12)x
4 00625 correct3 0125 correct2 025 correct1 1 correct0 1 correct-1 2 incorrect-2 4 incorrect-3 8 incorrect-4 16 incorrect
negative side of graph is incorrectpositive side of graph is correct
00625012502505124
8
16
-5 -4 -3 -2 -1 0 1 2 3 4 5P
osit
ive
negative positive
BS Math for (12)x
correctincorrect
Incorrect BS math and graph for
All graphs are symmetrical with SM
Two BS Math Equations that Produce Incorrect Answers
2 0x x starting equation 2
2
x x x x
x x
Add lsquoxrsquo to both sides
2(1) 1 let lsquoxrsquo = 1
2 1 This is an incorrect answer 2(0)-(0)=0 The only answer that BK math can give is x=0
1
a b starting equation ( )( ) ( )( )a a a b multiply both sides by ( )a
( )( ) ( )( ) ( )( ) ( )( )a a b b a b b b subtract ( )( )b b from both sides
( )( ) ( )( )a b a b a b b factor ( )a b from both sides
( )a b b divide both sides by ( )a b
( )b b b replace ( )a with ( )b 2b b simplify 2(1) (1) Let ( )b = 1
2 1 Again BS math yields and incorrect answer 1
This incorrect result is a problem with the distributive law SM explains how this incorrect answer is obtained and the problem with the distributive law that allows the error
Page 148 ndash149 Barry Mazur ndash Imagining Numbers
3
2
31
Do the computation carefully on paper using the rules we agreed to Then ponder your answer which should be something of a surprise to you donrsquot stop there Think of what your answer might possibly mean or might imply
Solve this equation using the distributive law
JK definitely a surprise the answer wasincorrect and illogical
Gerhard Gade University Professor doing math at Harvard University
Here is the BS math answer that Barry Mazur calculated
2
31
2
31
2
31
1 3 1 3 1 3 3 3 2 2 3 1 3
2 2 2 2 4 4 4 4 4 4 2 2
dash
1 3 1 3 1 31
2 2 2 2 4 4
3 3
4 4
When the distributive law is used with the BS Rule-of-Signs and one of the signs is a (-) dash sign the middle two terms always cancel each other This leads to the incorrect answer of (-1) in the above equation
Barry Mazur did not go back and check his answer with the original equation If he had he would have known that the answer was incorrect
3
1 31
2
check answer by setting equal
3
331 3
12
1 31
2
Take the Cube root of both sides
1 3 2 multiply both sides by 2
3 3 subtract 1 from both sides
2 2
3 3 square both sides
3 9 This is an incorrect answer A simple equation check could have shown that the distributive law using BS math provides an incorrect answer
1
BS Illogical and Incorrect Numbers Raised to Powers
nC D n DC
642 DDD
In BS math only a positive number can have an answer if ldquonrdquo is an even number (246---)
All negative numbers have no solutions
642 DDD
With SM just add a direction arrow and any number gives the correct answer
or n nC D C D
3 3( 3 8 8 or 8 2 or 2 )n D C
( 2 16 C= 16 or 16 4 or 4 )n D
Since a negative number has no square root answer in BS math BS mathematicians invented imaginary numbers
i2=-1
24414 iiC n
These equations are easily solved without the use of imaginary numbers with SM math
If 4 then 4 2x
If 4 then 4 2x
The 360-degree circle is a 4400 years old extremely outdated system based on a year of 12 months of 30 days each that is 360 days
SM will use a more logical system that will have a circle of 6283185307o
d
( )
(deg )
d r radians
d r rees
the arc distance around
the circumference of a circle
d
radiusr
2 2
2 2
d r
dv rd t t
v dda rd t t t
1
d
r
( )av tv
The average angular velocity of a body is the rate of its angular displacement
( )0
liminstt
vt
The instantaneous angular velocity is the limit of the ratio
t
as t approaches zero
( )d
av
d
tv
The average circumference velocity of the body is the rate of its distance traveled d divided by the time t
dvv
r
1 dd vdrv
t t t r r
The equation of the relationship between the angular velocity and the velocity of a point on that body
( )o
av
v v v
t ta
The average angular acceleration is the change in the angular velocity
( )0
liminstt
v
ta
The instantaneous angular acceleration is the
limit of the ration of v
t
as t approaches
zero
daa
r
1
1
d
d s d
vva
t r tv a
r t r
The equation of the relationship between the angular acceleration and the acceleration of a point on the body
21
2v a t
If a body with an initial angular velocity ov has a constant
angular acceleration a it will turn through an angle in a time t
1
Straight Line Symbol Units Rotational Symbol Units Distance d m Angular
Distance radians or degrees
Velocity dt
v m
s Angular Velocity
V
t
radians or degrees
s
Acceleration vat
2ms
Angular Acceleration
VA
t
2
radians or degrees
s
Mass M Kg Mass distribution
dM 2Kgm
Force actionA Ma
2
Kgm
sNewton
Torque
dT M A 2
2
Kgm
s
Momentum Mv Kgm
s
Angular Momentum
dM V 2Kgm
s
Work (Force)(distance)
Ad 2
Kgmm
sNm Joule
Work T
Using AS arrow math to solve and direction in space problems using the equation 1 2
2 2 2 2 2 2( ) amp and ( ) amp 3
4
2 2 2
2 2
( ) amp
(5 ) 4 3
(5 ) 16 9
(5 ) 25
(5 ) (5 )
2 2(5 ) 4 3
(5 ) 16 9
(5 ) 25
(5 ) (5 )
In SM math Pythagorasrsquo theorem provides both magnitude and direction
In BS math Pythagorasrsquo theorem provides magnitude only
time set equal to zero
23
( ) ( 0 )10 1
x t xx
3( ) ( 0)
10 1W t f
In BS math this is an equation for a traveling wave
must use a square term in order to make the curve symmetric
NO SQUARE TERM IS NECESSARY
In SM math when t is set to ldquo0rdquo
This is a major fault in BS math
x f(x)
-4 0019
-3 0033
-2 0073
-1 0273
0 3000
1 0273
2 0073
3 0033
4 0019 -4 -3 -2 -1 0 1 2 3 400
05
10
15
20
25
30
35
BS wave (must use square term)
snap shot of wave at x = 0
Am
plit
ude
lt to gt lt to gt f(ltgt)
lt -4 0073
lt -3 0097
lt -2 0143
lt -1 0273
origin 0 3000
gt 1 0273
gt 2 0143
gt 3 0097
gt 4 0073lt0073 lt0097 lt0143 lt0273
3000
0273gt 0143gt 0097gt 0073gt
Wave moving
snap shot of wave
Am
plit
ude
lt4 lt3 lt2 lt1 0 1gt 2gt 3gt 4gt
3
2
1
In SM NO squared terms are needed
SPRING
An Action is needed to stretch or compress a spring to a starting point where it can execute cycle motion
tsA (ts to start) is the work required to move a mass attached to a spring a distance (d1) from
its initial rest position RAK 0 the spring constant is the stiffness of the spring
o A is attraction o R is repulsion o
ARRA
00 is for the back and forth motion through the zero initial rest position A
stretched spring goes from ldquoArdquo to ldquo0rdquo to ldquoRrdquo to ldquoArdquo to ldquo0rdquo to ldquoRrdquo d1 is the distance a mass is moved from its initial rest position
0
0 1ts A RA K d
Once ( )mad
is performed to move the mass to position d1 (against the restraint of the spring) a of the ( )mad
is transferred into the spring as potential attractionrepulsion (PAR) There is always
more ( )mad
performed than is stored as PAR When the spring is released from the restraining Ats the PAR starts the oscillation Conservation of ( )mad
requires that the PAR and the motion ( )mad
( )( )madM be a constant
( )amp ( ) constantmadPAR M 02
0 0
1
2A R A RPAR K d
02
( ) 0 0
1( )
2mad A R A RM m v
0 0 02 2 2
0 0 0 1
2 2 21
2 2 21
1 1 1amp
2 2 2
1 1 1amp
2 2 2
amp
A R A R A RK d m v K d
Kd mv Kd
Kd mv Kd
0
1d is frac12 the height from the base to the peak of a cycle
An experiment of moving paper under a pendulum through a small angle filled with sand that can escape through a small hole in the bottom shows that a pendulum cycling back and forth follows a curve that is mathematically analogous to a lsquosinrsquo or lsquocosrsquo
The projection of the curved area onto the straight line exactly duplicate the back and forth motion of CM This allows for the math of lsquosinrsquo or lsquocosrsquo to be used
SM will use lsquosinrdquo math It could use lsquocosrsquo or a combination of both lsquosinrsquo and lsquocosrsquo Using just the lsquosinrsquo math makes the understanding and math easier
0 sin 0 0SM 0 sin157 1SM The diagram at the left shows a spring being stretched from its (zero rest) position-2 and the PAR being released at position-1 There are 4 sections (1 to 2 2 to 3 3 to 4 and 4 to5) that alternate between maximum acceleration and zero velocity to max velocity and zero acceleration
At position-1 there is maximum acceleration and zero velocity o The lsquosinrsquo values from 0o to 157o will be used for velocity (values from 0 to 1) o The lsquosinrsquo values from 157o to 0o will be used for acceleration (values from 1 to 0)
At position-2 there is maximum velocity and zero acceleration Acceleration changes to deceleration
o The lsquosinrsquo values from 157o to 0o will be used for deceleration (values from 1 to 0) o The lsquosinrsquo values from 0o to 157o will be used for acceleration (values from 0 to 1)
At position-3 there is maximum acceleration and zero velocity o The lsquosinrsquo values from 0o to 157o will be used for velocity (values from 0 to 1) o The lsquosinrsquo values from 157o to 0o will be used for acceleration (values from 1 to 0)
At position-4 there is maximum velocity and zero acceleration Acceleration changes to deceleration
o The lsquosinrsquo values from 157o to 0o will be used for deceleration (values from 1 to 0) o The lsquosinrsquo values from 0o to 157o will be used for acceleration (values from 0 to 1)
At position-5 the cycle starts again
2
)539(
cT
MK
spring constant (K) when mass (M) and cycle-time (T) are known
)(
))(89)((
111 md
uKgM
d
MG
d
FK sts
spring constant (K) when mass (M) and amplitude ( d1) are known
2 21( )
Kv d d
m
velocity (v) at any point during the cycle
velocity
dforM
Kdv
max
012 velocity (v2) maximum (when d = 0 at position-2 and 4)
acceleration (a1) maximum when spring constant (K) mass (M) and amplitude (d1) are known
acceleration (a1) maximum when amplitude (d1) and cycle-time (Tc) are known
acceleration (a) at any point during the cycle-time (T)
1
151
286
v
dTc
cycle-time (Tc) when amplitude (d1) and velocity (v1) are known
K
MTc 28651
cycle-time (Tc) when mass (M) and spring constant (K) are known
fTc
1
cycle-time (Tc) when the frequency is known
cTf
1
frequency (f) when the cycle-time (Tc) is known
of cycles
elapsed timef
frequency (f) when the number of cycles in an elapsed time is known
d1 amplitude (distance from rest to start position)
Also frac12 the distance from the base to the peak
3131 oror dM
Ka
251
3131 539
c
oror T
da
2539
cT
da
In December of 2006 I discovered the mathematical error in Einsteinrsquos Special Relativity Einstein used BS math and that math only works in an imaginary world
It cannot describe real symmetrical space
Einsteinrsquos Special Relativity should be abandoned and a theory of real space using SM should be used to replace it
Numerous articles on the Internet show Einsteinrsquos theory to be incorrect but no one in academia is willing to publish the information
Figure 1-10 A moving particle observed from two frames of reference in uniform translation
Einsteinrsquos Math Error in Special Relativity Transformation of Velocity (From Eisenberg- Fundamentals of Modern Physics pages 27-29) Consider the particle shown in figure (1-10) moving with velocity v as
seen in a frame of reference O
dx dy dz
Vx Vy Vzdt dt dt
dx dy dz
V x V y V zdy dt dt
We would like to evaluate the velocity v of the particle as seen in the frame of reference 0 which is itself moving relative to 0 with velocity v Measured in the 0 frame the velocity vector of the particle has com ponents
The same velocity vector as measured in the 0 frame has components
2
2
2
2
2
1
1
x vtx
v
cy y
z z
vxt
ctv
c
Now from equations (1-13) we know that the relation between the primed and the unprimed coordinates and times is
Take the differential of these equations remembering that v is a constant This gives
2
2
2
2
2
1
1
dx vdtdx
v
cdy dy
dz dz
vxdt
cdtv
c
2
2
2 2 2 2 2
2
2
2
11 ( )
1( ) 1
1
1
1
dx vdt
v dxdx vdt v
dx dx vdtc dt dtvdx vdx vdx v dxdt dt dt dt
dt dtc c c c
VxV x
Vx
VxV x
vv
vv
c
x
v
c
cV
( )
coordinate of particle relative to moving primed system
coordinate of particle relative to stationary system
velocity of primed coordina
V x
Vx
te system relative to stationaryv sys
tem
Page 28 Fundamentals of Modern Physics
(1-16)
According to Einstein this equation tells you how to transform the observed velocity from one frame of reference to another frame of reference First we note that as Vc and vc approachzero the equation approaches those which would be derived from the Galilean transformation
2 2
( ) 1 01 1 ( )
Vx v Vx vV
VxV x
Vx v
v cVxx
v
c
v
cc
Another property of this equation is that it is impossible to choose V and v such that Vrsquo the magnitude of the velocity which is seen in the new frame is greater than c
1
2
09 ( 09 ) 18009876543
( 09 ) 1811 (09 )
c c cV c c
cc
c
Example As seen by O particle 1 has velocity 09c in the direction of positive x and particle 2 has velocity 09c in the negative x direction
Einstein equation however predicts To evaluate the velocity of particle 1 with respect to particle 2 we transform from the O frame to the Orsquo frame moving in the negative x direction with velocity v = -09c using equation 1-16 We obtain
JK LOGIC will observe the particles separating at 18c
A confusing INCORRECT and ILLOGICAL answer
left is a dir rightsub of a
1
2
1
09 ( 09 ) 18009876543
181( 09 )1 ( 09 )
The multiplication of dir left time a dir right is ONLY a dir left
09
dir
direction leftdirection rightsubtraction
c c cV c c
cc
c
V
dir
22
22
09 18 18009876543
181( 081)1(09)(09)
1
left rightsub
sub
c c c cc c
ccc
c
Analyzing Einsteinrsquos BS math Equation
Using SM math a logical answer can be obtained
Cannot multiply a subtraction operator times a direction in space
Can only subtract and change its direction
Cannot multiply a direction to the left times a direction to the right
Cannot multiply opposite directions in space
Can only solve for the resultant
rightleft
subtraction
1
subtraction
2
sub of a dir to the left is a dir to the right
1 resultant 0 sub 2
2sub 2
2
09 09
09 amp 091
09 09 18 1
(0)1
(09) amp (09)1
c cV
c cc
c c cV
cc
cc
8
181
cc
SM math predicts a logical answer using logic
21st century scientists state in many articles that Einsteins General Relativity and Quantum Mechanics are incompatible
They are not only incompatible they are both mathematically incorrect
They both use illogical BS math
Any equations in math that use imaginary numbers should be abandoned
Imaginary numbers are only necessary in the current BS math system
This means that Schroumldingers equations should be abandoned
Any equation in quantum mechanics that uses imaginary number should be abandoned
Prof Kanarev has developed logical physics and chemistrytheories without the use of Quantum mechanics or relativity
A partial list of his theories coverbull Mechanics ndash replaces Newtonbull Photon - bull Electrons ndash Protons- Neutronsbull Spectroscopybull Nucleusbull Atoms amp Moleculesbull Electrodynamicsbull Heatbull Curvature of Space ndash replaces Schwarzcchildrsquos equation
Kanarevrsquos Hypothesis
bull Electron align linearly with proton They do NOT spin or form clouds around the nucleus
bull Neutrons are in the center of atom Proton are outside of Neutrons
bull The wavelength of all particles is the experimental radius
bull Double slit experiment explained correctly
bull Heisenbergrsquos inequality explained correctly
bull is the radius of approach of the magnetic lines around the torus of the electron
( 1 137)
bull Why water expands when it freezes
bull Electron clusters form lightning
rbull The meaning of Plancks constant is explained
22 2 kg m
h m f mr f consts
2210254 42m r kgm bull If the mr-constant describes a ring the equations
describe experiments bull If the mr-constant describes a wave the equations do
NOT describe experiments
Kanarevrsquos 5 Equations of Mechanics
Kanarevrsquos 1st Equation Acceleration ampmv maMma R R
Kanarevrsquos 2nd Equation Uniform motion amp mv maMma R R
maM mvR R
Kanarevrsquos 3rd Equation
Kanarevrsquos 4th Equation Equal and opposite reaction
1 21 21 2
12
a m
m a m ama
Kanarevrsquos 5th Equation 1 2 3 Mi M M M Mna R R R R R
The sum of the Actions working on a moving body is never equal to zero
Kanarevrsquos Photon model is a rolling ring
His equations accurately describe known experiments with photonsbull Electromagnetic radiation should be called photon radiationbull Diffraction and interference are the same phenomenabull Correctly explains the double slit experiment without quantum equationsbull Correctly explains the meaning of Heisenbergrsquos inequality
e
2 (2)(3142)(2817 15 ) 10007297352568
R 2426 12 137mag
xp
r m
m
Charge
Magnetic action
Proton
The intensity of the magnetic action near the geometrical center of the proton ring is so great that we theorize that it is the strong nuclear force
(9383 8)(1602 19)8476 14
4 (4)(3142)(1411 26)p
pp
EB T
M
(9395 8)(1602 19)1035 15
4 (4)(3142)(0966 26)N
NN
EB T
M
Neutron
1a Hydrogenavi
According to Professor Philip Kanarev there has not been a major discovery in physics or chemistry by ldquopeer-reviewing-academiardquo due to THEORY since the early 1900rsquos
There have been virtually no science-changing discoveries since the theories of quantum mechanics and relativity were introduced
One of the reasons is that the math being used is BS math and it operates on Broken-Symmetry and can only provide usable answers
BS math cannot be used to describe reality in space or time Space is symmetrical Space cannot depend on the direction one orientates a coordinate system
BS mathematicians should have discovered the problem Instead they invented symbols and definitions and bypassed the real problem ie [(imaginary numbers i2 = -1) and (absolutely values |-X|=+X)]
The Mathematical Principle of Error Joseph A Rybczyk 1996
Once an error enters a calculation all calculations after that point become an extension of the error
Studying the dash-cross codes of BS math has allowed me to break the codes and understand the reasons for the errors
Dash sign (-) CODES In Math used as
22
1 15 004255
bull (-) the number of zeros to the right of a decimal point 10-5=000001
bull (-) a subtraction operator
bull (-) a direction in space labeled negative (whatever that means)
bull (-) an exponent to mean divide
bull (-) describe the negative half of a sin or cos graph
bull (-) describe the negative half of an ldquoerdquo and ldquo1erdquo graph
bull (-) describe the negative side of all numbers raised to a power x
In Physics
bull (-) An Electron has been labeled as negative
bull (-) Anti-Particles have been labeled as negative
In Chemistry
bull (-) amp (+) Thermochemistry
Cross sign (+) CODES
(+) a direction in space labeled positive (whatever that means)
(+) an addition operator
(+) used to show the number of zeros to the left of a decimal point 10+5 = 100000
In Physics and Chemistry
bull (+) Protons have been Labeled as positive
bull (+) numerous other particles have been Labeled as positive
bull (+) amp (-) Thermochemistry
In Math used
The same dash symbol (-) is used for the numerous different math operations
The definitions established for the use of the dash symbol (-) do not distinguish between their different operations
In many math operations the dash symbol is changed to mean one of the other meanings
It is amazing that math has proceeded to its current level of use with this illogical use of a symbol
The same cross symbol (+) is used for numerous different math operations and produces the same illogical answers
A Summary of the BS Broken-Symmetry math
1 1 cosh tanhln
x xe x xe
bull Originated with broken-symmetry in the XYZ plane
bull Originated with mirror-broken-symmetry in the XZ plane
bull Created a Rule-of-Signs that cannot be proved (-)(-)=(+) (+)(+)=(+) (-)(+)=(-)
bull Created imaginary numbers to compensate for broken-symmetry
bull Created absolute values to change illogical negative number answers into positive number answers
bull Produces broken symmetry graphs for many functions
It took four years of study to discover why a (-)(-)=(+) only works in an imaginary world where space in the dash (-) direction is different from space in the cross (+) direction
Started working on the problem in August of 2001
Data points created using BS math produce graphs that are not symmetrical dash side different from cross side
However if a real problems data points follow a non symmetrical graph the graph can provide usable answers
This is why no one discovered the problem
Solved in January of 2005
Symmetry-Math(SM) vs Broken-Symmetry(BS) Math
In SM if we specify that the dash sign (-) means only subtraction and the cross sign (+) means only addition and an appropriate symbol is used for a direction in space space becomes symmetrical and math becomes logical
Math operators and directions in space are not the same and the same symbol should not be used to represent them
Letrsquos start with the (dash)multiplied by a (dash)=(cross) in the BS system
Usable answers may be obtained but it will be for illogical reasons
multiplied by a ( ) ( ) ( )subtraction Subtraction addition
rightleft left
dash dash cross
positivenegative negative
Symmetry-Math The subtraction of a direction is equal to the opposite direction The answers are correct using correct logic
opposite direction direction
SM
(negative)(negative) = (positive)
( ) ( ) ( )subtraction
BS
There is NO multiplication of a subtraction operator by a direction in space
There is just the subtraction of a direction in space
If instead of labeling the left side of a coordinate system as a negative (-) the same as a subtraction operator we label it with an arrow ( ) to represent the direction then a subtraction from that arrow direction will be in the opposite direction ( )
( ) mutiplied by a ( ) ( )
negative Positive negative
subtraction addition subtraction
dash cross dash
left right left
A subtraction operator multiplied by an addition operator is equal to a subtraction operator This is illogical
BS Math
A negative multiplied by a positive is equal to a negative This is illogical
A direction to the left multiplied by a direction to the right is a direction to the left This is illogical and Einsteins math error in special relativity
A dash times a cross is equal to a dash This is illogical
A vector moving to the left multiplied by a vector moving to the right is equal to ONLY to a vector moving to the left that is eight orders of magnitude faster thanThe speed of light
opposite direction subtraction direction
SM ( ) ( )
BS (negative)(positive) = (negative)
Symmetry Math The subtraction of a direction is equal to the opposite direction
There is no multiplication of a subtractionoperator times a direction in space
( ) multiplied by a ( ) ( )
positive positive positiveaddition addition additioncross cross cross
direction direction direction
What is the meaning of multiplying addition operators Addition operators are not multiplication operators
BS Math
() multiplied by a direction ( )=
number
() multiplied by a direction ( )=
number
A number times a direction maintains the same direction
( ) ()
dirdiradd
( ) ()
dirdiradd
The addition of a direction is in the same direction There is no multiplication by addition operators
Symmetry Math
Future math books need to eliminate the use of the cross sign (+) to represent something labeled a positive direction in space and the dash sign (-) to represent something labeled a negative direction in space Space does not have positive and negative directions
Symbols that are logical and have no illogical representation should be adopted SM uses either arrows or symbols
Symmetry-Math Number Line
Symmetry-Math Rules
The dash sign (-) will have only one use subtraction
The cross sign (+) will have only one use addition
Directions have an arrow and a number
All observers see the same direction and magnitude
bull There is no multiplication of arrows
amp ampamp ampampamp
( 3 ) ( 5 ) ( 6 ) ( 65 ) (17256 ) ( 32993 )
dir dirdir dirdirdir
Negative and Positive Directions in Space
If I asked you to point to a negative direction in space which way would you point
Hopefully you will realize that there is no such thing as a negative direction in space
If I asked you to point to a positive direction in space which way would you point
Again hopefully you will realize that there is no such thing as a positive direction in space
Objects on the right side of the x-axis are positive (+)(+) = (+)
Objects on the left side of the x-axis are negative (-)(-) = (+) negative math is different from positive math This is illogical
Objects on the top of the y-axis are positive
Objects on the bottom of the y-axis are negative
Objects in the front of the z-axis are positive
Objects in the back of the z-axis are negative
Again BS math of the positive direction (x right y up and z front) is different from the BS math of the negative direction (x left y down and z back)
BS math is Broken-Symmetry
Mirror image is broken symmetry in the x and z axis If the left amp right-axis are
reversed symmetry is broken Obs-1 math is different from obs-2
If the front amp back-axis is reversed symmetry is broken Obs-1 math is different from obs-2
If the top amp bottom-axis are
reversed symmetry is not broken Obs-1 math is the same as obs-2
Math answers cannot depend on which side of a number line an observer sit
Absolute values for Displacement
Even for a simple displacement of an object BS math invented a definition and absolute values to obtain something labeled a positive number answer
I----------I----------I----------I----------I----------I----------I----------I----------I
-4 -3 -2 -1 0 +1 +2 +3 +4 a gtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtb cltltltltltltltltltltltltltltltltltltltltltltltltltltltb
If an object starts at ( 3)ia X and moves to ( X 3) and the to ( 1)b fb c X
BS Math provides a usable answer without a definition or absolute values
However If an object starts at
( 1) and moves to ( X 3) and the to ( 3)i b fa X b c X
Then absolute values and a definition must be used
---------I----------I----------I----------I----------I----------I----------I---------- -4 -3 -2 -1 0 +1 +2 +3 +4
a gtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtb cltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltb
BS Math makes up a definition that all displacements are positive Since this example provides an answer with something labeled a negative the positive definition must be applied and (-2) is changed to (+2) A definition is needed to arrive at a useable answer This is Illogical
In SM absolute values and a meaningless definition of ldquopositiverdquo are removed In SM objects and directions in space are defined by the direction and magnitude of the resultant of arrows
With SM you get total distance traveled by the object and the final direction of the displacement
22222 2 bababbaababababa If w e su b stitu te p o s itiv e n u m b e rs in to th e eq u a tio n th e an sw er w ill b e co rrec t If (a = 5 ) an d (b = 3 )
2
2
8
5 3 5 3 5 3 5 5 5 3 3 5 3 3 25 15 15 9 64
8 8 8 8 8 8 8 8 8 8 64
W h en tw o p o sitiv e n u m b ers a re a d d ed an d sq u a re d th e B S m a th d is tr ib u tiv e law p ro v id es a co rre c t an sw er 1
Illogical and Incorrect BS math for the distributive law
222 2 bababbabbaaabababa B y d e fin itio n in B S m a th n o n u m b er sq u a red c a n ev e r b e n eg a tiv e T h e re fo re n o sq u a red n u m b er c an ev e r b e a v e c to r m o v in g in th e d a sh ( -) o r n eg a tiv e d ire c tio n if it is sq u a red In B S m a th an d ad d itio n o p e ra to r m u ltip lied b y a su b trac tio n o p e ra to r is A L W A Y S eq u a l to a su b tra c tio n o p e ra to r If th is h ad n o t b e en u sed fo r th e la s t fo u r h u n d red y e a rs y o u w o u ld d ie lau g h in g a t th e u tte r ab su rd ity o f th e la s t s ta tem e n t T h is is illo g ica l an d v io la te s sy m m etry
If a = 5 an d b = 3 42222352
22
4930253)3)(5(253)5)(3()3)(5(5353535 22222 A s lo n g a s ldquo ardquo is g re a te r th an ldquob rdquo th e B S m a th d is tr ib u tiv e law p ro v id es co rrec t an sw ers 1
B S m a th b re a k s d o w n w h e n ldquo b rdquo is g rea ter th a n ldquo a rdquo
If a = 3 an d b = 5
2 2
2
2
3 5 2 4
2 ( 2) ( 2) 4
2 ( )( 2) ( )( 2) 4
A n y co m b in a tio n ex cep t th e m id d le o n e is illo g ica l an d in co rre c t B y d e fin itio n th e an sw er can n o t b e -4 in B S m a th B S m a th is in co rrec t
4253095)5)(3(235)3)(5()5)(3(3535353 22222 T h is is w h ere th e d is tr ib u tiv e la w p ro v id es in co rrec t an sw ers I t p ro d u ces an an sw er o f + 4 b e cau se b y d e fin iti o n n o n u m b er sq u a red in B S m a th c an b e a n eg a tiv e n u m b er U sin g in co rre c t B S m a th th e d is tr ib u tiv e law w ill g iv e an in co rrec t an sw ers o f + 4 T h is is in co rrec t an d illo g ica l 1
Since there can be no negative direction from a squared term in BS math BS math should be modified or abandoned
The BS distributive law should be modified or abandoned
The numbers +1+2+3+4 and ndash1-2-3-4hellip should be abandoned
BS math gives incorrect answers because their Rule-of-Signs allows multiplying an arrow going in one direction by an arrow going in another direction
Clearly this is not logical
Arrows should be used that are specific for directions
W h at d o es it m e an to m u ltip ly ( 4) b y (4 ) A s w e k n o w 1
m u ltip lica tio n is ju s t ad d itio n W h a t w o u ld ( 4) tim e (4 ) b e 2
In BS math the answer is by definition -16 unless you square -4 and then it is +16
In SM you cannot multiply arrows going in opposite directions
The middle two terms are not logical You cannot multiply opposite directions
This is a MAJOR error in BS math They multiply a dash [(-) a direction to the left] by a cross [(+) a direction to the right] And then ldquoby definitionrdquo label the answer dash (-) This is illogical and produces incorrect answers
2
2 2
This is Einsteinrsquos math error
in Special relativity
Magnitude 4 Direction 1 Magnitude 3 Direction 2 Magnitude 5 Direction 3 Relative to an observer at 0 4
Magnitude 4 Direction 1 Magnitude 3 Direction 2 Magnitude 5 Direction 3 Relative to an observer at 0 4
Magnitude and Direction (MD) An arrow has both a magnitude andA direction relative to a 2nd objectobserver
bull Magnitude and velocity (MV) relative to a 2nd objectobserver
bull Magnitude And acceleration (MA) relative to a 2nd objectobserver2
v da
t t
Introducing a new symbol the ldquoamprdquo sign The ldquoamprdquo sign means resultant ADD all the Arrows for each direction and then SUBTRACT to find the final resultant
1 1amp 1 2 2amp 2 4 If =
1 amp1 2 2 amp2 4 If =
amp 5 amp 3 2 If gt
amp 3 amp 5= 2 If gt
amp 0 3 amp 3=0 if =
only the resultant equals zero
the vectors do not - they are still there
Does not exist There is no multiplication of arrows
There are two arrows One has a magnitude of 3 One has a magnitude of 5 The maximum these two arrows can be is 8
Only the resultant of arrows exist
An SM arrow has a direction and a magnitude
Arrows and magnitudes are still there even if the resultant is zero
There are no negative SM Arrows
There are no positive SM Arrows
There are no ZERO SM Arrows A zero would negate the definition of an SM Arrow something with no magnitude and no direction cannot be the definition of an arrow that is defined as having magnitude and direction
bull A resultant is the addition and subtraction of the differences in magnitude and direction arrows
A B and C are the arrows (a) A amp B exist(b) A amp B amp C exist(c) A-B Do Not Exist in Sm there are no negatives or positives(d) A+B-C Do Not Exist in SM there are no negatives or positives
There are only resultants when there are multiple arrows
SM Graphs versus BS Graphs
Numbers Raised to Powers
When numbers are raised to powers the negative sides of all BS math graphs are ILLOGICAL and INCORRECT
Incorrect BS Math for 2x
2x
4 16 correct3 8 correct2 4 correct1 2 correct0 1 correct-1 005 incorrect-2 025 incorrect-3 0125 incorrect-4 00625 incorrect
negative side of graph is incorrectpositive side of graph is correct
16
8
42105025012500625
-4 -3 -2 -1 0 1 2 3 4
po
sit
ive
negative positive
2x BS Math
correctincorrect
In BS math the dash sign (-) used in this example represents three different operations They arebull (-) Subtractionbull (-) A direction to the leftbull (-) As an exponent to mean divide by
All positive numbers produce correct results and graphs
All negative numbers produce incorrect results and graphs
A direction of three units to the left (-3) is placed in the equation as an exponent
A direction in space (3 units toward the left) changes a whole number to a fraction by using the dash (-) as an exponent rather than a direction in space
This is illogical and incorrect BS math should have performed this operation as
Unfortunately the answers to problems can come out correct because data points of a real problem fit the curve
However the answers will not be due to logical reasoning
Because the equation produced a curve that real problem data points follow it has been falsely believed that the equations and graphs are correct
1
2
x
Incorrect BK graph for (12)x
(12)x
4 00625 correct3 0125 correct2 025 correct1 1 correct0 1 correct-1 2 incorrect-2 4 incorrect-3 8 incorrect-4 16 incorrect
negative side of graph is incorrectpositive side of graph is correct
00625012502505124
8
16
-5 -4 -3 -2 -1 0 1 2 3 4 5P
osit
ive
negative positive
BS Math for (12)x
correctincorrect
Incorrect BS math and graph for
All graphs are symmetrical with SM
Two BS Math Equations that Produce Incorrect Answers
2 0x x starting equation 2
2
x x x x
x x
Add lsquoxrsquo to both sides
2(1) 1 let lsquoxrsquo = 1
2 1 This is an incorrect answer 2(0)-(0)=0 The only answer that BK math can give is x=0
1
a b starting equation ( )( ) ( )( )a a a b multiply both sides by ( )a
( )( ) ( )( ) ( )( ) ( )( )a a b b a b b b subtract ( )( )b b from both sides
( )( ) ( )( )a b a b a b b factor ( )a b from both sides
( )a b b divide both sides by ( )a b
( )b b b replace ( )a with ( )b 2b b simplify 2(1) (1) Let ( )b = 1
2 1 Again BS math yields and incorrect answer 1
This incorrect result is a problem with the distributive law SM explains how this incorrect answer is obtained and the problem with the distributive law that allows the error
Page 148 ndash149 Barry Mazur ndash Imagining Numbers
3
2
31
Do the computation carefully on paper using the rules we agreed to Then ponder your answer which should be something of a surprise to you donrsquot stop there Think of what your answer might possibly mean or might imply
Solve this equation using the distributive law
JK definitely a surprise the answer wasincorrect and illogical
Gerhard Gade University Professor doing math at Harvard University
Here is the BS math answer that Barry Mazur calculated
2
31
2
31
2
31
1 3 1 3 1 3 3 3 2 2 3 1 3
2 2 2 2 4 4 4 4 4 4 2 2
dash
1 3 1 3 1 31
2 2 2 2 4 4
3 3
4 4
When the distributive law is used with the BS Rule-of-Signs and one of the signs is a (-) dash sign the middle two terms always cancel each other This leads to the incorrect answer of (-1) in the above equation
Barry Mazur did not go back and check his answer with the original equation If he had he would have known that the answer was incorrect
3
1 31
2
check answer by setting equal
3
331 3
12
1 31
2
Take the Cube root of both sides
1 3 2 multiply both sides by 2
3 3 subtract 1 from both sides
2 2
3 3 square both sides
3 9 This is an incorrect answer A simple equation check could have shown that the distributive law using BS math provides an incorrect answer
1
BS Illogical and Incorrect Numbers Raised to Powers
nC D n DC
642 DDD
In BS math only a positive number can have an answer if ldquonrdquo is an even number (246---)
All negative numbers have no solutions
642 DDD
With SM just add a direction arrow and any number gives the correct answer
or n nC D C D
3 3( 3 8 8 or 8 2 or 2 )n D C
( 2 16 C= 16 or 16 4 or 4 )n D
Since a negative number has no square root answer in BS math BS mathematicians invented imaginary numbers
i2=-1
24414 iiC n
These equations are easily solved without the use of imaginary numbers with SM math
If 4 then 4 2x
If 4 then 4 2x
The 360-degree circle is a 4400 years old extremely outdated system based on a year of 12 months of 30 days each that is 360 days
SM will use a more logical system that will have a circle of 6283185307o
d
( )
(deg )
d r radians
d r rees
the arc distance around
the circumference of a circle
d
radiusr
2 2
2 2
d r
dv rd t t
v dda rd t t t
1
d
r
( )av tv
The average angular velocity of a body is the rate of its angular displacement
( )0
liminstt
vt
The instantaneous angular velocity is the limit of the ratio
t
as t approaches zero
( )d
av
d
tv
The average circumference velocity of the body is the rate of its distance traveled d divided by the time t
dvv
r
1 dd vdrv
t t t r r
The equation of the relationship between the angular velocity and the velocity of a point on that body
( )o
av
v v v
t ta
The average angular acceleration is the change in the angular velocity
( )0
liminstt
v
ta
The instantaneous angular acceleration is the
limit of the ration of v
t
as t approaches
zero
daa
r
1
1
d
d s d
vva
t r tv a
r t r
The equation of the relationship between the angular acceleration and the acceleration of a point on the body
21
2v a t
If a body with an initial angular velocity ov has a constant
angular acceleration a it will turn through an angle in a time t
1
Straight Line Symbol Units Rotational Symbol Units Distance d m Angular
Distance radians or degrees
Velocity dt
v m
s Angular Velocity
V
t
radians or degrees
s
Acceleration vat
2ms
Angular Acceleration
VA
t
2
radians or degrees
s
Mass M Kg Mass distribution
dM 2Kgm
Force actionA Ma
2
Kgm
sNewton
Torque
dT M A 2
2
Kgm
s
Momentum Mv Kgm
s
Angular Momentum
dM V 2Kgm
s
Work (Force)(distance)
Ad 2
Kgmm
sNm Joule
Work T
Using AS arrow math to solve and direction in space problems using the equation 1 2
2 2 2 2 2 2( ) amp and ( ) amp 3
4
2 2 2
2 2
( ) amp
(5 ) 4 3
(5 ) 16 9
(5 ) 25
(5 ) (5 )
2 2(5 ) 4 3
(5 ) 16 9
(5 ) 25
(5 ) (5 )
In SM math Pythagorasrsquo theorem provides both magnitude and direction
In BS math Pythagorasrsquo theorem provides magnitude only
time set equal to zero
23
( ) ( 0 )10 1
x t xx
3( ) ( 0)
10 1W t f
In BS math this is an equation for a traveling wave
must use a square term in order to make the curve symmetric
NO SQUARE TERM IS NECESSARY
In SM math when t is set to ldquo0rdquo
This is a major fault in BS math
x f(x)
-4 0019
-3 0033
-2 0073
-1 0273
0 3000
1 0273
2 0073
3 0033
4 0019 -4 -3 -2 -1 0 1 2 3 400
05
10
15
20
25
30
35
BS wave (must use square term)
snap shot of wave at x = 0
Am
plit
ude
lt to gt lt to gt f(ltgt)
lt -4 0073
lt -3 0097
lt -2 0143
lt -1 0273
origin 0 3000
gt 1 0273
gt 2 0143
gt 3 0097
gt 4 0073lt0073 lt0097 lt0143 lt0273
3000
0273gt 0143gt 0097gt 0073gt
Wave moving
snap shot of wave
Am
plit
ude
lt4 lt3 lt2 lt1 0 1gt 2gt 3gt 4gt
3
2
1
In SM NO squared terms are needed
SPRING
An Action is needed to stretch or compress a spring to a starting point where it can execute cycle motion
tsA (ts to start) is the work required to move a mass attached to a spring a distance (d1) from
its initial rest position RAK 0 the spring constant is the stiffness of the spring
o A is attraction o R is repulsion o
ARRA
00 is for the back and forth motion through the zero initial rest position A
stretched spring goes from ldquoArdquo to ldquo0rdquo to ldquoRrdquo to ldquoArdquo to ldquo0rdquo to ldquoRrdquo d1 is the distance a mass is moved from its initial rest position
0
0 1ts A RA K d
Once ( )mad
is performed to move the mass to position d1 (against the restraint of the spring) a of the ( )mad
is transferred into the spring as potential attractionrepulsion (PAR) There is always
more ( )mad
performed than is stored as PAR When the spring is released from the restraining Ats the PAR starts the oscillation Conservation of ( )mad
requires that the PAR and the motion ( )mad
( )( )madM be a constant
( )amp ( ) constantmadPAR M 02
0 0
1
2A R A RPAR K d
02
( ) 0 0
1( )
2mad A R A RM m v
0 0 02 2 2
0 0 0 1
2 2 21
2 2 21
1 1 1amp
2 2 2
1 1 1amp
2 2 2
amp
A R A R A RK d m v K d
Kd mv Kd
Kd mv Kd
0
1d is frac12 the height from the base to the peak of a cycle
An experiment of moving paper under a pendulum through a small angle filled with sand that can escape through a small hole in the bottom shows that a pendulum cycling back and forth follows a curve that is mathematically analogous to a lsquosinrsquo or lsquocosrsquo
The projection of the curved area onto the straight line exactly duplicate the back and forth motion of CM This allows for the math of lsquosinrsquo or lsquocosrsquo to be used
SM will use lsquosinrdquo math It could use lsquocosrsquo or a combination of both lsquosinrsquo and lsquocosrsquo Using just the lsquosinrsquo math makes the understanding and math easier
0 sin 0 0SM 0 sin157 1SM The diagram at the left shows a spring being stretched from its (zero rest) position-2 and the PAR being released at position-1 There are 4 sections (1 to 2 2 to 3 3 to 4 and 4 to5) that alternate between maximum acceleration and zero velocity to max velocity and zero acceleration
At position-1 there is maximum acceleration and zero velocity o The lsquosinrsquo values from 0o to 157o will be used for velocity (values from 0 to 1) o The lsquosinrsquo values from 157o to 0o will be used for acceleration (values from 1 to 0)
At position-2 there is maximum velocity and zero acceleration Acceleration changes to deceleration
o The lsquosinrsquo values from 157o to 0o will be used for deceleration (values from 1 to 0) o The lsquosinrsquo values from 0o to 157o will be used for acceleration (values from 0 to 1)
At position-3 there is maximum acceleration and zero velocity o The lsquosinrsquo values from 0o to 157o will be used for velocity (values from 0 to 1) o The lsquosinrsquo values from 157o to 0o will be used for acceleration (values from 1 to 0)
At position-4 there is maximum velocity and zero acceleration Acceleration changes to deceleration
o The lsquosinrsquo values from 157o to 0o will be used for deceleration (values from 1 to 0) o The lsquosinrsquo values from 0o to 157o will be used for acceleration (values from 0 to 1)
At position-5 the cycle starts again
2
)539(
cT
MK
spring constant (K) when mass (M) and cycle-time (T) are known
)(
))(89)((
111 md
uKgM
d
MG
d
FK sts
spring constant (K) when mass (M) and amplitude ( d1) are known
2 21( )
Kv d d
m
velocity (v) at any point during the cycle
velocity
dforM
Kdv
max
012 velocity (v2) maximum (when d = 0 at position-2 and 4)
acceleration (a1) maximum when spring constant (K) mass (M) and amplitude (d1) are known
acceleration (a1) maximum when amplitude (d1) and cycle-time (Tc) are known
acceleration (a) at any point during the cycle-time (T)
1
151
286
v
dTc
cycle-time (Tc) when amplitude (d1) and velocity (v1) are known
K
MTc 28651
cycle-time (Tc) when mass (M) and spring constant (K) are known
fTc
1
cycle-time (Tc) when the frequency is known
cTf
1
frequency (f) when the cycle-time (Tc) is known
of cycles
elapsed timef
frequency (f) when the number of cycles in an elapsed time is known
d1 amplitude (distance from rest to start position)
Also frac12 the distance from the base to the peak
3131 oror dM
Ka
251
3131 539
c
oror T
da
2539
cT
da
In December of 2006 I discovered the mathematical error in Einsteinrsquos Special Relativity Einstein used BS math and that math only works in an imaginary world
It cannot describe real symmetrical space
Einsteinrsquos Special Relativity should be abandoned and a theory of real space using SM should be used to replace it
Numerous articles on the Internet show Einsteinrsquos theory to be incorrect but no one in academia is willing to publish the information
Figure 1-10 A moving particle observed from two frames of reference in uniform translation
Einsteinrsquos Math Error in Special Relativity Transformation of Velocity (From Eisenberg- Fundamentals of Modern Physics pages 27-29) Consider the particle shown in figure (1-10) moving with velocity v as
seen in a frame of reference O
dx dy dz
Vx Vy Vzdt dt dt
dx dy dz
V x V y V zdy dt dt
We would like to evaluate the velocity v of the particle as seen in the frame of reference 0 which is itself moving relative to 0 with velocity v Measured in the 0 frame the velocity vector of the particle has com ponents
The same velocity vector as measured in the 0 frame has components
2
2
2
2
2
1
1
x vtx
v
cy y
z z
vxt
ctv
c
Now from equations (1-13) we know that the relation between the primed and the unprimed coordinates and times is
Take the differential of these equations remembering that v is a constant This gives
2
2
2
2
2
1
1
dx vdtdx
v
cdy dy
dz dz
vxdt
cdtv
c
2
2
2 2 2 2 2
2
2
2
11 ( )
1( ) 1
1
1
1
dx vdt
v dxdx vdt v
dx dx vdtc dt dtvdx vdx vdx v dxdt dt dt dt
dt dtc c c c
VxV x
Vx
VxV x
vv
vv
c
x
v
c
cV
( )
coordinate of particle relative to moving primed system
coordinate of particle relative to stationary system
velocity of primed coordina
V x
Vx
te system relative to stationaryv sys
tem
Page 28 Fundamentals of Modern Physics
(1-16)
According to Einstein this equation tells you how to transform the observed velocity from one frame of reference to another frame of reference First we note that as Vc and vc approachzero the equation approaches those which would be derived from the Galilean transformation
2 2
( ) 1 01 1 ( )
Vx v Vx vV
VxV x
Vx v
v cVxx
v
c
v
cc
Another property of this equation is that it is impossible to choose V and v such that Vrsquo the magnitude of the velocity which is seen in the new frame is greater than c
1
2
09 ( 09 ) 18009876543
( 09 ) 1811 (09 )
c c cV c c
cc
c
Example As seen by O particle 1 has velocity 09c in the direction of positive x and particle 2 has velocity 09c in the negative x direction
Einstein equation however predicts To evaluate the velocity of particle 1 with respect to particle 2 we transform from the O frame to the Orsquo frame moving in the negative x direction with velocity v = -09c using equation 1-16 We obtain
JK LOGIC will observe the particles separating at 18c
A confusing INCORRECT and ILLOGICAL answer
left is a dir rightsub of a
1
2
1
09 ( 09 ) 18009876543
181( 09 )1 ( 09 )
The multiplication of dir left time a dir right is ONLY a dir left
09
dir
direction leftdirection rightsubtraction
c c cV c c
cc
c
V
dir
22
22
09 18 18009876543
181( 081)1(09)(09)
1
left rightsub
sub
c c c cc c
ccc
c
Analyzing Einsteinrsquos BS math Equation
Using SM math a logical answer can be obtained
Cannot multiply a subtraction operator times a direction in space
Can only subtract and change its direction
Cannot multiply a direction to the left times a direction to the right
Cannot multiply opposite directions in space
Can only solve for the resultant
rightleft
subtraction
1
subtraction
2
sub of a dir to the left is a dir to the right
1 resultant 0 sub 2
2sub 2
2
09 09
09 amp 091
09 09 18 1
(0)1
(09) amp (09)1
c cV
c cc
c c cV
cc
cc
8
181
cc
SM math predicts a logical answer using logic
21st century scientists state in many articles that Einsteins General Relativity and Quantum Mechanics are incompatible
They are not only incompatible they are both mathematically incorrect
They both use illogical BS math
Any equations in math that use imaginary numbers should be abandoned
Imaginary numbers are only necessary in the current BS math system
This means that Schroumldingers equations should be abandoned
Any equation in quantum mechanics that uses imaginary number should be abandoned
Prof Kanarev has developed logical physics and chemistrytheories without the use of Quantum mechanics or relativity
A partial list of his theories coverbull Mechanics ndash replaces Newtonbull Photon - bull Electrons ndash Protons- Neutronsbull Spectroscopybull Nucleusbull Atoms amp Moleculesbull Electrodynamicsbull Heatbull Curvature of Space ndash replaces Schwarzcchildrsquos equation
Kanarevrsquos Hypothesis
bull Electron align linearly with proton They do NOT spin or form clouds around the nucleus
bull Neutrons are in the center of atom Proton are outside of Neutrons
bull The wavelength of all particles is the experimental radius
bull Double slit experiment explained correctly
bull Heisenbergrsquos inequality explained correctly
bull is the radius of approach of the magnetic lines around the torus of the electron
( 1 137)
bull Why water expands when it freezes
bull Electron clusters form lightning
rbull The meaning of Plancks constant is explained
22 2 kg m
h m f mr f consts
2210254 42m r kgm bull If the mr-constant describes a ring the equations
describe experiments bull If the mr-constant describes a wave the equations do
NOT describe experiments
Kanarevrsquos 5 Equations of Mechanics
Kanarevrsquos 1st Equation Acceleration ampmv maMma R R
Kanarevrsquos 2nd Equation Uniform motion amp mv maMma R R
maM mvR R
Kanarevrsquos 3rd Equation
Kanarevrsquos 4th Equation Equal and opposite reaction
1 21 21 2
12
a m
m a m ama
Kanarevrsquos 5th Equation 1 2 3 Mi M M M Mna R R R R R
The sum of the Actions working on a moving body is never equal to zero
Kanarevrsquos Photon model is a rolling ring
His equations accurately describe known experiments with photonsbull Electromagnetic radiation should be called photon radiationbull Diffraction and interference are the same phenomenabull Correctly explains the double slit experiment without quantum equationsbull Correctly explains the meaning of Heisenbergrsquos inequality
e
2 (2)(3142)(2817 15 ) 10007297352568
R 2426 12 137mag
xp
r m
m
Charge
Magnetic action
Proton
The intensity of the magnetic action near the geometrical center of the proton ring is so great that we theorize that it is the strong nuclear force
(9383 8)(1602 19)8476 14
4 (4)(3142)(1411 26)p
pp
EB T
M
(9395 8)(1602 19)1035 15
4 (4)(3142)(0966 26)N
NN
EB T
M
Neutron
1a Hydrogenavi
The Mathematical Principle of Error Joseph A Rybczyk 1996
Once an error enters a calculation all calculations after that point become an extension of the error
Studying the dash-cross codes of BS math has allowed me to break the codes and understand the reasons for the errors
Dash sign (-) CODES In Math used as
22
1 15 004255
bull (-) the number of zeros to the right of a decimal point 10-5=000001
bull (-) a subtraction operator
bull (-) a direction in space labeled negative (whatever that means)
bull (-) an exponent to mean divide
bull (-) describe the negative half of a sin or cos graph
bull (-) describe the negative half of an ldquoerdquo and ldquo1erdquo graph
bull (-) describe the negative side of all numbers raised to a power x
In Physics
bull (-) An Electron has been labeled as negative
bull (-) Anti-Particles have been labeled as negative
In Chemistry
bull (-) amp (+) Thermochemistry
Cross sign (+) CODES
(+) a direction in space labeled positive (whatever that means)
(+) an addition operator
(+) used to show the number of zeros to the left of a decimal point 10+5 = 100000
In Physics and Chemistry
bull (+) Protons have been Labeled as positive
bull (+) numerous other particles have been Labeled as positive
bull (+) amp (-) Thermochemistry
In Math used
The same dash symbol (-) is used for the numerous different math operations
The definitions established for the use of the dash symbol (-) do not distinguish between their different operations
In many math operations the dash symbol is changed to mean one of the other meanings
It is amazing that math has proceeded to its current level of use with this illogical use of a symbol
The same cross symbol (+) is used for numerous different math operations and produces the same illogical answers
A Summary of the BS Broken-Symmetry math
1 1 cosh tanhln
x xe x xe
bull Originated with broken-symmetry in the XYZ plane
bull Originated with mirror-broken-symmetry in the XZ plane
bull Created a Rule-of-Signs that cannot be proved (-)(-)=(+) (+)(+)=(+) (-)(+)=(-)
bull Created imaginary numbers to compensate for broken-symmetry
bull Created absolute values to change illogical negative number answers into positive number answers
bull Produces broken symmetry graphs for many functions
It took four years of study to discover why a (-)(-)=(+) only works in an imaginary world where space in the dash (-) direction is different from space in the cross (+) direction
Started working on the problem in August of 2001
Data points created using BS math produce graphs that are not symmetrical dash side different from cross side
However if a real problems data points follow a non symmetrical graph the graph can provide usable answers
This is why no one discovered the problem
Solved in January of 2005
Symmetry-Math(SM) vs Broken-Symmetry(BS) Math
In SM if we specify that the dash sign (-) means only subtraction and the cross sign (+) means only addition and an appropriate symbol is used for a direction in space space becomes symmetrical and math becomes logical
Math operators and directions in space are not the same and the same symbol should not be used to represent them
Letrsquos start with the (dash)multiplied by a (dash)=(cross) in the BS system
Usable answers may be obtained but it will be for illogical reasons
multiplied by a ( ) ( ) ( )subtraction Subtraction addition
rightleft left
dash dash cross
positivenegative negative
Symmetry-Math The subtraction of a direction is equal to the opposite direction The answers are correct using correct logic
opposite direction direction
SM
(negative)(negative) = (positive)
( ) ( ) ( )subtraction
BS
There is NO multiplication of a subtraction operator by a direction in space
There is just the subtraction of a direction in space
If instead of labeling the left side of a coordinate system as a negative (-) the same as a subtraction operator we label it with an arrow ( ) to represent the direction then a subtraction from that arrow direction will be in the opposite direction ( )
( ) mutiplied by a ( ) ( )
negative Positive negative
subtraction addition subtraction
dash cross dash
left right left
A subtraction operator multiplied by an addition operator is equal to a subtraction operator This is illogical
BS Math
A negative multiplied by a positive is equal to a negative This is illogical
A direction to the left multiplied by a direction to the right is a direction to the left This is illogical and Einsteins math error in special relativity
A dash times a cross is equal to a dash This is illogical
A vector moving to the left multiplied by a vector moving to the right is equal to ONLY to a vector moving to the left that is eight orders of magnitude faster thanThe speed of light
opposite direction subtraction direction
SM ( ) ( )
BS (negative)(positive) = (negative)
Symmetry Math The subtraction of a direction is equal to the opposite direction
There is no multiplication of a subtractionoperator times a direction in space
( ) multiplied by a ( ) ( )
positive positive positiveaddition addition additioncross cross cross
direction direction direction
What is the meaning of multiplying addition operators Addition operators are not multiplication operators
BS Math
() multiplied by a direction ( )=
number
() multiplied by a direction ( )=
number
A number times a direction maintains the same direction
( ) ()
dirdiradd
( ) ()
dirdiradd
The addition of a direction is in the same direction There is no multiplication by addition operators
Symmetry Math
Future math books need to eliminate the use of the cross sign (+) to represent something labeled a positive direction in space and the dash sign (-) to represent something labeled a negative direction in space Space does not have positive and negative directions
Symbols that are logical and have no illogical representation should be adopted SM uses either arrows or symbols
Symmetry-Math Number Line
Symmetry-Math Rules
The dash sign (-) will have only one use subtraction
The cross sign (+) will have only one use addition
Directions have an arrow and a number
All observers see the same direction and magnitude
bull There is no multiplication of arrows
amp ampamp ampampamp
( 3 ) ( 5 ) ( 6 ) ( 65 ) (17256 ) ( 32993 )
dir dirdir dirdirdir
Negative and Positive Directions in Space
If I asked you to point to a negative direction in space which way would you point
Hopefully you will realize that there is no such thing as a negative direction in space
If I asked you to point to a positive direction in space which way would you point
Again hopefully you will realize that there is no such thing as a positive direction in space
Objects on the right side of the x-axis are positive (+)(+) = (+)
Objects on the left side of the x-axis are negative (-)(-) = (+) negative math is different from positive math This is illogical
Objects on the top of the y-axis are positive
Objects on the bottom of the y-axis are negative
Objects in the front of the z-axis are positive
Objects in the back of the z-axis are negative
Again BS math of the positive direction (x right y up and z front) is different from the BS math of the negative direction (x left y down and z back)
BS math is Broken-Symmetry
Mirror image is broken symmetry in the x and z axis If the left amp right-axis are
reversed symmetry is broken Obs-1 math is different from obs-2
If the front amp back-axis is reversed symmetry is broken Obs-1 math is different from obs-2
If the top amp bottom-axis are
reversed symmetry is not broken Obs-1 math is the same as obs-2
Math answers cannot depend on which side of a number line an observer sit
Absolute values for Displacement
Even for a simple displacement of an object BS math invented a definition and absolute values to obtain something labeled a positive number answer
I----------I----------I----------I----------I----------I----------I----------I----------I
-4 -3 -2 -1 0 +1 +2 +3 +4 a gtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtb cltltltltltltltltltltltltltltltltltltltltltltltltltltltb
If an object starts at ( 3)ia X and moves to ( X 3) and the to ( 1)b fb c X
BS Math provides a usable answer without a definition or absolute values
However If an object starts at
( 1) and moves to ( X 3) and the to ( 3)i b fa X b c X
Then absolute values and a definition must be used
---------I----------I----------I----------I----------I----------I----------I---------- -4 -3 -2 -1 0 +1 +2 +3 +4
a gtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtb cltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltb
BS Math makes up a definition that all displacements are positive Since this example provides an answer with something labeled a negative the positive definition must be applied and (-2) is changed to (+2) A definition is needed to arrive at a useable answer This is Illogical
In SM absolute values and a meaningless definition of ldquopositiverdquo are removed In SM objects and directions in space are defined by the direction and magnitude of the resultant of arrows
With SM you get total distance traveled by the object and the final direction of the displacement
22222 2 bababbaababababa If w e su b stitu te p o s itiv e n u m b e rs in to th e eq u a tio n th e an sw er w ill b e co rrec t If (a = 5 ) an d (b = 3 )
2
2
8
5 3 5 3 5 3 5 5 5 3 3 5 3 3 25 15 15 9 64
8 8 8 8 8 8 8 8 8 8 64
W h en tw o p o sitiv e n u m b ers a re a d d ed an d sq u a re d th e B S m a th d is tr ib u tiv e law p ro v id es a co rre c t an sw er 1
Illogical and Incorrect BS math for the distributive law
222 2 bababbabbaaabababa B y d e fin itio n in B S m a th n o n u m b er sq u a red c a n ev e r b e n eg a tiv e T h e re fo re n o sq u a red n u m b er c an ev e r b e a v e c to r m o v in g in th e d a sh ( -) o r n eg a tiv e d ire c tio n if it is sq u a red In B S m a th an d ad d itio n o p e ra to r m u ltip lied b y a su b trac tio n o p e ra to r is A L W A Y S eq u a l to a su b tra c tio n o p e ra to r If th is h ad n o t b e en u sed fo r th e la s t fo u r h u n d red y e a rs y o u w o u ld d ie lau g h in g a t th e u tte r ab su rd ity o f th e la s t s ta tem e n t T h is is illo g ica l an d v io la te s sy m m etry
If a = 5 an d b = 3 42222352
22
4930253)3)(5(253)5)(3()3)(5(5353535 22222 A s lo n g a s ldquo ardquo is g re a te r th an ldquob rdquo th e B S m a th d is tr ib u tiv e law p ro v id es co rrec t an sw ers 1
B S m a th b re a k s d o w n w h e n ldquo b rdquo is g rea ter th a n ldquo a rdquo
If a = 3 an d b = 5
2 2
2
2
3 5 2 4
2 ( 2) ( 2) 4
2 ( )( 2) ( )( 2) 4
A n y co m b in a tio n ex cep t th e m id d le o n e is illo g ica l an d in co rre c t B y d e fin itio n th e an sw er can n o t b e -4 in B S m a th B S m a th is in co rrec t
4253095)5)(3(235)3)(5()5)(3(3535353 22222 T h is is w h ere th e d is tr ib u tiv e la w p ro v id es in co rrec t an sw ers I t p ro d u ces an an sw er o f + 4 b e cau se b y d e fin iti o n n o n u m b er sq u a red in B S m a th c an b e a n eg a tiv e n u m b er U sin g in co rre c t B S m a th th e d is tr ib u tiv e law w ill g iv e an in co rrec t an sw ers o f + 4 T h is is in co rrec t an d illo g ica l 1
Since there can be no negative direction from a squared term in BS math BS math should be modified or abandoned
The BS distributive law should be modified or abandoned
The numbers +1+2+3+4 and ndash1-2-3-4hellip should be abandoned
BS math gives incorrect answers because their Rule-of-Signs allows multiplying an arrow going in one direction by an arrow going in another direction
Clearly this is not logical
Arrows should be used that are specific for directions
W h at d o es it m e an to m u ltip ly ( 4) b y (4 ) A s w e k n o w 1
m u ltip lica tio n is ju s t ad d itio n W h a t w o u ld ( 4) tim e (4 ) b e 2
In BS math the answer is by definition -16 unless you square -4 and then it is +16
In SM you cannot multiply arrows going in opposite directions
The middle two terms are not logical You cannot multiply opposite directions
This is a MAJOR error in BS math They multiply a dash [(-) a direction to the left] by a cross [(+) a direction to the right] And then ldquoby definitionrdquo label the answer dash (-) This is illogical and produces incorrect answers
2
2 2
This is Einsteinrsquos math error
in Special relativity
Magnitude 4 Direction 1 Magnitude 3 Direction 2 Magnitude 5 Direction 3 Relative to an observer at 0 4
Magnitude 4 Direction 1 Magnitude 3 Direction 2 Magnitude 5 Direction 3 Relative to an observer at 0 4
Magnitude and Direction (MD) An arrow has both a magnitude andA direction relative to a 2nd objectobserver
bull Magnitude and velocity (MV) relative to a 2nd objectobserver
bull Magnitude And acceleration (MA) relative to a 2nd objectobserver2
v da
t t
Introducing a new symbol the ldquoamprdquo sign The ldquoamprdquo sign means resultant ADD all the Arrows for each direction and then SUBTRACT to find the final resultant
1 1amp 1 2 2amp 2 4 If =
1 amp1 2 2 amp2 4 If =
amp 5 amp 3 2 If gt
amp 3 amp 5= 2 If gt
amp 0 3 amp 3=0 if =
only the resultant equals zero
the vectors do not - they are still there
Does not exist There is no multiplication of arrows
There are two arrows One has a magnitude of 3 One has a magnitude of 5 The maximum these two arrows can be is 8
Only the resultant of arrows exist
An SM arrow has a direction and a magnitude
Arrows and magnitudes are still there even if the resultant is zero
There are no negative SM Arrows
There are no positive SM Arrows
There are no ZERO SM Arrows A zero would negate the definition of an SM Arrow something with no magnitude and no direction cannot be the definition of an arrow that is defined as having magnitude and direction
bull A resultant is the addition and subtraction of the differences in magnitude and direction arrows
A B and C are the arrows (a) A amp B exist(b) A amp B amp C exist(c) A-B Do Not Exist in Sm there are no negatives or positives(d) A+B-C Do Not Exist in SM there are no negatives or positives
There are only resultants when there are multiple arrows
SM Graphs versus BS Graphs
Numbers Raised to Powers
When numbers are raised to powers the negative sides of all BS math graphs are ILLOGICAL and INCORRECT
Incorrect BS Math for 2x
2x
4 16 correct3 8 correct2 4 correct1 2 correct0 1 correct-1 005 incorrect-2 025 incorrect-3 0125 incorrect-4 00625 incorrect
negative side of graph is incorrectpositive side of graph is correct
16
8
42105025012500625
-4 -3 -2 -1 0 1 2 3 4
po
sit
ive
negative positive
2x BS Math
correctincorrect
In BS math the dash sign (-) used in this example represents three different operations They arebull (-) Subtractionbull (-) A direction to the leftbull (-) As an exponent to mean divide by
All positive numbers produce correct results and graphs
All negative numbers produce incorrect results and graphs
A direction of three units to the left (-3) is placed in the equation as an exponent
A direction in space (3 units toward the left) changes a whole number to a fraction by using the dash (-) as an exponent rather than a direction in space
This is illogical and incorrect BS math should have performed this operation as
Unfortunately the answers to problems can come out correct because data points of a real problem fit the curve
However the answers will not be due to logical reasoning
Because the equation produced a curve that real problem data points follow it has been falsely believed that the equations and graphs are correct
1
2
x
Incorrect BK graph for (12)x
(12)x
4 00625 correct3 0125 correct2 025 correct1 1 correct0 1 correct-1 2 incorrect-2 4 incorrect-3 8 incorrect-4 16 incorrect
negative side of graph is incorrectpositive side of graph is correct
00625012502505124
8
16
-5 -4 -3 -2 -1 0 1 2 3 4 5P
osit
ive
negative positive
BS Math for (12)x
correctincorrect
Incorrect BS math and graph for
All graphs are symmetrical with SM
Two BS Math Equations that Produce Incorrect Answers
2 0x x starting equation 2
2
x x x x
x x
Add lsquoxrsquo to both sides
2(1) 1 let lsquoxrsquo = 1
2 1 This is an incorrect answer 2(0)-(0)=0 The only answer that BK math can give is x=0
1
a b starting equation ( )( ) ( )( )a a a b multiply both sides by ( )a
( )( ) ( )( ) ( )( ) ( )( )a a b b a b b b subtract ( )( )b b from both sides
( )( ) ( )( )a b a b a b b factor ( )a b from both sides
( )a b b divide both sides by ( )a b
( )b b b replace ( )a with ( )b 2b b simplify 2(1) (1) Let ( )b = 1
2 1 Again BS math yields and incorrect answer 1
This incorrect result is a problem with the distributive law SM explains how this incorrect answer is obtained and the problem with the distributive law that allows the error
Page 148 ndash149 Barry Mazur ndash Imagining Numbers
3
2
31
Do the computation carefully on paper using the rules we agreed to Then ponder your answer which should be something of a surprise to you donrsquot stop there Think of what your answer might possibly mean or might imply
Solve this equation using the distributive law
JK definitely a surprise the answer wasincorrect and illogical
Gerhard Gade University Professor doing math at Harvard University
Here is the BS math answer that Barry Mazur calculated
2
31
2
31
2
31
1 3 1 3 1 3 3 3 2 2 3 1 3
2 2 2 2 4 4 4 4 4 4 2 2
dash
1 3 1 3 1 31
2 2 2 2 4 4
3 3
4 4
When the distributive law is used with the BS Rule-of-Signs and one of the signs is a (-) dash sign the middle two terms always cancel each other This leads to the incorrect answer of (-1) in the above equation
Barry Mazur did not go back and check his answer with the original equation If he had he would have known that the answer was incorrect
3
1 31
2
check answer by setting equal
3
331 3
12
1 31
2
Take the Cube root of both sides
1 3 2 multiply both sides by 2
3 3 subtract 1 from both sides
2 2
3 3 square both sides
3 9 This is an incorrect answer A simple equation check could have shown that the distributive law using BS math provides an incorrect answer
1
BS Illogical and Incorrect Numbers Raised to Powers
nC D n DC
642 DDD
In BS math only a positive number can have an answer if ldquonrdquo is an even number (246---)
All negative numbers have no solutions
642 DDD
With SM just add a direction arrow and any number gives the correct answer
or n nC D C D
3 3( 3 8 8 or 8 2 or 2 )n D C
( 2 16 C= 16 or 16 4 or 4 )n D
Since a negative number has no square root answer in BS math BS mathematicians invented imaginary numbers
i2=-1
24414 iiC n
These equations are easily solved without the use of imaginary numbers with SM math
If 4 then 4 2x
If 4 then 4 2x
The 360-degree circle is a 4400 years old extremely outdated system based on a year of 12 months of 30 days each that is 360 days
SM will use a more logical system that will have a circle of 6283185307o
d
( )
(deg )
d r radians
d r rees
the arc distance around
the circumference of a circle
d
radiusr
2 2
2 2
d r
dv rd t t
v dda rd t t t
1
d
r
( )av tv
The average angular velocity of a body is the rate of its angular displacement
( )0
liminstt
vt
The instantaneous angular velocity is the limit of the ratio
t
as t approaches zero
( )d
av
d
tv
The average circumference velocity of the body is the rate of its distance traveled d divided by the time t
dvv
r
1 dd vdrv
t t t r r
The equation of the relationship between the angular velocity and the velocity of a point on that body
( )o
av
v v v
t ta
The average angular acceleration is the change in the angular velocity
( )0
liminstt
v
ta
The instantaneous angular acceleration is the
limit of the ration of v
t
as t approaches
zero
daa
r
1
1
d
d s d
vva
t r tv a
r t r
The equation of the relationship between the angular acceleration and the acceleration of a point on the body
21
2v a t
If a body with an initial angular velocity ov has a constant
angular acceleration a it will turn through an angle in a time t
1
Straight Line Symbol Units Rotational Symbol Units Distance d m Angular
Distance radians or degrees
Velocity dt
v m
s Angular Velocity
V
t
radians or degrees
s
Acceleration vat
2ms
Angular Acceleration
VA
t
2
radians or degrees
s
Mass M Kg Mass distribution
dM 2Kgm
Force actionA Ma
2
Kgm
sNewton
Torque
dT M A 2
2
Kgm
s
Momentum Mv Kgm
s
Angular Momentum
dM V 2Kgm
s
Work (Force)(distance)
Ad 2
Kgmm
sNm Joule
Work T
Using AS arrow math to solve and direction in space problems using the equation 1 2
2 2 2 2 2 2( ) amp and ( ) amp 3
4
2 2 2
2 2
( ) amp
(5 ) 4 3
(5 ) 16 9
(5 ) 25
(5 ) (5 )
2 2(5 ) 4 3
(5 ) 16 9
(5 ) 25
(5 ) (5 )
In SM math Pythagorasrsquo theorem provides both magnitude and direction
In BS math Pythagorasrsquo theorem provides magnitude only
time set equal to zero
23
( ) ( 0 )10 1
x t xx
3( ) ( 0)
10 1W t f
In BS math this is an equation for a traveling wave
must use a square term in order to make the curve symmetric
NO SQUARE TERM IS NECESSARY
In SM math when t is set to ldquo0rdquo
This is a major fault in BS math
x f(x)
-4 0019
-3 0033
-2 0073
-1 0273
0 3000
1 0273
2 0073
3 0033
4 0019 -4 -3 -2 -1 0 1 2 3 400
05
10
15
20
25
30
35
BS wave (must use square term)
snap shot of wave at x = 0
Am
plit
ude
lt to gt lt to gt f(ltgt)
lt -4 0073
lt -3 0097
lt -2 0143
lt -1 0273
origin 0 3000
gt 1 0273
gt 2 0143
gt 3 0097
gt 4 0073lt0073 lt0097 lt0143 lt0273
3000
0273gt 0143gt 0097gt 0073gt
Wave moving
snap shot of wave
Am
plit
ude
lt4 lt3 lt2 lt1 0 1gt 2gt 3gt 4gt
3
2
1
In SM NO squared terms are needed
SPRING
An Action is needed to stretch or compress a spring to a starting point where it can execute cycle motion
tsA (ts to start) is the work required to move a mass attached to a spring a distance (d1) from
its initial rest position RAK 0 the spring constant is the stiffness of the spring
o A is attraction o R is repulsion o
ARRA
00 is for the back and forth motion through the zero initial rest position A
stretched spring goes from ldquoArdquo to ldquo0rdquo to ldquoRrdquo to ldquoArdquo to ldquo0rdquo to ldquoRrdquo d1 is the distance a mass is moved from its initial rest position
0
0 1ts A RA K d
Once ( )mad
is performed to move the mass to position d1 (against the restraint of the spring) a of the ( )mad
is transferred into the spring as potential attractionrepulsion (PAR) There is always
more ( )mad
performed than is stored as PAR When the spring is released from the restraining Ats the PAR starts the oscillation Conservation of ( )mad
requires that the PAR and the motion ( )mad
( )( )madM be a constant
( )amp ( ) constantmadPAR M 02
0 0
1
2A R A RPAR K d
02
( ) 0 0
1( )
2mad A R A RM m v
0 0 02 2 2
0 0 0 1
2 2 21
2 2 21
1 1 1amp
2 2 2
1 1 1amp
2 2 2
amp
A R A R A RK d m v K d
Kd mv Kd
Kd mv Kd
0
1d is frac12 the height from the base to the peak of a cycle
An experiment of moving paper under a pendulum through a small angle filled with sand that can escape through a small hole in the bottom shows that a pendulum cycling back and forth follows a curve that is mathematically analogous to a lsquosinrsquo or lsquocosrsquo
The projection of the curved area onto the straight line exactly duplicate the back and forth motion of CM This allows for the math of lsquosinrsquo or lsquocosrsquo to be used
SM will use lsquosinrdquo math It could use lsquocosrsquo or a combination of both lsquosinrsquo and lsquocosrsquo Using just the lsquosinrsquo math makes the understanding and math easier
0 sin 0 0SM 0 sin157 1SM The diagram at the left shows a spring being stretched from its (zero rest) position-2 and the PAR being released at position-1 There are 4 sections (1 to 2 2 to 3 3 to 4 and 4 to5) that alternate between maximum acceleration and zero velocity to max velocity and zero acceleration
At position-1 there is maximum acceleration and zero velocity o The lsquosinrsquo values from 0o to 157o will be used for velocity (values from 0 to 1) o The lsquosinrsquo values from 157o to 0o will be used for acceleration (values from 1 to 0)
At position-2 there is maximum velocity and zero acceleration Acceleration changes to deceleration
o The lsquosinrsquo values from 157o to 0o will be used for deceleration (values from 1 to 0) o The lsquosinrsquo values from 0o to 157o will be used for acceleration (values from 0 to 1)
At position-3 there is maximum acceleration and zero velocity o The lsquosinrsquo values from 0o to 157o will be used for velocity (values from 0 to 1) o The lsquosinrsquo values from 157o to 0o will be used for acceleration (values from 1 to 0)
At position-4 there is maximum velocity and zero acceleration Acceleration changes to deceleration
o The lsquosinrsquo values from 157o to 0o will be used for deceleration (values from 1 to 0) o The lsquosinrsquo values from 0o to 157o will be used for acceleration (values from 0 to 1)
At position-5 the cycle starts again
2
)539(
cT
MK
spring constant (K) when mass (M) and cycle-time (T) are known
)(
))(89)((
111 md
uKgM
d
MG
d
FK sts
spring constant (K) when mass (M) and amplitude ( d1) are known
2 21( )
Kv d d
m
velocity (v) at any point during the cycle
velocity
dforM
Kdv
max
012 velocity (v2) maximum (when d = 0 at position-2 and 4)
acceleration (a1) maximum when spring constant (K) mass (M) and amplitude (d1) are known
acceleration (a1) maximum when amplitude (d1) and cycle-time (Tc) are known
acceleration (a) at any point during the cycle-time (T)
1
151
286
v
dTc
cycle-time (Tc) when amplitude (d1) and velocity (v1) are known
K
MTc 28651
cycle-time (Tc) when mass (M) and spring constant (K) are known
fTc
1
cycle-time (Tc) when the frequency is known
cTf
1
frequency (f) when the cycle-time (Tc) is known
of cycles
elapsed timef
frequency (f) when the number of cycles in an elapsed time is known
d1 amplitude (distance from rest to start position)
Also frac12 the distance from the base to the peak
3131 oror dM
Ka
251
3131 539
c
oror T
da
2539
cT
da
In December of 2006 I discovered the mathematical error in Einsteinrsquos Special Relativity Einstein used BS math and that math only works in an imaginary world
It cannot describe real symmetrical space
Einsteinrsquos Special Relativity should be abandoned and a theory of real space using SM should be used to replace it
Numerous articles on the Internet show Einsteinrsquos theory to be incorrect but no one in academia is willing to publish the information
Figure 1-10 A moving particle observed from two frames of reference in uniform translation
Einsteinrsquos Math Error in Special Relativity Transformation of Velocity (From Eisenberg- Fundamentals of Modern Physics pages 27-29) Consider the particle shown in figure (1-10) moving with velocity v as
seen in a frame of reference O
dx dy dz
Vx Vy Vzdt dt dt
dx dy dz
V x V y V zdy dt dt
We would like to evaluate the velocity v of the particle as seen in the frame of reference 0 which is itself moving relative to 0 with velocity v Measured in the 0 frame the velocity vector of the particle has com ponents
The same velocity vector as measured in the 0 frame has components
2
2
2
2
2
1
1
x vtx
v
cy y
z z
vxt
ctv
c
Now from equations (1-13) we know that the relation between the primed and the unprimed coordinates and times is
Take the differential of these equations remembering that v is a constant This gives
2
2
2
2
2
1
1
dx vdtdx
v
cdy dy
dz dz
vxdt
cdtv
c
2
2
2 2 2 2 2
2
2
2
11 ( )
1( ) 1
1
1
1
dx vdt
v dxdx vdt v
dx dx vdtc dt dtvdx vdx vdx v dxdt dt dt dt
dt dtc c c c
VxV x
Vx
VxV x
vv
vv
c
x
v
c
cV
( )
coordinate of particle relative to moving primed system
coordinate of particle relative to stationary system
velocity of primed coordina
V x
Vx
te system relative to stationaryv sys
tem
Page 28 Fundamentals of Modern Physics
(1-16)
According to Einstein this equation tells you how to transform the observed velocity from one frame of reference to another frame of reference First we note that as Vc and vc approachzero the equation approaches those which would be derived from the Galilean transformation
2 2
( ) 1 01 1 ( )
Vx v Vx vV
VxV x
Vx v
v cVxx
v
c
v
cc
Another property of this equation is that it is impossible to choose V and v such that Vrsquo the magnitude of the velocity which is seen in the new frame is greater than c
1
2
09 ( 09 ) 18009876543
( 09 ) 1811 (09 )
c c cV c c
cc
c
Example As seen by O particle 1 has velocity 09c in the direction of positive x and particle 2 has velocity 09c in the negative x direction
Einstein equation however predicts To evaluate the velocity of particle 1 with respect to particle 2 we transform from the O frame to the Orsquo frame moving in the negative x direction with velocity v = -09c using equation 1-16 We obtain
JK LOGIC will observe the particles separating at 18c
A confusing INCORRECT and ILLOGICAL answer
left is a dir rightsub of a
1
2
1
09 ( 09 ) 18009876543
181( 09 )1 ( 09 )
The multiplication of dir left time a dir right is ONLY a dir left
09
dir
direction leftdirection rightsubtraction
c c cV c c
cc
c
V
dir
22
22
09 18 18009876543
181( 081)1(09)(09)
1
left rightsub
sub
c c c cc c
ccc
c
Analyzing Einsteinrsquos BS math Equation
Using SM math a logical answer can be obtained
Cannot multiply a subtraction operator times a direction in space
Can only subtract and change its direction
Cannot multiply a direction to the left times a direction to the right
Cannot multiply opposite directions in space
Can only solve for the resultant
rightleft
subtraction
1
subtraction
2
sub of a dir to the left is a dir to the right
1 resultant 0 sub 2
2sub 2
2
09 09
09 amp 091
09 09 18 1
(0)1
(09) amp (09)1
c cV
c cc
c c cV
cc
cc
8
181
cc
SM math predicts a logical answer using logic
21st century scientists state in many articles that Einsteins General Relativity and Quantum Mechanics are incompatible
They are not only incompatible they are both mathematically incorrect
They both use illogical BS math
Any equations in math that use imaginary numbers should be abandoned
Imaginary numbers are only necessary in the current BS math system
This means that Schroumldingers equations should be abandoned
Any equation in quantum mechanics that uses imaginary number should be abandoned
Prof Kanarev has developed logical physics and chemistrytheories without the use of Quantum mechanics or relativity
A partial list of his theories coverbull Mechanics ndash replaces Newtonbull Photon - bull Electrons ndash Protons- Neutronsbull Spectroscopybull Nucleusbull Atoms amp Moleculesbull Electrodynamicsbull Heatbull Curvature of Space ndash replaces Schwarzcchildrsquos equation
Kanarevrsquos Hypothesis
bull Electron align linearly with proton They do NOT spin or form clouds around the nucleus
bull Neutrons are in the center of atom Proton are outside of Neutrons
bull The wavelength of all particles is the experimental radius
bull Double slit experiment explained correctly
bull Heisenbergrsquos inequality explained correctly
bull is the radius of approach of the magnetic lines around the torus of the electron
( 1 137)
bull Why water expands when it freezes
bull Electron clusters form lightning
rbull The meaning of Plancks constant is explained
22 2 kg m
h m f mr f consts
2210254 42m r kgm bull If the mr-constant describes a ring the equations
describe experiments bull If the mr-constant describes a wave the equations do
NOT describe experiments
Kanarevrsquos 5 Equations of Mechanics
Kanarevrsquos 1st Equation Acceleration ampmv maMma R R
Kanarevrsquos 2nd Equation Uniform motion amp mv maMma R R
maM mvR R
Kanarevrsquos 3rd Equation
Kanarevrsquos 4th Equation Equal and opposite reaction
1 21 21 2
12
a m
m a m ama
Kanarevrsquos 5th Equation 1 2 3 Mi M M M Mna R R R R R
The sum of the Actions working on a moving body is never equal to zero
Kanarevrsquos Photon model is a rolling ring
His equations accurately describe known experiments with photonsbull Electromagnetic radiation should be called photon radiationbull Diffraction and interference are the same phenomenabull Correctly explains the double slit experiment without quantum equationsbull Correctly explains the meaning of Heisenbergrsquos inequality
e
2 (2)(3142)(2817 15 ) 10007297352568
R 2426 12 137mag
xp
r m
m
Charge
Magnetic action
Proton
The intensity of the magnetic action near the geometrical center of the proton ring is so great that we theorize that it is the strong nuclear force
(9383 8)(1602 19)8476 14
4 (4)(3142)(1411 26)p
pp
EB T
M
(9395 8)(1602 19)1035 15
4 (4)(3142)(0966 26)N
NN
EB T
M
Neutron
1a Hydrogenavi
Studying the dash-cross codes of BS math has allowed me to break the codes and understand the reasons for the errors
Dash sign (-) CODES In Math used as
22
1 15 004255
bull (-) the number of zeros to the right of a decimal point 10-5=000001
bull (-) a subtraction operator
bull (-) a direction in space labeled negative (whatever that means)
bull (-) an exponent to mean divide
bull (-) describe the negative half of a sin or cos graph
bull (-) describe the negative half of an ldquoerdquo and ldquo1erdquo graph
bull (-) describe the negative side of all numbers raised to a power x
In Physics
bull (-) An Electron has been labeled as negative
bull (-) Anti-Particles have been labeled as negative
In Chemistry
bull (-) amp (+) Thermochemistry
Cross sign (+) CODES
(+) a direction in space labeled positive (whatever that means)
(+) an addition operator
(+) used to show the number of zeros to the left of a decimal point 10+5 = 100000
In Physics and Chemistry
bull (+) Protons have been Labeled as positive
bull (+) numerous other particles have been Labeled as positive
bull (+) amp (-) Thermochemistry
In Math used
The same dash symbol (-) is used for the numerous different math operations
The definitions established for the use of the dash symbol (-) do not distinguish between their different operations
In many math operations the dash symbol is changed to mean one of the other meanings
It is amazing that math has proceeded to its current level of use with this illogical use of a symbol
The same cross symbol (+) is used for numerous different math operations and produces the same illogical answers
A Summary of the BS Broken-Symmetry math
1 1 cosh tanhln
x xe x xe
bull Originated with broken-symmetry in the XYZ plane
bull Originated with mirror-broken-symmetry in the XZ plane
bull Created a Rule-of-Signs that cannot be proved (-)(-)=(+) (+)(+)=(+) (-)(+)=(-)
bull Created imaginary numbers to compensate for broken-symmetry
bull Created absolute values to change illogical negative number answers into positive number answers
bull Produces broken symmetry graphs for many functions
It took four years of study to discover why a (-)(-)=(+) only works in an imaginary world where space in the dash (-) direction is different from space in the cross (+) direction
Started working on the problem in August of 2001
Data points created using BS math produce graphs that are not symmetrical dash side different from cross side
However if a real problems data points follow a non symmetrical graph the graph can provide usable answers
This is why no one discovered the problem
Solved in January of 2005
Symmetry-Math(SM) vs Broken-Symmetry(BS) Math
In SM if we specify that the dash sign (-) means only subtraction and the cross sign (+) means only addition and an appropriate symbol is used for a direction in space space becomes symmetrical and math becomes logical
Math operators and directions in space are not the same and the same symbol should not be used to represent them
Letrsquos start with the (dash)multiplied by a (dash)=(cross) in the BS system
Usable answers may be obtained but it will be for illogical reasons
multiplied by a ( ) ( ) ( )subtraction Subtraction addition
rightleft left
dash dash cross
positivenegative negative
Symmetry-Math The subtraction of a direction is equal to the opposite direction The answers are correct using correct logic
opposite direction direction
SM
(negative)(negative) = (positive)
( ) ( ) ( )subtraction
BS
There is NO multiplication of a subtraction operator by a direction in space
There is just the subtraction of a direction in space
If instead of labeling the left side of a coordinate system as a negative (-) the same as a subtraction operator we label it with an arrow ( ) to represent the direction then a subtraction from that arrow direction will be in the opposite direction ( )
( ) mutiplied by a ( ) ( )
negative Positive negative
subtraction addition subtraction
dash cross dash
left right left
A subtraction operator multiplied by an addition operator is equal to a subtraction operator This is illogical
BS Math
A negative multiplied by a positive is equal to a negative This is illogical
A direction to the left multiplied by a direction to the right is a direction to the left This is illogical and Einsteins math error in special relativity
A dash times a cross is equal to a dash This is illogical
A vector moving to the left multiplied by a vector moving to the right is equal to ONLY to a vector moving to the left that is eight orders of magnitude faster thanThe speed of light
opposite direction subtraction direction
SM ( ) ( )
BS (negative)(positive) = (negative)
Symmetry Math The subtraction of a direction is equal to the opposite direction
There is no multiplication of a subtractionoperator times a direction in space
( ) multiplied by a ( ) ( )
positive positive positiveaddition addition additioncross cross cross
direction direction direction
What is the meaning of multiplying addition operators Addition operators are not multiplication operators
BS Math
() multiplied by a direction ( )=
number
() multiplied by a direction ( )=
number
A number times a direction maintains the same direction
( ) ()
dirdiradd
( ) ()
dirdiradd
The addition of a direction is in the same direction There is no multiplication by addition operators
Symmetry Math
Future math books need to eliminate the use of the cross sign (+) to represent something labeled a positive direction in space and the dash sign (-) to represent something labeled a negative direction in space Space does not have positive and negative directions
Symbols that are logical and have no illogical representation should be adopted SM uses either arrows or symbols
Symmetry-Math Number Line
Symmetry-Math Rules
The dash sign (-) will have only one use subtraction
The cross sign (+) will have only one use addition
Directions have an arrow and a number
All observers see the same direction and magnitude
bull There is no multiplication of arrows
amp ampamp ampampamp
( 3 ) ( 5 ) ( 6 ) ( 65 ) (17256 ) ( 32993 )
dir dirdir dirdirdir
Negative and Positive Directions in Space
If I asked you to point to a negative direction in space which way would you point
Hopefully you will realize that there is no such thing as a negative direction in space
If I asked you to point to a positive direction in space which way would you point
Again hopefully you will realize that there is no such thing as a positive direction in space
Objects on the right side of the x-axis are positive (+)(+) = (+)
Objects on the left side of the x-axis are negative (-)(-) = (+) negative math is different from positive math This is illogical
Objects on the top of the y-axis are positive
Objects on the bottom of the y-axis are negative
Objects in the front of the z-axis are positive
Objects in the back of the z-axis are negative
Again BS math of the positive direction (x right y up and z front) is different from the BS math of the negative direction (x left y down and z back)
BS math is Broken-Symmetry
Mirror image is broken symmetry in the x and z axis If the left amp right-axis are
reversed symmetry is broken Obs-1 math is different from obs-2
If the front amp back-axis is reversed symmetry is broken Obs-1 math is different from obs-2
If the top amp bottom-axis are
reversed symmetry is not broken Obs-1 math is the same as obs-2
Math answers cannot depend on which side of a number line an observer sit
Absolute values for Displacement
Even for a simple displacement of an object BS math invented a definition and absolute values to obtain something labeled a positive number answer
I----------I----------I----------I----------I----------I----------I----------I----------I
-4 -3 -2 -1 0 +1 +2 +3 +4 a gtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtb cltltltltltltltltltltltltltltltltltltltltltltltltltltltb
If an object starts at ( 3)ia X and moves to ( X 3) and the to ( 1)b fb c X
BS Math provides a usable answer without a definition or absolute values
However If an object starts at
( 1) and moves to ( X 3) and the to ( 3)i b fa X b c X
Then absolute values and a definition must be used
---------I----------I----------I----------I----------I----------I----------I---------- -4 -3 -2 -1 0 +1 +2 +3 +4
a gtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtb cltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltb
BS Math makes up a definition that all displacements are positive Since this example provides an answer with something labeled a negative the positive definition must be applied and (-2) is changed to (+2) A definition is needed to arrive at a useable answer This is Illogical
In SM absolute values and a meaningless definition of ldquopositiverdquo are removed In SM objects and directions in space are defined by the direction and magnitude of the resultant of arrows
With SM you get total distance traveled by the object and the final direction of the displacement
22222 2 bababbaababababa If w e su b stitu te p o s itiv e n u m b e rs in to th e eq u a tio n th e an sw er w ill b e co rrec t If (a = 5 ) an d (b = 3 )
2
2
8
5 3 5 3 5 3 5 5 5 3 3 5 3 3 25 15 15 9 64
8 8 8 8 8 8 8 8 8 8 64
W h en tw o p o sitiv e n u m b ers a re a d d ed an d sq u a re d th e B S m a th d is tr ib u tiv e law p ro v id es a co rre c t an sw er 1
Illogical and Incorrect BS math for the distributive law
222 2 bababbabbaaabababa B y d e fin itio n in B S m a th n o n u m b er sq u a red c a n ev e r b e n eg a tiv e T h e re fo re n o sq u a red n u m b er c an ev e r b e a v e c to r m o v in g in th e d a sh ( -) o r n eg a tiv e d ire c tio n if it is sq u a red In B S m a th an d ad d itio n o p e ra to r m u ltip lied b y a su b trac tio n o p e ra to r is A L W A Y S eq u a l to a su b tra c tio n o p e ra to r If th is h ad n o t b e en u sed fo r th e la s t fo u r h u n d red y e a rs y o u w o u ld d ie lau g h in g a t th e u tte r ab su rd ity o f th e la s t s ta tem e n t T h is is illo g ica l an d v io la te s sy m m etry
If a = 5 an d b = 3 42222352
22
4930253)3)(5(253)5)(3()3)(5(5353535 22222 A s lo n g a s ldquo ardquo is g re a te r th an ldquob rdquo th e B S m a th d is tr ib u tiv e law p ro v id es co rrec t an sw ers 1
B S m a th b re a k s d o w n w h e n ldquo b rdquo is g rea ter th a n ldquo a rdquo
If a = 3 an d b = 5
2 2
2
2
3 5 2 4
2 ( 2) ( 2) 4
2 ( )( 2) ( )( 2) 4
A n y co m b in a tio n ex cep t th e m id d le o n e is illo g ica l an d in co rre c t B y d e fin itio n th e an sw er can n o t b e -4 in B S m a th B S m a th is in co rrec t
4253095)5)(3(235)3)(5()5)(3(3535353 22222 T h is is w h ere th e d is tr ib u tiv e la w p ro v id es in co rrec t an sw ers I t p ro d u ces an an sw er o f + 4 b e cau se b y d e fin iti o n n o n u m b er sq u a red in B S m a th c an b e a n eg a tiv e n u m b er U sin g in co rre c t B S m a th th e d is tr ib u tiv e law w ill g iv e an in co rrec t an sw ers o f + 4 T h is is in co rrec t an d illo g ica l 1
Since there can be no negative direction from a squared term in BS math BS math should be modified or abandoned
The BS distributive law should be modified or abandoned
The numbers +1+2+3+4 and ndash1-2-3-4hellip should be abandoned
BS math gives incorrect answers because their Rule-of-Signs allows multiplying an arrow going in one direction by an arrow going in another direction
Clearly this is not logical
Arrows should be used that are specific for directions
W h at d o es it m e an to m u ltip ly ( 4) b y (4 ) A s w e k n o w 1
m u ltip lica tio n is ju s t ad d itio n W h a t w o u ld ( 4) tim e (4 ) b e 2
In BS math the answer is by definition -16 unless you square -4 and then it is +16
In SM you cannot multiply arrows going in opposite directions
The middle two terms are not logical You cannot multiply opposite directions
This is a MAJOR error in BS math They multiply a dash [(-) a direction to the left] by a cross [(+) a direction to the right] And then ldquoby definitionrdquo label the answer dash (-) This is illogical and produces incorrect answers
2
2 2
This is Einsteinrsquos math error
in Special relativity
Magnitude 4 Direction 1 Magnitude 3 Direction 2 Magnitude 5 Direction 3 Relative to an observer at 0 4
Magnitude 4 Direction 1 Magnitude 3 Direction 2 Magnitude 5 Direction 3 Relative to an observer at 0 4
Magnitude and Direction (MD) An arrow has both a magnitude andA direction relative to a 2nd objectobserver
bull Magnitude and velocity (MV) relative to a 2nd objectobserver
bull Magnitude And acceleration (MA) relative to a 2nd objectobserver2
v da
t t
Introducing a new symbol the ldquoamprdquo sign The ldquoamprdquo sign means resultant ADD all the Arrows for each direction and then SUBTRACT to find the final resultant
1 1amp 1 2 2amp 2 4 If =
1 amp1 2 2 amp2 4 If =
amp 5 amp 3 2 If gt
amp 3 amp 5= 2 If gt
amp 0 3 amp 3=0 if =
only the resultant equals zero
the vectors do not - they are still there
Does not exist There is no multiplication of arrows
There are two arrows One has a magnitude of 3 One has a magnitude of 5 The maximum these two arrows can be is 8
Only the resultant of arrows exist
An SM arrow has a direction and a magnitude
Arrows and magnitudes are still there even if the resultant is zero
There are no negative SM Arrows
There are no positive SM Arrows
There are no ZERO SM Arrows A zero would negate the definition of an SM Arrow something with no magnitude and no direction cannot be the definition of an arrow that is defined as having magnitude and direction
bull A resultant is the addition and subtraction of the differences in magnitude and direction arrows
A B and C are the arrows (a) A amp B exist(b) A amp B amp C exist(c) A-B Do Not Exist in Sm there are no negatives or positives(d) A+B-C Do Not Exist in SM there are no negatives or positives
There are only resultants when there are multiple arrows
SM Graphs versus BS Graphs
Numbers Raised to Powers
When numbers are raised to powers the negative sides of all BS math graphs are ILLOGICAL and INCORRECT
Incorrect BS Math for 2x
2x
4 16 correct3 8 correct2 4 correct1 2 correct0 1 correct-1 005 incorrect-2 025 incorrect-3 0125 incorrect-4 00625 incorrect
negative side of graph is incorrectpositive side of graph is correct
16
8
42105025012500625
-4 -3 -2 -1 0 1 2 3 4
po
sit
ive
negative positive
2x BS Math
correctincorrect
In BS math the dash sign (-) used in this example represents three different operations They arebull (-) Subtractionbull (-) A direction to the leftbull (-) As an exponent to mean divide by
All positive numbers produce correct results and graphs
All negative numbers produce incorrect results and graphs
A direction of three units to the left (-3) is placed in the equation as an exponent
A direction in space (3 units toward the left) changes a whole number to a fraction by using the dash (-) as an exponent rather than a direction in space
This is illogical and incorrect BS math should have performed this operation as
Unfortunately the answers to problems can come out correct because data points of a real problem fit the curve
However the answers will not be due to logical reasoning
Because the equation produced a curve that real problem data points follow it has been falsely believed that the equations and graphs are correct
1
2
x
Incorrect BK graph for (12)x
(12)x
4 00625 correct3 0125 correct2 025 correct1 1 correct0 1 correct-1 2 incorrect-2 4 incorrect-3 8 incorrect-4 16 incorrect
negative side of graph is incorrectpositive side of graph is correct
00625012502505124
8
16
-5 -4 -3 -2 -1 0 1 2 3 4 5P
osit
ive
negative positive
BS Math for (12)x
correctincorrect
Incorrect BS math and graph for
All graphs are symmetrical with SM
Two BS Math Equations that Produce Incorrect Answers
2 0x x starting equation 2
2
x x x x
x x
Add lsquoxrsquo to both sides
2(1) 1 let lsquoxrsquo = 1
2 1 This is an incorrect answer 2(0)-(0)=0 The only answer that BK math can give is x=0
1
a b starting equation ( )( ) ( )( )a a a b multiply both sides by ( )a
( )( ) ( )( ) ( )( ) ( )( )a a b b a b b b subtract ( )( )b b from both sides
( )( ) ( )( )a b a b a b b factor ( )a b from both sides
( )a b b divide both sides by ( )a b
( )b b b replace ( )a with ( )b 2b b simplify 2(1) (1) Let ( )b = 1
2 1 Again BS math yields and incorrect answer 1
This incorrect result is a problem with the distributive law SM explains how this incorrect answer is obtained and the problem with the distributive law that allows the error
Page 148 ndash149 Barry Mazur ndash Imagining Numbers
3
2
31
Do the computation carefully on paper using the rules we agreed to Then ponder your answer which should be something of a surprise to you donrsquot stop there Think of what your answer might possibly mean or might imply
Solve this equation using the distributive law
JK definitely a surprise the answer wasincorrect and illogical
Gerhard Gade University Professor doing math at Harvard University
Here is the BS math answer that Barry Mazur calculated
2
31
2
31
2
31
1 3 1 3 1 3 3 3 2 2 3 1 3
2 2 2 2 4 4 4 4 4 4 2 2
dash
1 3 1 3 1 31
2 2 2 2 4 4
3 3
4 4
When the distributive law is used with the BS Rule-of-Signs and one of the signs is a (-) dash sign the middle two terms always cancel each other This leads to the incorrect answer of (-1) in the above equation
Barry Mazur did not go back and check his answer with the original equation If he had he would have known that the answer was incorrect
3
1 31
2
check answer by setting equal
3
331 3
12
1 31
2
Take the Cube root of both sides
1 3 2 multiply both sides by 2
3 3 subtract 1 from both sides
2 2
3 3 square both sides
3 9 This is an incorrect answer A simple equation check could have shown that the distributive law using BS math provides an incorrect answer
1
BS Illogical and Incorrect Numbers Raised to Powers
nC D n DC
642 DDD
In BS math only a positive number can have an answer if ldquonrdquo is an even number (246---)
All negative numbers have no solutions
642 DDD
With SM just add a direction arrow and any number gives the correct answer
or n nC D C D
3 3( 3 8 8 or 8 2 or 2 )n D C
( 2 16 C= 16 or 16 4 or 4 )n D
Since a negative number has no square root answer in BS math BS mathematicians invented imaginary numbers
i2=-1
24414 iiC n
These equations are easily solved without the use of imaginary numbers with SM math
If 4 then 4 2x
If 4 then 4 2x
The 360-degree circle is a 4400 years old extremely outdated system based on a year of 12 months of 30 days each that is 360 days
SM will use a more logical system that will have a circle of 6283185307o
d
( )
(deg )
d r radians
d r rees
the arc distance around
the circumference of a circle
d
radiusr
2 2
2 2
d r
dv rd t t
v dda rd t t t
1
d
r
( )av tv
The average angular velocity of a body is the rate of its angular displacement
( )0
liminstt
vt
The instantaneous angular velocity is the limit of the ratio
t
as t approaches zero
( )d
av
d
tv
The average circumference velocity of the body is the rate of its distance traveled d divided by the time t
dvv
r
1 dd vdrv
t t t r r
The equation of the relationship between the angular velocity and the velocity of a point on that body
( )o
av
v v v
t ta
The average angular acceleration is the change in the angular velocity
( )0
liminstt
v
ta
The instantaneous angular acceleration is the
limit of the ration of v
t
as t approaches
zero
daa
r
1
1
d
d s d
vva
t r tv a
r t r
The equation of the relationship between the angular acceleration and the acceleration of a point on the body
21
2v a t
If a body with an initial angular velocity ov has a constant
angular acceleration a it will turn through an angle in a time t
1
Straight Line Symbol Units Rotational Symbol Units Distance d m Angular
Distance radians or degrees
Velocity dt
v m
s Angular Velocity
V
t
radians or degrees
s
Acceleration vat
2ms
Angular Acceleration
VA
t
2
radians or degrees
s
Mass M Kg Mass distribution
dM 2Kgm
Force actionA Ma
2
Kgm
sNewton
Torque
dT M A 2
2
Kgm
s
Momentum Mv Kgm
s
Angular Momentum
dM V 2Kgm
s
Work (Force)(distance)
Ad 2
Kgmm
sNm Joule
Work T
Using AS arrow math to solve and direction in space problems using the equation 1 2
2 2 2 2 2 2( ) amp and ( ) amp 3
4
2 2 2
2 2
( ) amp
(5 ) 4 3
(5 ) 16 9
(5 ) 25
(5 ) (5 )
2 2(5 ) 4 3
(5 ) 16 9
(5 ) 25
(5 ) (5 )
In SM math Pythagorasrsquo theorem provides both magnitude and direction
In BS math Pythagorasrsquo theorem provides magnitude only
time set equal to zero
23
( ) ( 0 )10 1
x t xx
3( ) ( 0)
10 1W t f
In BS math this is an equation for a traveling wave
must use a square term in order to make the curve symmetric
NO SQUARE TERM IS NECESSARY
In SM math when t is set to ldquo0rdquo
This is a major fault in BS math
x f(x)
-4 0019
-3 0033
-2 0073
-1 0273
0 3000
1 0273
2 0073
3 0033
4 0019 -4 -3 -2 -1 0 1 2 3 400
05
10
15
20
25
30
35
BS wave (must use square term)
snap shot of wave at x = 0
Am
plit
ude
lt to gt lt to gt f(ltgt)
lt -4 0073
lt -3 0097
lt -2 0143
lt -1 0273
origin 0 3000
gt 1 0273
gt 2 0143
gt 3 0097
gt 4 0073lt0073 lt0097 lt0143 lt0273
3000
0273gt 0143gt 0097gt 0073gt
Wave moving
snap shot of wave
Am
plit
ude
lt4 lt3 lt2 lt1 0 1gt 2gt 3gt 4gt
3
2
1
In SM NO squared terms are needed
SPRING
An Action is needed to stretch or compress a spring to a starting point where it can execute cycle motion
tsA (ts to start) is the work required to move a mass attached to a spring a distance (d1) from
its initial rest position RAK 0 the spring constant is the stiffness of the spring
o A is attraction o R is repulsion o
ARRA
00 is for the back and forth motion through the zero initial rest position A
stretched spring goes from ldquoArdquo to ldquo0rdquo to ldquoRrdquo to ldquoArdquo to ldquo0rdquo to ldquoRrdquo d1 is the distance a mass is moved from its initial rest position
0
0 1ts A RA K d
Once ( )mad
is performed to move the mass to position d1 (against the restraint of the spring) a of the ( )mad
is transferred into the spring as potential attractionrepulsion (PAR) There is always
more ( )mad
performed than is stored as PAR When the spring is released from the restraining Ats the PAR starts the oscillation Conservation of ( )mad
requires that the PAR and the motion ( )mad
( )( )madM be a constant
( )amp ( ) constantmadPAR M 02
0 0
1
2A R A RPAR K d
02
( ) 0 0
1( )
2mad A R A RM m v
0 0 02 2 2
0 0 0 1
2 2 21
2 2 21
1 1 1amp
2 2 2
1 1 1amp
2 2 2
amp
A R A R A RK d m v K d
Kd mv Kd
Kd mv Kd
0
1d is frac12 the height from the base to the peak of a cycle
An experiment of moving paper under a pendulum through a small angle filled with sand that can escape through a small hole in the bottom shows that a pendulum cycling back and forth follows a curve that is mathematically analogous to a lsquosinrsquo or lsquocosrsquo
The projection of the curved area onto the straight line exactly duplicate the back and forth motion of CM This allows for the math of lsquosinrsquo or lsquocosrsquo to be used
SM will use lsquosinrdquo math It could use lsquocosrsquo or a combination of both lsquosinrsquo and lsquocosrsquo Using just the lsquosinrsquo math makes the understanding and math easier
0 sin 0 0SM 0 sin157 1SM The diagram at the left shows a spring being stretched from its (zero rest) position-2 and the PAR being released at position-1 There are 4 sections (1 to 2 2 to 3 3 to 4 and 4 to5) that alternate between maximum acceleration and zero velocity to max velocity and zero acceleration
At position-1 there is maximum acceleration and zero velocity o The lsquosinrsquo values from 0o to 157o will be used for velocity (values from 0 to 1) o The lsquosinrsquo values from 157o to 0o will be used for acceleration (values from 1 to 0)
At position-2 there is maximum velocity and zero acceleration Acceleration changes to deceleration
o The lsquosinrsquo values from 157o to 0o will be used for deceleration (values from 1 to 0) o The lsquosinrsquo values from 0o to 157o will be used for acceleration (values from 0 to 1)
At position-3 there is maximum acceleration and zero velocity o The lsquosinrsquo values from 0o to 157o will be used for velocity (values from 0 to 1) o The lsquosinrsquo values from 157o to 0o will be used for acceleration (values from 1 to 0)
At position-4 there is maximum velocity and zero acceleration Acceleration changes to deceleration
o The lsquosinrsquo values from 157o to 0o will be used for deceleration (values from 1 to 0) o The lsquosinrsquo values from 0o to 157o will be used for acceleration (values from 0 to 1)
At position-5 the cycle starts again
2
)539(
cT
MK
spring constant (K) when mass (M) and cycle-time (T) are known
)(
))(89)((
111 md
uKgM
d
MG
d
FK sts
spring constant (K) when mass (M) and amplitude ( d1) are known
2 21( )
Kv d d
m
velocity (v) at any point during the cycle
velocity
dforM
Kdv
max
012 velocity (v2) maximum (when d = 0 at position-2 and 4)
acceleration (a1) maximum when spring constant (K) mass (M) and amplitude (d1) are known
acceleration (a1) maximum when amplitude (d1) and cycle-time (Tc) are known
acceleration (a) at any point during the cycle-time (T)
1
151
286
v
dTc
cycle-time (Tc) when amplitude (d1) and velocity (v1) are known
K
MTc 28651
cycle-time (Tc) when mass (M) and spring constant (K) are known
fTc
1
cycle-time (Tc) when the frequency is known
cTf
1
frequency (f) when the cycle-time (Tc) is known
of cycles
elapsed timef
frequency (f) when the number of cycles in an elapsed time is known
d1 amplitude (distance from rest to start position)
Also frac12 the distance from the base to the peak
3131 oror dM
Ka
251
3131 539
c
oror T
da
2539
cT
da
In December of 2006 I discovered the mathematical error in Einsteinrsquos Special Relativity Einstein used BS math and that math only works in an imaginary world
It cannot describe real symmetrical space
Einsteinrsquos Special Relativity should be abandoned and a theory of real space using SM should be used to replace it
Numerous articles on the Internet show Einsteinrsquos theory to be incorrect but no one in academia is willing to publish the information
Figure 1-10 A moving particle observed from two frames of reference in uniform translation
Einsteinrsquos Math Error in Special Relativity Transformation of Velocity (From Eisenberg- Fundamentals of Modern Physics pages 27-29) Consider the particle shown in figure (1-10) moving with velocity v as
seen in a frame of reference O
dx dy dz
Vx Vy Vzdt dt dt
dx dy dz
V x V y V zdy dt dt
We would like to evaluate the velocity v of the particle as seen in the frame of reference 0 which is itself moving relative to 0 with velocity v Measured in the 0 frame the velocity vector of the particle has com ponents
The same velocity vector as measured in the 0 frame has components
2
2
2
2
2
1
1
x vtx
v
cy y
z z
vxt
ctv
c
Now from equations (1-13) we know that the relation between the primed and the unprimed coordinates and times is
Take the differential of these equations remembering that v is a constant This gives
2
2
2
2
2
1
1
dx vdtdx
v
cdy dy
dz dz
vxdt
cdtv
c
2
2
2 2 2 2 2
2
2
2
11 ( )
1( ) 1
1
1
1
dx vdt
v dxdx vdt v
dx dx vdtc dt dtvdx vdx vdx v dxdt dt dt dt
dt dtc c c c
VxV x
Vx
VxV x
vv
vv
c
x
v
c
cV
( )
coordinate of particle relative to moving primed system
coordinate of particle relative to stationary system
velocity of primed coordina
V x
Vx
te system relative to stationaryv sys
tem
Page 28 Fundamentals of Modern Physics
(1-16)
According to Einstein this equation tells you how to transform the observed velocity from one frame of reference to another frame of reference First we note that as Vc and vc approachzero the equation approaches those which would be derived from the Galilean transformation
2 2
( ) 1 01 1 ( )
Vx v Vx vV
VxV x
Vx v
v cVxx
v
c
v
cc
Another property of this equation is that it is impossible to choose V and v such that Vrsquo the magnitude of the velocity which is seen in the new frame is greater than c
1
2
09 ( 09 ) 18009876543
( 09 ) 1811 (09 )
c c cV c c
cc
c
Example As seen by O particle 1 has velocity 09c in the direction of positive x and particle 2 has velocity 09c in the negative x direction
Einstein equation however predicts To evaluate the velocity of particle 1 with respect to particle 2 we transform from the O frame to the Orsquo frame moving in the negative x direction with velocity v = -09c using equation 1-16 We obtain
JK LOGIC will observe the particles separating at 18c
A confusing INCORRECT and ILLOGICAL answer
left is a dir rightsub of a
1
2
1
09 ( 09 ) 18009876543
181( 09 )1 ( 09 )
The multiplication of dir left time a dir right is ONLY a dir left
09
dir
direction leftdirection rightsubtraction
c c cV c c
cc
c
V
dir
22
22
09 18 18009876543
181( 081)1(09)(09)
1
left rightsub
sub
c c c cc c
ccc
c
Analyzing Einsteinrsquos BS math Equation
Using SM math a logical answer can be obtained
Cannot multiply a subtraction operator times a direction in space
Can only subtract and change its direction
Cannot multiply a direction to the left times a direction to the right
Cannot multiply opposite directions in space
Can only solve for the resultant
rightleft
subtraction
1
subtraction
2
sub of a dir to the left is a dir to the right
1 resultant 0 sub 2
2sub 2
2
09 09
09 amp 091
09 09 18 1
(0)1
(09) amp (09)1
c cV
c cc
c c cV
cc
cc
8
181
cc
SM math predicts a logical answer using logic
21st century scientists state in many articles that Einsteins General Relativity and Quantum Mechanics are incompatible
They are not only incompatible they are both mathematically incorrect
They both use illogical BS math
Any equations in math that use imaginary numbers should be abandoned
Imaginary numbers are only necessary in the current BS math system
This means that Schroumldingers equations should be abandoned
Any equation in quantum mechanics that uses imaginary number should be abandoned
Prof Kanarev has developed logical physics and chemistrytheories without the use of Quantum mechanics or relativity
A partial list of his theories coverbull Mechanics ndash replaces Newtonbull Photon - bull Electrons ndash Protons- Neutronsbull Spectroscopybull Nucleusbull Atoms amp Moleculesbull Electrodynamicsbull Heatbull Curvature of Space ndash replaces Schwarzcchildrsquos equation
Kanarevrsquos Hypothesis
bull Electron align linearly with proton They do NOT spin or form clouds around the nucleus
bull Neutrons are in the center of atom Proton are outside of Neutrons
bull The wavelength of all particles is the experimental radius
bull Double slit experiment explained correctly
bull Heisenbergrsquos inequality explained correctly
bull is the radius of approach of the magnetic lines around the torus of the electron
( 1 137)
bull Why water expands when it freezes
bull Electron clusters form lightning
rbull The meaning of Plancks constant is explained
22 2 kg m
h m f mr f consts
2210254 42m r kgm bull If the mr-constant describes a ring the equations
describe experiments bull If the mr-constant describes a wave the equations do
NOT describe experiments
Kanarevrsquos 5 Equations of Mechanics
Kanarevrsquos 1st Equation Acceleration ampmv maMma R R
Kanarevrsquos 2nd Equation Uniform motion amp mv maMma R R
maM mvR R
Kanarevrsquos 3rd Equation
Kanarevrsquos 4th Equation Equal and opposite reaction
1 21 21 2
12
a m
m a m ama
Kanarevrsquos 5th Equation 1 2 3 Mi M M M Mna R R R R R
The sum of the Actions working on a moving body is never equal to zero
Kanarevrsquos Photon model is a rolling ring
His equations accurately describe known experiments with photonsbull Electromagnetic radiation should be called photon radiationbull Diffraction and interference are the same phenomenabull Correctly explains the double slit experiment without quantum equationsbull Correctly explains the meaning of Heisenbergrsquos inequality
e
2 (2)(3142)(2817 15 ) 10007297352568
R 2426 12 137mag
xp
r m
m
Charge
Magnetic action
Proton
The intensity of the magnetic action near the geometrical center of the proton ring is so great that we theorize that it is the strong nuclear force
(9383 8)(1602 19)8476 14
4 (4)(3142)(1411 26)p
pp
EB T
M
(9395 8)(1602 19)1035 15
4 (4)(3142)(0966 26)N
NN
EB T
M
Neutron
1a Hydrogenavi
In Physics
bull (-) An Electron has been labeled as negative
bull (-) Anti-Particles have been labeled as negative
In Chemistry
bull (-) amp (+) Thermochemistry
Cross sign (+) CODES
(+) a direction in space labeled positive (whatever that means)
(+) an addition operator
(+) used to show the number of zeros to the left of a decimal point 10+5 = 100000
In Physics and Chemistry
bull (+) Protons have been Labeled as positive
bull (+) numerous other particles have been Labeled as positive
bull (+) amp (-) Thermochemistry
In Math used
The same dash symbol (-) is used for the numerous different math operations
The definitions established for the use of the dash symbol (-) do not distinguish between their different operations
In many math operations the dash symbol is changed to mean one of the other meanings
It is amazing that math has proceeded to its current level of use with this illogical use of a symbol
The same cross symbol (+) is used for numerous different math operations and produces the same illogical answers
A Summary of the BS Broken-Symmetry math
1 1 cosh tanhln
x xe x xe
bull Originated with broken-symmetry in the XYZ plane
bull Originated with mirror-broken-symmetry in the XZ plane
bull Created a Rule-of-Signs that cannot be proved (-)(-)=(+) (+)(+)=(+) (-)(+)=(-)
bull Created imaginary numbers to compensate for broken-symmetry
bull Created absolute values to change illogical negative number answers into positive number answers
bull Produces broken symmetry graphs for many functions
It took four years of study to discover why a (-)(-)=(+) only works in an imaginary world where space in the dash (-) direction is different from space in the cross (+) direction
Started working on the problem in August of 2001
Data points created using BS math produce graphs that are not symmetrical dash side different from cross side
However if a real problems data points follow a non symmetrical graph the graph can provide usable answers
This is why no one discovered the problem
Solved in January of 2005
Symmetry-Math(SM) vs Broken-Symmetry(BS) Math
In SM if we specify that the dash sign (-) means only subtraction and the cross sign (+) means only addition and an appropriate symbol is used for a direction in space space becomes symmetrical and math becomes logical
Math operators and directions in space are not the same and the same symbol should not be used to represent them
Letrsquos start with the (dash)multiplied by a (dash)=(cross) in the BS system
Usable answers may be obtained but it will be for illogical reasons
multiplied by a ( ) ( ) ( )subtraction Subtraction addition
rightleft left
dash dash cross
positivenegative negative
Symmetry-Math The subtraction of a direction is equal to the opposite direction The answers are correct using correct logic
opposite direction direction
SM
(negative)(negative) = (positive)
( ) ( ) ( )subtraction
BS
There is NO multiplication of a subtraction operator by a direction in space
There is just the subtraction of a direction in space
If instead of labeling the left side of a coordinate system as a negative (-) the same as a subtraction operator we label it with an arrow ( ) to represent the direction then a subtraction from that arrow direction will be in the opposite direction ( )
( ) mutiplied by a ( ) ( )
negative Positive negative
subtraction addition subtraction
dash cross dash
left right left
A subtraction operator multiplied by an addition operator is equal to a subtraction operator This is illogical
BS Math
A negative multiplied by a positive is equal to a negative This is illogical
A direction to the left multiplied by a direction to the right is a direction to the left This is illogical and Einsteins math error in special relativity
A dash times a cross is equal to a dash This is illogical
A vector moving to the left multiplied by a vector moving to the right is equal to ONLY to a vector moving to the left that is eight orders of magnitude faster thanThe speed of light
opposite direction subtraction direction
SM ( ) ( )
BS (negative)(positive) = (negative)
Symmetry Math The subtraction of a direction is equal to the opposite direction
There is no multiplication of a subtractionoperator times a direction in space
( ) multiplied by a ( ) ( )
positive positive positiveaddition addition additioncross cross cross
direction direction direction
What is the meaning of multiplying addition operators Addition operators are not multiplication operators
BS Math
() multiplied by a direction ( )=
number
() multiplied by a direction ( )=
number
A number times a direction maintains the same direction
( ) ()
dirdiradd
( ) ()
dirdiradd
The addition of a direction is in the same direction There is no multiplication by addition operators
Symmetry Math
Future math books need to eliminate the use of the cross sign (+) to represent something labeled a positive direction in space and the dash sign (-) to represent something labeled a negative direction in space Space does not have positive and negative directions
Symbols that are logical and have no illogical representation should be adopted SM uses either arrows or symbols
Symmetry-Math Number Line
Symmetry-Math Rules
The dash sign (-) will have only one use subtraction
The cross sign (+) will have only one use addition
Directions have an arrow and a number
All observers see the same direction and magnitude
bull There is no multiplication of arrows
amp ampamp ampampamp
( 3 ) ( 5 ) ( 6 ) ( 65 ) (17256 ) ( 32993 )
dir dirdir dirdirdir
Negative and Positive Directions in Space
If I asked you to point to a negative direction in space which way would you point
Hopefully you will realize that there is no such thing as a negative direction in space
If I asked you to point to a positive direction in space which way would you point
Again hopefully you will realize that there is no such thing as a positive direction in space
Objects on the right side of the x-axis are positive (+)(+) = (+)
Objects on the left side of the x-axis are negative (-)(-) = (+) negative math is different from positive math This is illogical
Objects on the top of the y-axis are positive
Objects on the bottom of the y-axis are negative
Objects in the front of the z-axis are positive
Objects in the back of the z-axis are negative
Again BS math of the positive direction (x right y up and z front) is different from the BS math of the negative direction (x left y down and z back)
BS math is Broken-Symmetry
Mirror image is broken symmetry in the x and z axis If the left amp right-axis are
reversed symmetry is broken Obs-1 math is different from obs-2
If the front amp back-axis is reversed symmetry is broken Obs-1 math is different from obs-2
If the top amp bottom-axis are
reversed symmetry is not broken Obs-1 math is the same as obs-2
Math answers cannot depend on which side of a number line an observer sit
Absolute values for Displacement
Even for a simple displacement of an object BS math invented a definition and absolute values to obtain something labeled a positive number answer
I----------I----------I----------I----------I----------I----------I----------I----------I
-4 -3 -2 -1 0 +1 +2 +3 +4 a gtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtb cltltltltltltltltltltltltltltltltltltltltltltltltltltltb
If an object starts at ( 3)ia X and moves to ( X 3) and the to ( 1)b fb c X
BS Math provides a usable answer without a definition or absolute values
However If an object starts at
( 1) and moves to ( X 3) and the to ( 3)i b fa X b c X
Then absolute values and a definition must be used
---------I----------I----------I----------I----------I----------I----------I---------- -4 -3 -2 -1 0 +1 +2 +3 +4
a gtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtb cltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltb
BS Math makes up a definition that all displacements are positive Since this example provides an answer with something labeled a negative the positive definition must be applied and (-2) is changed to (+2) A definition is needed to arrive at a useable answer This is Illogical
In SM absolute values and a meaningless definition of ldquopositiverdquo are removed In SM objects and directions in space are defined by the direction and magnitude of the resultant of arrows
With SM you get total distance traveled by the object and the final direction of the displacement
22222 2 bababbaababababa If w e su b stitu te p o s itiv e n u m b e rs in to th e eq u a tio n th e an sw er w ill b e co rrec t If (a = 5 ) an d (b = 3 )
2
2
8
5 3 5 3 5 3 5 5 5 3 3 5 3 3 25 15 15 9 64
8 8 8 8 8 8 8 8 8 8 64
W h en tw o p o sitiv e n u m b ers a re a d d ed an d sq u a re d th e B S m a th d is tr ib u tiv e law p ro v id es a co rre c t an sw er 1
Illogical and Incorrect BS math for the distributive law
222 2 bababbabbaaabababa B y d e fin itio n in B S m a th n o n u m b er sq u a red c a n ev e r b e n eg a tiv e T h e re fo re n o sq u a red n u m b er c an ev e r b e a v e c to r m o v in g in th e d a sh ( -) o r n eg a tiv e d ire c tio n if it is sq u a red In B S m a th an d ad d itio n o p e ra to r m u ltip lied b y a su b trac tio n o p e ra to r is A L W A Y S eq u a l to a su b tra c tio n o p e ra to r If th is h ad n o t b e en u sed fo r th e la s t fo u r h u n d red y e a rs y o u w o u ld d ie lau g h in g a t th e u tte r ab su rd ity o f th e la s t s ta tem e n t T h is is illo g ica l an d v io la te s sy m m etry
If a = 5 an d b = 3 42222352
22
4930253)3)(5(253)5)(3()3)(5(5353535 22222 A s lo n g a s ldquo ardquo is g re a te r th an ldquob rdquo th e B S m a th d is tr ib u tiv e law p ro v id es co rrec t an sw ers 1
B S m a th b re a k s d o w n w h e n ldquo b rdquo is g rea ter th a n ldquo a rdquo
If a = 3 an d b = 5
2 2
2
2
3 5 2 4
2 ( 2) ( 2) 4
2 ( )( 2) ( )( 2) 4
A n y co m b in a tio n ex cep t th e m id d le o n e is illo g ica l an d in co rre c t B y d e fin itio n th e an sw er can n o t b e -4 in B S m a th B S m a th is in co rrec t
4253095)5)(3(235)3)(5()5)(3(3535353 22222 T h is is w h ere th e d is tr ib u tiv e la w p ro v id es in co rrec t an sw ers I t p ro d u ces an an sw er o f + 4 b e cau se b y d e fin iti o n n o n u m b er sq u a red in B S m a th c an b e a n eg a tiv e n u m b er U sin g in co rre c t B S m a th th e d is tr ib u tiv e law w ill g iv e an in co rrec t an sw ers o f + 4 T h is is in co rrec t an d illo g ica l 1
Since there can be no negative direction from a squared term in BS math BS math should be modified or abandoned
The BS distributive law should be modified or abandoned
The numbers +1+2+3+4 and ndash1-2-3-4hellip should be abandoned
BS math gives incorrect answers because their Rule-of-Signs allows multiplying an arrow going in one direction by an arrow going in another direction
Clearly this is not logical
Arrows should be used that are specific for directions
W h at d o es it m e an to m u ltip ly ( 4) b y (4 ) A s w e k n o w 1
m u ltip lica tio n is ju s t ad d itio n W h a t w o u ld ( 4) tim e (4 ) b e 2
In BS math the answer is by definition -16 unless you square -4 and then it is +16
In SM you cannot multiply arrows going in opposite directions
The middle two terms are not logical You cannot multiply opposite directions
This is a MAJOR error in BS math They multiply a dash [(-) a direction to the left] by a cross [(+) a direction to the right] And then ldquoby definitionrdquo label the answer dash (-) This is illogical and produces incorrect answers
2
2 2
This is Einsteinrsquos math error
in Special relativity
Magnitude 4 Direction 1 Magnitude 3 Direction 2 Magnitude 5 Direction 3 Relative to an observer at 0 4
Magnitude 4 Direction 1 Magnitude 3 Direction 2 Magnitude 5 Direction 3 Relative to an observer at 0 4
Magnitude and Direction (MD) An arrow has both a magnitude andA direction relative to a 2nd objectobserver
bull Magnitude and velocity (MV) relative to a 2nd objectobserver
bull Magnitude And acceleration (MA) relative to a 2nd objectobserver2
v da
t t
Introducing a new symbol the ldquoamprdquo sign The ldquoamprdquo sign means resultant ADD all the Arrows for each direction and then SUBTRACT to find the final resultant
1 1amp 1 2 2amp 2 4 If =
1 amp1 2 2 amp2 4 If =
amp 5 amp 3 2 If gt
amp 3 amp 5= 2 If gt
amp 0 3 amp 3=0 if =
only the resultant equals zero
the vectors do not - they are still there
Does not exist There is no multiplication of arrows
There are two arrows One has a magnitude of 3 One has a magnitude of 5 The maximum these two arrows can be is 8
Only the resultant of arrows exist
An SM arrow has a direction and a magnitude
Arrows and magnitudes are still there even if the resultant is zero
There are no negative SM Arrows
There are no positive SM Arrows
There are no ZERO SM Arrows A zero would negate the definition of an SM Arrow something with no magnitude and no direction cannot be the definition of an arrow that is defined as having magnitude and direction
bull A resultant is the addition and subtraction of the differences in magnitude and direction arrows
A B and C are the arrows (a) A amp B exist(b) A amp B amp C exist(c) A-B Do Not Exist in Sm there are no negatives or positives(d) A+B-C Do Not Exist in SM there are no negatives or positives
There are only resultants when there are multiple arrows
SM Graphs versus BS Graphs
Numbers Raised to Powers
When numbers are raised to powers the negative sides of all BS math graphs are ILLOGICAL and INCORRECT
Incorrect BS Math for 2x
2x
4 16 correct3 8 correct2 4 correct1 2 correct0 1 correct-1 005 incorrect-2 025 incorrect-3 0125 incorrect-4 00625 incorrect
negative side of graph is incorrectpositive side of graph is correct
16
8
42105025012500625
-4 -3 -2 -1 0 1 2 3 4
po
sit
ive
negative positive
2x BS Math
correctincorrect
In BS math the dash sign (-) used in this example represents three different operations They arebull (-) Subtractionbull (-) A direction to the leftbull (-) As an exponent to mean divide by
All positive numbers produce correct results and graphs
All negative numbers produce incorrect results and graphs
A direction of three units to the left (-3) is placed in the equation as an exponent
A direction in space (3 units toward the left) changes a whole number to a fraction by using the dash (-) as an exponent rather than a direction in space
This is illogical and incorrect BS math should have performed this operation as
Unfortunately the answers to problems can come out correct because data points of a real problem fit the curve
However the answers will not be due to logical reasoning
Because the equation produced a curve that real problem data points follow it has been falsely believed that the equations and graphs are correct
1
2
x
Incorrect BK graph for (12)x
(12)x
4 00625 correct3 0125 correct2 025 correct1 1 correct0 1 correct-1 2 incorrect-2 4 incorrect-3 8 incorrect-4 16 incorrect
negative side of graph is incorrectpositive side of graph is correct
00625012502505124
8
16
-5 -4 -3 -2 -1 0 1 2 3 4 5P
osit
ive
negative positive
BS Math for (12)x
correctincorrect
Incorrect BS math and graph for
All graphs are symmetrical with SM
Two BS Math Equations that Produce Incorrect Answers
2 0x x starting equation 2
2
x x x x
x x
Add lsquoxrsquo to both sides
2(1) 1 let lsquoxrsquo = 1
2 1 This is an incorrect answer 2(0)-(0)=0 The only answer that BK math can give is x=0
1
a b starting equation ( )( ) ( )( )a a a b multiply both sides by ( )a
( )( ) ( )( ) ( )( ) ( )( )a a b b a b b b subtract ( )( )b b from both sides
( )( ) ( )( )a b a b a b b factor ( )a b from both sides
( )a b b divide both sides by ( )a b
( )b b b replace ( )a with ( )b 2b b simplify 2(1) (1) Let ( )b = 1
2 1 Again BS math yields and incorrect answer 1
This incorrect result is a problem with the distributive law SM explains how this incorrect answer is obtained and the problem with the distributive law that allows the error
Page 148 ndash149 Barry Mazur ndash Imagining Numbers
3
2
31
Do the computation carefully on paper using the rules we agreed to Then ponder your answer which should be something of a surprise to you donrsquot stop there Think of what your answer might possibly mean or might imply
Solve this equation using the distributive law
JK definitely a surprise the answer wasincorrect and illogical
Gerhard Gade University Professor doing math at Harvard University
Here is the BS math answer that Barry Mazur calculated
2
31
2
31
2
31
1 3 1 3 1 3 3 3 2 2 3 1 3
2 2 2 2 4 4 4 4 4 4 2 2
dash
1 3 1 3 1 31
2 2 2 2 4 4
3 3
4 4
When the distributive law is used with the BS Rule-of-Signs and one of the signs is a (-) dash sign the middle two terms always cancel each other This leads to the incorrect answer of (-1) in the above equation
Barry Mazur did not go back and check his answer with the original equation If he had he would have known that the answer was incorrect
3
1 31
2
check answer by setting equal
3
331 3
12
1 31
2
Take the Cube root of both sides
1 3 2 multiply both sides by 2
3 3 subtract 1 from both sides
2 2
3 3 square both sides
3 9 This is an incorrect answer A simple equation check could have shown that the distributive law using BS math provides an incorrect answer
1
BS Illogical and Incorrect Numbers Raised to Powers
nC D n DC
642 DDD
In BS math only a positive number can have an answer if ldquonrdquo is an even number (246---)
All negative numbers have no solutions
642 DDD
With SM just add a direction arrow and any number gives the correct answer
or n nC D C D
3 3( 3 8 8 or 8 2 or 2 )n D C
( 2 16 C= 16 or 16 4 or 4 )n D
Since a negative number has no square root answer in BS math BS mathematicians invented imaginary numbers
i2=-1
24414 iiC n
These equations are easily solved without the use of imaginary numbers with SM math
If 4 then 4 2x
If 4 then 4 2x
The 360-degree circle is a 4400 years old extremely outdated system based on a year of 12 months of 30 days each that is 360 days
SM will use a more logical system that will have a circle of 6283185307o
d
( )
(deg )
d r radians
d r rees
the arc distance around
the circumference of a circle
d
radiusr
2 2
2 2
d r
dv rd t t
v dda rd t t t
1
d
r
( )av tv
The average angular velocity of a body is the rate of its angular displacement
( )0
liminstt
vt
The instantaneous angular velocity is the limit of the ratio
t
as t approaches zero
( )d
av
d
tv
The average circumference velocity of the body is the rate of its distance traveled d divided by the time t
dvv
r
1 dd vdrv
t t t r r
The equation of the relationship between the angular velocity and the velocity of a point on that body
( )o
av
v v v
t ta
The average angular acceleration is the change in the angular velocity
( )0
liminstt
v
ta
The instantaneous angular acceleration is the
limit of the ration of v
t
as t approaches
zero
daa
r
1
1
d
d s d
vva
t r tv a
r t r
The equation of the relationship between the angular acceleration and the acceleration of a point on the body
21
2v a t
If a body with an initial angular velocity ov has a constant
angular acceleration a it will turn through an angle in a time t
1
Straight Line Symbol Units Rotational Symbol Units Distance d m Angular
Distance radians or degrees
Velocity dt
v m
s Angular Velocity
V
t
radians or degrees
s
Acceleration vat
2ms
Angular Acceleration
VA
t
2
radians or degrees
s
Mass M Kg Mass distribution
dM 2Kgm
Force actionA Ma
2
Kgm
sNewton
Torque
dT M A 2
2
Kgm
s
Momentum Mv Kgm
s
Angular Momentum
dM V 2Kgm
s
Work (Force)(distance)
Ad 2
Kgmm
sNm Joule
Work T
Using AS arrow math to solve and direction in space problems using the equation 1 2
2 2 2 2 2 2( ) amp and ( ) amp 3
4
2 2 2
2 2
( ) amp
(5 ) 4 3
(5 ) 16 9
(5 ) 25
(5 ) (5 )
2 2(5 ) 4 3
(5 ) 16 9
(5 ) 25
(5 ) (5 )
In SM math Pythagorasrsquo theorem provides both magnitude and direction
In BS math Pythagorasrsquo theorem provides magnitude only
time set equal to zero
23
( ) ( 0 )10 1
x t xx
3( ) ( 0)
10 1W t f
In BS math this is an equation for a traveling wave
must use a square term in order to make the curve symmetric
NO SQUARE TERM IS NECESSARY
In SM math when t is set to ldquo0rdquo
This is a major fault in BS math
x f(x)
-4 0019
-3 0033
-2 0073
-1 0273
0 3000
1 0273
2 0073
3 0033
4 0019 -4 -3 -2 -1 0 1 2 3 400
05
10
15
20
25
30
35
BS wave (must use square term)
snap shot of wave at x = 0
Am
plit
ude
lt to gt lt to gt f(ltgt)
lt -4 0073
lt -3 0097
lt -2 0143
lt -1 0273
origin 0 3000
gt 1 0273
gt 2 0143
gt 3 0097
gt 4 0073lt0073 lt0097 lt0143 lt0273
3000
0273gt 0143gt 0097gt 0073gt
Wave moving
snap shot of wave
Am
plit
ude
lt4 lt3 lt2 lt1 0 1gt 2gt 3gt 4gt
3
2
1
In SM NO squared terms are needed
SPRING
An Action is needed to stretch or compress a spring to a starting point where it can execute cycle motion
tsA (ts to start) is the work required to move a mass attached to a spring a distance (d1) from
its initial rest position RAK 0 the spring constant is the stiffness of the spring
o A is attraction o R is repulsion o
ARRA
00 is for the back and forth motion through the zero initial rest position A
stretched spring goes from ldquoArdquo to ldquo0rdquo to ldquoRrdquo to ldquoArdquo to ldquo0rdquo to ldquoRrdquo d1 is the distance a mass is moved from its initial rest position
0
0 1ts A RA K d
Once ( )mad
is performed to move the mass to position d1 (against the restraint of the spring) a of the ( )mad
is transferred into the spring as potential attractionrepulsion (PAR) There is always
more ( )mad
performed than is stored as PAR When the spring is released from the restraining Ats the PAR starts the oscillation Conservation of ( )mad
requires that the PAR and the motion ( )mad
( )( )madM be a constant
( )amp ( ) constantmadPAR M 02
0 0
1
2A R A RPAR K d
02
( ) 0 0
1( )
2mad A R A RM m v
0 0 02 2 2
0 0 0 1
2 2 21
2 2 21
1 1 1amp
2 2 2
1 1 1amp
2 2 2
amp
A R A R A RK d m v K d
Kd mv Kd
Kd mv Kd
0
1d is frac12 the height from the base to the peak of a cycle
An experiment of moving paper under a pendulum through a small angle filled with sand that can escape through a small hole in the bottom shows that a pendulum cycling back and forth follows a curve that is mathematically analogous to a lsquosinrsquo or lsquocosrsquo
The projection of the curved area onto the straight line exactly duplicate the back and forth motion of CM This allows for the math of lsquosinrsquo or lsquocosrsquo to be used
SM will use lsquosinrdquo math It could use lsquocosrsquo or a combination of both lsquosinrsquo and lsquocosrsquo Using just the lsquosinrsquo math makes the understanding and math easier
0 sin 0 0SM 0 sin157 1SM The diagram at the left shows a spring being stretched from its (zero rest) position-2 and the PAR being released at position-1 There are 4 sections (1 to 2 2 to 3 3 to 4 and 4 to5) that alternate between maximum acceleration and zero velocity to max velocity and zero acceleration
At position-1 there is maximum acceleration and zero velocity o The lsquosinrsquo values from 0o to 157o will be used for velocity (values from 0 to 1) o The lsquosinrsquo values from 157o to 0o will be used for acceleration (values from 1 to 0)
At position-2 there is maximum velocity and zero acceleration Acceleration changes to deceleration
o The lsquosinrsquo values from 157o to 0o will be used for deceleration (values from 1 to 0) o The lsquosinrsquo values from 0o to 157o will be used for acceleration (values from 0 to 1)
At position-3 there is maximum acceleration and zero velocity o The lsquosinrsquo values from 0o to 157o will be used for velocity (values from 0 to 1) o The lsquosinrsquo values from 157o to 0o will be used for acceleration (values from 1 to 0)
At position-4 there is maximum velocity and zero acceleration Acceleration changes to deceleration
o The lsquosinrsquo values from 157o to 0o will be used for deceleration (values from 1 to 0) o The lsquosinrsquo values from 0o to 157o will be used for acceleration (values from 0 to 1)
At position-5 the cycle starts again
2
)539(
cT
MK
spring constant (K) when mass (M) and cycle-time (T) are known
)(
))(89)((
111 md
uKgM
d
MG
d
FK sts
spring constant (K) when mass (M) and amplitude ( d1) are known
2 21( )
Kv d d
m
velocity (v) at any point during the cycle
velocity
dforM
Kdv
max
012 velocity (v2) maximum (when d = 0 at position-2 and 4)
acceleration (a1) maximum when spring constant (K) mass (M) and amplitude (d1) are known
acceleration (a1) maximum when amplitude (d1) and cycle-time (Tc) are known
acceleration (a) at any point during the cycle-time (T)
1
151
286
v
dTc
cycle-time (Tc) when amplitude (d1) and velocity (v1) are known
K
MTc 28651
cycle-time (Tc) when mass (M) and spring constant (K) are known
fTc
1
cycle-time (Tc) when the frequency is known
cTf
1
frequency (f) when the cycle-time (Tc) is known
of cycles
elapsed timef
frequency (f) when the number of cycles in an elapsed time is known
d1 amplitude (distance from rest to start position)
Also frac12 the distance from the base to the peak
3131 oror dM
Ka
251
3131 539
c
oror T
da
2539
cT
da
In December of 2006 I discovered the mathematical error in Einsteinrsquos Special Relativity Einstein used BS math and that math only works in an imaginary world
It cannot describe real symmetrical space
Einsteinrsquos Special Relativity should be abandoned and a theory of real space using SM should be used to replace it
Numerous articles on the Internet show Einsteinrsquos theory to be incorrect but no one in academia is willing to publish the information
Figure 1-10 A moving particle observed from two frames of reference in uniform translation
Einsteinrsquos Math Error in Special Relativity Transformation of Velocity (From Eisenberg- Fundamentals of Modern Physics pages 27-29) Consider the particle shown in figure (1-10) moving with velocity v as
seen in a frame of reference O
dx dy dz
Vx Vy Vzdt dt dt
dx dy dz
V x V y V zdy dt dt
We would like to evaluate the velocity v of the particle as seen in the frame of reference 0 which is itself moving relative to 0 with velocity v Measured in the 0 frame the velocity vector of the particle has com ponents
The same velocity vector as measured in the 0 frame has components
2
2
2
2
2
1
1
x vtx
v
cy y
z z
vxt
ctv
c
Now from equations (1-13) we know that the relation between the primed and the unprimed coordinates and times is
Take the differential of these equations remembering that v is a constant This gives
2
2
2
2
2
1
1
dx vdtdx
v
cdy dy
dz dz
vxdt
cdtv
c
2
2
2 2 2 2 2
2
2
2
11 ( )
1( ) 1
1
1
1
dx vdt
v dxdx vdt v
dx dx vdtc dt dtvdx vdx vdx v dxdt dt dt dt
dt dtc c c c
VxV x
Vx
VxV x
vv
vv
c
x
v
c
cV
( )
coordinate of particle relative to moving primed system
coordinate of particle relative to stationary system
velocity of primed coordina
V x
Vx
te system relative to stationaryv sys
tem
Page 28 Fundamentals of Modern Physics
(1-16)
According to Einstein this equation tells you how to transform the observed velocity from one frame of reference to another frame of reference First we note that as Vc and vc approachzero the equation approaches those which would be derived from the Galilean transformation
2 2
( ) 1 01 1 ( )
Vx v Vx vV
VxV x
Vx v
v cVxx
v
c
v
cc
Another property of this equation is that it is impossible to choose V and v such that Vrsquo the magnitude of the velocity which is seen in the new frame is greater than c
1
2
09 ( 09 ) 18009876543
( 09 ) 1811 (09 )
c c cV c c
cc
c
Example As seen by O particle 1 has velocity 09c in the direction of positive x and particle 2 has velocity 09c in the negative x direction
Einstein equation however predicts To evaluate the velocity of particle 1 with respect to particle 2 we transform from the O frame to the Orsquo frame moving in the negative x direction with velocity v = -09c using equation 1-16 We obtain
JK LOGIC will observe the particles separating at 18c
A confusing INCORRECT and ILLOGICAL answer
left is a dir rightsub of a
1
2
1
09 ( 09 ) 18009876543
181( 09 )1 ( 09 )
The multiplication of dir left time a dir right is ONLY a dir left
09
dir
direction leftdirection rightsubtraction
c c cV c c
cc
c
V
dir
22
22
09 18 18009876543
181( 081)1(09)(09)
1
left rightsub
sub
c c c cc c
ccc
c
Analyzing Einsteinrsquos BS math Equation
Using SM math a logical answer can be obtained
Cannot multiply a subtraction operator times a direction in space
Can only subtract and change its direction
Cannot multiply a direction to the left times a direction to the right
Cannot multiply opposite directions in space
Can only solve for the resultant
rightleft
subtraction
1
subtraction
2
sub of a dir to the left is a dir to the right
1 resultant 0 sub 2
2sub 2
2
09 09
09 amp 091
09 09 18 1
(0)1
(09) amp (09)1
c cV
c cc
c c cV
cc
cc
8
181
cc
SM math predicts a logical answer using logic
21st century scientists state in many articles that Einsteins General Relativity and Quantum Mechanics are incompatible
They are not only incompatible they are both mathematically incorrect
They both use illogical BS math
Any equations in math that use imaginary numbers should be abandoned
Imaginary numbers are only necessary in the current BS math system
This means that Schroumldingers equations should be abandoned
Any equation in quantum mechanics that uses imaginary number should be abandoned
Prof Kanarev has developed logical physics and chemistrytheories without the use of Quantum mechanics or relativity
A partial list of his theories coverbull Mechanics ndash replaces Newtonbull Photon - bull Electrons ndash Protons- Neutronsbull Spectroscopybull Nucleusbull Atoms amp Moleculesbull Electrodynamicsbull Heatbull Curvature of Space ndash replaces Schwarzcchildrsquos equation
Kanarevrsquos Hypothesis
bull Electron align linearly with proton They do NOT spin or form clouds around the nucleus
bull Neutrons are in the center of atom Proton are outside of Neutrons
bull The wavelength of all particles is the experimental radius
bull Double slit experiment explained correctly
bull Heisenbergrsquos inequality explained correctly
bull is the radius of approach of the magnetic lines around the torus of the electron
( 1 137)
bull Why water expands when it freezes
bull Electron clusters form lightning
rbull The meaning of Plancks constant is explained
22 2 kg m
h m f mr f consts
2210254 42m r kgm bull If the mr-constant describes a ring the equations
describe experiments bull If the mr-constant describes a wave the equations do
NOT describe experiments
Kanarevrsquos 5 Equations of Mechanics
Kanarevrsquos 1st Equation Acceleration ampmv maMma R R
Kanarevrsquos 2nd Equation Uniform motion amp mv maMma R R
maM mvR R
Kanarevrsquos 3rd Equation
Kanarevrsquos 4th Equation Equal and opposite reaction
1 21 21 2
12
a m
m a m ama
Kanarevrsquos 5th Equation 1 2 3 Mi M M M Mna R R R R R
The sum of the Actions working on a moving body is never equal to zero
Kanarevrsquos Photon model is a rolling ring
His equations accurately describe known experiments with photonsbull Electromagnetic radiation should be called photon radiationbull Diffraction and interference are the same phenomenabull Correctly explains the double slit experiment without quantum equationsbull Correctly explains the meaning of Heisenbergrsquos inequality
e
2 (2)(3142)(2817 15 ) 10007297352568
R 2426 12 137mag
xp
r m
m
Charge
Magnetic action
Proton
The intensity of the magnetic action near the geometrical center of the proton ring is so great that we theorize that it is the strong nuclear force
(9383 8)(1602 19)8476 14
4 (4)(3142)(1411 26)p
pp
EB T
M
(9395 8)(1602 19)1035 15
4 (4)(3142)(0966 26)N
NN
EB T
M
Neutron
1a Hydrogenavi
Cross sign (+) CODES
(+) a direction in space labeled positive (whatever that means)
(+) an addition operator
(+) used to show the number of zeros to the left of a decimal point 10+5 = 100000
In Physics and Chemistry
bull (+) Protons have been Labeled as positive
bull (+) numerous other particles have been Labeled as positive
bull (+) amp (-) Thermochemistry
In Math used
The same dash symbol (-) is used for the numerous different math operations
The definitions established for the use of the dash symbol (-) do not distinguish between their different operations
In many math operations the dash symbol is changed to mean one of the other meanings
It is amazing that math has proceeded to its current level of use with this illogical use of a symbol
The same cross symbol (+) is used for numerous different math operations and produces the same illogical answers
A Summary of the BS Broken-Symmetry math
1 1 cosh tanhln
x xe x xe
bull Originated with broken-symmetry in the XYZ plane
bull Originated with mirror-broken-symmetry in the XZ plane
bull Created a Rule-of-Signs that cannot be proved (-)(-)=(+) (+)(+)=(+) (-)(+)=(-)
bull Created imaginary numbers to compensate for broken-symmetry
bull Created absolute values to change illogical negative number answers into positive number answers
bull Produces broken symmetry graphs for many functions
It took four years of study to discover why a (-)(-)=(+) only works in an imaginary world where space in the dash (-) direction is different from space in the cross (+) direction
Started working on the problem in August of 2001
Data points created using BS math produce graphs that are not symmetrical dash side different from cross side
However if a real problems data points follow a non symmetrical graph the graph can provide usable answers
This is why no one discovered the problem
Solved in January of 2005
Symmetry-Math(SM) vs Broken-Symmetry(BS) Math
In SM if we specify that the dash sign (-) means only subtraction and the cross sign (+) means only addition and an appropriate symbol is used for a direction in space space becomes symmetrical and math becomes logical
Math operators and directions in space are not the same and the same symbol should not be used to represent them
Letrsquos start with the (dash)multiplied by a (dash)=(cross) in the BS system
Usable answers may be obtained but it will be for illogical reasons
multiplied by a ( ) ( ) ( )subtraction Subtraction addition
rightleft left
dash dash cross
positivenegative negative
Symmetry-Math The subtraction of a direction is equal to the opposite direction The answers are correct using correct logic
opposite direction direction
SM
(negative)(negative) = (positive)
( ) ( ) ( )subtraction
BS
There is NO multiplication of a subtraction operator by a direction in space
There is just the subtraction of a direction in space
If instead of labeling the left side of a coordinate system as a negative (-) the same as a subtraction operator we label it with an arrow ( ) to represent the direction then a subtraction from that arrow direction will be in the opposite direction ( )
( ) mutiplied by a ( ) ( )
negative Positive negative
subtraction addition subtraction
dash cross dash
left right left
A subtraction operator multiplied by an addition operator is equal to a subtraction operator This is illogical
BS Math
A negative multiplied by a positive is equal to a negative This is illogical
A direction to the left multiplied by a direction to the right is a direction to the left This is illogical and Einsteins math error in special relativity
A dash times a cross is equal to a dash This is illogical
A vector moving to the left multiplied by a vector moving to the right is equal to ONLY to a vector moving to the left that is eight orders of magnitude faster thanThe speed of light
opposite direction subtraction direction
SM ( ) ( )
BS (negative)(positive) = (negative)
Symmetry Math The subtraction of a direction is equal to the opposite direction
There is no multiplication of a subtractionoperator times a direction in space
( ) multiplied by a ( ) ( )
positive positive positiveaddition addition additioncross cross cross
direction direction direction
What is the meaning of multiplying addition operators Addition operators are not multiplication operators
BS Math
() multiplied by a direction ( )=
number
() multiplied by a direction ( )=
number
A number times a direction maintains the same direction
( ) ()
dirdiradd
( ) ()
dirdiradd
The addition of a direction is in the same direction There is no multiplication by addition operators
Symmetry Math
Future math books need to eliminate the use of the cross sign (+) to represent something labeled a positive direction in space and the dash sign (-) to represent something labeled a negative direction in space Space does not have positive and negative directions
Symbols that are logical and have no illogical representation should be adopted SM uses either arrows or symbols
Symmetry-Math Number Line
Symmetry-Math Rules
The dash sign (-) will have only one use subtraction
The cross sign (+) will have only one use addition
Directions have an arrow and a number
All observers see the same direction and magnitude
bull There is no multiplication of arrows
amp ampamp ampampamp
( 3 ) ( 5 ) ( 6 ) ( 65 ) (17256 ) ( 32993 )
dir dirdir dirdirdir
Negative and Positive Directions in Space
If I asked you to point to a negative direction in space which way would you point
Hopefully you will realize that there is no such thing as a negative direction in space
If I asked you to point to a positive direction in space which way would you point
Again hopefully you will realize that there is no such thing as a positive direction in space
Objects on the right side of the x-axis are positive (+)(+) = (+)
Objects on the left side of the x-axis are negative (-)(-) = (+) negative math is different from positive math This is illogical
Objects on the top of the y-axis are positive
Objects on the bottom of the y-axis are negative
Objects in the front of the z-axis are positive
Objects in the back of the z-axis are negative
Again BS math of the positive direction (x right y up and z front) is different from the BS math of the negative direction (x left y down and z back)
BS math is Broken-Symmetry
Mirror image is broken symmetry in the x and z axis If the left amp right-axis are
reversed symmetry is broken Obs-1 math is different from obs-2
If the front amp back-axis is reversed symmetry is broken Obs-1 math is different from obs-2
If the top amp bottom-axis are
reversed symmetry is not broken Obs-1 math is the same as obs-2
Math answers cannot depend on which side of a number line an observer sit
Absolute values for Displacement
Even for a simple displacement of an object BS math invented a definition and absolute values to obtain something labeled a positive number answer
I----------I----------I----------I----------I----------I----------I----------I----------I
-4 -3 -2 -1 0 +1 +2 +3 +4 a gtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtb cltltltltltltltltltltltltltltltltltltltltltltltltltltltb
If an object starts at ( 3)ia X and moves to ( X 3) and the to ( 1)b fb c X
BS Math provides a usable answer without a definition or absolute values
However If an object starts at
( 1) and moves to ( X 3) and the to ( 3)i b fa X b c X
Then absolute values and a definition must be used
---------I----------I----------I----------I----------I----------I----------I---------- -4 -3 -2 -1 0 +1 +2 +3 +4
a gtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtb cltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltb
BS Math makes up a definition that all displacements are positive Since this example provides an answer with something labeled a negative the positive definition must be applied and (-2) is changed to (+2) A definition is needed to arrive at a useable answer This is Illogical
In SM absolute values and a meaningless definition of ldquopositiverdquo are removed In SM objects and directions in space are defined by the direction and magnitude of the resultant of arrows
With SM you get total distance traveled by the object and the final direction of the displacement
22222 2 bababbaababababa If w e su b stitu te p o s itiv e n u m b e rs in to th e eq u a tio n th e an sw er w ill b e co rrec t If (a = 5 ) an d (b = 3 )
2
2
8
5 3 5 3 5 3 5 5 5 3 3 5 3 3 25 15 15 9 64
8 8 8 8 8 8 8 8 8 8 64
W h en tw o p o sitiv e n u m b ers a re a d d ed an d sq u a re d th e B S m a th d is tr ib u tiv e law p ro v id es a co rre c t an sw er 1
Illogical and Incorrect BS math for the distributive law
222 2 bababbabbaaabababa B y d e fin itio n in B S m a th n o n u m b er sq u a red c a n ev e r b e n eg a tiv e T h e re fo re n o sq u a red n u m b er c an ev e r b e a v e c to r m o v in g in th e d a sh ( -) o r n eg a tiv e d ire c tio n if it is sq u a red In B S m a th an d ad d itio n o p e ra to r m u ltip lied b y a su b trac tio n o p e ra to r is A L W A Y S eq u a l to a su b tra c tio n o p e ra to r If th is h ad n o t b e en u sed fo r th e la s t fo u r h u n d red y e a rs y o u w o u ld d ie lau g h in g a t th e u tte r ab su rd ity o f th e la s t s ta tem e n t T h is is illo g ica l an d v io la te s sy m m etry
If a = 5 an d b = 3 42222352
22
4930253)3)(5(253)5)(3()3)(5(5353535 22222 A s lo n g a s ldquo ardquo is g re a te r th an ldquob rdquo th e B S m a th d is tr ib u tiv e law p ro v id es co rrec t an sw ers 1
B S m a th b re a k s d o w n w h e n ldquo b rdquo is g rea ter th a n ldquo a rdquo
If a = 3 an d b = 5
2 2
2
2
3 5 2 4
2 ( 2) ( 2) 4
2 ( )( 2) ( )( 2) 4
A n y co m b in a tio n ex cep t th e m id d le o n e is illo g ica l an d in co rre c t B y d e fin itio n th e an sw er can n o t b e -4 in B S m a th B S m a th is in co rrec t
4253095)5)(3(235)3)(5()5)(3(3535353 22222 T h is is w h ere th e d is tr ib u tiv e la w p ro v id es in co rrec t an sw ers I t p ro d u ces an an sw er o f + 4 b e cau se b y d e fin iti o n n o n u m b er sq u a red in B S m a th c an b e a n eg a tiv e n u m b er U sin g in co rre c t B S m a th th e d is tr ib u tiv e law w ill g iv e an in co rrec t an sw ers o f + 4 T h is is in co rrec t an d illo g ica l 1
Since there can be no negative direction from a squared term in BS math BS math should be modified or abandoned
The BS distributive law should be modified or abandoned
The numbers +1+2+3+4 and ndash1-2-3-4hellip should be abandoned
BS math gives incorrect answers because their Rule-of-Signs allows multiplying an arrow going in one direction by an arrow going in another direction
Clearly this is not logical
Arrows should be used that are specific for directions
W h at d o es it m e an to m u ltip ly ( 4) b y (4 ) A s w e k n o w 1
m u ltip lica tio n is ju s t ad d itio n W h a t w o u ld ( 4) tim e (4 ) b e 2
In BS math the answer is by definition -16 unless you square -4 and then it is +16
In SM you cannot multiply arrows going in opposite directions
The middle two terms are not logical You cannot multiply opposite directions
This is a MAJOR error in BS math They multiply a dash [(-) a direction to the left] by a cross [(+) a direction to the right] And then ldquoby definitionrdquo label the answer dash (-) This is illogical and produces incorrect answers
2
2 2
This is Einsteinrsquos math error
in Special relativity
Magnitude 4 Direction 1 Magnitude 3 Direction 2 Magnitude 5 Direction 3 Relative to an observer at 0 4
Magnitude 4 Direction 1 Magnitude 3 Direction 2 Magnitude 5 Direction 3 Relative to an observer at 0 4
Magnitude and Direction (MD) An arrow has both a magnitude andA direction relative to a 2nd objectobserver
bull Magnitude and velocity (MV) relative to a 2nd objectobserver
bull Magnitude And acceleration (MA) relative to a 2nd objectobserver2
v da
t t
Introducing a new symbol the ldquoamprdquo sign The ldquoamprdquo sign means resultant ADD all the Arrows for each direction and then SUBTRACT to find the final resultant
1 1amp 1 2 2amp 2 4 If =
1 amp1 2 2 amp2 4 If =
amp 5 amp 3 2 If gt
amp 3 amp 5= 2 If gt
amp 0 3 amp 3=0 if =
only the resultant equals zero
the vectors do not - they are still there
Does not exist There is no multiplication of arrows
There are two arrows One has a magnitude of 3 One has a magnitude of 5 The maximum these two arrows can be is 8
Only the resultant of arrows exist
An SM arrow has a direction and a magnitude
Arrows and magnitudes are still there even if the resultant is zero
There are no negative SM Arrows
There are no positive SM Arrows
There are no ZERO SM Arrows A zero would negate the definition of an SM Arrow something with no magnitude and no direction cannot be the definition of an arrow that is defined as having magnitude and direction
bull A resultant is the addition and subtraction of the differences in magnitude and direction arrows
A B and C are the arrows (a) A amp B exist(b) A amp B amp C exist(c) A-B Do Not Exist in Sm there are no negatives or positives(d) A+B-C Do Not Exist in SM there are no negatives or positives
There are only resultants when there are multiple arrows
SM Graphs versus BS Graphs
Numbers Raised to Powers
When numbers are raised to powers the negative sides of all BS math graphs are ILLOGICAL and INCORRECT
Incorrect BS Math for 2x
2x
4 16 correct3 8 correct2 4 correct1 2 correct0 1 correct-1 005 incorrect-2 025 incorrect-3 0125 incorrect-4 00625 incorrect
negative side of graph is incorrectpositive side of graph is correct
16
8
42105025012500625
-4 -3 -2 -1 0 1 2 3 4
po
sit
ive
negative positive
2x BS Math
correctincorrect
In BS math the dash sign (-) used in this example represents three different operations They arebull (-) Subtractionbull (-) A direction to the leftbull (-) As an exponent to mean divide by
All positive numbers produce correct results and graphs
All negative numbers produce incorrect results and graphs
A direction of three units to the left (-3) is placed in the equation as an exponent
A direction in space (3 units toward the left) changes a whole number to a fraction by using the dash (-) as an exponent rather than a direction in space
This is illogical and incorrect BS math should have performed this operation as
Unfortunately the answers to problems can come out correct because data points of a real problem fit the curve
However the answers will not be due to logical reasoning
Because the equation produced a curve that real problem data points follow it has been falsely believed that the equations and graphs are correct
1
2
x
Incorrect BK graph for (12)x
(12)x
4 00625 correct3 0125 correct2 025 correct1 1 correct0 1 correct-1 2 incorrect-2 4 incorrect-3 8 incorrect-4 16 incorrect
negative side of graph is incorrectpositive side of graph is correct
00625012502505124
8
16
-5 -4 -3 -2 -1 0 1 2 3 4 5P
osit
ive
negative positive
BS Math for (12)x
correctincorrect
Incorrect BS math and graph for
All graphs are symmetrical with SM
Two BS Math Equations that Produce Incorrect Answers
2 0x x starting equation 2
2
x x x x
x x
Add lsquoxrsquo to both sides
2(1) 1 let lsquoxrsquo = 1
2 1 This is an incorrect answer 2(0)-(0)=0 The only answer that BK math can give is x=0
1
a b starting equation ( )( ) ( )( )a a a b multiply both sides by ( )a
( )( ) ( )( ) ( )( ) ( )( )a a b b a b b b subtract ( )( )b b from both sides
( )( ) ( )( )a b a b a b b factor ( )a b from both sides
( )a b b divide both sides by ( )a b
( )b b b replace ( )a with ( )b 2b b simplify 2(1) (1) Let ( )b = 1
2 1 Again BS math yields and incorrect answer 1
This incorrect result is a problem with the distributive law SM explains how this incorrect answer is obtained and the problem with the distributive law that allows the error
Page 148 ndash149 Barry Mazur ndash Imagining Numbers
3
2
31
Do the computation carefully on paper using the rules we agreed to Then ponder your answer which should be something of a surprise to you donrsquot stop there Think of what your answer might possibly mean or might imply
Solve this equation using the distributive law
JK definitely a surprise the answer wasincorrect and illogical
Gerhard Gade University Professor doing math at Harvard University
Here is the BS math answer that Barry Mazur calculated
2
31
2
31
2
31
1 3 1 3 1 3 3 3 2 2 3 1 3
2 2 2 2 4 4 4 4 4 4 2 2
dash
1 3 1 3 1 31
2 2 2 2 4 4
3 3
4 4
When the distributive law is used with the BS Rule-of-Signs and one of the signs is a (-) dash sign the middle two terms always cancel each other This leads to the incorrect answer of (-1) in the above equation
Barry Mazur did not go back and check his answer with the original equation If he had he would have known that the answer was incorrect
3
1 31
2
check answer by setting equal
3
331 3
12
1 31
2
Take the Cube root of both sides
1 3 2 multiply both sides by 2
3 3 subtract 1 from both sides
2 2
3 3 square both sides
3 9 This is an incorrect answer A simple equation check could have shown that the distributive law using BS math provides an incorrect answer
1
BS Illogical and Incorrect Numbers Raised to Powers
nC D n DC
642 DDD
In BS math only a positive number can have an answer if ldquonrdquo is an even number (246---)
All negative numbers have no solutions
642 DDD
With SM just add a direction arrow and any number gives the correct answer
or n nC D C D
3 3( 3 8 8 or 8 2 or 2 )n D C
( 2 16 C= 16 or 16 4 or 4 )n D
Since a negative number has no square root answer in BS math BS mathematicians invented imaginary numbers
i2=-1
24414 iiC n
These equations are easily solved without the use of imaginary numbers with SM math
If 4 then 4 2x
If 4 then 4 2x
The 360-degree circle is a 4400 years old extremely outdated system based on a year of 12 months of 30 days each that is 360 days
SM will use a more logical system that will have a circle of 6283185307o
d
( )
(deg )
d r radians
d r rees
the arc distance around
the circumference of a circle
d
radiusr
2 2
2 2
d r
dv rd t t
v dda rd t t t
1
d
r
( )av tv
The average angular velocity of a body is the rate of its angular displacement
( )0
liminstt
vt
The instantaneous angular velocity is the limit of the ratio
t
as t approaches zero
( )d
av
d
tv
The average circumference velocity of the body is the rate of its distance traveled d divided by the time t
dvv
r
1 dd vdrv
t t t r r
The equation of the relationship between the angular velocity and the velocity of a point on that body
( )o
av
v v v
t ta
The average angular acceleration is the change in the angular velocity
( )0
liminstt
v
ta
The instantaneous angular acceleration is the
limit of the ration of v
t
as t approaches
zero
daa
r
1
1
d
d s d
vva
t r tv a
r t r
The equation of the relationship between the angular acceleration and the acceleration of a point on the body
21
2v a t
If a body with an initial angular velocity ov has a constant
angular acceleration a it will turn through an angle in a time t
1
Straight Line Symbol Units Rotational Symbol Units Distance d m Angular
Distance radians or degrees
Velocity dt
v m
s Angular Velocity
V
t
radians or degrees
s
Acceleration vat
2ms
Angular Acceleration
VA
t
2
radians or degrees
s
Mass M Kg Mass distribution
dM 2Kgm
Force actionA Ma
2
Kgm
sNewton
Torque
dT M A 2
2
Kgm
s
Momentum Mv Kgm
s
Angular Momentum
dM V 2Kgm
s
Work (Force)(distance)
Ad 2
Kgmm
sNm Joule
Work T
Using AS arrow math to solve and direction in space problems using the equation 1 2
2 2 2 2 2 2( ) amp and ( ) amp 3
4
2 2 2
2 2
( ) amp
(5 ) 4 3
(5 ) 16 9
(5 ) 25
(5 ) (5 )
2 2(5 ) 4 3
(5 ) 16 9
(5 ) 25
(5 ) (5 )
In SM math Pythagorasrsquo theorem provides both magnitude and direction
In BS math Pythagorasrsquo theorem provides magnitude only
time set equal to zero
23
( ) ( 0 )10 1
x t xx
3( ) ( 0)
10 1W t f
In BS math this is an equation for a traveling wave
must use a square term in order to make the curve symmetric
NO SQUARE TERM IS NECESSARY
In SM math when t is set to ldquo0rdquo
This is a major fault in BS math
x f(x)
-4 0019
-3 0033
-2 0073
-1 0273
0 3000
1 0273
2 0073
3 0033
4 0019 -4 -3 -2 -1 0 1 2 3 400
05
10
15
20
25
30
35
BS wave (must use square term)
snap shot of wave at x = 0
Am
plit
ude
lt to gt lt to gt f(ltgt)
lt -4 0073
lt -3 0097
lt -2 0143
lt -1 0273
origin 0 3000
gt 1 0273
gt 2 0143
gt 3 0097
gt 4 0073lt0073 lt0097 lt0143 lt0273
3000
0273gt 0143gt 0097gt 0073gt
Wave moving
snap shot of wave
Am
plit
ude
lt4 lt3 lt2 lt1 0 1gt 2gt 3gt 4gt
3
2
1
In SM NO squared terms are needed
SPRING
An Action is needed to stretch or compress a spring to a starting point where it can execute cycle motion
tsA (ts to start) is the work required to move a mass attached to a spring a distance (d1) from
its initial rest position RAK 0 the spring constant is the stiffness of the spring
o A is attraction o R is repulsion o
ARRA
00 is for the back and forth motion through the zero initial rest position A
stretched spring goes from ldquoArdquo to ldquo0rdquo to ldquoRrdquo to ldquoArdquo to ldquo0rdquo to ldquoRrdquo d1 is the distance a mass is moved from its initial rest position
0
0 1ts A RA K d
Once ( )mad
is performed to move the mass to position d1 (against the restraint of the spring) a of the ( )mad
is transferred into the spring as potential attractionrepulsion (PAR) There is always
more ( )mad
performed than is stored as PAR When the spring is released from the restraining Ats the PAR starts the oscillation Conservation of ( )mad
requires that the PAR and the motion ( )mad
( )( )madM be a constant
( )amp ( ) constantmadPAR M 02
0 0
1
2A R A RPAR K d
02
( ) 0 0
1( )
2mad A R A RM m v
0 0 02 2 2
0 0 0 1
2 2 21
2 2 21
1 1 1amp
2 2 2
1 1 1amp
2 2 2
amp
A R A R A RK d m v K d
Kd mv Kd
Kd mv Kd
0
1d is frac12 the height from the base to the peak of a cycle
An experiment of moving paper under a pendulum through a small angle filled with sand that can escape through a small hole in the bottom shows that a pendulum cycling back and forth follows a curve that is mathematically analogous to a lsquosinrsquo or lsquocosrsquo
The projection of the curved area onto the straight line exactly duplicate the back and forth motion of CM This allows for the math of lsquosinrsquo or lsquocosrsquo to be used
SM will use lsquosinrdquo math It could use lsquocosrsquo or a combination of both lsquosinrsquo and lsquocosrsquo Using just the lsquosinrsquo math makes the understanding and math easier
0 sin 0 0SM 0 sin157 1SM The diagram at the left shows a spring being stretched from its (zero rest) position-2 and the PAR being released at position-1 There are 4 sections (1 to 2 2 to 3 3 to 4 and 4 to5) that alternate between maximum acceleration and zero velocity to max velocity and zero acceleration
At position-1 there is maximum acceleration and zero velocity o The lsquosinrsquo values from 0o to 157o will be used for velocity (values from 0 to 1) o The lsquosinrsquo values from 157o to 0o will be used for acceleration (values from 1 to 0)
At position-2 there is maximum velocity and zero acceleration Acceleration changes to deceleration
o The lsquosinrsquo values from 157o to 0o will be used for deceleration (values from 1 to 0) o The lsquosinrsquo values from 0o to 157o will be used for acceleration (values from 0 to 1)
At position-3 there is maximum acceleration and zero velocity o The lsquosinrsquo values from 0o to 157o will be used for velocity (values from 0 to 1) o The lsquosinrsquo values from 157o to 0o will be used for acceleration (values from 1 to 0)
At position-4 there is maximum velocity and zero acceleration Acceleration changes to deceleration
o The lsquosinrsquo values from 157o to 0o will be used for deceleration (values from 1 to 0) o The lsquosinrsquo values from 0o to 157o will be used for acceleration (values from 0 to 1)
At position-5 the cycle starts again
2
)539(
cT
MK
spring constant (K) when mass (M) and cycle-time (T) are known
)(
))(89)((
111 md
uKgM
d
MG
d
FK sts
spring constant (K) when mass (M) and amplitude ( d1) are known
2 21( )
Kv d d
m
velocity (v) at any point during the cycle
velocity
dforM
Kdv
max
012 velocity (v2) maximum (when d = 0 at position-2 and 4)
acceleration (a1) maximum when spring constant (K) mass (M) and amplitude (d1) are known
acceleration (a1) maximum when amplitude (d1) and cycle-time (Tc) are known
acceleration (a) at any point during the cycle-time (T)
1
151
286
v
dTc
cycle-time (Tc) when amplitude (d1) and velocity (v1) are known
K
MTc 28651
cycle-time (Tc) when mass (M) and spring constant (K) are known
fTc
1
cycle-time (Tc) when the frequency is known
cTf
1
frequency (f) when the cycle-time (Tc) is known
of cycles
elapsed timef
frequency (f) when the number of cycles in an elapsed time is known
d1 amplitude (distance from rest to start position)
Also frac12 the distance from the base to the peak
3131 oror dM
Ka
251
3131 539
c
oror T
da
2539
cT
da
In December of 2006 I discovered the mathematical error in Einsteinrsquos Special Relativity Einstein used BS math and that math only works in an imaginary world
It cannot describe real symmetrical space
Einsteinrsquos Special Relativity should be abandoned and a theory of real space using SM should be used to replace it
Numerous articles on the Internet show Einsteinrsquos theory to be incorrect but no one in academia is willing to publish the information
Figure 1-10 A moving particle observed from two frames of reference in uniform translation
Einsteinrsquos Math Error in Special Relativity Transformation of Velocity (From Eisenberg- Fundamentals of Modern Physics pages 27-29) Consider the particle shown in figure (1-10) moving with velocity v as
seen in a frame of reference O
dx dy dz
Vx Vy Vzdt dt dt
dx dy dz
V x V y V zdy dt dt
We would like to evaluate the velocity v of the particle as seen in the frame of reference 0 which is itself moving relative to 0 with velocity v Measured in the 0 frame the velocity vector of the particle has com ponents
The same velocity vector as measured in the 0 frame has components
2
2
2
2
2
1
1
x vtx
v
cy y
z z
vxt
ctv
c
Now from equations (1-13) we know that the relation between the primed and the unprimed coordinates and times is
Take the differential of these equations remembering that v is a constant This gives
2
2
2
2
2
1
1
dx vdtdx
v
cdy dy
dz dz
vxdt
cdtv
c
2
2
2 2 2 2 2
2
2
2
11 ( )
1( ) 1
1
1
1
dx vdt
v dxdx vdt v
dx dx vdtc dt dtvdx vdx vdx v dxdt dt dt dt
dt dtc c c c
VxV x
Vx
VxV x
vv
vv
c
x
v
c
cV
( )
coordinate of particle relative to moving primed system
coordinate of particle relative to stationary system
velocity of primed coordina
V x
Vx
te system relative to stationaryv sys
tem
Page 28 Fundamentals of Modern Physics
(1-16)
According to Einstein this equation tells you how to transform the observed velocity from one frame of reference to another frame of reference First we note that as Vc and vc approachzero the equation approaches those which would be derived from the Galilean transformation
2 2
( ) 1 01 1 ( )
Vx v Vx vV
VxV x
Vx v
v cVxx
v
c
v
cc
Another property of this equation is that it is impossible to choose V and v such that Vrsquo the magnitude of the velocity which is seen in the new frame is greater than c
1
2
09 ( 09 ) 18009876543
( 09 ) 1811 (09 )
c c cV c c
cc
c
Example As seen by O particle 1 has velocity 09c in the direction of positive x and particle 2 has velocity 09c in the negative x direction
Einstein equation however predicts To evaluate the velocity of particle 1 with respect to particle 2 we transform from the O frame to the Orsquo frame moving in the negative x direction with velocity v = -09c using equation 1-16 We obtain
JK LOGIC will observe the particles separating at 18c
A confusing INCORRECT and ILLOGICAL answer
left is a dir rightsub of a
1
2
1
09 ( 09 ) 18009876543
181( 09 )1 ( 09 )
The multiplication of dir left time a dir right is ONLY a dir left
09
dir
direction leftdirection rightsubtraction
c c cV c c
cc
c
V
dir
22
22
09 18 18009876543
181( 081)1(09)(09)
1
left rightsub
sub
c c c cc c
ccc
c
Analyzing Einsteinrsquos BS math Equation
Using SM math a logical answer can be obtained
Cannot multiply a subtraction operator times a direction in space
Can only subtract and change its direction
Cannot multiply a direction to the left times a direction to the right
Cannot multiply opposite directions in space
Can only solve for the resultant
rightleft
subtraction
1
subtraction
2
sub of a dir to the left is a dir to the right
1 resultant 0 sub 2
2sub 2
2
09 09
09 amp 091
09 09 18 1
(0)1
(09) amp (09)1
c cV
c cc
c c cV
cc
cc
8
181
cc
SM math predicts a logical answer using logic
21st century scientists state in many articles that Einsteins General Relativity and Quantum Mechanics are incompatible
They are not only incompatible they are both mathematically incorrect
They both use illogical BS math
Any equations in math that use imaginary numbers should be abandoned
Imaginary numbers are only necessary in the current BS math system
This means that Schroumldingers equations should be abandoned
Any equation in quantum mechanics that uses imaginary number should be abandoned
Prof Kanarev has developed logical physics and chemistrytheories without the use of Quantum mechanics or relativity
A partial list of his theories coverbull Mechanics ndash replaces Newtonbull Photon - bull Electrons ndash Protons- Neutronsbull Spectroscopybull Nucleusbull Atoms amp Moleculesbull Electrodynamicsbull Heatbull Curvature of Space ndash replaces Schwarzcchildrsquos equation
Kanarevrsquos Hypothesis
bull Electron align linearly with proton They do NOT spin or form clouds around the nucleus
bull Neutrons are in the center of atom Proton are outside of Neutrons
bull The wavelength of all particles is the experimental radius
bull Double slit experiment explained correctly
bull Heisenbergrsquos inequality explained correctly
bull is the radius of approach of the magnetic lines around the torus of the electron
( 1 137)
bull Why water expands when it freezes
bull Electron clusters form lightning
rbull The meaning of Plancks constant is explained
22 2 kg m
h m f mr f consts
2210254 42m r kgm bull If the mr-constant describes a ring the equations
describe experiments bull If the mr-constant describes a wave the equations do
NOT describe experiments
Kanarevrsquos 5 Equations of Mechanics
Kanarevrsquos 1st Equation Acceleration ampmv maMma R R
Kanarevrsquos 2nd Equation Uniform motion amp mv maMma R R
maM mvR R
Kanarevrsquos 3rd Equation
Kanarevrsquos 4th Equation Equal and opposite reaction
1 21 21 2
12
a m
m a m ama
Kanarevrsquos 5th Equation 1 2 3 Mi M M M Mna R R R R R
The sum of the Actions working on a moving body is never equal to zero
Kanarevrsquos Photon model is a rolling ring
His equations accurately describe known experiments with photonsbull Electromagnetic radiation should be called photon radiationbull Diffraction and interference are the same phenomenabull Correctly explains the double slit experiment without quantum equationsbull Correctly explains the meaning of Heisenbergrsquos inequality
e
2 (2)(3142)(2817 15 ) 10007297352568
R 2426 12 137mag
xp
r m
m
Charge
Magnetic action
Proton
The intensity of the magnetic action near the geometrical center of the proton ring is so great that we theorize that it is the strong nuclear force
(9383 8)(1602 19)8476 14
4 (4)(3142)(1411 26)p
pp
EB T
M
(9395 8)(1602 19)1035 15
4 (4)(3142)(0966 26)N
NN
EB T
M
Neutron
1a Hydrogenavi
The same dash symbol (-) is used for the numerous different math operations
The definitions established for the use of the dash symbol (-) do not distinguish between their different operations
In many math operations the dash symbol is changed to mean one of the other meanings
It is amazing that math has proceeded to its current level of use with this illogical use of a symbol
The same cross symbol (+) is used for numerous different math operations and produces the same illogical answers
A Summary of the BS Broken-Symmetry math
1 1 cosh tanhln
x xe x xe
bull Originated with broken-symmetry in the XYZ plane
bull Originated with mirror-broken-symmetry in the XZ plane
bull Created a Rule-of-Signs that cannot be proved (-)(-)=(+) (+)(+)=(+) (-)(+)=(-)
bull Created imaginary numbers to compensate for broken-symmetry
bull Created absolute values to change illogical negative number answers into positive number answers
bull Produces broken symmetry graphs for many functions
It took four years of study to discover why a (-)(-)=(+) only works in an imaginary world where space in the dash (-) direction is different from space in the cross (+) direction
Started working on the problem in August of 2001
Data points created using BS math produce graphs that are not symmetrical dash side different from cross side
However if a real problems data points follow a non symmetrical graph the graph can provide usable answers
This is why no one discovered the problem
Solved in January of 2005
Symmetry-Math(SM) vs Broken-Symmetry(BS) Math
In SM if we specify that the dash sign (-) means only subtraction and the cross sign (+) means only addition and an appropriate symbol is used for a direction in space space becomes symmetrical and math becomes logical
Math operators and directions in space are not the same and the same symbol should not be used to represent them
Letrsquos start with the (dash)multiplied by a (dash)=(cross) in the BS system
Usable answers may be obtained but it will be for illogical reasons
multiplied by a ( ) ( ) ( )subtraction Subtraction addition
rightleft left
dash dash cross
positivenegative negative
Symmetry-Math The subtraction of a direction is equal to the opposite direction The answers are correct using correct logic
opposite direction direction
SM
(negative)(negative) = (positive)
( ) ( ) ( )subtraction
BS
There is NO multiplication of a subtraction operator by a direction in space
There is just the subtraction of a direction in space
If instead of labeling the left side of a coordinate system as a negative (-) the same as a subtraction operator we label it with an arrow ( ) to represent the direction then a subtraction from that arrow direction will be in the opposite direction ( )
( ) mutiplied by a ( ) ( )
negative Positive negative
subtraction addition subtraction
dash cross dash
left right left
A subtraction operator multiplied by an addition operator is equal to a subtraction operator This is illogical
BS Math
A negative multiplied by a positive is equal to a negative This is illogical
A direction to the left multiplied by a direction to the right is a direction to the left This is illogical and Einsteins math error in special relativity
A dash times a cross is equal to a dash This is illogical
A vector moving to the left multiplied by a vector moving to the right is equal to ONLY to a vector moving to the left that is eight orders of magnitude faster thanThe speed of light
opposite direction subtraction direction
SM ( ) ( )
BS (negative)(positive) = (negative)
Symmetry Math The subtraction of a direction is equal to the opposite direction
There is no multiplication of a subtractionoperator times a direction in space
( ) multiplied by a ( ) ( )
positive positive positiveaddition addition additioncross cross cross
direction direction direction
What is the meaning of multiplying addition operators Addition operators are not multiplication operators
BS Math
() multiplied by a direction ( )=
number
() multiplied by a direction ( )=
number
A number times a direction maintains the same direction
( ) ()
dirdiradd
( ) ()
dirdiradd
The addition of a direction is in the same direction There is no multiplication by addition operators
Symmetry Math
Future math books need to eliminate the use of the cross sign (+) to represent something labeled a positive direction in space and the dash sign (-) to represent something labeled a negative direction in space Space does not have positive and negative directions
Symbols that are logical and have no illogical representation should be adopted SM uses either arrows or symbols
Symmetry-Math Number Line
Symmetry-Math Rules
The dash sign (-) will have only one use subtraction
The cross sign (+) will have only one use addition
Directions have an arrow and a number
All observers see the same direction and magnitude
bull There is no multiplication of arrows
amp ampamp ampampamp
( 3 ) ( 5 ) ( 6 ) ( 65 ) (17256 ) ( 32993 )
dir dirdir dirdirdir
Negative and Positive Directions in Space
If I asked you to point to a negative direction in space which way would you point
Hopefully you will realize that there is no such thing as a negative direction in space
If I asked you to point to a positive direction in space which way would you point
Again hopefully you will realize that there is no such thing as a positive direction in space
Objects on the right side of the x-axis are positive (+)(+) = (+)
Objects on the left side of the x-axis are negative (-)(-) = (+) negative math is different from positive math This is illogical
Objects on the top of the y-axis are positive
Objects on the bottom of the y-axis are negative
Objects in the front of the z-axis are positive
Objects in the back of the z-axis are negative
Again BS math of the positive direction (x right y up and z front) is different from the BS math of the negative direction (x left y down and z back)
BS math is Broken-Symmetry
Mirror image is broken symmetry in the x and z axis If the left amp right-axis are
reversed symmetry is broken Obs-1 math is different from obs-2
If the front amp back-axis is reversed symmetry is broken Obs-1 math is different from obs-2
If the top amp bottom-axis are
reversed symmetry is not broken Obs-1 math is the same as obs-2
Math answers cannot depend on which side of a number line an observer sit
Absolute values for Displacement
Even for a simple displacement of an object BS math invented a definition and absolute values to obtain something labeled a positive number answer
I----------I----------I----------I----------I----------I----------I----------I----------I
-4 -3 -2 -1 0 +1 +2 +3 +4 a gtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtb cltltltltltltltltltltltltltltltltltltltltltltltltltltltb
If an object starts at ( 3)ia X and moves to ( X 3) and the to ( 1)b fb c X
BS Math provides a usable answer without a definition or absolute values
However If an object starts at
( 1) and moves to ( X 3) and the to ( 3)i b fa X b c X
Then absolute values and a definition must be used
---------I----------I----------I----------I----------I----------I----------I---------- -4 -3 -2 -1 0 +1 +2 +3 +4
a gtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtb cltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltb
BS Math makes up a definition that all displacements are positive Since this example provides an answer with something labeled a negative the positive definition must be applied and (-2) is changed to (+2) A definition is needed to arrive at a useable answer This is Illogical
In SM absolute values and a meaningless definition of ldquopositiverdquo are removed In SM objects and directions in space are defined by the direction and magnitude of the resultant of arrows
With SM you get total distance traveled by the object and the final direction of the displacement
22222 2 bababbaababababa If w e su b stitu te p o s itiv e n u m b e rs in to th e eq u a tio n th e an sw er w ill b e co rrec t If (a = 5 ) an d (b = 3 )
2
2
8
5 3 5 3 5 3 5 5 5 3 3 5 3 3 25 15 15 9 64
8 8 8 8 8 8 8 8 8 8 64
W h en tw o p o sitiv e n u m b ers a re a d d ed an d sq u a re d th e B S m a th d is tr ib u tiv e law p ro v id es a co rre c t an sw er 1
Illogical and Incorrect BS math for the distributive law
222 2 bababbabbaaabababa B y d e fin itio n in B S m a th n o n u m b er sq u a red c a n ev e r b e n eg a tiv e T h e re fo re n o sq u a red n u m b er c an ev e r b e a v e c to r m o v in g in th e d a sh ( -) o r n eg a tiv e d ire c tio n if it is sq u a red In B S m a th an d ad d itio n o p e ra to r m u ltip lied b y a su b trac tio n o p e ra to r is A L W A Y S eq u a l to a su b tra c tio n o p e ra to r If th is h ad n o t b e en u sed fo r th e la s t fo u r h u n d red y e a rs y o u w o u ld d ie lau g h in g a t th e u tte r ab su rd ity o f th e la s t s ta tem e n t T h is is illo g ica l an d v io la te s sy m m etry
If a = 5 an d b = 3 42222352
22
4930253)3)(5(253)5)(3()3)(5(5353535 22222 A s lo n g a s ldquo ardquo is g re a te r th an ldquob rdquo th e B S m a th d is tr ib u tiv e law p ro v id es co rrec t an sw ers 1
B S m a th b re a k s d o w n w h e n ldquo b rdquo is g rea ter th a n ldquo a rdquo
If a = 3 an d b = 5
2 2
2
2
3 5 2 4
2 ( 2) ( 2) 4
2 ( )( 2) ( )( 2) 4
A n y co m b in a tio n ex cep t th e m id d le o n e is illo g ica l an d in co rre c t B y d e fin itio n th e an sw er can n o t b e -4 in B S m a th B S m a th is in co rrec t
4253095)5)(3(235)3)(5()5)(3(3535353 22222 T h is is w h ere th e d is tr ib u tiv e la w p ro v id es in co rrec t an sw ers I t p ro d u ces an an sw er o f + 4 b e cau se b y d e fin iti o n n o n u m b er sq u a red in B S m a th c an b e a n eg a tiv e n u m b er U sin g in co rre c t B S m a th th e d is tr ib u tiv e law w ill g iv e an in co rrec t an sw ers o f + 4 T h is is in co rrec t an d illo g ica l 1
Since there can be no negative direction from a squared term in BS math BS math should be modified or abandoned
The BS distributive law should be modified or abandoned
The numbers +1+2+3+4 and ndash1-2-3-4hellip should be abandoned
BS math gives incorrect answers because their Rule-of-Signs allows multiplying an arrow going in one direction by an arrow going in another direction
Clearly this is not logical
Arrows should be used that are specific for directions
W h at d o es it m e an to m u ltip ly ( 4) b y (4 ) A s w e k n o w 1
m u ltip lica tio n is ju s t ad d itio n W h a t w o u ld ( 4) tim e (4 ) b e 2
In BS math the answer is by definition -16 unless you square -4 and then it is +16
In SM you cannot multiply arrows going in opposite directions
The middle two terms are not logical You cannot multiply opposite directions
This is a MAJOR error in BS math They multiply a dash [(-) a direction to the left] by a cross [(+) a direction to the right] And then ldquoby definitionrdquo label the answer dash (-) This is illogical and produces incorrect answers
2
2 2
This is Einsteinrsquos math error
in Special relativity
Magnitude 4 Direction 1 Magnitude 3 Direction 2 Magnitude 5 Direction 3 Relative to an observer at 0 4
Magnitude 4 Direction 1 Magnitude 3 Direction 2 Magnitude 5 Direction 3 Relative to an observer at 0 4
Magnitude and Direction (MD) An arrow has both a magnitude andA direction relative to a 2nd objectobserver
bull Magnitude and velocity (MV) relative to a 2nd objectobserver
bull Magnitude And acceleration (MA) relative to a 2nd objectobserver2
v da
t t
Introducing a new symbol the ldquoamprdquo sign The ldquoamprdquo sign means resultant ADD all the Arrows for each direction and then SUBTRACT to find the final resultant
1 1amp 1 2 2amp 2 4 If =
1 amp1 2 2 amp2 4 If =
amp 5 amp 3 2 If gt
amp 3 amp 5= 2 If gt
amp 0 3 amp 3=0 if =
only the resultant equals zero
the vectors do not - they are still there
Does not exist There is no multiplication of arrows
There are two arrows One has a magnitude of 3 One has a magnitude of 5 The maximum these two arrows can be is 8
Only the resultant of arrows exist
An SM arrow has a direction and a magnitude
Arrows and magnitudes are still there even if the resultant is zero
There are no negative SM Arrows
There are no positive SM Arrows
There are no ZERO SM Arrows A zero would negate the definition of an SM Arrow something with no magnitude and no direction cannot be the definition of an arrow that is defined as having magnitude and direction
bull A resultant is the addition and subtraction of the differences in magnitude and direction arrows
A B and C are the arrows (a) A amp B exist(b) A amp B amp C exist(c) A-B Do Not Exist in Sm there are no negatives or positives(d) A+B-C Do Not Exist in SM there are no negatives or positives
There are only resultants when there are multiple arrows
SM Graphs versus BS Graphs
Numbers Raised to Powers
When numbers are raised to powers the negative sides of all BS math graphs are ILLOGICAL and INCORRECT
Incorrect BS Math for 2x
2x
4 16 correct3 8 correct2 4 correct1 2 correct0 1 correct-1 005 incorrect-2 025 incorrect-3 0125 incorrect-4 00625 incorrect
negative side of graph is incorrectpositive side of graph is correct
16
8
42105025012500625
-4 -3 -2 -1 0 1 2 3 4
po
sit
ive
negative positive
2x BS Math
correctincorrect
In BS math the dash sign (-) used in this example represents three different operations They arebull (-) Subtractionbull (-) A direction to the leftbull (-) As an exponent to mean divide by
All positive numbers produce correct results and graphs
All negative numbers produce incorrect results and graphs
A direction of three units to the left (-3) is placed in the equation as an exponent
A direction in space (3 units toward the left) changes a whole number to a fraction by using the dash (-) as an exponent rather than a direction in space
This is illogical and incorrect BS math should have performed this operation as
Unfortunately the answers to problems can come out correct because data points of a real problem fit the curve
However the answers will not be due to logical reasoning
Because the equation produced a curve that real problem data points follow it has been falsely believed that the equations and graphs are correct
1
2
x
Incorrect BK graph for (12)x
(12)x
4 00625 correct3 0125 correct2 025 correct1 1 correct0 1 correct-1 2 incorrect-2 4 incorrect-3 8 incorrect-4 16 incorrect
negative side of graph is incorrectpositive side of graph is correct
00625012502505124
8
16
-5 -4 -3 -2 -1 0 1 2 3 4 5P
osit
ive
negative positive
BS Math for (12)x
correctincorrect
Incorrect BS math and graph for
All graphs are symmetrical with SM
Two BS Math Equations that Produce Incorrect Answers
2 0x x starting equation 2
2
x x x x
x x
Add lsquoxrsquo to both sides
2(1) 1 let lsquoxrsquo = 1
2 1 This is an incorrect answer 2(0)-(0)=0 The only answer that BK math can give is x=0
1
a b starting equation ( )( ) ( )( )a a a b multiply both sides by ( )a
( )( ) ( )( ) ( )( ) ( )( )a a b b a b b b subtract ( )( )b b from both sides
( )( ) ( )( )a b a b a b b factor ( )a b from both sides
( )a b b divide both sides by ( )a b
( )b b b replace ( )a with ( )b 2b b simplify 2(1) (1) Let ( )b = 1
2 1 Again BS math yields and incorrect answer 1
This incorrect result is a problem with the distributive law SM explains how this incorrect answer is obtained and the problem with the distributive law that allows the error
Page 148 ndash149 Barry Mazur ndash Imagining Numbers
3
2
31
Do the computation carefully on paper using the rules we agreed to Then ponder your answer which should be something of a surprise to you donrsquot stop there Think of what your answer might possibly mean or might imply
Solve this equation using the distributive law
JK definitely a surprise the answer wasincorrect and illogical
Gerhard Gade University Professor doing math at Harvard University
Here is the BS math answer that Barry Mazur calculated
2
31
2
31
2
31
1 3 1 3 1 3 3 3 2 2 3 1 3
2 2 2 2 4 4 4 4 4 4 2 2
dash
1 3 1 3 1 31
2 2 2 2 4 4
3 3
4 4
When the distributive law is used with the BS Rule-of-Signs and one of the signs is a (-) dash sign the middle two terms always cancel each other This leads to the incorrect answer of (-1) in the above equation
Barry Mazur did not go back and check his answer with the original equation If he had he would have known that the answer was incorrect
3
1 31
2
check answer by setting equal
3
331 3
12
1 31
2
Take the Cube root of both sides
1 3 2 multiply both sides by 2
3 3 subtract 1 from both sides
2 2
3 3 square both sides
3 9 This is an incorrect answer A simple equation check could have shown that the distributive law using BS math provides an incorrect answer
1
BS Illogical and Incorrect Numbers Raised to Powers
nC D n DC
642 DDD
In BS math only a positive number can have an answer if ldquonrdquo is an even number (246---)
All negative numbers have no solutions
642 DDD
With SM just add a direction arrow and any number gives the correct answer
or n nC D C D
3 3( 3 8 8 or 8 2 or 2 )n D C
( 2 16 C= 16 or 16 4 or 4 )n D
Since a negative number has no square root answer in BS math BS mathematicians invented imaginary numbers
i2=-1
24414 iiC n
These equations are easily solved without the use of imaginary numbers with SM math
If 4 then 4 2x
If 4 then 4 2x
The 360-degree circle is a 4400 years old extremely outdated system based on a year of 12 months of 30 days each that is 360 days
SM will use a more logical system that will have a circle of 6283185307o
d
( )
(deg )
d r radians
d r rees
the arc distance around
the circumference of a circle
d
radiusr
2 2
2 2
d r
dv rd t t
v dda rd t t t
1
d
r
( )av tv
The average angular velocity of a body is the rate of its angular displacement
( )0
liminstt
vt
The instantaneous angular velocity is the limit of the ratio
t
as t approaches zero
( )d
av
d
tv
The average circumference velocity of the body is the rate of its distance traveled d divided by the time t
dvv
r
1 dd vdrv
t t t r r
The equation of the relationship between the angular velocity and the velocity of a point on that body
( )o
av
v v v
t ta
The average angular acceleration is the change in the angular velocity
( )0
liminstt
v
ta
The instantaneous angular acceleration is the
limit of the ration of v
t
as t approaches
zero
daa
r
1
1
d
d s d
vva
t r tv a
r t r
The equation of the relationship between the angular acceleration and the acceleration of a point on the body
21
2v a t
If a body with an initial angular velocity ov has a constant
angular acceleration a it will turn through an angle in a time t
1
Straight Line Symbol Units Rotational Symbol Units Distance d m Angular
Distance radians or degrees
Velocity dt
v m
s Angular Velocity
V
t
radians or degrees
s
Acceleration vat
2ms
Angular Acceleration
VA
t
2
radians or degrees
s
Mass M Kg Mass distribution
dM 2Kgm
Force actionA Ma
2
Kgm
sNewton
Torque
dT M A 2
2
Kgm
s
Momentum Mv Kgm
s
Angular Momentum
dM V 2Kgm
s
Work (Force)(distance)
Ad 2
Kgmm
sNm Joule
Work T
Using AS arrow math to solve and direction in space problems using the equation 1 2
2 2 2 2 2 2( ) amp and ( ) amp 3
4
2 2 2
2 2
( ) amp
(5 ) 4 3
(5 ) 16 9
(5 ) 25
(5 ) (5 )
2 2(5 ) 4 3
(5 ) 16 9
(5 ) 25
(5 ) (5 )
In SM math Pythagorasrsquo theorem provides both magnitude and direction
In BS math Pythagorasrsquo theorem provides magnitude only
time set equal to zero
23
( ) ( 0 )10 1
x t xx
3( ) ( 0)
10 1W t f
In BS math this is an equation for a traveling wave
must use a square term in order to make the curve symmetric
NO SQUARE TERM IS NECESSARY
In SM math when t is set to ldquo0rdquo
This is a major fault in BS math
x f(x)
-4 0019
-3 0033
-2 0073
-1 0273
0 3000
1 0273
2 0073
3 0033
4 0019 -4 -3 -2 -1 0 1 2 3 400
05
10
15
20
25
30
35
BS wave (must use square term)
snap shot of wave at x = 0
Am
plit
ude
lt to gt lt to gt f(ltgt)
lt -4 0073
lt -3 0097
lt -2 0143
lt -1 0273
origin 0 3000
gt 1 0273
gt 2 0143
gt 3 0097
gt 4 0073lt0073 lt0097 lt0143 lt0273
3000
0273gt 0143gt 0097gt 0073gt
Wave moving
snap shot of wave
Am
plit
ude
lt4 lt3 lt2 lt1 0 1gt 2gt 3gt 4gt
3
2
1
In SM NO squared terms are needed
SPRING
An Action is needed to stretch or compress a spring to a starting point where it can execute cycle motion
tsA (ts to start) is the work required to move a mass attached to a spring a distance (d1) from
its initial rest position RAK 0 the spring constant is the stiffness of the spring
o A is attraction o R is repulsion o
ARRA
00 is for the back and forth motion through the zero initial rest position A
stretched spring goes from ldquoArdquo to ldquo0rdquo to ldquoRrdquo to ldquoArdquo to ldquo0rdquo to ldquoRrdquo d1 is the distance a mass is moved from its initial rest position
0
0 1ts A RA K d
Once ( )mad
is performed to move the mass to position d1 (against the restraint of the spring) a of the ( )mad
is transferred into the spring as potential attractionrepulsion (PAR) There is always
more ( )mad
performed than is stored as PAR When the spring is released from the restraining Ats the PAR starts the oscillation Conservation of ( )mad
requires that the PAR and the motion ( )mad
( )( )madM be a constant
( )amp ( ) constantmadPAR M 02
0 0
1
2A R A RPAR K d
02
( ) 0 0
1( )
2mad A R A RM m v
0 0 02 2 2
0 0 0 1
2 2 21
2 2 21
1 1 1amp
2 2 2
1 1 1amp
2 2 2
amp
A R A R A RK d m v K d
Kd mv Kd
Kd mv Kd
0
1d is frac12 the height from the base to the peak of a cycle
An experiment of moving paper under a pendulum through a small angle filled with sand that can escape through a small hole in the bottom shows that a pendulum cycling back and forth follows a curve that is mathematically analogous to a lsquosinrsquo or lsquocosrsquo
The projection of the curved area onto the straight line exactly duplicate the back and forth motion of CM This allows for the math of lsquosinrsquo or lsquocosrsquo to be used
SM will use lsquosinrdquo math It could use lsquocosrsquo or a combination of both lsquosinrsquo and lsquocosrsquo Using just the lsquosinrsquo math makes the understanding and math easier
0 sin 0 0SM 0 sin157 1SM The diagram at the left shows a spring being stretched from its (zero rest) position-2 and the PAR being released at position-1 There are 4 sections (1 to 2 2 to 3 3 to 4 and 4 to5) that alternate between maximum acceleration and zero velocity to max velocity and zero acceleration
At position-1 there is maximum acceleration and zero velocity o The lsquosinrsquo values from 0o to 157o will be used for velocity (values from 0 to 1) o The lsquosinrsquo values from 157o to 0o will be used for acceleration (values from 1 to 0)
At position-2 there is maximum velocity and zero acceleration Acceleration changes to deceleration
o The lsquosinrsquo values from 157o to 0o will be used for deceleration (values from 1 to 0) o The lsquosinrsquo values from 0o to 157o will be used for acceleration (values from 0 to 1)
At position-3 there is maximum acceleration and zero velocity o The lsquosinrsquo values from 0o to 157o will be used for velocity (values from 0 to 1) o The lsquosinrsquo values from 157o to 0o will be used for acceleration (values from 1 to 0)
At position-4 there is maximum velocity and zero acceleration Acceleration changes to deceleration
o The lsquosinrsquo values from 157o to 0o will be used for deceleration (values from 1 to 0) o The lsquosinrsquo values from 0o to 157o will be used for acceleration (values from 0 to 1)
At position-5 the cycle starts again
2
)539(
cT
MK
spring constant (K) when mass (M) and cycle-time (T) are known
)(
))(89)((
111 md
uKgM
d
MG
d
FK sts
spring constant (K) when mass (M) and amplitude ( d1) are known
2 21( )
Kv d d
m
velocity (v) at any point during the cycle
velocity
dforM
Kdv
max
012 velocity (v2) maximum (when d = 0 at position-2 and 4)
acceleration (a1) maximum when spring constant (K) mass (M) and amplitude (d1) are known
acceleration (a1) maximum when amplitude (d1) and cycle-time (Tc) are known
acceleration (a) at any point during the cycle-time (T)
1
151
286
v
dTc
cycle-time (Tc) when amplitude (d1) and velocity (v1) are known
K
MTc 28651
cycle-time (Tc) when mass (M) and spring constant (K) are known
fTc
1
cycle-time (Tc) when the frequency is known
cTf
1
frequency (f) when the cycle-time (Tc) is known
of cycles
elapsed timef
frequency (f) when the number of cycles in an elapsed time is known
d1 amplitude (distance from rest to start position)
Also frac12 the distance from the base to the peak
3131 oror dM
Ka
251
3131 539
c
oror T
da
2539
cT
da
In December of 2006 I discovered the mathematical error in Einsteinrsquos Special Relativity Einstein used BS math and that math only works in an imaginary world
It cannot describe real symmetrical space
Einsteinrsquos Special Relativity should be abandoned and a theory of real space using SM should be used to replace it
Numerous articles on the Internet show Einsteinrsquos theory to be incorrect but no one in academia is willing to publish the information
Figure 1-10 A moving particle observed from two frames of reference in uniform translation
Einsteinrsquos Math Error in Special Relativity Transformation of Velocity (From Eisenberg- Fundamentals of Modern Physics pages 27-29) Consider the particle shown in figure (1-10) moving with velocity v as
seen in a frame of reference O
dx dy dz
Vx Vy Vzdt dt dt
dx dy dz
V x V y V zdy dt dt
We would like to evaluate the velocity v of the particle as seen in the frame of reference 0 which is itself moving relative to 0 with velocity v Measured in the 0 frame the velocity vector of the particle has com ponents
The same velocity vector as measured in the 0 frame has components
2
2
2
2
2
1
1
x vtx
v
cy y
z z
vxt
ctv
c
Now from equations (1-13) we know that the relation between the primed and the unprimed coordinates and times is
Take the differential of these equations remembering that v is a constant This gives
2
2
2
2
2
1
1
dx vdtdx
v
cdy dy
dz dz
vxdt
cdtv
c
2
2
2 2 2 2 2
2
2
2
11 ( )
1( ) 1
1
1
1
dx vdt
v dxdx vdt v
dx dx vdtc dt dtvdx vdx vdx v dxdt dt dt dt
dt dtc c c c
VxV x
Vx
VxV x
vv
vv
c
x
v
c
cV
( )
coordinate of particle relative to moving primed system
coordinate of particle relative to stationary system
velocity of primed coordina
V x
Vx
te system relative to stationaryv sys
tem
Page 28 Fundamentals of Modern Physics
(1-16)
According to Einstein this equation tells you how to transform the observed velocity from one frame of reference to another frame of reference First we note that as Vc and vc approachzero the equation approaches those which would be derived from the Galilean transformation
2 2
( ) 1 01 1 ( )
Vx v Vx vV
VxV x
Vx v
v cVxx
v
c
v
cc
Another property of this equation is that it is impossible to choose V and v such that Vrsquo the magnitude of the velocity which is seen in the new frame is greater than c
1
2
09 ( 09 ) 18009876543
( 09 ) 1811 (09 )
c c cV c c
cc
c
Example As seen by O particle 1 has velocity 09c in the direction of positive x and particle 2 has velocity 09c in the negative x direction
Einstein equation however predicts To evaluate the velocity of particle 1 with respect to particle 2 we transform from the O frame to the Orsquo frame moving in the negative x direction with velocity v = -09c using equation 1-16 We obtain
JK LOGIC will observe the particles separating at 18c
A confusing INCORRECT and ILLOGICAL answer
left is a dir rightsub of a
1
2
1
09 ( 09 ) 18009876543
181( 09 )1 ( 09 )
The multiplication of dir left time a dir right is ONLY a dir left
09
dir
direction leftdirection rightsubtraction
c c cV c c
cc
c
V
dir
22
22
09 18 18009876543
181( 081)1(09)(09)
1
left rightsub
sub
c c c cc c
ccc
c
Analyzing Einsteinrsquos BS math Equation
Using SM math a logical answer can be obtained
Cannot multiply a subtraction operator times a direction in space
Can only subtract and change its direction
Cannot multiply a direction to the left times a direction to the right
Cannot multiply opposite directions in space
Can only solve for the resultant
rightleft
subtraction
1
subtraction
2
sub of a dir to the left is a dir to the right
1 resultant 0 sub 2
2sub 2
2
09 09
09 amp 091
09 09 18 1
(0)1
(09) amp (09)1
c cV
c cc
c c cV
cc
cc
8
181
cc
SM math predicts a logical answer using logic
21st century scientists state in many articles that Einsteins General Relativity and Quantum Mechanics are incompatible
They are not only incompatible they are both mathematically incorrect
They both use illogical BS math
Any equations in math that use imaginary numbers should be abandoned
Imaginary numbers are only necessary in the current BS math system
This means that Schroumldingers equations should be abandoned
Any equation in quantum mechanics that uses imaginary number should be abandoned
Prof Kanarev has developed logical physics and chemistrytheories without the use of Quantum mechanics or relativity
A partial list of his theories coverbull Mechanics ndash replaces Newtonbull Photon - bull Electrons ndash Protons- Neutronsbull Spectroscopybull Nucleusbull Atoms amp Moleculesbull Electrodynamicsbull Heatbull Curvature of Space ndash replaces Schwarzcchildrsquos equation
Kanarevrsquos Hypothesis
bull Electron align linearly with proton They do NOT spin or form clouds around the nucleus
bull Neutrons are in the center of atom Proton are outside of Neutrons
bull The wavelength of all particles is the experimental radius
bull Double slit experiment explained correctly
bull Heisenbergrsquos inequality explained correctly
bull is the radius of approach of the magnetic lines around the torus of the electron
( 1 137)
bull Why water expands when it freezes
bull Electron clusters form lightning
rbull The meaning of Plancks constant is explained
22 2 kg m
h m f mr f consts
2210254 42m r kgm bull If the mr-constant describes a ring the equations
describe experiments bull If the mr-constant describes a wave the equations do
NOT describe experiments
Kanarevrsquos 5 Equations of Mechanics
Kanarevrsquos 1st Equation Acceleration ampmv maMma R R
Kanarevrsquos 2nd Equation Uniform motion amp mv maMma R R
maM mvR R
Kanarevrsquos 3rd Equation
Kanarevrsquos 4th Equation Equal and opposite reaction
1 21 21 2
12
a m
m a m ama
Kanarevrsquos 5th Equation 1 2 3 Mi M M M Mna R R R R R
The sum of the Actions working on a moving body is never equal to zero
Kanarevrsquos Photon model is a rolling ring
His equations accurately describe known experiments with photonsbull Electromagnetic radiation should be called photon radiationbull Diffraction and interference are the same phenomenabull Correctly explains the double slit experiment without quantum equationsbull Correctly explains the meaning of Heisenbergrsquos inequality
e
2 (2)(3142)(2817 15 ) 10007297352568
R 2426 12 137mag
xp
r m
m
Charge
Magnetic action
Proton
The intensity of the magnetic action near the geometrical center of the proton ring is so great that we theorize that it is the strong nuclear force
(9383 8)(1602 19)8476 14
4 (4)(3142)(1411 26)p
pp
EB T
M
(9395 8)(1602 19)1035 15
4 (4)(3142)(0966 26)N
NN
EB T
M
Neutron
1a Hydrogenavi
A Summary of the BS Broken-Symmetry math
1 1 cosh tanhln
x xe x xe
bull Originated with broken-symmetry in the XYZ plane
bull Originated with mirror-broken-symmetry in the XZ plane
bull Created a Rule-of-Signs that cannot be proved (-)(-)=(+) (+)(+)=(+) (-)(+)=(-)
bull Created imaginary numbers to compensate for broken-symmetry
bull Created absolute values to change illogical negative number answers into positive number answers
bull Produces broken symmetry graphs for many functions
It took four years of study to discover why a (-)(-)=(+) only works in an imaginary world where space in the dash (-) direction is different from space in the cross (+) direction
Started working on the problem in August of 2001
Data points created using BS math produce graphs that are not symmetrical dash side different from cross side
However if a real problems data points follow a non symmetrical graph the graph can provide usable answers
This is why no one discovered the problem
Solved in January of 2005
Symmetry-Math(SM) vs Broken-Symmetry(BS) Math
In SM if we specify that the dash sign (-) means only subtraction and the cross sign (+) means only addition and an appropriate symbol is used for a direction in space space becomes symmetrical and math becomes logical
Math operators and directions in space are not the same and the same symbol should not be used to represent them
Letrsquos start with the (dash)multiplied by a (dash)=(cross) in the BS system
Usable answers may be obtained but it will be for illogical reasons
multiplied by a ( ) ( ) ( )subtraction Subtraction addition
rightleft left
dash dash cross
positivenegative negative
Symmetry-Math The subtraction of a direction is equal to the opposite direction The answers are correct using correct logic
opposite direction direction
SM
(negative)(negative) = (positive)
( ) ( ) ( )subtraction
BS
There is NO multiplication of a subtraction operator by a direction in space
There is just the subtraction of a direction in space
If instead of labeling the left side of a coordinate system as a negative (-) the same as a subtraction operator we label it with an arrow ( ) to represent the direction then a subtraction from that arrow direction will be in the opposite direction ( )
( ) mutiplied by a ( ) ( )
negative Positive negative
subtraction addition subtraction
dash cross dash
left right left
A subtraction operator multiplied by an addition operator is equal to a subtraction operator This is illogical
BS Math
A negative multiplied by a positive is equal to a negative This is illogical
A direction to the left multiplied by a direction to the right is a direction to the left This is illogical and Einsteins math error in special relativity
A dash times a cross is equal to a dash This is illogical
A vector moving to the left multiplied by a vector moving to the right is equal to ONLY to a vector moving to the left that is eight orders of magnitude faster thanThe speed of light
opposite direction subtraction direction
SM ( ) ( )
BS (negative)(positive) = (negative)
Symmetry Math The subtraction of a direction is equal to the opposite direction
There is no multiplication of a subtractionoperator times a direction in space
( ) multiplied by a ( ) ( )
positive positive positiveaddition addition additioncross cross cross
direction direction direction
What is the meaning of multiplying addition operators Addition operators are not multiplication operators
BS Math
() multiplied by a direction ( )=
number
() multiplied by a direction ( )=
number
A number times a direction maintains the same direction
( ) ()
dirdiradd
( ) ()
dirdiradd
The addition of a direction is in the same direction There is no multiplication by addition operators
Symmetry Math
Future math books need to eliminate the use of the cross sign (+) to represent something labeled a positive direction in space and the dash sign (-) to represent something labeled a negative direction in space Space does not have positive and negative directions
Symbols that are logical and have no illogical representation should be adopted SM uses either arrows or symbols
Symmetry-Math Number Line
Symmetry-Math Rules
The dash sign (-) will have only one use subtraction
The cross sign (+) will have only one use addition
Directions have an arrow and a number
All observers see the same direction and magnitude
bull There is no multiplication of arrows
amp ampamp ampampamp
( 3 ) ( 5 ) ( 6 ) ( 65 ) (17256 ) ( 32993 )
dir dirdir dirdirdir
Negative and Positive Directions in Space
If I asked you to point to a negative direction in space which way would you point
Hopefully you will realize that there is no such thing as a negative direction in space
If I asked you to point to a positive direction in space which way would you point
Again hopefully you will realize that there is no such thing as a positive direction in space
Objects on the right side of the x-axis are positive (+)(+) = (+)
Objects on the left side of the x-axis are negative (-)(-) = (+) negative math is different from positive math This is illogical
Objects on the top of the y-axis are positive
Objects on the bottom of the y-axis are negative
Objects in the front of the z-axis are positive
Objects in the back of the z-axis are negative
Again BS math of the positive direction (x right y up and z front) is different from the BS math of the negative direction (x left y down and z back)
BS math is Broken-Symmetry
Mirror image is broken symmetry in the x and z axis If the left amp right-axis are
reversed symmetry is broken Obs-1 math is different from obs-2
If the front amp back-axis is reversed symmetry is broken Obs-1 math is different from obs-2
If the top amp bottom-axis are
reversed symmetry is not broken Obs-1 math is the same as obs-2
Math answers cannot depend on which side of a number line an observer sit
Absolute values for Displacement
Even for a simple displacement of an object BS math invented a definition and absolute values to obtain something labeled a positive number answer
I----------I----------I----------I----------I----------I----------I----------I----------I
-4 -3 -2 -1 0 +1 +2 +3 +4 a gtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtb cltltltltltltltltltltltltltltltltltltltltltltltltltltltb
If an object starts at ( 3)ia X and moves to ( X 3) and the to ( 1)b fb c X
BS Math provides a usable answer without a definition or absolute values
However If an object starts at
( 1) and moves to ( X 3) and the to ( 3)i b fa X b c X
Then absolute values and a definition must be used
---------I----------I----------I----------I----------I----------I----------I---------- -4 -3 -2 -1 0 +1 +2 +3 +4
a gtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtb cltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltb
BS Math makes up a definition that all displacements are positive Since this example provides an answer with something labeled a negative the positive definition must be applied and (-2) is changed to (+2) A definition is needed to arrive at a useable answer This is Illogical
In SM absolute values and a meaningless definition of ldquopositiverdquo are removed In SM objects and directions in space are defined by the direction and magnitude of the resultant of arrows
With SM you get total distance traveled by the object and the final direction of the displacement
22222 2 bababbaababababa If w e su b stitu te p o s itiv e n u m b e rs in to th e eq u a tio n th e an sw er w ill b e co rrec t If (a = 5 ) an d (b = 3 )
2
2
8
5 3 5 3 5 3 5 5 5 3 3 5 3 3 25 15 15 9 64
8 8 8 8 8 8 8 8 8 8 64
W h en tw o p o sitiv e n u m b ers a re a d d ed an d sq u a re d th e B S m a th d is tr ib u tiv e law p ro v id es a co rre c t an sw er 1
Illogical and Incorrect BS math for the distributive law
222 2 bababbabbaaabababa B y d e fin itio n in B S m a th n o n u m b er sq u a red c a n ev e r b e n eg a tiv e T h e re fo re n o sq u a red n u m b er c an ev e r b e a v e c to r m o v in g in th e d a sh ( -) o r n eg a tiv e d ire c tio n if it is sq u a red In B S m a th an d ad d itio n o p e ra to r m u ltip lied b y a su b trac tio n o p e ra to r is A L W A Y S eq u a l to a su b tra c tio n o p e ra to r If th is h ad n o t b e en u sed fo r th e la s t fo u r h u n d red y e a rs y o u w o u ld d ie lau g h in g a t th e u tte r ab su rd ity o f th e la s t s ta tem e n t T h is is illo g ica l an d v io la te s sy m m etry
If a = 5 an d b = 3 42222352
22
4930253)3)(5(253)5)(3()3)(5(5353535 22222 A s lo n g a s ldquo ardquo is g re a te r th an ldquob rdquo th e B S m a th d is tr ib u tiv e law p ro v id es co rrec t an sw ers 1
B S m a th b re a k s d o w n w h e n ldquo b rdquo is g rea ter th a n ldquo a rdquo
If a = 3 an d b = 5
2 2
2
2
3 5 2 4
2 ( 2) ( 2) 4
2 ( )( 2) ( )( 2) 4
A n y co m b in a tio n ex cep t th e m id d le o n e is illo g ica l an d in co rre c t B y d e fin itio n th e an sw er can n o t b e -4 in B S m a th B S m a th is in co rrec t
4253095)5)(3(235)3)(5()5)(3(3535353 22222 T h is is w h ere th e d is tr ib u tiv e la w p ro v id es in co rrec t an sw ers I t p ro d u ces an an sw er o f + 4 b e cau se b y d e fin iti o n n o n u m b er sq u a red in B S m a th c an b e a n eg a tiv e n u m b er U sin g in co rre c t B S m a th th e d is tr ib u tiv e law w ill g iv e an in co rrec t an sw ers o f + 4 T h is is in co rrec t an d illo g ica l 1
Since there can be no negative direction from a squared term in BS math BS math should be modified or abandoned
The BS distributive law should be modified or abandoned
The numbers +1+2+3+4 and ndash1-2-3-4hellip should be abandoned
BS math gives incorrect answers because their Rule-of-Signs allows multiplying an arrow going in one direction by an arrow going in another direction
Clearly this is not logical
Arrows should be used that are specific for directions
W h at d o es it m e an to m u ltip ly ( 4) b y (4 ) A s w e k n o w 1
m u ltip lica tio n is ju s t ad d itio n W h a t w o u ld ( 4) tim e (4 ) b e 2
In BS math the answer is by definition -16 unless you square -4 and then it is +16
In SM you cannot multiply arrows going in opposite directions
The middle two terms are not logical You cannot multiply opposite directions
This is a MAJOR error in BS math They multiply a dash [(-) a direction to the left] by a cross [(+) a direction to the right] And then ldquoby definitionrdquo label the answer dash (-) This is illogical and produces incorrect answers
2
2 2
This is Einsteinrsquos math error
in Special relativity
Magnitude 4 Direction 1 Magnitude 3 Direction 2 Magnitude 5 Direction 3 Relative to an observer at 0 4
Magnitude 4 Direction 1 Magnitude 3 Direction 2 Magnitude 5 Direction 3 Relative to an observer at 0 4
Magnitude and Direction (MD) An arrow has both a magnitude andA direction relative to a 2nd objectobserver
bull Magnitude and velocity (MV) relative to a 2nd objectobserver
bull Magnitude And acceleration (MA) relative to a 2nd objectobserver2
v da
t t
Introducing a new symbol the ldquoamprdquo sign The ldquoamprdquo sign means resultant ADD all the Arrows for each direction and then SUBTRACT to find the final resultant
1 1amp 1 2 2amp 2 4 If =
1 amp1 2 2 amp2 4 If =
amp 5 amp 3 2 If gt
amp 3 amp 5= 2 If gt
amp 0 3 amp 3=0 if =
only the resultant equals zero
the vectors do not - they are still there
Does not exist There is no multiplication of arrows
There are two arrows One has a magnitude of 3 One has a magnitude of 5 The maximum these two arrows can be is 8
Only the resultant of arrows exist
An SM arrow has a direction and a magnitude
Arrows and magnitudes are still there even if the resultant is zero
There are no negative SM Arrows
There are no positive SM Arrows
There are no ZERO SM Arrows A zero would negate the definition of an SM Arrow something with no magnitude and no direction cannot be the definition of an arrow that is defined as having magnitude and direction
bull A resultant is the addition and subtraction of the differences in magnitude and direction arrows
A B and C are the arrows (a) A amp B exist(b) A amp B amp C exist(c) A-B Do Not Exist in Sm there are no negatives or positives(d) A+B-C Do Not Exist in SM there are no negatives or positives
There are only resultants when there are multiple arrows
SM Graphs versus BS Graphs
Numbers Raised to Powers
When numbers are raised to powers the negative sides of all BS math graphs are ILLOGICAL and INCORRECT
Incorrect BS Math for 2x
2x
4 16 correct3 8 correct2 4 correct1 2 correct0 1 correct-1 005 incorrect-2 025 incorrect-3 0125 incorrect-4 00625 incorrect
negative side of graph is incorrectpositive side of graph is correct
16
8
42105025012500625
-4 -3 -2 -1 0 1 2 3 4
po
sit
ive
negative positive
2x BS Math
correctincorrect
In BS math the dash sign (-) used in this example represents three different operations They arebull (-) Subtractionbull (-) A direction to the leftbull (-) As an exponent to mean divide by
All positive numbers produce correct results and graphs
All negative numbers produce incorrect results and graphs
A direction of three units to the left (-3) is placed in the equation as an exponent
A direction in space (3 units toward the left) changes a whole number to a fraction by using the dash (-) as an exponent rather than a direction in space
This is illogical and incorrect BS math should have performed this operation as
Unfortunately the answers to problems can come out correct because data points of a real problem fit the curve
However the answers will not be due to logical reasoning
Because the equation produced a curve that real problem data points follow it has been falsely believed that the equations and graphs are correct
1
2
x
Incorrect BK graph for (12)x
(12)x
4 00625 correct3 0125 correct2 025 correct1 1 correct0 1 correct-1 2 incorrect-2 4 incorrect-3 8 incorrect-4 16 incorrect
negative side of graph is incorrectpositive side of graph is correct
00625012502505124
8
16
-5 -4 -3 -2 -1 0 1 2 3 4 5P
osit
ive
negative positive
BS Math for (12)x
correctincorrect
Incorrect BS math and graph for
All graphs are symmetrical with SM
Two BS Math Equations that Produce Incorrect Answers
2 0x x starting equation 2
2
x x x x
x x
Add lsquoxrsquo to both sides
2(1) 1 let lsquoxrsquo = 1
2 1 This is an incorrect answer 2(0)-(0)=0 The only answer that BK math can give is x=0
1
a b starting equation ( )( ) ( )( )a a a b multiply both sides by ( )a
( )( ) ( )( ) ( )( ) ( )( )a a b b a b b b subtract ( )( )b b from both sides
( )( ) ( )( )a b a b a b b factor ( )a b from both sides
( )a b b divide both sides by ( )a b
( )b b b replace ( )a with ( )b 2b b simplify 2(1) (1) Let ( )b = 1
2 1 Again BS math yields and incorrect answer 1
This incorrect result is a problem with the distributive law SM explains how this incorrect answer is obtained and the problem with the distributive law that allows the error
Page 148 ndash149 Barry Mazur ndash Imagining Numbers
3
2
31
Do the computation carefully on paper using the rules we agreed to Then ponder your answer which should be something of a surprise to you donrsquot stop there Think of what your answer might possibly mean or might imply
Solve this equation using the distributive law
JK definitely a surprise the answer wasincorrect and illogical
Gerhard Gade University Professor doing math at Harvard University
Here is the BS math answer that Barry Mazur calculated
2
31
2
31
2
31
1 3 1 3 1 3 3 3 2 2 3 1 3
2 2 2 2 4 4 4 4 4 4 2 2
dash
1 3 1 3 1 31
2 2 2 2 4 4
3 3
4 4
When the distributive law is used with the BS Rule-of-Signs and one of the signs is a (-) dash sign the middle two terms always cancel each other This leads to the incorrect answer of (-1) in the above equation
Barry Mazur did not go back and check his answer with the original equation If he had he would have known that the answer was incorrect
3
1 31
2
check answer by setting equal
3
331 3
12
1 31
2
Take the Cube root of both sides
1 3 2 multiply both sides by 2
3 3 subtract 1 from both sides
2 2
3 3 square both sides
3 9 This is an incorrect answer A simple equation check could have shown that the distributive law using BS math provides an incorrect answer
1
BS Illogical and Incorrect Numbers Raised to Powers
nC D n DC
642 DDD
In BS math only a positive number can have an answer if ldquonrdquo is an even number (246---)
All negative numbers have no solutions
642 DDD
With SM just add a direction arrow and any number gives the correct answer
or n nC D C D
3 3( 3 8 8 or 8 2 or 2 )n D C
( 2 16 C= 16 or 16 4 or 4 )n D
Since a negative number has no square root answer in BS math BS mathematicians invented imaginary numbers
i2=-1
24414 iiC n
These equations are easily solved without the use of imaginary numbers with SM math
If 4 then 4 2x
If 4 then 4 2x
The 360-degree circle is a 4400 years old extremely outdated system based on a year of 12 months of 30 days each that is 360 days
SM will use a more logical system that will have a circle of 6283185307o
d
( )
(deg )
d r radians
d r rees
the arc distance around
the circumference of a circle
d
radiusr
2 2
2 2
d r
dv rd t t
v dda rd t t t
1
d
r
( )av tv
The average angular velocity of a body is the rate of its angular displacement
( )0
liminstt
vt
The instantaneous angular velocity is the limit of the ratio
t
as t approaches zero
( )d
av
d
tv
The average circumference velocity of the body is the rate of its distance traveled d divided by the time t
dvv
r
1 dd vdrv
t t t r r
The equation of the relationship between the angular velocity and the velocity of a point on that body
( )o
av
v v v
t ta
The average angular acceleration is the change in the angular velocity
( )0
liminstt
v
ta
The instantaneous angular acceleration is the
limit of the ration of v
t
as t approaches
zero
daa
r
1
1
d
d s d
vva
t r tv a
r t r
The equation of the relationship between the angular acceleration and the acceleration of a point on the body
21
2v a t
If a body with an initial angular velocity ov has a constant
angular acceleration a it will turn through an angle in a time t
1
Straight Line Symbol Units Rotational Symbol Units Distance d m Angular
Distance radians or degrees
Velocity dt
v m
s Angular Velocity
V
t
radians or degrees
s
Acceleration vat
2ms
Angular Acceleration
VA
t
2
radians or degrees
s
Mass M Kg Mass distribution
dM 2Kgm
Force actionA Ma
2
Kgm
sNewton
Torque
dT M A 2
2
Kgm
s
Momentum Mv Kgm
s
Angular Momentum
dM V 2Kgm
s
Work (Force)(distance)
Ad 2
Kgmm
sNm Joule
Work T
Using AS arrow math to solve and direction in space problems using the equation 1 2
2 2 2 2 2 2( ) amp and ( ) amp 3
4
2 2 2
2 2
( ) amp
(5 ) 4 3
(5 ) 16 9
(5 ) 25
(5 ) (5 )
2 2(5 ) 4 3
(5 ) 16 9
(5 ) 25
(5 ) (5 )
In SM math Pythagorasrsquo theorem provides both magnitude and direction
In BS math Pythagorasrsquo theorem provides magnitude only
time set equal to zero
23
( ) ( 0 )10 1
x t xx
3( ) ( 0)
10 1W t f
In BS math this is an equation for a traveling wave
must use a square term in order to make the curve symmetric
NO SQUARE TERM IS NECESSARY
In SM math when t is set to ldquo0rdquo
This is a major fault in BS math
x f(x)
-4 0019
-3 0033
-2 0073
-1 0273
0 3000
1 0273
2 0073
3 0033
4 0019 -4 -3 -2 -1 0 1 2 3 400
05
10
15
20
25
30
35
BS wave (must use square term)
snap shot of wave at x = 0
Am
plit
ude
lt to gt lt to gt f(ltgt)
lt -4 0073
lt -3 0097
lt -2 0143
lt -1 0273
origin 0 3000
gt 1 0273
gt 2 0143
gt 3 0097
gt 4 0073lt0073 lt0097 lt0143 lt0273
3000
0273gt 0143gt 0097gt 0073gt
Wave moving
snap shot of wave
Am
plit
ude
lt4 lt3 lt2 lt1 0 1gt 2gt 3gt 4gt
3
2
1
In SM NO squared terms are needed
SPRING
An Action is needed to stretch or compress a spring to a starting point where it can execute cycle motion
tsA (ts to start) is the work required to move a mass attached to a spring a distance (d1) from
its initial rest position RAK 0 the spring constant is the stiffness of the spring
o A is attraction o R is repulsion o
ARRA
00 is for the back and forth motion through the zero initial rest position A
stretched spring goes from ldquoArdquo to ldquo0rdquo to ldquoRrdquo to ldquoArdquo to ldquo0rdquo to ldquoRrdquo d1 is the distance a mass is moved from its initial rest position
0
0 1ts A RA K d
Once ( )mad
is performed to move the mass to position d1 (against the restraint of the spring) a of the ( )mad
is transferred into the spring as potential attractionrepulsion (PAR) There is always
more ( )mad
performed than is stored as PAR When the spring is released from the restraining Ats the PAR starts the oscillation Conservation of ( )mad
requires that the PAR and the motion ( )mad
( )( )madM be a constant
( )amp ( ) constantmadPAR M 02
0 0
1
2A R A RPAR K d
02
( ) 0 0
1( )
2mad A R A RM m v
0 0 02 2 2
0 0 0 1
2 2 21
2 2 21
1 1 1amp
2 2 2
1 1 1amp
2 2 2
amp
A R A R A RK d m v K d
Kd mv Kd
Kd mv Kd
0
1d is frac12 the height from the base to the peak of a cycle
An experiment of moving paper under a pendulum through a small angle filled with sand that can escape through a small hole in the bottom shows that a pendulum cycling back and forth follows a curve that is mathematically analogous to a lsquosinrsquo or lsquocosrsquo
The projection of the curved area onto the straight line exactly duplicate the back and forth motion of CM This allows for the math of lsquosinrsquo or lsquocosrsquo to be used
SM will use lsquosinrdquo math It could use lsquocosrsquo or a combination of both lsquosinrsquo and lsquocosrsquo Using just the lsquosinrsquo math makes the understanding and math easier
0 sin 0 0SM 0 sin157 1SM The diagram at the left shows a spring being stretched from its (zero rest) position-2 and the PAR being released at position-1 There are 4 sections (1 to 2 2 to 3 3 to 4 and 4 to5) that alternate between maximum acceleration and zero velocity to max velocity and zero acceleration
At position-1 there is maximum acceleration and zero velocity o The lsquosinrsquo values from 0o to 157o will be used for velocity (values from 0 to 1) o The lsquosinrsquo values from 157o to 0o will be used for acceleration (values from 1 to 0)
At position-2 there is maximum velocity and zero acceleration Acceleration changes to deceleration
o The lsquosinrsquo values from 157o to 0o will be used for deceleration (values from 1 to 0) o The lsquosinrsquo values from 0o to 157o will be used for acceleration (values from 0 to 1)
At position-3 there is maximum acceleration and zero velocity o The lsquosinrsquo values from 0o to 157o will be used for velocity (values from 0 to 1) o The lsquosinrsquo values from 157o to 0o will be used for acceleration (values from 1 to 0)
At position-4 there is maximum velocity and zero acceleration Acceleration changes to deceleration
o The lsquosinrsquo values from 157o to 0o will be used for deceleration (values from 1 to 0) o The lsquosinrsquo values from 0o to 157o will be used for acceleration (values from 0 to 1)
At position-5 the cycle starts again
2
)539(
cT
MK
spring constant (K) when mass (M) and cycle-time (T) are known
)(
))(89)((
111 md
uKgM
d
MG
d
FK sts
spring constant (K) when mass (M) and amplitude ( d1) are known
2 21( )
Kv d d
m
velocity (v) at any point during the cycle
velocity
dforM
Kdv
max
012 velocity (v2) maximum (when d = 0 at position-2 and 4)
acceleration (a1) maximum when spring constant (K) mass (M) and amplitude (d1) are known
acceleration (a1) maximum when amplitude (d1) and cycle-time (Tc) are known
acceleration (a) at any point during the cycle-time (T)
1
151
286
v
dTc
cycle-time (Tc) when amplitude (d1) and velocity (v1) are known
K
MTc 28651
cycle-time (Tc) when mass (M) and spring constant (K) are known
fTc
1
cycle-time (Tc) when the frequency is known
cTf
1
frequency (f) when the cycle-time (Tc) is known
of cycles
elapsed timef
frequency (f) when the number of cycles in an elapsed time is known
d1 amplitude (distance from rest to start position)
Also frac12 the distance from the base to the peak
3131 oror dM
Ka
251
3131 539
c
oror T
da
2539
cT
da
In December of 2006 I discovered the mathematical error in Einsteinrsquos Special Relativity Einstein used BS math and that math only works in an imaginary world
It cannot describe real symmetrical space
Einsteinrsquos Special Relativity should be abandoned and a theory of real space using SM should be used to replace it
Numerous articles on the Internet show Einsteinrsquos theory to be incorrect but no one in academia is willing to publish the information
Figure 1-10 A moving particle observed from two frames of reference in uniform translation
Einsteinrsquos Math Error in Special Relativity Transformation of Velocity (From Eisenberg- Fundamentals of Modern Physics pages 27-29) Consider the particle shown in figure (1-10) moving with velocity v as
seen in a frame of reference O
dx dy dz
Vx Vy Vzdt dt dt
dx dy dz
V x V y V zdy dt dt
We would like to evaluate the velocity v of the particle as seen in the frame of reference 0 which is itself moving relative to 0 with velocity v Measured in the 0 frame the velocity vector of the particle has com ponents
The same velocity vector as measured in the 0 frame has components
2
2
2
2
2
1
1
x vtx
v
cy y
z z
vxt
ctv
c
Now from equations (1-13) we know that the relation between the primed and the unprimed coordinates and times is
Take the differential of these equations remembering that v is a constant This gives
2
2
2
2
2
1
1
dx vdtdx
v
cdy dy
dz dz
vxdt
cdtv
c
2
2
2 2 2 2 2
2
2
2
11 ( )
1( ) 1
1
1
1
dx vdt
v dxdx vdt v
dx dx vdtc dt dtvdx vdx vdx v dxdt dt dt dt
dt dtc c c c
VxV x
Vx
VxV x
vv
vv
c
x
v
c
cV
( )
coordinate of particle relative to moving primed system
coordinate of particle relative to stationary system
velocity of primed coordina
V x
Vx
te system relative to stationaryv sys
tem
Page 28 Fundamentals of Modern Physics
(1-16)
According to Einstein this equation tells you how to transform the observed velocity from one frame of reference to another frame of reference First we note that as Vc and vc approachzero the equation approaches those which would be derived from the Galilean transformation
2 2
( ) 1 01 1 ( )
Vx v Vx vV
VxV x
Vx v
v cVxx
v
c
v
cc
Another property of this equation is that it is impossible to choose V and v such that Vrsquo the magnitude of the velocity which is seen in the new frame is greater than c
1
2
09 ( 09 ) 18009876543
( 09 ) 1811 (09 )
c c cV c c
cc
c
Example As seen by O particle 1 has velocity 09c in the direction of positive x and particle 2 has velocity 09c in the negative x direction
Einstein equation however predicts To evaluate the velocity of particle 1 with respect to particle 2 we transform from the O frame to the Orsquo frame moving in the negative x direction with velocity v = -09c using equation 1-16 We obtain
JK LOGIC will observe the particles separating at 18c
A confusing INCORRECT and ILLOGICAL answer
left is a dir rightsub of a
1
2
1
09 ( 09 ) 18009876543
181( 09 )1 ( 09 )
The multiplication of dir left time a dir right is ONLY a dir left
09
dir
direction leftdirection rightsubtraction
c c cV c c
cc
c
V
dir
22
22
09 18 18009876543
181( 081)1(09)(09)
1
left rightsub
sub
c c c cc c
ccc
c
Analyzing Einsteinrsquos BS math Equation
Using SM math a logical answer can be obtained
Cannot multiply a subtraction operator times a direction in space
Can only subtract and change its direction
Cannot multiply a direction to the left times a direction to the right
Cannot multiply opposite directions in space
Can only solve for the resultant
rightleft
subtraction
1
subtraction
2
sub of a dir to the left is a dir to the right
1 resultant 0 sub 2
2sub 2
2
09 09
09 amp 091
09 09 18 1
(0)1
(09) amp (09)1
c cV
c cc
c c cV
cc
cc
8
181
cc
SM math predicts a logical answer using logic
21st century scientists state in many articles that Einsteins General Relativity and Quantum Mechanics are incompatible
They are not only incompatible they are both mathematically incorrect
They both use illogical BS math
Any equations in math that use imaginary numbers should be abandoned
Imaginary numbers are only necessary in the current BS math system
This means that Schroumldingers equations should be abandoned
Any equation in quantum mechanics that uses imaginary number should be abandoned
Prof Kanarev has developed logical physics and chemistrytheories without the use of Quantum mechanics or relativity
A partial list of his theories coverbull Mechanics ndash replaces Newtonbull Photon - bull Electrons ndash Protons- Neutronsbull Spectroscopybull Nucleusbull Atoms amp Moleculesbull Electrodynamicsbull Heatbull Curvature of Space ndash replaces Schwarzcchildrsquos equation
Kanarevrsquos Hypothesis
bull Electron align linearly with proton They do NOT spin or form clouds around the nucleus
bull Neutrons are in the center of atom Proton are outside of Neutrons
bull The wavelength of all particles is the experimental radius
bull Double slit experiment explained correctly
bull Heisenbergrsquos inequality explained correctly
bull is the radius of approach of the magnetic lines around the torus of the electron
( 1 137)
bull Why water expands when it freezes
bull Electron clusters form lightning
rbull The meaning of Plancks constant is explained
22 2 kg m
h m f mr f consts
2210254 42m r kgm bull If the mr-constant describes a ring the equations
describe experiments bull If the mr-constant describes a wave the equations do
NOT describe experiments
Kanarevrsquos 5 Equations of Mechanics
Kanarevrsquos 1st Equation Acceleration ampmv maMma R R
Kanarevrsquos 2nd Equation Uniform motion amp mv maMma R R
maM mvR R
Kanarevrsquos 3rd Equation
Kanarevrsquos 4th Equation Equal and opposite reaction
1 21 21 2
12
a m
m a m ama
Kanarevrsquos 5th Equation 1 2 3 Mi M M M Mna R R R R R
The sum of the Actions working on a moving body is never equal to zero
Kanarevrsquos Photon model is a rolling ring
His equations accurately describe known experiments with photonsbull Electromagnetic radiation should be called photon radiationbull Diffraction and interference are the same phenomenabull Correctly explains the double slit experiment without quantum equationsbull Correctly explains the meaning of Heisenbergrsquos inequality
e
2 (2)(3142)(2817 15 ) 10007297352568
R 2426 12 137mag
xp
r m
m
Charge
Magnetic action
Proton
The intensity of the magnetic action near the geometrical center of the proton ring is so great that we theorize that it is the strong nuclear force
(9383 8)(1602 19)8476 14
4 (4)(3142)(1411 26)p
pp
EB T
M
(9395 8)(1602 19)1035 15
4 (4)(3142)(0966 26)N
NN
EB T
M
Neutron
1a Hydrogenavi
It took four years of study to discover why a (-)(-)=(+) only works in an imaginary world where space in the dash (-) direction is different from space in the cross (+) direction
Started working on the problem in August of 2001
Data points created using BS math produce graphs that are not symmetrical dash side different from cross side
However if a real problems data points follow a non symmetrical graph the graph can provide usable answers
This is why no one discovered the problem
Solved in January of 2005
Symmetry-Math(SM) vs Broken-Symmetry(BS) Math
In SM if we specify that the dash sign (-) means only subtraction and the cross sign (+) means only addition and an appropriate symbol is used for a direction in space space becomes symmetrical and math becomes logical
Math operators and directions in space are not the same and the same symbol should not be used to represent them
Letrsquos start with the (dash)multiplied by a (dash)=(cross) in the BS system
Usable answers may be obtained but it will be for illogical reasons
multiplied by a ( ) ( ) ( )subtraction Subtraction addition
rightleft left
dash dash cross
positivenegative negative
Symmetry-Math The subtraction of a direction is equal to the opposite direction The answers are correct using correct logic
opposite direction direction
SM
(negative)(negative) = (positive)
( ) ( ) ( )subtraction
BS
There is NO multiplication of a subtraction operator by a direction in space
There is just the subtraction of a direction in space
If instead of labeling the left side of a coordinate system as a negative (-) the same as a subtraction operator we label it with an arrow ( ) to represent the direction then a subtraction from that arrow direction will be in the opposite direction ( )
( ) mutiplied by a ( ) ( )
negative Positive negative
subtraction addition subtraction
dash cross dash
left right left
A subtraction operator multiplied by an addition operator is equal to a subtraction operator This is illogical
BS Math
A negative multiplied by a positive is equal to a negative This is illogical
A direction to the left multiplied by a direction to the right is a direction to the left This is illogical and Einsteins math error in special relativity
A dash times a cross is equal to a dash This is illogical
A vector moving to the left multiplied by a vector moving to the right is equal to ONLY to a vector moving to the left that is eight orders of magnitude faster thanThe speed of light
opposite direction subtraction direction
SM ( ) ( )
BS (negative)(positive) = (negative)
Symmetry Math The subtraction of a direction is equal to the opposite direction
There is no multiplication of a subtractionoperator times a direction in space
( ) multiplied by a ( ) ( )
positive positive positiveaddition addition additioncross cross cross
direction direction direction
What is the meaning of multiplying addition operators Addition operators are not multiplication operators
BS Math
() multiplied by a direction ( )=
number
() multiplied by a direction ( )=
number
A number times a direction maintains the same direction
( ) ()
dirdiradd
( ) ()
dirdiradd
The addition of a direction is in the same direction There is no multiplication by addition operators
Symmetry Math
Future math books need to eliminate the use of the cross sign (+) to represent something labeled a positive direction in space and the dash sign (-) to represent something labeled a negative direction in space Space does not have positive and negative directions
Symbols that are logical and have no illogical representation should be adopted SM uses either arrows or symbols
Symmetry-Math Number Line
Symmetry-Math Rules
The dash sign (-) will have only one use subtraction
The cross sign (+) will have only one use addition
Directions have an arrow and a number
All observers see the same direction and magnitude
bull There is no multiplication of arrows
amp ampamp ampampamp
( 3 ) ( 5 ) ( 6 ) ( 65 ) (17256 ) ( 32993 )
dir dirdir dirdirdir
Negative and Positive Directions in Space
If I asked you to point to a negative direction in space which way would you point
Hopefully you will realize that there is no such thing as a negative direction in space
If I asked you to point to a positive direction in space which way would you point
Again hopefully you will realize that there is no such thing as a positive direction in space
Objects on the right side of the x-axis are positive (+)(+) = (+)
Objects on the left side of the x-axis are negative (-)(-) = (+) negative math is different from positive math This is illogical
Objects on the top of the y-axis are positive
Objects on the bottom of the y-axis are negative
Objects in the front of the z-axis are positive
Objects in the back of the z-axis are negative
Again BS math of the positive direction (x right y up and z front) is different from the BS math of the negative direction (x left y down and z back)
BS math is Broken-Symmetry
Mirror image is broken symmetry in the x and z axis If the left amp right-axis are
reversed symmetry is broken Obs-1 math is different from obs-2
If the front amp back-axis is reversed symmetry is broken Obs-1 math is different from obs-2
If the top amp bottom-axis are
reversed symmetry is not broken Obs-1 math is the same as obs-2
Math answers cannot depend on which side of a number line an observer sit
Absolute values for Displacement
Even for a simple displacement of an object BS math invented a definition and absolute values to obtain something labeled a positive number answer
I----------I----------I----------I----------I----------I----------I----------I----------I
-4 -3 -2 -1 0 +1 +2 +3 +4 a gtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtb cltltltltltltltltltltltltltltltltltltltltltltltltltltltb
If an object starts at ( 3)ia X and moves to ( X 3) and the to ( 1)b fb c X
BS Math provides a usable answer without a definition or absolute values
However If an object starts at
( 1) and moves to ( X 3) and the to ( 3)i b fa X b c X
Then absolute values and a definition must be used
---------I----------I----------I----------I----------I----------I----------I---------- -4 -3 -2 -1 0 +1 +2 +3 +4
a gtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtb cltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltb
BS Math makes up a definition that all displacements are positive Since this example provides an answer with something labeled a negative the positive definition must be applied and (-2) is changed to (+2) A definition is needed to arrive at a useable answer This is Illogical
In SM absolute values and a meaningless definition of ldquopositiverdquo are removed In SM objects and directions in space are defined by the direction and magnitude of the resultant of arrows
With SM you get total distance traveled by the object and the final direction of the displacement
22222 2 bababbaababababa If w e su b stitu te p o s itiv e n u m b e rs in to th e eq u a tio n th e an sw er w ill b e co rrec t If (a = 5 ) an d (b = 3 )
2
2
8
5 3 5 3 5 3 5 5 5 3 3 5 3 3 25 15 15 9 64
8 8 8 8 8 8 8 8 8 8 64
W h en tw o p o sitiv e n u m b ers a re a d d ed an d sq u a re d th e B S m a th d is tr ib u tiv e law p ro v id es a co rre c t an sw er 1
Illogical and Incorrect BS math for the distributive law
222 2 bababbabbaaabababa B y d e fin itio n in B S m a th n o n u m b er sq u a red c a n ev e r b e n eg a tiv e T h e re fo re n o sq u a red n u m b er c an ev e r b e a v e c to r m o v in g in th e d a sh ( -) o r n eg a tiv e d ire c tio n if it is sq u a red In B S m a th an d ad d itio n o p e ra to r m u ltip lied b y a su b trac tio n o p e ra to r is A L W A Y S eq u a l to a su b tra c tio n o p e ra to r If th is h ad n o t b e en u sed fo r th e la s t fo u r h u n d red y e a rs y o u w o u ld d ie lau g h in g a t th e u tte r ab su rd ity o f th e la s t s ta tem e n t T h is is illo g ica l an d v io la te s sy m m etry
If a = 5 an d b = 3 42222352
22
4930253)3)(5(253)5)(3()3)(5(5353535 22222 A s lo n g a s ldquo ardquo is g re a te r th an ldquob rdquo th e B S m a th d is tr ib u tiv e law p ro v id es co rrec t an sw ers 1
B S m a th b re a k s d o w n w h e n ldquo b rdquo is g rea ter th a n ldquo a rdquo
If a = 3 an d b = 5
2 2
2
2
3 5 2 4
2 ( 2) ( 2) 4
2 ( )( 2) ( )( 2) 4
A n y co m b in a tio n ex cep t th e m id d le o n e is illo g ica l an d in co rre c t B y d e fin itio n th e an sw er can n o t b e -4 in B S m a th B S m a th is in co rrec t
4253095)5)(3(235)3)(5()5)(3(3535353 22222 T h is is w h ere th e d is tr ib u tiv e la w p ro v id es in co rrec t an sw ers I t p ro d u ces an an sw er o f + 4 b e cau se b y d e fin iti o n n o n u m b er sq u a red in B S m a th c an b e a n eg a tiv e n u m b er U sin g in co rre c t B S m a th th e d is tr ib u tiv e law w ill g iv e an in co rrec t an sw ers o f + 4 T h is is in co rrec t an d illo g ica l 1
Since there can be no negative direction from a squared term in BS math BS math should be modified or abandoned
The BS distributive law should be modified or abandoned
The numbers +1+2+3+4 and ndash1-2-3-4hellip should be abandoned
BS math gives incorrect answers because their Rule-of-Signs allows multiplying an arrow going in one direction by an arrow going in another direction
Clearly this is not logical
Arrows should be used that are specific for directions
W h at d o es it m e an to m u ltip ly ( 4) b y (4 ) A s w e k n o w 1
m u ltip lica tio n is ju s t ad d itio n W h a t w o u ld ( 4) tim e (4 ) b e 2
In BS math the answer is by definition -16 unless you square -4 and then it is +16
In SM you cannot multiply arrows going in opposite directions
The middle two terms are not logical You cannot multiply opposite directions
This is a MAJOR error in BS math They multiply a dash [(-) a direction to the left] by a cross [(+) a direction to the right] And then ldquoby definitionrdquo label the answer dash (-) This is illogical and produces incorrect answers
2
2 2
This is Einsteinrsquos math error
in Special relativity
Magnitude 4 Direction 1 Magnitude 3 Direction 2 Magnitude 5 Direction 3 Relative to an observer at 0 4
Magnitude 4 Direction 1 Magnitude 3 Direction 2 Magnitude 5 Direction 3 Relative to an observer at 0 4
Magnitude and Direction (MD) An arrow has both a magnitude andA direction relative to a 2nd objectobserver
bull Magnitude and velocity (MV) relative to a 2nd objectobserver
bull Magnitude And acceleration (MA) relative to a 2nd objectobserver2
v da
t t
Introducing a new symbol the ldquoamprdquo sign The ldquoamprdquo sign means resultant ADD all the Arrows for each direction and then SUBTRACT to find the final resultant
1 1amp 1 2 2amp 2 4 If =
1 amp1 2 2 amp2 4 If =
amp 5 amp 3 2 If gt
amp 3 amp 5= 2 If gt
amp 0 3 amp 3=0 if =
only the resultant equals zero
the vectors do not - they are still there
Does not exist There is no multiplication of arrows
There are two arrows One has a magnitude of 3 One has a magnitude of 5 The maximum these two arrows can be is 8
Only the resultant of arrows exist
An SM arrow has a direction and a magnitude
Arrows and magnitudes are still there even if the resultant is zero
There are no negative SM Arrows
There are no positive SM Arrows
There are no ZERO SM Arrows A zero would negate the definition of an SM Arrow something with no magnitude and no direction cannot be the definition of an arrow that is defined as having magnitude and direction
bull A resultant is the addition and subtraction of the differences in magnitude and direction arrows
A B and C are the arrows (a) A amp B exist(b) A amp B amp C exist(c) A-B Do Not Exist in Sm there are no negatives or positives(d) A+B-C Do Not Exist in SM there are no negatives or positives
There are only resultants when there are multiple arrows
SM Graphs versus BS Graphs
Numbers Raised to Powers
When numbers are raised to powers the negative sides of all BS math graphs are ILLOGICAL and INCORRECT
Incorrect BS Math for 2x
2x
4 16 correct3 8 correct2 4 correct1 2 correct0 1 correct-1 005 incorrect-2 025 incorrect-3 0125 incorrect-4 00625 incorrect
negative side of graph is incorrectpositive side of graph is correct
16
8
42105025012500625
-4 -3 -2 -1 0 1 2 3 4
po
sit
ive
negative positive
2x BS Math
correctincorrect
In BS math the dash sign (-) used in this example represents three different operations They arebull (-) Subtractionbull (-) A direction to the leftbull (-) As an exponent to mean divide by
All positive numbers produce correct results and graphs
All negative numbers produce incorrect results and graphs
A direction of three units to the left (-3) is placed in the equation as an exponent
A direction in space (3 units toward the left) changes a whole number to a fraction by using the dash (-) as an exponent rather than a direction in space
This is illogical and incorrect BS math should have performed this operation as
Unfortunately the answers to problems can come out correct because data points of a real problem fit the curve
However the answers will not be due to logical reasoning
Because the equation produced a curve that real problem data points follow it has been falsely believed that the equations and graphs are correct
1
2
x
Incorrect BK graph for (12)x
(12)x
4 00625 correct3 0125 correct2 025 correct1 1 correct0 1 correct-1 2 incorrect-2 4 incorrect-3 8 incorrect-4 16 incorrect
negative side of graph is incorrectpositive side of graph is correct
00625012502505124
8
16
-5 -4 -3 -2 -1 0 1 2 3 4 5P
osit
ive
negative positive
BS Math for (12)x
correctincorrect
Incorrect BS math and graph for
All graphs are symmetrical with SM
Two BS Math Equations that Produce Incorrect Answers
2 0x x starting equation 2
2
x x x x
x x
Add lsquoxrsquo to both sides
2(1) 1 let lsquoxrsquo = 1
2 1 This is an incorrect answer 2(0)-(0)=0 The only answer that BK math can give is x=0
1
a b starting equation ( )( ) ( )( )a a a b multiply both sides by ( )a
( )( ) ( )( ) ( )( ) ( )( )a a b b a b b b subtract ( )( )b b from both sides
( )( ) ( )( )a b a b a b b factor ( )a b from both sides
( )a b b divide both sides by ( )a b
( )b b b replace ( )a with ( )b 2b b simplify 2(1) (1) Let ( )b = 1
2 1 Again BS math yields and incorrect answer 1
This incorrect result is a problem with the distributive law SM explains how this incorrect answer is obtained and the problem with the distributive law that allows the error
Page 148 ndash149 Barry Mazur ndash Imagining Numbers
3
2
31
Do the computation carefully on paper using the rules we agreed to Then ponder your answer which should be something of a surprise to you donrsquot stop there Think of what your answer might possibly mean or might imply
Solve this equation using the distributive law
JK definitely a surprise the answer wasincorrect and illogical
Gerhard Gade University Professor doing math at Harvard University
Here is the BS math answer that Barry Mazur calculated
2
31
2
31
2
31
1 3 1 3 1 3 3 3 2 2 3 1 3
2 2 2 2 4 4 4 4 4 4 2 2
dash
1 3 1 3 1 31
2 2 2 2 4 4
3 3
4 4
When the distributive law is used with the BS Rule-of-Signs and one of the signs is a (-) dash sign the middle two terms always cancel each other This leads to the incorrect answer of (-1) in the above equation
Barry Mazur did not go back and check his answer with the original equation If he had he would have known that the answer was incorrect
3
1 31
2
check answer by setting equal
3
331 3
12
1 31
2
Take the Cube root of both sides
1 3 2 multiply both sides by 2
3 3 subtract 1 from both sides
2 2
3 3 square both sides
3 9 This is an incorrect answer A simple equation check could have shown that the distributive law using BS math provides an incorrect answer
1
BS Illogical and Incorrect Numbers Raised to Powers
nC D n DC
642 DDD
In BS math only a positive number can have an answer if ldquonrdquo is an even number (246---)
All negative numbers have no solutions
642 DDD
With SM just add a direction arrow and any number gives the correct answer
or n nC D C D
3 3( 3 8 8 or 8 2 or 2 )n D C
( 2 16 C= 16 or 16 4 or 4 )n D
Since a negative number has no square root answer in BS math BS mathematicians invented imaginary numbers
i2=-1
24414 iiC n
These equations are easily solved without the use of imaginary numbers with SM math
If 4 then 4 2x
If 4 then 4 2x
The 360-degree circle is a 4400 years old extremely outdated system based on a year of 12 months of 30 days each that is 360 days
SM will use a more logical system that will have a circle of 6283185307o
d
( )
(deg )
d r radians
d r rees
the arc distance around
the circumference of a circle
d
radiusr
2 2
2 2
d r
dv rd t t
v dda rd t t t
1
d
r
( )av tv
The average angular velocity of a body is the rate of its angular displacement
( )0
liminstt
vt
The instantaneous angular velocity is the limit of the ratio
t
as t approaches zero
( )d
av
d
tv
The average circumference velocity of the body is the rate of its distance traveled d divided by the time t
dvv
r
1 dd vdrv
t t t r r
The equation of the relationship between the angular velocity and the velocity of a point on that body
( )o
av
v v v
t ta
The average angular acceleration is the change in the angular velocity
( )0
liminstt
v
ta
The instantaneous angular acceleration is the
limit of the ration of v
t
as t approaches
zero
daa
r
1
1
d
d s d
vva
t r tv a
r t r
The equation of the relationship between the angular acceleration and the acceleration of a point on the body
21
2v a t
If a body with an initial angular velocity ov has a constant
angular acceleration a it will turn through an angle in a time t
1
Straight Line Symbol Units Rotational Symbol Units Distance d m Angular
Distance radians or degrees
Velocity dt
v m
s Angular Velocity
V
t
radians or degrees
s
Acceleration vat
2ms
Angular Acceleration
VA
t
2
radians or degrees
s
Mass M Kg Mass distribution
dM 2Kgm
Force actionA Ma
2
Kgm
sNewton
Torque
dT M A 2
2
Kgm
s
Momentum Mv Kgm
s
Angular Momentum
dM V 2Kgm
s
Work (Force)(distance)
Ad 2
Kgmm
sNm Joule
Work T
Using AS arrow math to solve and direction in space problems using the equation 1 2
2 2 2 2 2 2( ) amp and ( ) amp 3
4
2 2 2
2 2
( ) amp
(5 ) 4 3
(5 ) 16 9
(5 ) 25
(5 ) (5 )
2 2(5 ) 4 3
(5 ) 16 9
(5 ) 25
(5 ) (5 )
In SM math Pythagorasrsquo theorem provides both magnitude and direction
In BS math Pythagorasrsquo theorem provides magnitude only
time set equal to zero
23
( ) ( 0 )10 1
x t xx
3( ) ( 0)
10 1W t f
In BS math this is an equation for a traveling wave
must use a square term in order to make the curve symmetric
NO SQUARE TERM IS NECESSARY
In SM math when t is set to ldquo0rdquo
This is a major fault in BS math
x f(x)
-4 0019
-3 0033
-2 0073
-1 0273
0 3000
1 0273
2 0073
3 0033
4 0019 -4 -3 -2 -1 0 1 2 3 400
05
10
15
20
25
30
35
BS wave (must use square term)
snap shot of wave at x = 0
Am
plit
ude
lt to gt lt to gt f(ltgt)
lt -4 0073
lt -3 0097
lt -2 0143
lt -1 0273
origin 0 3000
gt 1 0273
gt 2 0143
gt 3 0097
gt 4 0073lt0073 lt0097 lt0143 lt0273
3000
0273gt 0143gt 0097gt 0073gt
Wave moving
snap shot of wave
Am
plit
ude
lt4 lt3 lt2 lt1 0 1gt 2gt 3gt 4gt
3
2
1
In SM NO squared terms are needed
SPRING
An Action is needed to stretch or compress a spring to a starting point where it can execute cycle motion
tsA (ts to start) is the work required to move a mass attached to a spring a distance (d1) from
its initial rest position RAK 0 the spring constant is the stiffness of the spring
o A is attraction o R is repulsion o
ARRA
00 is for the back and forth motion through the zero initial rest position A
stretched spring goes from ldquoArdquo to ldquo0rdquo to ldquoRrdquo to ldquoArdquo to ldquo0rdquo to ldquoRrdquo d1 is the distance a mass is moved from its initial rest position
0
0 1ts A RA K d
Once ( )mad
is performed to move the mass to position d1 (against the restraint of the spring) a of the ( )mad
is transferred into the spring as potential attractionrepulsion (PAR) There is always
more ( )mad
performed than is stored as PAR When the spring is released from the restraining Ats the PAR starts the oscillation Conservation of ( )mad
requires that the PAR and the motion ( )mad
( )( )madM be a constant
( )amp ( ) constantmadPAR M 02
0 0
1
2A R A RPAR K d
02
( ) 0 0
1( )
2mad A R A RM m v
0 0 02 2 2
0 0 0 1
2 2 21
2 2 21
1 1 1amp
2 2 2
1 1 1amp
2 2 2
amp
A R A R A RK d m v K d
Kd mv Kd
Kd mv Kd
0
1d is frac12 the height from the base to the peak of a cycle
An experiment of moving paper under a pendulum through a small angle filled with sand that can escape through a small hole in the bottom shows that a pendulum cycling back and forth follows a curve that is mathematically analogous to a lsquosinrsquo or lsquocosrsquo
The projection of the curved area onto the straight line exactly duplicate the back and forth motion of CM This allows for the math of lsquosinrsquo or lsquocosrsquo to be used
SM will use lsquosinrdquo math It could use lsquocosrsquo or a combination of both lsquosinrsquo and lsquocosrsquo Using just the lsquosinrsquo math makes the understanding and math easier
0 sin 0 0SM 0 sin157 1SM The diagram at the left shows a spring being stretched from its (zero rest) position-2 and the PAR being released at position-1 There are 4 sections (1 to 2 2 to 3 3 to 4 and 4 to5) that alternate between maximum acceleration and zero velocity to max velocity and zero acceleration
At position-1 there is maximum acceleration and zero velocity o The lsquosinrsquo values from 0o to 157o will be used for velocity (values from 0 to 1) o The lsquosinrsquo values from 157o to 0o will be used for acceleration (values from 1 to 0)
At position-2 there is maximum velocity and zero acceleration Acceleration changes to deceleration
o The lsquosinrsquo values from 157o to 0o will be used for deceleration (values from 1 to 0) o The lsquosinrsquo values from 0o to 157o will be used for acceleration (values from 0 to 1)
At position-3 there is maximum acceleration and zero velocity o The lsquosinrsquo values from 0o to 157o will be used for velocity (values from 0 to 1) o The lsquosinrsquo values from 157o to 0o will be used for acceleration (values from 1 to 0)
At position-4 there is maximum velocity and zero acceleration Acceleration changes to deceleration
o The lsquosinrsquo values from 157o to 0o will be used for deceleration (values from 1 to 0) o The lsquosinrsquo values from 0o to 157o will be used for acceleration (values from 0 to 1)
At position-5 the cycle starts again
2
)539(
cT
MK
spring constant (K) when mass (M) and cycle-time (T) are known
)(
))(89)((
111 md
uKgM
d
MG
d
FK sts
spring constant (K) when mass (M) and amplitude ( d1) are known
2 21( )
Kv d d
m
velocity (v) at any point during the cycle
velocity
dforM
Kdv
max
012 velocity (v2) maximum (when d = 0 at position-2 and 4)
acceleration (a1) maximum when spring constant (K) mass (M) and amplitude (d1) are known
acceleration (a1) maximum when amplitude (d1) and cycle-time (Tc) are known
acceleration (a) at any point during the cycle-time (T)
1
151
286
v
dTc
cycle-time (Tc) when amplitude (d1) and velocity (v1) are known
K
MTc 28651
cycle-time (Tc) when mass (M) and spring constant (K) are known
fTc
1
cycle-time (Tc) when the frequency is known
cTf
1
frequency (f) when the cycle-time (Tc) is known
of cycles
elapsed timef
frequency (f) when the number of cycles in an elapsed time is known
d1 amplitude (distance from rest to start position)
Also frac12 the distance from the base to the peak
3131 oror dM
Ka
251
3131 539
c
oror T
da
2539
cT
da
In December of 2006 I discovered the mathematical error in Einsteinrsquos Special Relativity Einstein used BS math and that math only works in an imaginary world
It cannot describe real symmetrical space
Einsteinrsquos Special Relativity should be abandoned and a theory of real space using SM should be used to replace it
Numerous articles on the Internet show Einsteinrsquos theory to be incorrect but no one in academia is willing to publish the information
Figure 1-10 A moving particle observed from two frames of reference in uniform translation
Einsteinrsquos Math Error in Special Relativity Transformation of Velocity (From Eisenberg- Fundamentals of Modern Physics pages 27-29) Consider the particle shown in figure (1-10) moving with velocity v as
seen in a frame of reference O
dx dy dz
Vx Vy Vzdt dt dt
dx dy dz
V x V y V zdy dt dt
We would like to evaluate the velocity v of the particle as seen in the frame of reference 0 which is itself moving relative to 0 with velocity v Measured in the 0 frame the velocity vector of the particle has com ponents
The same velocity vector as measured in the 0 frame has components
2
2
2
2
2
1
1
x vtx
v
cy y
z z
vxt
ctv
c
Now from equations (1-13) we know that the relation between the primed and the unprimed coordinates and times is
Take the differential of these equations remembering that v is a constant This gives
2
2
2
2
2
1
1
dx vdtdx
v
cdy dy
dz dz
vxdt
cdtv
c
2
2
2 2 2 2 2
2
2
2
11 ( )
1( ) 1
1
1
1
dx vdt
v dxdx vdt v
dx dx vdtc dt dtvdx vdx vdx v dxdt dt dt dt
dt dtc c c c
VxV x
Vx
VxV x
vv
vv
c
x
v
c
cV
( )
coordinate of particle relative to moving primed system
coordinate of particle relative to stationary system
velocity of primed coordina
V x
Vx
te system relative to stationaryv sys
tem
Page 28 Fundamentals of Modern Physics
(1-16)
According to Einstein this equation tells you how to transform the observed velocity from one frame of reference to another frame of reference First we note that as Vc and vc approachzero the equation approaches those which would be derived from the Galilean transformation
2 2
( ) 1 01 1 ( )
Vx v Vx vV
VxV x
Vx v
v cVxx
v
c
v
cc
Another property of this equation is that it is impossible to choose V and v such that Vrsquo the magnitude of the velocity which is seen in the new frame is greater than c
1
2
09 ( 09 ) 18009876543
( 09 ) 1811 (09 )
c c cV c c
cc
c
Example As seen by O particle 1 has velocity 09c in the direction of positive x and particle 2 has velocity 09c in the negative x direction
Einstein equation however predicts To evaluate the velocity of particle 1 with respect to particle 2 we transform from the O frame to the Orsquo frame moving in the negative x direction with velocity v = -09c using equation 1-16 We obtain
JK LOGIC will observe the particles separating at 18c
A confusing INCORRECT and ILLOGICAL answer
left is a dir rightsub of a
1
2
1
09 ( 09 ) 18009876543
181( 09 )1 ( 09 )
The multiplication of dir left time a dir right is ONLY a dir left
09
dir
direction leftdirection rightsubtraction
c c cV c c
cc
c
V
dir
22
22
09 18 18009876543
181( 081)1(09)(09)
1
left rightsub
sub
c c c cc c
ccc
c
Analyzing Einsteinrsquos BS math Equation
Using SM math a logical answer can be obtained
Cannot multiply a subtraction operator times a direction in space
Can only subtract and change its direction
Cannot multiply a direction to the left times a direction to the right
Cannot multiply opposite directions in space
Can only solve for the resultant
rightleft
subtraction
1
subtraction
2
sub of a dir to the left is a dir to the right
1 resultant 0 sub 2
2sub 2
2
09 09
09 amp 091
09 09 18 1
(0)1
(09) amp (09)1
c cV
c cc
c c cV
cc
cc
8
181
cc
SM math predicts a logical answer using logic
21st century scientists state in many articles that Einsteins General Relativity and Quantum Mechanics are incompatible
They are not only incompatible they are both mathematically incorrect
They both use illogical BS math
Any equations in math that use imaginary numbers should be abandoned
Imaginary numbers are only necessary in the current BS math system
This means that Schroumldingers equations should be abandoned
Any equation in quantum mechanics that uses imaginary number should be abandoned
Prof Kanarev has developed logical physics and chemistrytheories without the use of Quantum mechanics or relativity
A partial list of his theories coverbull Mechanics ndash replaces Newtonbull Photon - bull Electrons ndash Protons- Neutronsbull Spectroscopybull Nucleusbull Atoms amp Moleculesbull Electrodynamicsbull Heatbull Curvature of Space ndash replaces Schwarzcchildrsquos equation
Kanarevrsquos Hypothesis
bull Electron align linearly with proton They do NOT spin or form clouds around the nucleus
bull Neutrons are in the center of atom Proton are outside of Neutrons
bull The wavelength of all particles is the experimental radius
bull Double slit experiment explained correctly
bull Heisenbergrsquos inequality explained correctly
bull is the radius of approach of the magnetic lines around the torus of the electron
( 1 137)
bull Why water expands when it freezes
bull Electron clusters form lightning
rbull The meaning of Plancks constant is explained
22 2 kg m
h m f mr f consts
2210254 42m r kgm bull If the mr-constant describes a ring the equations
describe experiments bull If the mr-constant describes a wave the equations do
NOT describe experiments
Kanarevrsquos 5 Equations of Mechanics
Kanarevrsquos 1st Equation Acceleration ampmv maMma R R
Kanarevrsquos 2nd Equation Uniform motion amp mv maMma R R
maM mvR R
Kanarevrsquos 3rd Equation
Kanarevrsquos 4th Equation Equal and opposite reaction
1 21 21 2
12
a m
m a m ama
Kanarevrsquos 5th Equation 1 2 3 Mi M M M Mna R R R R R
The sum of the Actions working on a moving body is never equal to zero
Kanarevrsquos Photon model is a rolling ring
His equations accurately describe known experiments with photonsbull Electromagnetic radiation should be called photon radiationbull Diffraction and interference are the same phenomenabull Correctly explains the double slit experiment without quantum equationsbull Correctly explains the meaning of Heisenbergrsquos inequality
e
2 (2)(3142)(2817 15 ) 10007297352568
R 2426 12 137mag
xp
r m
m
Charge
Magnetic action
Proton
The intensity of the magnetic action near the geometrical center of the proton ring is so great that we theorize that it is the strong nuclear force
(9383 8)(1602 19)8476 14
4 (4)(3142)(1411 26)p
pp
EB T
M
(9395 8)(1602 19)1035 15
4 (4)(3142)(0966 26)N
NN
EB T
M
Neutron
1a Hydrogenavi
Symmetry-Math(SM) vs Broken-Symmetry(BS) Math
In SM if we specify that the dash sign (-) means only subtraction and the cross sign (+) means only addition and an appropriate symbol is used for a direction in space space becomes symmetrical and math becomes logical
Math operators and directions in space are not the same and the same symbol should not be used to represent them
Letrsquos start with the (dash)multiplied by a (dash)=(cross) in the BS system
Usable answers may be obtained but it will be for illogical reasons
multiplied by a ( ) ( ) ( )subtraction Subtraction addition
rightleft left
dash dash cross
positivenegative negative
Symmetry-Math The subtraction of a direction is equal to the opposite direction The answers are correct using correct logic
opposite direction direction
SM
(negative)(negative) = (positive)
( ) ( ) ( )subtraction
BS
There is NO multiplication of a subtraction operator by a direction in space
There is just the subtraction of a direction in space
If instead of labeling the left side of a coordinate system as a negative (-) the same as a subtraction operator we label it with an arrow ( ) to represent the direction then a subtraction from that arrow direction will be in the opposite direction ( )
( ) mutiplied by a ( ) ( )
negative Positive negative
subtraction addition subtraction
dash cross dash
left right left
A subtraction operator multiplied by an addition operator is equal to a subtraction operator This is illogical
BS Math
A negative multiplied by a positive is equal to a negative This is illogical
A direction to the left multiplied by a direction to the right is a direction to the left This is illogical and Einsteins math error in special relativity
A dash times a cross is equal to a dash This is illogical
A vector moving to the left multiplied by a vector moving to the right is equal to ONLY to a vector moving to the left that is eight orders of magnitude faster thanThe speed of light
opposite direction subtraction direction
SM ( ) ( )
BS (negative)(positive) = (negative)
Symmetry Math The subtraction of a direction is equal to the opposite direction
There is no multiplication of a subtractionoperator times a direction in space
( ) multiplied by a ( ) ( )
positive positive positiveaddition addition additioncross cross cross
direction direction direction
What is the meaning of multiplying addition operators Addition operators are not multiplication operators
BS Math
() multiplied by a direction ( )=
number
() multiplied by a direction ( )=
number
A number times a direction maintains the same direction
( ) ()
dirdiradd
( ) ()
dirdiradd
The addition of a direction is in the same direction There is no multiplication by addition operators
Symmetry Math
Future math books need to eliminate the use of the cross sign (+) to represent something labeled a positive direction in space and the dash sign (-) to represent something labeled a negative direction in space Space does not have positive and negative directions
Symbols that are logical and have no illogical representation should be adopted SM uses either arrows or symbols
Symmetry-Math Number Line
Symmetry-Math Rules
The dash sign (-) will have only one use subtraction
The cross sign (+) will have only one use addition
Directions have an arrow and a number
All observers see the same direction and magnitude
bull There is no multiplication of arrows
amp ampamp ampampamp
( 3 ) ( 5 ) ( 6 ) ( 65 ) (17256 ) ( 32993 )
dir dirdir dirdirdir
Negative and Positive Directions in Space
If I asked you to point to a negative direction in space which way would you point
Hopefully you will realize that there is no such thing as a negative direction in space
If I asked you to point to a positive direction in space which way would you point
Again hopefully you will realize that there is no such thing as a positive direction in space
Objects on the right side of the x-axis are positive (+)(+) = (+)
Objects on the left side of the x-axis are negative (-)(-) = (+) negative math is different from positive math This is illogical
Objects on the top of the y-axis are positive
Objects on the bottom of the y-axis are negative
Objects in the front of the z-axis are positive
Objects in the back of the z-axis are negative
Again BS math of the positive direction (x right y up and z front) is different from the BS math of the negative direction (x left y down and z back)
BS math is Broken-Symmetry
Mirror image is broken symmetry in the x and z axis If the left amp right-axis are
reversed symmetry is broken Obs-1 math is different from obs-2
If the front amp back-axis is reversed symmetry is broken Obs-1 math is different from obs-2
If the top amp bottom-axis are
reversed symmetry is not broken Obs-1 math is the same as obs-2
Math answers cannot depend on which side of a number line an observer sit
Absolute values for Displacement
Even for a simple displacement of an object BS math invented a definition and absolute values to obtain something labeled a positive number answer
I----------I----------I----------I----------I----------I----------I----------I----------I
-4 -3 -2 -1 0 +1 +2 +3 +4 a gtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtb cltltltltltltltltltltltltltltltltltltltltltltltltltltltb
If an object starts at ( 3)ia X and moves to ( X 3) and the to ( 1)b fb c X
BS Math provides a usable answer without a definition or absolute values
However If an object starts at
( 1) and moves to ( X 3) and the to ( 3)i b fa X b c X
Then absolute values and a definition must be used
---------I----------I----------I----------I----------I----------I----------I---------- -4 -3 -2 -1 0 +1 +2 +3 +4
a gtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtb cltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltb
BS Math makes up a definition that all displacements are positive Since this example provides an answer with something labeled a negative the positive definition must be applied and (-2) is changed to (+2) A definition is needed to arrive at a useable answer This is Illogical
In SM absolute values and a meaningless definition of ldquopositiverdquo are removed In SM objects and directions in space are defined by the direction and magnitude of the resultant of arrows
With SM you get total distance traveled by the object and the final direction of the displacement
22222 2 bababbaababababa If w e su b stitu te p o s itiv e n u m b e rs in to th e eq u a tio n th e an sw er w ill b e co rrec t If (a = 5 ) an d (b = 3 )
2
2
8
5 3 5 3 5 3 5 5 5 3 3 5 3 3 25 15 15 9 64
8 8 8 8 8 8 8 8 8 8 64
W h en tw o p o sitiv e n u m b ers a re a d d ed an d sq u a re d th e B S m a th d is tr ib u tiv e law p ro v id es a co rre c t an sw er 1
Illogical and Incorrect BS math for the distributive law
222 2 bababbabbaaabababa B y d e fin itio n in B S m a th n o n u m b er sq u a red c a n ev e r b e n eg a tiv e T h e re fo re n o sq u a red n u m b er c an ev e r b e a v e c to r m o v in g in th e d a sh ( -) o r n eg a tiv e d ire c tio n if it is sq u a red In B S m a th an d ad d itio n o p e ra to r m u ltip lied b y a su b trac tio n o p e ra to r is A L W A Y S eq u a l to a su b tra c tio n o p e ra to r If th is h ad n o t b e en u sed fo r th e la s t fo u r h u n d red y e a rs y o u w o u ld d ie lau g h in g a t th e u tte r ab su rd ity o f th e la s t s ta tem e n t T h is is illo g ica l an d v io la te s sy m m etry
If a = 5 an d b = 3 42222352
22
4930253)3)(5(253)5)(3()3)(5(5353535 22222 A s lo n g a s ldquo ardquo is g re a te r th an ldquob rdquo th e B S m a th d is tr ib u tiv e law p ro v id es co rrec t an sw ers 1
B S m a th b re a k s d o w n w h e n ldquo b rdquo is g rea ter th a n ldquo a rdquo
If a = 3 an d b = 5
2 2
2
2
3 5 2 4
2 ( 2) ( 2) 4
2 ( )( 2) ( )( 2) 4
A n y co m b in a tio n ex cep t th e m id d le o n e is illo g ica l an d in co rre c t B y d e fin itio n th e an sw er can n o t b e -4 in B S m a th B S m a th is in co rrec t
4253095)5)(3(235)3)(5()5)(3(3535353 22222 T h is is w h ere th e d is tr ib u tiv e la w p ro v id es in co rrec t an sw ers I t p ro d u ces an an sw er o f + 4 b e cau se b y d e fin iti o n n o n u m b er sq u a red in B S m a th c an b e a n eg a tiv e n u m b er U sin g in co rre c t B S m a th th e d is tr ib u tiv e law w ill g iv e an in co rrec t an sw ers o f + 4 T h is is in co rrec t an d illo g ica l 1
Since there can be no negative direction from a squared term in BS math BS math should be modified or abandoned
The BS distributive law should be modified or abandoned
The numbers +1+2+3+4 and ndash1-2-3-4hellip should be abandoned
BS math gives incorrect answers because their Rule-of-Signs allows multiplying an arrow going in one direction by an arrow going in another direction
Clearly this is not logical
Arrows should be used that are specific for directions
W h at d o es it m e an to m u ltip ly ( 4) b y (4 ) A s w e k n o w 1
m u ltip lica tio n is ju s t ad d itio n W h a t w o u ld ( 4) tim e (4 ) b e 2
In BS math the answer is by definition -16 unless you square -4 and then it is +16
In SM you cannot multiply arrows going in opposite directions
The middle two terms are not logical You cannot multiply opposite directions
This is a MAJOR error in BS math They multiply a dash [(-) a direction to the left] by a cross [(+) a direction to the right] And then ldquoby definitionrdquo label the answer dash (-) This is illogical and produces incorrect answers
2
2 2
This is Einsteinrsquos math error
in Special relativity
Magnitude 4 Direction 1 Magnitude 3 Direction 2 Magnitude 5 Direction 3 Relative to an observer at 0 4
Magnitude 4 Direction 1 Magnitude 3 Direction 2 Magnitude 5 Direction 3 Relative to an observer at 0 4
Magnitude and Direction (MD) An arrow has both a magnitude andA direction relative to a 2nd objectobserver
bull Magnitude and velocity (MV) relative to a 2nd objectobserver
bull Magnitude And acceleration (MA) relative to a 2nd objectobserver2
v da
t t
Introducing a new symbol the ldquoamprdquo sign The ldquoamprdquo sign means resultant ADD all the Arrows for each direction and then SUBTRACT to find the final resultant
1 1amp 1 2 2amp 2 4 If =
1 amp1 2 2 amp2 4 If =
amp 5 amp 3 2 If gt
amp 3 amp 5= 2 If gt
amp 0 3 amp 3=0 if =
only the resultant equals zero
the vectors do not - they are still there
Does not exist There is no multiplication of arrows
There are two arrows One has a magnitude of 3 One has a magnitude of 5 The maximum these two arrows can be is 8
Only the resultant of arrows exist
An SM arrow has a direction and a magnitude
Arrows and magnitudes are still there even if the resultant is zero
There are no negative SM Arrows
There are no positive SM Arrows
There are no ZERO SM Arrows A zero would negate the definition of an SM Arrow something with no magnitude and no direction cannot be the definition of an arrow that is defined as having magnitude and direction
bull A resultant is the addition and subtraction of the differences in magnitude and direction arrows
A B and C are the arrows (a) A amp B exist(b) A amp B amp C exist(c) A-B Do Not Exist in Sm there are no negatives or positives(d) A+B-C Do Not Exist in SM there are no negatives or positives
There are only resultants when there are multiple arrows
SM Graphs versus BS Graphs
Numbers Raised to Powers
When numbers are raised to powers the negative sides of all BS math graphs are ILLOGICAL and INCORRECT
Incorrect BS Math for 2x
2x
4 16 correct3 8 correct2 4 correct1 2 correct0 1 correct-1 005 incorrect-2 025 incorrect-3 0125 incorrect-4 00625 incorrect
negative side of graph is incorrectpositive side of graph is correct
16
8
42105025012500625
-4 -3 -2 -1 0 1 2 3 4
po
sit
ive
negative positive
2x BS Math
correctincorrect
In BS math the dash sign (-) used in this example represents three different operations They arebull (-) Subtractionbull (-) A direction to the leftbull (-) As an exponent to mean divide by
All positive numbers produce correct results and graphs
All negative numbers produce incorrect results and graphs
A direction of three units to the left (-3) is placed in the equation as an exponent
A direction in space (3 units toward the left) changes a whole number to a fraction by using the dash (-) as an exponent rather than a direction in space
This is illogical and incorrect BS math should have performed this operation as
Unfortunately the answers to problems can come out correct because data points of a real problem fit the curve
However the answers will not be due to logical reasoning
Because the equation produced a curve that real problem data points follow it has been falsely believed that the equations and graphs are correct
1
2
x
Incorrect BK graph for (12)x
(12)x
4 00625 correct3 0125 correct2 025 correct1 1 correct0 1 correct-1 2 incorrect-2 4 incorrect-3 8 incorrect-4 16 incorrect
negative side of graph is incorrectpositive side of graph is correct
00625012502505124
8
16
-5 -4 -3 -2 -1 0 1 2 3 4 5P
osit
ive
negative positive
BS Math for (12)x
correctincorrect
Incorrect BS math and graph for
All graphs are symmetrical with SM
Two BS Math Equations that Produce Incorrect Answers
2 0x x starting equation 2
2
x x x x
x x
Add lsquoxrsquo to both sides
2(1) 1 let lsquoxrsquo = 1
2 1 This is an incorrect answer 2(0)-(0)=0 The only answer that BK math can give is x=0
1
a b starting equation ( )( ) ( )( )a a a b multiply both sides by ( )a
( )( ) ( )( ) ( )( ) ( )( )a a b b a b b b subtract ( )( )b b from both sides
( )( ) ( )( )a b a b a b b factor ( )a b from both sides
( )a b b divide both sides by ( )a b
( )b b b replace ( )a with ( )b 2b b simplify 2(1) (1) Let ( )b = 1
2 1 Again BS math yields and incorrect answer 1
This incorrect result is a problem with the distributive law SM explains how this incorrect answer is obtained and the problem with the distributive law that allows the error
Page 148 ndash149 Barry Mazur ndash Imagining Numbers
3
2
31
Do the computation carefully on paper using the rules we agreed to Then ponder your answer which should be something of a surprise to you donrsquot stop there Think of what your answer might possibly mean or might imply
Solve this equation using the distributive law
JK definitely a surprise the answer wasincorrect and illogical
Gerhard Gade University Professor doing math at Harvard University
Here is the BS math answer that Barry Mazur calculated
2
31
2
31
2
31
1 3 1 3 1 3 3 3 2 2 3 1 3
2 2 2 2 4 4 4 4 4 4 2 2
dash
1 3 1 3 1 31
2 2 2 2 4 4
3 3
4 4
When the distributive law is used with the BS Rule-of-Signs and one of the signs is a (-) dash sign the middle two terms always cancel each other This leads to the incorrect answer of (-1) in the above equation
Barry Mazur did not go back and check his answer with the original equation If he had he would have known that the answer was incorrect
3
1 31
2
check answer by setting equal
3
331 3
12
1 31
2
Take the Cube root of both sides
1 3 2 multiply both sides by 2
3 3 subtract 1 from both sides
2 2
3 3 square both sides
3 9 This is an incorrect answer A simple equation check could have shown that the distributive law using BS math provides an incorrect answer
1
BS Illogical and Incorrect Numbers Raised to Powers
nC D n DC
642 DDD
In BS math only a positive number can have an answer if ldquonrdquo is an even number (246---)
All negative numbers have no solutions
642 DDD
With SM just add a direction arrow and any number gives the correct answer
or n nC D C D
3 3( 3 8 8 or 8 2 or 2 )n D C
( 2 16 C= 16 or 16 4 or 4 )n D
Since a negative number has no square root answer in BS math BS mathematicians invented imaginary numbers
i2=-1
24414 iiC n
These equations are easily solved without the use of imaginary numbers with SM math
If 4 then 4 2x
If 4 then 4 2x
The 360-degree circle is a 4400 years old extremely outdated system based on a year of 12 months of 30 days each that is 360 days
SM will use a more logical system that will have a circle of 6283185307o
d
( )
(deg )
d r radians
d r rees
the arc distance around
the circumference of a circle
d
radiusr
2 2
2 2
d r
dv rd t t
v dda rd t t t
1
d
r
( )av tv
The average angular velocity of a body is the rate of its angular displacement
( )0
liminstt
vt
The instantaneous angular velocity is the limit of the ratio
t
as t approaches zero
( )d
av
d
tv
The average circumference velocity of the body is the rate of its distance traveled d divided by the time t
dvv
r
1 dd vdrv
t t t r r
The equation of the relationship between the angular velocity and the velocity of a point on that body
( )o
av
v v v
t ta
The average angular acceleration is the change in the angular velocity
( )0
liminstt
v
ta
The instantaneous angular acceleration is the
limit of the ration of v
t
as t approaches
zero
daa
r
1
1
d
d s d
vva
t r tv a
r t r
The equation of the relationship between the angular acceleration and the acceleration of a point on the body
21
2v a t
If a body with an initial angular velocity ov has a constant
angular acceleration a it will turn through an angle in a time t
1
Straight Line Symbol Units Rotational Symbol Units Distance d m Angular
Distance radians or degrees
Velocity dt
v m
s Angular Velocity
V
t
radians or degrees
s
Acceleration vat
2ms
Angular Acceleration
VA
t
2
radians or degrees
s
Mass M Kg Mass distribution
dM 2Kgm
Force actionA Ma
2
Kgm
sNewton
Torque
dT M A 2
2
Kgm
s
Momentum Mv Kgm
s
Angular Momentum
dM V 2Kgm
s
Work (Force)(distance)
Ad 2
Kgmm
sNm Joule
Work T
Using AS arrow math to solve and direction in space problems using the equation 1 2
2 2 2 2 2 2( ) amp and ( ) amp 3
4
2 2 2
2 2
( ) amp
(5 ) 4 3
(5 ) 16 9
(5 ) 25
(5 ) (5 )
2 2(5 ) 4 3
(5 ) 16 9
(5 ) 25
(5 ) (5 )
In SM math Pythagorasrsquo theorem provides both magnitude and direction
In BS math Pythagorasrsquo theorem provides magnitude only
time set equal to zero
23
( ) ( 0 )10 1
x t xx
3( ) ( 0)
10 1W t f
In BS math this is an equation for a traveling wave
must use a square term in order to make the curve symmetric
NO SQUARE TERM IS NECESSARY
In SM math when t is set to ldquo0rdquo
This is a major fault in BS math
x f(x)
-4 0019
-3 0033
-2 0073
-1 0273
0 3000
1 0273
2 0073
3 0033
4 0019 -4 -3 -2 -1 0 1 2 3 400
05
10
15
20
25
30
35
BS wave (must use square term)
snap shot of wave at x = 0
Am
plit
ude
lt to gt lt to gt f(ltgt)
lt -4 0073
lt -3 0097
lt -2 0143
lt -1 0273
origin 0 3000
gt 1 0273
gt 2 0143
gt 3 0097
gt 4 0073lt0073 lt0097 lt0143 lt0273
3000
0273gt 0143gt 0097gt 0073gt
Wave moving
snap shot of wave
Am
plit
ude
lt4 lt3 lt2 lt1 0 1gt 2gt 3gt 4gt
3
2
1
In SM NO squared terms are needed
SPRING
An Action is needed to stretch or compress a spring to a starting point where it can execute cycle motion
tsA (ts to start) is the work required to move a mass attached to a spring a distance (d1) from
its initial rest position RAK 0 the spring constant is the stiffness of the spring
o A is attraction o R is repulsion o
ARRA
00 is for the back and forth motion through the zero initial rest position A
stretched spring goes from ldquoArdquo to ldquo0rdquo to ldquoRrdquo to ldquoArdquo to ldquo0rdquo to ldquoRrdquo d1 is the distance a mass is moved from its initial rest position
0
0 1ts A RA K d
Once ( )mad
is performed to move the mass to position d1 (against the restraint of the spring) a of the ( )mad
is transferred into the spring as potential attractionrepulsion (PAR) There is always
more ( )mad
performed than is stored as PAR When the spring is released from the restraining Ats the PAR starts the oscillation Conservation of ( )mad
requires that the PAR and the motion ( )mad
( )( )madM be a constant
( )amp ( ) constantmadPAR M 02
0 0
1
2A R A RPAR K d
02
( ) 0 0
1( )
2mad A R A RM m v
0 0 02 2 2
0 0 0 1
2 2 21
2 2 21
1 1 1amp
2 2 2
1 1 1amp
2 2 2
amp
A R A R A RK d m v K d
Kd mv Kd
Kd mv Kd
0
1d is frac12 the height from the base to the peak of a cycle
An experiment of moving paper under a pendulum through a small angle filled with sand that can escape through a small hole in the bottom shows that a pendulum cycling back and forth follows a curve that is mathematically analogous to a lsquosinrsquo or lsquocosrsquo
The projection of the curved area onto the straight line exactly duplicate the back and forth motion of CM This allows for the math of lsquosinrsquo or lsquocosrsquo to be used
SM will use lsquosinrdquo math It could use lsquocosrsquo or a combination of both lsquosinrsquo and lsquocosrsquo Using just the lsquosinrsquo math makes the understanding and math easier
0 sin 0 0SM 0 sin157 1SM The diagram at the left shows a spring being stretched from its (zero rest) position-2 and the PAR being released at position-1 There are 4 sections (1 to 2 2 to 3 3 to 4 and 4 to5) that alternate between maximum acceleration and zero velocity to max velocity and zero acceleration
At position-1 there is maximum acceleration and zero velocity o The lsquosinrsquo values from 0o to 157o will be used for velocity (values from 0 to 1) o The lsquosinrsquo values from 157o to 0o will be used for acceleration (values from 1 to 0)
At position-2 there is maximum velocity and zero acceleration Acceleration changes to deceleration
o The lsquosinrsquo values from 157o to 0o will be used for deceleration (values from 1 to 0) o The lsquosinrsquo values from 0o to 157o will be used for acceleration (values from 0 to 1)
At position-3 there is maximum acceleration and zero velocity o The lsquosinrsquo values from 0o to 157o will be used for velocity (values from 0 to 1) o The lsquosinrsquo values from 157o to 0o will be used for acceleration (values from 1 to 0)
At position-4 there is maximum velocity and zero acceleration Acceleration changes to deceleration
o The lsquosinrsquo values from 157o to 0o will be used for deceleration (values from 1 to 0) o The lsquosinrsquo values from 0o to 157o will be used for acceleration (values from 0 to 1)
At position-5 the cycle starts again
2
)539(
cT
MK
spring constant (K) when mass (M) and cycle-time (T) are known
)(
))(89)((
111 md
uKgM
d
MG
d
FK sts
spring constant (K) when mass (M) and amplitude ( d1) are known
2 21( )
Kv d d
m
velocity (v) at any point during the cycle
velocity
dforM
Kdv
max
012 velocity (v2) maximum (when d = 0 at position-2 and 4)
acceleration (a1) maximum when spring constant (K) mass (M) and amplitude (d1) are known
acceleration (a1) maximum when amplitude (d1) and cycle-time (Tc) are known
acceleration (a) at any point during the cycle-time (T)
1
151
286
v
dTc
cycle-time (Tc) when amplitude (d1) and velocity (v1) are known
K
MTc 28651
cycle-time (Tc) when mass (M) and spring constant (K) are known
fTc
1
cycle-time (Tc) when the frequency is known
cTf
1
frequency (f) when the cycle-time (Tc) is known
of cycles
elapsed timef
frequency (f) when the number of cycles in an elapsed time is known
d1 amplitude (distance from rest to start position)
Also frac12 the distance from the base to the peak
3131 oror dM
Ka
251
3131 539
c
oror T
da
2539
cT
da
In December of 2006 I discovered the mathematical error in Einsteinrsquos Special Relativity Einstein used BS math and that math only works in an imaginary world
It cannot describe real symmetrical space
Einsteinrsquos Special Relativity should be abandoned and a theory of real space using SM should be used to replace it
Numerous articles on the Internet show Einsteinrsquos theory to be incorrect but no one in academia is willing to publish the information
Figure 1-10 A moving particle observed from two frames of reference in uniform translation
Einsteinrsquos Math Error in Special Relativity Transformation of Velocity (From Eisenberg- Fundamentals of Modern Physics pages 27-29) Consider the particle shown in figure (1-10) moving with velocity v as
seen in a frame of reference O
dx dy dz
Vx Vy Vzdt dt dt
dx dy dz
V x V y V zdy dt dt
We would like to evaluate the velocity v of the particle as seen in the frame of reference 0 which is itself moving relative to 0 with velocity v Measured in the 0 frame the velocity vector of the particle has com ponents
The same velocity vector as measured in the 0 frame has components
2
2
2
2
2
1
1
x vtx
v
cy y
z z
vxt
ctv
c
Now from equations (1-13) we know that the relation between the primed and the unprimed coordinates and times is
Take the differential of these equations remembering that v is a constant This gives
2
2
2
2
2
1
1
dx vdtdx
v
cdy dy
dz dz
vxdt
cdtv
c
2
2
2 2 2 2 2
2
2
2
11 ( )
1( ) 1
1
1
1
dx vdt
v dxdx vdt v
dx dx vdtc dt dtvdx vdx vdx v dxdt dt dt dt
dt dtc c c c
VxV x
Vx
VxV x
vv
vv
c
x
v
c
cV
( )
coordinate of particle relative to moving primed system
coordinate of particle relative to stationary system
velocity of primed coordina
V x
Vx
te system relative to stationaryv sys
tem
Page 28 Fundamentals of Modern Physics
(1-16)
According to Einstein this equation tells you how to transform the observed velocity from one frame of reference to another frame of reference First we note that as Vc and vc approachzero the equation approaches those which would be derived from the Galilean transformation
2 2
( ) 1 01 1 ( )
Vx v Vx vV
VxV x
Vx v
v cVxx
v
c
v
cc
Another property of this equation is that it is impossible to choose V and v such that Vrsquo the magnitude of the velocity which is seen in the new frame is greater than c
1
2
09 ( 09 ) 18009876543
( 09 ) 1811 (09 )
c c cV c c
cc
c
Example As seen by O particle 1 has velocity 09c in the direction of positive x and particle 2 has velocity 09c in the negative x direction
Einstein equation however predicts To evaluate the velocity of particle 1 with respect to particle 2 we transform from the O frame to the Orsquo frame moving in the negative x direction with velocity v = -09c using equation 1-16 We obtain
JK LOGIC will observe the particles separating at 18c
A confusing INCORRECT and ILLOGICAL answer
left is a dir rightsub of a
1
2
1
09 ( 09 ) 18009876543
181( 09 )1 ( 09 )
The multiplication of dir left time a dir right is ONLY a dir left
09
dir
direction leftdirection rightsubtraction
c c cV c c
cc
c
V
dir
22
22
09 18 18009876543
181( 081)1(09)(09)
1
left rightsub
sub
c c c cc c
ccc
c
Analyzing Einsteinrsquos BS math Equation
Using SM math a logical answer can be obtained
Cannot multiply a subtraction operator times a direction in space
Can only subtract and change its direction
Cannot multiply a direction to the left times a direction to the right
Cannot multiply opposite directions in space
Can only solve for the resultant
rightleft
subtraction
1
subtraction
2
sub of a dir to the left is a dir to the right
1 resultant 0 sub 2
2sub 2
2
09 09
09 amp 091
09 09 18 1
(0)1
(09) amp (09)1
c cV
c cc
c c cV
cc
cc
8
181
cc
SM math predicts a logical answer using logic
21st century scientists state in many articles that Einsteins General Relativity and Quantum Mechanics are incompatible
They are not only incompatible they are both mathematically incorrect
They both use illogical BS math
Any equations in math that use imaginary numbers should be abandoned
Imaginary numbers are only necessary in the current BS math system
This means that Schroumldingers equations should be abandoned
Any equation in quantum mechanics that uses imaginary number should be abandoned
Prof Kanarev has developed logical physics and chemistrytheories without the use of Quantum mechanics or relativity
A partial list of his theories coverbull Mechanics ndash replaces Newtonbull Photon - bull Electrons ndash Protons- Neutronsbull Spectroscopybull Nucleusbull Atoms amp Moleculesbull Electrodynamicsbull Heatbull Curvature of Space ndash replaces Schwarzcchildrsquos equation
Kanarevrsquos Hypothesis
bull Electron align linearly with proton They do NOT spin or form clouds around the nucleus
bull Neutrons are in the center of atom Proton are outside of Neutrons
bull The wavelength of all particles is the experimental radius
bull Double slit experiment explained correctly
bull Heisenbergrsquos inequality explained correctly
bull is the radius of approach of the magnetic lines around the torus of the electron
( 1 137)
bull Why water expands when it freezes
bull Electron clusters form lightning
rbull The meaning of Plancks constant is explained
22 2 kg m
h m f mr f consts
2210254 42m r kgm bull If the mr-constant describes a ring the equations
describe experiments bull If the mr-constant describes a wave the equations do
NOT describe experiments
Kanarevrsquos 5 Equations of Mechanics
Kanarevrsquos 1st Equation Acceleration ampmv maMma R R
Kanarevrsquos 2nd Equation Uniform motion amp mv maMma R R
maM mvR R
Kanarevrsquos 3rd Equation
Kanarevrsquos 4th Equation Equal and opposite reaction
1 21 21 2
12
a m
m a m ama
Kanarevrsquos 5th Equation 1 2 3 Mi M M M Mna R R R R R
The sum of the Actions working on a moving body is never equal to zero
Kanarevrsquos Photon model is a rolling ring
His equations accurately describe known experiments with photonsbull Electromagnetic radiation should be called photon radiationbull Diffraction and interference are the same phenomenabull Correctly explains the double slit experiment without quantum equationsbull Correctly explains the meaning of Heisenbergrsquos inequality
e
2 (2)(3142)(2817 15 ) 10007297352568
R 2426 12 137mag
xp
r m
m
Charge
Magnetic action
Proton
The intensity of the magnetic action near the geometrical center of the proton ring is so great that we theorize that it is the strong nuclear force
(9383 8)(1602 19)8476 14
4 (4)(3142)(1411 26)p
pp
EB T
M
(9395 8)(1602 19)1035 15
4 (4)(3142)(0966 26)N
NN
EB T
M
Neutron
1a Hydrogenavi
Symmetry-Math The subtraction of a direction is equal to the opposite direction The answers are correct using correct logic
opposite direction direction
SM
(negative)(negative) = (positive)
( ) ( ) ( )subtraction
BS
There is NO multiplication of a subtraction operator by a direction in space
There is just the subtraction of a direction in space
If instead of labeling the left side of a coordinate system as a negative (-) the same as a subtraction operator we label it with an arrow ( ) to represent the direction then a subtraction from that arrow direction will be in the opposite direction ( )
( ) mutiplied by a ( ) ( )
negative Positive negative
subtraction addition subtraction
dash cross dash
left right left
A subtraction operator multiplied by an addition operator is equal to a subtraction operator This is illogical
BS Math
A negative multiplied by a positive is equal to a negative This is illogical
A direction to the left multiplied by a direction to the right is a direction to the left This is illogical and Einsteins math error in special relativity
A dash times a cross is equal to a dash This is illogical
A vector moving to the left multiplied by a vector moving to the right is equal to ONLY to a vector moving to the left that is eight orders of magnitude faster thanThe speed of light
opposite direction subtraction direction
SM ( ) ( )
BS (negative)(positive) = (negative)
Symmetry Math The subtraction of a direction is equal to the opposite direction
There is no multiplication of a subtractionoperator times a direction in space
( ) multiplied by a ( ) ( )
positive positive positiveaddition addition additioncross cross cross
direction direction direction
What is the meaning of multiplying addition operators Addition operators are not multiplication operators
BS Math
() multiplied by a direction ( )=
number
() multiplied by a direction ( )=
number
A number times a direction maintains the same direction
( ) ()
dirdiradd
( ) ()
dirdiradd
The addition of a direction is in the same direction There is no multiplication by addition operators
Symmetry Math
Future math books need to eliminate the use of the cross sign (+) to represent something labeled a positive direction in space and the dash sign (-) to represent something labeled a negative direction in space Space does not have positive and negative directions
Symbols that are logical and have no illogical representation should be adopted SM uses either arrows or symbols
Symmetry-Math Number Line
Symmetry-Math Rules
The dash sign (-) will have only one use subtraction
The cross sign (+) will have only one use addition
Directions have an arrow and a number
All observers see the same direction and magnitude
bull There is no multiplication of arrows
amp ampamp ampampamp
( 3 ) ( 5 ) ( 6 ) ( 65 ) (17256 ) ( 32993 )
dir dirdir dirdirdir
Negative and Positive Directions in Space
If I asked you to point to a negative direction in space which way would you point
Hopefully you will realize that there is no such thing as a negative direction in space
If I asked you to point to a positive direction in space which way would you point
Again hopefully you will realize that there is no such thing as a positive direction in space
Objects on the right side of the x-axis are positive (+)(+) = (+)
Objects on the left side of the x-axis are negative (-)(-) = (+) negative math is different from positive math This is illogical
Objects on the top of the y-axis are positive
Objects on the bottom of the y-axis are negative
Objects in the front of the z-axis are positive
Objects in the back of the z-axis are negative
Again BS math of the positive direction (x right y up and z front) is different from the BS math of the negative direction (x left y down and z back)
BS math is Broken-Symmetry
Mirror image is broken symmetry in the x and z axis If the left amp right-axis are
reversed symmetry is broken Obs-1 math is different from obs-2
If the front amp back-axis is reversed symmetry is broken Obs-1 math is different from obs-2
If the top amp bottom-axis are
reversed symmetry is not broken Obs-1 math is the same as obs-2
Math answers cannot depend on which side of a number line an observer sit
Absolute values for Displacement
Even for a simple displacement of an object BS math invented a definition and absolute values to obtain something labeled a positive number answer
I----------I----------I----------I----------I----------I----------I----------I----------I
-4 -3 -2 -1 0 +1 +2 +3 +4 a gtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtb cltltltltltltltltltltltltltltltltltltltltltltltltltltltb
If an object starts at ( 3)ia X and moves to ( X 3) and the to ( 1)b fb c X
BS Math provides a usable answer without a definition or absolute values
However If an object starts at
( 1) and moves to ( X 3) and the to ( 3)i b fa X b c X
Then absolute values and a definition must be used
---------I----------I----------I----------I----------I----------I----------I---------- -4 -3 -2 -1 0 +1 +2 +3 +4
a gtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtb cltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltb
BS Math makes up a definition that all displacements are positive Since this example provides an answer with something labeled a negative the positive definition must be applied and (-2) is changed to (+2) A definition is needed to arrive at a useable answer This is Illogical
In SM absolute values and a meaningless definition of ldquopositiverdquo are removed In SM objects and directions in space are defined by the direction and magnitude of the resultant of arrows
With SM you get total distance traveled by the object and the final direction of the displacement
22222 2 bababbaababababa If w e su b stitu te p o s itiv e n u m b e rs in to th e eq u a tio n th e an sw er w ill b e co rrec t If (a = 5 ) an d (b = 3 )
2
2
8
5 3 5 3 5 3 5 5 5 3 3 5 3 3 25 15 15 9 64
8 8 8 8 8 8 8 8 8 8 64
W h en tw o p o sitiv e n u m b ers a re a d d ed an d sq u a re d th e B S m a th d is tr ib u tiv e law p ro v id es a co rre c t an sw er 1
Illogical and Incorrect BS math for the distributive law
222 2 bababbabbaaabababa B y d e fin itio n in B S m a th n o n u m b er sq u a red c a n ev e r b e n eg a tiv e T h e re fo re n o sq u a red n u m b er c an ev e r b e a v e c to r m o v in g in th e d a sh ( -) o r n eg a tiv e d ire c tio n if it is sq u a red In B S m a th an d ad d itio n o p e ra to r m u ltip lied b y a su b trac tio n o p e ra to r is A L W A Y S eq u a l to a su b tra c tio n o p e ra to r If th is h ad n o t b e en u sed fo r th e la s t fo u r h u n d red y e a rs y o u w o u ld d ie lau g h in g a t th e u tte r ab su rd ity o f th e la s t s ta tem e n t T h is is illo g ica l an d v io la te s sy m m etry
If a = 5 an d b = 3 42222352
22
4930253)3)(5(253)5)(3()3)(5(5353535 22222 A s lo n g a s ldquo ardquo is g re a te r th an ldquob rdquo th e B S m a th d is tr ib u tiv e law p ro v id es co rrec t an sw ers 1
B S m a th b re a k s d o w n w h e n ldquo b rdquo is g rea ter th a n ldquo a rdquo
If a = 3 an d b = 5
2 2
2
2
3 5 2 4
2 ( 2) ( 2) 4
2 ( )( 2) ( )( 2) 4
A n y co m b in a tio n ex cep t th e m id d le o n e is illo g ica l an d in co rre c t B y d e fin itio n th e an sw er can n o t b e -4 in B S m a th B S m a th is in co rrec t
4253095)5)(3(235)3)(5()5)(3(3535353 22222 T h is is w h ere th e d is tr ib u tiv e la w p ro v id es in co rrec t an sw ers I t p ro d u ces an an sw er o f + 4 b e cau se b y d e fin iti o n n o n u m b er sq u a red in B S m a th c an b e a n eg a tiv e n u m b er U sin g in co rre c t B S m a th th e d is tr ib u tiv e law w ill g iv e an in co rrec t an sw ers o f + 4 T h is is in co rrec t an d illo g ica l 1
Since there can be no negative direction from a squared term in BS math BS math should be modified or abandoned
The BS distributive law should be modified or abandoned
The numbers +1+2+3+4 and ndash1-2-3-4hellip should be abandoned
BS math gives incorrect answers because their Rule-of-Signs allows multiplying an arrow going in one direction by an arrow going in another direction
Clearly this is not logical
Arrows should be used that are specific for directions
W h at d o es it m e an to m u ltip ly ( 4) b y (4 ) A s w e k n o w 1
m u ltip lica tio n is ju s t ad d itio n W h a t w o u ld ( 4) tim e (4 ) b e 2
In BS math the answer is by definition -16 unless you square -4 and then it is +16
In SM you cannot multiply arrows going in opposite directions
The middle two terms are not logical You cannot multiply opposite directions
This is a MAJOR error in BS math They multiply a dash [(-) a direction to the left] by a cross [(+) a direction to the right] And then ldquoby definitionrdquo label the answer dash (-) This is illogical and produces incorrect answers
2
2 2
This is Einsteinrsquos math error
in Special relativity
Magnitude 4 Direction 1 Magnitude 3 Direction 2 Magnitude 5 Direction 3 Relative to an observer at 0 4
Magnitude 4 Direction 1 Magnitude 3 Direction 2 Magnitude 5 Direction 3 Relative to an observer at 0 4
Magnitude and Direction (MD) An arrow has both a magnitude andA direction relative to a 2nd objectobserver
bull Magnitude and velocity (MV) relative to a 2nd objectobserver
bull Magnitude And acceleration (MA) relative to a 2nd objectobserver2
v da
t t
Introducing a new symbol the ldquoamprdquo sign The ldquoamprdquo sign means resultant ADD all the Arrows for each direction and then SUBTRACT to find the final resultant
1 1amp 1 2 2amp 2 4 If =
1 amp1 2 2 amp2 4 If =
amp 5 amp 3 2 If gt
amp 3 amp 5= 2 If gt
amp 0 3 amp 3=0 if =
only the resultant equals zero
the vectors do not - they are still there
Does not exist There is no multiplication of arrows
There are two arrows One has a magnitude of 3 One has a magnitude of 5 The maximum these two arrows can be is 8
Only the resultant of arrows exist
An SM arrow has a direction and a magnitude
Arrows and magnitudes are still there even if the resultant is zero
There are no negative SM Arrows
There are no positive SM Arrows
There are no ZERO SM Arrows A zero would negate the definition of an SM Arrow something with no magnitude and no direction cannot be the definition of an arrow that is defined as having magnitude and direction
bull A resultant is the addition and subtraction of the differences in magnitude and direction arrows
A B and C are the arrows (a) A amp B exist(b) A amp B amp C exist(c) A-B Do Not Exist in Sm there are no negatives or positives(d) A+B-C Do Not Exist in SM there are no negatives or positives
There are only resultants when there are multiple arrows
SM Graphs versus BS Graphs
Numbers Raised to Powers
When numbers are raised to powers the negative sides of all BS math graphs are ILLOGICAL and INCORRECT
Incorrect BS Math for 2x
2x
4 16 correct3 8 correct2 4 correct1 2 correct0 1 correct-1 005 incorrect-2 025 incorrect-3 0125 incorrect-4 00625 incorrect
negative side of graph is incorrectpositive side of graph is correct
16
8
42105025012500625
-4 -3 -2 -1 0 1 2 3 4
po
sit
ive
negative positive
2x BS Math
correctincorrect
In BS math the dash sign (-) used in this example represents three different operations They arebull (-) Subtractionbull (-) A direction to the leftbull (-) As an exponent to mean divide by
All positive numbers produce correct results and graphs
All negative numbers produce incorrect results and graphs
A direction of three units to the left (-3) is placed in the equation as an exponent
A direction in space (3 units toward the left) changes a whole number to a fraction by using the dash (-) as an exponent rather than a direction in space
This is illogical and incorrect BS math should have performed this operation as
Unfortunately the answers to problems can come out correct because data points of a real problem fit the curve
However the answers will not be due to logical reasoning
Because the equation produced a curve that real problem data points follow it has been falsely believed that the equations and graphs are correct
1
2
x
Incorrect BK graph for (12)x
(12)x
4 00625 correct3 0125 correct2 025 correct1 1 correct0 1 correct-1 2 incorrect-2 4 incorrect-3 8 incorrect-4 16 incorrect
negative side of graph is incorrectpositive side of graph is correct
00625012502505124
8
16
-5 -4 -3 -2 -1 0 1 2 3 4 5P
osit
ive
negative positive
BS Math for (12)x
correctincorrect
Incorrect BS math and graph for
All graphs are symmetrical with SM
Two BS Math Equations that Produce Incorrect Answers
2 0x x starting equation 2
2
x x x x
x x
Add lsquoxrsquo to both sides
2(1) 1 let lsquoxrsquo = 1
2 1 This is an incorrect answer 2(0)-(0)=0 The only answer that BK math can give is x=0
1
a b starting equation ( )( ) ( )( )a a a b multiply both sides by ( )a
( )( ) ( )( ) ( )( ) ( )( )a a b b a b b b subtract ( )( )b b from both sides
( )( ) ( )( )a b a b a b b factor ( )a b from both sides
( )a b b divide both sides by ( )a b
( )b b b replace ( )a with ( )b 2b b simplify 2(1) (1) Let ( )b = 1
2 1 Again BS math yields and incorrect answer 1
This incorrect result is a problem with the distributive law SM explains how this incorrect answer is obtained and the problem with the distributive law that allows the error
Page 148 ndash149 Barry Mazur ndash Imagining Numbers
3
2
31
Do the computation carefully on paper using the rules we agreed to Then ponder your answer which should be something of a surprise to you donrsquot stop there Think of what your answer might possibly mean or might imply
Solve this equation using the distributive law
JK definitely a surprise the answer wasincorrect and illogical
Gerhard Gade University Professor doing math at Harvard University
Here is the BS math answer that Barry Mazur calculated
2
31
2
31
2
31
1 3 1 3 1 3 3 3 2 2 3 1 3
2 2 2 2 4 4 4 4 4 4 2 2
dash
1 3 1 3 1 31
2 2 2 2 4 4
3 3
4 4
When the distributive law is used with the BS Rule-of-Signs and one of the signs is a (-) dash sign the middle two terms always cancel each other This leads to the incorrect answer of (-1) in the above equation
Barry Mazur did not go back and check his answer with the original equation If he had he would have known that the answer was incorrect
3
1 31
2
check answer by setting equal
3
331 3
12
1 31
2
Take the Cube root of both sides
1 3 2 multiply both sides by 2
3 3 subtract 1 from both sides
2 2
3 3 square both sides
3 9 This is an incorrect answer A simple equation check could have shown that the distributive law using BS math provides an incorrect answer
1
BS Illogical and Incorrect Numbers Raised to Powers
nC D n DC
642 DDD
In BS math only a positive number can have an answer if ldquonrdquo is an even number (246---)
All negative numbers have no solutions
642 DDD
With SM just add a direction arrow and any number gives the correct answer
or n nC D C D
3 3( 3 8 8 or 8 2 or 2 )n D C
( 2 16 C= 16 or 16 4 or 4 )n D
Since a negative number has no square root answer in BS math BS mathematicians invented imaginary numbers
i2=-1
24414 iiC n
These equations are easily solved without the use of imaginary numbers with SM math
If 4 then 4 2x
If 4 then 4 2x
The 360-degree circle is a 4400 years old extremely outdated system based on a year of 12 months of 30 days each that is 360 days
SM will use a more logical system that will have a circle of 6283185307o
d
( )
(deg )
d r radians
d r rees
the arc distance around
the circumference of a circle
d
radiusr
2 2
2 2
d r
dv rd t t
v dda rd t t t
1
d
r
( )av tv
The average angular velocity of a body is the rate of its angular displacement
( )0
liminstt
vt
The instantaneous angular velocity is the limit of the ratio
t
as t approaches zero
( )d
av
d
tv
The average circumference velocity of the body is the rate of its distance traveled d divided by the time t
dvv
r
1 dd vdrv
t t t r r
The equation of the relationship between the angular velocity and the velocity of a point on that body
( )o
av
v v v
t ta
The average angular acceleration is the change in the angular velocity
( )0
liminstt
v
ta
The instantaneous angular acceleration is the
limit of the ration of v
t
as t approaches
zero
daa
r
1
1
d
d s d
vva
t r tv a
r t r
The equation of the relationship between the angular acceleration and the acceleration of a point on the body
21
2v a t
If a body with an initial angular velocity ov has a constant
angular acceleration a it will turn through an angle in a time t
1
Straight Line Symbol Units Rotational Symbol Units Distance d m Angular
Distance radians or degrees
Velocity dt
v m
s Angular Velocity
V
t
radians or degrees
s
Acceleration vat
2ms
Angular Acceleration
VA
t
2
radians or degrees
s
Mass M Kg Mass distribution
dM 2Kgm
Force actionA Ma
2
Kgm
sNewton
Torque
dT M A 2
2
Kgm
s
Momentum Mv Kgm
s
Angular Momentum
dM V 2Kgm
s
Work (Force)(distance)
Ad 2
Kgmm
sNm Joule
Work T
Using AS arrow math to solve and direction in space problems using the equation 1 2
2 2 2 2 2 2( ) amp and ( ) amp 3
4
2 2 2
2 2
( ) amp
(5 ) 4 3
(5 ) 16 9
(5 ) 25
(5 ) (5 )
2 2(5 ) 4 3
(5 ) 16 9
(5 ) 25
(5 ) (5 )
In SM math Pythagorasrsquo theorem provides both magnitude and direction
In BS math Pythagorasrsquo theorem provides magnitude only
time set equal to zero
23
( ) ( 0 )10 1
x t xx
3( ) ( 0)
10 1W t f
In BS math this is an equation for a traveling wave
must use a square term in order to make the curve symmetric
NO SQUARE TERM IS NECESSARY
In SM math when t is set to ldquo0rdquo
This is a major fault in BS math
x f(x)
-4 0019
-3 0033
-2 0073
-1 0273
0 3000
1 0273
2 0073
3 0033
4 0019 -4 -3 -2 -1 0 1 2 3 400
05
10
15
20
25
30
35
BS wave (must use square term)
snap shot of wave at x = 0
Am
plit
ude
lt to gt lt to gt f(ltgt)
lt -4 0073
lt -3 0097
lt -2 0143
lt -1 0273
origin 0 3000
gt 1 0273
gt 2 0143
gt 3 0097
gt 4 0073lt0073 lt0097 lt0143 lt0273
3000
0273gt 0143gt 0097gt 0073gt
Wave moving
snap shot of wave
Am
plit
ude
lt4 lt3 lt2 lt1 0 1gt 2gt 3gt 4gt
3
2
1
In SM NO squared terms are needed
SPRING
An Action is needed to stretch or compress a spring to a starting point where it can execute cycle motion
tsA (ts to start) is the work required to move a mass attached to a spring a distance (d1) from
its initial rest position RAK 0 the spring constant is the stiffness of the spring
o A is attraction o R is repulsion o
ARRA
00 is for the back and forth motion through the zero initial rest position A
stretched spring goes from ldquoArdquo to ldquo0rdquo to ldquoRrdquo to ldquoArdquo to ldquo0rdquo to ldquoRrdquo d1 is the distance a mass is moved from its initial rest position
0
0 1ts A RA K d
Once ( )mad
is performed to move the mass to position d1 (against the restraint of the spring) a of the ( )mad
is transferred into the spring as potential attractionrepulsion (PAR) There is always
more ( )mad
performed than is stored as PAR When the spring is released from the restraining Ats the PAR starts the oscillation Conservation of ( )mad
requires that the PAR and the motion ( )mad
( )( )madM be a constant
( )amp ( ) constantmadPAR M 02
0 0
1
2A R A RPAR K d
02
( ) 0 0
1( )
2mad A R A RM m v
0 0 02 2 2
0 0 0 1
2 2 21
2 2 21
1 1 1amp
2 2 2
1 1 1amp
2 2 2
amp
A R A R A RK d m v K d
Kd mv Kd
Kd mv Kd
0
1d is frac12 the height from the base to the peak of a cycle
An experiment of moving paper under a pendulum through a small angle filled with sand that can escape through a small hole in the bottom shows that a pendulum cycling back and forth follows a curve that is mathematically analogous to a lsquosinrsquo or lsquocosrsquo
The projection of the curved area onto the straight line exactly duplicate the back and forth motion of CM This allows for the math of lsquosinrsquo or lsquocosrsquo to be used
SM will use lsquosinrdquo math It could use lsquocosrsquo or a combination of both lsquosinrsquo and lsquocosrsquo Using just the lsquosinrsquo math makes the understanding and math easier
0 sin 0 0SM 0 sin157 1SM The diagram at the left shows a spring being stretched from its (zero rest) position-2 and the PAR being released at position-1 There are 4 sections (1 to 2 2 to 3 3 to 4 and 4 to5) that alternate between maximum acceleration and zero velocity to max velocity and zero acceleration
At position-1 there is maximum acceleration and zero velocity o The lsquosinrsquo values from 0o to 157o will be used for velocity (values from 0 to 1) o The lsquosinrsquo values from 157o to 0o will be used for acceleration (values from 1 to 0)
At position-2 there is maximum velocity and zero acceleration Acceleration changes to deceleration
o The lsquosinrsquo values from 157o to 0o will be used for deceleration (values from 1 to 0) o The lsquosinrsquo values from 0o to 157o will be used for acceleration (values from 0 to 1)
At position-3 there is maximum acceleration and zero velocity o The lsquosinrsquo values from 0o to 157o will be used for velocity (values from 0 to 1) o The lsquosinrsquo values from 157o to 0o will be used for acceleration (values from 1 to 0)
At position-4 there is maximum velocity and zero acceleration Acceleration changes to deceleration
o The lsquosinrsquo values from 157o to 0o will be used for deceleration (values from 1 to 0) o The lsquosinrsquo values from 0o to 157o will be used for acceleration (values from 0 to 1)
At position-5 the cycle starts again
2
)539(
cT
MK
spring constant (K) when mass (M) and cycle-time (T) are known
)(
))(89)((
111 md
uKgM
d
MG
d
FK sts
spring constant (K) when mass (M) and amplitude ( d1) are known
2 21( )
Kv d d
m
velocity (v) at any point during the cycle
velocity
dforM
Kdv
max
012 velocity (v2) maximum (when d = 0 at position-2 and 4)
acceleration (a1) maximum when spring constant (K) mass (M) and amplitude (d1) are known
acceleration (a1) maximum when amplitude (d1) and cycle-time (Tc) are known
acceleration (a) at any point during the cycle-time (T)
1
151
286
v
dTc
cycle-time (Tc) when amplitude (d1) and velocity (v1) are known
K
MTc 28651
cycle-time (Tc) when mass (M) and spring constant (K) are known
fTc
1
cycle-time (Tc) when the frequency is known
cTf
1
frequency (f) when the cycle-time (Tc) is known
of cycles
elapsed timef
frequency (f) when the number of cycles in an elapsed time is known
d1 amplitude (distance from rest to start position)
Also frac12 the distance from the base to the peak
3131 oror dM
Ka
251
3131 539
c
oror T
da
2539
cT
da
In December of 2006 I discovered the mathematical error in Einsteinrsquos Special Relativity Einstein used BS math and that math only works in an imaginary world
It cannot describe real symmetrical space
Einsteinrsquos Special Relativity should be abandoned and a theory of real space using SM should be used to replace it
Numerous articles on the Internet show Einsteinrsquos theory to be incorrect but no one in academia is willing to publish the information
Figure 1-10 A moving particle observed from two frames of reference in uniform translation
Einsteinrsquos Math Error in Special Relativity Transformation of Velocity (From Eisenberg- Fundamentals of Modern Physics pages 27-29) Consider the particle shown in figure (1-10) moving with velocity v as
seen in a frame of reference O
dx dy dz
Vx Vy Vzdt dt dt
dx dy dz
V x V y V zdy dt dt
We would like to evaluate the velocity v of the particle as seen in the frame of reference 0 which is itself moving relative to 0 with velocity v Measured in the 0 frame the velocity vector of the particle has com ponents
The same velocity vector as measured in the 0 frame has components
2
2
2
2
2
1
1
x vtx
v
cy y
z z
vxt
ctv
c
Now from equations (1-13) we know that the relation between the primed and the unprimed coordinates and times is
Take the differential of these equations remembering that v is a constant This gives
2
2
2
2
2
1
1
dx vdtdx
v
cdy dy
dz dz
vxdt
cdtv
c
2
2
2 2 2 2 2
2
2
2
11 ( )
1( ) 1
1
1
1
dx vdt
v dxdx vdt v
dx dx vdtc dt dtvdx vdx vdx v dxdt dt dt dt
dt dtc c c c
VxV x
Vx
VxV x
vv
vv
c
x
v
c
cV
( )
coordinate of particle relative to moving primed system
coordinate of particle relative to stationary system
velocity of primed coordina
V x
Vx
te system relative to stationaryv sys
tem
Page 28 Fundamentals of Modern Physics
(1-16)
According to Einstein this equation tells you how to transform the observed velocity from one frame of reference to another frame of reference First we note that as Vc and vc approachzero the equation approaches those which would be derived from the Galilean transformation
2 2
( ) 1 01 1 ( )
Vx v Vx vV
VxV x
Vx v
v cVxx
v
c
v
cc
Another property of this equation is that it is impossible to choose V and v such that Vrsquo the magnitude of the velocity which is seen in the new frame is greater than c
1
2
09 ( 09 ) 18009876543
( 09 ) 1811 (09 )
c c cV c c
cc
c
Example As seen by O particle 1 has velocity 09c in the direction of positive x and particle 2 has velocity 09c in the negative x direction
Einstein equation however predicts To evaluate the velocity of particle 1 with respect to particle 2 we transform from the O frame to the Orsquo frame moving in the negative x direction with velocity v = -09c using equation 1-16 We obtain
JK LOGIC will observe the particles separating at 18c
A confusing INCORRECT and ILLOGICAL answer
left is a dir rightsub of a
1
2
1
09 ( 09 ) 18009876543
181( 09 )1 ( 09 )
The multiplication of dir left time a dir right is ONLY a dir left
09
dir
direction leftdirection rightsubtraction
c c cV c c
cc
c
V
dir
22
22
09 18 18009876543
181( 081)1(09)(09)
1
left rightsub
sub
c c c cc c
ccc
c
Analyzing Einsteinrsquos BS math Equation
Using SM math a logical answer can be obtained
Cannot multiply a subtraction operator times a direction in space
Can only subtract and change its direction
Cannot multiply a direction to the left times a direction to the right
Cannot multiply opposite directions in space
Can only solve for the resultant
rightleft
subtraction
1
subtraction
2
sub of a dir to the left is a dir to the right
1 resultant 0 sub 2
2sub 2
2
09 09
09 amp 091
09 09 18 1
(0)1
(09) amp (09)1
c cV
c cc
c c cV
cc
cc
8
181
cc
SM math predicts a logical answer using logic
21st century scientists state in many articles that Einsteins General Relativity and Quantum Mechanics are incompatible
They are not only incompatible they are both mathematically incorrect
They both use illogical BS math
Any equations in math that use imaginary numbers should be abandoned
Imaginary numbers are only necessary in the current BS math system
This means that Schroumldingers equations should be abandoned
Any equation in quantum mechanics that uses imaginary number should be abandoned
Prof Kanarev has developed logical physics and chemistrytheories without the use of Quantum mechanics or relativity
A partial list of his theories coverbull Mechanics ndash replaces Newtonbull Photon - bull Electrons ndash Protons- Neutronsbull Spectroscopybull Nucleusbull Atoms amp Moleculesbull Electrodynamicsbull Heatbull Curvature of Space ndash replaces Schwarzcchildrsquos equation
Kanarevrsquos Hypothesis
bull Electron align linearly with proton They do NOT spin or form clouds around the nucleus
bull Neutrons are in the center of atom Proton are outside of Neutrons
bull The wavelength of all particles is the experimental radius
bull Double slit experiment explained correctly
bull Heisenbergrsquos inequality explained correctly
bull is the radius of approach of the magnetic lines around the torus of the electron
( 1 137)
bull Why water expands when it freezes
bull Electron clusters form lightning
rbull The meaning of Plancks constant is explained
22 2 kg m
h m f mr f consts
2210254 42m r kgm bull If the mr-constant describes a ring the equations
describe experiments bull If the mr-constant describes a wave the equations do
NOT describe experiments
Kanarevrsquos 5 Equations of Mechanics
Kanarevrsquos 1st Equation Acceleration ampmv maMma R R
Kanarevrsquos 2nd Equation Uniform motion amp mv maMma R R
maM mvR R
Kanarevrsquos 3rd Equation
Kanarevrsquos 4th Equation Equal and opposite reaction
1 21 21 2
12
a m
m a m ama
Kanarevrsquos 5th Equation 1 2 3 Mi M M M Mna R R R R R
The sum of the Actions working on a moving body is never equal to zero
Kanarevrsquos Photon model is a rolling ring
His equations accurately describe known experiments with photonsbull Electromagnetic radiation should be called photon radiationbull Diffraction and interference are the same phenomenabull Correctly explains the double slit experiment without quantum equationsbull Correctly explains the meaning of Heisenbergrsquos inequality
e
2 (2)(3142)(2817 15 ) 10007297352568
R 2426 12 137mag
xp
r m
m
Charge
Magnetic action
Proton
The intensity of the magnetic action near the geometrical center of the proton ring is so great that we theorize that it is the strong nuclear force
(9383 8)(1602 19)8476 14
4 (4)(3142)(1411 26)p
pp
EB T
M
(9395 8)(1602 19)1035 15
4 (4)(3142)(0966 26)N
NN
EB T
M
Neutron
1a Hydrogenavi
( ) mutiplied by a ( ) ( )
negative Positive negative
subtraction addition subtraction
dash cross dash
left right left
A subtraction operator multiplied by an addition operator is equal to a subtraction operator This is illogical
BS Math
A negative multiplied by a positive is equal to a negative This is illogical
A direction to the left multiplied by a direction to the right is a direction to the left This is illogical and Einsteins math error in special relativity
A dash times a cross is equal to a dash This is illogical
A vector moving to the left multiplied by a vector moving to the right is equal to ONLY to a vector moving to the left that is eight orders of magnitude faster thanThe speed of light
opposite direction subtraction direction
SM ( ) ( )
BS (negative)(positive) = (negative)
Symmetry Math The subtraction of a direction is equal to the opposite direction
There is no multiplication of a subtractionoperator times a direction in space
( ) multiplied by a ( ) ( )
positive positive positiveaddition addition additioncross cross cross
direction direction direction
What is the meaning of multiplying addition operators Addition operators are not multiplication operators
BS Math
() multiplied by a direction ( )=
number
() multiplied by a direction ( )=
number
A number times a direction maintains the same direction
( ) ()
dirdiradd
( ) ()
dirdiradd
The addition of a direction is in the same direction There is no multiplication by addition operators
Symmetry Math
Future math books need to eliminate the use of the cross sign (+) to represent something labeled a positive direction in space and the dash sign (-) to represent something labeled a negative direction in space Space does not have positive and negative directions
Symbols that are logical and have no illogical representation should be adopted SM uses either arrows or symbols
Symmetry-Math Number Line
Symmetry-Math Rules
The dash sign (-) will have only one use subtraction
The cross sign (+) will have only one use addition
Directions have an arrow and a number
All observers see the same direction and magnitude
bull There is no multiplication of arrows
amp ampamp ampampamp
( 3 ) ( 5 ) ( 6 ) ( 65 ) (17256 ) ( 32993 )
dir dirdir dirdirdir
Negative and Positive Directions in Space
If I asked you to point to a negative direction in space which way would you point
Hopefully you will realize that there is no such thing as a negative direction in space
If I asked you to point to a positive direction in space which way would you point
Again hopefully you will realize that there is no such thing as a positive direction in space
Objects on the right side of the x-axis are positive (+)(+) = (+)
Objects on the left side of the x-axis are negative (-)(-) = (+) negative math is different from positive math This is illogical
Objects on the top of the y-axis are positive
Objects on the bottom of the y-axis are negative
Objects in the front of the z-axis are positive
Objects in the back of the z-axis are negative
Again BS math of the positive direction (x right y up and z front) is different from the BS math of the negative direction (x left y down and z back)
BS math is Broken-Symmetry
Mirror image is broken symmetry in the x and z axis If the left amp right-axis are
reversed symmetry is broken Obs-1 math is different from obs-2
If the front amp back-axis is reversed symmetry is broken Obs-1 math is different from obs-2
If the top amp bottom-axis are
reversed symmetry is not broken Obs-1 math is the same as obs-2
Math answers cannot depend on which side of a number line an observer sit
Absolute values for Displacement
Even for a simple displacement of an object BS math invented a definition and absolute values to obtain something labeled a positive number answer
I----------I----------I----------I----------I----------I----------I----------I----------I
-4 -3 -2 -1 0 +1 +2 +3 +4 a gtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtb cltltltltltltltltltltltltltltltltltltltltltltltltltltltb
If an object starts at ( 3)ia X and moves to ( X 3) and the to ( 1)b fb c X
BS Math provides a usable answer without a definition or absolute values
However If an object starts at
( 1) and moves to ( X 3) and the to ( 3)i b fa X b c X
Then absolute values and a definition must be used
---------I----------I----------I----------I----------I----------I----------I---------- -4 -3 -2 -1 0 +1 +2 +3 +4
a gtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtb cltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltb
BS Math makes up a definition that all displacements are positive Since this example provides an answer with something labeled a negative the positive definition must be applied and (-2) is changed to (+2) A definition is needed to arrive at a useable answer This is Illogical
In SM absolute values and a meaningless definition of ldquopositiverdquo are removed In SM objects and directions in space are defined by the direction and magnitude of the resultant of arrows
With SM you get total distance traveled by the object and the final direction of the displacement
22222 2 bababbaababababa If w e su b stitu te p o s itiv e n u m b e rs in to th e eq u a tio n th e an sw er w ill b e co rrec t If (a = 5 ) an d (b = 3 )
2
2
8
5 3 5 3 5 3 5 5 5 3 3 5 3 3 25 15 15 9 64
8 8 8 8 8 8 8 8 8 8 64
W h en tw o p o sitiv e n u m b ers a re a d d ed an d sq u a re d th e B S m a th d is tr ib u tiv e law p ro v id es a co rre c t an sw er 1
Illogical and Incorrect BS math for the distributive law
222 2 bababbabbaaabababa B y d e fin itio n in B S m a th n o n u m b er sq u a red c a n ev e r b e n eg a tiv e T h e re fo re n o sq u a red n u m b er c an ev e r b e a v e c to r m o v in g in th e d a sh ( -) o r n eg a tiv e d ire c tio n if it is sq u a red In B S m a th an d ad d itio n o p e ra to r m u ltip lied b y a su b trac tio n o p e ra to r is A L W A Y S eq u a l to a su b tra c tio n o p e ra to r If th is h ad n o t b e en u sed fo r th e la s t fo u r h u n d red y e a rs y o u w o u ld d ie lau g h in g a t th e u tte r ab su rd ity o f th e la s t s ta tem e n t T h is is illo g ica l an d v io la te s sy m m etry
If a = 5 an d b = 3 42222352
22
4930253)3)(5(253)5)(3()3)(5(5353535 22222 A s lo n g a s ldquo ardquo is g re a te r th an ldquob rdquo th e B S m a th d is tr ib u tiv e law p ro v id es co rrec t an sw ers 1
B S m a th b re a k s d o w n w h e n ldquo b rdquo is g rea ter th a n ldquo a rdquo
If a = 3 an d b = 5
2 2
2
2
3 5 2 4
2 ( 2) ( 2) 4
2 ( )( 2) ( )( 2) 4
A n y co m b in a tio n ex cep t th e m id d le o n e is illo g ica l an d in co rre c t B y d e fin itio n th e an sw er can n o t b e -4 in B S m a th B S m a th is in co rrec t
4253095)5)(3(235)3)(5()5)(3(3535353 22222 T h is is w h ere th e d is tr ib u tiv e la w p ro v id es in co rrec t an sw ers I t p ro d u ces an an sw er o f + 4 b e cau se b y d e fin iti o n n o n u m b er sq u a red in B S m a th c an b e a n eg a tiv e n u m b er U sin g in co rre c t B S m a th th e d is tr ib u tiv e law w ill g iv e an in co rrec t an sw ers o f + 4 T h is is in co rrec t an d illo g ica l 1
Since there can be no negative direction from a squared term in BS math BS math should be modified or abandoned
The BS distributive law should be modified or abandoned
The numbers +1+2+3+4 and ndash1-2-3-4hellip should be abandoned
BS math gives incorrect answers because their Rule-of-Signs allows multiplying an arrow going in one direction by an arrow going in another direction
Clearly this is not logical
Arrows should be used that are specific for directions
W h at d o es it m e an to m u ltip ly ( 4) b y (4 ) A s w e k n o w 1
m u ltip lica tio n is ju s t ad d itio n W h a t w o u ld ( 4) tim e (4 ) b e 2
In BS math the answer is by definition -16 unless you square -4 and then it is +16
In SM you cannot multiply arrows going in opposite directions
The middle two terms are not logical You cannot multiply opposite directions
This is a MAJOR error in BS math They multiply a dash [(-) a direction to the left] by a cross [(+) a direction to the right] And then ldquoby definitionrdquo label the answer dash (-) This is illogical and produces incorrect answers
2
2 2
This is Einsteinrsquos math error
in Special relativity
Magnitude 4 Direction 1 Magnitude 3 Direction 2 Magnitude 5 Direction 3 Relative to an observer at 0 4
Magnitude 4 Direction 1 Magnitude 3 Direction 2 Magnitude 5 Direction 3 Relative to an observer at 0 4
Magnitude and Direction (MD) An arrow has both a magnitude andA direction relative to a 2nd objectobserver
bull Magnitude and velocity (MV) relative to a 2nd objectobserver
bull Magnitude And acceleration (MA) relative to a 2nd objectobserver2
v da
t t
Introducing a new symbol the ldquoamprdquo sign The ldquoamprdquo sign means resultant ADD all the Arrows for each direction and then SUBTRACT to find the final resultant
1 1amp 1 2 2amp 2 4 If =
1 amp1 2 2 amp2 4 If =
amp 5 amp 3 2 If gt
amp 3 amp 5= 2 If gt
amp 0 3 amp 3=0 if =
only the resultant equals zero
the vectors do not - they are still there
Does not exist There is no multiplication of arrows
There are two arrows One has a magnitude of 3 One has a magnitude of 5 The maximum these two arrows can be is 8
Only the resultant of arrows exist
An SM arrow has a direction and a magnitude
Arrows and magnitudes are still there even if the resultant is zero
There are no negative SM Arrows
There are no positive SM Arrows
There are no ZERO SM Arrows A zero would negate the definition of an SM Arrow something with no magnitude and no direction cannot be the definition of an arrow that is defined as having magnitude and direction
bull A resultant is the addition and subtraction of the differences in magnitude and direction arrows
A B and C are the arrows (a) A amp B exist(b) A amp B amp C exist(c) A-B Do Not Exist in Sm there are no negatives or positives(d) A+B-C Do Not Exist in SM there are no negatives or positives
There are only resultants when there are multiple arrows
SM Graphs versus BS Graphs
Numbers Raised to Powers
When numbers are raised to powers the negative sides of all BS math graphs are ILLOGICAL and INCORRECT
Incorrect BS Math for 2x
2x
4 16 correct3 8 correct2 4 correct1 2 correct0 1 correct-1 005 incorrect-2 025 incorrect-3 0125 incorrect-4 00625 incorrect
negative side of graph is incorrectpositive side of graph is correct
16
8
42105025012500625
-4 -3 -2 -1 0 1 2 3 4
po
sit
ive
negative positive
2x BS Math
correctincorrect
In BS math the dash sign (-) used in this example represents three different operations They arebull (-) Subtractionbull (-) A direction to the leftbull (-) As an exponent to mean divide by
All positive numbers produce correct results and graphs
All negative numbers produce incorrect results and graphs
A direction of three units to the left (-3) is placed in the equation as an exponent
A direction in space (3 units toward the left) changes a whole number to a fraction by using the dash (-) as an exponent rather than a direction in space
This is illogical and incorrect BS math should have performed this operation as
Unfortunately the answers to problems can come out correct because data points of a real problem fit the curve
However the answers will not be due to logical reasoning
Because the equation produced a curve that real problem data points follow it has been falsely believed that the equations and graphs are correct
1
2
x
Incorrect BK graph for (12)x
(12)x
4 00625 correct3 0125 correct2 025 correct1 1 correct0 1 correct-1 2 incorrect-2 4 incorrect-3 8 incorrect-4 16 incorrect
negative side of graph is incorrectpositive side of graph is correct
00625012502505124
8
16
-5 -4 -3 -2 -1 0 1 2 3 4 5P
osit
ive
negative positive
BS Math for (12)x
correctincorrect
Incorrect BS math and graph for
All graphs are symmetrical with SM
Two BS Math Equations that Produce Incorrect Answers
2 0x x starting equation 2
2
x x x x
x x
Add lsquoxrsquo to both sides
2(1) 1 let lsquoxrsquo = 1
2 1 This is an incorrect answer 2(0)-(0)=0 The only answer that BK math can give is x=0
1
a b starting equation ( )( ) ( )( )a a a b multiply both sides by ( )a
( )( ) ( )( ) ( )( ) ( )( )a a b b a b b b subtract ( )( )b b from both sides
( )( ) ( )( )a b a b a b b factor ( )a b from both sides
( )a b b divide both sides by ( )a b
( )b b b replace ( )a with ( )b 2b b simplify 2(1) (1) Let ( )b = 1
2 1 Again BS math yields and incorrect answer 1
This incorrect result is a problem with the distributive law SM explains how this incorrect answer is obtained and the problem with the distributive law that allows the error
Page 148 ndash149 Barry Mazur ndash Imagining Numbers
3
2
31
Do the computation carefully on paper using the rules we agreed to Then ponder your answer which should be something of a surprise to you donrsquot stop there Think of what your answer might possibly mean or might imply
Solve this equation using the distributive law
JK definitely a surprise the answer wasincorrect and illogical
Gerhard Gade University Professor doing math at Harvard University
Here is the BS math answer that Barry Mazur calculated
2
31
2
31
2
31
1 3 1 3 1 3 3 3 2 2 3 1 3
2 2 2 2 4 4 4 4 4 4 2 2
dash
1 3 1 3 1 31
2 2 2 2 4 4
3 3
4 4
When the distributive law is used with the BS Rule-of-Signs and one of the signs is a (-) dash sign the middle two terms always cancel each other This leads to the incorrect answer of (-1) in the above equation
Barry Mazur did not go back and check his answer with the original equation If he had he would have known that the answer was incorrect
3
1 31
2
check answer by setting equal
3
331 3
12
1 31
2
Take the Cube root of both sides
1 3 2 multiply both sides by 2
3 3 subtract 1 from both sides
2 2
3 3 square both sides
3 9 This is an incorrect answer A simple equation check could have shown that the distributive law using BS math provides an incorrect answer
1
BS Illogical and Incorrect Numbers Raised to Powers
nC D n DC
642 DDD
In BS math only a positive number can have an answer if ldquonrdquo is an even number (246---)
All negative numbers have no solutions
642 DDD
With SM just add a direction arrow and any number gives the correct answer
or n nC D C D
3 3( 3 8 8 or 8 2 or 2 )n D C
( 2 16 C= 16 or 16 4 or 4 )n D
Since a negative number has no square root answer in BS math BS mathematicians invented imaginary numbers
i2=-1
24414 iiC n
These equations are easily solved without the use of imaginary numbers with SM math
If 4 then 4 2x
If 4 then 4 2x
The 360-degree circle is a 4400 years old extremely outdated system based on a year of 12 months of 30 days each that is 360 days
SM will use a more logical system that will have a circle of 6283185307o
d
( )
(deg )
d r radians
d r rees
the arc distance around
the circumference of a circle
d
radiusr
2 2
2 2
d r
dv rd t t
v dda rd t t t
1
d
r
( )av tv
The average angular velocity of a body is the rate of its angular displacement
( )0
liminstt
vt
The instantaneous angular velocity is the limit of the ratio
t
as t approaches zero
( )d
av
d
tv
The average circumference velocity of the body is the rate of its distance traveled d divided by the time t
dvv
r
1 dd vdrv
t t t r r
The equation of the relationship between the angular velocity and the velocity of a point on that body
( )o
av
v v v
t ta
The average angular acceleration is the change in the angular velocity
( )0
liminstt
v
ta
The instantaneous angular acceleration is the
limit of the ration of v
t
as t approaches
zero
daa
r
1
1
d
d s d
vva
t r tv a
r t r
The equation of the relationship between the angular acceleration and the acceleration of a point on the body
21
2v a t
If a body with an initial angular velocity ov has a constant
angular acceleration a it will turn through an angle in a time t
1
Straight Line Symbol Units Rotational Symbol Units Distance d m Angular
Distance radians or degrees
Velocity dt
v m
s Angular Velocity
V
t
radians or degrees
s
Acceleration vat
2ms
Angular Acceleration
VA
t
2
radians or degrees
s
Mass M Kg Mass distribution
dM 2Kgm
Force actionA Ma
2
Kgm
sNewton
Torque
dT M A 2
2
Kgm
s
Momentum Mv Kgm
s
Angular Momentum
dM V 2Kgm
s
Work (Force)(distance)
Ad 2
Kgmm
sNm Joule
Work T
Using AS arrow math to solve and direction in space problems using the equation 1 2
2 2 2 2 2 2( ) amp and ( ) amp 3
4
2 2 2
2 2
( ) amp
(5 ) 4 3
(5 ) 16 9
(5 ) 25
(5 ) (5 )
2 2(5 ) 4 3
(5 ) 16 9
(5 ) 25
(5 ) (5 )
In SM math Pythagorasrsquo theorem provides both magnitude and direction
In BS math Pythagorasrsquo theorem provides magnitude only
time set equal to zero
23
( ) ( 0 )10 1
x t xx
3( ) ( 0)
10 1W t f
In BS math this is an equation for a traveling wave
must use a square term in order to make the curve symmetric
NO SQUARE TERM IS NECESSARY
In SM math when t is set to ldquo0rdquo
This is a major fault in BS math
x f(x)
-4 0019
-3 0033
-2 0073
-1 0273
0 3000
1 0273
2 0073
3 0033
4 0019 -4 -3 -2 -1 0 1 2 3 400
05
10
15
20
25
30
35
BS wave (must use square term)
snap shot of wave at x = 0
Am
plit
ude
lt to gt lt to gt f(ltgt)
lt -4 0073
lt -3 0097
lt -2 0143
lt -1 0273
origin 0 3000
gt 1 0273
gt 2 0143
gt 3 0097
gt 4 0073lt0073 lt0097 lt0143 lt0273
3000
0273gt 0143gt 0097gt 0073gt
Wave moving
snap shot of wave
Am
plit
ude
lt4 lt3 lt2 lt1 0 1gt 2gt 3gt 4gt
3
2
1
In SM NO squared terms are needed
SPRING
An Action is needed to stretch or compress a spring to a starting point where it can execute cycle motion
tsA (ts to start) is the work required to move a mass attached to a spring a distance (d1) from
its initial rest position RAK 0 the spring constant is the stiffness of the spring
o A is attraction o R is repulsion o
ARRA
00 is for the back and forth motion through the zero initial rest position A
stretched spring goes from ldquoArdquo to ldquo0rdquo to ldquoRrdquo to ldquoArdquo to ldquo0rdquo to ldquoRrdquo d1 is the distance a mass is moved from its initial rest position
0
0 1ts A RA K d
Once ( )mad
is performed to move the mass to position d1 (against the restraint of the spring) a of the ( )mad
is transferred into the spring as potential attractionrepulsion (PAR) There is always
more ( )mad
performed than is stored as PAR When the spring is released from the restraining Ats the PAR starts the oscillation Conservation of ( )mad
requires that the PAR and the motion ( )mad
( )( )madM be a constant
( )amp ( ) constantmadPAR M 02
0 0
1
2A R A RPAR K d
02
( ) 0 0
1( )
2mad A R A RM m v
0 0 02 2 2
0 0 0 1
2 2 21
2 2 21
1 1 1amp
2 2 2
1 1 1amp
2 2 2
amp
A R A R A RK d m v K d
Kd mv Kd
Kd mv Kd
0
1d is frac12 the height from the base to the peak of a cycle
An experiment of moving paper under a pendulum through a small angle filled with sand that can escape through a small hole in the bottom shows that a pendulum cycling back and forth follows a curve that is mathematically analogous to a lsquosinrsquo or lsquocosrsquo
The projection of the curved area onto the straight line exactly duplicate the back and forth motion of CM This allows for the math of lsquosinrsquo or lsquocosrsquo to be used
SM will use lsquosinrdquo math It could use lsquocosrsquo or a combination of both lsquosinrsquo and lsquocosrsquo Using just the lsquosinrsquo math makes the understanding and math easier
0 sin 0 0SM 0 sin157 1SM The diagram at the left shows a spring being stretched from its (zero rest) position-2 and the PAR being released at position-1 There are 4 sections (1 to 2 2 to 3 3 to 4 and 4 to5) that alternate between maximum acceleration and zero velocity to max velocity and zero acceleration
At position-1 there is maximum acceleration and zero velocity o The lsquosinrsquo values from 0o to 157o will be used for velocity (values from 0 to 1) o The lsquosinrsquo values from 157o to 0o will be used for acceleration (values from 1 to 0)
At position-2 there is maximum velocity and zero acceleration Acceleration changes to deceleration
o The lsquosinrsquo values from 157o to 0o will be used for deceleration (values from 1 to 0) o The lsquosinrsquo values from 0o to 157o will be used for acceleration (values from 0 to 1)
At position-3 there is maximum acceleration and zero velocity o The lsquosinrsquo values from 0o to 157o will be used for velocity (values from 0 to 1) o The lsquosinrsquo values from 157o to 0o will be used for acceleration (values from 1 to 0)
At position-4 there is maximum velocity and zero acceleration Acceleration changes to deceleration
o The lsquosinrsquo values from 157o to 0o will be used for deceleration (values from 1 to 0) o The lsquosinrsquo values from 0o to 157o will be used for acceleration (values from 0 to 1)
At position-5 the cycle starts again
2
)539(
cT
MK
spring constant (K) when mass (M) and cycle-time (T) are known
)(
))(89)((
111 md
uKgM
d
MG
d
FK sts
spring constant (K) when mass (M) and amplitude ( d1) are known
2 21( )
Kv d d
m
velocity (v) at any point during the cycle
velocity
dforM
Kdv
max
012 velocity (v2) maximum (when d = 0 at position-2 and 4)
acceleration (a1) maximum when spring constant (K) mass (M) and amplitude (d1) are known
acceleration (a1) maximum when amplitude (d1) and cycle-time (Tc) are known
acceleration (a) at any point during the cycle-time (T)
1
151
286
v
dTc
cycle-time (Tc) when amplitude (d1) and velocity (v1) are known
K
MTc 28651
cycle-time (Tc) when mass (M) and spring constant (K) are known
fTc
1
cycle-time (Tc) when the frequency is known
cTf
1
frequency (f) when the cycle-time (Tc) is known
of cycles
elapsed timef
frequency (f) when the number of cycles in an elapsed time is known
d1 amplitude (distance from rest to start position)
Also frac12 the distance from the base to the peak
3131 oror dM
Ka
251
3131 539
c
oror T
da
2539
cT
da
In December of 2006 I discovered the mathematical error in Einsteinrsquos Special Relativity Einstein used BS math and that math only works in an imaginary world
It cannot describe real symmetrical space
Einsteinrsquos Special Relativity should be abandoned and a theory of real space using SM should be used to replace it
Numerous articles on the Internet show Einsteinrsquos theory to be incorrect but no one in academia is willing to publish the information
Figure 1-10 A moving particle observed from two frames of reference in uniform translation
Einsteinrsquos Math Error in Special Relativity Transformation of Velocity (From Eisenberg- Fundamentals of Modern Physics pages 27-29) Consider the particle shown in figure (1-10) moving with velocity v as
seen in a frame of reference O
dx dy dz
Vx Vy Vzdt dt dt
dx dy dz
V x V y V zdy dt dt
We would like to evaluate the velocity v of the particle as seen in the frame of reference 0 which is itself moving relative to 0 with velocity v Measured in the 0 frame the velocity vector of the particle has com ponents
The same velocity vector as measured in the 0 frame has components
2
2
2
2
2
1
1
x vtx
v
cy y
z z
vxt
ctv
c
Now from equations (1-13) we know that the relation between the primed and the unprimed coordinates and times is
Take the differential of these equations remembering that v is a constant This gives
2
2
2
2
2
1
1
dx vdtdx
v
cdy dy
dz dz
vxdt
cdtv
c
2
2
2 2 2 2 2
2
2
2
11 ( )
1( ) 1
1
1
1
dx vdt
v dxdx vdt v
dx dx vdtc dt dtvdx vdx vdx v dxdt dt dt dt
dt dtc c c c
VxV x
Vx
VxV x
vv
vv
c
x
v
c
cV
( )
coordinate of particle relative to moving primed system
coordinate of particle relative to stationary system
velocity of primed coordina
V x
Vx
te system relative to stationaryv sys
tem
Page 28 Fundamentals of Modern Physics
(1-16)
According to Einstein this equation tells you how to transform the observed velocity from one frame of reference to another frame of reference First we note that as Vc and vc approachzero the equation approaches those which would be derived from the Galilean transformation
2 2
( ) 1 01 1 ( )
Vx v Vx vV
VxV x
Vx v
v cVxx
v
c
v
cc
Another property of this equation is that it is impossible to choose V and v such that Vrsquo the magnitude of the velocity which is seen in the new frame is greater than c
1
2
09 ( 09 ) 18009876543
( 09 ) 1811 (09 )
c c cV c c
cc
c
Example As seen by O particle 1 has velocity 09c in the direction of positive x and particle 2 has velocity 09c in the negative x direction
Einstein equation however predicts To evaluate the velocity of particle 1 with respect to particle 2 we transform from the O frame to the Orsquo frame moving in the negative x direction with velocity v = -09c using equation 1-16 We obtain
JK LOGIC will observe the particles separating at 18c
A confusing INCORRECT and ILLOGICAL answer
left is a dir rightsub of a
1
2
1
09 ( 09 ) 18009876543
181( 09 )1 ( 09 )
The multiplication of dir left time a dir right is ONLY a dir left
09
dir
direction leftdirection rightsubtraction
c c cV c c
cc
c
V
dir
22
22
09 18 18009876543
181( 081)1(09)(09)
1
left rightsub
sub
c c c cc c
ccc
c
Analyzing Einsteinrsquos BS math Equation
Using SM math a logical answer can be obtained
Cannot multiply a subtraction operator times a direction in space
Can only subtract and change its direction
Cannot multiply a direction to the left times a direction to the right
Cannot multiply opposite directions in space
Can only solve for the resultant
rightleft
subtraction
1
subtraction
2
sub of a dir to the left is a dir to the right
1 resultant 0 sub 2
2sub 2
2
09 09
09 amp 091
09 09 18 1
(0)1
(09) amp (09)1
c cV
c cc
c c cV
cc
cc
8
181
cc
SM math predicts a logical answer using logic
21st century scientists state in many articles that Einsteins General Relativity and Quantum Mechanics are incompatible
They are not only incompatible they are both mathematically incorrect
They both use illogical BS math
Any equations in math that use imaginary numbers should be abandoned
Imaginary numbers are only necessary in the current BS math system
This means that Schroumldingers equations should be abandoned
Any equation in quantum mechanics that uses imaginary number should be abandoned
Prof Kanarev has developed logical physics and chemistrytheories without the use of Quantum mechanics or relativity
A partial list of his theories coverbull Mechanics ndash replaces Newtonbull Photon - bull Electrons ndash Protons- Neutronsbull Spectroscopybull Nucleusbull Atoms amp Moleculesbull Electrodynamicsbull Heatbull Curvature of Space ndash replaces Schwarzcchildrsquos equation
Kanarevrsquos Hypothesis
bull Electron align linearly with proton They do NOT spin or form clouds around the nucleus
bull Neutrons are in the center of atom Proton are outside of Neutrons
bull The wavelength of all particles is the experimental radius
bull Double slit experiment explained correctly
bull Heisenbergrsquos inequality explained correctly
bull is the radius of approach of the magnetic lines around the torus of the electron
( 1 137)
bull Why water expands when it freezes
bull Electron clusters form lightning
rbull The meaning of Plancks constant is explained
22 2 kg m
h m f mr f consts
2210254 42m r kgm bull If the mr-constant describes a ring the equations
describe experiments bull If the mr-constant describes a wave the equations do
NOT describe experiments
Kanarevrsquos 5 Equations of Mechanics
Kanarevrsquos 1st Equation Acceleration ampmv maMma R R
Kanarevrsquos 2nd Equation Uniform motion amp mv maMma R R
maM mvR R
Kanarevrsquos 3rd Equation
Kanarevrsquos 4th Equation Equal and opposite reaction
1 21 21 2
12
a m
m a m ama
Kanarevrsquos 5th Equation 1 2 3 Mi M M M Mna R R R R R
The sum of the Actions working on a moving body is never equal to zero
Kanarevrsquos Photon model is a rolling ring
His equations accurately describe known experiments with photonsbull Electromagnetic radiation should be called photon radiationbull Diffraction and interference are the same phenomenabull Correctly explains the double slit experiment without quantum equationsbull Correctly explains the meaning of Heisenbergrsquos inequality
e
2 (2)(3142)(2817 15 ) 10007297352568
R 2426 12 137mag
xp
r m
m
Charge
Magnetic action
Proton
The intensity of the magnetic action near the geometrical center of the proton ring is so great that we theorize that it is the strong nuclear force
(9383 8)(1602 19)8476 14
4 (4)(3142)(1411 26)p
pp
EB T
M
(9395 8)(1602 19)1035 15
4 (4)(3142)(0966 26)N
NN
EB T
M
Neutron
1a Hydrogenavi
A vector moving to the left multiplied by a vector moving to the right is equal to ONLY to a vector moving to the left that is eight orders of magnitude faster thanThe speed of light
opposite direction subtraction direction
SM ( ) ( )
BS (negative)(positive) = (negative)
Symmetry Math The subtraction of a direction is equal to the opposite direction
There is no multiplication of a subtractionoperator times a direction in space
( ) multiplied by a ( ) ( )
positive positive positiveaddition addition additioncross cross cross
direction direction direction
What is the meaning of multiplying addition operators Addition operators are not multiplication operators
BS Math
() multiplied by a direction ( )=
number
() multiplied by a direction ( )=
number
A number times a direction maintains the same direction
( ) ()
dirdiradd
( ) ()
dirdiradd
The addition of a direction is in the same direction There is no multiplication by addition operators
Symmetry Math
Future math books need to eliminate the use of the cross sign (+) to represent something labeled a positive direction in space and the dash sign (-) to represent something labeled a negative direction in space Space does not have positive and negative directions
Symbols that are logical and have no illogical representation should be adopted SM uses either arrows or symbols
Symmetry-Math Number Line
Symmetry-Math Rules
The dash sign (-) will have only one use subtraction
The cross sign (+) will have only one use addition
Directions have an arrow and a number
All observers see the same direction and magnitude
bull There is no multiplication of arrows
amp ampamp ampampamp
( 3 ) ( 5 ) ( 6 ) ( 65 ) (17256 ) ( 32993 )
dir dirdir dirdirdir
Negative and Positive Directions in Space
If I asked you to point to a negative direction in space which way would you point
Hopefully you will realize that there is no such thing as a negative direction in space
If I asked you to point to a positive direction in space which way would you point
Again hopefully you will realize that there is no such thing as a positive direction in space
Objects on the right side of the x-axis are positive (+)(+) = (+)
Objects on the left side of the x-axis are negative (-)(-) = (+) negative math is different from positive math This is illogical
Objects on the top of the y-axis are positive
Objects on the bottom of the y-axis are negative
Objects in the front of the z-axis are positive
Objects in the back of the z-axis are negative
Again BS math of the positive direction (x right y up and z front) is different from the BS math of the negative direction (x left y down and z back)
BS math is Broken-Symmetry
Mirror image is broken symmetry in the x and z axis If the left amp right-axis are
reversed symmetry is broken Obs-1 math is different from obs-2
If the front amp back-axis is reversed symmetry is broken Obs-1 math is different from obs-2
If the top amp bottom-axis are
reversed symmetry is not broken Obs-1 math is the same as obs-2
Math answers cannot depend on which side of a number line an observer sit
Absolute values for Displacement
Even for a simple displacement of an object BS math invented a definition and absolute values to obtain something labeled a positive number answer
I----------I----------I----------I----------I----------I----------I----------I----------I
-4 -3 -2 -1 0 +1 +2 +3 +4 a gtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtb cltltltltltltltltltltltltltltltltltltltltltltltltltltltb
If an object starts at ( 3)ia X and moves to ( X 3) and the to ( 1)b fb c X
BS Math provides a usable answer without a definition or absolute values
However If an object starts at
( 1) and moves to ( X 3) and the to ( 3)i b fa X b c X
Then absolute values and a definition must be used
---------I----------I----------I----------I----------I----------I----------I---------- -4 -3 -2 -1 0 +1 +2 +3 +4
a gtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtb cltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltb
BS Math makes up a definition that all displacements are positive Since this example provides an answer with something labeled a negative the positive definition must be applied and (-2) is changed to (+2) A definition is needed to arrive at a useable answer This is Illogical
In SM absolute values and a meaningless definition of ldquopositiverdquo are removed In SM objects and directions in space are defined by the direction and magnitude of the resultant of arrows
With SM you get total distance traveled by the object and the final direction of the displacement
22222 2 bababbaababababa If w e su b stitu te p o s itiv e n u m b e rs in to th e eq u a tio n th e an sw er w ill b e co rrec t If (a = 5 ) an d (b = 3 )
2
2
8
5 3 5 3 5 3 5 5 5 3 3 5 3 3 25 15 15 9 64
8 8 8 8 8 8 8 8 8 8 64
W h en tw o p o sitiv e n u m b ers a re a d d ed an d sq u a re d th e B S m a th d is tr ib u tiv e law p ro v id es a co rre c t an sw er 1
Illogical and Incorrect BS math for the distributive law
222 2 bababbabbaaabababa B y d e fin itio n in B S m a th n o n u m b er sq u a red c a n ev e r b e n eg a tiv e T h e re fo re n o sq u a red n u m b er c an ev e r b e a v e c to r m o v in g in th e d a sh ( -) o r n eg a tiv e d ire c tio n if it is sq u a red In B S m a th an d ad d itio n o p e ra to r m u ltip lied b y a su b trac tio n o p e ra to r is A L W A Y S eq u a l to a su b tra c tio n o p e ra to r If th is h ad n o t b e en u sed fo r th e la s t fo u r h u n d red y e a rs y o u w o u ld d ie lau g h in g a t th e u tte r ab su rd ity o f th e la s t s ta tem e n t T h is is illo g ica l an d v io la te s sy m m etry
If a = 5 an d b = 3 42222352
22
4930253)3)(5(253)5)(3()3)(5(5353535 22222 A s lo n g a s ldquo ardquo is g re a te r th an ldquob rdquo th e B S m a th d is tr ib u tiv e law p ro v id es co rrec t an sw ers 1
B S m a th b re a k s d o w n w h e n ldquo b rdquo is g rea ter th a n ldquo a rdquo
If a = 3 an d b = 5
2 2
2
2
3 5 2 4
2 ( 2) ( 2) 4
2 ( )( 2) ( )( 2) 4
A n y co m b in a tio n ex cep t th e m id d le o n e is illo g ica l an d in co rre c t B y d e fin itio n th e an sw er can n o t b e -4 in B S m a th B S m a th is in co rrec t
4253095)5)(3(235)3)(5()5)(3(3535353 22222 T h is is w h ere th e d is tr ib u tiv e la w p ro v id es in co rrec t an sw ers I t p ro d u ces an an sw er o f + 4 b e cau se b y d e fin iti o n n o n u m b er sq u a red in B S m a th c an b e a n eg a tiv e n u m b er U sin g in co rre c t B S m a th th e d is tr ib u tiv e law w ill g iv e an in co rrec t an sw ers o f + 4 T h is is in co rrec t an d illo g ica l 1
Since there can be no negative direction from a squared term in BS math BS math should be modified or abandoned
The BS distributive law should be modified or abandoned
The numbers +1+2+3+4 and ndash1-2-3-4hellip should be abandoned
BS math gives incorrect answers because their Rule-of-Signs allows multiplying an arrow going in one direction by an arrow going in another direction
Clearly this is not logical
Arrows should be used that are specific for directions
W h at d o es it m e an to m u ltip ly ( 4) b y (4 ) A s w e k n o w 1
m u ltip lica tio n is ju s t ad d itio n W h a t w o u ld ( 4) tim e (4 ) b e 2
In BS math the answer is by definition -16 unless you square -4 and then it is +16
In SM you cannot multiply arrows going in opposite directions
The middle two terms are not logical You cannot multiply opposite directions
This is a MAJOR error in BS math They multiply a dash [(-) a direction to the left] by a cross [(+) a direction to the right] And then ldquoby definitionrdquo label the answer dash (-) This is illogical and produces incorrect answers
2
2 2
This is Einsteinrsquos math error
in Special relativity
Magnitude 4 Direction 1 Magnitude 3 Direction 2 Magnitude 5 Direction 3 Relative to an observer at 0 4
Magnitude 4 Direction 1 Magnitude 3 Direction 2 Magnitude 5 Direction 3 Relative to an observer at 0 4
Magnitude and Direction (MD) An arrow has both a magnitude andA direction relative to a 2nd objectobserver
bull Magnitude and velocity (MV) relative to a 2nd objectobserver
bull Magnitude And acceleration (MA) relative to a 2nd objectobserver2
v da
t t
Introducing a new symbol the ldquoamprdquo sign The ldquoamprdquo sign means resultant ADD all the Arrows for each direction and then SUBTRACT to find the final resultant
1 1amp 1 2 2amp 2 4 If =
1 amp1 2 2 amp2 4 If =
amp 5 amp 3 2 If gt
amp 3 amp 5= 2 If gt
amp 0 3 amp 3=0 if =
only the resultant equals zero
the vectors do not - they are still there
Does not exist There is no multiplication of arrows
There are two arrows One has a magnitude of 3 One has a magnitude of 5 The maximum these two arrows can be is 8
Only the resultant of arrows exist
An SM arrow has a direction and a magnitude
Arrows and magnitudes are still there even if the resultant is zero
There are no negative SM Arrows
There are no positive SM Arrows
There are no ZERO SM Arrows A zero would negate the definition of an SM Arrow something with no magnitude and no direction cannot be the definition of an arrow that is defined as having magnitude and direction
bull A resultant is the addition and subtraction of the differences in magnitude and direction arrows
A B and C are the arrows (a) A amp B exist(b) A amp B amp C exist(c) A-B Do Not Exist in Sm there are no negatives or positives(d) A+B-C Do Not Exist in SM there are no negatives or positives
There are only resultants when there are multiple arrows
SM Graphs versus BS Graphs
Numbers Raised to Powers
When numbers are raised to powers the negative sides of all BS math graphs are ILLOGICAL and INCORRECT
Incorrect BS Math for 2x
2x
4 16 correct3 8 correct2 4 correct1 2 correct0 1 correct-1 005 incorrect-2 025 incorrect-3 0125 incorrect-4 00625 incorrect
negative side of graph is incorrectpositive side of graph is correct
16
8
42105025012500625
-4 -3 -2 -1 0 1 2 3 4
po
sit
ive
negative positive
2x BS Math
correctincorrect
In BS math the dash sign (-) used in this example represents three different operations They arebull (-) Subtractionbull (-) A direction to the leftbull (-) As an exponent to mean divide by
All positive numbers produce correct results and graphs
All negative numbers produce incorrect results and graphs
A direction of three units to the left (-3) is placed in the equation as an exponent
A direction in space (3 units toward the left) changes a whole number to a fraction by using the dash (-) as an exponent rather than a direction in space
This is illogical and incorrect BS math should have performed this operation as
Unfortunately the answers to problems can come out correct because data points of a real problem fit the curve
However the answers will not be due to logical reasoning
Because the equation produced a curve that real problem data points follow it has been falsely believed that the equations and graphs are correct
1
2
x
Incorrect BK graph for (12)x
(12)x
4 00625 correct3 0125 correct2 025 correct1 1 correct0 1 correct-1 2 incorrect-2 4 incorrect-3 8 incorrect-4 16 incorrect
negative side of graph is incorrectpositive side of graph is correct
00625012502505124
8
16
-5 -4 -3 -2 -1 0 1 2 3 4 5P
osit
ive
negative positive
BS Math for (12)x
correctincorrect
Incorrect BS math and graph for
All graphs are symmetrical with SM
Two BS Math Equations that Produce Incorrect Answers
2 0x x starting equation 2
2
x x x x
x x
Add lsquoxrsquo to both sides
2(1) 1 let lsquoxrsquo = 1
2 1 This is an incorrect answer 2(0)-(0)=0 The only answer that BK math can give is x=0
1
a b starting equation ( )( ) ( )( )a a a b multiply both sides by ( )a
( )( ) ( )( ) ( )( ) ( )( )a a b b a b b b subtract ( )( )b b from both sides
( )( ) ( )( )a b a b a b b factor ( )a b from both sides
( )a b b divide both sides by ( )a b
( )b b b replace ( )a with ( )b 2b b simplify 2(1) (1) Let ( )b = 1
2 1 Again BS math yields and incorrect answer 1
This incorrect result is a problem with the distributive law SM explains how this incorrect answer is obtained and the problem with the distributive law that allows the error
Page 148 ndash149 Barry Mazur ndash Imagining Numbers
3
2
31
Do the computation carefully on paper using the rules we agreed to Then ponder your answer which should be something of a surprise to you donrsquot stop there Think of what your answer might possibly mean or might imply
Solve this equation using the distributive law
JK definitely a surprise the answer wasincorrect and illogical
Gerhard Gade University Professor doing math at Harvard University
Here is the BS math answer that Barry Mazur calculated
2
31
2
31
2
31
1 3 1 3 1 3 3 3 2 2 3 1 3
2 2 2 2 4 4 4 4 4 4 2 2
dash
1 3 1 3 1 31
2 2 2 2 4 4
3 3
4 4
When the distributive law is used with the BS Rule-of-Signs and one of the signs is a (-) dash sign the middle two terms always cancel each other This leads to the incorrect answer of (-1) in the above equation
Barry Mazur did not go back and check his answer with the original equation If he had he would have known that the answer was incorrect
3
1 31
2
check answer by setting equal
3
331 3
12
1 31
2
Take the Cube root of both sides
1 3 2 multiply both sides by 2
3 3 subtract 1 from both sides
2 2
3 3 square both sides
3 9 This is an incorrect answer A simple equation check could have shown that the distributive law using BS math provides an incorrect answer
1
BS Illogical and Incorrect Numbers Raised to Powers
nC D n DC
642 DDD
In BS math only a positive number can have an answer if ldquonrdquo is an even number (246---)
All negative numbers have no solutions
642 DDD
With SM just add a direction arrow and any number gives the correct answer
or n nC D C D
3 3( 3 8 8 or 8 2 or 2 )n D C
( 2 16 C= 16 or 16 4 or 4 )n D
Since a negative number has no square root answer in BS math BS mathematicians invented imaginary numbers
i2=-1
24414 iiC n
These equations are easily solved without the use of imaginary numbers with SM math
If 4 then 4 2x
If 4 then 4 2x
The 360-degree circle is a 4400 years old extremely outdated system based on a year of 12 months of 30 days each that is 360 days
SM will use a more logical system that will have a circle of 6283185307o
d
( )
(deg )
d r radians
d r rees
the arc distance around
the circumference of a circle
d
radiusr
2 2
2 2
d r
dv rd t t
v dda rd t t t
1
d
r
( )av tv
The average angular velocity of a body is the rate of its angular displacement
( )0
liminstt
vt
The instantaneous angular velocity is the limit of the ratio
t
as t approaches zero
( )d
av
d
tv
The average circumference velocity of the body is the rate of its distance traveled d divided by the time t
dvv
r
1 dd vdrv
t t t r r
The equation of the relationship between the angular velocity and the velocity of a point on that body
( )o
av
v v v
t ta
The average angular acceleration is the change in the angular velocity
( )0
liminstt
v
ta
The instantaneous angular acceleration is the
limit of the ration of v
t
as t approaches
zero
daa
r
1
1
d
d s d
vva
t r tv a
r t r
The equation of the relationship between the angular acceleration and the acceleration of a point on the body
21
2v a t
If a body with an initial angular velocity ov has a constant
angular acceleration a it will turn through an angle in a time t
1
Straight Line Symbol Units Rotational Symbol Units Distance d m Angular
Distance radians or degrees
Velocity dt
v m
s Angular Velocity
V
t
radians or degrees
s
Acceleration vat
2ms
Angular Acceleration
VA
t
2
radians or degrees
s
Mass M Kg Mass distribution
dM 2Kgm
Force actionA Ma
2
Kgm
sNewton
Torque
dT M A 2
2
Kgm
s
Momentum Mv Kgm
s
Angular Momentum
dM V 2Kgm
s
Work (Force)(distance)
Ad 2
Kgmm
sNm Joule
Work T
Using AS arrow math to solve and direction in space problems using the equation 1 2
2 2 2 2 2 2( ) amp and ( ) amp 3
4
2 2 2
2 2
( ) amp
(5 ) 4 3
(5 ) 16 9
(5 ) 25
(5 ) (5 )
2 2(5 ) 4 3
(5 ) 16 9
(5 ) 25
(5 ) (5 )
In SM math Pythagorasrsquo theorem provides both magnitude and direction
In BS math Pythagorasrsquo theorem provides magnitude only
time set equal to zero
23
( ) ( 0 )10 1
x t xx
3( ) ( 0)
10 1W t f
In BS math this is an equation for a traveling wave
must use a square term in order to make the curve symmetric
NO SQUARE TERM IS NECESSARY
In SM math when t is set to ldquo0rdquo
This is a major fault in BS math
x f(x)
-4 0019
-3 0033
-2 0073
-1 0273
0 3000
1 0273
2 0073
3 0033
4 0019 -4 -3 -2 -1 0 1 2 3 400
05
10
15
20
25
30
35
BS wave (must use square term)
snap shot of wave at x = 0
Am
plit
ude
lt to gt lt to gt f(ltgt)
lt -4 0073
lt -3 0097
lt -2 0143
lt -1 0273
origin 0 3000
gt 1 0273
gt 2 0143
gt 3 0097
gt 4 0073lt0073 lt0097 lt0143 lt0273
3000
0273gt 0143gt 0097gt 0073gt
Wave moving
snap shot of wave
Am
plit
ude
lt4 lt3 lt2 lt1 0 1gt 2gt 3gt 4gt
3
2
1
In SM NO squared terms are needed
SPRING
An Action is needed to stretch or compress a spring to a starting point where it can execute cycle motion
tsA (ts to start) is the work required to move a mass attached to a spring a distance (d1) from
its initial rest position RAK 0 the spring constant is the stiffness of the spring
o A is attraction o R is repulsion o
ARRA
00 is for the back and forth motion through the zero initial rest position A
stretched spring goes from ldquoArdquo to ldquo0rdquo to ldquoRrdquo to ldquoArdquo to ldquo0rdquo to ldquoRrdquo d1 is the distance a mass is moved from its initial rest position
0
0 1ts A RA K d
Once ( )mad
is performed to move the mass to position d1 (against the restraint of the spring) a of the ( )mad
is transferred into the spring as potential attractionrepulsion (PAR) There is always
more ( )mad
performed than is stored as PAR When the spring is released from the restraining Ats the PAR starts the oscillation Conservation of ( )mad
requires that the PAR and the motion ( )mad
( )( )madM be a constant
( )amp ( ) constantmadPAR M 02
0 0
1
2A R A RPAR K d
02
( ) 0 0
1( )
2mad A R A RM m v
0 0 02 2 2
0 0 0 1
2 2 21
2 2 21
1 1 1amp
2 2 2
1 1 1amp
2 2 2
amp
A R A R A RK d m v K d
Kd mv Kd
Kd mv Kd
0
1d is frac12 the height from the base to the peak of a cycle
An experiment of moving paper under a pendulum through a small angle filled with sand that can escape through a small hole in the bottom shows that a pendulum cycling back and forth follows a curve that is mathematically analogous to a lsquosinrsquo or lsquocosrsquo
The projection of the curved area onto the straight line exactly duplicate the back and forth motion of CM This allows for the math of lsquosinrsquo or lsquocosrsquo to be used
SM will use lsquosinrdquo math It could use lsquocosrsquo or a combination of both lsquosinrsquo and lsquocosrsquo Using just the lsquosinrsquo math makes the understanding and math easier
0 sin 0 0SM 0 sin157 1SM The diagram at the left shows a spring being stretched from its (zero rest) position-2 and the PAR being released at position-1 There are 4 sections (1 to 2 2 to 3 3 to 4 and 4 to5) that alternate between maximum acceleration and zero velocity to max velocity and zero acceleration
At position-1 there is maximum acceleration and zero velocity o The lsquosinrsquo values from 0o to 157o will be used for velocity (values from 0 to 1) o The lsquosinrsquo values from 157o to 0o will be used for acceleration (values from 1 to 0)
At position-2 there is maximum velocity and zero acceleration Acceleration changes to deceleration
o The lsquosinrsquo values from 157o to 0o will be used for deceleration (values from 1 to 0) o The lsquosinrsquo values from 0o to 157o will be used for acceleration (values from 0 to 1)
At position-3 there is maximum acceleration and zero velocity o The lsquosinrsquo values from 0o to 157o will be used for velocity (values from 0 to 1) o The lsquosinrsquo values from 157o to 0o will be used for acceleration (values from 1 to 0)
At position-4 there is maximum velocity and zero acceleration Acceleration changes to deceleration
o The lsquosinrsquo values from 157o to 0o will be used for deceleration (values from 1 to 0) o The lsquosinrsquo values from 0o to 157o will be used for acceleration (values from 0 to 1)
At position-5 the cycle starts again
2
)539(
cT
MK
spring constant (K) when mass (M) and cycle-time (T) are known
)(
))(89)((
111 md
uKgM
d
MG
d
FK sts
spring constant (K) when mass (M) and amplitude ( d1) are known
2 21( )
Kv d d
m
velocity (v) at any point during the cycle
velocity
dforM
Kdv
max
012 velocity (v2) maximum (when d = 0 at position-2 and 4)
acceleration (a1) maximum when spring constant (K) mass (M) and amplitude (d1) are known
acceleration (a1) maximum when amplitude (d1) and cycle-time (Tc) are known
acceleration (a) at any point during the cycle-time (T)
1
151
286
v
dTc
cycle-time (Tc) when amplitude (d1) and velocity (v1) are known
K
MTc 28651
cycle-time (Tc) when mass (M) and spring constant (K) are known
fTc
1
cycle-time (Tc) when the frequency is known
cTf
1
frequency (f) when the cycle-time (Tc) is known
of cycles
elapsed timef
frequency (f) when the number of cycles in an elapsed time is known
d1 amplitude (distance from rest to start position)
Also frac12 the distance from the base to the peak
3131 oror dM
Ka
251
3131 539
c
oror T
da
2539
cT
da
In December of 2006 I discovered the mathematical error in Einsteinrsquos Special Relativity Einstein used BS math and that math only works in an imaginary world
It cannot describe real symmetrical space
Einsteinrsquos Special Relativity should be abandoned and a theory of real space using SM should be used to replace it
Numerous articles on the Internet show Einsteinrsquos theory to be incorrect but no one in academia is willing to publish the information
Figure 1-10 A moving particle observed from two frames of reference in uniform translation
Einsteinrsquos Math Error in Special Relativity Transformation of Velocity (From Eisenberg- Fundamentals of Modern Physics pages 27-29) Consider the particle shown in figure (1-10) moving with velocity v as
seen in a frame of reference O
dx dy dz
Vx Vy Vzdt dt dt
dx dy dz
V x V y V zdy dt dt
We would like to evaluate the velocity v of the particle as seen in the frame of reference 0 which is itself moving relative to 0 with velocity v Measured in the 0 frame the velocity vector of the particle has com ponents
The same velocity vector as measured in the 0 frame has components
2
2
2
2
2
1
1
x vtx
v
cy y
z z
vxt
ctv
c
Now from equations (1-13) we know that the relation between the primed and the unprimed coordinates and times is
Take the differential of these equations remembering that v is a constant This gives
2
2
2
2
2
1
1
dx vdtdx
v
cdy dy
dz dz
vxdt
cdtv
c
2
2
2 2 2 2 2
2
2
2
11 ( )
1( ) 1
1
1
1
dx vdt
v dxdx vdt v
dx dx vdtc dt dtvdx vdx vdx v dxdt dt dt dt
dt dtc c c c
VxV x
Vx
VxV x
vv
vv
c
x
v
c
cV
( )
coordinate of particle relative to moving primed system
coordinate of particle relative to stationary system
velocity of primed coordina
V x
Vx
te system relative to stationaryv sys
tem
Page 28 Fundamentals of Modern Physics
(1-16)
According to Einstein this equation tells you how to transform the observed velocity from one frame of reference to another frame of reference First we note that as Vc and vc approachzero the equation approaches those which would be derived from the Galilean transformation
2 2
( ) 1 01 1 ( )
Vx v Vx vV
VxV x
Vx v
v cVxx
v
c
v
cc
Another property of this equation is that it is impossible to choose V and v such that Vrsquo the magnitude of the velocity which is seen in the new frame is greater than c
1
2
09 ( 09 ) 18009876543
( 09 ) 1811 (09 )
c c cV c c
cc
c
Example As seen by O particle 1 has velocity 09c in the direction of positive x and particle 2 has velocity 09c in the negative x direction
Einstein equation however predicts To evaluate the velocity of particle 1 with respect to particle 2 we transform from the O frame to the Orsquo frame moving in the negative x direction with velocity v = -09c using equation 1-16 We obtain
JK LOGIC will observe the particles separating at 18c
A confusing INCORRECT and ILLOGICAL answer
left is a dir rightsub of a
1
2
1
09 ( 09 ) 18009876543
181( 09 )1 ( 09 )
The multiplication of dir left time a dir right is ONLY a dir left
09
dir
direction leftdirection rightsubtraction
c c cV c c
cc
c
V
dir
22
22
09 18 18009876543
181( 081)1(09)(09)
1
left rightsub
sub
c c c cc c
ccc
c
Analyzing Einsteinrsquos BS math Equation
Using SM math a logical answer can be obtained
Cannot multiply a subtraction operator times a direction in space
Can only subtract and change its direction
Cannot multiply a direction to the left times a direction to the right
Cannot multiply opposite directions in space
Can only solve for the resultant
rightleft
subtraction
1
subtraction
2
sub of a dir to the left is a dir to the right
1 resultant 0 sub 2
2sub 2
2
09 09
09 amp 091
09 09 18 1
(0)1
(09) amp (09)1
c cV
c cc
c c cV
cc
cc
8
181
cc
SM math predicts a logical answer using logic
21st century scientists state in many articles that Einsteins General Relativity and Quantum Mechanics are incompatible
They are not only incompatible they are both mathematically incorrect
They both use illogical BS math
Any equations in math that use imaginary numbers should be abandoned
Imaginary numbers are only necessary in the current BS math system
This means that Schroumldingers equations should be abandoned
Any equation in quantum mechanics that uses imaginary number should be abandoned
Prof Kanarev has developed logical physics and chemistrytheories without the use of Quantum mechanics or relativity
A partial list of his theories coverbull Mechanics ndash replaces Newtonbull Photon - bull Electrons ndash Protons- Neutronsbull Spectroscopybull Nucleusbull Atoms amp Moleculesbull Electrodynamicsbull Heatbull Curvature of Space ndash replaces Schwarzcchildrsquos equation
Kanarevrsquos Hypothesis
bull Electron align linearly with proton They do NOT spin or form clouds around the nucleus
bull Neutrons are in the center of atom Proton are outside of Neutrons
bull The wavelength of all particles is the experimental radius
bull Double slit experiment explained correctly
bull Heisenbergrsquos inequality explained correctly
bull is the radius of approach of the magnetic lines around the torus of the electron
( 1 137)
bull Why water expands when it freezes
bull Electron clusters form lightning
rbull The meaning of Plancks constant is explained
22 2 kg m
h m f mr f consts
2210254 42m r kgm bull If the mr-constant describes a ring the equations
describe experiments bull If the mr-constant describes a wave the equations do
NOT describe experiments
Kanarevrsquos 5 Equations of Mechanics
Kanarevrsquos 1st Equation Acceleration ampmv maMma R R
Kanarevrsquos 2nd Equation Uniform motion amp mv maMma R R
maM mvR R
Kanarevrsquos 3rd Equation
Kanarevrsquos 4th Equation Equal and opposite reaction
1 21 21 2
12
a m
m a m ama
Kanarevrsquos 5th Equation 1 2 3 Mi M M M Mna R R R R R
The sum of the Actions working on a moving body is never equal to zero
Kanarevrsquos Photon model is a rolling ring
His equations accurately describe known experiments with photonsbull Electromagnetic radiation should be called photon radiationbull Diffraction and interference are the same phenomenabull Correctly explains the double slit experiment without quantum equationsbull Correctly explains the meaning of Heisenbergrsquos inequality
e
2 (2)(3142)(2817 15 ) 10007297352568
R 2426 12 137mag
xp
r m
m
Charge
Magnetic action
Proton
The intensity of the magnetic action near the geometrical center of the proton ring is so great that we theorize that it is the strong nuclear force
(9383 8)(1602 19)8476 14
4 (4)(3142)(1411 26)p
pp
EB T
M
(9395 8)(1602 19)1035 15
4 (4)(3142)(0966 26)N
NN
EB T
M
Neutron
1a Hydrogenavi
opposite direction subtraction direction
SM ( ) ( )
BS (negative)(positive) = (negative)
Symmetry Math The subtraction of a direction is equal to the opposite direction
There is no multiplication of a subtractionoperator times a direction in space
( ) multiplied by a ( ) ( )
positive positive positiveaddition addition additioncross cross cross
direction direction direction
What is the meaning of multiplying addition operators Addition operators are not multiplication operators
BS Math
() multiplied by a direction ( )=
number
() multiplied by a direction ( )=
number
A number times a direction maintains the same direction
( ) ()
dirdiradd
( ) ()
dirdiradd
The addition of a direction is in the same direction There is no multiplication by addition operators
Symmetry Math
Future math books need to eliminate the use of the cross sign (+) to represent something labeled a positive direction in space and the dash sign (-) to represent something labeled a negative direction in space Space does not have positive and negative directions
Symbols that are logical and have no illogical representation should be adopted SM uses either arrows or symbols
Symmetry-Math Number Line
Symmetry-Math Rules
The dash sign (-) will have only one use subtraction
The cross sign (+) will have only one use addition
Directions have an arrow and a number
All observers see the same direction and magnitude
bull There is no multiplication of arrows
amp ampamp ampampamp
( 3 ) ( 5 ) ( 6 ) ( 65 ) (17256 ) ( 32993 )
dir dirdir dirdirdir
Negative and Positive Directions in Space
If I asked you to point to a negative direction in space which way would you point
Hopefully you will realize that there is no such thing as a negative direction in space
If I asked you to point to a positive direction in space which way would you point
Again hopefully you will realize that there is no such thing as a positive direction in space
Objects on the right side of the x-axis are positive (+)(+) = (+)
Objects on the left side of the x-axis are negative (-)(-) = (+) negative math is different from positive math This is illogical
Objects on the top of the y-axis are positive
Objects on the bottom of the y-axis are negative
Objects in the front of the z-axis are positive
Objects in the back of the z-axis are negative
Again BS math of the positive direction (x right y up and z front) is different from the BS math of the negative direction (x left y down and z back)
BS math is Broken-Symmetry
Mirror image is broken symmetry in the x and z axis If the left amp right-axis are
reversed symmetry is broken Obs-1 math is different from obs-2
If the front amp back-axis is reversed symmetry is broken Obs-1 math is different from obs-2
If the top amp bottom-axis are
reversed symmetry is not broken Obs-1 math is the same as obs-2
Math answers cannot depend on which side of a number line an observer sit
Absolute values for Displacement
Even for a simple displacement of an object BS math invented a definition and absolute values to obtain something labeled a positive number answer
I----------I----------I----------I----------I----------I----------I----------I----------I
-4 -3 -2 -1 0 +1 +2 +3 +4 a gtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtb cltltltltltltltltltltltltltltltltltltltltltltltltltltltb
If an object starts at ( 3)ia X and moves to ( X 3) and the to ( 1)b fb c X
BS Math provides a usable answer without a definition or absolute values
However If an object starts at
( 1) and moves to ( X 3) and the to ( 3)i b fa X b c X
Then absolute values and a definition must be used
---------I----------I----------I----------I----------I----------I----------I---------- -4 -3 -2 -1 0 +1 +2 +3 +4
a gtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtb cltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltb
BS Math makes up a definition that all displacements are positive Since this example provides an answer with something labeled a negative the positive definition must be applied and (-2) is changed to (+2) A definition is needed to arrive at a useable answer This is Illogical
In SM absolute values and a meaningless definition of ldquopositiverdquo are removed In SM objects and directions in space are defined by the direction and magnitude of the resultant of arrows
With SM you get total distance traveled by the object and the final direction of the displacement
22222 2 bababbaababababa If w e su b stitu te p o s itiv e n u m b e rs in to th e eq u a tio n th e an sw er w ill b e co rrec t If (a = 5 ) an d (b = 3 )
2
2
8
5 3 5 3 5 3 5 5 5 3 3 5 3 3 25 15 15 9 64
8 8 8 8 8 8 8 8 8 8 64
W h en tw o p o sitiv e n u m b ers a re a d d ed an d sq u a re d th e B S m a th d is tr ib u tiv e law p ro v id es a co rre c t an sw er 1
Illogical and Incorrect BS math for the distributive law
222 2 bababbabbaaabababa B y d e fin itio n in B S m a th n o n u m b er sq u a red c a n ev e r b e n eg a tiv e T h e re fo re n o sq u a red n u m b er c an ev e r b e a v e c to r m o v in g in th e d a sh ( -) o r n eg a tiv e d ire c tio n if it is sq u a red In B S m a th an d ad d itio n o p e ra to r m u ltip lied b y a su b trac tio n o p e ra to r is A L W A Y S eq u a l to a su b tra c tio n o p e ra to r If th is h ad n o t b e en u sed fo r th e la s t fo u r h u n d red y e a rs y o u w o u ld d ie lau g h in g a t th e u tte r ab su rd ity o f th e la s t s ta tem e n t T h is is illo g ica l an d v io la te s sy m m etry
If a = 5 an d b = 3 42222352
22
4930253)3)(5(253)5)(3()3)(5(5353535 22222 A s lo n g a s ldquo ardquo is g re a te r th an ldquob rdquo th e B S m a th d is tr ib u tiv e law p ro v id es co rrec t an sw ers 1
B S m a th b re a k s d o w n w h e n ldquo b rdquo is g rea ter th a n ldquo a rdquo
If a = 3 an d b = 5
2 2
2
2
3 5 2 4
2 ( 2) ( 2) 4
2 ( )( 2) ( )( 2) 4
A n y co m b in a tio n ex cep t th e m id d le o n e is illo g ica l an d in co rre c t B y d e fin itio n th e an sw er can n o t b e -4 in B S m a th B S m a th is in co rrec t
4253095)5)(3(235)3)(5()5)(3(3535353 22222 T h is is w h ere th e d is tr ib u tiv e la w p ro v id es in co rrec t an sw ers I t p ro d u ces an an sw er o f + 4 b e cau se b y d e fin iti o n n o n u m b er sq u a red in B S m a th c an b e a n eg a tiv e n u m b er U sin g in co rre c t B S m a th th e d is tr ib u tiv e law w ill g iv e an in co rrec t an sw ers o f + 4 T h is is in co rrec t an d illo g ica l 1
Since there can be no negative direction from a squared term in BS math BS math should be modified or abandoned
The BS distributive law should be modified or abandoned
The numbers +1+2+3+4 and ndash1-2-3-4hellip should be abandoned
BS math gives incorrect answers because their Rule-of-Signs allows multiplying an arrow going in one direction by an arrow going in another direction
Clearly this is not logical
Arrows should be used that are specific for directions
W h at d o es it m e an to m u ltip ly ( 4) b y (4 ) A s w e k n o w 1
m u ltip lica tio n is ju s t ad d itio n W h a t w o u ld ( 4) tim e (4 ) b e 2
In BS math the answer is by definition -16 unless you square -4 and then it is +16
In SM you cannot multiply arrows going in opposite directions
The middle two terms are not logical You cannot multiply opposite directions
This is a MAJOR error in BS math They multiply a dash [(-) a direction to the left] by a cross [(+) a direction to the right] And then ldquoby definitionrdquo label the answer dash (-) This is illogical and produces incorrect answers
2
2 2
This is Einsteinrsquos math error
in Special relativity
Magnitude 4 Direction 1 Magnitude 3 Direction 2 Magnitude 5 Direction 3 Relative to an observer at 0 4
Magnitude 4 Direction 1 Magnitude 3 Direction 2 Magnitude 5 Direction 3 Relative to an observer at 0 4
Magnitude and Direction (MD) An arrow has both a magnitude andA direction relative to a 2nd objectobserver
bull Magnitude and velocity (MV) relative to a 2nd objectobserver
bull Magnitude And acceleration (MA) relative to a 2nd objectobserver2
v da
t t
Introducing a new symbol the ldquoamprdquo sign The ldquoamprdquo sign means resultant ADD all the Arrows for each direction and then SUBTRACT to find the final resultant
1 1amp 1 2 2amp 2 4 If =
1 amp1 2 2 amp2 4 If =
amp 5 amp 3 2 If gt
amp 3 amp 5= 2 If gt
amp 0 3 amp 3=0 if =
only the resultant equals zero
the vectors do not - they are still there
Does not exist There is no multiplication of arrows
There are two arrows One has a magnitude of 3 One has a magnitude of 5 The maximum these two arrows can be is 8
Only the resultant of arrows exist
An SM arrow has a direction and a magnitude
Arrows and magnitudes are still there even if the resultant is zero
There are no negative SM Arrows
There are no positive SM Arrows
There are no ZERO SM Arrows A zero would negate the definition of an SM Arrow something with no magnitude and no direction cannot be the definition of an arrow that is defined as having magnitude and direction
bull A resultant is the addition and subtraction of the differences in magnitude and direction arrows
A B and C are the arrows (a) A amp B exist(b) A amp B amp C exist(c) A-B Do Not Exist in Sm there are no negatives or positives(d) A+B-C Do Not Exist in SM there are no negatives or positives
There are only resultants when there are multiple arrows
SM Graphs versus BS Graphs
Numbers Raised to Powers
When numbers are raised to powers the negative sides of all BS math graphs are ILLOGICAL and INCORRECT
Incorrect BS Math for 2x
2x
4 16 correct3 8 correct2 4 correct1 2 correct0 1 correct-1 005 incorrect-2 025 incorrect-3 0125 incorrect-4 00625 incorrect
negative side of graph is incorrectpositive side of graph is correct
16
8
42105025012500625
-4 -3 -2 -1 0 1 2 3 4
po
sit
ive
negative positive
2x BS Math
correctincorrect
In BS math the dash sign (-) used in this example represents three different operations They arebull (-) Subtractionbull (-) A direction to the leftbull (-) As an exponent to mean divide by
All positive numbers produce correct results and graphs
All negative numbers produce incorrect results and graphs
A direction of three units to the left (-3) is placed in the equation as an exponent
A direction in space (3 units toward the left) changes a whole number to a fraction by using the dash (-) as an exponent rather than a direction in space
This is illogical and incorrect BS math should have performed this operation as
Unfortunately the answers to problems can come out correct because data points of a real problem fit the curve
However the answers will not be due to logical reasoning
Because the equation produced a curve that real problem data points follow it has been falsely believed that the equations and graphs are correct
1
2
x
Incorrect BK graph for (12)x
(12)x
4 00625 correct3 0125 correct2 025 correct1 1 correct0 1 correct-1 2 incorrect-2 4 incorrect-3 8 incorrect-4 16 incorrect
negative side of graph is incorrectpositive side of graph is correct
00625012502505124
8
16
-5 -4 -3 -2 -1 0 1 2 3 4 5P
osit
ive
negative positive
BS Math for (12)x
correctincorrect
Incorrect BS math and graph for
All graphs are symmetrical with SM
Two BS Math Equations that Produce Incorrect Answers
2 0x x starting equation 2
2
x x x x
x x
Add lsquoxrsquo to both sides
2(1) 1 let lsquoxrsquo = 1
2 1 This is an incorrect answer 2(0)-(0)=0 The only answer that BK math can give is x=0
1
a b starting equation ( )( ) ( )( )a a a b multiply both sides by ( )a
( )( ) ( )( ) ( )( ) ( )( )a a b b a b b b subtract ( )( )b b from both sides
( )( ) ( )( )a b a b a b b factor ( )a b from both sides
( )a b b divide both sides by ( )a b
( )b b b replace ( )a with ( )b 2b b simplify 2(1) (1) Let ( )b = 1
2 1 Again BS math yields and incorrect answer 1
This incorrect result is a problem with the distributive law SM explains how this incorrect answer is obtained and the problem with the distributive law that allows the error
Page 148 ndash149 Barry Mazur ndash Imagining Numbers
3
2
31
Do the computation carefully on paper using the rules we agreed to Then ponder your answer which should be something of a surprise to you donrsquot stop there Think of what your answer might possibly mean or might imply
Solve this equation using the distributive law
JK definitely a surprise the answer wasincorrect and illogical
Gerhard Gade University Professor doing math at Harvard University
Here is the BS math answer that Barry Mazur calculated
2
31
2
31
2
31
1 3 1 3 1 3 3 3 2 2 3 1 3
2 2 2 2 4 4 4 4 4 4 2 2
dash
1 3 1 3 1 31
2 2 2 2 4 4
3 3
4 4
When the distributive law is used with the BS Rule-of-Signs and one of the signs is a (-) dash sign the middle two terms always cancel each other This leads to the incorrect answer of (-1) in the above equation
Barry Mazur did not go back and check his answer with the original equation If he had he would have known that the answer was incorrect
3
1 31
2
check answer by setting equal
3
331 3
12
1 31
2
Take the Cube root of both sides
1 3 2 multiply both sides by 2
3 3 subtract 1 from both sides
2 2
3 3 square both sides
3 9 This is an incorrect answer A simple equation check could have shown that the distributive law using BS math provides an incorrect answer
1
BS Illogical and Incorrect Numbers Raised to Powers
nC D n DC
642 DDD
In BS math only a positive number can have an answer if ldquonrdquo is an even number (246---)
All negative numbers have no solutions
642 DDD
With SM just add a direction arrow and any number gives the correct answer
or n nC D C D
3 3( 3 8 8 or 8 2 or 2 )n D C
( 2 16 C= 16 or 16 4 or 4 )n D
Since a negative number has no square root answer in BS math BS mathematicians invented imaginary numbers
i2=-1
24414 iiC n
These equations are easily solved without the use of imaginary numbers with SM math
If 4 then 4 2x
If 4 then 4 2x
The 360-degree circle is a 4400 years old extremely outdated system based on a year of 12 months of 30 days each that is 360 days
SM will use a more logical system that will have a circle of 6283185307o
d
( )
(deg )
d r radians
d r rees
the arc distance around
the circumference of a circle
d
radiusr
2 2
2 2
d r
dv rd t t
v dda rd t t t
1
d
r
( )av tv
The average angular velocity of a body is the rate of its angular displacement
( )0
liminstt
vt
The instantaneous angular velocity is the limit of the ratio
t
as t approaches zero
( )d
av
d
tv
The average circumference velocity of the body is the rate of its distance traveled d divided by the time t
dvv
r
1 dd vdrv
t t t r r
The equation of the relationship between the angular velocity and the velocity of a point on that body
( )o
av
v v v
t ta
The average angular acceleration is the change in the angular velocity
( )0
liminstt
v
ta
The instantaneous angular acceleration is the
limit of the ration of v
t
as t approaches
zero
daa
r
1
1
d
d s d
vva
t r tv a
r t r
The equation of the relationship between the angular acceleration and the acceleration of a point on the body
21
2v a t
If a body with an initial angular velocity ov has a constant
angular acceleration a it will turn through an angle in a time t
1
Straight Line Symbol Units Rotational Symbol Units Distance d m Angular
Distance radians or degrees
Velocity dt
v m
s Angular Velocity
V
t
radians or degrees
s
Acceleration vat
2ms
Angular Acceleration
VA
t
2
radians or degrees
s
Mass M Kg Mass distribution
dM 2Kgm
Force actionA Ma
2
Kgm
sNewton
Torque
dT M A 2
2
Kgm
s
Momentum Mv Kgm
s
Angular Momentum
dM V 2Kgm
s
Work (Force)(distance)
Ad 2
Kgmm
sNm Joule
Work T
Using AS arrow math to solve and direction in space problems using the equation 1 2
2 2 2 2 2 2( ) amp and ( ) amp 3
4
2 2 2
2 2
( ) amp
(5 ) 4 3
(5 ) 16 9
(5 ) 25
(5 ) (5 )
2 2(5 ) 4 3
(5 ) 16 9
(5 ) 25
(5 ) (5 )
In SM math Pythagorasrsquo theorem provides both magnitude and direction
In BS math Pythagorasrsquo theorem provides magnitude only
time set equal to zero
23
( ) ( 0 )10 1
x t xx
3( ) ( 0)
10 1W t f
In BS math this is an equation for a traveling wave
must use a square term in order to make the curve symmetric
NO SQUARE TERM IS NECESSARY
In SM math when t is set to ldquo0rdquo
This is a major fault in BS math
x f(x)
-4 0019
-3 0033
-2 0073
-1 0273
0 3000
1 0273
2 0073
3 0033
4 0019 -4 -3 -2 -1 0 1 2 3 400
05
10
15
20
25
30
35
BS wave (must use square term)
snap shot of wave at x = 0
Am
plit
ude
lt to gt lt to gt f(ltgt)
lt -4 0073
lt -3 0097
lt -2 0143
lt -1 0273
origin 0 3000
gt 1 0273
gt 2 0143
gt 3 0097
gt 4 0073lt0073 lt0097 lt0143 lt0273
3000
0273gt 0143gt 0097gt 0073gt
Wave moving
snap shot of wave
Am
plit
ude
lt4 lt3 lt2 lt1 0 1gt 2gt 3gt 4gt
3
2
1
In SM NO squared terms are needed
SPRING
An Action is needed to stretch or compress a spring to a starting point where it can execute cycle motion
tsA (ts to start) is the work required to move a mass attached to a spring a distance (d1) from
its initial rest position RAK 0 the spring constant is the stiffness of the spring
o A is attraction o R is repulsion o
ARRA
00 is for the back and forth motion through the zero initial rest position A
stretched spring goes from ldquoArdquo to ldquo0rdquo to ldquoRrdquo to ldquoArdquo to ldquo0rdquo to ldquoRrdquo d1 is the distance a mass is moved from its initial rest position
0
0 1ts A RA K d
Once ( )mad
is performed to move the mass to position d1 (against the restraint of the spring) a of the ( )mad
is transferred into the spring as potential attractionrepulsion (PAR) There is always
more ( )mad
performed than is stored as PAR When the spring is released from the restraining Ats the PAR starts the oscillation Conservation of ( )mad
requires that the PAR and the motion ( )mad
( )( )madM be a constant
( )amp ( ) constantmadPAR M 02
0 0
1
2A R A RPAR K d
02
( ) 0 0
1( )
2mad A R A RM m v
0 0 02 2 2
0 0 0 1
2 2 21
2 2 21
1 1 1amp
2 2 2
1 1 1amp
2 2 2
amp
A R A R A RK d m v K d
Kd mv Kd
Kd mv Kd
0
1d is frac12 the height from the base to the peak of a cycle
An experiment of moving paper under a pendulum through a small angle filled with sand that can escape through a small hole in the bottom shows that a pendulum cycling back and forth follows a curve that is mathematically analogous to a lsquosinrsquo or lsquocosrsquo
The projection of the curved area onto the straight line exactly duplicate the back and forth motion of CM This allows for the math of lsquosinrsquo or lsquocosrsquo to be used
SM will use lsquosinrdquo math It could use lsquocosrsquo or a combination of both lsquosinrsquo and lsquocosrsquo Using just the lsquosinrsquo math makes the understanding and math easier
0 sin 0 0SM 0 sin157 1SM The diagram at the left shows a spring being stretched from its (zero rest) position-2 and the PAR being released at position-1 There are 4 sections (1 to 2 2 to 3 3 to 4 and 4 to5) that alternate between maximum acceleration and zero velocity to max velocity and zero acceleration
At position-1 there is maximum acceleration and zero velocity o The lsquosinrsquo values from 0o to 157o will be used for velocity (values from 0 to 1) o The lsquosinrsquo values from 157o to 0o will be used for acceleration (values from 1 to 0)
At position-2 there is maximum velocity and zero acceleration Acceleration changes to deceleration
o The lsquosinrsquo values from 157o to 0o will be used for deceleration (values from 1 to 0) o The lsquosinrsquo values from 0o to 157o will be used for acceleration (values from 0 to 1)
At position-3 there is maximum acceleration and zero velocity o The lsquosinrsquo values from 0o to 157o will be used for velocity (values from 0 to 1) o The lsquosinrsquo values from 157o to 0o will be used for acceleration (values from 1 to 0)
At position-4 there is maximum velocity and zero acceleration Acceleration changes to deceleration
o The lsquosinrsquo values from 157o to 0o will be used for deceleration (values from 1 to 0) o The lsquosinrsquo values from 0o to 157o will be used for acceleration (values from 0 to 1)
At position-5 the cycle starts again
2
)539(
cT
MK
spring constant (K) when mass (M) and cycle-time (T) are known
)(
))(89)((
111 md
uKgM
d
MG
d
FK sts
spring constant (K) when mass (M) and amplitude ( d1) are known
2 21( )
Kv d d
m
velocity (v) at any point during the cycle
velocity
dforM
Kdv
max
012 velocity (v2) maximum (when d = 0 at position-2 and 4)
acceleration (a1) maximum when spring constant (K) mass (M) and amplitude (d1) are known
acceleration (a1) maximum when amplitude (d1) and cycle-time (Tc) are known
acceleration (a) at any point during the cycle-time (T)
1
151
286
v
dTc
cycle-time (Tc) when amplitude (d1) and velocity (v1) are known
K
MTc 28651
cycle-time (Tc) when mass (M) and spring constant (K) are known
fTc
1
cycle-time (Tc) when the frequency is known
cTf
1
frequency (f) when the cycle-time (Tc) is known
of cycles
elapsed timef
frequency (f) when the number of cycles in an elapsed time is known
d1 amplitude (distance from rest to start position)
Also frac12 the distance from the base to the peak
3131 oror dM
Ka
251
3131 539
c
oror T
da
2539
cT
da
In December of 2006 I discovered the mathematical error in Einsteinrsquos Special Relativity Einstein used BS math and that math only works in an imaginary world
It cannot describe real symmetrical space
Einsteinrsquos Special Relativity should be abandoned and a theory of real space using SM should be used to replace it
Numerous articles on the Internet show Einsteinrsquos theory to be incorrect but no one in academia is willing to publish the information
Figure 1-10 A moving particle observed from two frames of reference in uniform translation
Einsteinrsquos Math Error in Special Relativity Transformation of Velocity (From Eisenberg- Fundamentals of Modern Physics pages 27-29) Consider the particle shown in figure (1-10) moving with velocity v as
seen in a frame of reference O
dx dy dz
Vx Vy Vzdt dt dt
dx dy dz
V x V y V zdy dt dt
We would like to evaluate the velocity v of the particle as seen in the frame of reference 0 which is itself moving relative to 0 with velocity v Measured in the 0 frame the velocity vector of the particle has com ponents
The same velocity vector as measured in the 0 frame has components
2
2
2
2
2
1
1
x vtx
v
cy y
z z
vxt
ctv
c
Now from equations (1-13) we know that the relation between the primed and the unprimed coordinates and times is
Take the differential of these equations remembering that v is a constant This gives
2
2
2
2
2
1
1
dx vdtdx
v
cdy dy
dz dz
vxdt
cdtv
c
2
2
2 2 2 2 2
2
2
2
11 ( )
1( ) 1
1
1
1
dx vdt
v dxdx vdt v
dx dx vdtc dt dtvdx vdx vdx v dxdt dt dt dt
dt dtc c c c
VxV x
Vx
VxV x
vv
vv
c
x
v
c
cV
( )
coordinate of particle relative to moving primed system
coordinate of particle relative to stationary system
velocity of primed coordina
V x
Vx
te system relative to stationaryv sys
tem
Page 28 Fundamentals of Modern Physics
(1-16)
According to Einstein this equation tells you how to transform the observed velocity from one frame of reference to another frame of reference First we note that as Vc and vc approachzero the equation approaches those which would be derived from the Galilean transformation
2 2
( ) 1 01 1 ( )
Vx v Vx vV
VxV x
Vx v
v cVxx
v
c
v
cc
Another property of this equation is that it is impossible to choose V and v such that Vrsquo the magnitude of the velocity which is seen in the new frame is greater than c
1
2
09 ( 09 ) 18009876543
( 09 ) 1811 (09 )
c c cV c c
cc
c
Example As seen by O particle 1 has velocity 09c in the direction of positive x and particle 2 has velocity 09c in the negative x direction
Einstein equation however predicts To evaluate the velocity of particle 1 with respect to particle 2 we transform from the O frame to the Orsquo frame moving in the negative x direction with velocity v = -09c using equation 1-16 We obtain
JK LOGIC will observe the particles separating at 18c
A confusing INCORRECT and ILLOGICAL answer
left is a dir rightsub of a
1
2
1
09 ( 09 ) 18009876543
181( 09 )1 ( 09 )
The multiplication of dir left time a dir right is ONLY a dir left
09
dir
direction leftdirection rightsubtraction
c c cV c c
cc
c
V
dir
22
22
09 18 18009876543
181( 081)1(09)(09)
1
left rightsub
sub
c c c cc c
ccc
c
Analyzing Einsteinrsquos BS math Equation
Using SM math a logical answer can be obtained
Cannot multiply a subtraction operator times a direction in space
Can only subtract and change its direction
Cannot multiply a direction to the left times a direction to the right
Cannot multiply opposite directions in space
Can only solve for the resultant
rightleft
subtraction
1
subtraction
2
sub of a dir to the left is a dir to the right
1 resultant 0 sub 2
2sub 2
2
09 09
09 amp 091
09 09 18 1
(0)1
(09) amp (09)1
c cV
c cc
c c cV
cc
cc
8
181
cc
SM math predicts a logical answer using logic
21st century scientists state in many articles that Einsteins General Relativity and Quantum Mechanics are incompatible
They are not only incompatible they are both mathematically incorrect
They both use illogical BS math
Any equations in math that use imaginary numbers should be abandoned
Imaginary numbers are only necessary in the current BS math system
This means that Schroumldingers equations should be abandoned
Any equation in quantum mechanics that uses imaginary number should be abandoned
Prof Kanarev has developed logical physics and chemistrytheories without the use of Quantum mechanics or relativity
A partial list of his theories coverbull Mechanics ndash replaces Newtonbull Photon - bull Electrons ndash Protons- Neutronsbull Spectroscopybull Nucleusbull Atoms amp Moleculesbull Electrodynamicsbull Heatbull Curvature of Space ndash replaces Schwarzcchildrsquos equation
Kanarevrsquos Hypothesis
bull Electron align linearly with proton They do NOT spin or form clouds around the nucleus
bull Neutrons are in the center of atom Proton are outside of Neutrons
bull The wavelength of all particles is the experimental radius
bull Double slit experiment explained correctly
bull Heisenbergrsquos inequality explained correctly
bull is the radius of approach of the magnetic lines around the torus of the electron
( 1 137)
bull Why water expands when it freezes
bull Electron clusters form lightning
rbull The meaning of Plancks constant is explained
22 2 kg m
h m f mr f consts
2210254 42m r kgm bull If the mr-constant describes a ring the equations
describe experiments bull If the mr-constant describes a wave the equations do
NOT describe experiments
Kanarevrsquos 5 Equations of Mechanics
Kanarevrsquos 1st Equation Acceleration ampmv maMma R R
Kanarevrsquos 2nd Equation Uniform motion amp mv maMma R R
maM mvR R
Kanarevrsquos 3rd Equation
Kanarevrsquos 4th Equation Equal and opposite reaction
1 21 21 2
12
a m
m a m ama
Kanarevrsquos 5th Equation 1 2 3 Mi M M M Mna R R R R R
The sum of the Actions working on a moving body is never equal to zero
Kanarevrsquos Photon model is a rolling ring
His equations accurately describe known experiments with photonsbull Electromagnetic radiation should be called photon radiationbull Diffraction and interference are the same phenomenabull Correctly explains the double slit experiment without quantum equationsbull Correctly explains the meaning of Heisenbergrsquos inequality
e
2 (2)(3142)(2817 15 ) 10007297352568
R 2426 12 137mag
xp
r m
m
Charge
Magnetic action
Proton
The intensity of the magnetic action near the geometrical center of the proton ring is so great that we theorize that it is the strong nuclear force
(9383 8)(1602 19)8476 14
4 (4)(3142)(1411 26)p
pp
EB T
M
(9395 8)(1602 19)1035 15
4 (4)(3142)(0966 26)N
NN
EB T
M
Neutron
1a Hydrogenavi
( ) multiplied by a ( ) ( )
positive positive positiveaddition addition additioncross cross cross
direction direction direction
What is the meaning of multiplying addition operators Addition operators are not multiplication operators
BS Math
() multiplied by a direction ( )=
number
() multiplied by a direction ( )=
number
A number times a direction maintains the same direction
( ) ()
dirdiradd
( ) ()
dirdiradd
The addition of a direction is in the same direction There is no multiplication by addition operators
Symmetry Math
Future math books need to eliminate the use of the cross sign (+) to represent something labeled a positive direction in space and the dash sign (-) to represent something labeled a negative direction in space Space does not have positive and negative directions
Symbols that are logical and have no illogical representation should be adopted SM uses either arrows or symbols
Symmetry-Math Number Line
Symmetry-Math Rules
The dash sign (-) will have only one use subtraction
The cross sign (+) will have only one use addition
Directions have an arrow and a number
All observers see the same direction and magnitude
bull There is no multiplication of arrows
amp ampamp ampampamp
( 3 ) ( 5 ) ( 6 ) ( 65 ) (17256 ) ( 32993 )
dir dirdir dirdirdir
Negative and Positive Directions in Space
If I asked you to point to a negative direction in space which way would you point
Hopefully you will realize that there is no such thing as a negative direction in space
If I asked you to point to a positive direction in space which way would you point
Again hopefully you will realize that there is no such thing as a positive direction in space
Objects on the right side of the x-axis are positive (+)(+) = (+)
Objects on the left side of the x-axis are negative (-)(-) = (+) negative math is different from positive math This is illogical
Objects on the top of the y-axis are positive
Objects on the bottom of the y-axis are negative
Objects in the front of the z-axis are positive
Objects in the back of the z-axis are negative
Again BS math of the positive direction (x right y up and z front) is different from the BS math of the negative direction (x left y down and z back)
BS math is Broken-Symmetry
Mirror image is broken symmetry in the x and z axis If the left amp right-axis are
reversed symmetry is broken Obs-1 math is different from obs-2
If the front amp back-axis is reversed symmetry is broken Obs-1 math is different from obs-2
If the top amp bottom-axis are
reversed symmetry is not broken Obs-1 math is the same as obs-2
Math answers cannot depend on which side of a number line an observer sit
Absolute values for Displacement
Even for a simple displacement of an object BS math invented a definition and absolute values to obtain something labeled a positive number answer
I----------I----------I----------I----------I----------I----------I----------I----------I
-4 -3 -2 -1 0 +1 +2 +3 +4 a gtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtb cltltltltltltltltltltltltltltltltltltltltltltltltltltltb
If an object starts at ( 3)ia X and moves to ( X 3) and the to ( 1)b fb c X
BS Math provides a usable answer without a definition or absolute values
However If an object starts at
( 1) and moves to ( X 3) and the to ( 3)i b fa X b c X
Then absolute values and a definition must be used
---------I----------I----------I----------I----------I----------I----------I---------- -4 -3 -2 -1 0 +1 +2 +3 +4
a gtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtb cltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltb
BS Math makes up a definition that all displacements are positive Since this example provides an answer with something labeled a negative the positive definition must be applied and (-2) is changed to (+2) A definition is needed to arrive at a useable answer This is Illogical
In SM absolute values and a meaningless definition of ldquopositiverdquo are removed In SM objects and directions in space are defined by the direction and magnitude of the resultant of arrows
With SM you get total distance traveled by the object and the final direction of the displacement
22222 2 bababbaababababa If w e su b stitu te p o s itiv e n u m b e rs in to th e eq u a tio n th e an sw er w ill b e co rrec t If (a = 5 ) an d (b = 3 )
2
2
8
5 3 5 3 5 3 5 5 5 3 3 5 3 3 25 15 15 9 64
8 8 8 8 8 8 8 8 8 8 64
W h en tw o p o sitiv e n u m b ers a re a d d ed an d sq u a re d th e B S m a th d is tr ib u tiv e law p ro v id es a co rre c t an sw er 1
Illogical and Incorrect BS math for the distributive law
222 2 bababbabbaaabababa B y d e fin itio n in B S m a th n o n u m b er sq u a red c a n ev e r b e n eg a tiv e T h e re fo re n o sq u a red n u m b er c an ev e r b e a v e c to r m o v in g in th e d a sh ( -) o r n eg a tiv e d ire c tio n if it is sq u a red In B S m a th an d ad d itio n o p e ra to r m u ltip lied b y a su b trac tio n o p e ra to r is A L W A Y S eq u a l to a su b tra c tio n o p e ra to r If th is h ad n o t b e en u sed fo r th e la s t fo u r h u n d red y e a rs y o u w o u ld d ie lau g h in g a t th e u tte r ab su rd ity o f th e la s t s ta tem e n t T h is is illo g ica l an d v io la te s sy m m etry
If a = 5 an d b = 3 42222352
22
4930253)3)(5(253)5)(3()3)(5(5353535 22222 A s lo n g a s ldquo ardquo is g re a te r th an ldquob rdquo th e B S m a th d is tr ib u tiv e law p ro v id es co rrec t an sw ers 1
B S m a th b re a k s d o w n w h e n ldquo b rdquo is g rea ter th a n ldquo a rdquo
If a = 3 an d b = 5
2 2
2
2
3 5 2 4
2 ( 2) ( 2) 4
2 ( )( 2) ( )( 2) 4
A n y co m b in a tio n ex cep t th e m id d le o n e is illo g ica l an d in co rre c t B y d e fin itio n th e an sw er can n o t b e -4 in B S m a th B S m a th is in co rrec t
4253095)5)(3(235)3)(5()5)(3(3535353 22222 T h is is w h ere th e d is tr ib u tiv e la w p ro v id es in co rrec t an sw ers I t p ro d u ces an an sw er o f + 4 b e cau se b y d e fin iti o n n o n u m b er sq u a red in B S m a th c an b e a n eg a tiv e n u m b er U sin g in co rre c t B S m a th th e d is tr ib u tiv e law w ill g iv e an in co rrec t an sw ers o f + 4 T h is is in co rrec t an d illo g ica l 1
Since there can be no negative direction from a squared term in BS math BS math should be modified or abandoned
The BS distributive law should be modified or abandoned
The numbers +1+2+3+4 and ndash1-2-3-4hellip should be abandoned
BS math gives incorrect answers because their Rule-of-Signs allows multiplying an arrow going in one direction by an arrow going in another direction
Clearly this is not logical
Arrows should be used that are specific for directions
W h at d o es it m e an to m u ltip ly ( 4) b y (4 ) A s w e k n o w 1
m u ltip lica tio n is ju s t ad d itio n W h a t w o u ld ( 4) tim e (4 ) b e 2
In BS math the answer is by definition -16 unless you square -4 and then it is +16
In SM you cannot multiply arrows going in opposite directions
The middle two terms are not logical You cannot multiply opposite directions
This is a MAJOR error in BS math They multiply a dash [(-) a direction to the left] by a cross [(+) a direction to the right] And then ldquoby definitionrdquo label the answer dash (-) This is illogical and produces incorrect answers
2
2 2
This is Einsteinrsquos math error
in Special relativity
Magnitude 4 Direction 1 Magnitude 3 Direction 2 Magnitude 5 Direction 3 Relative to an observer at 0 4
Magnitude 4 Direction 1 Magnitude 3 Direction 2 Magnitude 5 Direction 3 Relative to an observer at 0 4
Magnitude and Direction (MD) An arrow has both a magnitude andA direction relative to a 2nd objectobserver
bull Magnitude and velocity (MV) relative to a 2nd objectobserver
bull Magnitude And acceleration (MA) relative to a 2nd objectobserver2
v da
t t
Introducing a new symbol the ldquoamprdquo sign The ldquoamprdquo sign means resultant ADD all the Arrows for each direction and then SUBTRACT to find the final resultant
1 1amp 1 2 2amp 2 4 If =
1 amp1 2 2 amp2 4 If =
amp 5 amp 3 2 If gt
amp 3 amp 5= 2 If gt
amp 0 3 amp 3=0 if =
only the resultant equals zero
the vectors do not - they are still there
Does not exist There is no multiplication of arrows
There are two arrows One has a magnitude of 3 One has a magnitude of 5 The maximum these two arrows can be is 8
Only the resultant of arrows exist
An SM arrow has a direction and a magnitude
Arrows and magnitudes are still there even if the resultant is zero
There are no negative SM Arrows
There are no positive SM Arrows
There are no ZERO SM Arrows A zero would negate the definition of an SM Arrow something with no magnitude and no direction cannot be the definition of an arrow that is defined as having magnitude and direction
bull A resultant is the addition and subtraction of the differences in magnitude and direction arrows
A B and C are the arrows (a) A amp B exist(b) A amp B amp C exist(c) A-B Do Not Exist in Sm there are no negatives or positives(d) A+B-C Do Not Exist in SM there are no negatives or positives
There are only resultants when there are multiple arrows
SM Graphs versus BS Graphs
Numbers Raised to Powers
When numbers are raised to powers the negative sides of all BS math graphs are ILLOGICAL and INCORRECT
Incorrect BS Math for 2x
2x
4 16 correct3 8 correct2 4 correct1 2 correct0 1 correct-1 005 incorrect-2 025 incorrect-3 0125 incorrect-4 00625 incorrect
negative side of graph is incorrectpositive side of graph is correct
16
8
42105025012500625
-4 -3 -2 -1 0 1 2 3 4
po
sit
ive
negative positive
2x BS Math
correctincorrect
In BS math the dash sign (-) used in this example represents three different operations They arebull (-) Subtractionbull (-) A direction to the leftbull (-) As an exponent to mean divide by
All positive numbers produce correct results and graphs
All negative numbers produce incorrect results and graphs
A direction of three units to the left (-3) is placed in the equation as an exponent
A direction in space (3 units toward the left) changes a whole number to a fraction by using the dash (-) as an exponent rather than a direction in space
This is illogical and incorrect BS math should have performed this operation as
Unfortunately the answers to problems can come out correct because data points of a real problem fit the curve
However the answers will not be due to logical reasoning
Because the equation produced a curve that real problem data points follow it has been falsely believed that the equations and graphs are correct
1
2
x
Incorrect BK graph for (12)x
(12)x
4 00625 correct3 0125 correct2 025 correct1 1 correct0 1 correct-1 2 incorrect-2 4 incorrect-3 8 incorrect-4 16 incorrect
negative side of graph is incorrectpositive side of graph is correct
00625012502505124
8
16
-5 -4 -3 -2 -1 0 1 2 3 4 5P
osit
ive
negative positive
BS Math for (12)x
correctincorrect
Incorrect BS math and graph for
All graphs are symmetrical with SM
Two BS Math Equations that Produce Incorrect Answers
2 0x x starting equation 2
2
x x x x
x x
Add lsquoxrsquo to both sides
2(1) 1 let lsquoxrsquo = 1
2 1 This is an incorrect answer 2(0)-(0)=0 The only answer that BK math can give is x=0
1
a b starting equation ( )( ) ( )( )a a a b multiply both sides by ( )a
( )( ) ( )( ) ( )( ) ( )( )a a b b a b b b subtract ( )( )b b from both sides
( )( ) ( )( )a b a b a b b factor ( )a b from both sides
( )a b b divide both sides by ( )a b
( )b b b replace ( )a with ( )b 2b b simplify 2(1) (1) Let ( )b = 1
2 1 Again BS math yields and incorrect answer 1
This incorrect result is a problem with the distributive law SM explains how this incorrect answer is obtained and the problem with the distributive law that allows the error
Page 148 ndash149 Barry Mazur ndash Imagining Numbers
3
2
31
Do the computation carefully on paper using the rules we agreed to Then ponder your answer which should be something of a surprise to you donrsquot stop there Think of what your answer might possibly mean or might imply
Solve this equation using the distributive law
JK definitely a surprise the answer wasincorrect and illogical
Gerhard Gade University Professor doing math at Harvard University
Here is the BS math answer that Barry Mazur calculated
2
31
2
31
2
31
1 3 1 3 1 3 3 3 2 2 3 1 3
2 2 2 2 4 4 4 4 4 4 2 2
dash
1 3 1 3 1 31
2 2 2 2 4 4
3 3
4 4
When the distributive law is used with the BS Rule-of-Signs and one of the signs is a (-) dash sign the middle two terms always cancel each other This leads to the incorrect answer of (-1) in the above equation
Barry Mazur did not go back and check his answer with the original equation If he had he would have known that the answer was incorrect
3
1 31
2
check answer by setting equal
3
331 3
12
1 31
2
Take the Cube root of both sides
1 3 2 multiply both sides by 2
3 3 subtract 1 from both sides
2 2
3 3 square both sides
3 9 This is an incorrect answer A simple equation check could have shown that the distributive law using BS math provides an incorrect answer
1
BS Illogical and Incorrect Numbers Raised to Powers
nC D n DC
642 DDD
In BS math only a positive number can have an answer if ldquonrdquo is an even number (246---)
All negative numbers have no solutions
642 DDD
With SM just add a direction arrow and any number gives the correct answer
or n nC D C D
3 3( 3 8 8 or 8 2 or 2 )n D C
( 2 16 C= 16 or 16 4 or 4 )n D
Since a negative number has no square root answer in BS math BS mathematicians invented imaginary numbers
i2=-1
24414 iiC n
These equations are easily solved without the use of imaginary numbers with SM math
If 4 then 4 2x
If 4 then 4 2x
The 360-degree circle is a 4400 years old extremely outdated system based on a year of 12 months of 30 days each that is 360 days
SM will use a more logical system that will have a circle of 6283185307o
d
( )
(deg )
d r radians
d r rees
the arc distance around
the circumference of a circle
d
radiusr
2 2
2 2
d r
dv rd t t
v dda rd t t t
1
d
r
( )av tv
The average angular velocity of a body is the rate of its angular displacement
( )0
liminstt
vt
The instantaneous angular velocity is the limit of the ratio
t
as t approaches zero
( )d
av
d
tv
The average circumference velocity of the body is the rate of its distance traveled d divided by the time t
dvv
r
1 dd vdrv
t t t r r
The equation of the relationship between the angular velocity and the velocity of a point on that body
( )o
av
v v v
t ta
The average angular acceleration is the change in the angular velocity
( )0
liminstt
v
ta
The instantaneous angular acceleration is the
limit of the ration of v
t
as t approaches
zero
daa
r
1
1
d
d s d
vva
t r tv a
r t r
The equation of the relationship between the angular acceleration and the acceleration of a point on the body
21
2v a t
If a body with an initial angular velocity ov has a constant
angular acceleration a it will turn through an angle in a time t
1
Straight Line Symbol Units Rotational Symbol Units Distance d m Angular
Distance radians or degrees
Velocity dt
v m
s Angular Velocity
V
t
radians or degrees
s
Acceleration vat
2ms
Angular Acceleration
VA
t
2
radians or degrees
s
Mass M Kg Mass distribution
dM 2Kgm
Force actionA Ma
2
Kgm
sNewton
Torque
dT M A 2
2
Kgm
s
Momentum Mv Kgm
s
Angular Momentum
dM V 2Kgm
s
Work (Force)(distance)
Ad 2
Kgmm
sNm Joule
Work T
Using AS arrow math to solve and direction in space problems using the equation 1 2
2 2 2 2 2 2( ) amp and ( ) amp 3
4
2 2 2
2 2
( ) amp
(5 ) 4 3
(5 ) 16 9
(5 ) 25
(5 ) (5 )
2 2(5 ) 4 3
(5 ) 16 9
(5 ) 25
(5 ) (5 )
In SM math Pythagorasrsquo theorem provides both magnitude and direction
In BS math Pythagorasrsquo theorem provides magnitude only
time set equal to zero
23
( ) ( 0 )10 1
x t xx
3( ) ( 0)
10 1W t f
In BS math this is an equation for a traveling wave
must use a square term in order to make the curve symmetric
NO SQUARE TERM IS NECESSARY
In SM math when t is set to ldquo0rdquo
This is a major fault in BS math
x f(x)
-4 0019
-3 0033
-2 0073
-1 0273
0 3000
1 0273
2 0073
3 0033
4 0019 -4 -3 -2 -1 0 1 2 3 400
05
10
15
20
25
30
35
BS wave (must use square term)
snap shot of wave at x = 0
Am
plit
ude
lt to gt lt to gt f(ltgt)
lt -4 0073
lt -3 0097
lt -2 0143
lt -1 0273
origin 0 3000
gt 1 0273
gt 2 0143
gt 3 0097
gt 4 0073lt0073 lt0097 lt0143 lt0273
3000
0273gt 0143gt 0097gt 0073gt
Wave moving
snap shot of wave
Am
plit
ude
lt4 lt3 lt2 lt1 0 1gt 2gt 3gt 4gt
3
2
1
In SM NO squared terms are needed
SPRING
An Action is needed to stretch or compress a spring to a starting point where it can execute cycle motion
tsA (ts to start) is the work required to move a mass attached to a spring a distance (d1) from
its initial rest position RAK 0 the spring constant is the stiffness of the spring
o A is attraction o R is repulsion o
ARRA
00 is for the back and forth motion through the zero initial rest position A
stretched spring goes from ldquoArdquo to ldquo0rdquo to ldquoRrdquo to ldquoArdquo to ldquo0rdquo to ldquoRrdquo d1 is the distance a mass is moved from its initial rest position
0
0 1ts A RA K d
Once ( )mad
is performed to move the mass to position d1 (against the restraint of the spring) a of the ( )mad
is transferred into the spring as potential attractionrepulsion (PAR) There is always
more ( )mad
performed than is stored as PAR When the spring is released from the restraining Ats the PAR starts the oscillation Conservation of ( )mad
requires that the PAR and the motion ( )mad
( )( )madM be a constant
( )amp ( ) constantmadPAR M 02
0 0
1
2A R A RPAR K d
02
( ) 0 0
1( )
2mad A R A RM m v
0 0 02 2 2
0 0 0 1
2 2 21
2 2 21
1 1 1amp
2 2 2
1 1 1amp
2 2 2
amp
A R A R A RK d m v K d
Kd mv Kd
Kd mv Kd
0
1d is frac12 the height from the base to the peak of a cycle
An experiment of moving paper under a pendulum through a small angle filled with sand that can escape through a small hole in the bottom shows that a pendulum cycling back and forth follows a curve that is mathematically analogous to a lsquosinrsquo or lsquocosrsquo
The projection of the curved area onto the straight line exactly duplicate the back and forth motion of CM This allows for the math of lsquosinrsquo or lsquocosrsquo to be used
SM will use lsquosinrdquo math It could use lsquocosrsquo or a combination of both lsquosinrsquo and lsquocosrsquo Using just the lsquosinrsquo math makes the understanding and math easier
0 sin 0 0SM 0 sin157 1SM The diagram at the left shows a spring being stretched from its (zero rest) position-2 and the PAR being released at position-1 There are 4 sections (1 to 2 2 to 3 3 to 4 and 4 to5) that alternate between maximum acceleration and zero velocity to max velocity and zero acceleration
At position-1 there is maximum acceleration and zero velocity o The lsquosinrsquo values from 0o to 157o will be used for velocity (values from 0 to 1) o The lsquosinrsquo values from 157o to 0o will be used for acceleration (values from 1 to 0)
At position-2 there is maximum velocity and zero acceleration Acceleration changes to deceleration
o The lsquosinrsquo values from 157o to 0o will be used for deceleration (values from 1 to 0) o The lsquosinrsquo values from 0o to 157o will be used for acceleration (values from 0 to 1)
At position-3 there is maximum acceleration and zero velocity o The lsquosinrsquo values from 0o to 157o will be used for velocity (values from 0 to 1) o The lsquosinrsquo values from 157o to 0o will be used for acceleration (values from 1 to 0)
At position-4 there is maximum velocity and zero acceleration Acceleration changes to deceleration
o The lsquosinrsquo values from 157o to 0o will be used for deceleration (values from 1 to 0) o The lsquosinrsquo values from 0o to 157o will be used for acceleration (values from 0 to 1)
At position-5 the cycle starts again
2
)539(
cT
MK
spring constant (K) when mass (M) and cycle-time (T) are known
)(
))(89)((
111 md
uKgM
d
MG
d
FK sts
spring constant (K) when mass (M) and amplitude ( d1) are known
2 21( )
Kv d d
m
velocity (v) at any point during the cycle
velocity
dforM
Kdv
max
012 velocity (v2) maximum (when d = 0 at position-2 and 4)
acceleration (a1) maximum when spring constant (K) mass (M) and amplitude (d1) are known
acceleration (a1) maximum when amplitude (d1) and cycle-time (Tc) are known
acceleration (a) at any point during the cycle-time (T)
1
151
286
v
dTc
cycle-time (Tc) when amplitude (d1) and velocity (v1) are known
K
MTc 28651
cycle-time (Tc) when mass (M) and spring constant (K) are known
fTc
1
cycle-time (Tc) when the frequency is known
cTf
1
frequency (f) when the cycle-time (Tc) is known
of cycles
elapsed timef
frequency (f) when the number of cycles in an elapsed time is known
d1 amplitude (distance from rest to start position)
Also frac12 the distance from the base to the peak
3131 oror dM
Ka
251
3131 539
c
oror T
da
2539
cT
da
In December of 2006 I discovered the mathematical error in Einsteinrsquos Special Relativity Einstein used BS math and that math only works in an imaginary world
It cannot describe real symmetrical space
Einsteinrsquos Special Relativity should be abandoned and a theory of real space using SM should be used to replace it
Numerous articles on the Internet show Einsteinrsquos theory to be incorrect but no one in academia is willing to publish the information
Figure 1-10 A moving particle observed from two frames of reference in uniform translation
Einsteinrsquos Math Error in Special Relativity Transformation of Velocity (From Eisenberg- Fundamentals of Modern Physics pages 27-29) Consider the particle shown in figure (1-10) moving with velocity v as
seen in a frame of reference O
dx dy dz
Vx Vy Vzdt dt dt
dx dy dz
V x V y V zdy dt dt
We would like to evaluate the velocity v of the particle as seen in the frame of reference 0 which is itself moving relative to 0 with velocity v Measured in the 0 frame the velocity vector of the particle has com ponents
The same velocity vector as measured in the 0 frame has components
2
2
2
2
2
1
1
x vtx
v
cy y
z z
vxt
ctv
c
Now from equations (1-13) we know that the relation between the primed and the unprimed coordinates and times is
Take the differential of these equations remembering that v is a constant This gives
2
2
2
2
2
1
1
dx vdtdx
v
cdy dy
dz dz
vxdt
cdtv
c
2
2
2 2 2 2 2
2
2
2
11 ( )
1( ) 1
1
1
1
dx vdt
v dxdx vdt v
dx dx vdtc dt dtvdx vdx vdx v dxdt dt dt dt
dt dtc c c c
VxV x
Vx
VxV x
vv
vv
c
x
v
c
cV
( )
coordinate of particle relative to moving primed system
coordinate of particle relative to stationary system
velocity of primed coordina
V x
Vx
te system relative to stationaryv sys
tem
Page 28 Fundamentals of Modern Physics
(1-16)
According to Einstein this equation tells you how to transform the observed velocity from one frame of reference to another frame of reference First we note that as Vc and vc approachzero the equation approaches those which would be derived from the Galilean transformation
2 2
( ) 1 01 1 ( )
Vx v Vx vV
VxV x
Vx v
v cVxx
v
c
v
cc
Another property of this equation is that it is impossible to choose V and v such that Vrsquo the magnitude of the velocity which is seen in the new frame is greater than c
1
2
09 ( 09 ) 18009876543
( 09 ) 1811 (09 )
c c cV c c
cc
c
Example As seen by O particle 1 has velocity 09c in the direction of positive x and particle 2 has velocity 09c in the negative x direction
Einstein equation however predicts To evaluate the velocity of particle 1 with respect to particle 2 we transform from the O frame to the Orsquo frame moving in the negative x direction with velocity v = -09c using equation 1-16 We obtain
JK LOGIC will observe the particles separating at 18c
A confusing INCORRECT and ILLOGICAL answer
left is a dir rightsub of a
1
2
1
09 ( 09 ) 18009876543
181( 09 )1 ( 09 )
The multiplication of dir left time a dir right is ONLY a dir left
09
dir
direction leftdirection rightsubtraction
c c cV c c
cc
c
V
dir
22
22
09 18 18009876543
181( 081)1(09)(09)
1
left rightsub
sub
c c c cc c
ccc
c
Analyzing Einsteinrsquos BS math Equation
Using SM math a logical answer can be obtained
Cannot multiply a subtraction operator times a direction in space
Can only subtract and change its direction
Cannot multiply a direction to the left times a direction to the right
Cannot multiply opposite directions in space
Can only solve for the resultant
rightleft
subtraction
1
subtraction
2
sub of a dir to the left is a dir to the right
1 resultant 0 sub 2
2sub 2
2
09 09
09 amp 091
09 09 18 1
(0)1
(09) amp (09)1
c cV
c cc
c c cV
cc
cc
8
181
cc
SM math predicts a logical answer using logic
21st century scientists state in many articles that Einsteins General Relativity and Quantum Mechanics are incompatible
They are not only incompatible they are both mathematically incorrect
They both use illogical BS math
Any equations in math that use imaginary numbers should be abandoned
Imaginary numbers are only necessary in the current BS math system
This means that Schroumldingers equations should be abandoned
Any equation in quantum mechanics that uses imaginary number should be abandoned
Prof Kanarev has developed logical physics and chemistrytheories without the use of Quantum mechanics or relativity
A partial list of his theories coverbull Mechanics ndash replaces Newtonbull Photon - bull Electrons ndash Protons- Neutronsbull Spectroscopybull Nucleusbull Atoms amp Moleculesbull Electrodynamicsbull Heatbull Curvature of Space ndash replaces Schwarzcchildrsquos equation
Kanarevrsquos Hypothesis
bull Electron align linearly with proton They do NOT spin or form clouds around the nucleus
bull Neutrons are in the center of atom Proton are outside of Neutrons
bull The wavelength of all particles is the experimental radius
bull Double slit experiment explained correctly
bull Heisenbergrsquos inequality explained correctly
bull is the radius of approach of the magnetic lines around the torus of the electron
( 1 137)
bull Why water expands when it freezes
bull Electron clusters form lightning
rbull The meaning of Plancks constant is explained
22 2 kg m
h m f mr f consts
2210254 42m r kgm bull If the mr-constant describes a ring the equations
describe experiments bull If the mr-constant describes a wave the equations do
NOT describe experiments
Kanarevrsquos 5 Equations of Mechanics
Kanarevrsquos 1st Equation Acceleration ampmv maMma R R
Kanarevrsquos 2nd Equation Uniform motion amp mv maMma R R
maM mvR R
Kanarevrsquos 3rd Equation
Kanarevrsquos 4th Equation Equal and opposite reaction
1 21 21 2
12
a m
m a m ama
Kanarevrsquos 5th Equation 1 2 3 Mi M M M Mna R R R R R
The sum of the Actions working on a moving body is never equal to zero
Kanarevrsquos Photon model is a rolling ring
His equations accurately describe known experiments with photonsbull Electromagnetic radiation should be called photon radiationbull Diffraction and interference are the same phenomenabull Correctly explains the double slit experiment without quantum equationsbull Correctly explains the meaning of Heisenbergrsquos inequality
e
2 (2)(3142)(2817 15 ) 10007297352568
R 2426 12 137mag
xp
r m
m
Charge
Magnetic action
Proton
The intensity of the magnetic action near the geometrical center of the proton ring is so great that we theorize that it is the strong nuclear force
(9383 8)(1602 19)8476 14
4 (4)(3142)(1411 26)p
pp
EB T
M
(9395 8)(1602 19)1035 15
4 (4)(3142)(0966 26)N
NN
EB T
M
Neutron
1a Hydrogenavi
() multiplied by a direction ( )=
number
() multiplied by a direction ( )=
number
A number times a direction maintains the same direction
( ) ()
dirdiradd
( ) ()
dirdiradd
The addition of a direction is in the same direction There is no multiplication by addition operators
Symmetry Math
Future math books need to eliminate the use of the cross sign (+) to represent something labeled a positive direction in space and the dash sign (-) to represent something labeled a negative direction in space Space does not have positive and negative directions
Symbols that are logical and have no illogical representation should be adopted SM uses either arrows or symbols
Symmetry-Math Number Line
Symmetry-Math Rules
The dash sign (-) will have only one use subtraction
The cross sign (+) will have only one use addition
Directions have an arrow and a number
All observers see the same direction and magnitude
bull There is no multiplication of arrows
amp ampamp ampampamp
( 3 ) ( 5 ) ( 6 ) ( 65 ) (17256 ) ( 32993 )
dir dirdir dirdirdir
Negative and Positive Directions in Space
If I asked you to point to a negative direction in space which way would you point
Hopefully you will realize that there is no such thing as a negative direction in space
If I asked you to point to a positive direction in space which way would you point
Again hopefully you will realize that there is no such thing as a positive direction in space
Objects on the right side of the x-axis are positive (+)(+) = (+)
Objects on the left side of the x-axis are negative (-)(-) = (+) negative math is different from positive math This is illogical
Objects on the top of the y-axis are positive
Objects on the bottom of the y-axis are negative
Objects in the front of the z-axis are positive
Objects in the back of the z-axis are negative
Again BS math of the positive direction (x right y up and z front) is different from the BS math of the negative direction (x left y down and z back)
BS math is Broken-Symmetry
Mirror image is broken symmetry in the x and z axis If the left amp right-axis are
reversed symmetry is broken Obs-1 math is different from obs-2
If the front amp back-axis is reversed symmetry is broken Obs-1 math is different from obs-2
If the top amp bottom-axis are
reversed symmetry is not broken Obs-1 math is the same as obs-2
Math answers cannot depend on which side of a number line an observer sit
Absolute values for Displacement
Even for a simple displacement of an object BS math invented a definition and absolute values to obtain something labeled a positive number answer
I----------I----------I----------I----------I----------I----------I----------I----------I
-4 -3 -2 -1 0 +1 +2 +3 +4 a gtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtb cltltltltltltltltltltltltltltltltltltltltltltltltltltltb
If an object starts at ( 3)ia X and moves to ( X 3) and the to ( 1)b fb c X
BS Math provides a usable answer without a definition or absolute values
However If an object starts at
( 1) and moves to ( X 3) and the to ( 3)i b fa X b c X
Then absolute values and a definition must be used
---------I----------I----------I----------I----------I----------I----------I---------- -4 -3 -2 -1 0 +1 +2 +3 +4
a gtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtb cltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltb
BS Math makes up a definition that all displacements are positive Since this example provides an answer with something labeled a negative the positive definition must be applied and (-2) is changed to (+2) A definition is needed to arrive at a useable answer This is Illogical
In SM absolute values and a meaningless definition of ldquopositiverdquo are removed In SM objects and directions in space are defined by the direction and magnitude of the resultant of arrows
With SM you get total distance traveled by the object and the final direction of the displacement
22222 2 bababbaababababa If w e su b stitu te p o s itiv e n u m b e rs in to th e eq u a tio n th e an sw er w ill b e co rrec t If (a = 5 ) an d (b = 3 )
2
2
8
5 3 5 3 5 3 5 5 5 3 3 5 3 3 25 15 15 9 64
8 8 8 8 8 8 8 8 8 8 64
W h en tw o p o sitiv e n u m b ers a re a d d ed an d sq u a re d th e B S m a th d is tr ib u tiv e law p ro v id es a co rre c t an sw er 1
Illogical and Incorrect BS math for the distributive law
222 2 bababbabbaaabababa B y d e fin itio n in B S m a th n o n u m b er sq u a red c a n ev e r b e n eg a tiv e T h e re fo re n o sq u a red n u m b er c an ev e r b e a v e c to r m o v in g in th e d a sh ( -) o r n eg a tiv e d ire c tio n if it is sq u a red In B S m a th an d ad d itio n o p e ra to r m u ltip lied b y a su b trac tio n o p e ra to r is A L W A Y S eq u a l to a su b tra c tio n o p e ra to r If th is h ad n o t b e en u sed fo r th e la s t fo u r h u n d red y e a rs y o u w o u ld d ie lau g h in g a t th e u tte r ab su rd ity o f th e la s t s ta tem e n t T h is is illo g ica l an d v io la te s sy m m etry
If a = 5 an d b = 3 42222352
22
4930253)3)(5(253)5)(3()3)(5(5353535 22222 A s lo n g a s ldquo ardquo is g re a te r th an ldquob rdquo th e B S m a th d is tr ib u tiv e law p ro v id es co rrec t an sw ers 1
B S m a th b re a k s d o w n w h e n ldquo b rdquo is g rea ter th a n ldquo a rdquo
If a = 3 an d b = 5
2 2
2
2
3 5 2 4
2 ( 2) ( 2) 4
2 ( )( 2) ( )( 2) 4
A n y co m b in a tio n ex cep t th e m id d le o n e is illo g ica l an d in co rre c t B y d e fin itio n th e an sw er can n o t b e -4 in B S m a th B S m a th is in co rrec t
4253095)5)(3(235)3)(5()5)(3(3535353 22222 T h is is w h ere th e d is tr ib u tiv e la w p ro v id es in co rrec t an sw ers I t p ro d u ces an an sw er o f + 4 b e cau se b y d e fin iti o n n o n u m b er sq u a red in B S m a th c an b e a n eg a tiv e n u m b er U sin g in co rre c t B S m a th th e d is tr ib u tiv e law w ill g iv e an in co rrec t an sw ers o f + 4 T h is is in co rrec t an d illo g ica l 1
Since there can be no negative direction from a squared term in BS math BS math should be modified or abandoned
The BS distributive law should be modified or abandoned
The numbers +1+2+3+4 and ndash1-2-3-4hellip should be abandoned
BS math gives incorrect answers because their Rule-of-Signs allows multiplying an arrow going in one direction by an arrow going in another direction
Clearly this is not logical
Arrows should be used that are specific for directions
W h at d o es it m e an to m u ltip ly ( 4) b y (4 ) A s w e k n o w 1
m u ltip lica tio n is ju s t ad d itio n W h a t w o u ld ( 4) tim e (4 ) b e 2
In BS math the answer is by definition -16 unless you square -4 and then it is +16
In SM you cannot multiply arrows going in opposite directions
The middle two terms are not logical You cannot multiply opposite directions
This is a MAJOR error in BS math They multiply a dash [(-) a direction to the left] by a cross [(+) a direction to the right] And then ldquoby definitionrdquo label the answer dash (-) This is illogical and produces incorrect answers
2
2 2
This is Einsteinrsquos math error
in Special relativity
Magnitude 4 Direction 1 Magnitude 3 Direction 2 Magnitude 5 Direction 3 Relative to an observer at 0 4
Magnitude 4 Direction 1 Magnitude 3 Direction 2 Magnitude 5 Direction 3 Relative to an observer at 0 4
Magnitude and Direction (MD) An arrow has both a magnitude andA direction relative to a 2nd objectobserver
bull Magnitude and velocity (MV) relative to a 2nd objectobserver
bull Magnitude And acceleration (MA) relative to a 2nd objectobserver2
v da
t t
Introducing a new symbol the ldquoamprdquo sign The ldquoamprdquo sign means resultant ADD all the Arrows for each direction and then SUBTRACT to find the final resultant
1 1amp 1 2 2amp 2 4 If =
1 amp1 2 2 amp2 4 If =
amp 5 amp 3 2 If gt
amp 3 amp 5= 2 If gt
amp 0 3 amp 3=0 if =
only the resultant equals zero
the vectors do not - they are still there
Does not exist There is no multiplication of arrows
There are two arrows One has a magnitude of 3 One has a magnitude of 5 The maximum these two arrows can be is 8
Only the resultant of arrows exist
An SM arrow has a direction and a magnitude
Arrows and magnitudes are still there even if the resultant is zero
There are no negative SM Arrows
There are no positive SM Arrows
There are no ZERO SM Arrows A zero would negate the definition of an SM Arrow something with no magnitude and no direction cannot be the definition of an arrow that is defined as having magnitude and direction
bull A resultant is the addition and subtraction of the differences in magnitude and direction arrows
A B and C are the arrows (a) A amp B exist(b) A amp B amp C exist(c) A-B Do Not Exist in Sm there are no negatives or positives(d) A+B-C Do Not Exist in SM there are no negatives or positives
There are only resultants when there are multiple arrows
SM Graphs versus BS Graphs
Numbers Raised to Powers
When numbers are raised to powers the negative sides of all BS math graphs are ILLOGICAL and INCORRECT
Incorrect BS Math for 2x
2x
4 16 correct3 8 correct2 4 correct1 2 correct0 1 correct-1 005 incorrect-2 025 incorrect-3 0125 incorrect-4 00625 incorrect
negative side of graph is incorrectpositive side of graph is correct
16
8
42105025012500625
-4 -3 -2 -1 0 1 2 3 4
po
sit
ive
negative positive
2x BS Math
correctincorrect
In BS math the dash sign (-) used in this example represents three different operations They arebull (-) Subtractionbull (-) A direction to the leftbull (-) As an exponent to mean divide by
All positive numbers produce correct results and graphs
All negative numbers produce incorrect results and graphs
A direction of three units to the left (-3) is placed in the equation as an exponent
A direction in space (3 units toward the left) changes a whole number to a fraction by using the dash (-) as an exponent rather than a direction in space
This is illogical and incorrect BS math should have performed this operation as
Unfortunately the answers to problems can come out correct because data points of a real problem fit the curve
However the answers will not be due to logical reasoning
Because the equation produced a curve that real problem data points follow it has been falsely believed that the equations and graphs are correct
1
2
x
Incorrect BK graph for (12)x
(12)x
4 00625 correct3 0125 correct2 025 correct1 1 correct0 1 correct-1 2 incorrect-2 4 incorrect-3 8 incorrect-4 16 incorrect
negative side of graph is incorrectpositive side of graph is correct
00625012502505124
8
16
-5 -4 -3 -2 -1 0 1 2 3 4 5P
osit
ive
negative positive
BS Math for (12)x
correctincorrect
Incorrect BS math and graph for
All graphs are symmetrical with SM
Two BS Math Equations that Produce Incorrect Answers
2 0x x starting equation 2
2
x x x x
x x
Add lsquoxrsquo to both sides
2(1) 1 let lsquoxrsquo = 1
2 1 This is an incorrect answer 2(0)-(0)=0 The only answer that BK math can give is x=0
1
a b starting equation ( )( ) ( )( )a a a b multiply both sides by ( )a
( )( ) ( )( ) ( )( ) ( )( )a a b b a b b b subtract ( )( )b b from both sides
( )( ) ( )( )a b a b a b b factor ( )a b from both sides
( )a b b divide both sides by ( )a b
( )b b b replace ( )a with ( )b 2b b simplify 2(1) (1) Let ( )b = 1
2 1 Again BS math yields and incorrect answer 1
This incorrect result is a problem with the distributive law SM explains how this incorrect answer is obtained and the problem with the distributive law that allows the error
Page 148 ndash149 Barry Mazur ndash Imagining Numbers
3
2
31
Do the computation carefully on paper using the rules we agreed to Then ponder your answer which should be something of a surprise to you donrsquot stop there Think of what your answer might possibly mean or might imply
Solve this equation using the distributive law
JK definitely a surprise the answer wasincorrect and illogical
Gerhard Gade University Professor doing math at Harvard University
Here is the BS math answer that Barry Mazur calculated
2
31
2
31
2
31
1 3 1 3 1 3 3 3 2 2 3 1 3
2 2 2 2 4 4 4 4 4 4 2 2
dash
1 3 1 3 1 31
2 2 2 2 4 4
3 3
4 4
When the distributive law is used with the BS Rule-of-Signs and one of the signs is a (-) dash sign the middle two terms always cancel each other This leads to the incorrect answer of (-1) in the above equation
Barry Mazur did not go back and check his answer with the original equation If he had he would have known that the answer was incorrect
3
1 31
2
check answer by setting equal
3
331 3
12
1 31
2
Take the Cube root of both sides
1 3 2 multiply both sides by 2
3 3 subtract 1 from both sides
2 2
3 3 square both sides
3 9 This is an incorrect answer A simple equation check could have shown that the distributive law using BS math provides an incorrect answer
1
BS Illogical and Incorrect Numbers Raised to Powers
nC D n DC
642 DDD
In BS math only a positive number can have an answer if ldquonrdquo is an even number (246---)
All negative numbers have no solutions
642 DDD
With SM just add a direction arrow and any number gives the correct answer
or n nC D C D
3 3( 3 8 8 or 8 2 or 2 )n D C
( 2 16 C= 16 or 16 4 or 4 )n D
Since a negative number has no square root answer in BS math BS mathematicians invented imaginary numbers
i2=-1
24414 iiC n
These equations are easily solved without the use of imaginary numbers with SM math
If 4 then 4 2x
If 4 then 4 2x
The 360-degree circle is a 4400 years old extremely outdated system based on a year of 12 months of 30 days each that is 360 days
SM will use a more logical system that will have a circle of 6283185307o
d
( )
(deg )
d r radians
d r rees
the arc distance around
the circumference of a circle
d
radiusr
2 2
2 2
d r
dv rd t t
v dda rd t t t
1
d
r
( )av tv
The average angular velocity of a body is the rate of its angular displacement
( )0
liminstt
vt
The instantaneous angular velocity is the limit of the ratio
t
as t approaches zero
( )d
av
d
tv
The average circumference velocity of the body is the rate of its distance traveled d divided by the time t
dvv
r
1 dd vdrv
t t t r r
The equation of the relationship between the angular velocity and the velocity of a point on that body
( )o
av
v v v
t ta
The average angular acceleration is the change in the angular velocity
( )0
liminstt
v
ta
The instantaneous angular acceleration is the
limit of the ration of v
t
as t approaches
zero
daa
r
1
1
d
d s d
vva
t r tv a
r t r
The equation of the relationship between the angular acceleration and the acceleration of a point on the body
21
2v a t
If a body with an initial angular velocity ov has a constant
angular acceleration a it will turn through an angle in a time t
1
Straight Line Symbol Units Rotational Symbol Units Distance d m Angular
Distance radians or degrees
Velocity dt
v m
s Angular Velocity
V
t
radians or degrees
s
Acceleration vat
2ms
Angular Acceleration
VA
t
2
radians or degrees
s
Mass M Kg Mass distribution
dM 2Kgm
Force actionA Ma
2
Kgm
sNewton
Torque
dT M A 2
2
Kgm
s
Momentum Mv Kgm
s
Angular Momentum
dM V 2Kgm
s
Work (Force)(distance)
Ad 2
Kgmm
sNm Joule
Work T
Using AS arrow math to solve and direction in space problems using the equation 1 2
2 2 2 2 2 2( ) amp and ( ) amp 3
4
2 2 2
2 2
( ) amp
(5 ) 4 3
(5 ) 16 9
(5 ) 25
(5 ) (5 )
2 2(5 ) 4 3
(5 ) 16 9
(5 ) 25
(5 ) (5 )
In SM math Pythagorasrsquo theorem provides both magnitude and direction
In BS math Pythagorasrsquo theorem provides magnitude only
time set equal to zero
23
( ) ( 0 )10 1
x t xx
3( ) ( 0)
10 1W t f
In BS math this is an equation for a traveling wave
must use a square term in order to make the curve symmetric
NO SQUARE TERM IS NECESSARY
In SM math when t is set to ldquo0rdquo
This is a major fault in BS math
x f(x)
-4 0019
-3 0033
-2 0073
-1 0273
0 3000
1 0273
2 0073
3 0033
4 0019 -4 -3 -2 -1 0 1 2 3 400
05
10
15
20
25
30
35
BS wave (must use square term)
snap shot of wave at x = 0
Am
plit
ude
lt to gt lt to gt f(ltgt)
lt -4 0073
lt -3 0097
lt -2 0143
lt -1 0273
origin 0 3000
gt 1 0273
gt 2 0143
gt 3 0097
gt 4 0073lt0073 lt0097 lt0143 lt0273
3000
0273gt 0143gt 0097gt 0073gt
Wave moving
snap shot of wave
Am
plit
ude
lt4 lt3 lt2 lt1 0 1gt 2gt 3gt 4gt
3
2
1
In SM NO squared terms are needed
SPRING
An Action is needed to stretch or compress a spring to a starting point where it can execute cycle motion
tsA (ts to start) is the work required to move a mass attached to a spring a distance (d1) from
its initial rest position RAK 0 the spring constant is the stiffness of the spring
o A is attraction o R is repulsion o
ARRA
00 is for the back and forth motion through the zero initial rest position A
stretched spring goes from ldquoArdquo to ldquo0rdquo to ldquoRrdquo to ldquoArdquo to ldquo0rdquo to ldquoRrdquo d1 is the distance a mass is moved from its initial rest position
0
0 1ts A RA K d
Once ( )mad
is performed to move the mass to position d1 (against the restraint of the spring) a of the ( )mad
is transferred into the spring as potential attractionrepulsion (PAR) There is always
more ( )mad
performed than is stored as PAR When the spring is released from the restraining Ats the PAR starts the oscillation Conservation of ( )mad
requires that the PAR and the motion ( )mad
( )( )madM be a constant
( )amp ( ) constantmadPAR M 02
0 0
1
2A R A RPAR K d
02
( ) 0 0
1( )
2mad A R A RM m v
0 0 02 2 2
0 0 0 1
2 2 21
2 2 21
1 1 1amp
2 2 2
1 1 1amp
2 2 2
amp
A R A R A RK d m v K d
Kd mv Kd
Kd mv Kd
0
1d is frac12 the height from the base to the peak of a cycle
An experiment of moving paper under a pendulum through a small angle filled with sand that can escape through a small hole in the bottom shows that a pendulum cycling back and forth follows a curve that is mathematically analogous to a lsquosinrsquo or lsquocosrsquo
The projection of the curved area onto the straight line exactly duplicate the back and forth motion of CM This allows for the math of lsquosinrsquo or lsquocosrsquo to be used
SM will use lsquosinrdquo math It could use lsquocosrsquo or a combination of both lsquosinrsquo and lsquocosrsquo Using just the lsquosinrsquo math makes the understanding and math easier
0 sin 0 0SM 0 sin157 1SM The diagram at the left shows a spring being stretched from its (zero rest) position-2 and the PAR being released at position-1 There are 4 sections (1 to 2 2 to 3 3 to 4 and 4 to5) that alternate between maximum acceleration and zero velocity to max velocity and zero acceleration
At position-1 there is maximum acceleration and zero velocity o The lsquosinrsquo values from 0o to 157o will be used for velocity (values from 0 to 1) o The lsquosinrsquo values from 157o to 0o will be used for acceleration (values from 1 to 0)
At position-2 there is maximum velocity and zero acceleration Acceleration changes to deceleration
o The lsquosinrsquo values from 157o to 0o will be used for deceleration (values from 1 to 0) o The lsquosinrsquo values from 0o to 157o will be used for acceleration (values from 0 to 1)
At position-3 there is maximum acceleration and zero velocity o The lsquosinrsquo values from 0o to 157o will be used for velocity (values from 0 to 1) o The lsquosinrsquo values from 157o to 0o will be used for acceleration (values from 1 to 0)
At position-4 there is maximum velocity and zero acceleration Acceleration changes to deceleration
o The lsquosinrsquo values from 157o to 0o will be used for deceleration (values from 1 to 0) o The lsquosinrsquo values from 0o to 157o will be used for acceleration (values from 0 to 1)
At position-5 the cycle starts again
2
)539(
cT
MK
spring constant (K) when mass (M) and cycle-time (T) are known
)(
))(89)((
111 md
uKgM
d
MG
d
FK sts
spring constant (K) when mass (M) and amplitude ( d1) are known
2 21( )
Kv d d
m
velocity (v) at any point during the cycle
velocity
dforM
Kdv
max
012 velocity (v2) maximum (when d = 0 at position-2 and 4)
acceleration (a1) maximum when spring constant (K) mass (M) and amplitude (d1) are known
acceleration (a1) maximum when amplitude (d1) and cycle-time (Tc) are known
acceleration (a) at any point during the cycle-time (T)
1
151
286
v
dTc
cycle-time (Tc) when amplitude (d1) and velocity (v1) are known
K
MTc 28651
cycle-time (Tc) when mass (M) and spring constant (K) are known
fTc
1
cycle-time (Tc) when the frequency is known
cTf
1
frequency (f) when the cycle-time (Tc) is known
of cycles
elapsed timef
frequency (f) when the number of cycles in an elapsed time is known
d1 amplitude (distance from rest to start position)
Also frac12 the distance from the base to the peak
3131 oror dM
Ka
251
3131 539
c
oror T
da
2539
cT
da
In December of 2006 I discovered the mathematical error in Einsteinrsquos Special Relativity Einstein used BS math and that math only works in an imaginary world
It cannot describe real symmetrical space
Einsteinrsquos Special Relativity should be abandoned and a theory of real space using SM should be used to replace it
Numerous articles on the Internet show Einsteinrsquos theory to be incorrect but no one in academia is willing to publish the information
Figure 1-10 A moving particle observed from two frames of reference in uniform translation
Einsteinrsquos Math Error in Special Relativity Transformation of Velocity (From Eisenberg- Fundamentals of Modern Physics pages 27-29) Consider the particle shown in figure (1-10) moving with velocity v as
seen in a frame of reference O
dx dy dz
Vx Vy Vzdt dt dt
dx dy dz
V x V y V zdy dt dt
We would like to evaluate the velocity v of the particle as seen in the frame of reference 0 which is itself moving relative to 0 with velocity v Measured in the 0 frame the velocity vector of the particle has com ponents
The same velocity vector as measured in the 0 frame has components
2
2
2
2
2
1
1
x vtx
v
cy y
z z
vxt
ctv
c
Now from equations (1-13) we know that the relation between the primed and the unprimed coordinates and times is
Take the differential of these equations remembering that v is a constant This gives
2
2
2
2
2
1
1
dx vdtdx
v
cdy dy
dz dz
vxdt
cdtv
c
2
2
2 2 2 2 2
2
2
2
11 ( )
1( ) 1
1
1
1
dx vdt
v dxdx vdt v
dx dx vdtc dt dtvdx vdx vdx v dxdt dt dt dt
dt dtc c c c
VxV x
Vx
VxV x
vv
vv
c
x
v
c
cV
( )
coordinate of particle relative to moving primed system
coordinate of particle relative to stationary system
velocity of primed coordina
V x
Vx
te system relative to stationaryv sys
tem
Page 28 Fundamentals of Modern Physics
(1-16)
According to Einstein this equation tells you how to transform the observed velocity from one frame of reference to another frame of reference First we note that as Vc and vc approachzero the equation approaches those which would be derived from the Galilean transformation
2 2
( ) 1 01 1 ( )
Vx v Vx vV
VxV x
Vx v
v cVxx
v
c
v
cc
Another property of this equation is that it is impossible to choose V and v such that Vrsquo the magnitude of the velocity which is seen in the new frame is greater than c
1
2
09 ( 09 ) 18009876543
( 09 ) 1811 (09 )
c c cV c c
cc
c
Example As seen by O particle 1 has velocity 09c in the direction of positive x and particle 2 has velocity 09c in the negative x direction
Einstein equation however predicts To evaluate the velocity of particle 1 with respect to particle 2 we transform from the O frame to the Orsquo frame moving in the negative x direction with velocity v = -09c using equation 1-16 We obtain
JK LOGIC will observe the particles separating at 18c
A confusing INCORRECT and ILLOGICAL answer
left is a dir rightsub of a
1
2
1
09 ( 09 ) 18009876543
181( 09 )1 ( 09 )
The multiplication of dir left time a dir right is ONLY a dir left
09
dir
direction leftdirection rightsubtraction
c c cV c c
cc
c
V
dir
22
22
09 18 18009876543
181( 081)1(09)(09)
1
left rightsub
sub
c c c cc c
ccc
c
Analyzing Einsteinrsquos BS math Equation
Using SM math a logical answer can be obtained
Cannot multiply a subtraction operator times a direction in space
Can only subtract and change its direction
Cannot multiply a direction to the left times a direction to the right
Cannot multiply opposite directions in space
Can only solve for the resultant
rightleft
subtraction
1
subtraction
2
sub of a dir to the left is a dir to the right
1 resultant 0 sub 2
2sub 2
2
09 09
09 amp 091
09 09 18 1
(0)1
(09) amp (09)1
c cV
c cc
c c cV
cc
cc
8
181
cc
SM math predicts a logical answer using logic
21st century scientists state in many articles that Einsteins General Relativity and Quantum Mechanics are incompatible
They are not only incompatible they are both mathematically incorrect
They both use illogical BS math
Any equations in math that use imaginary numbers should be abandoned
Imaginary numbers are only necessary in the current BS math system
This means that Schroumldingers equations should be abandoned
Any equation in quantum mechanics that uses imaginary number should be abandoned
Prof Kanarev has developed logical physics and chemistrytheories without the use of Quantum mechanics or relativity
A partial list of his theories coverbull Mechanics ndash replaces Newtonbull Photon - bull Electrons ndash Protons- Neutronsbull Spectroscopybull Nucleusbull Atoms amp Moleculesbull Electrodynamicsbull Heatbull Curvature of Space ndash replaces Schwarzcchildrsquos equation
Kanarevrsquos Hypothesis
bull Electron align linearly with proton They do NOT spin or form clouds around the nucleus
bull Neutrons are in the center of atom Proton are outside of Neutrons
bull The wavelength of all particles is the experimental radius
bull Double slit experiment explained correctly
bull Heisenbergrsquos inequality explained correctly
bull is the radius of approach of the magnetic lines around the torus of the electron
( 1 137)
bull Why water expands when it freezes
bull Electron clusters form lightning
rbull The meaning of Plancks constant is explained
22 2 kg m
h m f mr f consts
2210254 42m r kgm bull If the mr-constant describes a ring the equations
describe experiments bull If the mr-constant describes a wave the equations do
NOT describe experiments
Kanarevrsquos 5 Equations of Mechanics
Kanarevrsquos 1st Equation Acceleration ampmv maMma R R
Kanarevrsquos 2nd Equation Uniform motion amp mv maMma R R
maM mvR R
Kanarevrsquos 3rd Equation
Kanarevrsquos 4th Equation Equal and opposite reaction
1 21 21 2
12
a m
m a m ama
Kanarevrsquos 5th Equation 1 2 3 Mi M M M Mna R R R R R
The sum of the Actions working on a moving body is never equal to zero
Kanarevrsquos Photon model is a rolling ring
His equations accurately describe known experiments with photonsbull Electromagnetic radiation should be called photon radiationbull Diffraction and interference are the same phenomenabull Correctly explains the double slit experiment without quantum equationsbull Correctly explains the meaning of Heisenbergrsquos inequality
e
2 (2)(3142)(2817 15 ) 10007297352568
R 2426 12 137mag
xp
r m
m
Charge
Magnetic action
Proton
The intensity of the magnetic action near the geometrical center of the proton ring is so great that we theorize that it is the strong nuclear force
(9383 8)(1602 19)8476 14
4 (4)(3142)(1411 26)p
pp
EB T
M
(9395 8)(1602 19)1035 15
4 (4)(3142)(0966 26)N
NN
EB T
M
Neutron
1a Hydrogenavi
Future math books need to eliminate the use of the cross sign (+) to represent something labeled a positive direction in space and the dash sign (-) to represent something labeled a negative direction in space Space does not have positive and negative directions
Symbols that are logical and have no illogical representation should be adopted SM uses either arrows or symbols
Symmetry-Math Number Line
Symmetry-Math Rules
The dash sign (-) will have only one use subtraction
The cross sign (+) will have only one use addition
Directions have an arrow and a number
All observers see the same direction and magnitude
bull There is no multiplication of arrows
amp ampamp ampampamp
( 3 ) ( 5 ) ( 6 ) ( 65 ) (17256 ) ( 32993 )
dir dirdir dirdirdir
Negative and Positive Directions in Space
If I asked you to point to a negative direction in space which way would you point
Hopefully you will realize that there is no such thing as a negative direction in space
If I asked you to point to a positive direction in space which way would you point
Again hopefully you will realize that there is no such thing as a positive direction in space
Objects on the right side of the x-axis are positive (+)(+) = (+)
Objects on the left side of the x-axis are negative (-)(-) = (+) negative math is different from positive math This is illogical
Objects on the top of the y-axis are positive
Objects on the bottom of the y-axis are negative
Objects in the front of the z-axis are positive
Objects in the back of the z-axis are negative
Again BS math of the positive direction (x right y up and z front) is different from the BS math of the negative direction (x left y down and z back)
BS math is Broken-Symmetry
Mirror image is broken symmetry in the x and z axis If the left amp right-axis are
reversed symmetry is broken Obs-1 math is different from obs-2
If the front amp back-axis is reversed symmetry is broken Obs-1 math is different from obs-2
If the top amp bottom-axis are
reversed symmetry is not broken Obs-1 math is the same as obs-2
Math answers cannot depend on which side of a number line an observer sit
Absolute values for Displacement
Even for a simple displacement of an object BS math invented a definition and absolute values to obtain something labeled a positive number answer
I----------I----------I----------I----------I----------I----------I----------I----------I
-4 -3 -2 -1 0 +1 +2 +3 +4 a gtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtb cltltltltltltltltltltltltltltltltltltltltltltltltltltltb
If an object starts at ( 3)ia X and moves to ( X 3) and the to ( 1)b fb c X
BS Math provides a usable answer without a definition or absolute values
However If an object starts at
( 1) and moves to ( X 3) and the to ( 3)i b fa X b c X
Then absolute values and a definition must be used
---------I----------I----------I----------I----------I----------I----------I---------- -4 -3 -2 -1 0 +1 +2 +3 +4
a gtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtb cltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltb
BS Math makes up a definition that all displacements are positive Since this example provides an answer with something labeled a negative the positive definition must be applied and (-2) is changed to (+2) A definition is needed to arrive at a useable answer This is Illogical
In SM absolute values and a meaningless definition of ldquopositiverdquo are removed In SM objects and directions in space are defined by the direction and magnitude of the resultant of arrows
With SM you get total distance traveled by the object and the final direction of the displacement
22222 2 bababbaababababa If w e su b stitu te p o s itiv e n u m b e rs in to th e eq u a tio n th e an sw er w ill b e co rrec t If (a = 5 ) an d (b = 3 )
2
2
8
5 3 5 3 5 3 5 5 5 3 3 5 3 3 25 15 15 9 64
8 8 8 8 8 8 8 8 8 8 64
W h en tw o p o sitiv e n u m b ers a re a d d ed an d sq u a re d th e B S m a th d is tr ib u tiv e law p ro v id es a co rre c t an sw er 1
Illogical and Incorrect BS math for the distributive law
222 2 bababbabbaaabababa B y d e fin itio n in B S m a th n o n u m b er sq u a red c a n ev e r b e n eg a tiv e T h e re fo re n o sq u a red n u m b er c an ev e r b e a v e c to r m o v in g in th e d a sh ( -) o r n eg a tiv e d ire c tio n if it is sq u a red In B S m a th an d ad d itio n o p e ra to r m u ltip lied b y a su b trac tio n o p e ra to r is A L W A Y S eq u a l to a su b tra c tio n o p e ra to r If th is h ad n o t b e en u sed fo r th e la s t fo u r h u n d red y e a rs y o u w o u ld d ie lau g h in g a t th e u tte r ab su rd ity o f th e la s t s ta tem e n t T h is is illo g ica l an d v io la te s sy m m etry
If a = 5 an d b = 3 42222352
22
4930253)3)(5(253)5)(3()3)(5(5353535 22222 A s lo n g a s ldquo ardquo is g re a te r th an ldquob rdquo th e B S m a th d is tr ib u tiv e law p ro v id es co rrec t an sw ers 1
B S m a th b re a k s d o w n w h e n ldquo b rdquo is g rea ter th a n ldquo a rdquo
If a = 3 an d b = 5
2 2
2
2
3 5 2 4
2 ( 2) ( 2) 4
2 ( )( 2) ( )( 2) 4
A n y co m b in a tio n ex cep t th e m id d le o n e is illo g ica l an d in co rre c t B y d e fin itio n th e an sw er can n o t b e -4 in B S m a th B S m a th is in co rrec t
4253095)5)(3(235)3)(5()5)(3(3535353 22222 T h is is w h ere th e d is tr ib u tiv e la w p ro v id es in co rrec t an sw ers I t p ro d u ces an an sw er o f + 4 b e cau se b y d e fin iti o n n o n u m b er sq u a red in B S m a th c an b e a n eg a tiv e n u m b er U sin g in co rre c t B S m a th th e d is tr ib u tiv e law w ill g iv e an in co rrec t an sw ers o f + 4 T h is is in co rrec t an d illo g ica l 1
Since there can be no negative direction from a squared term in BS math BS math should be modified or abandoned
The BS distributive law should be modified or abandoned
The numbers +1+2+3+4 and ndash1-2-3-4hellip should be abandoned
BS math gives incorrect answers because their Rule-of-Signs allows multiplying an arrow going in one direction by an arrow going in another direction
Clearly this is not logical
Arrows should be used that are specific for directions
W h at d o es it m e an to m u ltip ly ( 4) b y (4 ) A s w e k n o w 1
m u ltip lica tio n is ju s t ad d itio n W h a t w o u ld ( 4) tim e (4 ) b e 2
In BS math the answer is by definition -16 unless you square -4 and then it is +16
In SM you cannot multiply arrows going in opposite directions
The middle two terms are not logical You cannot multiply opposite directions
This is a MAJOR error in BS math They multiply a dash [(-) a direction to the left] by a cross [(+) a direction to the right] And then ldquoby definitionrdquo label the answer dash (-) This is illogical and produces incorrect answers
2
2 2
This is Einsteinrsquos math error
in Special relativity
Magnitude 4 Direction 1 Magnitude 3 Direction 2 Magnitude 5 Direction 3 Relative to an observer at 0 4
Magnitude 4 Direction 1 Magnitude 3 Direction 2 Magnitude 5 Direction 3 Relative to an observer at 0 4
Magnitude and Direction (MD) An arrow has both a magnitude andA direction relative to a 2nd objectobserver
bull Magnitude and velocity (MV) relative to a 2nd objectobserver
bull Magnitude And acceleration (MA) relative to a 2nd objectobserver2
v da
t t
Introducing a new symbol the ldquoamprdquo sign The ldquoamprdquo sign means resultant ADD all the Arrows for each direction and then SUBTRACT to find the final resultant
1 1amp 1 2 2amp 2 4 If =
1 amp1 2 2 amp2 4 If =
amp 5 amp 3 2 If gt
amp 3 amp 5= 2 If gt
amp 0 3 amp 3=0 if =
only the resultant equals zero
the vectors do not - they are still there
Does not exist There is no multiplication of arrows
There are two arrows One has a magnitude of 3 One has a magnitude of 5 The maximum these two arrows can be is 8
Only the resultant of arrows exist
An SM arrow has a direction and a magnitude
Arrows and magnitudes are still there even if the resultant is zero
There are no negative SM Arrows
There are no positive SM Arrows
There are no ZERO SM Arrows A zero would negate the definition of an SM Arrow something with no magnitude and no direction cannot be the definition of an arrow that is defined as having magnitude and direction
bull A resultant is the addition and subtraction of the differences in magnitude and direction arrows
A B and C are the arrows (a) A amp B exist(b) A amp B amp C exist(c) A-B Do Not Exist in Sm there are no negatives or positives(d) A+B-C Do Not Exist in SM there are no negatives or positives
There are only resultants when there are multiple arrows
SM Graphs versus BS Graphs
Numbers Raised to Powers
When numbers are raised to powers the negative sides of all BS math graphs are ILLOGICAL and INCORRECT
Incorrect BS Math for 2x
2x
4 16 correct3 8 correct2 4 correct1 2 correct0 1 correct-1 005 incorrect-2 025 incorrect-3 0125 incorrect-4 00625 incorrect
negative side of graph is incorrectpositive side of graph is correct
16
8
42105025012500625
-4 -3 -2 -1 0 1 2 3 4
po
sit
ive
negative positive
2x BS Math
correctincorrect
In BS math the dash sign (-) used in this example represents three different operations They arebull (-) Subtractionbull (-) A direction to the leftbull (-) As an exponent to mean divide by
All positive numbers produce correct results and graphs
All negative numbers produce incorrect results and graphs
A direction of three units to the left (-3) is placed in the equation as an exponent
A direction in space (3 units toward the left) changes a whole number to a fraction by using the dash (-) as an exponent rather than a direction in space
This is illogical and incorrect BS math should have performed this operation as
Unfortunately the answers to problems can come out correct because data points of a real problem fit the curve
However the answers will not be due to logical reasoning
Because the equation produced a curve that real problem data points follow it has been falsely believed that the equations and graphs are correct
1
2
x
Incorrect BK graph for (12)x
(12)x
4 00625 correct3 0125 correct2 025 correct1 1 correct0 1 correct-1 2 incorrect-2 4 incorrect-3 8 incorrect-4 16 incorrect
negative side of graph is incorrectpositive side of graph is correct
00625012502505124
8
16
-5 -4 -3 -2 -1 0 1 2 3 4 5P
osit
ive
negative positive
BS Math for (12)x
correctincorrect
Incorrect BS math and graph for
All graphs are symmetrical with SM
Two BS Math Equations that Produce Incorrect Answers
2 0x x starting equation 2
2
x x x x
x x
Add lsquoxrsquo to both sides
2(1) 1 let lsquoxrsquo = 1
2 1 This is an incorrect answer 2(0)-(0)=0 The only answer that BK math can give is x=0
1
a b starting equation ( )( ) ( )( )a a a b multiply both sides by ( )a
( )( ) ( )( ) ( )( ) ( )( )a a b b a b b b subtract ( )( )b b from both sides
( )( ) ( )( )a b a b a b b factor ( )a b from both sides
( )a b b divide both sides by ( )a b
( )b b b replace ( )a with ( )b 2b b simplify 2(1) (1) Let ( )b = 1
2 1 Again BS math yields and incorrect answer 1
This incorrect result is a problem with the distributive law SM explains how this incorrect answer is obtained and the problem with the distributive law that allows the error
Page 148 ndash149 Barry Mazur ndash Imagining Numbers
3
2
31
Do the computation carefully on paper using the rules we agreed to Then ponder your answer which should be something of a surprise to you donrsquot stop there Think of what your answer might possibly mean or might imply
Solve this equation using the distributive law
JK definitely a surprise the answer wasincorrect and illogical
Gerhard Gade University Professor doing math at Harvard University
Here is the BS math answer that Barry Mazur calculated
2
31
2
31
2
31
1 3 1 3 1 3 3 3 2 2 3 1 3
2 2 2 2 4 4 4 4 4 4 2 2
dash
1 3 1 3 1 31
2 2 2 2 4 4
3 3
4 4
When the distributive law is used with the BS Rule-of-Signs and one of the signs is a (-) dash sign the middle two terms always cancel each other This leads to the incorrect answer of (-1) in the above equation
Barry Mazur did not go back and check his answer with the original equation If he had he would have known that the answer was incorrect
3
1 31
2
check answer by setting equal
3
331 3
12
1 31
2
Take the Cube root of both sides
1 3 2 multiply both sides by 2
3 3 subtract 1 from both sides
2 2
3 3 square both sides
3 9 This is an incorrect answer A simple equation check could have shown that the distributive law using BS math provides an incorrect answer
1
BS Illogical and Incorrect Numbers Raised to Powers
nC D n DC
642 DDD
In BS math only a positive number can have an answer if ldquonrdquo is an even number (246---)
All negative numbers have no solutions
642 DDD
With SM just add a direction arrow and any number gives the correct answer
or n nC D C D
3 3( 3 8 8 or 8 2 or 2 )n D C
( 2 16 C= 16 or 16 4 or 4 )n D
Since a negative number has no square root answer in BS math BS mathematicians invented imaginary numbers
i2=-1
24414 iiC n
These equations are easily solved without the use of imaginary numbers with SM math
If 4 then 4 2x
If 4 then 4 2x
The 360-degree circle is a 4400 years old extremely outdated system based on a year of 12 months of 30 days each that is 360 days
SM will use a more logical system that will have a circle of 6283185307o
d
( )
(deg )
d r radians
d r rees
the arc distance around
the circumference of a circle
d
radiusr
2 2
2 2
d r
dv rd t t
v dda rd t t t
1
d
r
( )av tv
The average angular velocity of a body is the rate of its angular displacement
( )0
liminstt
vt
The instantaneous angular velocity is the limit of the ratio
t
as t approaches zero
( )d
av
d
tv
The average circumference velocity of the body is the rate of its distance traveled d divided by the time t
dvv
r
1 dd vdrv
t t t r r
The equation of the relationship between the angular velocity and the velocity of a point on that body
( )o
av
v v v
t ta
The average angular acceleration is the change in the angular velocity
( )0
liminstt
v
ta
The instantaneous angular acceleration is the
limit of the ration of v
t
as t approaches
zero
daa
r
1
1
d
d s d
vva
t r tv a
r t r
The equation of the relationship between the angular acceleration and the acceleration of a point on the body
21
2v a t
If a body with an initial angular velocity ov has a constant
angular acceleration a it will turn through an angle in a time t
1
Straight Line Symbol Units Rotational Symbol Units Distance d m Angular
Distance radians or degrees
Velocity dt
v m
s Angular Velocity
V
t
radians or degrees
s
Acceleration vat
2ms
Angular Acceleration
VA
t
2
radians or degrees
s
Mass M Kg Mass distribution
dM 2Kgm
Force actionA Ma
2
Kgm
sNewton
Torque
dT M A 2
2
Kgm
s
Momentum Mv Kgm
s
Angular Momentum
dM V 2Kgm
s
Work (Force)(distance)
Ad 2
Kgmm
sNm Joule
Work T
Using AS arrow math to solve and direction in space problems using the equation 1 2
2 2 2 2 2 2( ) amp and ( ) amp 3
4
2 2 2
2 2
( ) amp
(5 ) 4 3
(5 ) 16 9
(5 ) 25
(5 ) (5 )
2 2(5 ) 4 3
(5 ) 16 9
(5 ) 25
(5 ) (5 )
In SM math Pythagorasrsquo theorem provides both magnitude and direction
In BS math Pythagorasrsquo theorem provides magnitude only
time set equal to zero
23
( ) ( 0 )10 1
x t xx
3( ) ( 0)
10 1W t f
In BS math this is an equation for a traveling wave
must use a square term in order to make the curve symmetric
NO SQUARE TERM IS NECESSARY
In SM math when t is set to ldquo0rdquo
This is a major fault in BS math
x f(x)
-4 0019
-3 0033
-2 0073
-1 0273
0 3000
1 0273
2 0073
3 0033
4 0019 -4 -3 -2 -1 0 1 2 3 400
05
10
15
20
25
30
35
BS wave (must use square term)
snap shot of wave at x = 0
Am
plit
ude
lt to gt lt to gt f(ltgt)
lt -4 0073
lt -3 0097
lt -2 0143
lt -1 0273
origin 0 3000
gt 1 0273
gt 2 0143
gt 3 0097
gt 4 0073lt0073 lt0097 lt0143 lt0273
3000
0273gt 0143gt 0097gt 0073gt
Wave moving
snap shot of wave
Am
plit
ude
lt4 lt3 lt2 lt1 0 1gt 2gt 3gt 4gt
3
2
1
In SM NO squared terms are needed
SPRING
An Action is needed to stretch or compress a spring to a starting point where it can execute cycle motion
tsA (ts to start) is the work required to move a mass attached to a spring a distance (d1) from
its initial rest position RAK 0 the spring constant is the stiffness of the spring
o A is attraction o R is repulsion o
ARRA
00 is for the back and forth motion through the zero initial rest position A
stretched spring goes from ldquoArdquo to ldquo0rdquo to ldquoRrdquo to ldquoArdquo to ldquo0rdquo to ldquoRrdquo d1 is the distance a mass is moved from its initial rest position
0
0 1ts A RA K d
Once ( )mad
is performed to move the mass to position d1 (against the restraint of the spring) a of the ( )mad
is transferred into the spring as potential attractionrepulsion (PAR) There is always
more ( )mad
performed than is stored as PAR When the spring is released from the restraining Ats the PAR starts the oscillation Conservation of ( )mad
requires that the PAR and the motion ( )mad
( )( )madM be a constant
( )amp ( ) constantmadPAR M 02
0 0
1
2A R A RPAR K d
02
( ) 0 0
1( )
2mad A R A RM m v
0 0 02 2 2
0 0 0 1
2 2 21
2 2 21
1 1 1amp
2 2 2
1 1 1amp
2 2 2
amp
A R A R A RK d m v K d
Kd mv Kd
Kd mv Kd
0
1d is frac12 the height from the base to the peak of a cycle
An experiment of moving paper under a pendulum through a small angle filled with sand that can escape through a small hole in the bottom shows that a pendulum cycling back and forth follows a curve that is mathematically analogous to a lsquosinrsquo or lsquocosrsquo
The projection of the curved area onto the straight line exactly duplicate the back and forth motion of CM This allows for the math of lsquosinrsquo or lsquocosrsquo to be used
SM will use lsquosinrdquo math It could use lsquocosrsquo or a combination of both lsquosinrsquo and lsquocosrsquo Using just the lsquosinrsquo math makes the understanding and math easier
0 sin 0 0SM 0 sin157 1SM The diagram at the left shows a spring being stretched from its (zero rest) position-2 and the PAR being released at position-1 There are 4 sections (1 to 2 2 to 3 3 to 4 and 4 to5) that alternate between maximum acceleration and zero velocity to max velocity and zero acceleration
At position-1 there is maximum acceleration and zero velocity o The lsquosinrsquo values from 0o to 157o will be used for velocity (values from 0 to 1) o The lsquosinrsquo values from 157o to 0o will be used for acceleration (values from 1 to 0)
At position-2 there is maximum velocity and zero acceleration Acceleration changes to deceleration
o The lsquosinrsquo values from 157o to 0o will be used for deceleration (values from 1 to 0) o The lsquosinrsquo values from 0o to 157o will be used for acceleration (values from 0 to 1)
At position-3 there is maximum acceleration and zero velocity o The lsquosinrsquo values from 0o to 157o will be used for velocity (values from 0 to 1) o The lsquosinrsquo values from 157o to 0o will be used for acceleration (values from 1 to 0)
At position-4 there is maximum velocity and zero acceleration Acceleration changes to deceleration
o The lsquosinrsquo values from 157o to 0o will be used for deceleration (values from 1 to 0) o The lsquosinrsquo values from 0o to 157o will be used for acceleration (values from 0 to 1)
At position-5 the cycle starts again
2
)539(
cT
MK
spring constant (K) when mass (M) and cycle-time (T) are known
)(
))(89)((
111 md
uKgM
d
MG
d
FK sts
spring constant (K) when mass (M) and amplitude ( d1) are known
2 21( )
Kv d d
m
velocity (v) at any point during the cycle
velocity
dforM
Kdv
max
012 velocity (v2) maximum (when d = 0 at position-2 and 4)
acceleration (a1) maximum when spring constant (K) mass (M) and amplitude (d1) are known
acceleration (a1) maximum when amplitude (d1) and cycle-time (Tc) are known
acceleration (a) at any point during the cycle-time (T)
1
151
286
v
dTc
cycle-time (Tc) when amplitude (d1) and velocity (v1) are known
K
MTc 28651
cycle-time (Tc) when mass (M) and spring constant (K) are known
fTc
1
cycle-time (Tc) when the frequency is known
cTf
1
frequency (f) when the cycle-time (Tc) is known
of cycles
elapsed timef
frequency (f) when the number of cycles in an elapsed time is known
d1 amplitude (distance from rest to start position)
Also frac12 the distance from the base to the peak
3131 oror dM
Ka
251
3131 539
c
oror T
da
2539
cT
da
In December of 2006 I discovered the mathematical error in Einsteinrsquos Special Relativity Einstein used BS math and that math only works in an imaginary world
It cannot describe real symmetrical space
Einsteinrsquos Special Relativity should be abandoned and a theory of real space using SM should be used to replace it
Numerous articles on the Internet show Einsteinrsquos theory to be incorrect but no one in academia is willing to publish the information
Figure 1-10 A moving particle observed from two frames of reference in uniform translation
Einsteinrsquos Math Error in Special Relativity Transformation of Velocity (From Eisenberg- Fundamentals of Modern Physics pages 27-29) Consider the particle shown in figure (1-10) moving with velocity v as
seen in a frame of reference O
dx dy dz
Vx Vy Vzdt dt dt
dx dy dz
V x V y V zdy dt dt
We would like to evaluate the velocity v of the particle as seen in the frame of reference 0 which is itself moving relative to 0 with velocity v Measured in the 0 frame the velocity vector of the particle has com ponents
The same velocity vector as measured in the 0 frame has components
2
2
2
2
2
1
1
x vtx
v
cy y
z z
vxt
ctv
c
Now from equations (1-13) we know that the relation between the primed and the unprimed coordinates and times is
Take the differential of these equations remembering that v is a constant This gives
2
2
2
2
2
1
1
dx vdtdx
v
cdy dy
dz dz
vxdt
cdtv
c
2
2
2 2 2 2 2
2
2
2
11 ( )
1( ) 1
1
1
1
dx vdt
v dxdx vdt v
dx dx vdtc dt dtvdx vdx vdx v dxdt dt dt dt
dt dtc c c c
VxV x
Vx
VxV x
vv
vv
c
x
v
c
cV
( )
coordinate of particle relative to moving primed system
coordinate of particle relative to stationary system
velocity of primed coordina
V x
Vx
te system relative to stationaryv sys
tem
Page 28 Fundamentals of Modern Physics
(1-16)
According to Einstein this equation tells you how to transform the observed velocity from one frame of reference to another frame of reference First we note that as Vc and vc approachzero the equation approaches those which would be derived from the Galilean transformation
2 2
( ) 1 01 1 ( )
Vx v Vx vV
VxV x
Vx v
v cVxx
v
c
v
cc
Another property of this equation is that it is impossible to choose V and v such that Vrsquo the magnitude of the velocity which is seen in the new frame is greater than c
1
2
09 ( 09 ) 18009876543
( 09 ) 1811 (09 )
c c cV c c
cc
c
Example As seen by O particle 1 has velocity 09c in the direction of positive x and particle 2 has velocity 09c in the negative x direction
Einstein equation however predicts To evaluate the velocity of particle 1 with respect to particle 2 we transform from the O frame to the Orsquo frame moving in the negative x direction with velocity v = -09c using equation 1-16 We obtain
JK LOGIC will observe the particles separating at 18c
A confusing INCORRECT and ILLOGICAL answer
left is a dir rightsub of a
1
2
1
09 ( 09 ) 18009876543
181( 09 )1 ( 09 )
The multiplication of dir left time a dir right is ONLY a dir left
09
dir
direction leftdirection rightsubtraction
c c cV c c
cc
c
V
dir
22
22
09 18 18009876543
181( 081)1(09)(09)
1
left rightsub
sub
c c c cc c
ccc
c
Analyzing Einsteinrsquos BS math Equation
Using SM math a logical answer can be obtained
Cannot multiply a subtraction operator times a direction in space
Can only subtract and change its direction
Cannot multiply a direction to the left times a direction to the right
Cannot multiply opposite directions in space
Can only solve for the resultant
rightleft
subtraction
1
subtraction
2
sub of a dir to the left is a dir to the right
1 resultant 0 sub 2
2sub 2
2
09 09
09 amp 091
09 09 18 1
(0)1
(09) amp (09)1
c cV
c cc
c c cV
cc
cc
8
181
cc
SM math predicts a logical answer using logic
21st century scientists state in many articles that Einsteins General Relativity and Quantum Mechanics are incompatible
They are not only incompatible they are both mathematically incorrect
They both use illogical BS math
Any equations in math that use imaginary numbers should be abandoned
Imaginary numbers are only necessary in the current BS math system
This means that Schroumldingers equations should be abandoned
Any equation in quantum mechanics that uses imaginary number should be abandoned
Prof Kanarev has developed logical physics and chemistrytheories without the use of Quantum mechanics or relativity
A partial list of his theories coverbull Mechanics ndash replaces Newtonbull Photon - bull Electrons ndash Protons- Neutronsbull Spectroscopybull Nucleusbull Atoms amp Moleculesbull Electrodynamicsbull Heatbull Curvature of Space ndash replaces Schwarzcchildrsquos equation
Kanarevrsquos Hypothesis
bull Electron align linearly with proton They do NOT spin or form clouds around the nucleus
bull Neutrons are in the center of atom Proton are outside of Neutrons
bull The wavelength of all particles is the experimental radius
bull Double slit experiment explained correctly
bull Heisenbergrsquos inequality explained correctly
bull is the radius of approach of the magnetic lines around the torus of the electron
( 1 137)
bull Why water expands when it freezes
bull Electron clusters form lightning
rbull The meaning of Plancks constant is explained
22 2 kg m
h m f mr f consts
2210254 42m r kgm bull If the mr-constant describes a ring the equations
describe experiments bull If the mr-constant describes a wave the equations do
NOT describe experiments
Kanarevrsquos 5 Equations of Mechanics
Kanarevrsquos 1st Equation Acceleration ampmv maMma R R
Kanarevrsquos 2nd Equation Uniform motion amp mv maMma R R
maM mvR R
Kanarevrsquos 3rd Equation
Kanarevrsquos 4th Equation Equal and opposite reaction
1 21 21 2
12
a m
m a m ama
Kanarevrsquos 5th Equation 1 2 3 Mi M M M Mna R R R R R
The sum of the Actions working on a moving body is never equal to zero
Kanarevrsquos Photon model is a rolling ring
His equations accurately describe known experiments with photonsbull Electromagnetic radiation should be called photon radiationbull Diffraction and interference are the same phenomenabull Correctly explains the double slit experiment without quantum equationsbull Correctly explains the meaning of Heisenbergrsquos inequality
e
2 (2)(3142)(2817 15 ) 10007297352568
R 2426 12 137mag
xp
r m
m
Charge
Magnetic action
Proton
The intensity of the magnetic action near the geometrical center of the proton ring is so great that we theorize that it is the strong nuclear force
(9383 8)(1602 19)8476 14
4 (4)(3142)(1411 26)p
pp
EB T
M
(9395 8)(1602 19)1035 15
4 (4)(3142)(0966 26)N
NN
EB T
M
Neutron
1a Hydrogenavi
Symmetry-Math Number Line
Symmetry-Math Rules
The dash sign (-) will have only one use subtraction
The cross sign (+) will have only one use addition
Directions have an arrow and a number
All observers see the same direction and magnitude
bull There is no multiplication of arrows
amp ampamp ampampamp
( 3 ) ( 5 ) ( 6 ) ( 65 ) (17256 ) ( 32993 )
dir dirdir dirdirdir
Negative and Positive Directions in Space
If I asked you to point to a negative direction in space which way would you point
Hopefully you will realize that there is no such thing as a negative direction in space
If I asked you to point to a positive direction in space which way would you point
Again hopefully you will realize that there is no such thing as a positive direction in space
Objects on the right side of the x-axis are positive (+)(+) = (+)
Objects on the left side of the x-axis are negative (-)(-) = (+) negative math is different from positive math This is illogical
Objects on the top of the y-axis are positive
Objects on the bottom of the y-axis are negative
Objects in the front of the z-axis are positive
Objects in the back of the z-axis are negative
Again BS math of the positive direction (x right y up and z front) is different from the BS math of the negative direction (x left y down and z back)
BS math is Broken-Symmetry
Mirror image is broken symmetry in the x and z axis If the left amp right-axis are
reversed symmetry is broken Obs-1 math is different from obs-2
If the front amp back-axis is reversed symmetry is broken Obs-1 math is different from obs-2
If the top amp bottom-axis are
reversed symmetry is not broken Obs-1 math is the same as obs-2
Math answers cannot depend on which side of a number line an observer sit
Absolute values for Displacement
Even for a simple displacement of an object BS math invented a definition and absolute values to obtain something labeled a positive number answer
I----------I----------I----------I----------I----------I----------I----------I----------I
-4 -3 -2 -1 0 +1 +2 +3 +4 a gtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtb cltltltltltltltltltltltltltltltltltltltltltltltltltltltb
If an object starts at ( 3)ia X and moves to ( X 3) and the to ( 1)b fb c X
BS Math provides a usable answer without a definition or absolute values
However If an object starts at
( 1) and moves to ( X 3) and the to ( 3)i b fa X b c X
Then absolute values and a definition must be used
---------I----------I----------I----------I----------I----------I----------I---------- -4 -3 -2 -1 0 +1 +2 +3 +4
a gtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtb cltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltb
BS Math makes up a definition that all displacements are positive Since this example provides an answer with something labeled a negative the positive definition must be applied and (-2) is changed to (+2) A definition is needed to arrive at a useable answer This is Illogical
In SM absolute values and a meaningless definition of ldquopositiverdquo are removed In SM objects and directions in space are defined by the direction and magnitude of the resultant of arrows
With SM you get total distance traveled by the object and the final direction of the displacement
22222 2 bababbaababababa If w e su b stitu te p o s itiv e n u m b e rs in to th e eq u a tio n th e an sw er w ill b e co rrec t If (a = 5 ) an d (b = 3 )
2
2
8
5 3 5 3 5 3 5 5 5 3 3 5 3 3 25 15 15 9 64
8 8 8 8 8 8 8 8 8 8 64
W h en tw o p o sitiv e n u m b ers a re a d d ed an d sq u a re d th e B S m a th d is tr ib u tiv e law p ro v id es a co rre c t an sw er 1
Illogical and Incorrect BS math for the distributive law
222 2 bababbabbaaabababa B y d e fin itio n in B S m a th n o n u m b er sq u a red c a n ev e r b e n eg a tiv e T h e re fo re n o sq u a red n u m b er c an ev e r b e a v e c to r m o v in g in th e d a sh ( -) o r n eg a tiv e d ire c tio n if it is sq u a red In B S m a th an d ad d itio n o p e ra to r m u ltip lied b y a su b trac tio n o p e ra to r is A L W A Y S eq u a l to a su b tra c tio n o p e ra to r If th is h ad n o t b e en u sed fo r th e la s t fo u r h u n d red y e a rs y o u w o u ld d ie lau g h in g a t th e u tte r ab su rd ity o f th e la s t s ta tem e n t T h is is illo g ica l an d v io la te s sy m m etry
If a = 5 an d b = 3 42222352
22
4930253)3)(5(253)5)(3()3)(5(5353535 22222 A s lo n g a s ldquo ardquo is g re a te r th an ldquob rdquo th e B S m a th d is tr ib u tiv e law p ro v id es co rrec t an sw ers 1
B S m a th b re a k s d o w n w h e n ldquo b rdquo is g rea ter th a n ldquo a rdquo
If a = 3 an d b = 5
2 2
2
2
3 5 2 4
2 ( 2) ( 2) 4
2 ( )( 2) ( )( 2) 4
A n y co m b in a tio n ex cep t th e m id d le o n e is illo g ica l an d in co rre c t B y d e fin itio n th e an sw er can n o t b e -4 in B S m a th B S m a th is in co rrec t
4253095)5)(3(235)3)(5()5)(3(3535353 22222 T h is is w h ere th e d is tr ib u tiv e la w p ro v id es in co rrec t an sw ers I t p ro d u ces an an sw er o f + 4 b e cau se b y d e fin iti o n n o n u m b er sq u a red in B S m a th c an b e a n eg a tiv e n u m b er U sin g in co rre c t B S m a th th e d is tr ib u tiv e law w ill g iv e an in co rrec t an sw ers o f + 4 T h is is in co rrec t an d illo g ica l 1
Since there can be no negative direction from a squared term in BS math BS math should be modified or abandoned
The BS distributive law should be modified or abandoned
The numbers +1+2+3+4 and ndash1-2-3-4hellip should be abandoned
BS math gives incorrect answers because their Rule-of-Signs allows multiplying an arrow going in one direction by an arrow going in another direction
Clearly this is not logical
Arrows should be used that are specific for directions
W h at d o es it m e an to m u ltip ly ( 4) b y (4 ) A s w e k n o w 1
m u ltip lica tio n is ju s t ad d itio n W h a t w o u ld ( 4) tim e (4 ) b e 2
In BS math the answer is by definition -16 unless you square -4 and then it is +16
In SM you cannot multiply arrows going in opposite directions
The middle two terms are not logical You cannot multiply opposite directions
This is a MAJOR error in BS math They multiply a dash [(-) a direction to the left] by a cross [(+) a direction to the right] And then ldquoby definitionrdquo label the answer dash (-) This is illogical and produces incorrect answers
2
2 2
This is Einsteinrsquos math error
in Special relativity
Magnitude 4 Direction 1 Magnitude 3 Direction 2 Magnitude 5 Direction 3 Relative to an observer at 0 4
Magnitude 4 Direction 1 Magnitude 3 Direction 2 Magnitude 5 Direction 3 Relative to an observer at 0 4
Magnitude and Direction (MD) An arrow has both a magnitude andA direction relative to a 2nd objectobserver
bull Magnitude and velocity (MV) relative to a 2nd objectobserver
bull Magnitude And acceleration (MA) relative to a 2nd objectobserver2
v da
t t
Introducing a new symbol the ldquoamprdquo sign The ldquoamprdquo sign means resultant ADD all the Arrows for each direction and then SUBTRACT to find the final resultant
1 1amp 1 2 2amp 2 4 If =
1 amp1 2 2 amp2 4 If =
amp 5 amp 3 2 If gt
amp 3 amp 5= 2 If gt
amp 0 3 amp 3=0 if =
only the resultant equals zero
the vectors do not - they are still there
Does not exist There is no multiplication of arrows
There are two arrows One has a magnitude of 3 One has a magnitude of 5 The maximum these two arrows can be is 8
Only the resultant of arrows exist
An SM arrow has a direction and a magnitude
Arrows and magnitudes are still there even if the resultant is zero
There are no negative SM Arrows
There are no positive SM Arrows
There are no ZERO SM Arrows A zero would negate the definition of an SM Arrow something with no magnitude and no direction cannot be the definition of an arrow that is defined as having magnitude and direction
bull A resultant is the addition and subtraction of the differences in magnitude and direction arrows
A B and C are the arrows (a) A amp B exist(b) A amp B amp C exist(c) A-B Do Not Exist in Sm there are no negatives or positives(d) A+B-C Do Not Exist in SM there are no negatives or positives
There are only resultants when there are multiple arrows
SM Graphs versus BS Graphs
Numbers Raised to Powers
When numbers are raised to powers the negative sides of all BS math graphs are ILLOGICAL and INCORRECT
Incorrect BS Math for 2x
2x
4 16 correct3 8 correct2 4 correct1 2 correct0 1 correct-1 005 incorrect-2 025 incorrect-3 0125 incorrect-4 00625 incorrect
negative side of graph is incorrectpositive side of graph is correct
16
8
42105025012500625
-4 -3 -2 -1 0 1 2 3 4
po
sit
ive
negative positive
2x BS Math
correctincorrect
In BS math the dash sign (-) used in this example represents three different operations They arebull (-) Subtractionbull (-) A direction to the leftbull (-) As an exponent to mean divide by
All positive numbers produce correct results and graphs
All negative numbers produce incorrect results and graphs
A direction of three units to the left (-3) is placed in the equation as an exponent
A direction in space (3 units toward the left) changes a whole number to a fraction by using the dash (-) as an exponent rather than a direction in space
This is illogical and incorrect BS math should have performed this operation as
Unfortunately the answers to problems can come out correct because data points of a real problem fit the curve
However the answers will not be due to logical reasoning
Because the equation produced a curve that real problem data points follow it has been falsely believed that the equations and graphs are correct
1
2
x
Incorrect BK graph for (12)x
(12)x
4 00625 correct3 0125 correct2 025 correct1 1 correct0 1 correct-1 2 incorrect-2 4 incorrect-3 8 incorrect-4 16 incorrect
negative side of graph is incorrectpositive side of graph is correct
00625012502505124
8
16
-5 -4 -3 -2 -1 0 1 2 3 4 5P
osit
ive
negative positive
BS Math for (12)x
correctincorrect
Incorrect BS math and graph for
All graphs are symmetrical with SM
Two BS Math Equations that Produce Incorrect Answers
2 0x x starting equation 2
2
x x x x
x x
Add lsquoxrsquo to both sides
2(1) 1 let lsquoxrsquo = 1
2 1 This is an incorrect answer 2(0)-(0)=0 The only answer that BK math can give is x=0
1
a b starting equation ( )( ) ( )( )a a a b multiply both sides by ( )a
( )( ) ( )( ) ( )( ) ( )( )a a b b a b b b subtract ( )( )b b from both sides
( )( ) ( )( )a b a b a b b factor ( )a b from both sides
( )a b b divide both sides by ( )a b
( )b b b replace ( )a with ( )b 2b b simplify 2(1) (1) Let ( )b = 1
2 1 Again BS math yields and incorrect answer 1
This incorrect result is a problem with the distributive law SM explains how this incorrect answer is obtained and the problem with the distributive law that allows the error
Page 148 ndash149 Barry Mazur ndash Imagining Numbers
3
2
31
Do the computation carefully on paper using the rules we agreed to Then ponder your answer which should be something of a surprise to you donrsquot stop there Think of what your answer might possibly mean or might imply
Solve this equation using the distributive law
JK definitely a surprise the answer wasincorrect and illogical
Gerhard Gade University Professor doing math at Harvard University
Here is the BS math answer that Barry Mazur calculated
2
31
2
31
2
31
1 3 1 3 1 3 3 3 2 2 3 1 3
2 2 2 2 4 4 4 4 4 4 2 2
dash
1 3 1 3 1 31
2 2 2 2 4 4
3 3
4 4
When the distributive law is used with the BS Rule-of-Signs and one of the signs is a (-) dash sign the middle two terms always cancel each other This leads to the incorrect answer of (-1) in the above equation
Barry Mazur did not go back and check his answer with the original equation If he had he would have known that the answer was incorrect
3
1 31
2
check answer by setting equal
3
331 3
12
1 31
2
Take the Cube root of both sides
1 3 2 multiply both sides by 2
3 3 subtract 1 from both sides
2 2
3 3 square both sides
3 9 This is an incorrect answer A simple equation check could have shown that the distributive law using BS math provides an incorrect answer
1
BS Illogical and Incorrect Numbers Raised to Powers
nC D n DC
642 DDD
In BS math only a positive number can have an answer if ldquonrdquo is an even number (246---)
All negative numbers have no solutions
642 DDD
With SM just add a direction arrow and any number gives the correct answer
or n nC D C D
3 3( 3 8 8 or 8 2 or 2 )n D C
( 2 16 C= 16 or 16 4 or 4 )n D
Since a negative number has no square root answer in BS math BS mathematicians invented imaginary numbers
i2=-1
24414 iiC n
These equations are easily solved without the use of imaginary numbers with SM math
If 4 then 4 2x
If 4 then 4 2x
The 360-degree circle is a 4400 years old extremely outdated system based on a year of 12 months of 30 days each that is 360 days
SM will use a more logical system that will have a circle of 6283185307o
d
( )
(deg )
d r radians
d r rees
the arc distance around
the circumference of a circle
d
radiusr
2 2
2 2
d r
dv rd t t
v dda rd t t t
1
d
r
( )av tv
The average angular velocity of a body is the rate of its angular displacement
( )0
liminstt
vt
The instantaneous angular velocity is the limit of the ratio
t
as t approaches zero
( )d
av
d
tv
The average circumference velocity of the body is the rate of its distance traveled d divided by the time t
dvv
r
1 dd vdrv
t t t r r
The equation of the relationship between the angular velocity and the velocity of a point on that body
( )o
av
v v v
t ta
The average angular acceleration is the change in the angular velocity
( )0
liminstt
v
ta
The instantaneous angular acceleration is the
limit of the ration of v
t
as t approaches
zero
daa
r
1
1
d
d s d
vva
t r tv a
r t r
The equation of the relationship between the angular acceleration and the acceleration of a point on the body
21
2v a t
If a body with an initial angular velocity ov has a constant
angular acceleration a it will turn through an angle in a time t
1
Straight Line Symbol Units Rotational Symbol Units Distance d m Angular
Distance radians or degrees
Velocity dt
v m
s Angular Velocity
V
t
radians or degrees
s
Acceleration vat
2ms
Angular Acceleration
VA
t
2
radians or degrees
s
Mass M Kg Mass distribution
dM 2Kgm
Force actionA Ma
2
Kgm
sNewton
Torque
dT M A 2
2
Kgm
s
Momentum Mv Kgm
s
Angular Momentum
dM V 2Kgm
s
Work (Force)(distance)
Ad 2
Kgmm
sNm Joule
Work T
Using AS arrow math to solve and direction in space problems using the equation 1 2
2 2 2 2 2 2( ) amp and ( ) amp 3
4
2 2 2
2 2
( ) amp
(5 ) 4 3
(5 ) 16 9
(5 ) 25
(5 ) (5 )
2 2(5 ) 4 3
(5 ) 16 9
(5 ) 25
(5 ) (5 )
In SM math Pythagorasrsquo theorem provides both magnitude and direction
In BS math Pythagorasrsquo theorem provides magnitude only
time set equal to zero
23
( ) ( 0 )10 1
x t xx
3( ) ( 0)
10 1W t f
In BS math this is an equation for a traveling wave
must use a square term in order to make the curve symmetric
NO SQUARE TERM IS NECESSARY
In SM math when t is set to ldquo0rdquo
This is a major fault in BS math
x f(x)
-4 0019
-3 0033
-2 0073
-1 0273
0 3000
1 0273
2 0073
3 0033
4 0019 -4 -3 -2 -1 0 1 2 3 400
05
10
15
20
25
30
35
BS wave (must use square term)
snap shot of wave at x = 0
Am
plit
ude
lt to gt lt to gt f(ltgt)
lt -4 0073
lt -3 0097
lt -2 0143
lt -1 0273
origin 0 3000
gt 1 0273
gt 2 0143
gt 3 0097
gt 4 0073lt0073 lt0097 lt0143 lt0273
3000
0273gt 0143gt 0097gt 0073gt
Wave moving
snap shot of wave
Am
plit
ude
lt4 lt3 lt2 lt1 0 1gt 2gt 3gt 4gt
3
2
1
In SM NO squared terms are needed
SPRING
An Action is needed to stretch or compress a spring to a starting point where it can execute cycle motion
tsA (ts to start) is the work required to move a mass attached to a spring a distance (d1) from
its initial rest position RAK 0 the spring constant is the stiffness of the spring
o A is attraction o R is repulsion o
ARRA
00 is for the back and forth motion through the zero initial rest position A
stretched spring goes from ldquoArdquo to ldquo0rdquo to ldquoRrdquo to ldquoArdquo to ldquo0rdquo to ldquoRrdquo d1 is the distance a mass is moved from its initial rest position
0
0 1ts A RA K d
Once ( )mad
is performed to move the mass to position d1 (against the restraint of the spring) a of the ( )mad
is transferred into the spring as potential attractionrepulsion (PAR) There is always
more ( )mad
performed than is stored as PAR When the spring is released from the restraining Ats the PAR starts the oscillation Conservation of ( )mad
requires that the PAR and the motion ( )mad
( )( )madM be a constant
( )amp ( ) constantmadPAR M 02
0 0
1
2A R A RPAR K d
02
( ) 0 0
1( )
2mad A R A RM m v
0 0 02 2 2
0 0 0 1
2 2 21
2 2 21
1 1 1amp
2 2 2
1 1 1amp
2 2 2
amp
A R A R A RK d m v K d
Kd mv Kd
Kd mv Kd
0
1d is frac12 the height from the base to the peak of a cycle
An experiment of moving paper under a pendulum through a small angle filled with sand that can escape through a small hole in the bottom shows that a pendulum cycling back and forth follows a curve that is mathematically analogous to a lsquosinrsquo or lsquocosrsquo
The projection of the curved area onto the straight line exactly duplicate the back and forth motion of CM This allows for the math of lsquosinrsquo or lsquocosrsquo to be used
SM will use lsquosinrdquo math It could use lsquocosrsquo or a combination of both lsquosinrsquo and lsquocosrsquo Using just the lsquosinrsquo math makes the understanding and math easier
0 sin 0 0SM 0 sin157 1SM The diagram at the left shows a spring being stretched from its (zero rest) position-2 and the PAR being released at position-1 There are 4 sections (1 to 2 2 to 3 3 to 4 and 4 to5) that alternate between maximum acceleration and zero velocity to max velocity and zero acceleration
At position-1 there is maximum acceleration and zero velocity o The lsquosinrsquo values from 0o to 157o will be used for velocity (values from 0 to 1) o The lsquosinrsquo values from 157o to 0o will be used for acceleration (values from 1 to 0)
At position-2 there is maximum velocity and zero acceleration Acceleration changes to deceleration
o The lsquosinrsquo values from 157o to 0o will be used for deceleration (values from 1 to 0) o The lsquosinrsquo values from 0o to 157o will be used for acceleration (values from 0 to 1)
At position-3 there is maximum acceleration and zero velocity o The lsquosinrsquo values from 0o to 157o will be used for velocity (values from 0 to 1) o The lsquosinrsquo values from 157o to 0o will be used for acceleration (values from 1 to 0)
At position-4 there is maximum velocity and zero acceleration Acceleration changes to deceleration
o The lsquosinrsquo values from 157o to 0o will be used for deceleration (values from 1 to 0) o The lsquosinrsquo values from 0o to 157o will be used for acceleration (values from 0 to 1)
At position-5 the cycle starts again
2
)539(
cT
MK
spring constant (K) when mass (M) and cycle-time (T) are known
)(
))(89)((
111 md
uKgM
d
MG
d
FK sts
spring constant (K) when mass (M) and amplitude ( d1) are known
2 21( )
Kv d d
m
velocity (v) at any point during the cycle
velocity
dforM
Kdv
max
012 velocity (v2) maximum (when d = 0 at position-2 and 4)
acceleration (a1) maximum when spring constant (K) mass (M) and amplitude (d1) are known
acceleration (a1) maximum when amplitude (d1) and cycle-time (Tc) are known
acceleration (a) at any point during the cycle-time (T)
1
151
286
v
dTc
cycle-time (Tc) when amplitude (d1) and velocity (v1) are known
K
MTc 28651
cycle-time (Tc) when mass (M) and spring constant (K) are known
fTc
1
cycle-time (Tc) when the frequency is known
cTf
1
frequency (f) when the cycle-time (Tc) is known
of cycles
elapsed timef
frequency (f) when the number of cycles in an elapsed time is known
d1 amplitude (distance from rest to start position)
Also frac12 the distance from the base to the peak
3131 oror dM
Ka
251
3131 539
c
oror T
da
2539
cT
da
In December of 2006 I discovered the mathematical error in Einsteinrsquos Special Relativity Einstein used BS math and that math only works in an imaginary world
It cannot describe real symmetrical space
Einsteinrsquos Special Relativity should be abandoned and a theory of real space using SM should be used to replace it
Numerous articles on the Internet show Einsteinrsquos theory to be incorrect but no one in academia is willing to publish the information
Figure 1-10 A moving particle observed from two frames of reference in uniform translation
Einsteinrsquos Math Error in Special Relativity Transformation of Velocity (From Eisenberg- Fundamentals of Modern Physics pages 27-29) Consider the particle shown in figure (1-10) moving with velocity v as
seen in a frame of reference O
dx dy dz
Vx Vy Vzdt dt dt
dx dy dz
V x V y V zdy dt dt
We would like to evaluate the velocity v of the particle as seen in the frame of reference 0 which is itself moving relative to 0 with velocity v Measured in the 0 frame the velocity vector of the particle has com ponents
The same velocity vector as measured in the 0 frame has components
2
2
2
2
2
1
1
x vtx
v
cy y
z z
vxt
ctv
c
Now from equations (1-13) we know that the relation between the primed and the unprimed coordinates and times is
Take the differential of these equations remembering that v is a constant This gives
2
2
2
2
2
1
1
dx vdtdx
v
cdy dy
dz dz
vxdt
cdtv
c
2
2
2 2 2 2 2
2
2
2
11 ( )
1( ) 1
1
1
1
dx vdt
v dxdx vdt v
dx dx vdtc dt dtvdx vdx vdx v dxdt dt dt dt
dt dtc c c c
VxV x
Vx
VxV x
vv
vv
c
x
v
c
cV
( )
coordinate of particle relative to moving primed system
coordinate of particle relative to stationary system
velocity of primed coordina
V x
Vx
te system relative to stationaryv sys
tem
Page 28 Fundamentals of Modern Physics
(1-16)
According to Einstein this equation tells you how to transform the observed velocity from one frame of reference to another frame of reference First we note that as Vc and vc approachzero the equation approaches those which would be derived from the Galilean transformation
2 2
( ) 1 01 1 ( )
Vx v Vx vV
VxV x
Vx v
v cVxx
v
c
v
cc
Another property of this equation is that it is impossible to choose V and v such that Vrsquo the magnitude of the velocity which is seen in the new frame is greater than c
1
2
09 ( 09 ) 18009876543
( 09 ) 1811 (09 )
c c cV c c
cc
c
Example As seen by O particle 1 has velocity 09c in the direction of positive x and particle 2 has velocity 09c in the negative x direction
Einstein equation however predicts To evaluate the velocity of particle 1 with respect to particle 2 we transform from the O frame to the Orsquo frame moving in the negative x direction with velocity v = -09c using equation 1-16 We obtain
JK LOGIC will observe the particles separating at 18c
A confusing INCORRECT and ILLOGICAL answer
left is a dir rightsub of a
1
2
1
09 ( 09 ) 18009876543
181( 09 )1 ( 09 )
The multiplication of dir left time a dir right is ONLY a dir left
09
dir
direction leftdirection rightsubtraction
c c cV c c
cc
c
V
dir
22
22
09 18 18009876543
181( 081)1(09)(09)
1
left rightsub
sub
c c c cc c
ccc
c
Analyzing Einsteinrsquos BS math Equation
Using SM math a logical answer can be obtained
Cannot multiply a subtraction operator times a direction in space
Can only subtract and change its direction
Cannot multiply a direction to the left times a direction to the right
Cannot multiply opposite directions in space
Can only solve for the resultant
rightleft
subtraction
1
subtraction
2
sub of a dir to the left is a dir to the right
1 resultant 0 sub 2
2sub 2
2
09 09
09 amp 091
09 09 18 1
(0)1
(09) amp (09)1
c cV
c cc
c c cV
cc
cc
8
181
cc
SM math predicts a logical answer using logic
21st century scientists state in many articles that Einsteins General Relativity and Quantum Mechanics are incompatible
They are not only incompatible they are both mathematically incorrect
They both use illogical BS math
Any equations in math that use imaginary numbers should be abandoned
Imaginary numbers are only necessary in the current BS math system
This means that Schroumldingers equations should be abandoned
Any equation in quantum mechanics that uses imaginary number should be abandoned
Prof Kanarev has developed logical physics and chemistrytheories without the use of Quantum mechanics or relativity
A partial list of his theories coverbull Mechanics ndash replaces Newtonbull Photon - bull Electrons ndash Protons- Neutronsbull Spectroscopybull Nucleusbull Atoms amp Moleculesbull Electrodynamicsbull Heatbull Curvature of Space ndash replaces Schwarzcchildrsquos equation
Kanarevrsquos Hypothesis
bull Electron align linearly with proton They do NOT spin or form clouds around the nucleus
bull Neutrons are in the center of atom Proton are outside of Neutrons
bull The wavelength of all particles is the experimental radius
bull Double slit experiment explained correctly
bull Heisenbergrsquos inequality explained correctly
bull is the radius of approach of the magnetic lines around the torus of the electron
( 1 137)
bull Why water expands when it freezes
bull Electron clusters form lightning
rbull The meaning of Plancks constant is explained
22 2 kg m
h m f mr f consts
2210254 42m r kgm bull If the mr-constant describes a ring the equations
describe experiments bull If the mr-constant describes a wave the equations do
NOT describe experiments
Kanarevrsquos 5 Equations of Mechanics
Kanarevrsquos 1st Equation Acceleration ampmv maMma R R
Kanarevrsquos 2nd Equation Uniform motion amp mv maMma R R
maM mvR R
Kanarevrsquos 3rd Equation
Kanarevrsquos 4th Equation Equal and opposite reaction
1 21 21 2
12
a m
m a m ama
Kanarevrsquos 5th Equation 1 2 3 Mi M M M Mna R R R R R
The sum of the Actions working on a moving body is never equal to zero
Kanarevrsquos Photon model is a rolling ring
His equations accurately describe known experiments with photonsbull Electromagnetic radiation should be called photon radiationbull Diffraction and interference are the same phenomenabull Correctly explains the double slit experiment without quantum equationsbull Correctly explains the meaning of Heisenbergrsquos inequality
e
2 (2)(3142)(2817 15 ) 10007297352568
R 2426 12 137mag
xp
r m
m
Charge
Magnetic action
Proton
The intensity of the magnetic action near the geometrical center of the proton ring is so great that we theorize that it is the strong nuclear force
(9383 8)(1602 19)8476 14
4 (4)(3142)(1411 26)p
pp
EB T
M
(9395 8)(1602 19)1035 15
4 (4)(3142)(0966 26)N
NN
EB T
M
Neutron
1a Hydrogenavi
Symmetry-Math Rules
The dash sign (-) will have only one use subtraction
The cross sign (+) will have only one use addition
Directions have an arrow and a number
All observers see the same direction and magnitude
bull There is no multiplication of arrows
amp ampamp ampampamp
( 3 ) ( 5 ) ( 6 ) ( 65 ) (17256 ) ( 32993 )
dir dirdir dirdirdir
Negative and Positive Directions in Space
If I asked you to point to a negative direction in space which way would you point
Hopefully you will realize that there is no such thing as a negative direction in space
If I asked you to point to a positive direction in space which way would you point
Again hopefully you will realize that there is no such thing as a positive direction in space
Objects on the right side of the x-axis are positive (+)(+) = (+)
Objects on the left side of the x-axis are negative (-)(-) = (+) negative math is different from positive math This is illogical
Objects on the top of the y-axis are positive
Objects on the bottom of the y-axis are negative
Objects in the front of the z-axis are positive
Objects in the back of the z-axis are negative
Again BS math of the positive direction (x right y up and z front) is different from the BS math of the negative direction (x left y down and z back)
BS math is Broken-Symmetry
Mirror image is broken symmetry in the x and z axis If the left amp right-axis are
reversed symmetry is broken Obs-1 math is different from obs-2
If the front amp back-axis is reversed symmetry is broken Obs-1 math is different from obs-2
If the top amp bottom-axis are
reversed symmetry is not broken Obs-1 math is the same as obs-2
Math answers cannot depend on which side of a number line an observer sit
Absolute values for Displacement
Even for a simple displacement of an object BS math invented a definition and absolute values to obtain something labeled a positive number answer
I----------I----------I----------I----------I----------I----------I----------I----------I
-4 -3 -2 -1 0 +1 +2 +3 +4 a gtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtb cltltltltltltltltltltltltltltltltltltltltltltltltltltltb
If an object starts at ( 3)ia X and moves to ( X 3) and the to ( 1)b fb c X
BS Math provides a usable answer without a definition or absolute values
However If an object starts at
( 1) and moves to ( X 3) and the to ( 3)i b fa X b c X
Then absolute values and a definition must be used
---------I----------I----------I----------I----------I----------I----------I---------- -4 -3 -2 -1 0 +1 +2 +3 +4
a gtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtb cltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltb
BS Math makes up a definition that all displacements are positive Since this example provides an answer with something labeled a negative the positive definition must be applied and (-2) is changed to (+2) A definition is needed to arrive at a useable answer This is Illogical
In SM absolute values and a meaningless definition of ldquopositiverdquo are removed In SM objects and directions in space are defined by the direction and magnitude of the resultant of arrows
With SM you get total distance traveled by the object and the final direction of the displacement
22222 2 bababbaababababa If w e su b stitu te p o s itiv e n u m b e rs in to th e eq u a tio n th e an sw er w ill b e co rrec t If (a = 5 ) an d (b = 3 )
2
2
8
5 3 5 3 5 3 5 5 5 3 3 5 3 3 25 15 15 9 64
8 8 8 8 8 8 8 8 8 8 64
W h en tw o p o sitiv e n u m b ers a re a d d ed an d sq u a re d th e B S m a th d is tr ib u tiv e law p ro v id es a co rre c t an sw er 1
Illogical and Incorrect BS math for the distributive law
222 2 bababbabbaaabababa B y d e fin itio n in B S m a th n o n u m b er sq u a red c a n ev e r b e n eg a tiv e T h e re fo re n o sq u a red n u m b er c an ev e r b e a v e c to r m o v in g in th e d a sh ( -) o r n eg a tiv e d ire c tio n if it is sq u a red In B S m a th an d ad d itio n o p e ra to r m u ltip lied b y a su b trac tio n o p e ra to r is A L W A Y S eq u a l to a su b tra c tio n o p e ra to r If th is h ad n o t b e en u sed fo r th e la s t fo u r h u n d red y e a rs y o u w o u ld d ie lau g h in g a t th e u tte r ab su rd ity o f th e la s t s ta tem e n t T h is is illo g ica l an d v io la te s sy m m etry
If a = 5 an d b = 3 42222352
22
4930253)3)(5(253)5)(3()3)(5(5353535 22222 A s lo n g a s ldquo ardquo is g re a te r th an ldquob rdquo th e B S m a th d is tr ib u tiv e law p ro v id es co rrec t an sw ers 1
B S m a th b re a k s d o w n w h e n ldquo b rdquo is g rea ter th a n ldquo a rdquo
If a = 3 an d b = 5
2 2
2
2
3 5 2 4
2 ( 2) ( 2) 4
2 ( )( 2) ( )( 2) 4
A n y co m b in a tio n ex cep t th e m id d le o n e is illo g ica l an d in co rre c t B y d e fin itio n th e an sw er can n o t b e -4 in B S m a th B S m a th is in co rrec t
4253095)5)(3(235)3)(5()5)(3(3535353 22222 T h is is w h ere th e d is tr ib u tiv e la w p ro v id es in co rrec t an sw ers I t p ro d u ces an an sw er o f + 4 b e cau se b y d e fin iti o n n o n u m b er sq u a red in B S m a th c an b e a n eg a tiv e n u m b er U sin g in co rre c t B S m a th th e d is tr ib u tiv e law w ill g iv e an in co rrec t an sw ers o f + 4 T h is is in co rrec t an d illo g ica l 1
Since there can be no negative direction from a squared term in BS math BS math should be modified or abandoned
The BS distributive law should be modified or abandoned
The numbers +1+2+3+4 and ndash1-2-3-4hellip should be abandoned
BS math gives incorrect answers because their Rule-of-Signs allows multiplying an arrow going in one direction by an arrow going in another direction
Clearly this is not logical
Arrows should be used that are specific for directions
W h at d o es it m e an to m u ltip ly ( 4) b y (4 ) A s w e k n o w 1
m u ltip lica tio n is ju s t ad d itio n W h a t w o u ld ( 4) tim e (4 ) b e 2
In BS math the answer is by definition -16 unless you square -4 and then it is +16
In SM you cannot multiply arrows going in opposite directions
The middle two terms are not logical You cannot multiply opposite directions
This is a MAJOR error in BS math They multiply a dash [(-) a direction to the left] by a cross [(+) a direction to the right] And then ldquoby definitionrdquo label the answer dash (-) This is illogical and produces incorrect answers
2
2 2
This is Einsteinrsquos math error
in Special relativity
Magnitude 4 Direction 1 Magnitude 3 Direction 2 Magnitude 5 Direction 3 Relative to an observer at 0 4
Magnitude 4 Direction 1 Magnitude 3 Direction 2 Magnitude 5 Direction 3 Relative to an observer at 0 4
Magnitude and Direction (MD) An arrow has both a magnitude andA direction relative to a 2nd objectobserver
bull Magnitude and velocity (MV) relative to a 2nd objectobserver
bull Magnitude And acceleration (MA) relative to a 2nd objectobserver2
v da
t t
Introducing a new symbol the ldquoamprdquo sign The ldquoamprdquo sign means resultant ADD all the Arrows for each direction and then SUBTRACT to find the final resultant
1 1amp 1 2 2amp 2 4 If =
1 amp1 2 2 amp2 4 If =
amp 5 amp 3 2 If gt
amp 3 amp 5= 2 If gt
amp 0 3 amp 3=0 if =
only the resultant equals zero
the vectors do not - they are still there
Does not exist There is no multiplication of arrows
There are two arrows One has a magnitude of 3 One has a magnitude of 5 The maximum these two arrows can be is 8
Only the resultant of arrows exist
An SM arrow has a direction and a magnitude
Arrows and magnitudes are still there even if the resultant is zero
There are no negative SM Arrows
There are no positive SM Arrows
There are no ZERO SM Arrows A zero would negate the definition of an SM Arrow something with no magnitude and no direction cannot be the definition of an arrow that is defined as having magnitude and direction
bull A resultant is the addition and subtraction of the differences in magnitude and direction arrows
A B and C are the arrows (a) A amp B exist(b) A amp B amp C exist(c) A-B Do Not Exist in Sm there are no negatives or positives(d) A+B-C Do Not Exist in SM there are no negatives or positives
There are only resultants when there are multiple arrows
SM Graphs versus BS Graphs
Numbers Raised to Powers
When numbers are raised to powers the negative sides of all BS math graphs are ILLOGICAL and INCORRECT
Incorrect BS Math for 2x
2x
4 16 correct3 8 correct2 4 correct1 2 correct0 1 correct-1 005 incorrect-2 025 incorrect-3 0125 incorrect-4 00625 incorrect
negative side of graph is incorrectpositive side of graph is correct
16
8
42105025012500625
-4 -3 -2 -1 0 1 2 3 4
po
sit
ive
negative positive
2x BS Math
correctincorrect
In BS math the dash sign (-) used in this example represents three different operations They arebull (-) Subtractionbull (-) A direction to the leftbull (-) As an exponent to mean divide by
All positive numbers produce correct results and graphs
All negative numbers produce incorrect results and graphs
A direction of three units to the left (-3) is placed in the equation as an exponent
A direction in space (3 units toward the left) changes a whole number to a fraction by using the dash (-) as an exponent rather than a direction in space
This is illogical and incorrect BS math should have performed this operation as
Unfortunately the answers to problems can come out correct because data points of a real problem fit the curve
However the answers will not be due to logical reasoning
Because the equation produced a curve that real problem data points follow it has been falsely believed that the equations and graphs are correct
1
2
x
Incorrect BK graph for (12)x
(12)x
4 00625 correct3 0125 correct2 025 correct1 1 correct0 1 correct-1 2 incorrect-2 4 incorrect-3 8 incorrect-4 16 incorrect
negative side of graph is incorrectpositive side of graph is correct
00625012502505124
8
16
-5 -4 -3 -2 -1 0 1 2 3 4 5P
osit
ive
negative positive
BS Math for (12)x
correctincorrect
Incorrect BS math and graph for
All graphs are symmetrical with SM
Two BS Math Equations that Produce Incorrect Answers
2 0x x starting equation 2
2
x x x x
x x
Add lsquoxrsquo to both sides
2(1) 1 let lsquoxrsquo = 1
2 1 This is an incorrect answer 2(0)-(0)=0 The only answer that BK math can give is x=0
1
a b starting equation ( )( ) ( )( )a a a b multiply both sides by ( )a
( )( ) ( )( ) ( )( ) ( )( )a a b b a b b b subtract ( )( )b b from both sides
( )( ) ( )( )a b a b a b b factor ( )a b from both sides
( )a b b divide both sides by ( )a b
( )b b b replace ( )a with ( )b 2b b simplify 2(1) (1) Let ( )b = 1
2 1 Again BS math yields and incorrect answer 1
This incorrect result is a problem with the distributive law SM explains how this incorrect answer is obtained and the problem with the distributive law that allows the error
Page 148 ndash149 Barry Mazur ndash Imagining Numbers
3
2
31
Do the computation carefully on paper using the rules we agreed to Then ponder your answer which should be something of a surprise to you donrsquot stop there Think of what your answer might possibly mean or might imply
Solve this equation using the distributive law
JK definitely a surprise the answer wasincorrect and illogical
Gerhard Gade University Professor doing math at Harvard University
Here is the BS math answer that Barry Mazur calculated
2
31
2
31
2
31
1 3 1 3 1 3 3 3 2 2 3 1 3
2 2 2 2 4 4 4 4 4 4 2 2
dash
1 3 1 3 1 31
2 2 2 2 4 4
3 3
4 4
When the distributive law is used with the BS Rule-of-Signs and one of the signs is a (-) dash sign the middle two terms always cancel each other This leads to the incorrect answer of (-1) in the above equation
Barry Mazur did not go back and check his answer with the original equation If he had he would have known that the answer was incorrect
3
1 31
2
check answer by setting equal
3
331 3
12
1 31
2
Take the Cube root of both sides
1 3 2 multiply both sides by 2
3 3 subtract 1 from both sides
2 2
3 3 square both sides
3 9 This is an incorrect answer A simple equation check could have shown that the distributive law using BS math provides an incorrect answer
1
BS Illogical and Incorrect Numbers Raised to Powers
nC D n DC
642 DDD
In BS math only a positive number can have an answer if ldquonrdquo is an even number (246---)
All negative numbers have no solutions
642 DDD
With SM just add a direction arrow and any number gives the correct answer
or n nC D C D
3 3( 3 8 8 or 8 2 or 2 )n D C
( 2 16 C= 16 or 16 4 or 4 )n D
Since a negative number has no square root answer in BS math BS mathematicians invented imaginary numbers
i2=-1
24414 iiC n
These equations are easily solved without the use of imaginary numbers with SM math
If 4 then 4 2x
If 4 then 4 2x
The 360-degree circle is a 4400 years old extremely outdated system based on a year of 12 months of 30 days each that is 360 days
SM will use a more logical system that will have a circle of 6283185307o
d
( )
(deg )
d r radians
d r rees
the arc distance around
the circumference of a circle
d
radiusr
2 2
2 2
d r
dv rd t t
v dda rd t t t
1
d
r
( )av tv
The average angular velocity of a body is the rate of its angular displacement
( )0
liminstt
vt
The instantaneous angular velocity is the limit of the ratio
t
as t approaches zero
( )d
av
d
tv
The average circumference velocity of the body is the rate of its distance traveled d divided by the time t
dvv
r
1 dd vdrv
t t t r r
The equation of the relationship between the angular velocity and the velocity of a point on that body
( )o
av
v v v
t ta
The average angular acceleration is the change in the angular velocity
( )0
liminstt
v
ta
The instantaneous angular acceleration is the
limit of the ration of v
t
as t approaches
zero
daa
r
1
1
d
d s d
vva
t r tv a
r t r
The equation of the relationship between the angular acceleration and the acceleration of a point on the body
21
2v a t
If a body with an initial angular velocity ov has a constant
angular acceleration a it will turn through an angle in a time t
1
Straight Line Symbol Units Rotational Symbol Units Distance d m Angular
Distance radians or degrees
Velocity dt
v m
s Angular Velocity
V
t
radians or degrees
s
Acceleration vat
2ms
Angular Acceleration
VA
t
2
radians or degrees
s
Mass M Kg Mass distribution
dM 2Kgm
Force actionA Ma
2
Kgm
sNewton
Torque
dT M A 2
2
Kgm
s
Momentum Mv Kgm
s
Angular Momentum
dM V 2Kgm
s
Work (Force)(distance)
Ad 2
Kgmm
sNm Joule
Work T
Using AS arrow math to solve and direction in space problems using the equation 1 2
2 2 2 2 2 2( ) amp and ( ) amp 3
4
2 2 2
2 2
( ) amp
(5 ) 4 3
(5 ) 16 9
(5 ) 25
(5 ) (5 )
2 2(5 ) 4 3
(5 ) 16 9
(5 ) 25
(5 ) (5 )
In SM math Pythagorasrsquo theorem provides both magnitude and direction
In BS math Pythagorasrsquo theorem provides magnitude only
time set equal to zero
23
( ) ( 0 )10 1
x t xx
3( ) ( 0)
10 1W t f
In BS math this is an equation for a traveling wave
must use a square term in order to make the curve symmetric
NO SQUARE TERM IS NECESSARY
In SM math when t is set to ldquo0rdquo
This is a major fault in BS math
x f(x)
-4 0019
-3 0033
-2 0073
-1 0273
0 3000
1 0273
2 0073
3 0033
4 0019 -4 -3 -2 -1 0 1 2 3 400
05
10
15
20
25
30
35
BS wave (must use square term)
snap shot of wave at x = 0
Am
plit
ude
lt to gt lt to gt f(ltgt)
lt -4 0073
lt -3 0097
lt -2 0143
lt -1 0273
origin 0 3000
gt 1 0273
gt 2 0143
gt 3 0097
gt 4 0073lt0073 lt0097 lt0143 lt0273
3000
0273gt 0143gt 0097gt 0073gt
Wave moving
snap shot of wave
Am
plit
ude
lt4 lt3 lt2 lt1 0 1gt 2gt 3gt 4gt
3
2
1
In SM NO squared terms are needed
SPRING
An Action is needed to stretch or compress a spring to a starting point where it can execute cycle motion
tsA (ts to start) is the work required to move a mass attached to a spring a distance (d1) from
its initial rest position RAK 0 the spring constant is the stiffness of the spring
o A is attraction o R is repulsion o
ARRA
00 is for the back and forth motion through the zero initial rest position A
stretched spring goes from ldquoArdquo to ldquo0rdquo to ldquoRrdquo to ldquoArdquo to ldquo0rdquo to ldquoRrdquo d1 is the distance a mass is moved from its initial rest position
0
0 1ts A RA K d
Once ( )mad
is performed to move the mass to position d1 (against the restraint of the spring) a of the ( )mad
is transferred into the spring as potential attractionrepulsion (PAR) There is always
more ( )mad
performed than is stored as PAR When the spring is released from the restraining Ats the PAR starts the oscillation Conservation of ( )mad
requires that the PAR and the motion ( )mad
( )( )madM be a constant
( )amp ( ) constantmadPAR M 02
0 0
1
2A R A RPAR K d
02
( ) 0 0
1( )
2mad A R A RM m v
0 0 02 2 2
0 0 0 1
2 2 21
2 2 21
1 1 1amp
2 2 2
1 1 1amp
2 2 2
amp
A R A R A RK d m v K d
Kd mv Kd
Kd mv Kd
0
1d is frac12 the height from the base to the peak of a cycle
An experiment of moving paper under a pendulum through a small angle filled with sand that can escape through a small hole in the bottom shows that a pendulum cycling back and forth follows a curve that is mathematically analogous to a lsquosinrsquo or lsquocosrsquo
The projection of the curved area onto the straight line exactly duplicate the back and forth motion of CM This allows for the math of lsquosinrsquo or lsquocosrsquo to be used
SM will use lsquosinrdquo math It could use lsquocosrsquo or a combination of both lsquosinrsquo and lsquocosrsquo Using just the lsquosinrsquo math makes the understanding and math easier
0 sin 0 0SM 0 sin157 1SM The diagram at the left shows a spring being stretched from its (zero rest) position-2 and the PAR being released at position-1 There are 4 sections (1 to 2 2 to 3 3 to 4 and 4 to5) that alternate between maximum acceleration and zero velocity to max velocity and zero acceleration
At position-1 there is maximum acceleration and zero velocity o The lsquosinrsquo values from 0o to 157o will be used for velocity (values from 0 to 1) o The lsquosinrsquo values from 157o to 0o will be used for acceleration (values from 1 to 0)
At position-2 there is maximum velocity and zero acceleration Acceleration changes to deceleration
o The lsquosinrsquo values from 157o to 0o will be used for deceleration (values from 1 to 0) o The lsquosinrsquo values from 0o to 157o will be used for acceleration (values from 0 to 1)
At position-3 there is maximum acceleration and zero velocity o The lsquosinrsquo values from 0o to 157o will be used for velocity (values from 0 to 1) o The lsquosinrsquo values from 157o to 0o will be used for acceleration (values from 1 to 0)
At position-4 there is maximum velocity and zero acceleration Acceleration changes to deceleration
o The lsquosinrsquo values from 157o to 0o will be used for deceleration (values from 1 to 0) o The lsquosinrsquo values from 0o to 157o will be used for acceleration (values from 0 to 1)
At position-5 the cycle starts again
2
)539(
cT
MK
spring constant (K) when mass (M) and cycle-time (T) are known
)(
))(89)((
111 md
uKgM
d
MG
d
FK sts
spring constant (K) when mass (M) and amplitude ( d1) are known
2 21( )
Kv d d
m
velocity (v) at any point during the cycle
velocity
dforM
Kdv
max
012 velocity (v2) maximum (when d = 0 at position-2 and 4)
acceleration (a1) maximum when spring constant (K) mass (M) and amplitude (d1) are known
acceleration (a1) maximum when amplitude (d1) and cycle-time (Tc) are known
acceleration (a) at any point during the cycle-time (T)
1
151
286
v
dTc
cycle-time (Tc) when amplitude (d1) and velocity (v1) are known
K
MTc 28651
cycle-time (Tc) when mass (M) and spring constant (K) are known
fTc
1
cycle-time (Tc) when the frequency is known
cTf
1
frequency (f) when the cycle-time (Tc) is known
of cycles
elapsed timef
frequency (f) when the number of cycles in an elapsed time is known
d1 amplitude (distance from rest to start position)
Also frac12 the distance from the base to the peak
3131 oror dM
Ka
251
3131 539
c
oror T
da
2539
cT
da
In December of 2006 I discovered the mathematical error in Einsteinrsquos Special Relativity Einstein used BS math and that math only works in an imaginary world
It cannot describe real symmetrical space
Einsteinrsquos Special Relativity should be abandoned and a theory of real space using SM should be used to replace it
Numerous articles on the Internet show Einsteinrsquos theory to be incorrect but no one in academia is willing to publish the information
Figure 1-10 A moving particle observed from two frames of reference in uniform translation
Einsteinrsquos Math Error in Special Relativity Transformation of Velocity (From Eisenberg- Fundamentals of Modern Physics pages 27-29) Consider the particle shown in figure (1-10) moving with velocity v as
seen in a frame of reference O
dx dy dz
Vx Vy Vzdt dt dt
dx dy dz
V x V y V zdy dt dt
We would like to evaluate the velocity v of the particle as seen in the frame of reference 0 which is itself moving relative to 0 with velocity v Measured in the 0 frame the velocity vector of the particle has com ponents
The same velocity vector as measured in the 0 frame has components
2
2
2
2
2
1
1
x vtx
v
cy y
z z
vxt
ctv
c
Now from equations (1-13) we know that the relation between the primed and the unprimed coordinates and times is
Take the differential of these equations remembering that v is a constant This gives
2
2
2
2
2
1
1
dx vdtdx
v
cdy dy
dz dz
vxdt
cdtv
c
2
2
2 2 2 2 2
2
2
2
11 ( )
1( ) 1
1
1
1
dx vdt
v dxdx vdt v
dx dx vdtc dt dtvdx vdx vdx v dxdt dt dt dt
dt dtc c c c
VxV x
Vx
VxV x
vv
vv
c
x
v
c
cV
( )
coordinate of particle relative to moving primed system
coordinate of particle relative to stationary system
velocity of primed coordina
V x
Vx
te system relative to stationaryv sys
tem
Page 28 Fundamentals of Modern Physics
(1-16)
According to Einstein this equation tells you how to transform the observed velocity from one frame of reference to another frame of reference First we note that as Vc and vc approachzero the equation approaches those which would be derived from the Galilean transformation
2 2
( ) 1 01 1 ( )
Vx v Vx vV
VxV x
Vx v
v cVxx
v
c
v
cc
Another property of this equation is that it is impossible to choose V and v such that Vrsquo the magnitude of the velocity which is seen in the new frame is greater than c
1
2
09 ( 09 ) 18009876543
( 09 ) 1811 (09 )
c c cV c c
cc
c
Example As seen by O particle 1 has velocity 09c in the direction of positive x and particle 2 has velocity 09c in the negative x direction
Einstein equation however predicts To evaluate the velocity of particle 1 with respect to particle 2 we transform from the O frame to the Orsquo frame moving in the negative x direction with velocity v = -09c using equation 1-16 We obtain
JK LOGIC will observe the particles separating at 18c
A confusing INCORRECT and ILLOGICAL answer
left is a dir rightsub of a
1
2
1
09 ( 09 ) 18009876543
181( 09 )1 ( 09 )
The multiplication of dir left time a dir right is ONLY a dir left
09
dir
direction leftdirection rightsubtraction
c c cV c c
cc
c
V
dir
22
22
09 18 18009876543
181( 081)1(09)(09)
1
left rightsub
sub
c c c cc c
ccc
c
Analyzing Einsteinrsquos BS math Equation
Using SM math a logical answer can be obtained
Cannot multiply a subtraction operator times a direction in space
Can only subtract and change its direction
Cannot multiply a direction to the left times a direction to the right
Cannot multiply opposite directions in space
Can only solve for the resultant
rightleft
subtraction
1
subtraction
2
sub of a dir to the left is a dir to the right
1 resultant 0 sub 2
2sub 2
2
09 09
09 amp 091
09 09 18 1
(0)1
(09) amp (09)1
c cV
c cc
c c cV
cc
cc
8
181
cc
SM math predicts a logical answer using logic
21st century scientists state in many articles that Einsteins General Relativity and Quantum Mechanics are incompatible
They are not only incompatible they are both mathematically incorrect
They both use illogical BS math
Any equations in math that use imaginary numbers should be abandoned
Imaginary numbers are only necessary in the current BS math system
This means that Schroumldingers equations should be abandoned
Any equation in quantum mechanics that uses imaginary number should be abandoned
Prof Kanarev has developed logical physics and chemistrytheories without the use of Quantum mechanics or relativity
A partial list of his theories coverbull Mechanics ndash replaces Newtonbull Photon - bull Electrons ndash Protons- Neutronsbull Spectroscopybull Nucleusbull Atoms amp Moleculesbull Electrodynamicsbull Heatbull Curvature of Space ndash replaces Schwarzcchildrsquos equation
Kanarevrsquos Hypothesis
bull Electron align linearly with proton They do NOT spin or form clouds around the nucleus
bull Neutrons are in the center of atom Proton are outside of Neutrons
bull The wavelength of all particles is the experimental radius
bull Double slit experiment explained correctly
bull Heisenbergrsquos inequality explained correctly
bull is the radius of approach of the magnetic lines around the torus of the electron
( 1 137)
bull Why water expands when it freezes
bull Electron clusters form lightning
rbull The meaning of Plancks constant is explained
22 2 kg m
h m f mr f consts
2210254 42m r kgm bull If the mr-constant describes a ring the equations
describe experiments bull If the mr-constant describes a wave the equations do
NOT describe experiments
Kanarevrsquos 5 Equations of Mechanics
Kanarevrsquos 1st Equation Acceleration ampmv maMma R R
Kanarevrsquos 2nd Equation Uniform motion amp mv maMma R R
maM mvR R
Kanarevrsquos 3rd Equation
Kanarevrsquos 4th Equation Equal and opposite reaction
1 21 21 2
12
a m
m a m ama
Kanarevrsquos 5th Equation 1 2 3 Mi M M M Mna R R R R R
The sum of the Actions working on a moving body is never equal to zero
Kanarevrsquos Photon model is a rolling ring
His equations accurately describe known experiments with photonsbull Electromagnetic radiation should be called photon radiationbull Diffraction and interference are the same phenomenabull Correctly explains the double slit experiment without quantum equationsbull Correctly explains the meaning of Heisenbergrsquos inequality
e
2 (2)(3142)(2817 15 ) 10007297352568
R 2426 12 137mag
xp
r m
m
Charge
Magnetic action
Proton
The intensity of the magnetic action near the geometrical center of the proton ring is so great that we theorize that it is the strong nuclear force
(9383 8)(1602 19)8476 14
4 (4)(3142)(1411 26)p
pp
EB T
M
(9395 8)(1602 19)1035 15
4 (4)(3142)(0966 26)N
NN
EB T
M
Neutron
1a Hydrogenavi
Negative and Positive Directions in Space
If I asked you to point to a negative direction in space which way would you point
Hopefully you will realize that there is no such thing as a negative direction in space
If I asked you to point to a positive direction in space which way would you point
Again hopefully you will realize that there is no such thing as a positive direction in space
Objects on the right side of the x-axis are positive (+)(+) = (+)
Objects on the left side of the x-axis are negative (-)(-) = (+) negative math is different from positive math This is illogical
Objects on the top of the y-axis are positive
Objects on the bottom of the y-axis are negative
Objects in the front of the z-axis are positive
Objects in the back of the z-axis are negative
Again BS math of the positive direction (x right y up and z front) is different from the BS math of the negative direction (x left y down and z back)
BS math is Broken-Symmetry
Mirror image is broken symmetry in the x and z axis If the left amp right-axis are
reversed symmetry is broken Obs-1 math is different from obs-2
If the front amp back-axis is reversed symmetry is broken Obs-1 math is different from obs-2
If the top amp bottom-axis are
reversed symmetry is not broken Obs-1 math is the same as obs-2
Math answers cannot depend on which side of a number line an observer sit
Absolute values for Displacement
Even for a simple displacement of an object BS math invented a definition and absolute values to obtain something labeled a positive number answer
I----------I----------I----------I----------I----------I----------I----------I----------I
-4 -3 -2 -1 0 +1 +2 +3 +4 a gtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtb cltltltltltltltltltltltltltltltltltltltltltltltltltltltb
If an object starts at ( 3)ia X and moves to ( X 3) and the to ( 1)b fb c X
BS Math provides a usable answer without a definition or absolute values
However If an object starts at
( 1) and moves to ( X 3) and the to ( 3)i b fa X b c X
Then absolute values and a definition must be used
---------I----------I----------I----------I----------I----------I----------I---------- -4 -3 -2 -1 0 +1 +2 +3 +4
a gtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtb cltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltb
BS Math makes up a definition that all displacements are positive Since this example provides an answer with something labeled a negative the positive definition must be applied and (-2) is changed to (+2) A definition is needed to arrive at a useable answer This is Illogical
In SM absolute values and a meaningless definition of ldquopositiverdquo are removed In SM objects and directions in space are defined by the direction and magnitude of the resultant of arrows
With SM you get total distance traveled by the object and the final direction of the displacement
22222 2 bababbaababababa If w e su b stitu te p o s itiv e n u m b e rs in to th e eq u a tio n th e an sw er w ill b e co rrec t If (a = 5 ) an d (b = 3 )
2
2
8
5 3 5 3 5 3 5 5 5 3 3 5 3 3 25 15 15 9 64
8 8 8 8 8 8 8 8 8 8 64
W h en tw o p o sitiv e n u m b ers a re a d d ed an d sq u a re d th e B S m a th d is tr ib u tiv e law p ro v id es a co rre c t an sw er 1
Illogical and Incorrect BS math for the distributive law
222 2 bababbabbaaabababa B y d e fin itio n in B S m a th n o n u m b er sq u a red c a n ev e r b e n eg a tiv e T h e re fo re n o sq u a red n u m b er c an ev e r b e a v e c to r m o v in g in th e d a sh ( -) o r n eg a tiv e d ire c tio n if it is sq u a red In B S m a th an d ad d itio n o p e ra to r m u ltip lied b y a su b trac tio n o p e ra to r is A L W A Y S eq u a l to a su b tra c tio n o p e ra to r If th is h ad n o t b e en u sed fo r th e la s t fo u r h u n d red y e a rs y o u w o u ld d ie lau g h in g a t th e u tte r ab su rd ity o f th e la s t s ta tem e n t T h is is illo g ica l an d v io la te s sy m m etry
If a = 5 an d b = 3 42222352
22
4930253)3)(5(253)5)(3()3)(5(5353535 22222 A s lo n g a s ldquo ardquo is g re a te r th an ldquob rdquo th e B S m a th d is tr ib u tiv e law p ro v id es co rrec t an sw ers 1
B S m a th b re a k s d o w n w h e n ldquo b rdquo is g rea ter th a n ldquo a rdquo
If a = 3 an d b = 5
2 2
2
2
3 5 2 4
2 ( 2) ( 2) 4
2 ( )( 2) ( )( 2) 4
A n y co m b in a tio n ex cep t th e m id d le o n e is illo g ica l an d in co rre c t B y d e fin itio n th e an sw er can n o t b e -4 in B S m a th B S m a th is in co rrec t
4253095)5)(3(235)3)(5()5)(3(3535353 22222 T h is is w h ere th e d is tr ib u tiv e la w p ro v id es in co rrec t an sw ers I t p ro d u ces an an sw er o f + 4 b e cau se b y d e fin iti o n n o n u m b er sq u a red in B S m a th c an b e a n eg a tiv e n u m b er U sin g in co rre c t B S m a th th e d is tr ib u tiv e law w ill g iv e an in co rrec t an sw ers o f + 4 T h is is in co rrec t an d illo g ica l 1
Since there can be no negative direction from a squared term in BS math BS math should be modified or abandoned
The BS distributive law should be modified or abandoned
The numbers +1+2+3+4 and ndash1-2-3-4hellip should be abandoned
BS math gives incorrect answers because their Rule-of-Signs allows multiplying an arrow going in one direction by an arrow going in another direction
Clearly this is not logical
Arrows should be used that are specific for directions
W h at d o es it m e an to m u ltip ly ( 4) b y (4 ) A s w e k n o w 1
m u ltip lica tio n is ju s t ad d itio n W h a t w o u ld ( 4) tim e (4 ) b e 2
In BS math the answer is by definition -16 unless you square -4 and then it is +16
In SM you cannot multiply arrows going in opposite directions
The middle two terms are not logical You cannot multiply opposite directions
This is a MAJOR error in BS math They multiply a dash [(-) a direction to the left] by a cross [(+) a direction to the right] And then ldquoby definitionrdquo label the answer dash (-) This is illogical and produces incorrect answers
2
2 2
This is Einsteinrsquos math error
in Special relativity
Magnitude 4 Direction 1 Magnitude 3 Direction 2 Magnitude 5 Direction 3 Relative to an observer at 0 4
Magnitude 4 Direction 1 Magnitude 3 Direction 2 Magnitude 5 Direction 3 Relative to an observer at 0 4
Magnitude and Direction (MD) An arrow has both a magnitude andA direction relative to a 2nd objectobserver
bull Magnitude and velocity (MV) relative to a 2nd objectobserver
bull Magnitude And acceleration (MA) relative to a 2nd objectobserver2
v da
t t
Introducing a new symbol the ldquoamprdquo sign The ldquoamprdquo sign means resultant ADD all the Arrows for each direction and then SUBTRACT to find the final resultant
1 1amp 1 2 2amp 2 4 If =
1 amp1 2 2 amp2 4 If =
amp 5 amp 3 2 If gt
amp 3 amp 5= 2 If gt
amp 0 3 amp 3=0 if =
only the resultant equals zero
the vectors do not - they are still there
Does not exist There is no multiplication of arrows
There are two arrows One has a magnitude of 3 One has a magnitude of 5 The maximum these two arrows can be is 8
Only the resultant of arrows exist
An SM arrow has a direction and a magnitude
Arrows and magnitudes are still there even if the resultant is zero
There are no negative SM Arrows
There are no positive SM Arrows
There are no ZERO SM Arrows A zero would negate the definition of an SM Arrow something with no magnitude and no direction cannot be the definition of an arrow that is defined as having magnitude and direction
bull A resultant is the addition and subtraction of the differences in magnitude and direction arrows
A B and C are the arrows (a) A amp B exist(b) A amp B amp C exist(c) A-B Do Not Exist in Sm there are no negatives or positives(d) A+B-C Do Not Exist in SM there are no negatives or positives
There are only resultants when there are multiple arrows
SM Graphs versus BS Graphs
Numbers Raised to Powers
When numbers are raised to powers the negative sides of all BS math graphs are ILLOGICAL and INCORRECT
Incorrect BS Math for 2x
2x
4 16 correct3 8 correct2 4 correct1 2 correct0 1 correct-1 005 incorrect-2 025 incorrect-3 0125 incorrect-4 00625 incorrect
negative side of graph is incorrectpositive side of graph is correct
16
8
42105025012500625
-4 -3 -2 -1 0 1 2 3 4
po
sit
ive
negative positive
2x BS Math
correctincorrect
In BS math the dash sign (-) used in this example represents three different operations They arebull (-) Subtractionbull (-) A direction to the leftbull (-) As an exponent to mean divide by
All positive numbers produce correct results and graphs
All negative numbers produce incorrect results and graphs
A direction of three units to the left (-3) is placed in the equation as an exponent
A direction in space (3 units toward the left) changes a whole number to a fraction by using the dash (-) as an exponent rather than a direction in space
This is illogical and incorrect BS math should have performed this operation as
Unfortunately the answers to problems can come out correct because data points of a real problem fit the curve
However the answers will not be due to logical reasoning
Because the equation produced a curve that real problem data points follow it has been falsely believed that the equations and graphs are correct
1
2
x
Incorrect BK graph for (12)x
(12)x
4 00625 correct3 0125 correct2 025 correct1 1 correct0 1 correct-1 2 incorrect-2 4 incorrect-3 8 incorrect-4 16 incorrect
negative side of graph is incorrectpositive side of graph is correct
00625012502505124
8
16
-5 -4 -3 -2 -1 0 1 2 3 4 5P
osit
ive
negative positive
BS Math for (12)x
correctincorrect
Incorrect BS math and graph for
All graphs are symmetrical with SM
Two BS Math Equations that Produce Incorrect Answers
2 0x x starting equation 2
2
x x x x
x x
Add lsquoxrsquo to both sides
2(1) 1 let lsquoxrsquo = 1
2 1 This is an incorrect answer 2(0)-(0)=0 The only answer that BK math can give is x=0
1
a b starting equation ( )( ) ( )( )a a a b multiply both sides by ( )a
( )( ) ( )( ) ( )( ) ( )( )a a b b a b b b subtract ( )( )b b from both sides
( )( ) ( )( )a b a b a b b factor ( )a b from both sides
( )a b b divide both sides by ( )a b
( )b b b replace ( )a with ( )b 2b b simplify 2(1) (1) Let ( )b = 1
2 1 Again BS math yields and incorrect answer 1
This incorrect result is a problem with the distributive law SM explains how this incorrect answer is obtained and the problem with the distributive law that allows the error
Page 148 ndash149 Barry Mazur ndash Imagining Numbers
3
2
31
Do the computation carefully on paper using the rules we agreed to Then ponder your answer which should be something of a surprise to you donrsquot stop there Think of what your answer might possibly mean or might imply
Solve this equation using the distributive law
JK definitely a surprise the answer wasincorrect and illogical
Gerhard Gade University Professor doing math at Harvard University
Here is the BS math answer that Barry Mazur calculated
2
31
2
31
2
31
1 3 1 3 1 3 3 3 2 2 3 1 3
2 2 2 2 4 4 4 4 4 4 2 2
dash
1 3 1 3 1 31
2 2 2 2 4 4
3 3
4 4
When the distributive law is used with the BS Rule-of-Signs and one of the signs is a (-) dash sign the middle two terms always cancel each other This leads to the incorrect answer of (-1) in the above equation
Barry Mazur did not go back and check his answer with the original equation If he had he would have known that the answer was incorrect
3
1 31
2
check answer by setting equal
3
331 3
12
1 31
2
Take the Cube root of both sides
1 3 2 multiply both sides by 2
3 3 subtract 1 from both sides
2 2
3 3 square both sides
3 9 This is an incorrect answer A simple equation check could have shown that the distributive law using BS math provides an incorrect answer
1
BS Illogical and Incorrect Numbers Raised to Powers
nC D n DC
642 DDD
In BS math only a positive number can have an answer if ldquonrdquo is an even number (246---)
All negative numbers have no solutions
642 DDD
With SM just add a direction arrow and any number gives the correct answer
or n nC D C D
3 3( 3 8 8 or 8 2 or 2 )n D C
( 2 16 C= 16 or 16 4 or 4 )n D
Since a negative number has no square root answer in BS math BS mathematicians invented imaginary numbers
i2=-1
24414 iiC n
These equations are easily solved without the use of imaginary numbers with SM math
If 4 then 4 2x
If 4 then 4 2x
The 360-degree circle is a 4400 years old extremely outdated system based on a year of 12 months of 30 days each that is 360 days
SM will use a more logical system that will have a circle of 6283185307o
d
( )
(deg )
d r radians
d r rees
the arc distance around
the circumference of a circle
d
radiusr
2 2
2 2
d r
dv rd t t
v dda rd t t t
1
d
r
( )av tv
The average angular velocity of a body is the rate of its angular displacement
( )0
liminstt
vt
The instantaneous angular velocity is the limit of the ratio
t
as t approaches zero
( )d
av
d
tv
The average circumference velocity of the body is the rate of its distance traveled d divided by the time t
dvv
r
1 dd vdrv
t t t r r
The equation of the relationship between the angular velocity and the velocity of a point on that body
( )o
av
v v v
t ta
The average angular acceleration is the change in the angular velocity
( )0
liminstt
v
ta
The instantaneous angular acceleration is the
limit of the ration of v
t
as t approaches
zero
daa
r
1
1
d
d s d
vva
t r tv a
r t r
The equation of the relationship between the angular acceleration and the acceleration of a point on the body
21
2v a t
If a body with an initial angular velocity ov has a constant
angular acceleration a it will turn through an angle in a time t
1
Straight Line Symbol Units Rotational Symbol Units Distance d m Angular
Distance radians or degrees
Velocity dt
v m
s Angular Velocity
V
t
radians or degrees
s
Acceleration vat
2ms
Angular Acceleration
VA
t
2
radians or degrees
s
Mass M Kg Mass distribution
dM 2Kgm
Force actionA Ma
2
Kgm
sNewton
Torque
dT M A 2
2
Kgm
s
Momentum Mv Kgm
s
Angular Momentum
dM V 2Kgm
s
Work (Force)(distance)
Ad 2
Kgmm
sNm Joule
Work T
Using AS arrow math to solve and direction in space problems using the equation 1 2
2 2 2 2 2 2( ) amp and ( ) amp 3
4
2 2 2
2 2
( ) amp
(5 ) 4 3
(5 ) 16 9
(5 ) 25
(5 ) (5 )
2 2(5 ) 4 3
(5 ) 16 9
(5 ) 25
(5 ) (5 )
In SM math Pythagorasrsquo theorem provides both magnitude and direction
In BS math Pythagorasrsquo theorem provides magnitude only
time set equal to zero
23
( ) ( 0 )10 1
x t xx
3( ) ( 0)
10 1W t f
In BS math this is an equation for a traveling wave
must use a square term in order to make the curve symmetric
NO SQUARE TERM IS NECESSARY
In SM math when t is set to ldquo0rdquo
This is a major fault in BS math
x f(x)
-4 0019
-3 0033
-2 0073
-1 0273
0 3000
1 0273
2 0073
3 0033
4 0019 -4 -3 -2 -1 0 1 2 3 400
05
10
15
20
25
30
35
BS wave (must use square term)
snap shot of wave at x = 0
Am
plit
ude
lt to gt lt to gt f(ltgt)
lt -4 0073
lt -3 0097
lt -2 0143
lt -1 0273
origin 0 3000
gt 1 0273
gt 2 0143
gt 3 0097
gt 4 0073lt0073 lt0097 lt0143 lt0273
3000
0273gt 0143gt 0097gt 0073gt
Wave moving
snap shot of wave
Am
plit
ude
lt4 lt3 lt2 lt1 0 1gt 2gt 3gt 4gt
3
2
1
In SM NO squared terms are needed
SPRING
An Action is needed to stretch or compress a spring to a starting point where it can execute cycle motion
tsA (ts to start) is the work required to move a mass attached to a spring a distance (d1) from
its initial rest position RAK 0 the spring constant is the stiffness of the spring
o A is attraction o R is repulsion o
ARRA
00 is for the back and forth motion through the zero initial rest position A
stretched spring goes from ldquoArdquo to ldquo0rdquo to ldquoRrdquo to ldquoArdquo to ldquo0rdquo to ldquoRrdquo d1 is the distance a mass is moved from its initial rest position
0
0 1ts A RA K d
Once ( )mad
is performed to move the mass to position d1 (against the restraint of the spring) a of the ( )mad
is transferred into the spring as potential attractionrepulsion (PAR) There is always
more ( )mad
performed than is stored as PAR When the spring is released from the restraining Ats the PAR starts the oscillation Conservation of ( )mad
requires that the PAR and the motion ( )mad
( )( )madM be a constant
( )amp ( ) constantmadPAR M 02
0 0
1
2A R A RPAR K d
02
( ) 0 0
1( )
2mad A R A RM m v
0 0 02 2 2
0 0 0 1
2 2 21
2 2 21
1 1 1amp
2 2 2
1 1 1amp
2 2 2
amp
A R A R A RK d m v K d
Kd mv Kd
Kd mv Kd
0
1d is frac12 the height from the base to the peak of a cycle
An experiment of moving paper under a pendulum through a small angle filled with sand that can escape through a small hole in the bottom shows that a pendulum cycling back and forth follows a curve that is mathematically analogous to a lsquosinrsquo or lsquocosrsquo
The projection of the curved area onto the straight line exactly duplicate the back and forth motion of CM This allows for the math of lsquosinrsquo or lsquocosrsquo to be used
SM will use lsquosinrdquo math It could use lsquocosrsquo or a combination of both lsquosinrsquo and lsquocosrsquo Using just the lsquosinrsquo math makes the understanding and math easier
0 sin 0 0SM 0 sin157 1SM The diagram at the left shows a spring being stretched from its (zero rest) position-2 and the PAR being released at position-1 There are 4 sections (1 to 2 2 to 3 3 to 4 and 4 to5) that alternate between maximum acceleration and zero velocity to max velocity and zero acceleration
At position-1 there is maximum acceleration and zero velocity o The lsquosinrsquo values from 0o to 157o will be used for velocity (values from 0 to 1) o The lsquosinrsquo values from 157o to 0o will be used for acceleration (values from 1 to 0)
At position-2 there is maximum velocity and zero acceleration Acceleration changes to deceleration
o The lsquosinrsquo values from 157o to 0o will be used for deceleration (values from 1 to 0) o The lsquosinrsquo values from 0o to 157o will be used for acceleration (values from 0 to 1)
At position-3 there is maximum acceleration and zero velocity o The lsquosinrsquo values from 0o to 157o will be used for velocity (values from 0 to 1) o The lsquosinrsquo values from 157o to 0o will be used for acceleration (values from 1 to 0)
At position-4 there is maximum velocity and zero acceleration Acceleration changes to deceleration
o The lsquosinrsquo values from 157o to 0o will be used for deceleration (values from 1 to 0) o The lsquosinrsquo values from 0o to 157o will be used for acceleration (values from 0 to 1)
At position-5 the cycle starts again
2
)539(
cT
MK
spring constant (K) when mass (M) and cycle-time (T) are known
)(
))(89)((
111 md
uKgM
d
MG
d
FK sts
spring constant (K) when mass (M) and amplitude ( d1) are known
2 21( )
Kv d d
m
velocity (v) at any point during the cycle
velocity
dforM
Kdv
max
012 velocity (v2) maximum (when d = 0 at position-2 and 4)
acceleration (a1) maximum when spring constant (K) mass (M) and amplitude (d1) are known
acceleration (a1) maximum when amplitude (d1) and cycle-time (Tc) are known
acceleration (a) at any point during the cycle-time (T)
1
151
286
v
dTc
cycle-time (Tc) when amplitude (d1) and velocity (v1) are known
K
MTc 28651
cycle-time (Tc) when mass (M) and spring constant (K) are known
fTc
1
cycle-time (Tc) when the frequency is known
cTf
1
frequency (f) when the cycle-time (Tc) is known
of cycles
elapsed timef
frequency (f) when the number of cycles in an elapsed time is known
d1 amplitude (distance from rest to start position)
Also frac12 the distance from the base to the peak
3131 oror dM
Ka
251
3131 539
c
oror T
da
2539
cT
da
In December of 2006 I discovered the mathematical error in Einsteinrsquos Special Relativity Einstein used BS math and that math only works in an imaginary world
It cannot describe real symmetrical space
Einsteinrsquos Special Relativity should be abandoned and a theory of real space using SM should be used to replace it
Numerous articles on the Internet show Einsteinrsquos theory to be incorrect but no one in academia is willing to publish the information
Figure 1-10 A moving particle observed from two frames of reference in uniform translation
Einsteinrsquos Math Error in Special Relativity Transformation of Velocity (From Eisenberg- Fundamentals of Modern Physics pages 27-29) Consider the particle shown in figure (1-10) moving with velocity v as
seen in a frame of reference O
dx dy dz
Vx Vy Vzdt dt dt
dx dy dz
V x V y V zdy dt dt
We would like to evaluate the velocity v of the particle as seen in the frame of reference 0 which is itself moving relative to 0 with velocity v Measured in the 0 frame the velocity vector of the particle has com ponents
The same velocity vector as measured in the 0 frame has components
2
2
2
2
2
1
1
x vtx
v
cy y
z z
vxt
ctv
c
Now from equations (1-13) we know that the relation between the primed and the unprimed coordinates and times is
Take the differential of these equations remembering that v is a constant This gives
2
2
2
2
2
1
1
dx vdtdx
v
cdy dy
dz dz
vxdt
cdtv
c
2
2
2 2 2 2 2
2
2
2
11 ( )
1( ) 1
1
1
1
dx vdt
v dxdx vdt v
dx dx vdtc dt dtvdx vdx vdx v dxdt dt dt dt
dt dtc c c c
VxV x
Vx
VxV x
vv
vv
c
x
v
c
cV
( )
coordinate of particle relative to moving primed system
coordinate of particle relative to stationary system
velocity of primed coordina
V x
Vx
te system relative to stationaryv sys
tem
Page 28 Fundamentals of Modern Physics
(1-16)
According to Einstein this equation tells you how to transform the observed velocity from one frame of reference to another frame of reference First we note that as Vc and vc approachzero the equation approaches those which would be derived from the Galilean transformation
2 2
( ) 1 01 1 ( )
Vx v Vx vV
VxV x
Vx v
v cVxx
v
c
v
cc
Another property of this equation is that it is impossible to choose V and v such that Vrsquo the magnitude of the velocity which is seen in the new frame is greater than c
1
2
09 ( 09 ) 18009876543
( 09 ) 1811 (09 )
c c cV c c
cc
c
Example As seen by O particle 1 has velocity 09c in the direction of positive x and particle 2 has velocity 09c in the negative x direction
Einstein equation however predicts To evaluate the velocity of particle 1 with respect to particle 2 we transform from the O frame to the Orsquo frame moving in the negative x direction with velocity v = -09c using equation 1-16 We obtain
JK LOGIC will observe the particles separating at 18c
A confusing INCORRECT and ILLOGICAL answer
left is a dir rightsub of a
1
2
1
09 ( 09 ) 18009876543
181( 09 )1 ( 09 )
The multiplication of dir left time a dir right is ONLY a dir left
09
dir
direction leftdirection rightsubtraction
c c cV c c
cc
c
V
dir
22
22
09 18 18009876543
181( 081)1(09)(09)
1
left rightsub
sub
c c c cc c
ccc
c
Analyzing Einsteinrsquos BS math Equation
Using SM math a logical answer can be obtained
Cannot multiply a subtraction operator times a direction in space
Can only subtract and change its direction
Cannot multiply a direction to the left times a direction to the right
Cannot multiply opposite directions in space
Can only solve for the resultant
rightleft
subtraction
1
subtraction
2
sub of a dir to the left is a dir to the right
1 resultant 0 sub 2
2sub 2
2
09 09
09 amp 091
09 09 18 1
(0)1
(09) amp (09)1
c cV
c cc
c c cV
cc
cc
8
181
cc
SM math predicts a logical answer using logic
21st century scientists state in many articles that Einsteins General Relativity and Quantum Mechanics are incompatible
They are not only incompatible they are both mathematically incorrect
They both use illogical BS math
Any equations in math that use imaginary numbers should be abandoned
Imaginary numbers are only necessary in the current BS math system
This means that Schroumldingers equations should be abandoned
Any equation in quantum mechanics that uses imaginary number should be abandoned
Prof Kanarev has developed logical physics and chemistrytheories without the use of Quantum mechanics or relativity
A partial list of his theories coverbull Mechanics ndash replaces Newtonbull Photon - bull Electrons ndash Protons- Neutronsbull Spectroscopybull Nucleusbull Atoms amp Moleculesbull Electrodynamicsbull Heatbull Curvature of Space ndash replaces Schwarzcchildrsquos equation
Kanarevrsquos Hypothesis
bull Electron align linearly with proton They do NOT spin or form clouds around the nucleus
bull Neutrons are in the center of atom Proton are outside of Neutrons
bull The wavelength of all particles is the experimental radius
bull Double slit experiment explained correctly
bull Heisenbergrsquos inequality explained correctly
bull is the radius of approach of the magnetic lines around the torus of the electron
( 1 137)
bull Why water expands when it freezes
bull Electron clusters form lightning
rbull The meaning of Plancks constant is explained
22 2 kg m
h m f mr f consts
2210254 42m r kgm bull If the mr-constant describes a ring the equations
describe experiments bull If the mr-constant describes a wave the equations do
NOT describe experiments
Kanarevrsquos 5 Equations of Mechanics
Kanarevrsquos 1st Equation Acceleration ampmv maMma R R
Kanarevrsquos 2nd Equation Uniform motion amp mv maMma R R
maM mvR R
Kanarevrsquos 3rd Equation
Kanarevrsquos 4th Equation Equal and opposite reaction
1 21 21 2
12
a m
m a m ama
Kanarevrsquos 5th Equation 1 2 3 Mi M M M Mna R R R R R
The sum of the Actions working on a moving body is never equal to zero
Kanarevrsquos Photon model is a rolling ring
His equations accurately describe known experiments with photonsbull Electromagnetic radiation should be called photon radiationbull Diffraction and interference are the same phenomenabull Correctly explains the double slit experiment without quantum equationsbull Correctly explains the meaning of Heisenbergrsquos inequality
e
2 (2)(3142)(2817 15 ) 10007297352568
R 2426 12 137mag
xp
r m
m
Charge
Magnetic action
Proton
The intensity of the magnetic action near the geometrical center of the proton ring is so great that we theorize that it is the strong nuclear force
(9383 8)(1602 19)8476 14
4 (4)(3142)(1411 26)p
pp
EB T
M
(9395 8)(1602 19)1035 15
4 (4)(3142)(0966 26)N
NN
EB T
M
Neutron
1a Hydrogenavi
Objects on the right side of the x-axis are positive (+)(+) = (+)
Objects on the left side of the x-axis are negative (-)(-) = (+) negative math is different from positive math This is illogical
Objects on the top of the y-axis are positive
Objects on the bottom of the y-axis are negative
Objects in the front of the z-axis are positive
Objects in the back of the z-axis are negative
Again BS math of the positive direction (x right y up and z front) is different from the BS math of the negative direction (x left y down and z back)
BS math is Broken-Symmetry
Mirror image is broken symmetry in the x and z axis If the left amp right-axis are
reversed symmetry is broken Obs-1 math is different from obs-2
If the front amp back-axis is reversed symmetry is broken Obs-1 math is different from obs-2
If the top amp bottom-axis are
reversed symmetry is not broken Obs-1 math is the same as obs-2
Math answers cannot depend on which side of a number line an observer sit
Absolute values for Displacement
Even for a simple displacement of an object BS math invented a definition and absolute values to obtain something labeled a positive number answer
I----------I----------I----------I----------I----------I----------I----------I----------I
-4 -3 -2 -1 0 +1 +2 +3 +4 a gtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtb cltltltltltltltltltltltltltltltltltltltltltltltltltltltb
If an object starts at ( 3)ia X and moves to ( X 3) and the to ( 1)b fb c X
BS Math provides a usable answer without a definition or absolute values
However If an object starts at
( 1) and moves to ( X 3) and the to ( 3)i b fa X b c X
Then absolute values and a definition must be used
---------I----------I----------I----------I----------I----------I----------I---------- -4 -3 -2 -1 0 +1 +2 +3 +4
a gtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtb cltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltb
BS Math makes up a definition that all displacements are positive Since this example provides an answer with something labeled a negative the positive definition must be applied and (-2) is changed to (+2) A definition is needed to arrive at a useable answer This is Illogical
In SM absolute values and a meaningless definition of ldquopositiverdquo are removed In SM objects and directions in space are defined by the direction and magnitude of the resultant of arrows
With SM you get total distance traveled by the object and the final direction of the displacement
22222 2 bababbaababababa If w e su b stitu te p o s itiv e n u m b e rs in to th e eq u a tio n th e an sw er w ill b e co rrec t If (a = 5 ) an d (b = 3 )
2
2
8
5 3 5 3 5 3 5 5 5 3 3 5 3 3 25 15 15 9 64
8 8 8 8 8 8 8 8 8 8 64
W h en tw o p o sitiv e n u m b ers a re a d d ed an d sq u a re d th e B S m a th d is tr ib u tiv e law p ro v id es a co rre c t an sw er 1
Illogical and Incorrect BS math for the distributive law
222 2 bababbabbaaabababa B y d e fin itio n in B S m a th n o n u m b er sq u a red c a n ev e r b e n eg a tiv e T h e re fo re n o sq u a red n u m b er c an ev e r b e a v e c to r m o v in g in th e d a sh ( -) o r n eg a tiv e d ire c tio n if it is sq u a red In B S m a th an d ad d itio n o p e ra to r m u ltip lied b y a su b trac tio n o p e ra to r is A L W A Y S eq u a l to a su b tra c tio n o p e ra to r If th is h ad n o t b e en u sed fo r th e la s t fo u r h u n d red y e a rs y o u w o u ld d ie lau g h in g a t th e u tte r ab su rd ity o f th e la s t s ta tem e n t T h is is illo g ica l an d v io la te s sy m m etry
If a = 5 an d b = 3 42222352
22
4930253)3)(5(253)5)(3()3)(5(5353535 22222 A s lo n g a s ldquo ardquo is g re a te r th an ldquob rdquo th e B S m a th d is tr ib u tiv e law p ro v id es co rrec t an sw ers 1
B S m a th b re a k s d o w n w h e n ldquo b rdquo is g rea ter th a n ldquo a rdquo
If a = 3 an d b = 5
2 2
2
2
3 5 2 4
2 ( 2) ( 2) 4
2 ( )( 2) ( )( 2) 4
A n y co m b in a tio n ex cep t th e m id d le o n e is illo g ica l an d in co rre c t B y d e fin itio n th e an sw er can n o t b e -4 in B S m a th B S m a th is in co rrec t
4253095)5)(3(235)3)(5()5)(3(3535353 22222 T h is is w h ere th e d is tr ib u tiv e la w p ro v id es in co rrec t an sw ers I t p ro d u ces an an sw er o f + 4 b e cau se b y d e fin iti o n n o n u m b er sq u a red in B S m a th c an b e a n eg a tiv e n u m b er U sin g in co rre c t B S m a th th e d is tr ib u tiv e law w ill g iv e an in co rrec t an sw ers o f + 4 T h is is in co rrec t an d illo g ica l 1
Since there can be no negative direction from a squared term in BS math BS math should be modified or abandoned
The BS distributive law should be modified or abandoned
The numbers +1+2+3+4 and ndash1-2-3-4hellip should be abandoned
BS math gives incorrect answers because their Rule-of-Signs allows multiplying an arrow going in one direction by an arrow going in another direction
Clearly this is not logical
Arrows should be used that are specific for directions
W h at d o es it m e an to m u ltip ly ( 4) b y (4 ) A s w e k n o w 1
m u ltip lica tio n is ju s t ad d itio n W h a t w o u ld ( 4) tim e (4 ) b e 2
In BS math the answer is by definition -16 unless you square -4 and then it is +16
In SM you cannot multiply arrows going in opposite directions
The middle two terms are not logical You cannot multiply opposite directions
This is a MAJOR error in BS math They multiply a dash [(-) a direction to the left] by a cross [(+) a direction to the right] And then ldquoby definitionrdquo label the answer dash (-) This is illogical and produces incorrect answers
2
2 2
This is Einsteinrsquos math error
in Special relativity
Magnitude 4 Direction 1 Magnitude 3 Direction 2 Magnitude 5 Direction 3 Relative to an observer at 0 4
Magnitude 4 Direction 1 Magnitude 3 Direction 2 Magnitude 5 Direction 3 Relative to an observer at 0 4
Magnitude and Direction (MD) An arrow has both a magnitude andA direction relative to a 2nd objectobserver
bull Magnitude and velocity (MV) relative to a 2nd objectobserver
bull Magnitude And acceleration (MA) relative to a 2nd objectobserver2
v da
t t
Introducing a new symbol the ldquoamprdquo sign The ldquoamprdquo sign means resultant ADD all the Arrows for each direction and then SUBTRACT to find the final resultant
1 1amp 1 2 2amp 2 4 If =
1 amp1 2 2 amp2 4 If =
amp 5 amp 3 2 If gt
amp 3 amp 5= 2 If gt
amp 0 3 amp 3=0 if =
only the resultant equals zero
the vectors do not - they are still there
Does not exist There is no multiplication of arrows
There are two arrows One has a magnitude of 3 One has a magnitude of 5 The maximum these two arrows can be is 8
Only the resultant of arrows exist
An SM arrow has a direction and a magnitude
Arrows and magnitudes are still there even if the resultant is zero
There are no negative SM Arrows
There are no positive SM Arrows
There are no ZERO SM Arrows A zero would negate the definition of an SM Arrow something with no magnitude and no direction cannot be the definition of an arrow that is defined as having magnitude and direction
bull A resultant is the addition and subtraction of the differences in magnitude and direction arrows
A B and C are the arrows (a) A amp B exist(b) A amp B amp C exist(c) A-B Do Not Exist in Sm there are no negatives or positives(d) A+B-C Do Not Exist in SM there are no negatives or positives
There are only resultants when there are multiple arrows
SM Graphs versus BS Graphs
Numbers Raised to Powers
When numbers are raised to powers the negative sides of all BS math graphs are ILLOGICAL and INCORRECT
Incorrect BS Math for 2x
2x
4 16 correct3 8 correct2 4 correct1 2 correct0 1 correct-1 005 incorrect-2 025 incorrect-3 0125 incorrect-4 00625 incorrect
negative side of graph is incorrectpositive side of graph is correct
16
8
42105025012500625
-4 -3 -2 -1 0 1 2 3 4
po
sit
ive
negative positive
2x BS Math
correctincorrect
In BS math the dash sign (-) used in this example represents three different operations They arebull (-) Subtractionbull (-) A direction to the leftbull (-) As an exponent to mean divide by
All positive numbers produce correct results and graphs
All negative numbers produce incorrect results and graphs
A direction of three units to the left (-3) is placed in the equation as an exponent
A direction in space (3 units toward the left) changes a whole number to a fraction by using the dash (-) as an exponent rather than a direction in space
This is illogical and incorrect BS math should have performed this operation as
Unfortunately the answers to problems can come out correct because data points of a real problem fit the curve
However the answers will not be due to logical reasoning
Because the equation produced a curve that real problem data points follow it has been falsely believed that the equations and graphs are correct
1
2
x
Incorrect BK graph for (12)x
(12)x
4 00625 correct3 0125 correct2 025 correct1 1 correct0 1 correct-1 2 incorrect-2 4 incorrect-3 8 incorrect-4 16 incorrect
negative side of graph is incorrectpositive side of graph is correct
00625012502505124
8
16
-5 -4 -3 -2 -1 0 1 2 3 4 5P
osit
ive
negative positive
BS Math for (12)x
correctincorrect
Incorrect BS math and graph for
All graphs are symmetrical with SM
Two BS Math Equations that Produce Incorrect Answers
2 0x x starting equation 2
2
x x x x
x x
Add lsquoxrsquo to both sides
2(1) 1 let lsquoxrsquo = 1
2 1 This is an incorrect answer 2(0)-(0)=0 The only answer that BK math can give is x=0
1
a b starting equation ( )( ) ( )( )a a a b multiply both sides by ( )a
( )( ) ( )( ) ( )( ) ( )( )a a b b a b b b subtract ( )( )b b from both sides
( )( ) ( )( )a b a b a b b factor ( )a b from both sides
( )a b b divide both sides by ( )a b
( )b b b replace ( )a with ( )b 2b b simplify 2(1) (1) Let ( )b = 1
2 1 Again BS math yields and incorrect answer 1
This incorrect result is a problem with the distributive law SM explains how this incorrect answer is obtained and the problem with the distributive law that allows the error
Page 148 ndash149 Barry Mazur ndash Imagining Numbers
3
2
31
Do the computation carefully on paper using the rules we agreed to Then ponder your answer which should be something of a surprise to you donrsquot stop there Think of what your answer might possibly mean or might imply
Solve this equation using the distributive law
JK definitely a surprise the answer wasincorrect and illogical
Gerhard Gade University Professor doing math at Harvard University
Here is the BS math answer that Barry Mazur calculated
2
31
2
31
2
31
1 3 1 3 1 3 3 3 2 2 3 1 3
2 2 2 2 4 4 4 4 4 4 2 2
dash
1 3 1 3 1 31
2 2 2 2 4 4
3 3
4 4
When the distributive law is used with the BS Rule-of-Signs and one of the signs is a (-) dash sign the middle two terms always cancel each other This leads to the incorrect answer of (-1) in the above equation
Barry Mazur did not go back and check his answer with the original equation If he had he would have known that the answer was incorrect
3
1 31
2
check answer by setting equal
3
331 3
12
1 31
2
Take the Cube root of both sides
1 3 2 multiply both sides by 2
3 3 subtract 1 from both sides
2 2
3 3 square both sides
3 9 This is an incorrect answer A simple equation check could have shown that the distributive law using BS math provides an incorrect answer
1
BS Illogical and Incorrect Numbers Raised to Powers
nC D n DC
642 DDD
In BS math only a positive number can have an answer if ldquonrdquo is an even number (246---)
All negative numbers have no solutions
642 DDD
With SM just add a direction arrow and any number gives the correct answer
or n nC D C D
3 3( 3 8 8 or 8 2 or 2 )n D C
( 2 16 C= 16 or 16 4 or 4 )n D
Since a negative number has no square root answer in BS math BS mathematicians invented imaginary numbers
i2=-1
24414 iiC n
These equations are easily solved without the use of imaginary numbers with SM math
If 4 then 4 2x
If 4 then 4 2x
The 360-degree circle is a 4400 years old extremely outdated system based on a year of 12 months of 30 days each that is 360 days
SM will use a more logical system that will have a circle of 6283185307o
d
( )
(deg )
d r radians
d r rees
the arc distance around
the circumference of a circle
d
radiusr
2 2
2 2
d r
dv rd t t
v dda rd t t t
1
d
r
( )av tv
The average angular velocity of a body is the rate of its angular displacement
( )0
liminstt
vt
The instantaneous angular velocity is the limit of the ratio
t
as t approaches zero
( )d
av
d
tv
The average circumference velocity of the body is the rate of its distance traveled d divided by the time t
dvv
r
1 dd vdrv
t t t r r
The equation of the relationship between the angular velocity and the velocity of a point on that body
( )o
av
v v v
t ta
The average angular acceleration is the change in the angular velocity
( )0
liminstt
v
ta
The instantaneous angular acceleration is the
limit of the ration of v
t
as t approaches
zero
daa
r
1
1
d
d s d
vva
t r tv a
r t r
The equation of the relationship between the angular acceleration and the acceleration of a point on the body
21
2v a t
If a body with an initial angular velocity ov has a constant
angular acceleration a it will turn through an angle in a time t
1
Straight Line Symbol Units Rotational Symbol Units Distance d m Angular
Distance radians or degrees
Velocity dt
v m
s Angular Velocity
V
t
radians or degrees
s
Acceleration vat
2ms
Angular Acceleration
VA
t
2
radians or degrees
s
Mass M Kg Mass distribution
dM 2Kgm
Force actionA Ma
2
Kgm
sNewton
Torque
dT M A 2
2
Kgm
s
Momentum Mv Kgm
s
Angular Momentum
dM V 2Kgm
s
Work (Force)(distance)
Ad 2
Kgmm
sNm Joule
Work T
Using AS arrow math to solve and direction in space problems using the equation 1 2
2 2 2 2 2 2( ) amp and ( ) amp 3
4
2 2 2
2 2
( ) amp
(5 ) 4 3
(5 ) 16 9
(5 ) 25
(5 ) (5 )
2 2(5 ) 4 3
(5 ) 16 9
(5 ) 25
(5 ) (5 )
In SM math Pythagorasrsquo theorem provides both magnitude and direction
In BS math Pythagorasrsquo theorem provides magnitude only
time set equal to zero
23
( ) ( 0 )10 1
x t xx
3( ) ( 0)
10 1W t f
In BS math this is an equation for a traveling wave
must use a square term in order to make the curve symmetric
NO SQUARE TERM IS NECESSARY
In SM math when t is set to ldquo0rdquo
This is a major fault in BS math
x f(x)
-4 0019
-3 0033
-2 0073
-1 0273
0 3000
1 0273
2 0073
3 0033
4 0019 -4 -3 -2 -1 0 1 2 3 400
05
10
15
20
25
30
35
BS wave (must use square term)
snap shot of wave at x = 0
Am
plit
ude
lt to gt lt to gt f(ltgt)
lt -4 0073
lt -3 0097
lt -2 0143
lt -1 0273
origin 0 3000
gt 1 0273
gt 2 0143
gt 3 0097
gt 4 0073lt0073 lt0097 lt0143 lt0273
3000
0273gt 0143gt 0097gt 0073gt
Wave moving
snap shot of wave
Am
plit
ude
lt4 lt3 lt2 lt1 0 1gt 2gt 3gt 4gt
3
2
1
In SM NO squared terms are needed
SPRING
An Action is needed to stretch or compress a spring to a starting point where it can execute cycle motion
tsA (ts to start) is the work required to move a mass attached to a spring a distance (d1) from
its initial rest position RAK 0 the spring constant is the stiffness of the spring
o A is attraction o R is repulsion o
ARRA
00 is for the back and forth motion through the zero initial rest position A
stretched spring goes from ldquoArdquo to ldquo0rdquo to ldquoRrdquo to ldquoArdquo to ldquo0rdquo to ldquoRrdquo d1 is the distance a mass is moved from its initial rest position
0
0 1ts A RA K d
Once ( )mad
is performed to move the mass to position d1 (against the restraint of the spring) a of the ( )mad
is transferred into the spring as potential attractionrepulsion (PAR) There is always
more ( )mad
performed than is stored as PAR When the spring is released from the restraining Ats the PAR starts the oscillation Conservation of ( )mad
requires that the PAR and the motion ( )mad
( )( )madM be a constant
( )amp ( ) constantmadPAR M 02
0 0
1
2A R A RPAR K d
02
( ) 0 0
1( )
2mad A R A RM m v
0 0 02 2 2
0 0 0 1
2 2 21
2 2 21
1 1 1amp
2 2 2
1 1 1amp
2 2 2
amp
A R A R A RK d m v K d
Kd mv Kd
Kd mv Kd
0
1d is frac12 the height from the base to the peak of a cycle
An experiment of moving paper under a pendulum through a small angle filled with sand that can escape through a small hole in the bottom shows that a pendulum cycling back and forth follows a curve that is mathematically analogous to a lsquosinrsquo or lsquocosrsquo
The projection of the curved area onto the straight line exactly duplicate the back and forth motion of CM This allows for the math of lsquosinrsquo or lsquocosrsquo to be used
SM will use lsquosinrdquo math It could use lsquocosrsquo or a combination of both lsquosinrsquo and lsquocosrsquo Using just the lsquosinrsquo math makes the understanding and math easier
0 sin 0 0SM 0 sin157 1SM The diagram at the left shows a spring being stretched from its (zero rest) position-2 and the PAR being released at position-1 There are 4 sections (1 to 2 2 to 3 3 to 4 and 4 to5) that alternate between maximum acceleration and zero velocity to max velocity and zero acceleration
At position-1 there is maximum acceleration and zero velocity o The lsquosinrsquo values from 0o to 157o will be used for velocity (values from 0 to 1) o The lsquosinrsquo values from 157o to 0o will be used for acceleration (values from 1 to 0)
At position-2 there is maximum velocity and zero acceleration Acceleration changes to deceleration
o The lsquosinrsquo values from 157o to 0o will be used for deceleration (values from 1 to 0) o The lsquosinrsquo values from 0o to 157o will be used for acceleration (values from 0 to 1)
At position-3 there is maximum acceleration and zero velocity o The lsquosinrsquo values from 0o to 157o will be used for velocity (values from 0 to 1) o The lsquosinrsquo values from 157o to 0o will be used for acceleration (values from 1 to 0)
At position-4 there is maximum velocity and zero acceleration Acceleration changes to deceleration
o The lsquosinrsquo values from 157o to 0o will be used for deceleration (values from 1 to 0) o The lsquosinrsquo values from 0o to 157o will be used for acceleration (values from 0 to 1)
At position-5 the cycle starts again
2
)539(
cT
MK
spring constant (K) when mass (M) and cycle-time (T) are known
)(
))(89)((
111 md
uKgM
d
MG
d
FK sts
spring constant (K) when mass (M) and amplitude ( d1) are known
2 21( )
Kv d d
m
velocity (v) at any point during the cycle
velocity
dforM
Kdv
max
012 velocity (v2) maximum (when d = 0 at position-2 and 4)
acceleration (a1) maximum when spring constant (K) mass (M) and amplitude (d1) are known
acceleration (a1) maximum when amplitude (d1) and cycle-time (Tc) are known
acceleration (a) at any point during the cycle-time (T)
1
151
286
v
dTc
cycle-time (Tc) when amplitude (d1) and velocity (v1) are known
K
MTc 28651
cycle-time (Tc) when mass (M) and spring constant (K) are known
fTc
1
cycle-time (Tc) when the frequency is known
cTf
1
frequency (f) when the cycle-time (Tc) is known
of cycles
elapsed timef
frequency (f) when the number of cycles in an elapsed time is known
d1 amplitude (distance from rest to start position)
Also frac12 the distance from the base to the peak
3131 oror dM
Ka
251
3131 539
c
oror T
da
2539
cT
da
In December of 2006 I discovered the mathematical error in Einsteinrsquos Special Relativity Einstein used BS math and that math only works in an imaginary world
It cannot describe real symmetrical space
Einsteinrsquos Special Relativity should be abandoned and a theory of real space using SM should be used to replace it
Numerous articles on the Internet show Einsteinrsquos theory to be incorrect but no one in academia is willing to publish the information
Figure 1-10 A moving particle observed from two frames of reference in uniform translation
Einsteinrsquos Math Error in Special Relativity Transformation of Velocity (From Eisenberg- Fundamentals of Modern Physics pages 27-29) Consider the particle shown in figure (1-10) moving with velocity v as
seen in a frame of reference O
dx dy dz
Vx Vy Vzdt dt dt
dx dy dz
V x V y V zdy dt dt
We would like to evaluate the velocity v of the particle as seen in the frame of reference 0 which is itself moving relative to 0 with velocity v Measured in the 0 frame the velocity vector of the particle has com ponents
The same velocity vector as measured in the 0 frame has components
2
2
2
2
2
1
1
x vtx
v
cy y
z z
vxt
ctv
c
Now from equations (1-13) we know that the relation between the primed and the unprimed coordinates and times is
Take the differential of these equations remembering that v is a constant This gives
2
2
2
2
2
1
1
dx vdtdx
v
cdy dy
dz dz
vxdt
cdtv
c
2
2
2 2 2 2 2
2
2
2
11 ( )
1( ) 1
1
1
1
dx vdt
v dxdx vdt v
dx dx vdtc dt dtvdx vdx vdx v dxdt dt dt dt
dt dtc c c c
VxV x
Vx
VxV x
vv
vv
c
x
v
c
cV
( )
coordinate of particle relative to moving primed system
coordinate of particle relative to stationary system
velocity of primed coordina
V x
Vx
te system relative to stationaryv sys
tem
Page 28 Fundamentals of Modern Physics
(1-16)
According to Einstein this equation tells you how to transform the observed velocity from one frame of reference to another frame of reference First we note that as Vc and vc approachzero the equation approaches those which would be derived from the Galilean transformation
2 2
( ) 1 01 1 ( )
Vx v Vx vV
VxV x
Vx v
v cVxx
v
c
v
cc
Another property of this equation is that it is impossible to choose V and v such that Vrsquo the magnitude of the velocity which is seen in the new frame is greater than c
1
2
09 ( 09 ) 18009876543
( 09 ) 1811 (09 )
c c cV c c
cc
c
Example As seen by O particle 1 has velocity 09c in the direction of positive x and particle 2 has velocity 09c in the negative x direction
Einstein equation however predicts To evaluate the velocity of particle 1 with respect to particle 2 we transform from the O frame to the Orsquo frame moving in the negative x direction with velocity v = -09c using equation 1-16 We obtain
JK LOGIC will observe the particles separating at 18c
A confusing INCORRECT and ILLOGICAL answer
left is a dir rightsub of a
1
2
1
09 ( 09 ) 18009876543
181( 09 )1 ( 09 )
The multiplication of dir left time a dir right is ONLY a dir left
09
dir
direction leftdirection rightsubtraction
c c cV c c
cc
c
V
dir
22
22
09 18 18009876543
181( 081)1(09)(09)
1
left rightsub
sub
c c c cc c
ccc
c
Analyzing Einsteinrsquos BS math Equation
Using SM math a logical answer can be obtained
Cannot multiply a subtraction operator times a direction in space
Can only subtract and change its direction
Cannot multiply a direction to the left times a direction to the right
Cannot multiply opposite directions in space
Can only solve for the resultant
rightleft
subtraction
1
subtraction
2
sub of a dir to the left is a dir to the right
1 resultant 0 sub 2
2sub 2
2
09 09
09 amp 091
09 09 18 1
(0)1
(09) amp (09)1
c cV
c cc
c c cV
cc
cc
8
181
cc
SM math predicts a logical answer using logic
21st century scientists state in many articles that Einsteins General Relativity and Quantum Mechanics are incompatible
They are not only incompatible they are both mathematically incorrect
They both use illogical BS math
Any equations in math that use imaginary numbers should be abandoned
Imaginary numbers are only necessary in the current BS math system
This means that Schroumldingers equations should be abandoned
Any equation in quantum mechanics that uses imaginary number should be abandoned
Prof Kanarev has developed logical physics and chemistrytheories without the use of Quantum mechanics or relativity
A partial list of his theories coverbull Mechanics ndash replaces Newtonbull Photon - bull Electrons ndash Protons- Neutronsbull Spectroscopybull Nucleusbull Atoms amp Moleculesbull Electrodynamicsbull Heatbull Curvature of Space ndash replaces Schwarzcchildrsquos equation
Kanarevrsquos Hypothesis
bull Electron align linearly with proton They do NOT spin or form clouds around the nucleus
bull Neutrons are in the center of atom Proton are outside of Neutrons
bull The wavelength of all particles is the experimental radius
bull Double slit experiment explained correctly
bull Heisenbergrsquos inequality explained correctly
bull is the radius of approach of the magnetic lines around the torus of the electron
( 1 137)
bull Why water expands when it freezes
bull Electron clusters form lightning
rbull The meaning of Plancks constant is explained
22 2 kg m
h m f mr f consts
2210254 42m r kgm bull If the mr-constant describes a ring the equations
describe experiments bull If the mr-constant describes a wave the equations do
NOT describe experiments
Kanarevrsquos 5 Equations of Mechanics
Kanarevrsquos 1st Equation Acceleration ampmv maMma R R
Kanarevrsquos 2nd Equation Uniform motion amp mv maMma R R
maM mvR R
Kanarevrsquos 3rd Equation
Kanarevrsquos 4th Equation Equal and opposite reaction
1 21 21 2
12
a m
m a m ama
Kanarevrsquos 5th Equation 1 2 3 Mi M M M Mna R R R R R
The sum of the Actions working on a moving body is never equal to zero
Kanarevrsquos Photon model is a rolling ring
His equations accurately describe known experiments with photonsbull Electromagnetic radiation should be called photon radiationbull Diffraction and interference are the same phenomenabull Correctly explains the double slit experiment without quantum equationsbull Correctly explains the meaning of Heisenbergrsquos inequality
e
2 (2)(3142)(2817 15 ) 10007297352568
R 2426 12 137mag
xp
r m
m
Charge
Magnetic action
Proton
The intensity of the magnetic action near the geometrical center of the proton ring is so great that we theorize that it is the strong nuclear force
(9383 8)(1602 19)8476 14
4 (4)(3142)(1411 26)p
pp
EB T
M
(9395 8)(1602 19)1035 15
4 (4)(3142)(0966 26)N
NN
EB T
M
Neutron
1a Hydrogenavi
Mirror image is broken symmetry in the x and z axis If the left amp right-axis are
reversed symmetry is broken Obs-1 math is different from obs-2
If the front amp back-axis is reversed symmetry is broken Obs-1 math is different from obs-2
If the top amp bottom-axis are
reversed symmetry is not broken Obs-1 math is the same as obs-2
Math answers cannot depend on which side of a number line an observer sit
Absolute values for Displacement
Even for a simple displacement of an object BS math invented a definition and absolute values to obtain something labeled a positive number answer
I----------I----------I----------I----------I----------I----------I----------I----------I
-4 -3 -2 -1 0 +1 +2 +3 +4 a gtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtb cltltltltltltltltltltltltltltltltltltltltltltltltltltltb
If an object starts at ( 3)ia X and moves to ( X 3) and the to ( 1)b fb c X
BS Math provides a usable answer without a definition or absolute values
However If an object starts at
( 1) and moves to ( X 3) and the to ( 3)i b fa X b c X
Then absolute values and a definition must be used
---------I----------I----------I----------I----------I----------I----------I---------- -4 -3 -2 -1 0 +1 +2 +3 +4
a gtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtb cltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltb
BS Math makes up a definition that all displacements are positive Since this example provides an answer with something labeled a negative the positive definition must be applied and (-2) is changed to (+2) A definition is needed to arrive at a useable answer This is Illogical
In SM absolute values and a meaningless definition of ldquopositiverdquo are removed In SM objects and directions in space are defined by the direction and magnitude of the resultant of arrows
With SM you get total distance traveled by the object and the final direction of the displacement
22222 2 bababbaababababa If w e su b stitu te p o s itiv e n u m b e rs in to th e eq u a tio n th e an sw er w ill b e co rrec t If (a = 5 ) an d (b = 3 )
2
2
8
5 3 5 3 5 3 5 5 5 3 3 5 3 3 25 15 15 9 64
8 8 8 8 8 8 8 8 8 8 64
W h en tw o p o sitiv e n u m b ers a re a d d ed an d sq u a re d th e B S m a th d is tr ib u tiv e law p ro v id es a co rre c t an sw er 1
Illogical and Incorrect BS math for the distributive law
222 2 bababbabbaaabababa B y d e fin itio n in B S m a th n o n u m b er sq u a red c a n ev e r b e n eg a tiv e T h e re fo re n o sq u a red n u m b er c an ev e r b e a v e c to r m o v in g in th e d a sh ( -) o r n eg a tiv e d ire c tio n if it is sq u a red In B S m a th an d ad d itio n o p e ra to r m u ltip lied b y a su b trac tio n o p e ra to r is A L W A Y S eq u a l to a su b tra c tio n o p e ra to r If th is h ad n o t b e en u sed fo r th e la s t fo u r h u n d red y e a rs y o u w o u ld d ie lau g h in g a t th e u tte r ab su rd ity o f th e la s t s ta tem e n t T h is is illo g ica l an d v io la te s sy m m etry
If a = 5 an d b = 3 42222352
22
4930253)3)(5(253)5)(3()3)(5(5353535 22222 A s lo n g a s ldquo ardquo is g re a te r th an ldquob rdquo th e B S m a th d is tr ib u tiv e law p ro v id es co rrec t an sw ers 1
B S m a th b re a k s d o w n w h e n ldquo b rdquo is g rea ter th a n ldquo a rdquo
If a = 3 an d b = 5
2 2
2
2
3 5 2 4
2 ( 2) ( 2) 4
2 ( )( 2) ( )( 2) 4
A n y co m b in a tio n ex cep t th e m id d le o n e is illo g ica l an d in co rre c t B y d e fin itio n th e an sw er can n o t b e -4 in B S m a th B S m a th is in co rrec t
4253095)5)(3(235)3)(5()5)(3(3535353 22222 T h is is w h ere th e d is tr ib u tiv e la w p ro v id es in co rrec t an sw ers I t p ro d u ces an an sw er o f + 4 b e cau se b y d e fin iti o n n o n u m b er sq u a red in B S m a th c an b e a n eg a tiv e n u m b er U sin g in co rre c t B S m a th th e d is tr ib u tiv e law w ill g iv e an in co rrec t an sw ers o f + 4 T h is is in co rrec t an d illo g ica l 1
Since there can be no negative direction from a squared term in BS math BS math should be modified or abandoned
The BS distributive law should be modified or abandoned
The numbers +1+2+3+4 and ndash1-2-3-4hellip should be abandoned
BS math gives incorrect answers because their Rule-of-Signs allows multiplying an arrow going in one direction by an arrow going in another direction
Clearly this is not logical
Arrows should be used that are specific for directions
W h at d o es it m e an to m u ltip ly ( 4) b y (4 ) A s w e k n o w 1
m u ltip lica tio n is ju s t ad d itio n W h a t w o u ld ( 4) tim e (4 ) b e 2
In BS math the answer is by definition -16 unless you square -4 and then it is +16
In SM you cannot multiply arrows going in opposite directions
The middle two terms are not logical You cannot multiply opposite directions
This is a MAJOR error in BS math They multiply a dash [(-) a direction to the left] by a cross [(+) a direction to the right] And then ldquoby definitionrdquo label the answer dash (-) This is illogical and produces incorrect answers
2
2 2
This is Einsteinrsquos math error
in Special relativity
Magnitude 4 Direction 1 Magnitude 3 Direction 2 Magnitude 5 Direction 3 Relative to an observer at 0 4
Magnitude 4 Direction 1 Magnitude 3 Direction 2 Magnitude 5 Direction 3 Relative to an observer at 0 4
Magnitude and Direction (MD) An arrow has both a magnitude andA direction relative to a 2nd objectobserver
bull Magnitude and velocity (MV) relative to a 2nd objectobserver
bull Magnitude And acceleration (MA) relative to a 2nd objectobserver2
v da
t t
Introducing a new symbol the ldquoamprdquo sign The ldquoamprdquo sign means resultant ADD all the Arrows for each direction and then SUBTRACT to find the final resultant
1 1amp 1 2 2amp 2 4 If =
1 amp1 2 2 amp2 4 If =
amp 5 amp 3 2 If gt
amp 3 amp 5= 2 If gt
amp 0 3 amp 3=0 if =
only the resultant equals zero
the vectors do not - they are still there
Does not exist There is no multiplication of arrows
There are two arrows One has a magnitude of 3 One has a magnitude of 5 The maximum these two arrows can be is 8
Only the resultant of arrows exist
An SM arrow has a direction and a magnitude
Arrows and magnitudes are still there even if the resultant is zero
There are no negative SM Arrows
There are no positive SM Arrows
There are no ZERO SM Arrows A zero would negate the definition of an SM Arrow something with no magnitude and no direction cannot be the definition of an arrow that is defined as having magnitude and direction
bull A resultant is the addition and subtraction of the differences in magnitude and direction arrows
A B and C are the arrows (a) A amp B exist(b) A amp B amp C exist(c) A-B Do Not Exist in Sm there are no negatives or positives(d) A+B-C Do Not Exist in SM there are no negatives or positives
There are only resultants when there are multiple arrows
SM Graphs versus BS Graphs
Numbers Raised to Powers
When numbers are raised to powers the negative sides of all BS math graphs are ILLOGICAL and INCORRECT
Incorrect BS Math for 2x
2x
4 16 correct3 8 correct2 4 correct1 2 correct0 1 correct-1 005 incorrect-2 025 incorrect-3 0125 incorrect-4 00625 incorrect
negative side of graph is incorrectpositive side of graph is correct
16
8
42105025012500625
-4 -3 -2 -1 0 1 2 3 4
po
sit
ive
negative positive
2x BS Math
correctincorrect
In BS math the dash sign (-) used in this example represents three different operations They arebull (-) Subtractionbull (-) A direction to the leftbull (-) As an exponent to mean divide by
All positive numbers produce correct results and graphs
All negative numbers produce incorrect results and graphs
A direction of three units to the left (-3) is placed in the equation as an exponent
A direction in space (3 units toward the left) changes a whole number to a fraction by using the dash (-) as an exponent rather than a direction in space
This is illogical and incorrect BS math should have performed this operation as
Unfortunately the answers to problems can come out correct because data points of a real problem fit the curve
However the answers will not be due to logical reasoning
Because the equation produced a curve that real problem data points follow it has been falsely believed that the equations and graphs are correct
1
2
x
Incorrect BK graph for (12)x
(12)x
4 00625 correct3 0125 correct2 025 correct1 1 correct0 1 correct-1 2 incorrect-2 4 incorrect-3 8 incorrect-4 16 incorrect
negative side of graph is incorrectpositive side of graph is correct
00625012502505124
8
16
-5 -4 -3 -2 -1 0 1 2 3 4 5P
osit
ive
negative positive
BS Math for (12)x
correctincorrect
Incorrect BS math and graph for
All graphs are symmetrical with SM
Two BS Math Equations that Produce Incorrect Answers
2 0x x starting equation 2
2
x x x x
x x
Add lsquoxrsquo to both sides
2(1) 1 let lsquoxrsquo = 1
2 1 This is an incorrect answer 2(0)-(0)=0 The only answer that BK math can give is x=0
1
a b starting equation ( )( ) ( )( )a a a b multiply both sides by ( )a
( )( ) ( )( ) ( )( ) ( )( )a a b b a b b b subtract ( )( )b b from both sides
( )( ) ( )( )a b a b a b b factor ( )a b from both sides
( )a b b divide both sides by ( )a b
( )b b b replace ( )a with ( )b 2b b simplify 2(1) (1) Let ( )b = 1
2 1 Again BS math yields and incorrect answer 1
This incorrect result is a problem with the distributive law SM explains how this incorrect answer is obtained and the problem with the distributive law that allows the error
Page 148 ndash149 Barry Mazur ndash Imagining Numbers
3
2
31
Do the computation carefully on paper using the rules we agreed to Then ponder your answer which should be something of a surprise to you donrsquot stop there Think of what your answer might possibly mean or might imply
Solve this equation using the distributive law
JK definitely a surprise the answer wasincorrect and illogical
Gerhard Gade University Professor doing math at Harvard University
Here is the BS math answer that Barry Mazur calculated
2
31
2
31
2
31
1 3 1 3 1 3 3 3 2 2 3 1 3
2 2 2 2 4 4 4 4 4 4 2 2
dash
1 3 1 3 1 31
2 2 2 2 4 4
3 3
4 4
When the distributive law is used with the BS Rule-of-Signs and one of the signs is a (-) dash sign the middle two terms always cancel each other This leads to the incorrect answer of (-1) in the above equation
Barry Mazur did not go back and check his answer with the original equation If he had he would have known that the answer was incorrect
3
1 31
2
check answer by setting equal
3
331 3
12
1 31
2
Take the Cube root of both sides
1 3 2 multiply both sides by 2
3 3 subtract 1 from both sides
2 2
3 3 square both sides
3 9 This is an incorrect answer A simple equation check could have shown that the distributive law using BS math provides an incorrect answer
1
BS Illogical and Incorrect Numbers Raised to Powers
nC D n DC
642 DDD
In BS math only a positive number can have an answer if ldquonrdquo is an even number (246---)
All negative numbers have no solutions
642 DDD
With SM just add a direction arrow and any number gives the correct answer
or n nC D C D
3 3( 3 8 8 or 8 2 or 2 )n D C
( 2 16 C= 16 or 16 4 or 4 )n D
Since a negative number has no square root answer in BS math BS mathematicians invented imaginary numbers
i2=-1
24414 iiC n
These equations are easily solved without the use of imaginary numbers with SM math
If 4 then 4 2x
If 4 then 4 2x
The 360-degree circle is a 4400 years old extremely outdated system based on a year of 12 months of 30 days each that is 360 days
SM will use a more logical system that will have a circle of 6283185307o
d
( )
(deg )
d r radians
d r rees
the arc distance around
the circumference of a circle
d
radiusr
2 2
2 2
d r
dv rd t t
v dda rd t t t
1
d
r
( )av tv
The average angular velocity of a body is the rate of its angular displacement
( )0
liminstt
vt
The instantaneous angular velocity is the limit of the ratio
t
as t approaches zero
( )d
av
d
tv
The average circumference velocity of the body is the rate of its distance traveled d divided by the time t
dvv
r
1 dd vdrv
t t t r r
The equation of the relationship between the angular velocity and the velocity of a point on that body
( )o
av
v v v
t ta
The average angular acceleration is the change in the angular velocity
( )0
liminstt
v
ta
The instantaneous angular acceleration is the
limit of the ration of v
t
as t approaches
zero
daa
r
1
1
d
d s d
vva
t r tv a
r t r
The equation of the relationship between the angular acceleration and the acceleration of a point on the body
21
2v a t
If a body with an initial angular velocity ov has a constant
angular acceleration a it will turn through an angle in a time t
1
Straight Line Symbol Units Rotational Symbol Units Distance d m Angular
Distance radians or degrees
Velocity dt
v m
s Angular Velocity
V
t
radians or degrees
s
Acceleration vat
2ms
Angular Acceleration
VA
t
2
radians or degrees
s
Mass M Kg Mass distribution
dM 2Kgm
Force actionA Ma
2
Kgm
sNewton
Torque
dT M A 2
2
Kgm
s
Momentum Mv Kgm
s
Angular Momentum
dM V 2Kgm
s
Work (Force)(distance)
Ad 2
Kgmm
sNm Joule
Work T
Using AS arrow math to solve and direction in space problems using the equation 1 2
2 2 2 2 2 2( ) amp and ( ) amp 3
4
2 2 2
2 2
( ) amp
(5 ) 4 3
(5 ) 16 9
(5 ) 25
(5 ) (5 )
2 2(5 ) 4 3
(5 ) 16 9
(5 ) 25
(5 ) (5 )
In SM math Pythagorasrsquo theorem provides both magnitude and direction
In BS math Pythagorasrsquo theorem provides magnitude only
time set equal to zero
23
( ) ( 0 )10 1
x t xx
3( ) ( 0)
10 1W t f
In BS math this is an equation for a traveling wave
must use a square term in order to make the curve symmetric
NO SQUARE TERM IS NECESSARY
In SM math when t is set to ldquo0rdquo
This is a major fault in BS math
x f(x)
-4 0019
-3 0033
-2 0073
-1 0273
0 3000
1 0273
2 0073
3 0033
4 0019 -4 -3 -2 -1 0 1 2 3 400
05
10
15
20
25
30
35
BS wave (must use square term)
snap shot of wave at x = 0
Am
plit
ude
lt to gt lt to gt f(ltgt)
lt -4 0073
lt -3 0097
lt -2 0143
lt -1 0273
origin 0 3000
gt 1 0273
gt 2 0143
gt 3 0097
gt 4 0073lt0073 lt0097 lt0143 lt0273
3000
0273gt 0143gt 0097gt 0073gt
Wave moving
snap shot of wave
Am
plit
ude
lt4 lt3 lt2 lt1 0 1gt 2gt 3gt 4gt
3
2
1
In SM NO squared terms are needed
SPRING
An Action is needed to stretch or compress a spring to a starting point where it can execute cycle motion
tsA (ts to start) is the work required to move a mass attached to a spring a distance (d1) from
its initial rest position RAK 0 the spring constant is the stiffness of the spring
o A is attraction o R is repulsion o
ARRA
00 is for the back and forth motion through the zero initial rest position A
stretched spring goes from ldquoArdquo to ldquo0rdquo to ldquoRrdquo to ldquoArdquo to ldquo0rdquo to ldquoRrdquo d1 is the distance a mass is moved from its initial rest position
0
0 1ts A RA K d
Once ( )mad
is performed to move the mass to position d1 (against the restraint of the spring) a of the ( )mad
is transferred into the spring as potential attractionrepulsion (PAR) There is always
more ( )mad
performed than is stored as PAR When the spring is released from the restraining Ats the PAR starts the oscillation Conservation of ( )mad
requires that the PAR and the motion ( )mad
( )( )madM be a constant
( )amp ( ) constantmadPAR M 02
0 0
1
2A R A RPAR K d
02
( ) 0 0
1( )
2mad A R A RM m v
0 0 02 2 2
0 0 0 1
2 2 21
2 2 21
1 1 1amp
2 2 2
1 1 1amp
2 2 2
amp
A R A R A RK d m v K d
Kd mv Kd
Kd mv Kd
0
1d is frac12 the height from the base to the peak of a cycle
An experiment of moving paper under a pendulum through a small angle filled with sand that can escape through a small hole in the bottom shows that a pendulum cycling back and forth follows a curve that is mathematically analogous to a lsquosinrsquo or lsquocosrsquo
The projection of the curved area onto the straight line exactly duplicate the back and forth motion of CM This allows for the math of lsquosinrsquo or lsquocosrsquo to be used
SM will use lsquosinrdquo math It could use lsquocosrsquo or a combination of both lsquosinrsquo and lsquocosrsquo Using just the lsquosinrsquo math makes the understanding and math easier
0 sin 0 0SM 0 sin157 1SM The diagram at the left shows a spring being stretched from its (zero rest) position-2 and the PAR being released at position-1 There are 4 sections (1 to 2 2 to 3 3 to 4 and 4 to5) that alternate between maximum acceleration and zero velocity to max velocity and zero acceleration
At position-1 there is maximum acceleration and zero velocity o The lsquosinrsquo values from 0o to 157o will be used for velocity (values from 0 to 1) o The lsquosinrsquo values from 157o to 0o will be used for acceleration (values from 1 to 0)
At position-2 there is maximum velocity and zero acceleration Acceleration changes to deceleration
o The lsquosinrsquo values from 157o to 0o will be used for deceleration (values from 1 to 0) o The lsquosinrsquo values from 0o to 157o will be used for acceleration (values from 0 to 1)
At position-3 there is maximum acceleration and zero velocity o The lsquosinrsquo values from 0o to 157o will be used for velocity (values from 0 to 1) o The lsquosinrsquo values from 157o to 0o will be used for acceleration (values from 1 to 0)
At position-4 there is maximum velocity and zero acceleration Acceleration changes to deceleration
o The lsquosinrsquo values from 157o to 0o will be used for deceleration (values from 1 to 0) o The lsquosinrsquo values from 0o to 157o will be used for acceleration (values from 0 to 1)
At position-5 the cycle starts again
2
)539(
cT
MK
spring constant (K) when mass (M) and cycle-time (T) are known
)(
))(89)((
111 md
uKgM
d
MG
d
FK sts
spring constant (K) when mass (M) and amplitude ( d1) are known
2 21( )
Kv d d
m
velocity (v) at any point during the cycle
velocity
dforM
Kdv
max
012 velocity (v2) maximum (when d = 0 at position-2 and 4)
acceleration (a1) maximum when spring constant (K) mass (M) and amplitude (d1) are known
acceleration (a1) maximum when amplitude (d1) and cycle-time (Tc) are known
acceleration (a) at any point during the cycle-time (T)
1
151
286
v
dTc
cycle-time (Tc) when amplitude (d1) and velocity (v1) are known
K
MTc 28651
cycle-time (Tc) when mass (M) and spring constant (K) are known
fTc
1
cycle-time (Tc) when the frequency is known
cTf
1
frequency (f) when the cycle-time (Tc) is known
of cycles
elapsed timef
frequency (f) when the number of cycles in an elapsed time is known
d1 amplitude (distance from rest to start position)
Also frac12 the distance from the base to the peak
3131 oror dM
Ka
251
3131 539
c
oror T
da
2539
cT
da
In December of 2006 I discovered the mathematical error in Einsteinrsquos Special Relativity Einstein used BS math and that math only works in an imaginary world
It cannot describe real symmetrical space
Einsteinrsquos Special Relativity should be abandoned and a theory of real space using SM should be used to replace it
Numerous articles on the Internet show Einsteinrsquos theory to be incorrect but no one in academia is willing to publish the information
Figure 1-10 A moving particle observed from two frames of reference in uniform translation
Einsteinrsquos Math Error in Special Relativity Transformation of Velocity (From Eisenberg- Fundamentals of Modern Physics pages 27-29) Consider the particle shown in figure (1-10) moving with velocity v as
seen in a frame of reference O
dx dy dz
Vx Vy Vzdt dt dt
dx dy dz
V x V y V zdy dt dt
We would like to evaluate the velocity v of the particle as seen in the frame of reference 0 which is itself moving relative to 0 with velocity v Measured in the 0 frame the velocity vector of the particle has com ponents
The same velocity vector as measured in the 0 frame has components
2
2
2
2
2
1
1
x vtx
v
cy y
z z
vxt
ctv
c
Now from equations (1-13) we know that the relation between the primed and the unprimed coordinates and times is
Take the differential of these equations remembering that v is a constant This gives
2
2
2
2
2
1
1
dx vdtdx
v
cdy dy
dz dz
vxdt
cdtv
c
2
2
2 2 2 2 2
2
2
2
11 ( )
1( ) 1
1
1
1
dx vdt
v dxdx vdt v
dx dx vdtc dt dtvdx vdx vdx v dxdt dt dt dt
dt dtc c c c
VxV x
Vx
VxV x
vv
vv
c
x
v
c
cV
( )
coordinate of particle relative to moving primed system
coordinate of particle relative to stationary system
velocity of primed coordina
V x
Vx
te system relative to stationaryv sys
tem
Page 28 Fundamentals of Modern Physics
(1-16)
According to Einstein this equation tells you how to transform the observed velocity from one frame of reference to another frame of reference First we note that as Vc and vc approachzero the equation approaches those which would be derived from the Galilean transformation
2 2
( ) 1 01 1 ( )
Vx v Vx vV
VxV x
Vx v
v cVxx
v
c
v
cc
Another property of this equation is that it is impossible to choose V and v such that Vrsquo the magnitude of the velocity which is seen in the new frame is greater than c
1
2
09 ( 09 ) 18009876543
( 09 ) 1811 (09 )
c c cV c c
cc
c
Example As seen by O particle 1 has velocity 09c in the direction of positive x and particle 2 has velocity 09c in the negative x direction
Einstein equation however predicts To evaluate the velocity of particle 1 with respect to particle 2 we transform from the O frame to the Orsquo frame moving in the negative x direction with velocity v = -09c using equation 1-16 We obtain
JK LOGIC will observe the particles separating at 18c
A confusing INCORRECT and ILLOGICAL answer
left is a dir rightsub of a
1
2
1
09 ( 09 ) 18009876543
181( 09 )1 ( 09 )
The multiplication of dir left time a dir right is ONLY a dir left
09
dir
direction leftdirection rightsubtraction
c c cV c c
cc
c
V
dir
22
22
09 18 18009876543
181( 081)1(09)(09)
1
left rightsub
sub
c c c cc c
ccc
c
Analyzing Einsteinrsquos BS math Equation
Using SM math a logical answer can be obtained
Cannot multiply a subtraction operator times a direction in space
Can only subtract and change its direction
Cannot multiply a direction to the left times a direction to the right
Cannot multiply opposite directions in space
Can only solve for the resultant
rightleft
subtraction
1
subtraction
2
sub of a dir to the left is a dir to the right
1 resultant 0 sub 2
2sub 2
2
09 09
09 amp 091
09 09 18 1
(0)1
(09) amp (09)1
c cV
c cc
c c cV
cc
cc
8
181
cc
SM math predicts a logical answer using logic
21st century scientists state in many articles that Einsteins General Relativity and Quantum Mechanics are incompatible
They are not only incompatible they are both mathematically incorrect
They both use illogical BS math
Any equations in math that use imaginary numbers should be abandoned
Imaginary numbers are only necessary in the current BS math system
This means that Schroumldingers equations should be abandoned
Any equation in quantum mechanics that uses imaginary number should be abandoned
Prof Kanarev has developed logical physics and chemistrytheories without the use of Quantum mechanics or relativity
A partial list of his theories coverbull Mechanics ndash replaces Newtonbull Photon - bull Electrons ndash Protons- Neutronsbull Spectroscopybull Nucleusbull Atoms amp Moleculesbull Electrodynamicsbull Heatbull Curvature of Space ndash replaces Schwarzcchildrsquos equation
Kanarevrsquos Hypothesis
bull Electron align linearly with proton They do NOT spin or form clouds around the nucleus
bull Neutrons are in the center of atom Proton are outside of Neutrons
bull The wavelength of all particles is the experimental radius
bull Double slit experiment explained correctly
bull Heisenbergrsquos inequality explained correctly
bull is the radius of approach of the magnetic lines around the torus of the electron
( 1 137)
bull Why water expands when it freezes
bull Electron clusters form lightning
rbull The meaning of Plancks constant is explained
22 2 kg m
h m f mr f consts
2210254 42m r kgm bull If the mr-constant describes a ring the equations
describe experiments bull If the mr-constant describes a wave the equations do
NOT describe experiments
Kanarevrsquos 5 Equations of Mechanics
Kanarevrsquos 1st Equation Acceleration ampmv maMma R R
Kanarevrsquos 2nd Equation Uniform motion amp mv maMma R R
maM mvR R
Kanarevrsquos 3rd Equation
Kanarevrsquos 4th Equation Equal and opposite reaction
1 21 21 2
12
a m
m a m ama
Kanarevrsquos 5th Equation 1 2 3 Mi M M M Mna R R R R R
The sum of the Actions working on a moving body is never equal to zero
Kanarevrsquos Photon model is a rolling ring
His equations accurately describe known experiments with photonsbull Electromagnetic radiation should be called photon radiationbull Diffraction and interference are the same phenomenabull Correctly explains the double slit experiment without quantum equationsbull Correctly explains the meaning of Heisenbergrsquos inequality
e
2 (2)(3142)(2817 15 ) 10007297352568
R 2426 12 137mag
xp
r m
m
Charge
Magnetic action
Proton
The intensity of the magnetic action near the geometrical center of the proton ring is so great that we theorize that it is the strong nuclear force
(9383 8)(1602 19)8476 14
4 (4)(3142)(1411 26)p
pp
EB T
M
(9395 8)(1602 19)1035 15
4 (4)(3142)(0966 26)N
NN
EB T
M
Neutron
1a Hydrogenavi
Absolute values for Displacement
Even for a simple displacement of an object BS math invented a definition and absolute values to obtain something labeled a positive number answer
I----------I----------I----------I----------I----------I----------I----------I----------I
-4 -3 -2 -1 0 +1 +2 +3 +4 a gtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtb cltltltltltltltltltltltltltltltltltltltltltltltltltltltb
If an object starts at ( 3)ia X and moves to ( X 3) and the to ( 1)b fb c X
BS Math provides a usable answer without a definition or absolute values
However If an object starts at
( 1) and moves to ( X 3) and the to ( 3)i b fa X b c X
Then absolute values and a definition must be used
---------I----------I----------I----------I----------I----------I----------I---------- -4 -3 -2 -1 0 +1 +2 +3 +4
a gtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtb cltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltb
BS Math makes up a definition that all displacements are positive Since this example provides an answer with something labeled a negative the positive definition must be applied and (-2) is changed to (+2) A definition is needed to arrive at a useable answer This is Illogical
In SM absolute values and a meaningless definition of ldquopositiverdquo are removed In SM objects and directions in space are defined by the direction and magnitude of the resultant of arrows
With SM you get total distance traveled by the object and the final direction of the displacement
22222 2 bababbaababababa If w e su b stitu te p o s itiv e n u m b e rs in to th e eq u a tio n th e an sw er w ill b e co rrec t If (a = 5 ) an d (b = 3 )
2
2
8
5 3 5 3 5 3 5 5 5 3 3 5 3 3 25 15 15 9 64
8 8 8 8 8 8 8 8 8 8 64
W h en tw o p o sitiv e n u m b ers a re a d d ed an d sq u a re d th e B S m a th d is tr ib u tiv e law p ro v id es a co rre c t an sw er 1
Illogical and Incorrect BS math for the distributive law
222 2 bababbabbaaabababa B y d e fin itio n in B S m a th n o n u m b er sq u a red c a n ev e r b e n eg a tiv e T h e re fo re n o sq u a red n u m b er c an ev e r b e a v e c to r m o v in g in th e d a sh ( -) o r n eg a tiv e d ire c tio n if it is sq u a red In B S m a th an d ad d itio n o p e ra to r m u ltip lied b y a su b trac tio n o p e ra to r is A L W A Y S eq u a l to a su b tra c tio n o p e ra to r If th is h ad n o t b e en u sed fo r th e la s t fo u r h u n d red y e a rs y o u w o u ld d ie lau g h in g a t th e u tte r ab su rd ity o f th e la s t s ta tem e n t T h is is illo g ica l an d v io la te s sy m m etry
If a = 5 an d b = 3 42222352
22
4930253)3)(5(253)5)(3()3)(5(5353535 22222 A s lo n g a s ldquo ardquo is g re a te r th an ldquob rdquo th e B S m a th d is tr ib u tiv e law p ro v id es co rrec t an sw ers 1
B S m a th b re a k s d o w n w h e n ldquo b rdquo is g rea ter th a n ldquo a rdquo
If a = 3 an d b = 5
2 2
2
2
3 5 2 4
2 ( 2) ( 2) 4
2 ( )( 2) ( )( 2) 4
A n y co m b in a tio n ex cep t th e m id d le o n e is illo g ica l an d in co rre c t B y d e fin itio n th e an sw er can n o t b e -4 in B S m a th B S m a th is in co rrec t
4253095)5)(3(235)3)(5()5)(3(3535353 22222 T h is is w h ere th e d is tr ib u tiv e la w p ro v id es in co rrec t an sw ers I t p ro d u ces an an sw er o f + 4 b e cau se b y d e fin iti o n n o n u m b er sq u a red in B S m a th c an b e a n eg a tiv e n u m b er U sin g in co rre c t B S m a th th e d is tr ib u tiv e law w ill g iv e an in co rrec t an sw ers o f + 4 T h is is in co rrec t an d illo g ica l 1
Since there can be no negative direction from a squared term in BS math BS math should be modified or abandoned
The BS distributive law should be modified or abandoned
The numbers +1+2+3+4 and ndash1-2-3-4hellip should be abandoned
BS math gives incorrect answers because their Rule-of-Signs allows multiplying an arrow going in one direction by an arrow going in another direction
Clearly this is not logical
Arrows should be used that are specific for directions
W h at d o es it m e an to m u ltip ly ( 4) b y (4 ) A s w e k n o w 1
m u ltip lica tio n is ju s t ad d itio n W h a t w o u ld ( 4) tim e (4 ) b e 2
In BS math the answer is by definition -16 unless you square -4 and then it is +16
In SM you cannot multiply arrows going in opposite directions
The middle two terms are not logical You cannot multiply opposite directions
This is a MAJOR error in BS math They multiply a dash [(-) a direction to the left] by a cross [(+) a direction to the right] And then ldquoby definitionrdquo label the answer dash (-) This is illogical and produces incorrect answers
2
2 2
This is Einsteinrsquos math error
in Special relativity
Magnitude 4 Direction 1 Magnitude 3 Direction 2 Magnitude 5 Direction 3 Relative to an observer at 0 4
Magnitude 4 Direction 1 Magnitude 3 Direction 2 Magnitude 5 Direction 3 Relative to an observer at 0 4
Magnitude and Direction (MD) An arrow has both a magnitude andA direction relative to a 2nd objectobserver
bull Magnitude and velocity (MV) relative to a 2nd objectobserver
bull Magnitude And acceleration (MA) relative to a 2nd objectobserver2
v da
t t
Introducing a new symbol the ldquoamprdquo sign The ldquoamprdquo sign means resultant ADD all the Arrows for each direction and then SUBTRACT to find the final resultant
1 1amp 1 2 2amp 2 4 If =
1 amp1 2 2 amp2 4 If =
amp 5 amp 3 2 If gt
amp 3 amp 5= 2 If gt
amp 0 3 amp 3=0 if =
only the resultant equals zero
the vectors do not - they are still there
Does not exist There is no multiplication of arrows
There are two arrows One has a magnitude of 3 One has a magnitude of 5 The maximum these two arrows can be is 8
Only the resultant of arrows exist
An SM arrow has a direction and a magnitude
Arrows and magnitudes are still there even if the resultant is zero
There are no negative SM Arrows
There are no positive SM Arrows
There are no ZERO SM Arrows A zero would negate the definition of an SM Arrow something with no magnitude and no direction cannot be the definition of an arrow that is defined as having magnitude and direction
bull A resultant is the addition and subtraction of the differences in magnitude and direction arrows
A B and C are the arrows (a) A amp B exist(b) A amp B amp C exist(c) A-B Do Not Exist in Sm there are no negatives or positives(d) A+B-C Do Not Exist in SM there are no negatives or positives
There are only resultants when there are multiple arrows
SM Graphs versus BS Graphs
Numbers Raised to Powers
When numbers are raised to powers the negative sides of all BS math graphs are ILLOGICAL and INCORRECT
Incorrect BS Math for 2x
2x
4 16 correct3 8 correct2 4 correct1 2 correct0 1 correct-1 005 incorrect-2 025 incorrect-3 0125 incorrect-4 00625 incorrect
negative side of graph is incorrectpositive side of graph is correct
16
8
42105025012500625
-4 -3 -2 -1 0 1 2 3 4
po
sit
ive
negative positive
2x BS Math
correctincorrect
In BS math the dash sign (-) used in this example represents three different operations They arebull (-) Subtractionbull (-) A direction to the leftbull (-) As an exponent to mean divide by
All positive numbers produce correct results and graphs
All negative numbers produce incorrect results and graphs
A direction of three units to the left (-3) is placed in the equation as an exponent
A direction in space (3 units toward the left) changes a whole number to a fraction by using the dash (-) as an exponent rather than a direction in space
This is illogical and incorrect BS math should have performed this operation as
Unfortunately the answers to problems can come out correct because data points of a real problem fit the curve
However the answers will not be due to logical reasoning
Because the equation produced a curve that real problem data points follow it has been falsely believed that the equations and graphs are correct
1
2
x
Incorrect BK graph for (12)x
(12)x
4 00625 correct3 0125 correct2 025 correct1 1 correct0 1 correct-1 2 incorrect-2 4 incorrect-3 8 incorrect-4 16 incorrect
negative side of graph is incorrectpositive side of graph is correct
00625012502505124
8
16
-5 -4 -3 -2 -1 0 1 2 3 4 5P
osit
ive
negative positive
BS Math for (12)x
correctincorrect
Incorrect BS math and graph for
All graphs are symmetrical with SM
Two BS Math Equations that Produce Incorrect Answers
2 0x x starting equation 2
2
x x x x
x x
Add lsquoxrsquo to both sides
2(1) 1 let lsquoxrsquo = 1
2 1 This is an incorrect answer 2(0)-(0)=0 The only answer that BK math can give is x=0
1
a b starting equation ( )( ) ( )( )a a a b multiply both sides by ( )a
( )( ) ( )( ) ( )( ) ( )( )a a b b a b b b subtract ( )( )b b from both sides
( )( ) ( )( )a b a b a b b factor ( )a b from both sides
( )a b b divide both sides by ( )a b
( )b b b replace ( )a with ( )b 2b b simplify 2(1) (1) Let ( )b = 1
2 1 Again BS math yields and incorrect answer 1
This incorrect result is a problem with the distributive law SM explains how this incorrect answer is obtained and the problem with the distributive law that allows the error
Page 148 ndash149 Barry Mazur ndash Imagining Numbers
3
2
31
Do the computation carefully on paper using the rules we agreed to Then ponder your answer which should be something of a surprise to you donrsquot stop there Think of what your answer might possibly mean or might imply
Solve this equation using the distributive law
JK definitely a surprise the answer wasincorrect and illogical
Gerhard Gade University Professor doing math at Harvard University
Here is the BS math answer that Barry Mazur calculated
2
31
2
31
2
31
1 3 1 3 1 3 3 3 2 2 3 1 3
2 2 2 2 4 4 4 4 4 4 2 2
dash
1 3 1 3 1 31
2 2 2 2 4 4
3 3
4 4
When the distributive law is used with the BS Rule-of-Signs and one of the signs is a (-) dash sign the middle two terms always cancel each other This leads to the incorrect answer of (-1) in the above equation
Barry Mazur did not go back and check his answer with the original equation If he had he would have known that the answer was incorrect
3
1 31
2
check answer by setting equal
3
331 3
12
1 31
2
Take the Cube root of both sides
1 3 2 multiply both sides by 2
3 3 subtract 1 from both sides
2 2
3 3 square both sides
3 9 This is an incorrect answer A simple equation check could have shown that the distributive law using BS math provides an incorrect answer
1
BS Illogical and Incorrect Numbers Raised to Powers
nC D n DC
642 DDD
In BS math only a positive number can have an answer if ldquonrdquo is an even number (246---)
All negative numbers have no solutions
642 DDD
With SM just add a direction arrow and any number gives the correct answer
or n nC D C D
3 3( 3 8 8 or 8 2 or 2 )n D C
( 2 16 C= 16 or 16 4 or 4 )n D
Since a negative number has no square root answer in BS math BS mathematicians invented imaginary numbers
i2=-1
24414 iiC n
These equations are easily solved without the use of imaginary numbers with SM math
If 4 then 4 2x
If 4 then 4 2x
The 360-degree circle is a 4400 years old extremely outdated system based on a year of 12 months of 30 days each that is 360 days
SM will use a more logical system that will have a circle of 6283185307o
d
( )
(deg )
d r radians
d r rees
the arc distance around
the circumference of a circle
d
radiusr
2 2
2 2
d r
dv rd t t
v dda rd t t t
1
d
r
( )av tv
The average angular velocity of a body is the rate of its angular displacement
( )0
liminstt
vt
The instantaneous angular velocity is the limit of the ratio
t
as t approaches zero
( )d
av
d
tv
The average circumference velocity of the body is the rate of its distance traveled d divided by the time t
dvv
r
1 dd vdrv
t t t r r
The equation of the relationship between the angular velocity and the velocity of a point on that body
( )o
av
v v v
t ta
The average angular acceleration is the change in the angular velocity
( )0
liminstt
v
ta
The instantaneous angular acceleration is the
limit of the ration of v
t
as t approaches
zero
daa
r
1
1
d
d s d
vva
t r tv a
r t r
The equation of the relationship between the angular acceleration and the acceleration of a point on the body
21
2v a t
If a body with an initial angular velocity ov has a constant
angular acceleration a it will turn through an angle in a time t
1
Straight Line Symbol Units Rotational Symbol Units Distance d m Angular
Distance radians or degrees
Velocity dt
v m
s Angular Velocity
V
t
radians or degrees
s
Acceleration vat
2ms
Angular Acceleration
VA
t
2
radians or degrees
s
Mass M Kg Mass distribution
dM 2Kgm
Force actionA Ma
2
Kgm
sNewton
Torque
dT M A 2
2
Kgm
s
Momentum Mv Kgm
s
Angular Momentum
dM V 2Kgm
s
Work (Force)(distance)
Ad 2
Kgmm
sNm Joule
Work T
Using AS arrow math to solve and direction in space problems using the equation 1 2
2 2 2 2 2 2( ) amp and ( ) amp 3
4
2 2 2
2 2
( ) amp
(5 ) 4 3
(5 ) 16 9
(5 ) 25
(5 ) (5 )
2 2(5 ) 4 3
(5 ) 16 9
(5 ) 25
(5 ) (5 )
In SM math Pythagorasrsquo theorem provides both magnitude and direction
In BS math Pythagorasrsquo theorem provides magnitude only
time set equal to zero
23
( ) ( 0 )10 1
x t xx
3( ) ( 0)
10 1W t f
In BS math this is an equation for a traveling wave
must use a square term in order to make the curve symmetric
NO SQUARE TERM IS NECESSARY
In SM math when t is set to ldquo0rdquo
This is a major fault in BS math
x f(x)
-4 0019
-3 0033
-2 0073
-1 0273
0 3000
1 0273
2 0073
3 0033
4 0019 -4 -3 -2 -1 0 1 2 3 400
05
10
15
20
25
30
35
BS wave (must use square term)
snap shot of wave at x = 0
Am
plit
ude
lt to gt lt to gt f(ltgt)
lt -4 0073
lt -3 0097
lt -2 0143
lt -1 0273
origin 0 3000
gt 1 0273
gt 2 0143
gt 3 0097
gt 4 0073lt0073 lt0097 lt0143 lt0273
3000
0273gt 0143gt 0097gt 0073gt
Wave moving
snap shot of wave
Am
plit
ude
lt4 lt3 lt2 lt1 0 1gt 2gt 3gt 4gt
3
2
1
In SM NO squared terms are needed
SPRING
An Action is needed to stretch or compress a spring to a starting point where it can execute cycle motion
tsA (ts to start) is the work required to move a mass attached to a spring a distance (d1) from
its initial rest position RAK 0 the spring constant is the stiffness of the spring
o A is attraction o R is repulsion o
ARRA
00 is for the back and forth motion through the zero initial rest position A
stretched spring goes from ldquoArdquo to ldquo0rdquo to ldquoRrdquo to ldquoArdquo to ldquo0rdquo to ldquoRrdquo d1 is the distance a mass is moved from its initial rest position
0
0 1ts A RA K d
Once ( )mad
is performed to move the mass to position d1 (against the restraint of the spring) a of the ( )mad
is transferred into the spring as potential attractionrepulsion (PAR) There is always
more ( )mad
performed than is stored as PAR When the spring is released from the restraining Ats the PAR starts the oscillation Conservation of ( )mad
requires that the PAR and the motion ( )mad
( )( )madM be a constant
( )amp ( ) constantmadPAR M 02
0 0
1
2A R A RPAR K d
02
( ) 0 0
1( )
2mad A R A RM m v
0 0 02 2 2
0 0 0 1
2 2 21
2 2 21
1 1 1amp
2 2 2
1 1 1amp
2 2 2
amp
A R A R A RK d m v K d
Kd mv Kd
Kd mv Kd
0
1d is frac12 the height from the base to the peak of a cycle
An experiment of moving paper under a pendulum through a small angle filled with sand that can escape through a small hole in the bottom shows that a pendulum cycling back and forth follows a curve that is mathematically analogous to a lsquosinrsquo or lsquocosrsquo
The projection of the curved area onto the straight line exactly duplicate the back and forth motion of CM This allows for the math of lsquosinrsquo or lsquocosrsquo to be used
SM will use lsquosinrdquo math It could use lsquocosrsquo or a combination of both lsquosinrsquo and lsquocosrsquo Using just the lsquosinrsquo math makes the understanding and math easier
0 sin 0 0SM 0 sin157 1SM The diagram at the left shows a spring being stretched from its (zero rest) position-2 and the PAR being released at position-1 There are 4 sections (1 to 2 2 to 3 3 to 4 and 4 to5) that alternate between maximum acceleration and zero velocity to max velocity and zero acceleration
At position-1 there is maximum acceleration and zero velocity o The lsquosinrsquo values from 0o to 157o will be used for velocity (values from 0 to 1) o The lsquosinrsquo values from 157o to 0o will be used for acceleration (values from 1 to 0)
At position-2 there is maximum velocity and zero acceleration Acceleration changes to deceleration
o The lsquosinrsquo values from 157o to 0o will be used for deceleration (values from 1 to 0) o The lsquosinrsquo values from 0o to 157o will be used for acceleration (values from 0 to 1)
At position-3 there is maximum acceleration and zero velocity o The lsquosinrsquo values from 0o to 157o will be used for velocity (values from 0 to 1) o The lsquosinrsquo values from 157o to 0o will be used for acceleration (values from 1 to 0)
At position-4 there is maximum velocity and zero acceleration Acceleration changes to deceleration
o The lsquosinrsquo values from 157o to 0o will be used for deceleration (values from 1 to 0) o The lsquosinrsquo values from 0o to 157o will be used for acceleration (values from 0 to 1)
At position-5 the cycle starts again
2
)539(
cT
MK
spring constant (K) when mass (M) and cycle-time (T) are known
)(
))(89)((
111 md
uKgM
d
MG
d
FK sts
spring constant (K) when mass (M) and amplitude ( d1) are known
2 21( )
Kv d d
m
velocity (v) at any point during the cycle
velocity
dforM
Kdv
max
012 velocity (v2) maximum (when d = 0 at position-2 and 4)
acceleration (a1) maximum when spring constant (K) mass (M) and amplitude (d1) are known
acceleration (a1) maximum when amplitude (d1) and cycle-time (Tc) are known
acceleration (a) at any point during the cycle-time (T)
1
151
286
v
dTc
cycle-time (Tc) when amplitude (d1) and velocity (v1) are known
K
MTc 28651
cycle-time (Tc) when mass (M) and spring constant (K) are known
fTc
1
cycle-time (Tc) when the frequency is known
cTf
1
frequency (f) when the cycle-time (Tc) is known
of cycles
elapsed timef
frequency (f) when the number of cycles in an elapsed time is known
d1 amplitude (distance from rest to start position)
Also frac12 the distance from the base to the peak
3131 oror dM
Ka
251
3131 539
c
oror T
da
2539
cT
da
In December of 2006 I discovered the mathematical error in Einsteinrsquos Special Relativity Einstein used BS math and that math only works in an imaginary world
It cannot describe real symmetrical space
Einsteinrsquos Special Relativity should be abandoned and a theory of real space using SM should be used to replace it
Numerous articles on the Internet show Einsteinrsquos theory to be incorrect but no one in academia is willing to publish the information
Figure 1-10 A moving particle observed from two frames of reference in uniform translation
Einsteinrsquos Math Error in Special Relativity Transformation of Velocity (From Eisenberg- Fundamentals of Modern Physics pages 27-29) Consider the particle shown in figure (1-10) moving with velocity v as
seen in a frame of reference O
dx dy dz
Vx Vy Vzdt dt dt
dx dy dz
V x V y V zdy dt dt
We would like to evaluate the velocity v of the particle as seen in the frame of reference 0 which is itself moving relative to 0 with velocity v Measured in the 0 frame the velocity vector of the particle has com ponents
The same velocity vector as measured in the 0 frame has components
2
2
2
2
2
1
1
x vtx
v
cy y
z z
vxt
ctv
c
Now from equations (1-13) we know that the relation between the primed and the unprimed coordinates and times is
Take the differential of these equations remembering that v is a constant This gives
2
2
2
2
2
1
1
dx vdtdx
v
cdy dy
dz dz
vxdt
cdtv
c
2
2
2 2 2 2 2
2
2
2
11 ( )
1( ) 1
1
1
1
dx vdt
v dxdx vdt v
dx dx vdtc dt dtvdx vdx vdx v dxdt dt dt dt
dt dtc c c c
VxV x
Vx
VxV x
vv
vv
c
x
v
c
cV
( )
coordinate of particle relative to moving primed system
coordinate of particle relative to stationary system
velocity of primed coordina
V x
Vx
te system relative to stationaryv sys
tem
Page 28 Fundamentals of Modern Physics
(1-16)
According to Einstein this equation tells you how to transform the observed velocity from one frame of reference to another frame of reference First we note that as Vc and vc approachzero the equation approaches those which would be derived from the Galilean transformation
2 2
( ) 1 01 1 ( )
Vx v Vx vV
VxV x
Vx v
v cVxx
v
c
v
cc
Another property of this equation is that it is impossible to choose V and v such that Vrsquo the magnitude of the velocity which is seen in the new frame is greater than c
1
2
09 ( 09 ) 18009876543
( 09 ) 1811 (09 )
c c cV c c
cc
c
Example As seen by O particle 1 has velocity 09c in the direction of positive x and particle 2 has velocity 09c in the negative x direction
Einstein equation however predicts To evaluate the velocity of particle 1 with respect to particle 2 we transform from the O frame to the Orsquo frame moving in the negative x direction with velocity v = -09c using equation 1-16 We obtain
JK LOGIC will observe the particles separating at 18c
A confusing INCORRECT and ILLOGICAL answer
left is a dir rightsub of a
1
2
1
09 ( 09 ) 18009876543
181( 09 )1 ( 09 )
The multiplication of dir left time a dir right is ONLY a dir left
09
dir
direction leftdirection rightsubtraction
c c cV c c
cc
c
V
dir
22
22
09 18 18009876543
181( 081)1(09)(09)
1
left rightsub
sub
c c c cc c
ccc
c
Analyzing Einsteinrsquos BS math Equation
Using SM math a logical answer can be obtained
Cannot multiply a subtraction operator times a direction in space
Can only subtract and change its direction
Cannot multiply a direction to the left times a direction to the right
Cannot multiply opposite directions in space
Can only solve for the resultant
rightleft
subtraction
1
subtraction
2
sub of a dir to the left is a dir to the right
1 resultant 0 sub 2
2sub 2
2
09 09
09 amp 091
09 09 18 1
(0)1
(09) amp (09)1
c cV
c cc
c c cV
cc
cc
8
181
cc
SM math predicts a logical answer using logic
21st century scientists state in many articles that Einsteins General Relativity and Quantum Mechanics are incompatible
They are not only incompatible they are both mathematically incorrect
They both use illogical BS math
Any equations in math that use imaginary numbers should be abandoned
Imaginary numbers are only necessary in the current BS math system
This means that Schroumldingers equations should be abandoned
Any equation in quantum mechanics that uses imaginary number should be abandoned
Prof Kanarev has developed logical physics and chemistrytheories without the use of Quantum mechanics or relativity
A partial list of his theories coverbull Mechanics ndash replaces Newtonbull Photon - bull Electrons ndash Protons- Neutronsbull Spectroscopybull Nucleusbull Atoms amp Moleculesbull Electrodynamicsbull Heatbull Curvature of Space ndash replaces Schwarzcchildrsquos equation
Kanarevrsquos Hypothesis
bull Electron align linearly with proton They do NOT spin or form clouds around the nucleus
bull Neutrons are in the center of atom Proton are outside of Neutrons
bull The wavelength of all particles is the experimental radius
bull Double slit experiment explained correctly
bull Heisenbergrsquos inequality explained correctly
bull is the radius of approach of the magnetic lines around the torus of the electron
( 1 137)
bull Why water expands when it freezes
bull Electron clusters form lightning
rbull The meaning of Plancks constant is explained
22 2 kg m
h m f mr f consts
2210254 42m r kgm bull If the mr-constant describes a ring the equations
describe experiments bull If the mr-constant describes a wave the equations do
NOT describe experiments
Kanarevrsquos 5 Equations of Mechanics
Kanarevrsquos 1st Equation Acceleration ampmv maMma R R
Kanarevrsquos 2nd Equation Uniform motion amp mv maMma R R
maM mvR R
Kanarevrsquos 3rd Equation
Kanarevrsquos 4th Equation Equal and opposite reaction
1 21 21 2
12
a m
m a m ama
Kanarevrsquos 5th Equation 1 2 3 Mi M M M Mna R R R R R
The sum of the Actions working on a moving body is never equal to zero
Kanarevrsquos Photon model is a rolling ring
His equations accurately describe known experiments with photonsbull Electromagnetic radiation should be called photon radiationbull Diffraction and interference are the same phenomenabull Correctly explains the double slit experiment without quantum equationsbull Correctly explains the meaning of Heisenbergrsquos inequality
e
2 (2)(3142)(2817 15 ) 10007297352568
R 2426 12 137mag
xp
r m
m
Charge
Magnetic action
Proton
The intensity of the magnetic action near the geometrical center of the proton ring is so great that we theorize that it is the strong nuclear force
(9383 8)(1602 19)8476 14
4 (4)(3142)(1411 26)p
pp
EB T
M
(9395 8)(1602 19)1035 15
4 (4)(3142)(0966 26)N
NN
EB T
M
Neutron
1a Hydrogenavi
However If an object starts at
( 1) and moves to ( X 3) and the to ( 3)i b fa X b c X
Then absolute values and a definition must be used
---------I----------I----------I----------I----------I----------I----------I---------- -4 -3 -2 -1 0 +1 +2 +3 +4
a gtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtgtb cltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltltb
BS Math makes up a definition that all displacements are positive Since this example provides an answer with something labeled a negative the positive definition must be applied and (-2) is changed to (+2) A definition is needed to arrive at a useable answer This is Illogical
In SM absolute values and a meaningless definition of ldquopositiverdquo are removed In SM objects and directions in space are defined by the direction and magnitude of the resultant of arrows
With SM you get total distance traveled by the object and the final direction of the displacement
22222 2 bababbaababababa If w e su b stitu te p o s itiv e n u m b e rs in to th e eq u a tio n th e an sw er w ill b e co rrec t If (a = 5 ) an d (b = 3 )
2
2
8
5 3 5 3 5 3 5 5 5 3 3 5 3 3 25 15 15 9 64
8 8 8 8 8 8 8 8 8 8 64
W h en tw o p o sitiv e n u m b ers a re a d d ed an d sq u a re d th e B S m a th d is tr ib u tiv e law p ro v id es a co rre c t an sw er 1
Illogical and Incorrect BS math for the distributive law
222 2 bababbabbaaabababa B y d e fin itio n in B S m a th n o n u m b er sq u a red c a n ev e r b e n eg a tiv e T h e re fo re n o sq u a red n u m b er c an ev e r b e a v e c to r m o v in g in th e d a sh ( -) o r n eg a tiv e d ire c tio n if it is sq u a red In B S m a th an d ad d itio n o p e ra to r m u ltip lied b y a su b trac tio n o p e ra to r is A L W A Y S eq u a l to a su b tra c tio n o p e ra to r If th is h ad n o t b e en u sed fo r th e la s t fo u r h u n d red y e a rs y o u w o u ld d ie lau g h in g a t th e u tte r ab su rd ity o f th e la s t s ta tem e n t T h is is illo g ica l an d v io la te s sy m m etry
If a = 5 an d b = 3 42222352
22
4930253)3)(5(253)5)(3()3)(5(5353535 22222 A s lo n g a s ldquo ardquo is g re a te r th an ldquob rdquo th e B S m a th d is tr ib u tiv e law p ro v id es co rrec t an sw ers 1
B S m a th b re a k s d o w n w h e n ldquo b rdquo is g rea ter th a n ldquo a rdquo
If a = 3 an d b = 5
2 2
2
2
3 5 2 4
2 ( 2) ( 2) 4
2 ( )( 2) ( )( 2) 4
A n y co m b in a tio n ex cep t th e m id d le o n e is illo g ica l an d in co rre c t B y d e fin itio n th e an sw er can n o t b e -4 in B S m a th B S m a th is in co rrec t
4253095)5)(3(235)3)(5()5)(3(3535353 22222 T h is is w h ere th e d is tr ib u tiv e la w p ro v id es in co rrec t an sw ers I t p ro d u ces an an sw er o f + 4 b e cau se b y d e fin iti o n n o n u m b er sq u a red in B S m a th c an b e a n eg a tiv e n u m b er U sin g in co rre c t B S m a th th e d is tr ib u tiv e law w ill g iv e an in co rrec t an sw ers o f + 4 T h is is in co rrec t an d illo g ica l 1
Since there can be no negative direction from a squared term in BS math BS math should be modified or abandoned
The BS distributive law should be modified or abandoned
The numbers +1+2+3+4 and ndash1-2-3-4hellip should be abandoned
BS math gives incorrect answers because their Rule-of-Signs allows multiplying an arrow going in one direction by an arrow going in another direction
Clearly this is not logical
Arrows should be used that are specific for directions
W h at d o es it m e an to m u ltip ly ( 4) b y (4 ) A s w e k n o w 1
m u ltip lica tio n is ju s t ad d itio n W h a t w o u ld ( 4) tim e (4 ) b e 2
In BS math the answer is by definition -16 unless you square -4 and then it is +16
In SM you cannot multiply arrows going in opposite directions
The middle two terms are not logical You cannot multiply opposite directions
This is a MAJOR error in BS math They multiply a dash [(-) a direction to the left] by a cross [(+) a direction to the right] And then ldquoby definitionrdquo label the answer dash (-) This is illogical and produces incorrect answers
2
2 2
This is Einsteinrsquos math error
in Special relativity
Magnitude 4 Direction 1 Magnitude 3 Direction 2 Magnitude 5 Direction 3 Relative to an observer at 0 4
Magnitude 4 Direction 1 Magnitude 3 Direction 2 Magnitude 5 Direction 3 Relative to an observer at 0 4
Magnitude and Direction (MD) An arrow has both a magnitude andA direction relative to a 2nd objectobserver
bull Magnitude and velocity (MV) relative to a 2nd objectobserver
bull Magnitude And acceleration (MA) relative to a 2nd objectobserver2
v da
t t
Introducing a new symbol the ldquoamprdquo sign The ldquoamprdquo sign means resultant ADD all the Arrows for each direction and then SUBTRACT to find the final resultant
1 1amp 1 2 2amp 2 4 If =
1 amp1 2 2 amp2 4 If =
amp 5 amp 3 2 If gt
amp 3 amp 5= 2 If gt
amp 0 3 amp 3=0 if =
only the resultant equals zero
the vectors do not - they are still there
Does not exist There is no multiplication of arrows
There are two arrows One has a magnitude of 3 One has a magnitude of 5 The maximum these two arrows can be is 8
Only the resultant of arrows exist
An SM arrow has a direction and a magnitude
Arrows and magnitudes are still there even if the resultant is zero
There are no negative SM Arrows
There are no positive SM Arrows
There are no ZERO SM Arrows A zero would negate the definition of an SM Arrow something with no magnitude and no direction cannot be the definition of an arrow that is defined as having magnitude and direction
bull A resultant is the addition and subtraction of the differences in magnitude and direction arrows
A B and C are the arrows (a) A amp B exist(b) A amp B amp C exist(c) A-B Do Not Exist in Sm there are no negatives or positives(d) A+B-C Do Not Exist in SM there are no negatives or positives
There are only resultants when there are multiple arrows
SM Graphs versus BS Graphs
Numbers Raised to Powers
When numbers are raised to powers the negative sides of all BS math graphs are ILLOGICAL and INCORRECT
Incorrect BS Math for 2x
2x
4 16 correct3 8 correct2 4 correct1 2 correct0 1 correct-1 005 incorrect-2 025 incorrect-3 0125 incorrect-4 00625 incorrect
negative side of graph is incorrectpositive side of graph is correct
16
8
42105025012500625
-4 -3 -2 -1 0 1 2 3 4
po
sit
ive
negative positive
2x BS Math
correctincorrect
In BS math the dash sign (-) used in this example represents three different operations They arebull (-) Subtractionbull (-) A direction to the leftbull (-) As an exponent to mean divide by
All positive numbers produce correct results and graphs
All negative numbers produce incorrect results and graphs
A direction of three units to the left (-3) is placed in the equation as an exponent
A direction in space (3 units toward the left) changes a whole number to a fraction by using the dash (-) as an exponent rather than a direction in space
This is illogical and incorrect BS math should have performed this operation as
Unfortunately the answers to problems can come out correct because data points of a real problem fit the curve
However the answers will not be due to logical reasoning
Because the equation produced a curve that real problem data points follow it has been falsely believed that the equations and graphs are correct
1
2
x
Incorrect BK graph for (12)x
(12)x
4 00625 correct3 0125 correct2 025 correct1 1 correct0 1 correct-1 2 incorrect-2 4 incorrect-3 8 incorrect-4 16 incorrect
negative side of graph is incorrectpositive side of graph is correct
00625012502505124
8
16
-5 -4 -3 -2 -1 0 1 2 3 4 5P
osit
ive
negative positive
BS Math for (12)x
correctincorrect
Incorrect BS math and graph for
All graphs are symmetrical with SM
Two BS Math Equations that Produce Incorrect Answers
2 0x x starting equation 2
2
x x x x
x x
Add lsquoxrsquo to both sides
2(1) 1 let lsquoxrsquo = 1
2 1 This is an incorrect answer 2(0)-(0)=0 The only answer that BK math can give is x=0
1
a b starting equation ( )( ) ( )( )a a a b multiply both sides by ( )a
( )( ) ( )( ) ( )( ) ( )( )a a b b a b b b subtract ( )( )b b from both sides
( )( ) ( )( )a b a b a b b factor ( )a b from both sides
( )a b b divide both sides by ( )a b
( )b b b replace ( )a with ( )b 2b b simplify 2(1) (1) Let ( )b = 1
2 1 Again BS math yields and incorrect answer 1
This incorrect result is a problem with the distributive law SM explains how this incorrect answer is obtained and the problem with the distributive law that allows the error
Page 148 ndash149 Barry Mazur ndash Imagining Numbers
3
2
31
Do the computation carefully on paper using the rules we agreed to Then ponder your answer which should be something of a surprise to you donrsquot stop there Think of what your answer might possibly mean or might imply
Solve this equation using the distributive law
JK definitely a surprise the answer wasincorrect and illogical
Gerhard Gade University Professor doing math at Harvard University
Here is the BS math answer that Barry Mazur calculated
2
31
2
31
2
31
1 3 1 3 1 3 3 3 2 2 3 1 3
2 2 2 2 4 4 4 4 4 4 2 2
dash
1 3 1 3 1 31
2 2 2 2 4 4
3 3
4 4
When the distributive law is used with the BS Rule-of-Signs and one of the signs is a (-) dash sign the middle two terms always cancel each other This leads to the incorrect answer of (-1) in the above equation
Barry Mazur did not go back and check his answer with the original equation If he had he would have known that the answer was incorrect
3
1 31
2
check answer by setting equal
3
331 3
12
1 31
2
Take the Cube root of both sides
1 3 2 multiply both sides by 2
3 3 subtract 1 from both sides
2 2
3 3 square both sides
3 9 This is an incorrect answer A simple equation check could have shown that the distributive law using BS math provides an incorrect answer
1
BS Illogical and Incorrect Numbers Raised to Powers
nC D n DC
642 DDD
In BS math only a positive number can have an answer if ldquonrdquo is an even number (246---)
All negative numbers have no solutions
642 DDD
With SM just add a direction arrow and any number gives the correct answer
or n nC D C D
3 3( 3 8 8 or 8 2 or 2 )n D C
( 2 16 C= 16 or 16 4 or 4 )n D
Since a negative number has no square root answer in BS math BS mathematicians invented imaginary numbers
i2=-1
24414 iiC n
These equations are easily solved without the use of imaginary numbers with SM math
If 4 then 4 2x
If 4 then 4 2x
The 360-degree circle is a 4400 years old extremely outdated system based on a year of 12 months of 30 days each that is 360 days
SM will use a more logical system that will have a circle of 6283185307o
d
( )
(deg )
d r radians
d r rees
the arc distance around
the circumference of a circle
d
radiusr
2 2
2 2
d r
dv rd t t
v dda rd t t t
1
d
r
( )av tv
The average angular velocity of a body is the rate of its angular displacement
( )0
liminstt
vt
The instantaneous angular velocity is the limit of the ratio
t
as t approaches zero
( )d
av
d
tv
The average circumference velocity of the body is the rate of its distance traveled d divided by the time t
dvv
r
1 dd vdrv
t t t r r
The equation of the relationship between the angular velocity and the velocity of a point on that body
( )o
av
v v v
t ta
The average angular acceleration is the change in the angular velocity
( )0
liminstt
v
ta
The instantaneous angular acceleration is the
limit of the ration of v
t
as t approaches
zero
daa
r
1
1
d
d s d
vva
t r tv a
r t r
The equation of the relationship between the angular acceleration and the acceleration of a point on the body
21
2v a t
If a body with an initial angular velocity ov has a constant
angular acceleration a it will turn through an angle in a time t
1
Straight Line Symbol Units Rotational Symbol Units Distance d m Angular
Distance radians or degrees
Velocity dt
v m
s Angular Velocity
V
t
radians or degrees
s
Acceleration vat
2ms
Angular Acceleration
VA
t
2
radians or degrees
s
Mass M Kg Mass distribution
dM 2Kgm
Force actionA Ma
2
Kgm
sNewton
Torque
dT M A 2
2
Kgm
s
Momentum Mv Kgm
s
Angular Momentum
dM V 2Kgm
s
Work (Force)(distance)
Ad 2
Kgmm
sNm Joule
Work T
Using AS arrow math to solve and direction in space problems using the equation 1 2
2 2 2 2 2 2( ) amp and ( ) amp 3
4
2 2 2
2 2
( ) amp
(5 ) 4 3
(5 ) 16 9
(5 ) 25
(5 ) (5 )
2 2(5 ) 4 3
(5 ) 16 9
(5 ) 25
(5 ) (5 )
In SM math Pythagorasrsquo theorem provides both magnitude and direction
In BS math Pythagorasrsquo theorem provides magnitude only
time set equal to zero
23
( ) ( 0 )10 1
x t xx
3( ) ( 0)
10 1W t f
In BS math this is an equation for a traveling wave
must use a square term in order to make the curve symmetric
NO SQUARE TERM IS NECESSARY
In SM math when t is set to ldquo0rdquo
This is a major fault in BS math
x f(x)
-4 0019
-3 0033
-2 0073
-1 0273
0 3000
1 0273
2 0073
3 0033
4 0019 -4 -3 -2 -1 0 1 2 3 400
05
10
15
20
25
30
35
BS wave (must use square term)
snap shot of wave at x = 0
Am
plit
ude
lt to gt lt to gt f(ltgt)
lt -4 0073
lt -3 0097
lt -2 0143
lt -1 0273
origin 0 3000
gt 1 0273
gt 2 0143
gt 3 0097
gt 4 0073lt0073 lt0097 lt0143 lt0273
3000
0273gt 0143gt 0097gt 0073gt
Wave moving
snap shot of wave
Am
plit
ude
lt4 lt3 lt2 lt1 0 1gt 2gt 3gt 4gt
3
2
1
In SM NO squared terms are needed
SPRING
An Action is needed to stretch or compress a spring to a starting point where it can execute cycle motion
tsA (ts to start) is the work required to move a mass attached to a spring a distance (d1) from
its initial rest position RAK 0 the spring constant is the stiffness of the spring
o A is attraction o R is repulsion o
ARRA
00 is for the back and forth motion through the zero initial rest position A
stretched spring goes from ldquoArdquo to ldquo0rdquo to ldquoRrdquo to ldquoArdquo to ldquo0rdquo to ldquoRrdquo d1 is the distance a mass is moved from its initial rest position
0
0 1ts A RA K d
Once ( )mad
is performed to move the mass to position d1 (against the restraint of the spring) a of the ( )mad
is transferred into the spring as potential attractionrepulsion (PAR) There is always
more ( )mad
performed than is stored as PAR When the spring is released from the restraining Ats the PAR starts the oscillation Conservation of ( )mad
requires that the PAR and the motion ( )mad
( )( )madM be a constant
( )amp ( ) constantmadPAR M 02
0 0
1
2A R A RPAR K d
02
( ) 0 0
1( )
2mad A R A RM m v
0 0 02 2 2
0 0 0 1
2 2 21
2 2 21
1 1 1amp
2 2 2
1 1 1amp
2 2 2
amp
A R A R A RK d m v K d
Kd mv Kd
Kd mv Kd
0
1d is frac12 the height from the base to the peak of a cycle
An experiment of moving paper under a pendulum through a small angle filled with sand that can escape through a small hole in the bottom shows that a pendulum cycling back and forth follows a curve that is mathematically analogous to a lsquosinrsquo or lsquocosrsquo
The projection of the curved area onto the straight line exactly duplicate the back and forth motion of CM This allows for the math of lsquosinrsquo or lsquocosrsquo to be used
SM will use lsquosinrdquo math It could use lsquocosrsquo or a combination of both lsquosinrsquo and lsquocosrsquo Using just the lsquosinrsquo math makes the understanding and math easier
0 sin 0 0SM 0 sin157 1SM The diagram at the left shows a spring being stretched from its (zero rest) position-2 and the PAR being released at position-1 There are 4 sections (1 to 2 2 to 3 3 to 4 and 4 to5) that alternate between maximum acceleration and zero velocity to max velocity and zero acceleration
At position-1 there is maximum acceleration and zero velocity o The lsquosinrsquo values from 0o to 157o will be used for velocity (values from 0 to 1) o The lsquosinrsquo values from 157o to 0o will be used for acceleration (values from 1 to 0)
At position-2 there is maximum velocity and zero acceleration Acceleration changes to deceleration
o The lsquosinrsquo values from 157o to 0o will be used for deceleration (values from 1 to 0) o The lsquosinrsquo values from 0o to 157o will be used for acceleration (values from 0 to 1)
At position-3 there is maximum acceleration and zero velocity o The lsquosinrsquo values from 0o to 157o will be used for velocity (values from 0 to 1) o The lsquosinrsquo values from 157o to 0o will be used for acceleration (values from 1 to 0)
At position-4 there is maximum velocity and zero acceleration Acceleration changes to deceleration
o The lsquosinrsquo values from 157o to 0o will be used for deceleration (values from 1 to 0) o The lsquosinrsquo values from 0o to 157o will be used for acceleration (values from 0 to 1)
At position-5 the cycle starts again
2
)539(
cT
MK
spring constant (K) when mass (M) and cycle-time (T) are known
)(
))(89)((
111 md
uKgM
d
MG
d
FK sts
spring constant (K) when mass (M) and amplitude ( d1) are known
2 21( )
Kv d d
m
velocity (v) at any point during the cycle
velocity
dforM
Kdv
max
012 velocity (v2) maximum (when d = 0 at position-2 and 4)
acceleration (a1) maximum when spring constant (K) mass (M) and amplitude (d1) are known
acceleration (a1) maximum when amplitude (d1) and cycle-time (Tc) are known
acceleration (a) at any point during the cycle-time (T)
1
151
286
v
dTc
cycle-time (Tc) when amplitude (d1) and velocity (v1) are known
K
MTc 28651
cycle-time (Tc) when mass (M) and spring constant (K) are known
fTc
1
cycle-time (Tc) when the frequency is known
cTf
1
frequency (f) when the cycle-time (Tc) is known
of cycles
elapsed timef
frequency (f) when the number of cycles in an elapsed time is known
d1 amplitude (distance from rest to start position)
Also frac12 the distance from the base to the peak
3131 oror dM
Ka
251
3131 539
c
oror T
da
2539
cT
da
In December of 2006 I discovered the mathematical error in Einsteinrsquos Special Relativity Einstein used BS math and that math only works in an imaginary world
It cannot describe real symmetrical space
Einsteinrsquos Special Relativity should be abandoned and a theory of real space using SM should be used to replace it
Numerous articles on the Internet show Einsteinrsquos theory to be incorrect but no one in academia is willing to publish the information
Figure 1-10 A moving particle observed from two frames of reference in uniform translation
Einsteinrsquos Math Error in Special Relativity Transformation of Velocity (From Eisenberg- Fundamentals of Modern Physics pages 27-29) Consider the particle shown in figure (1-10) moving with velocity v as
seen in a frame of reference O
dx dy dz
Vx Vy Vzdt dt dt
dx dy dz
V x V y V zdy dt dt
We would like to evaluate the velocity v of the particle as seen in the frame of reference 0 which is itself moving relative to 0 with velocity v Measured in the 0 frame the velocity vector of the particle has com ponents
The same velocity vector as measured in the 0 frame has components
2
2
2
2
2
1
1
x vtx
v
cy y
z z
vxt
ctv
c
Now from equations (1-13) we know that the relation between the primed and the unprimed coordinates and times is
Take the differential of these equations remembering that v is a constant This gives
2
2
2
2
2
1
1
dx vdtdx
v
cdy dy
dz dz
vxdt
cdtv
c
2
2
2 2 2 2 2
2
2
2
11 ( )
1( ) 1
1
1
1
dx vdt
v dxdx vdt v
dx dx vdtc dt dtvdx vdx vdx v dxdt dt dt dt
dt dtc c c c
VxV x
Vx
VxV x
vv
vv
c
x
v
c
cV
( )
coordinate of particle relative to moving primed system
coordinate of particle relative to stationary system
velocity of primed coordina
V x
Vx
te system relative to stationaryv sys
tem
Page 28 Fundamentals of Modern Physics
(1-16)
According to Einstein this equation tells you how to transform the observed velocity from one frame of reference to another frame of reference First we note that as Vc and vc approachzero the equation approaches those which would be derived from the Galilean transformation
2 2
( ) 1 01 1 ( )
Vx v Vx vV
VxV x
Vx v
v cVxx
v
c
v
cc
Another property of this equation is that it is impossible to choose V and v such that Vrsquo the magnitude of the velocity which is seen in the new frame is greater than c
1
2
09 ( 09 ) 18009876543
( 09 ) 1811 (09 )
c c cV c c
cc
c
Example As seen by O particle 1 has velocity 09c in the direction of positive x and particle 2 has velocity 09c in the negative x direction
Einstein equation however predicts To evaluate the velocity of particle 1 with respect to particle 2 we transform from the O frame to the Orsquo frame moving in the negative x direction with velocity v = -09c using equation 1-16 We obtain
JK LOGIC will observe the particles separating at 18c
A confusing INCORRECT and ILLOGICAL answer
left is a dir rightsub of a
1
2
1
09 ( 09 ) 18009876543
181( 09 )1 ( 09 )
The multiplication of dir left time a dir right is ONLY a dir left
09
dir
direction leftdirection rightsubtraction
c c cV c c
cc
c
V
dir
22
22
09 18 18009876543
181( 081)1(09)(09)
1
left rightsub
sub
c c c cc c
ccc
c
Analyzing Einsteinrsquos BS math Equation
Using SM math a logical answer can be obtained
Cannot multiply a subtraction operator times a direction in space
Can only subtract and change its direction
Cannot multiply a direction to the left times a direction to the right
Cannot multiply opposite directions in space
Can only solve for the resultant
rightleft
subtraction
1
subtraction
2
sub of a dir to the left is a dir to the right
1 resultant 0 sub 2
2sub 2
2
09 09
09 amp 091
09 09 18 1
(0)1
(09) amp (09)1
c cV
c cc
c c cV
cc
cc
8
181
cc
SM math predicts a logical answer using logic
21st century scientists state in many articles that Einsteins General Relativity and Quantum Mechanics are incompatible
They are not only incompatible they are both mathematically incorrect
They both use illogical BS math
Any equations in math that use imaginary numbers should be abandoned
Imaginary numbers are only necessary in the current BS math system
This means that Schroumldingers equations should be abandoned
Any equation in quantum mechanics that uses imaginary number should be abandoned
Prof Kanarev has developed logical physics and chemistrytheories without the use of Quantum mechanics or relativity
A partial list of his theories coverbull Mechanics ndash replaces Newtonbull Photon - bull Electrons ndash Protons- Neutronsbull Spectroscopybull Nucleusbull Atoms amp Moleculesbull Electrodynamicsbull Heatbull Curvature of Space ndash replaces Schwarzcchildrsquos equation
Kanarevrsquos Hypothesis
bull Electron align linearly with proton They do NOT spin or form clouds around the nucleus
bull Neutrons are in the center of atom Proton are outside of Neutrons
bull The wavelength of all particles is the experimental radius
bull Double slit experiment explained correctly
bull Heisenbergrsquos inequality explained correctly
bull is the radius of approach of the magnetic lines around the torus of the electron
( 1 137)
bull Why water expands when it freezes
bull Electron clusters form lightning
rbull The meaning of Plancks constant is explained
22 2 kg m
h m f mr f consts
2210254 42m r kgm bull If the mr-constant describes a ring the equations
describe experiments bull If the mr-constant describes a wave the equations do
NOT describe experiments
Kanarevrsquos 5 Equations of Mechanics
Kanarevrsquos 1st Equation Acceleration ampmv maMma R R
Kanarevrsquos 2nd Equation Uniform motion amp mv maMma R R
maM mvR R
Kanarevrsquos 3rd Equation
Kanarevrsquos 4th Equation Equal and opposite reaction
1 21 21 2
12
a m
m a m ama
Kanarevrsquos 5th Equation 1 2 3 Mi M M M Mna R R R R R
The sum of the Actions working on a moving body is never equal to zero
Kanarevrsquos Photon model is a rolling ring
His equations accurately describe known experiments with photonsbull Electromagnetic radiation should be called photon radiationbull Diffraction and interference are the same phenomenabull Correctly explains the double slit experiment without quantum equationsbull Correctly explains the meaning of Heisenbergrsquos inequality
e
2 (2)(3142)(2817 15 ) 10007297352568
R 2426 12 137mag
xp
r m
m
Charge
Magnetic action
Proton
The intensity of the magnetic action near the geometrical center of the proton ring is so great that we theorize that it is the strong nuclear force
(9383 8)(1602 19)8476 14
4 (4)(3142)(1411 26)p
pp
EB T
M
(9395 8)(1602 19)1035 15
4 (4)(3142)(0966 26)N
NN
EB T
M
Neutron
1a Hydrogenavi
In SM absolute values and a meaningless definition of ldquopositiverdquo are removed In SM objects and directions in space are defined by the direction and magnitude of the resultant of arrows
With SM you get total distance traveled by the object and the final direction of the displacement
22222 2 bababbaababababa If w e su b stitu te p o s itiv e n u m b e rs in to th e eq u a tio n th e an sw er w ill b e co rrec t If (a = 5 ) an d (b = 3 )
2
2
8
5 3 5 3 5 3 5 5 5 3 3 5 3 3 25 15 15 9 64
8 8 8 8 8 8 8 8 8 8 64
W h en tw o p o sitiv e n u m b ers a re a d d ed an d sq u a re d th e B S m a th d is tr ib u tiv e law p ro v id es a co rre c t an sw er 1
Illogical and Incorrect BS math for the distributive law
222 2 bababbabbaaabababa B y d e fin itio n in B S m a th n o n u m b er sq u a red c a n ev e r b e n eg a tiv e T h e re fo re n o sq u a red n u m b er c an ev e r b e a v e c to r m o v in g in th e d a sh ( -) o r n eg a tiv e d ire c tio n if it is sq u a red In B S m a th an d ad d itio n o p e ra to r m u ltip lied b y a su b trac tio n o p e ra to r is A L W A Y S eq u a l to a su b tra c tio n o p e ra to r If th is h ad n o t b e en u sed fo r th e la s t fo u r h u n d red y e a rs y o u w o u ld d ie lau g h in g a t th e u tte r ab su rd ity o f th e la s t s ta tem e n t T h is is illo g ica l an d v io la te s sy m m etry
If a = 5 an d b = 3 42222352
22
4930253)3)(5(253)5)(3()3)(5(5353535 22222 A s lo n g a s ldquo ardquo is g re a te r th an ldquob rdquo th e B S m a th d is tr ib u tiv e law p ro v id es co rrec t an sw ers 1
B S m a th b re a k s d o w n w h e n ldquo b rdquo is g rea ter th a n ldquo a rdquo
If a = 3 an d b = 5
2 2
2
2
3 5 2 4
2 ( 2) ( 2) 4
2 ( )( 2) ( )( 2) 4
A n y co m b in a tio n ex cep t th e m id d le o n e is illo g ica l an d in co rre c t B y d e fin itio n th e an sw er can n o t b e -4 in B S m a th B S m a th is in co rrec t
4253095)5)(3(235)3)(5()5)(3(3535353 22222 T h is is w h ere th e d is tr ib u tiv e la w p ro v id es in co rrec t an sw ers I t p ro d u ces an an sw er o f + 4 b e cau se b y d e fin iti o n n o n u m b er sq u a red in B S m a th c an b e a n eg a tiv e n u m b er U sin g in co rre c t B S m a th th e d is tr ib u tiv e law w ill g iv e an in co rrec t an sw ers o f + 4 T h is is in co rrec t an d illo g ica l 1
Since there can be no negative direction from a squared term in BS math BS math should be modified or abandoned
The BS distributive law should be modified or abandoned
The numbers +1+2+3+4 and ndash1-2-3-4hellip should be abandoned
BS math gives incorrect answers because their Rule-of-Signs allows multiplying an arrow going in one direction by an arrow going in another direction
Clearly this is not logical
Arrows should be used that are specific for directions
W h at d o es it m e an to m u ltip ly ( 4) b y (4 ) A s w e k n o w 1
m u ltip lica tio n is ju s t ad d itio n W h a t w o u ld ( 4) tim e (4 ) b e 2
In BS math the answer is by definition -16 unless you square -4 and then it is +16
In SM you cannot multiply arrows going in opposite directions
The middle two terms are not logical You cannot multiply opposite directions
This is a MAJOR error in BS math They multiply a dash [(-) a direction to the left] by a cross [(+) a direction to the right] And then ldquoby definitionrdquo label the answer dash (-) This is illogical and produces incorrect answers
2
2 2
This is Einsteinrsquos math error
in Special relativity
Magnitude 4 Direction 1 Magnitude 3 Direction 2 Magnitude 5 Direction 3 Relative to an observer at 0 4
Magnitude 4 Direction 1 Magnitude 3 Direction 2 Magnitude 5 Direction 3 Relative to an observer at 0 4
Magnitude and Direction (MD) An arrow has both a magnitude andA direction relative to a 2nd objectobserver
bull Magnitude and velocity (MV) relative to a 2nd objectobserver
bull Magnitude And acceleration (MA) relative to a 2nd objectobserver2
v da
t t
Introducing a new symbol the ldquoamprdquo sign The ldquoamprdquo sign means resultant ADD all the Arrows for each direction and then SUBTRACT to find the final resultant
1 1amp 1 2 2amp 2 4 If =
1 amp1 2 2 amp2 4 If =
amp 5 amp 3 2 If gt
amp 3 amp 5= 2 If gt
amp 0 3 amp 3=0 if =
only the resultant equals zero
the vectors do not - they are still there
Does not exist There is no multiplication of arrows
There are two arrows One has a magnitude of 3 One has a magnitude of 5 The maximum these two arrows can be is 8
Only the resultant of arrows exist
An SM arrow has a direction and a magnitude
Arrows and magnitudes are still there even if the resultant is zero
There are no negative SM Arrows
There are no positive SM Arrows
There are no ZERO SM Arrows A zero would negate the definition of an SM Arrow something with no magnitude and no direction cannot be the definition of an arrow that is defined as having magnitude and direction
bull A resultant is the addition and subtraction of the differences in magnitude and direction arrows
A B and C are the arrows (a) A amp B exist(b) A amp B amp C exist(c) A-B Do Not Exist in Sm there are no negatives or positives(d) A+B-C Do Not Exist in SM there are no negatives or positives
There are only resultants when there are multiple arrows
SM Graphs versus BS Graphs
Numbers Raised to Powers
When numbers are raised to powers the negative sides of all BS math graphs are ILLOGICAL and INCORRECT
Incorrect BS Math for 2x
2x
4 16 correct3 8 correct2 4 correct1 2 correct0 1 correct-1 005 incorrect-2 025 incorrect-3 0125 incorrect-4 00625 incorrect
negative side of graph is incorrectpositive side of graph is correct
16
8
42105025012500625
-4 -3 -2 -1 0 1 2 3 4
po
sit
ive
negative positive
2x BS Math
correctincorrect
In BS math the dash sign (-) used in this example represents three different operations They arebull (-) Subtractionbull (-) A direction to the leftbull (-) As an exponent to mean divide by
All positive numbers produce correct results and graphs
All negative numbers produce incorrect results and graphs
A direction of three units to the left (-3) is placed in the equation as an exponent
A direction in space (3 units toward the left) changes a whole number to a fraction by using the dash (-) as an exponent rather than a direction in space
This is illogical and incorrect BS math should have performed this operation as
Unfortunately the answers to problems can come out correct because data points of a real problem fit the curve
However the answers will not be due to logical reasoning
Because the equation produced a curve that real problem data points follow it has been falsely believed that the equations and graphs are correct
1
2
x
Incorrect BK graph for (12)x
(12)x
4 00625 correct3 0125 correct2 025 correct1 1 correct0 1 correct-1 2 incorrect-2 4 incorrect-3 8 incorrect-4 16 incorrect
negative side of graph is incorrectpositive side of graph is correct
00625012502505124
8
16
-5 -4 -3 -2 -1 0 1 2 3 4 5P
osit
ive
negative positive
BS Math for (12)x
correctincorrect
Incorrect BS math and graph for
All graphs are symmetrical with SM
Two BS Math Equations that Produce Incorrect Answers
2 0x x starting equation 2
2
x x x x
x x
Add lsquoxrsquo to both sides
2(1) 1 let lsquoxrsquo = 1
2 1 This is an incorrect answer 2(0)-(0)=0 The only answer that BK math can give is x=0
1
a b starting equation ( )( ) ( )( )a a a b multiply both sides by ( )a
( )( ) ( )( ) ( )( ) ( )( )a a b b a b b b subtract ( )( )b b from both sides
( )( ) ( )( )a b a b a b b factor ( )a b from both sides
( )a b b divide both sides by ( )a b
( )b b b replace ( )a with ( )b 2b b simplify 2(1) (1) Let ( )b = 1
2 1 Again BS math yields and incorrect answer 1
This incorrect result is a problem with the distributive law SM explains how this incorrect answer is obtained and the problem with the distributive law that allows the error
Page 148 ndash149 Barry Mazur ndash Imagining Numbers
3
2
31
Do the computation carefully on paper using the rules we agreed to Then ponder your answer which should be something of a surprise to you donrsquot stop there Think of what your answer might possibly mean or might imply
Solve this equation using the distributive law
JK definitely a surprise the answer wasincorrect and illogical
Gerhard Gade University Professor doing math at Harvard University
Here is the BS math answer that Barry Mazur calculated
2
31
2
31
2
31
1 3 1 3 1 3 3 3 2 2 3 1 3
2 2 2 2 4 4 4 4 4 4 2 2
dash
1 3 1 3 1 31
2 2 2 2 4 4
3 3
4 4
When the distributive law is used with the BS Rule-of-Signs and one of the signs is a (-) dash sign the middle two terms always cancel each other This leads to the incorrect answer of (-1) in the above equation
Barry Mazur did not go back and check his answer with the original equation If he had he would have known that the answer was incorrect
3
1 31
2
check answer by setting equal
3
331 3
12
1 31
2
Take the Cube root of both sides
1 3 2 multiply both sides by 2
3 3 subtract 1 from both sides
2 2
3 3 square both sides
3 9 This is an incorrect answer A simple equation check could have shown that the distributive law using BS math provides an incorrect answer
1
BS Illogical and Incorrect Numbers Raised to Powers
nC D n DC
642 DDD
In BS math only a positive number can have an answer if ldquonrdquo is an even number (246---)
All negative numbers have no solutions
642 DDD
With SM just add a direction arrow and any number gives the correct answer
or n nC D C D
3 3( 3 8 8 or 8 2 or 2 )n D C
( 2 16 C= 16 or 16 4 or 4 )n D
Since a negative number has no square root answer in BS math BS mathematicians invented imaginary numbers
i2=-1
24414 iiC n
These equations are easily solved without the use of imaginary numbers with SM math
If 4 then 4 2x
If 4 then 4 2x
The 360-degree circle is a 4400 years old extremely outdated system based on a year of 12 months of 30 days each that is 360 days
SM will use a more logical system that will have a circle of 6283185307o
d
( )
(deg )
d r radians
d r rees
the arc distance around
the circumference of a circle
d
radiusr
2 2
2 2
d r
dv rd t t
v dda rd t t t
1
d
r
( )av tv
The average angular velocity of a body is the rate of its angular displacement
( )0
liminstt
vt
The instantaneous angular velocity is the limit of the ratio
t
as t approaches zero
( )d
av
d
tv
The average circumference velocity of the body is the rate of its distance traveled d divided by the time t
dvv
r
1 dd vdrv
t t t r r
The equation of the relationship between the angular velocity and the velocity of a point on that body
( )o
av
v v v
t ta
The average angular acceleration is the change in the angular velocity
( )0
liminstt
v
ta
The instantaneous angular acceleration is the
limit of the ration of v
t
as t approaches
zero
daa
r
1
1
d
d s d
vva
t r tv a
r t r
The equation of the relationship between the angular acceleration and the acceleration of a point on the body
21
2v a t
If a body with an initial angular velocity ov has a constant
angular acceleration a it will turn through an angle in a time t
1
Straight Line Symbol Units Rotational Symbol Units Distance d m Angular
Distance radians or degrees
Velocity dt
v m
s Angular Velocity
V
t
radians or degrees
s
Acceleration vat
2ms
Angular Acceleration
VA
t
2
radians or degrees
s
Mass M Kg Mass distribution
dM 2Kgm
Force actionA Ma
2
Kgm
sNewton
Torque
dT M A 2
2
Kgm
s
Momentum Mv Kgm
s
Angular Momentum
dM V 2Kgm
s
Work (Force)(distance)
Ad 2
Kgmm
sNm Joule
Work T
Using AS arrow math to solve and direction in space problems using the equation 1 2
2 2 2 2 2 2( ) amp and ( ) amp 3
4
2 2 2
2 2
( ) amp
(5 ) 4 3
(5 ) 16 9
(5 ) 25
(5 ) (5 )
2 2(5 ) 4 3
(5 ) 16 9
(5 ) 25
(5 ) (5 )
In SM math Pythagorasrsquo theorem provides both magnitude and direction
In BS math Pythagorasrsquo theorem provides magnitude only
time set equal to zero
23
( ) ( 0 )10 1
x t xx
3( ) ( 0)
10 1W t f
In BS math this is an equation for a traveling wave
must use a square term in order to make the curve symmetric
NO SQUARE TERM IS NECESSARY
In SM math when t is set to ldquo0rdquo
This is a major fault in BS math
x f(x)
-4 0019
-3 0033
-2 0073
-1 0273
0 3000
1 0273
2 0073
3 0033
4 0019 -4 -3 -2 -1 0 1 2 3 400
05
10
15
20
25
30
35
BS wave (must use square term)
snap shot of wave at x = 0
Am
plit
ude
lt to gt lt to gt f(ltgt)
lt -4 0073
lt -3 0097
lt -2 0143
lt -1 0273
origin 0 3000
gt 1 0273
gt 2 0143
gt 3 0097
gt 4 0073lt0073 lt0097 lt0143 lt0273
3000
0273gt 0143gt 0097gt 0073gt
Wave moving
snap shot of wave
Am
plit
ude
lt4 lt3 lt2 lt1 0 1gt 2gt 3gt 4gt
3
2
1
In SM NO squared terms are needed
SPRING
An Action is needed to stretch or compress a spring to a starting point where it can execute cycle motion
tsA (ts to start) is the work required to move a mass attached to a spring a distance (d1) from
its initial rest position RAK 0 the spring constant is the stiffness of the spring
o A is attraction o R is repulsion o
ARRA
00 is for the back and forth motion through the zero initial rest position A
stretched spring goes from ldquoArdquo to ldquo0rdquo to ldquoRrdquo to ldquoArdquo to ldquo0rdquo to ldquoRrdquo d1 is the distance a mass is moved from its initial rest position
0
0 1ts A RA K d
Once ( )mad
is performed to move the mass to position d1 (against the restraint of the spring) a of the ( )mad
is transferred into the spring as potential attractionrepulsion (PAR) There is always
more ( )mad
performed than is stored as PAR When the spring is released from the restraining Ats the PAR starts the oscillation Conservation of ( )mad
requires that the PAR and the motion ( )mad
( )( )madM be a constant
( )amp ( ) constantmadPAR M 02
0 0
1
2A R A RPAR K d
02
( ) 0 0
1( )
2mad A R A RM m v
0 0 02 2 2
0 0 0 1
2 2 21
2 2 21
1 1 1amp
2 2 2
1 1 1amp
2 2 2
amp
A R A R A RK d m v K d
Kd mv Kd
Kd mv Kd
0
1d is frac12 the height from the base to the peak of a cycle
An experiment of moving paper under a pendulum through a small angle filled with sand that can escape through a small hole in the bottom shows that a pendulum cycling back and forth follows a curve that is mathematically analogous to a lsquosinrsquo or lsquocosrsquo
The projection of the curved area onto the straight line exactly duplicate the back and forth motion of CM This allows for the math of lsquosinrsquo or lsquocosrsquo to be used
SM will use lsquosinrdquo math It could use lsquocosrsquo or a combination of both lsquosinrsquo and lsquocosrsquo Using just the lsquosinrsquo math makes the understanding and math easier
0 sin 0 0SM 0 sin157 1SM The diagram at the left shows a spring being stretched from its (zero rest) position-2 and the PAR being released at position-1 There are 4 sections (1 to 2 2 to 3 3 to 4 and 4 to5) that alternate between maximum acceleration and zero velocity to max velocity and zero acceleration
At position-1 there is maximum acceleration and zero velocity o The lsquosinrsquo values from 0o to 157o will be used for velocity (values from 0 to 1) o The lsquosinrsquo values from 157o to 0o will be used for acceleration (values from 1 to 0)
At position-2 there is maximum velocity and zero acceleration Acceleration changes to deceleration
o The lsquosinrsquo values from 157o to 0o will be used for deceleration (values from 1 to 0) o The lsquosinrsquo values from 0o to 157o will be used for acceleration (values from 0 to 1)
At position-3 there is maximum acceleration and zero velocity o The lsquosinrsquo values from 0o to 157o will be used for velocity (values from 0 to 1) o The lsquosinrsquo values from 157o to 0o will be used for acceleration (values from 1 to 0)
At position-4 there is maximum velocity and zero acceleration Acceleration changes to deceleration
o The lsquosinrsquo values from 157o to 0o will be used for deceleration (values from 1 to 0) o The lsquosinrsquo values from 0o to 157o will be used for acceleration (values from 0 to 1)
At position-5 the cycle starts again
2
)539(
cT
MK
spring constant (K) when mass (M) and cycle-time (T) are known
)(
))(89)((
111 md
uKgM
d
MG
d
FK sts
spring constant (K) when mass (M) and amplitude ( d1) are known
2 21( )
Kv d d
m
velocity (v) at any point during the cycle
velocity
dforM
Kdv
max
012 velocity (v2) maximum (when d = 0 at position-2 and 4)
acceleration (a1) maximum when spring constant (K) mass (M) and amplitude (d1) are known
acceleration (a1) maximum when amplitude (d1) and cycle-time (Tc) are known
acceleration (a) at any point during the cycle-time (T)
1
151
286
v
dTc
cycle-time (Tc) when amplitude (d1) and velocity (v1) are known
K
MTc 28651
cycle-time (Tc) when mass (M) and spring constant (K) are known
fTc
1
cycle-time (Tc) when the frequency is known
cTf
1
frequency (f) when the cycle-time (Tc) is known
of cycles
elapsed timef
frequency (f) when the number of cycles in an elapsed time is known
d1 amplitude (distance from rest to start position)
Also frac12 the distance from the base to the peak
3131 oror dM
Ka
251
3131 539
c
oror T
da
2539
cT
da
In December of 2006 I discovered the mathematical error in Einsteinrsquos Special Relativity Einstein used BS math and that math only works in an imaginary world
It cannot describe real symmetrical space
Einsteinrsquos Special Relativity should be abandoned and a theory of real space using SM should be used to replace it
Numerous articles on the Internet show Einsteinrsquos theory to be incorrect but no one in academia is willing to publish the information
Figure 1-10 A moving particle observed from two frames of reference in uniform translation
Einsteinrsquos Math Error in Special Relativity Transformation of Velocity (From Eisenberg- Fundamentals of Modern Physics pages 27-29) Consider the particle shown in figure (1-10) moving with velocity v as
seen in a frame of reference O
dx dy dz
Vx Vy Vzdt dt dt
dx dy dz
V x V y V zdy dt dt
We would like to evaluate the velocity v of the particle as seen in the frame of reference 0 which is itself moving relative to 0 with velocity v Measured in the 0 frame the velocity vector of the particle has com ponents
The same velocity vector as measured in the 0 frame has components
2
2
2
2
2
1
1
x vtx
v
cy y
z z
vxt
ctv
c
Now from equations (1-13) we know that the relation between the primed and the unprimed coordinates and times is
Take the differential of these equations remembering that v is a constant This gives
2
2
2
2
2
1
1
dx vdtdx
v
cdy dy
dz dz
vxdt
cdtv
c
2
2
2 2 2 2 2
2
2
2
11 ( )
1( ) 1
1
1
1
dx vdt
v dxdx vdt v
dx dx vdtc dt dtvdx vdx vdx v dxdt dt dt dt
dt dtc c c c
VxV x
Vx
VxV x
vv
vv
c
x
v
c
cV
( )
coordinate of particle relative to moving primed system
coordinate of particle relative to stationary system
velocity of primed coordina
V x
Vx
te system relative to stationaryv sys
tem
Page 28 Fundamentals of Modern Physics
(1-16)
According to Einstein this equation tells you how to transform the observed velocity from one frame of reference to another frame of reference First we note that as Vc and vc approachzero the equation approaches those which would be derived from the Galilean transformation
2 2
( ) 1 01 1 ( )
Vx v Vx vV
VxV x
Vx v
v cVxx
v
c
v
cc
Another property of this equation is that it is impossible to choose V and v such that Vrsquo the magnitude of the velocity which is seen in the new frame is greater than c
1
2
09 ( 09 ) 18009876543
( 09 ) 1811 (09 )
c c cV c c
cc
c
Example As seen by O particle 1 has velocity 09c in the direction of positive x and particle 2 has velocity 09c in the negative x direction
Einstein equation however predicts To evaluate the velocity of particle 1 with respect to particle 2 we transform from the O frame to the Orsquo frame moving in the negative x direction with velocity v = -09c using equation 1-16 We obtain
JK LOGIC will observe the particles separating at 18c
A confusing INCORRECT and ILLOGICAL answer
left is a dir rightsub of a
1
2
1
09 ( 09 ) 18009876543
181( 09 )1 ( 09 )
The multiplication of dir left time a dir right is ONLY a dir left
09
dir
direction leftdirection rightsubtraction
c c cV c c
cc
c
V
dir
22
22
09 18 18009876543
181( 081)1(09)(09)
1
left rightsub
sub
c c c cc c
ccc
c
Analyzing Einsteinrsquos BS math Equation
Using SM math a logical answer can be obtained
Cannot multiply a subtraction operator times a direction in space
Can only subtract and change its direction
Cannot multiply a direction to the left times a direction to the right
Cannot multiply opposite directions in space
Can only solve for the resultant
rightleft
subtraction
1
subtraction
2
sub of a dir to the left is a dir to the right
1 resultant 0 sub 2
2sub 2
2
09 09
09 amp 091
09 09 18 1
(0)1
(09) amp (09)1
c cV
c cc
c c cV
cc
cc
8
181
cc
SM math predicts a logical answer using logic
21st century scientists state in many articles that Einsteins General Relativity and Quantum Mechanics are incompatible
They are not only incompatible they are both mathematically incorrect
They both use illogical BS math
Any equations in math that use imaginary numbers should be abandoned
Imaginary numbers are only necessary in the current BS math system
This means that Schroumldingers equations should be abandoned
Any equation in quantum mechanics that uses imaginary number should be abandoned
Prof Kanarev has developed logical physics and chemistrytheories without the use of Quantum mechanics or relativity
A partial list of his theories coverbull Mechanics ndash replaces Newtonbull Photon - bull Electrons ndash Protons- Neutronsbull Spectroscopybull Nucleusbull Atoms amp Moleculesbull Electrodynamicsbull Heatbull Curvature of Space ndash replaces Schwarzcchildrsquos equation
Kanarevrsquos Hypothesis
bull Electron align linearly with proton They do NOT spin or form clouds around the nucleus
bull Neutrons are in the center of atom Proton are outside of Neutrons
bull The wavelength of all particles is the experimental radius
bull Double slit experiment explained correctly
bull Heisenbergrsquos inequality explained correctly
bull is the radius of approach of the magnetic lines around the torus of the electron
( 1 137)
bull Why water expands when it freezes
bull Electron clusters form lightning
rbull The meaning of Plancks constant is explained
22 2 kg m
h m f mr f consts
2210254 42m r kgm bull If the mr-constant describes a ring the equations
describe experiments bull If the mr-constant describes a wave the equations do
NOT describe experiments
Kanarevrsquos 5 Equations of Mechanics
Kanarevrsquos 1st Equation Acceleration ampmv maMma R R
Kanarevrsquos 2nd Equation Uniform motion amp mv maMma R R
maM mvR R
Kanarevrsquos 3rd Equation
Kanarevrsquos 4th Equation Equal and opposite reaction
1 21 21 2
12
a m
m a m ama
Kanarevrsquos 5th Equation 1 2 3 Mi M M M Mna R R R R R
The sum of the Actions working on a moving body is never equal to zero
Kanarevrsquos Photon model is a rolling ring
His equations accurately describe known experiments with photonsbull Electromagnetic radiation should be called photon radiationbull Diffraction and interference are the same phenomenabull Correctly explains the double slit experiment without quantum equationsbull Correctly explains the meaning of Heisenbergrsquos inequality
e
2 (2)(3142)(2817 15 ) 10007297352568
R 2426 12 137mag
xp
r m
m
Charge
Magnetic action
Proton
The intensity of the magnetic action near the geometrical center of the proton ring is so great that we theorize that it is the strong nuclear force
(9383 8)(1602 19)8476 14
4 (4)(3142)(1411 26)p
pp
EB T
M
(9395 8)(1602 19)1035 15
4 (4)(3142)(0966 26)N
NN
EB T
M
Neutron
1a Hydrogenavi
22222 2 bababbaababababa If w e su b stitu te p o s itiv e n u m b e rs in to th e eq u a tio n th e an sw er w ill b e co rrec t If (a = 5 ) an d (b = 3 )
2
2
8
5 3 5 3 5 3 5 5 5 3 3 5 3 3 25 15 15 9 64
8 8 8 8 8 8 8 8 8 8 64
W h en tw o p o sitiv e n u m b ers a re a d d ed an d sq u a re d th e B S m a th d is tr ib u tiv e law p ro v id es a co rre c t an sw er 1
Illogical and Incorrect BS math for the distributive law
222 2 bababbabbaaabababa B y d e fin itio n in B S m a th n o n u m b er sq u a red c a n ev e r b e n eg a tiv e T h e re fo re n o sq u a red n u m b er c an ev e r b e a v e c to r m o v in g in th e d a sh ( -) o r n eg a tiv e d ire c tio n if it is sq u a red In B S m a th an d ad d itio n o p e ra to r m u ltip lied b y a su b trac tio n o p e ra to r is A L W A Y S eq u a l to a su b tra c tio n o p e ra to r If th is h ad n o t b e en u sed fo r th e la s t fo u r h u n d red y e a rs y o u w o u ld d ie lau g h in g a t th e u tte r ab su rd ity o f th e la s t s ta tem e n t T h is is illo g ica l an d v io la te s sy m m etry
If a = 5 an d b = 3 42222352
22
4930253)3)(5(253)5)(3()3)(5(5353535 22222 A s lo n g a s ldquo ardquo is g re a te r th an ldquob rdquo th e B S m a th d is tr ib u tiv e law p ro v id es co rrec t an sw ers 1
B S m a th b re a k s d o w n w h e n ldquo b rdquo is g rea ter th a n ldquo a rdquo
If a = 3 an d b = 5
2 2
2
2
3 5 2 4
2 ( 2) ( 2) 4
2 ( )( 2) ( )( 2) 4
A n y co m b in a tio n ex cep t th e m id d le o n e is illo g ica l an d in co rre c t B y d e fin itio n th e an sw er can n o t b e -4 in B S m a th B S m a th is in co rrec t
4253095)5)(3(235)3)(5()5)(3(3535353 22222 T h is is w h ere th e d is tr ib u tiv e la w p ro v id es in co rrec t an sw ers I t p ro d u ces an an sw er o f + 4 b e cau se b y d e fin iti o n n o n u m b er sq u a red in B S m a th c an b e a n eg a tiv e n u m b er U sin g in co rre c t B S m a th th e d is tr ib u tiv e law w ill g iv e an in co rrec t an sw ers o f + 4 T h is is in co rrec t an d illo g ica l 1
Since there can be no negative direction from a squared term in BS math BS math should be modified or abandoned
The BS distributive law should be modified or abandoned
The numbers +1+2+3+4 and ndash1-2-3-4hellip should be abandoned
BS math gives incorrect answers because their Rule-of-Signs allows multiplying an arrow going in one direction by an arrow going in another direction
Clearly this is not logical
Arrows should be used that are specific for directions
W h at d o es it m e an to m u ltip ly ( 4) b y (4 ) A s w e k n o w 1
m u ltip lica tio n is ju s t ad d itio n W h a t w o u ld ( 4) tim e (4 ) b e 2
In BS math the answer is by definition -16 unless you square -4 and then it is +16
In SM you cannot multiply arrows going in opposite directions
The middle two terms are not logical You cannot multiply opposite directions
This is a MAJOR error in BS math They multiply a dash [(-) a direction to the left] by a cross [(+) a direction to the right] And then ldquoby definitionrdquo label the answer dash (-) This is illogical and produces incorrect answers
2
2 2
This is Einsteinrsquos math error
in Special relativity
Magnitude 4 Direction 1 Magnitude 3 Direction 2 Magnitude 5 Direction 3 Relative to an observer at 0 4
Magnitude 4 Direction 1 Magnitude 3 Direction 2 Magnitude 5 Direction 3 Relative to an observer at 0 4
Magnitude and Direction (MD) An arrow has both a magnitude andA direction relative to a 2nd objectobserver
bull Magnitude and velocity (MV) relative to a 2nd objectobserver
bull Magnitude And acceleration (MA) relative to a 2nd objectobserver2
v da
t t
Introducing a new symbol the ldquoamprdquo sign The ldquoamprdquo sign means resultant ADD all the Arrows for each direction and then SUBTRACT to find the final resultant
1 1amp 1 2 2amp 2 4 If =
1 amp1 2 2 amp2 4 If =
amp 5 amp 3 2 If gt
amp 3 amp 5= 2 If gt
amp 0 3 amp 3=0 if =
only the resultant equals zero
the vectors do not - they are still there
Does not exist There is no multiplication of arrows
There are two arrows One has a magnitude of 3 One has a magnitude of 5 The maximum these two arrows can be is 8
Only the resultant of arrows exist
An SM arrow has a direction and a magnitude
Arrows and magnitudes are still there even if the resultant is zero
There are no negative SM Arrows
There are no positive SM Arrows
There are no ZERO SM Arrows A zero would negate the definition of an SM Arrow something with no magnitude and no direction cannot be the definition of an arrow that is defined as having magnitude and direction
bull A resultant is the addition and subtraction of the differences in magnitude and direction arrows
A B and C are the arrows (a) A amp B exist(b) A amp B amp C exist(c) A-B Do Not Exist in Sm there are no negatives or positives(d) A+B-C Do Not Exist in SM there are no negatives or positives
There are only resultants when there are multiple arrows
SM Graphs versus BS Graphs
Numbers Raised to Powers
When numbers are raised to powers the negative sides of all BS math graphs are ILLOGICAL and INCORRECT
Incorrect BS Math for 2x
2x
4 16 correct3 8 correct2 4 correct1 2 correct0 1 correct-1 005 incorrect-2 025 incorrect-3 0125 incorrect-4 00625 incorrect
negative side of graph is incorrectpositive side of graph is correct
16
8
42105025012500625
-4 -3 -2 -1 0 1 2 3 4
po
sit
ive
negative positive
2x BS Math
correctincorrect
In BS math the dash sign (-) used in this example represents three different operations They arebull (-) Subtractionbull (-) A direction to the leftbull (-) As an exponent to mean divide by
All positive numbers produce correct results and graphs
All negative numbers produce incorrect results and graphs
A direction of three units to the left (-3) is placed in the equation as an exponent
A direction in space (3 units toward the left) changes a whole number to a fraction by using the dash (-) as an exponent rather than a direction in space
This is illogical and incorrect BS math should have performed this operation as
Unfortunately the answers to problems can come out correct because data points of a real problem fit the curve
However the answers will not be due to logical reasoning
Because the equation produced a curve that real problem data points follow it has been falsely believed that the equations and graphs are correct
1
2
x
Incorrect BK graph for (12)x
(12)x
4 00625 correct3 0125 correct2 025 correct1 1 correct0 1 correct-1 2 incorrect-2 4 incorrect-3 8 incorrect-4 16 incorrect
negative side of graph is incorrectpositive side of graph is correct
00625012502505124
8
16
-5 -4 -3 -2 -1 0 1 2 3 4 5P
osit
ive
negative positive
BS Math for (12)x
correctincorrect
Incorrect BS math and graph for
All graphs are symmetrical with SM
Two BS Math Equations that Produce Incorrect Answers
2 0x x starting equation 2
2
x x x x
x x
Add lsquoxrsquo to both sides
2(1) 1 let lsquoxrsquo = 1
2 1 This is an incorrect answer 2(0)-(0)=0 The only answer that BK math can give is x=0
1
a b starting equation ( )( ) ( )( )a a a b multiply both sides by ( )a
( )( ) ( )( ) ( )( ) ( )( )a a b b a b b b subtract ( )( )b b from both sides
( )( ) ( )( )a b a b a b b factor ( )a b from both sides
( )a b b divide both sides by ( )a b
( )b b b replace ( )a with ( )b 2b b simplify 2(1) (1) Let ( )b = 1
2 1 Again BS math yields and incorrect answer 1
This incorrect result is a problem with the distributive law SM explains how this incorrect answer is obtained and the problem with the distributive law that allows the error
Page 148 ndash149 Barry Mazur ndash Imagining Numbers
3
2
31
Do the computation carefully on paper using the rules we agreed to Then ponder your answer which should be something of a surprise to you donrsquot stop there Think of what your answer might possibly mean or might imply
Solve this equation using the distributive law
JK definitely a surprise the answer wasincorrect and illogical
Gerhard Gade University Professor doing math at Harvard University
Here is the BS math answer that Barry Mazur calculated
2
31
2
31
2
31
1 3 1 3 1 3 3 3 2 2 3 1 3
2 2 2 2 4 4 4 4 4 4 2 2
dash
1 3 1 3 1 31
2 2 2 2 4 4
3 3
4 4
When the distributive law is used with the BS Rule-of-Signs and one of the signs is a (-) dash sign the middle two terms always cancel each other This leads to the incorrect answer of (-1) in the above equation
Barry Mazur did not go back and check his answer with the original equation If he had he would have known that the answer was incorrect
3
1 31
2
check answer by setting equal
3
331 3
12
1 31
2
Take the Cube root of both sides
1 3 2 multiply both sides by 2
3 3 subtract 1 from both sides
2 2
3 3 square both sides
3 9 This is an incorrect answer A simple equation check could have shown that the distributive law using BS math provides an incorrect answer
1
BS Illogical and Incorrect Numbers Raised to Powers
nC D n DC
642 DDD
In BS math only a positive number can have an answer if ldquonrdquo is an even number (246---)
All negative numbers have no solutions
642 DDD
With SM just add a direction arrow and any number gives the correct answer
or n nC D C D
3 3( 3 8 8 or 8 2 or 2 )n D C
( 2 16 C= 16 or 16 4 or 4 )n D
Since a negative number has no square root answer in BS math BS mathematicians invented imaginary numbers
i2=-1
24414 iiC n
These equations are easily solved without the use of imaginary numbers with SM math
If 4 then 4 2x
If 4 then 4 2x
The 360-degree circle is a 4400 years old extremely outdated system based on a year of 12 months of 30 days each that is 360 days
SM will use a more logical system that will have a circle of 6283185307o
d
( )
(deg )
d r radians
d r rees
the arc distance around
the circumference of a circle
d
radiusr
2 2
2 2
d r
dv rd t t
v dda rd t t t
1
d
r
( )av tv
The average angular velocity of a body is the rate of its angular displacement
( )0
liminstt
vt
The instantaneous angular velocity is the limit of the ratio
t
as t approaches zero
( )d
av
d
tv
The average circumference velocity of the body is the rate of its distance traveled d divided by the time t
dvv
r
1 dd vdrv
t t t r r
The equation of the relationship between the angular velocity and the velocity of a point on that body
( )o
av
v v v
t ta
The average angular acceleration is the change in the angular velocity
( )0
liminstt
v
ta
The instantaneous angular acceleration is the
limit of the ration of v
t
as t approaches
zero
daa
r
1
1
d
d s d
vva
t r tv a
r t r
The equation of the relationship between the angular acceleration and the acceleration of a point on the body
21
2v a t
If a body with an initial angular velocity ov has a constant
angular acceleration a it will turn through an angle in a time t
1
Straight Line Symbol Units Rotational Symbol Units Distance d m Angular
Distance radians or degrees
Velocity dt
v m
s Angular Velocity
V
t
radians or degrees
s
Acceleration vat
2ms
Angular Acceleration
VA
t
2
radians or degrees
s
Mass M Kg Mass distribution
dM 2Kgm
Force actionA Ma
2
Kgm
sNewton
Torque
dT M A 2
2
Kgm
s
Momentum Mv Kgm
s
Angular Momentum
dM V 2Kgm
s
Work (Force)(distance)
Ad 2
Kgmm
sNm Joule
Work T
Using AS arrow math to solve and direction in space problems using the equation 1 2
2 2 2 2 2 2( ) amp and ( ) amp 3
4
2 2 2
2 2
( ) amp
(5 ) 4 3
(5 ) 16 9
(5 ) 25
(5 ) (5 )
2 2(5 ) 4 3
(5 ) 16 9
(5 ) 25
(5 ) (5 )
In SM math Pythagorasrsquo theorem provides both magnitude and direction
In BS math Pythagorasrsquo theorem provides magnitude only
time set equal to zero
23
( ) ( 0 )10 1
x t xx
3( ) ( 0)
10 1W t f
In BS math this is an equation for a traveling wave
must use a square term in order to make the curve symmetric
NO SQUARE TERM IS NECESSARY
In SM math when t is set to ldquo0rdquo
This is a major fault in BS math
x f(x)
-4 0019
-3 0033
-2 0073
-1 0273
0 3000
1 0273
2 0073
3 0033
4 0019 -4 -3 -2 -1 0 1 2 3 400
05
10
15
20
25
30
35
BS wave (must use square term)
snap shot of wave at x = 0
Am
plit
ude
lt to gt lt to gt f(ltgt)
lt -4 0073
lt -3 0097
lt -2 0143
lt -1 0273
origin 0 3000
gt 1 0273
gt 2 0143
gt 3 0097
gt 4 0073lt0073 lt0097 lt0143 lt0273
3000
0273gt 0143gt 0097gt 0073gt
Wave moving
snap shot of wave
Am
plit
ude
lt4 lt3 lt2 lt1 0 1gt 2gt 3gt 4gt
3
2
1
In SM NO squared terms are needed
SPRING
An Action is needed to stretch or compress a spring to a starting point where it can execute cycle motion
tsA (ts to start) is the work required to move a mass attached to a spring a distance (d1) from
its initial rest position RAK 0 the spring constant is the stiffness of the spring
o A is attraction o R is repulsion o
ARRA
00 is for the back and forth motion through the zero initial rest position A
stretched spring goes from ldquoArdquo to ldquo0rdquo to ldquoRrdquo to ldquoArdquo to ldquo0rdquo to ldquoRrdquo d1 is the distance a mass is moved from its initial rest position
0
0 1ts A RA K d
Once ( )mad
is performed to move the mass to position d1 (against the restraint of the spring) a of the ( )mad
is transferred into the spring as potential attractionrepulsion (PAR) There is always
more ( )mad
performed than is stored as PAR When the spring is released from the restraining Ats the PAR starts the oscillation Conservation of ( )mad
requires that the PAR and the motion ( )mad
( )( )madM be a constant
( )amp ( ) constantmadPAR M 02
0 0
1
2A R A RPAR K d
02
( ) 0 0
1( )
2mad A R A RM m v
0 0 02 2 2
0 0 0 1
2 2 21
2 2 21
1 1 1amp
2 2 2
1 1 1amp
2 2 2
amp
A R A R A RK d m v K d
Kd mv Kd
Kd mv Kd
0
1d is frac12 the height from the base to the peak of a cycle
An experiment of moving paper under a pendulum through a small angle filled with sand that can escape through a small hole in the bottom shows that a pendulum cycling back and forth follows a curve that is mathematically analogous to a lsquosinrsquo or lsquocosrsquo
The projection of the curved area onto the straight line exactly duplicate the back and forth motion of CM This allows for the math of lsquosinrsquo or lsquocosrsquo to be used
SM will use lsquosinrdquo math It could use lsquocosrsquo or a combination of both lsquosinrsquo and lsquocosrsquo Using just the lsquosinrsquo math makes the understanding and math easier
0 sin 0 0SM 0 sin157 1SM The diagram at the left shows a spring being stretched from its (zero rest) position-2 and the PAR being released at position-1 There are 4 sections (1 to 2 2 to 3 3 to 4 and 4 to5) that alternate between maximum acceleration and zero velocity to max velocity and zero acceleration
At position-1 there is maximum acceleration and zero velocity o The lsquosinrsquo values from 0o to 157o will be used for velocity (values from 0 to 1) o The lsquosinrsquo values from 157o to 0o will be used for acceleration (values from 1 to 0)
At position-2 there is maximum velocity and zero acceleration Acceleration changes to deceleration
o The lsquosinrsquo values from 157o to 0o will be used for deceleration (values from 1 to 0) o The lsquosinrsquo values from 0o to 157o will be used for acceleration (values from 0 to 1)
At position-3 there is maximum acceleration and zero velocity o The lsquosinrsquo values from 0o to 157o will be used for velocity (values from 0 to 1) o The lsquosinrsquo values from 157o to 0o will be used for acceleration (values from 1 to 0)
At position-4 there is maximum velocity and zero acceleration Acceleration changes to deceleration
o The lsquosinrsquo values from 157o to 0o will be used for deceleration (values from 1 to 0) o The lsquosinrsquo values from 0o to 157o will be used for acceleration (values from 0 to 1)
At position-5 the cycle starts again
2
)539(
cT
MK
spring constant (K) when mass (M) and cycle-time (T) are known
)(
))(89)((
111 md
uKgM
d
MG
d
FK sts
spring constant (K) when mass (M) and amplitude ( d1) are known
2 21( )
Kv d d
m
velocity (v) at any point during the cycle
velocity
dforM
Kdv
max
012 velocity (v2) maximum (when d = 0 at position-2 and 4)
acceleration (a1) maximum when spring constant (K) mass (M) and amplitude (d1) are known
acceleration (a1) maximum when amplitude (d1) and cycle-time (Tc) are known
acceleration (a) at any point during the cycle-time (T)
1
151
286
v
dTc
cycle-time (Tc) when amplitude (d1) and velocity (v1) are known
K
MTc 28651
cycle-time (Tc) when mass (M) and spring constant (K) are known
fTc
1
cycle-time (Tc) when the frequency is known
cTf
1
frequency (f) when the cycle-time (Tc) is known
of cycles
elapsed timef
frequency (f) when the number of cycles in an elapsed time is known
d1 amplitude (distance from rest to start position)
Also frac12 the distance from the base to the peak
3131 oror dM
Ka
251
3131 539
c
oror T
da
2539
cT
da
In December of 2006 I discovered the mathematical error in Einsteinrsquos Special Relativity Einstein used BS math and that math only works in an imaginary world
It cannot describe real symmetrical space
Einsteinrsquos Special Relativity should be abandoned and a theory of real space using SM should be used to replace it
Numerous articles on the Internet show Einsteinrsquos theory to be incorrect but no one in academia is willing to publish the information
Figure 1-10 A moving particle observed from two frames of reference in uniform translation
Einsteinrsquos Math Error in Special Relativity Transformation of Velocity (From Eisenberg- Fundamentals of Modern Physics pages 27-29) Consider the particle shown in figure (1-10) moving with velocity v as
seen in a frame of reference O
dx dy dz
Vx Vy Vzdt dt dt
dx dy dz
V x V y V zdy dt dt
We would like to evaluate the velocity v of the particle as seen in the frame of reference 0 which is itself moving relative to 0 with velocity v Measured in the 0 frame the velocity vector of the particle has com ponents
The same velocity vector as measured in the 0 frame has components
2
2
2
2
2
1
1
x vtx
v
cy y
z z
vxt
ctv
c
Now from equations (1-13) we know that the relation between the primed and the unprimed coordinates and times is
Take the differential of these equations remembering that v is a constant This gives
2
2
2
2
2
1
1
dx vdtdx
v
cdy dy
dz dz
vxdt
cdtv
c
2
2
2 2 2 2 2
2
2
2
11 ( )
1( ) 1
1
1
1
dx vdt
v dxdx vdt v
dx dx vdtc dt dtvdx vdx vdx v dxdt dt dt dt
dt dtc c c c
VxV x
Vx
VxV x
vv
vv
c
x
v
c
cV
( )
coordinate of particle relative to moving primed system
coordinate of particle relative to stationary system
velocity of primed coordina
V x
Vx
te system relative to stationaryv sys
tem
Page 28 Fundamentals of Modern Physics
(1-16)
According to Einstein this equation tells you how to transform the observed velocity from one frame of reference to another frame of reference First we note that as Vc and vc approachzero the equation approaches those which would be derived from the Galilean transformation
2 2
( ) 1 01 1 ( )
Vx v Vx vV
VxV x
Vx v
v cVxx
v
c
v
cc
Another property of this equation is that it is impossible to choose V and v such that Vrsquo the magnitude of the velocity which is seen in the new frame is greater than c
1
2
09 ( 09 ) 18009876543
( 09 ) 1811 (09 )
c c cV c c
cc
c
Example As seen by O particle 1 has velocity 09c in the direction of positive x and particle 2 has velocity 09c in the negative x direction
Einstein equation however predicts To evaluate the velocity of particle 1 with respect to particle 2 we transform from the O frame to the Orsquo frame moving in the negative x direction with velocity v = -09c using equation 1-16 We obtain
JK LOGIC will observe the particles separating at 18c
A confusing INCORRECT and ILLOGICAL answer
left is a dir rightsub of a
1
2
1
09 ( 09 ) 18009876543
181( 09 )1 ( 09 )
The multiplication of dir left time a dir right is ONLY a dir left
09
dir
direction leftdirection rightsubtraction
c c cV c c
cc
c
V
dir
22
22
09 18 18009876543
181( 081)1(09)(09)
1
left rightsub
sub
c c c cc c
ccc
c
Analyzing Einsteinrsquos BS math Equation
Using SM math a logical answer can be obtained
Cannot multiply a subtraction operator times a direction in space
Can only subtract and change its direction
Cannot multiply a direction to the left times a direction to the right
Cannot multiply opposite directions in space
Can only solve for the resultant
rightleft
subtraction
1
subtraction
2
sub of a dir to the left is a dir to the right
1 resultant 0 sub 2
2sub 2
2
09 09
09 amp 091
09 09 18 1
(0)1
(09) amp (09)1
c cV
c cc
c c cV
cc
cc
8
181
cc
SM math predicts a logical answer using logic
21st century scientists state in many articles that Einsteins General Relativity and Quantum Mechanics are incompatible
They are not only incompatible they are both mathematically incorrect
They both use illogical BS math
Any equations in math that use imaginary numbers should be abandoned
Imaginary numbers are only necessary in the current BS math system
This means that Schroumldingers equations should be abandoned
Any equation in quantum mechanics that uses imaginary number should be abandoned
Prof Kanarev has developed logical physics and chemistrytheories without the use of Quantum mechanics or relativity
A partial list of his theories coverbull Mechanics ndash replaces Newtonbull Photon - bull Electrons ndash Protons- Neutronsbull Spectroscopybull Nucleusbull Atoms amp Moleculesbull Electrodynamicsbull Heatbull Curvature of Space ndash replaces Schwarzcchildrsquos equation
Kanarevrsquos Hypothesis
bull Electron align linearly with proton They do NOT spin or form clouds around the nucleus
bull Neutrons are in the center of atom Proton are outside of Neutrons
bull The wavelength of all particles is the experimental radius
bull Double slit experiment explained correctly
bull Heisenbergrsquos inequality explained correctly
bull is the radius of approach of the magnetic lines around the torus of the electron
( 1 137)
bull Why water expands when it freezes
bull Electron clusters form lightning
rbull The meaning of Plancks constant is explained
22 2 kg m
h m f mr f consts
2210254 42m r kgm bull If the mr-constant describes a ring the equations
describe experiments bull If the mr-constant describes a wave the equations do
NOT describe experiments
Kanarevrsquos 5 Equations of Mechanics
Kanarevrsquos 1st Equation Acceleration ampmv maMma R R
Kanarevrsquos 2nd Equation Uniform motion amp mv maMma R R
maM mvR R
Kanarevrsquos 3rd Equation
Kanarevrsquos 4th Equation Equal and opposite reaction
1 21 21 2
12
a m
m a m ama
Kanarevrsquos 5th Equation 1 2 3 Mi M M M Mna R R R R R
The sum of the Actions working on a moving body is never equal to zero
Kanarevrsquos Photon model is a rolling ring
His equations accurately describe known experiments with photonsbull Electromagnetic radiation should be called photon radiationbull Diffraction and interference are the same phenomenabull Correctly explains the double slit experiment without quantum equationsbull Correctly explains the meaning of Heisenbergrsquos inequality
e
2 (2)(3142)(2817 15 ) 10007297352568
R 2426 12 137mag
xp
r m
m
Charge
Magnetic action
Proton
The intensity of the magnetic action near the geometrical center of the proton ring is so great that we theorize that it is the strong nuclear force
(9383 8)(1602 19)8476 14
4 (4)(3142)(1411 26)p
pp
EB T
M
(9395 8)(1602 19)1035 15
4 (4)(3142)(0966 26)N
NN
EB T
M
Neutron
1a Hydrogenavi
222 2 bababbabbaaabababa B y d e fin itio n in B S m a th n o n u m b er sq u a red c a n ev e r b e n eg a tiv e T h e re fo re n o sq u a red n u m b er c an ev e r b e a v e c to r m o v in g in th e d a sh ( -) o r n eg a tiv e d ire c tio n if it is sq u a red In B S m a th an d ad d itio n o p e ra to r m u ltip lied b y a su b trac tio n o p e ra to r is A L W A Y S eq u a l to a su b tra c tio n o p e ra to r If th is h ad n o t b e en u sed fo r th e la s t fo u r h u n d red y e a rs y o u w o u ld d ie lau g h in g a t th e u tte r ab su rd ity o f th e la s t s ta tem e n t T h is is illo g ica l an d v io la te s sy m m etry
If a = 5 an d b = 3 42222352
22
4930253)3)(5(253)5)(3()3)(5(5353535 22222 A s lo n g a s ldquo ardquo is g re a te r th an ldquob rdquo th e B S m a th d is tr ib u tiv e law p ro v id es co rrec t an sw ers 1
B S m a th b re a k s d o w n w h e n ldquo b rdquo is g rea ter th a n ldquo a rdquo
If a = 3 an d b = 5
2 2
2
2
3 5 2 4
2 ( 2) ( 2) 4
2 ( )( 2) ( )( 2) 4
A n y co m b in a tio n ex cep t th e m id d le o n e is illo g ica l an d in co rre c t B y d e fin itio n th e an sw er can n o t b e -4 in B S m a th B S m a th is in co rrec t
4253095)5)(3(235)3)(5()5)(3(3535353 22222 T h is is w h ere th e d is tr ib u tiv e la w p ro v id es in co rrec t an sw ers I t p ro d u ces an an sw er o f + 4 b e cau se b y d e fin iti o n n o n u m b er sq u a red in B S m a th c an b e a n eg a tiv e n u m b er U sin g in co rre c t B S m a th th e d is tr ib u tiv e law w ill g iv e an in co rrec t an sw ers o f + 4 T h is is in co rrec t an d illo g ica l 1
Since there can be no negative direction from a squared term in BS math BS math should be modified or abandoned
The BS distributive law should be modified or abandoned
The numbers +1+2+3+4 and ndash1-2-3-4hellip should be abandoned
BS math gives incorrect answers because their Rule-of-Signs allows multiplying an arrow going in one direction by an arrow going in another direction
Clearly this is not logical
Arrows should be used that are specific for directions
W h at d o es it m e an to m u ltip ly ( 4) b y (4 ) A s w e k n o w 1
m u ltip lica tio n is ju s t ad d itio n W h a t w o u ld ( 4) tim e (4 ) b e 2
In BS math the answer is by definition -16 unless you square -4 and then it is +16
In SM you cannot multiply arrows going in opposite directions
The middle two terms are not logical You cannot multiply opposite directions
This is a MAJOR error in BS math They multiply a dash [(-) a direction to the left] by a cross [(+) a direction to the right] And then ldquoby definitionrdquo label the answer dash (-) This is illogical and produces incorrect answers
2
2 2
This is Einsteinrsquos math error
in Special relativity
Magnitude 4 Direction 1 Magnitude 3 Direction 2 Magnitude 5 Direction 3 Relative to an observer at 0 4
Magnitude 4 Direction 1 Magnitude 3 Direction 2 Magnitude 5 Direction 3 Relative to an observer at 0 4
Magnitude and Direction (MD) An arrow has both a magnitude andA direction relative to a 2nd objectobserver
bull Magnitude and velocity (MV) relative to a 2nd objectobserver
bull Magnitude And acceleration (MA) relative to a 2nd objectobserver2
v da
t t
Introducing a new symbol the ldquoamprdquo sign The ldquoamprdquo sign means resultant ADD all the Arrows for each direction and then SUBTRACT to find the final resultant
1 1amp 1 2 2amp 2 4 If =
1 amp1 2 2 amp2 4 If =
amp 5 amp 3 2 If gt
amp 3 amp 5= 2 If gt
amp 0 3 amp 3=0 if =
only the resultant equals zero
the vectors do not - they are still there
Does not exist There is no multiplication of arrows
There are two arrows One has a magnitude of 3 One has a magnitude of 5 The maximum these two arrows can be is 8
Only the resultant of arrows exist
An SM arrow has a direction and a magnitude
Arrows and magnitudes are still there even if the resultant is zero
There are no negative SM Arrows
There are no positive SM Arrows
There are no ZERO SM Arrows A zero would negate the definition of an SM Arrow something with no magnitude and no direction cannot be the definition of an arrow that is defined as having magnitude and direction
bull A resultant is the addition and subtraction of the differences in magnitude and direction arrows
A B and C are the arrows (a) A amp B exist(b) A amp B amp C exist(c) A-B Do Not Exist in Sm there are no negatives or positives(d) A+B-C Do Not Exist in SM there are no negatives or positives
There are only resultants when there are multiple arrows
SM Graphs versus BS Graphs
Numbers Raised to Powers
When numbers are raised to powers the negative sides of all BS math graphs are ILLOGICAL and INCORRECT
Incorrect BS Math for 2x
2x
4 16 correct3 8 correct2 4 correct1 2 correct0 1 correct-1 005 incorrect-2 025 incorrect-3 0125 incorrect-4 00625 incorrect
negative side of graph is incorrectpositive side of graph is correct
16
8
42105025012500625
-4 -3 -2 -1 0 1 2 3 4
po
sit
ive
negative positive
2x BS Math
correctincorrect
In BS math the dash sign (-) used in this example represents three different operations They arebull (-) Subtractionbull (-) A direction to the leftbull (-) As an exponent to mean divide by
All positive numbers produce correct results and graphs
All negative numbers produce incorrect results and graphs
A direction of three units to the left (-3) is placed in the equation as an exponent
A direction in space (3 units toward the left) changes a whole number to a fraction by using the dash (-) as an exponent rather than a direction in space
This is illogical and incorrect BS math should have performed this operation as
Unfortunately the answers to problems can come out correct because data points of a real problem fit the curve
However the answers will not be due to logical reasoning
Because the equation produced a curve that real problem data points follow it has been falsely believed that the equations and graphs are correct
1
2
x
Incorrect BK graph for (12)x
(12)x
4 00625 correct3 0125 correct2 025 correct1 1 correct0 1 correct-1 2 incorrect-2 4 incorrect-3 8 incorrect-4 16 incorrect
negative side of graph is incorrectpositive side of graph is correct
00625012502505124
8
16
-5 -4 -3 -2 -1 0 1 2 3 4 5P
osit
ive
negative positive
BS Math for (12)x
correctincorrect
Incorrect BS math and graph for
All graphs are symmetrical with SM
Two BS Math Equations that Produce Incorrect Answers
2 0x x starting equation 2
2
x x x x
x x
Add lsquoxrsquo to both sides
2(1) 1 let lsquoxrsquo = 1
2 1 This is an incorrect answer 2(0)-(0)=0 The only answer that BK math can give is x=0
1
a b starting equation ( )( ) ( )( )a a a b multiply both sides by ( )a
( )( ) ( )( ) ( )( ) ( )( )a a b b a b b b subtract ( )( )b b from both sides
( )( ) ( )( )a b a b a b b factor ( )a b from both sides
( )a b b divide both sides by ( )a b
( )b b b replace ( )a with ( )b 2b b simplify 2(1) (1) Let ( )b = 1
2 1 Again BS math yields and incorrect answer 1
This incorrect result is a problem with the distributive law SM explains how this incorrect answer is obtained and the problem with the distributive law that allows the error
Page 148 ndash149 Barry Mazur ndash Imagining Numbers
3
2
31
Do the computation carefully on paper using the rules we agreed to Then ponder your answer which should be something of a surprise to you donrsquot stop there Think of what your answer might possibly mean or might imply
Solve this equation using the distributive law
JK definitely a surprise the answer wasincorrect and illogical
Gerhard Gade University Professor doing math at Harvard University
Here is the BS math answer that Barry Mazur calculated
2
31
2
31
2
31
1 3 1 3 1 3 3 3 2 2 3 1 3
2 2 2 2 4 4 4 4 4 4 2 2
dash
1 3 1 3 1 31
2 2 2 2 4 4
3 3
4 4
When the distributive law is used with the BS Rule-of-Signs and one of the signs is a (-) dash sign the middle two terms always cancel each other This leads to the incorrect answer of (-1) in the above equation
Barry Mazur did not go back and check his answer with the original equation If he had he would have known that the answer was incorrect
3
1 31
2
check answer by setting equal
3
331 3
12
1 31
2
Take the Cube root of both sides
1 3 2 multiply both sides by 2
3 3 subtract 1 from both sides
2 2
3 3 square both sides
3 9 This is an incorrect answer A simple equation check could have shown that the distributive law using BS math provides an incorrect answer
1
BS Illogical and Incorrect Numbers Raised to Powers
nC D n DC
642 DDD
In BS math only a positive number can have an answer if ldquonrdquo is an even number (246---)
All negative numbers have no solutions
642 DDD
With SM just add a direction arrow and any number gives the correct answer
or n nC D C D
3 3( 3 8 8 or 8 2 or 2 )n D C
( 2 16 C= 16 or 16 4 or 4 )n D
Since a negative number has no square root answer in BS math BS mathematicians invented imaginary numbers
i2=-1
24414 iiC n
These equations are easily solved without the use of imaginary numbers with SM math
If 4 then 4 2x
If 4 then 4 2x
The 360-degree circle is a 4400 years old extremely outdated system based on a year of 12 months of 30 days each that is 360 days
SM will use a more logical system that will have a circle of 6283185307o
d
( )
(deg )
d r radians
d r rees
the arc distance around
the circumference of a circle
d
radiusr
2 2
2 2
d r
dv rd t t
v dda rd t t t
1
d
r
( )av tv
The average angular velocity of a body is the rate of its angular displacement
( )0
liminstt
vt
The instantaneous angular velocity is the limit of the ratio
t
as t approaches zero
( )d
av
d
tv
The average circumference velocity of the body is the rate of its distance traveled d divided by the time t
dvv
r
1 dd vdrv
t t t r r
The equation of the relationship between the angular velocity and the velocity of a point on that body
( )o
av
v v v
t ta
The average angular acceleration is the change in the angular velocity
( )0
liminstt
v
ta
The instantaneous angular acceleration is the
limit of the ration of v
t
as t approaches
zero
daa
r
1
1
d
d s d
vva
t r tv a
r t r
The equation of the relationship between the angular acceleration and the acceleration of a point on the body
21
2v a t
If a body with an initial angular velocity ov has a constant
angular acceleration a it will turn through an angle in a time t
1
Straight Line Symbol Units Rotational Symbol Units Distance d m Angular
Distance radians or degrees
Velocity dt
v m
s Angular Velocity
V
t
radians or degrees
s
Acceleration vat
2ms
Angular Acceleration
VA
t
2
radians or degrees
s
Mass M Kg Mass distribution
dM 2Kgm
Force actionA Ma
2
Kgm
sNewton
Torque
dT M A 2
2
Kgm
s
Momentum Mv Kgm
s
Angular Momentum
dM V 2Kgm
s
Work (Force)(distance)
Ad 2
Kgmm
sNm Joule
Work T
Using AS arrow math to solve and direction in space problems using the equation 1 2
2 2 2 2 2 2( ) amp and ( ) amp 3
4
2 2 2
2 2
( ) amp
(5 ) 4 3
(5 ) 16 9
(5 ) 25
(5 ) (5 )
2 2(5 ) 4 3
(5 ) 16 9
(5 ) 25
(5 ) (5 )
In SM math Pythagorasrsquo theorem provides both magnitude and direction
In BS math Pythagorasrsquo theorem provides magnitude only
time set equal to zero
23
( ) ( 0 )10 1
x t xx
3( ) ( 0)
10 1W t f
In BS math this is an equation for a traveling wave
must use a square term in order to make the curve symmetric
NO SQUARE TERM IS NECESSARY
In SM math when t is set to ldquo0rdquo
This is a major fault in BS math
x f(x)
-4 0019
-3 0033
-2 0073
-1 0273
0 3000
1 0273
2 0073
3 0033
4 0019 -4 -3 -2 -1 0 1 2 3 400
05
10
15
20
25
30
35
BS wave (must use square term)
snap shot of wave at x = 0
Am
plit
ude
lt to gt lt to gt f(ltgt)
lt -4 0073
lt -3 0097
lt -2 0143
lt -1 0273
origin 0 3000
gt 1 0273
gt 2 0143
gt 3 0097
gt 4 0073lt0073 lt0097 lt0143 lt0273
3000
0273gt 0143gt 0097gt 0073gt
Wave moving
snap shot of wave
Am
plit
ude
lt4 lt3 lt2 lt1 0 1gt 2gt 3gt 4gt
3
2
1
In SM NO squared terms are needed
SPRING
An Action is needed to stretch or compress a spring to a starting point where it can execute cycle motion
tsA (ts to start) is the work required to move a mass attached to a spring a distance (d1) from
its initial rest position RAK 0 the spring constant is the stiffness of the spring
o A is attraction o R is repulsion o
ARRA
00 is for the back and forth motion through the zero initial rest position A
stretched spring goes from ldquoArdquo to ldquo0rdquo to ldquoRrdquo to ldquoArdquo to ldquo0rdquo to ldquoRrdquo d1 is the distance a mass is moved from its initial rest position
0
0 1ts A RA K d
Once ( )mad
is performed to move the mass to position d1 (against the restraint of the spring) a of the ( )mad
is transferred into the spring as potential attractionrepulsion (PAR) There is always
more ( )mad
performed than is stored as PAR When the spring is released from the restraining Ats the PAR starts the oscillation Conservation of ( )mad
requires that the PAR and the motion ( )mad
( )( )madM be a constant
( )amp ( ) constantmadPAR M 02
0 0
1
2A R A RPAR K d
02
( ) 0 0
1( )
2mad A R A RM m v
0 0 02 2 2
0 0 0 1
2 2 21
2 2 21
1 1 1amp
2 2 2
1 1 1amp
2 2 2
amp
A R A R A RK d m v K d
Kd mv Kd
Kd mv Kd
0
1d is frac12 the height from the base to the peak of a cycle
An experiment of moving paper under a pendulum through a small angle filled with sand that can escape through a small hole in the bottom shows that a pendulum cycling back and forth follows a curve that is mathematically analogous to a lsquosinrsquo or lsquocosrsquo
The projection of the curved area onto the straight line exactly duplicate the back and forth motion of CM This allows for the math of lsquosinrsquo or lsquocosrsquo to be used
SM will use lsquosinrdquo math It could use lsquocosrsquo or a combination of both lsquosinrsquo and lsquocosrsquo Using just the lsquosinrsquo math makes the understanding and math easier
0 sin 0 0SM 0 sin157 1SM The diagram at the left shows a spring being stretched from its (zero rest) position-2 and the PAR being released at position-1 There are 4 sections (1 to 2 2 to 3 3 to 4 and 4 to5) that alternate between maximum acceleration and zero velocity to max velocity and zero acceleration
At position-1 there is maximum acceleration and zero velocity o The lsquosinrsquo values from 0o to 157o will be used for velocity (values from 0 to 1) o The lsquosinrsquo values from 157o to 0o will be used for acceleration (values from 1 to 0)
At position-2 there is maximum velocity and zero acceleration Acceleration changes to deceleration
o The lsquosinrsquo values from 157o to 0o will be used for deceleration (values from 1 to 0) o The lsquosinrsquo values from 0o to 157o will be used for acceleration (values from 0 to 1)
At position-3 there is maximum acceleration and zero velocity o The lsquosinrsquo values from 0o to 157o will be used for velocity (values from 0 to 1) o The lsquosinrsquo values from 157o to 0o will be used for acceleration (values from 1 to 0)
At position-4 there is maximum velocity and zero acceleration Acceleration changes to deceleration
o The lsquosinrsquo values from 157o to 0o will be used for deceleration (values from 1 to 0) o The lsquosinrsquo values from 0o to 157o will be used for acceleration (values from 0 to 1)
At position-5 the cycle starts again
2
)539(
cT
MK
spring constant (K) when mass (M) and cycle-time (T) are known
)(
))(89)((
111 md
uKgM
d
MG
d
FK sts
spring constant (K) when mass (M) and amplitude ( d1) are known
2 21( )
Kv d d
m
velocity (v) at any point during the cycle
velocity
dforM
Kdv
max
012 velocity (v2) maximum (when d = 0 at position-2 and 4)
acceleration (a1) maximum when spring constant (K) mass (M) and amplitude (d1) are known
acceleration (a1) maximum when amplitude (d1) and cycle-time (Tc) are known
acceleration (a) at any point during the cycle-time (T)
1
151
286
v
dTc
cycle-time (Tc) when amplitude (d1) and velocity (v1) are known
K
MTc 28651
cycle-time (Tc) when mass (M) and spring constant (K) are known
fTc
1
cycle-time (Tc) when the frequency is known
cTf
1
frequency (f) when the cycle-time (Tc) is known
of cycles
elapsed timef
frequency (f) when the number of cycles in an elapsed time is known
d1 amplitude (distance from rest to start position)
Also frac12 the distance from the base to the peak
3131 oror dM
Ka
251
3131 539
c
oror T
da
2539
cT
da
In December of 2006 I discovered the mathematical error in Einsteinrsquos Special Relativity Einstein used BS math and that math only works in an imaginary world
It cannot describe real symmetrical space
Einsteinrsquos Special Relativity should be abandoned and a theory of real space using SM should be used to replace it
Numerous articles on the Internet show Einsteinrsquos theory to be incorrect but no one in academia is willing to publish the information
Figure 1-10 A moving particle observed from two frames of reference in uniform translation
Einsteinrsquos Math Error in Special Relativity Transformation of Velocity (From Eisenberg- Fundamentals of Modern Physics pages 27-29) Consider the particle shown in figure (1-10) moving with velocity v as
seen in a frame of reference O
dx dy dz
Vx Vy Vzdt dt dt
dx dy dz
V x V y V zdy dt dt
We would like to evaluate the velocity v of the particle as seen in the frame of reference 0 which is itself moving relative to 0 with velocity v Measured in the 0 frame the velocity vector of the particle has com ponents
The same velocity vector as measured in the 0 frame has components
2
2
2
2
2
1
1
x vtx
v
cy y
z z
vxt
ctv
c
Now from equations (1-13) we know that the relation between the primed and the unprimed coordinates and times is
Take the differential of these equations remembering that v is a constant This gives
2
2
2
2
2
1
1
dx vdtdx
v
cdy dy
dz dz
vxdt
cdtv
c
2
2
2 2 2 2 2
2
2
2
11 ( )
1( ) 1
1
1
1
dx vdt
v dxdx vdt v
dx dx vdtc dt dtvdx vdx vdx v dxdt dt dt dt
dt dtc c c c
VxV x
Vx
VxV x
vv
vv
c
x
v
c
cV
( )
coordinate of particle relative to moving primed system
coordinate of particle relative to stationary system
velocity of primed coordina
V x
Vx
te system relative to stationaryv sys
tem
Page 28 Fundamentals of Modern Physics
(1-16)
According to Einstein this equation tells you how to transform the observed velocity from one frame of reference to another frame of reference First we note that as Vc and vc approachzero the equation approaches those which would be derived from the Galilean transformation
2 2
( ) 1 01 1 ( )
Vx v Vx vV
VxV x
Vx v
v cVxx
v
c
v
cc
Another property of this equation is that it is impossible to choose V and v such that Vrsquo the magnitude of the velocity which is seen in the new frame is greater than c
1
2
09 ( 09 ) 18009876543
( 09 ) 1811 (09 )
c c cV c c
cc
c
Example As seen by O particle 1 has velocity 09c in the direction of positive x and particle 2 has velocity 09c in the negative x direction
Einstein equation however predicts To evaluate the velocity of particle 1 with respect to particle 2 we transform from the O frame to the Orsquo frame moving in the negative x direction with velocity v = -09c using equation 1-16 We obtain
JK LOGIC will observe the particles separating at 18c
A confusing INCORRECT and ILLOGICAL answer
left is a dir rightsub of a
1
2
1
09 ( 09 ) 18009876543
181( 09 )1 ( 09 )
The multiplication of dir left time a dir right is ONLY a dir left
09
dir
direction leftdirection rightsubtraction
c c cV c c
cc
c
V
dir
22
22
09 18 18009876543
181( 081)1(09)(09)
1
left rightsub
sub
c c c cc c
ccc
c
Analyzing Einsteinrsquos BS math Equation
Using SM math a logical answer can be obtained
Cannot multiply a subtraction operator times a direction in space
Can only subtract and change its direction
Cannot multiply a direction to the left times a direction to the right
Cannot multiply opposite directions in space
Can only solve for the resultant
rightleft
subtraction
1
subtraction
2
sub of a dir to the left is a dir to the right
1 resultant 0 sub 2
2sub 2
2
09 09
09 amp 091
09 09 18 1
(0)1
(09) amp (09)1
c cV
c cc
c c cV
cc
cc
8
181
cc
SM math predicts a logical answer using logic
21st century scientists state in many articles that Einsteins General Relativity and Quantum Mechanics are incompatible
They are not only incompatible they are both mathematically incorrect
They both use illogical BS math
Any equations in math that use imaginary numbers should be abandoned
Imaginary numbers are only necessary in the current BS math system
This means that Schroumldingers equations should be abandoned
Any equation in quantum mechanics that uses imaginary number should be abandoned
Prof Kanarev has developed logical physics and chemistrytheories without the use of Quantum mechanics or relativity
A partial list of his theories coverbull Mechanics ndash replaces Newtonbull Photon - bull Electrons ndash Protons- Neutronsbull Spectroscopybull Nucleusbull Atoms amp Moleculesbull Electrodynamicsbull Heatbull Curvature of Space ndash replaces Schwarzcchildrsquos equation
Kanarevrsquos Hypothesis
bull Electron align linearly with proton They do NOT spin or form clouds around the nucleus
bull Neutrons are in the center of atom Proton are outside of Neutrons
bull The wavelength of all particles is the experimental radius
bull Double slit experiment explained correctly
bull Heisenbergrsquos inequality explained correctly
bull is the radius of approach of the magnetic lines around the torus of the electron
( 1 137)
bull Why water expands when it freezes
bull Electron clusters form lightning
rbull The meaning of Plancks constant is explained
22 2 kg m
h m f mr f consts
2210254 42m r kgm bull If the mr-constant describes a ring the equations
describe experiments bull If the mr-constant describes a wave the equations do
NOT describe experiments
Kanarevrsquos 5 Equations of Mechanics
Kanarevrsquos 1st Equation Acceleration ampmv maMma R R
Kanarevrsquos 2nd Equation Uniform motion amp mv maMma R R
maM mvR R
Kanarevrsquos 3rd Equation
Kanarevrsquos 4th Equation Equal and opposite reaction
1 21 21 2
12
a m
m a m ama
Kanarevrsquos 5th Equation 1 2 3 Mi M M M Mna R R R R R
The sum of the Actions working on a moving body is never equal to zero
Kanarevrsquos Photon model is a rolling ring
His equations accurately describe known experiments with photonsbull Electromagnetic radiation should be called photon radiationbull Diffraction and interference are the same phenomenabull Correctly explains the double slit experiment without quantum equationsbull Correctly explains the meaning of Heisenbergrsquos inequality
e
2 (2)(3142)(2817 15 ) 10007297352568
R 2426 12 137mag
xp
r m
m
Charge
Magnetic action
Proton
The intensity of the magnetic action near the geometrical center of the proton ring is so great that we theorize that it is the strong nuclear force
(9383 8)(1602 19)8476 14
4 (4)(3142)(1411 26)p
pp
EB T
M
(9395 8)(1602 19)1035 15
4 (4)(3142)(0966 26)N
NN
EB T
M
Neutron
1a Hydrogenavi
B S m a th b re a k s d o w n w h e n ldquo b rdquo is g rea ter th a n ldquo a rdquo
If a = 3 an d b = 5
2 2
2
2
3 5 2 4
2 ( 2) ( 2) 4
2 ( )( 2) ( )( 2) 4
A n y co m b in a tio n ex cep t th e m id d le o n e is illo g ica l an d in co rre c t B y d e fin itio n th e an sw er can n o t b e -4 in B S m a th B S m a th is in co rrec t
4253095)5)(3(235)3)(5()5)(3(3535353 22222 T h is is w h ere th e d is tr ib u tiv e la w p ro v id es in co rrec t an sw ers I t p ro d u ces an an sw er o f + 4 b e cau se b y d e fin iti o n n o n u m b er sq u a red in B S m a th c an b e a n eg a tiv e n u m b er U sin g in co rre c t B S m a th th e d is tr ib u tiv e law w ill g iv e an in co rrec t an sw ers o f + 4 T h is is in co rrec t an d illo g ica l 1
Since there can be no negative direction from a squared term in BS math BS math should be modified or abandoned
The BS distributive law should be modified or abandoned
The numbers +1+2+3+4 and ndash1-2-3-4hellip should be abandoned
BS math gives incorrect answers because their Rule-of-Signs allows multiplying an arrow going in one direction by an arrow going in another direction
Clearly this is not logical
Arrows should be used that are specific for directions
W h at d o es it m e an to m u ltip ly ( 4) b y (4 ) A s w e k n o w 1
m u ltip lica tio n is ju s t ad d itio n W h a t w o u ld ( 4) tim e (4 ) b e 2
In BS math the answer is by definition -16 unless you square -4 and then it is +16
In SM you cannot multiply arrows going in opposite directions
The middle two terms are not logical You cannot multiply opposite directions
This is a MAJOR error in BS math They multiply a dash [(-) a direction to the left] by a cross [(+) a direction to the right] And then ldquoby definitionrdquo label the answer dash (-) This is illogical and produces incorrect answers
2
2 2
This is Einsteinrsquos math error
in Special relativity
Magnitude 4 Direction 1 Magnitude 3 Direction 2 Magnitude 5 Direction 3 Relative to an observer at 0 4
Magnitude 4 Direction 1 Magnitude 3 Direction 2 Magnitude 5 Direction 3 Relative to an observer at 0 4
Magnitude and Direction (MD) An arrow has both a magnitude andA direction relative to a 2nd objectobserver
bull Magnitude and velocity (MV) relative to a 2nd objectobserver
bull Magnitude And acceleration (MA) relative to a 2nd objectobserver2
v da
t t
Introducing a new symbol the ldquoamprdquo sign The ldquoamprdquo sign means resultant ADD all the Arrows for each direction and then SUBTRACT to find the final resultant
1 1amp 1 2 2amp 2 4 If =
1 amp1 2 2 amp2 4 If =
amp 5 amp 3 2 If gt
amp 3 amp 5= 2 If gt
amp 0 3 amp 3=0 if =
only the resultant equals zero
the vectors do not - they are still there
Does not exist There is no multiplication of arrows
There are two arrows One has a magnitude of 3 One has a magnitude of 5 The maximum these two arrows can be is 8
Only the resultant of arrows exist
An SM arrow has a direction and a magnitude
Arrows and magnitudes are still there even if the resultant is zero
There are no negative SM Arrows
There are no positive SM Arrows
There are no ZERO SM Arrows A zero would negate the definition of an SM Arrow something with no magnitude and no direction cannot be the definition of an arrow that is defined as having magnitude and direction
bull A resultant is the addition and subtraction of the differences in magnitude and direction arrows
A B and C are the arrows (a) A amp B exist(b) A amp B amp C exist(c) A-B Do Not Exist in Sm there are no negatives or positives(d) A+B-C Do Not Exist in SM there are no negatives or positives
There are only resultants when there are multiple arrows
SM Graphs versus BS Graphs
Numbers Raised to Powers
When numbers are raised to powers the negative sides of all BS math graphs are ILLOGICAL and INCORRECT
Incorrect BS Math for 2x
2x
4 16 correct3 8 correct2 4 correct1 2 correct0 1 correct-1 005 incorrect-2 025 incorrect-3 0125 incorrect-4 00625 incorrect
negative side of graph is incorrectpositive side of graph is correct
16
8
42105025012500625
-4 -3 -2 -1 0 1 2 3 4
po
sit
ive
negative positive
2x BS Math
correctincorrect
In BS math the dash sign (-) used in this example represents three different operations They arebull (-) Subtractionbull (-) A direction to the leftbull (-) As an exponent to mean divide by
All positive numbers produce correct results and graphs
All negative numbers produce incorrect results and graphs
A direction of three units to the left (-3) is placed in the equation as an exponent
A direction in space (3 units toward the left) changes a whole number to a fraction by using the dash (-) as an exponent rather than a direction in space
This is illogical and incorrect BS math should have performed this operation as
Unfortunately the answers to problems can come out correct because data points of a real problem fit the curve
However the answers will not be due to logical reasoning
Because the equation produced a curve that real problem data points follow it has been falsely believed that the equations and graphs are correct
1
2
x
Incorrect BK graph for (12)x
(12)x
4 00625 correct3 0125 correct2 025 correct1 1 correct0 1 correct-1 2 incorrect-2 4 incorrect-3 8 incorrect-4 16 incorrect
negative side of graph is incorrectpositive side of graph is correct
00625012502505124
8
16
-5 -4 -3 -2 -1 0 1 2 3 4 5P
osit
ive
negative positive
BS Math for (12)x
correctincorrect
Incorrect BS math and graph for
All graphs are symmetrical with SM
Two BS Math Equations that Produce Incorrect Answers
2 0x x starting equation 2
2
x x x x
x x
Add lsquoxrsquo to both sides
2(1) 1 let lsquoxrsquo = 1
2 1 This is an incorrect answer 2(0)-(0)=0 The only answer that BK math can give is x=0
1
a b starting equation ( )( ) ( )( )a a a b multiply both sides by ( )a
( )( ) ( )( ) ( )( ) ( )( )a a b b a b b b subtract ( )( )b b from both sides
( )( ) ( )( )a b a b a b b factor ( )a b from both sides
( )a b b divide both sides by ( )a b
( )b b b replace ( )a with ( )b 2b b simplify 2(1) (1) Let ( )b = 1
2 1 Again BS math yields and incorrect answer 1
This incorrect result is a problem with the distributive law SM explains how this incorrect answer is obtained and the problem with the distributive law that allows the error
Page 148 ndash149 Barry Mazur ndash Imagining Numbers
3
2
31
Do the computation carefully on paper using the rules we agreed to Then ponder your answer which should be something of a surprise to you donrsquot stop there Think of what your answer might possibly mean or might imply
Solve this equation using the distributive law
JK definitely a surprise the answer wasincorrect and illogical
Gerhard Gade University Professor doing math at Harvard University
Here is the BS math answer that Barry Mazur calculated
2
31
2
31
2
31
1 3 1 3 1 3 3 3 2 2 3 1 3
2 2 2 2 4 4 4 4 4 4 2 2
dash
1 3 1 3 1 31
2 2 2 2 4 4
3 3
4 4
When the distributive law is used with the BS Rule-of-Signs and one of the signs is a (-) dash sign the middle two terms always cancel each other This leads to the incorrect answer of (-1) in the above equation
Barry Mazur did not go back and check his answer with the original equation If he had he would have known that the answer was incorrect
3
1 31
2
check answer by setting equal
3
331 3
12
1 31
2
Take the Cube root of both sides
1 3 2 multiply both sides by 2
3 3 subtract 1 from both sides
2 2
3 3 square both sides
3 9 This is an incorrect answer A simple equation check could have shown that the distributive law using BS math provides an incorrect answer
1
BS Illogical and Incorrect Numbers Raised to Powers
nC D n DC
642 DDD
In BS math only a positive number can have an answer if ldquonrdquo is an even number (246---)
All negative numbers have no solutions
642 DDD
With SM just add a direction arrow and any number gives the correct answer
or n nC D C D
3 3( 3 8 8 or 8 2 or 2 )n D C
( 2 16 C= 16 or 16 4 or 4 )n D
Since a negative number has no square root answer in BS math BS mathematicians invented imaginary numbers
i2=-1
24414 iiC n
These equations are easily solved without the use of imaginary numbers with SM math
If 4 then 4 2x
If 4 then 4 2x
The 360-degree circle is a 4400 years old extremely outdated system based on a year of 12 months of 30 days each that is 360 days
SM will use a more logical system that will have a circle of 6283185307o
d
( )
(deg )
d r radians
d r rees
the arc distance around
the circumference of a circle
d
radiusr
2 2
2 2
d r
dv rd t t
v dda rd t t t
1
d
r
( )av tv
The average angular velocity of a body is the rate of its angular displacement
( )0
liminstt
vt
The instantaneous angular velocity is the limit of the ratio
t
as t approaches zero
( )d
av
d
tv
The average circumference velocity of the body is the rate of its distance traveled d divided by the time t
dvv
r
1 dd vdrv
t t t r r
The equation of the relationship between the angular velocity and the velocity of a point on that body
( )o
av
v v v
t ta
The average angular acceleration is the change in the angular velocity
( )0
liminstt
v
ta
The instantaneous angular acceleration is the
limit of the ration of v
t
as t approaches
zero
daa
r
1
1
d
d s d
vva
t r tv a
r t r
The equation of the relationship between the angular acceleration and the acceleration of a point on the body
21
2v a t
If a body with an initial angular velocity ov has a constant
angular acceleration a it will turn through an angle in a time t
1
Straight Line Symbol Units Rotational Symbol Units Distance d m Angular
Distance radians or degrees
Velocity dt
v m
s Angular Velocity
V
t
radians or degrees
s
Acceleration vat
2ms
Angular Acceleration
VA
t
2
radians or degrees
s
Mass M Kg Mass distribution
dM 2Kgm
Force actionA Ma
2
Kgm
sNewton
Torque
dT M A 2
2
Kgm
s
Momentum Mv Kgm
s
Angular Momentum
dM V 2Kgm
s
Work (Force)(distance)
Ad 2
Kgmm
sNm Joule
Work T
Using AS arrow math to solve and direction in space problems using the equation 1 2
2 2 2 2 2 2( ) amp and ( ) amp 3
4
2 2 2
2 2
( ) amp
(5 ) 4 3
(5 ) 16 9
(5 ) 25
(5 ) (5 )
2 2(5 ) 4 3
(5 ) 16 9
(5 ) 25
(5 ) (5 )
In SM math Pythagorasrsquo theorem provides both magnitude and direction
In BS math Pythagorasrsquo theorem provides magnitude only
time set equal to zero
23
( ) ( 0 )10 1
x t xx
3( ) ( 0)
10 1W t f
In BS math this is an equation for a traveling wave
must use a square term in order to make the curve symmetric
NO SQUARE TERM IS NECESSARY
In SM math when t is set to ldquo0rdquo
This is a major fault in BS math
x f(x)
-4 0019
-3 0033
-2 0073
-1 0273
0 3000
1 0273
2 0073
3 0033
4 0019 -4 -3 -2 -1 0 1 2 3 400
05
10
15
20
25
30
35
BS wave (must use square term)
snap shot of wave at x = 0
Am
plit
ude
lt to gt lt to gt f(ltgt)
lt -4 0073
lt -3 0097
lt -2 0143
lt -1 0273
origin 0 3000
gt 1 0273
gt 2 0143
gt 3 0097
gt 4 0073lt0073 lt0097 lt0143 lt0273
3000
0273gt 0143gt 0097gt 0073gt
Wave moving
snap shot of wave
Am
plit
ude
lt4 lt3 lt2 lt1 0 1gt 2gt 3gt 4gt
3
2
1
In SM NO squared terms are needed
SPRING
An Action is needed to stretch or compress a spring to a starting point where it can execute cycle motion
tsA (ts to start) is the work required to move a mass attached to a spring a distance (d1) from
its initial rest position RAK 0 the spring constant is the stiffness of the spring
o A is attraction o R is repulsion o
ARRA
00 is for the back and forth motion through the zero initial rest position A
stretched spring goes from ldquoArdquo to ldquo0rdquo to ldquoRrdquo to ldquoArdquo to ldquo0rdquo to ldquoRrdquo d1 is the distance a mass is moved from its initial rest position
0
0 1ts A RA K d
Once ( )mad
is performed to move the mass to position d1 (against the restraint of the spring) a of the ( )mad
is transferred into the spring as potential attractionrepulsion (PAR) There is always
more ( )mad
performed than is stored as PAR When the spring is released from the restraining Ats the PAR starts the oscillation Conservation of ( )mad
requires that the PAR and the motion ( )mad
( )( )madM be a constant
( )amp ( ) constantmadPAR M 02
0 0
1
2A R A RPAR K d
02
( ) 0 0
1( )
2mad A R A RM m v
0 0 02 2 2
0 0 0 1
2 2 21
2 2 21
1 1 1amp
2 2 2
1 1 1amp
2 2 2
amp
A R A R A RK d m v K d
Kd mv Kd
Kd mv Kd
0
1d is frac12 the height from the base to the peak of a cycle
An experiment of moving paper under a pendulum through a small angle filled with sand that can escape through a small hole in the bottom shows that a pendulum cycling back and forth follows a curve that is mathematically analogous to a lsquosinrsquo or lsquocosrsquo
The projection of the curved area onto the straight line exactly duplicate the back and forth motion of CM This allows for the math of lsquosinrsquo or lsquocosrsquo to be used
SM will use lsquosinrdquo math It could use lsquocosrsquo or a combination of both lsquosinrsquo and lsquocosrsquo Using just the lsquosinrsquo math makes the understanding and math easier
0 sin 0 0SM 0 sin157 1SM The diagram at the left shows a spring being stretched from its (zero rest) position-2 and the PAR being released at position-1 There are 4 sections (1 to 2 2 to 3 3 to 4 and 4 to5) that alternate between maximum acceleration and zero velocity to max velocity and zero acceleration
At position-1 there is maximum acceleration and zero velocity o The lsquosinrsquo values from 0o to 157o will be used for velocity (values from 0 to 1) o The lsquosinrsquo values from 157o to 0o will be used for acceleration (values from 1 to 0)
At position-2 there is maximum velocity and zero acceleration Acceleration changes to deceleration
o The lsquosinrsquo values from 157o to 0o will be used for deceleration (values from 1 to 0) o The lsquosinrsquo values from 0o to 157o will be used for acceleration (values from 0 to 1)
At position-3 there is maximum acceleration and zero velocity o The lsquosinrsquo values from 0o to 157o will be used for velocity (values from 0 to 1) o The lsquosinrsquo values from 157o to 0o will be used for acceleration (values from 1 to 0)
At position-4 there is maximum velocity and zero acceleration Acceleration changes to deceleration
o The lsquosinrsquo values from 157o to 0o will be used for deceleration (values from 1 to 0) o The lsquosinrsquo values from 0o to 157o will be used for acceleration (values from 0 to 1)
At position-5 the cycle starts again
2
)539(
cT
MK
spring constant (K) when mass (M) and cycle-time (T) are known
)(
))(89)((
111 md
uKgM
d
MG
d
FK sts
spring constant (K) when mass (M) and amplitude ( d1) are known
2 21( )
Kv d d
m
velocity (v) at any point during the cycle
velocity
dforM
Kdv
max
012 velocity (v2) maximum (when d = 0 at position-2 and 4)
acceleration (a1) maximum when spring constant (K) mass (M) and amplitude (d1) are known
acceleration (a1) maximum when amplitude (d1) and cycle-time (Tc) are known
acceleration (a) at any point during the cycle-time (T)
1
151
286
v
dTc
cycle-time (Tc) when amplitude (d1) and velocity (v1) are known
K
MTc 28651
cycle-time (Tc) when mass (M) and spring constant (K) are known
fTc
1
cycle-time (Tc) when the frequency is known
cTf
1
frequency (f) when the cycle-time (Tc) is known
of cycles
elapsed timef
frequency (f) when the number of cycles in an elapsed time is known
d1 amplitude (distance from rest to start position)
Also frac12 the distance from the base to the peak
3131 oror dM
Ka
251
3131 539
c
oror T
da
2539
cT
da
In December of 2006 I discovered the mathematical error in Einsteinrsquos Special Relativity Einstein used BS math and that math only works in an imaginary world
It cannot describe real symmetrical space
Einsteinrsquos Special Relativity should be abandoned and a theory of real space using SM should be used to replace it
Numerous articles on the Internet show Einsteinrsquos theory to be incorrect but no one in academia is willing to publish the information
Figure 1-10 A moving particle observed from two frames of reference in uniform translation
Einsteinrsquos Math Error in Special Relativity Transformation of Velocity (From Eisenberg- Fundamentals of Modern Physics pages 27-29) Consider the particle shown in figure (1-10) moving with velocity v as
seen in a frame of reference O
dx dy dz
Vx Vy Vzdt dt dt
dx dy dz
V x V y V zdy dt dt
We would like to evaluate the velocity v of the particle as seen in the frame of reference 0 which is itself moving relative to 0 with velocity v Measured in the 0 frame the velocity vector of the particle has com ponents
The same velocity vector as measured in the 0 frame has components
2
2
2
2
2
1
1
x vtx
v
cy y
z z
vxt
ctv
c
Now from equations (1-13) we know that the relation between the primed and the unprimed coordinates and times is
Take the differential of these equations remembering that v is a constant This gives
2
2
2
2
2
1
1
dx vdtdx
v
cdy dy
dz dz
vxdt
cdtv
c
2
2
2 2 2 2 2
2
2
2
11 ( )
1( ) 1
1
1
1
dx vdt
v dxdx vdt v
dx dx vdtc dt dtvdx vdx vdx v dxdt dt dt dt
dt dtc c c c
VxV x
Vx
VxV x
vv
vv
c
x
v
c
cV
( )
coordinate of particle relative to moving primed system
coordinate of particle relative to stationary system
velocity of primed coordina
V x
Vx
te system relative to stationaryv sys
tem
Page 28 Fundamentals of Modern Physics
(1-16)
According to Einstein this equation tells you how to transform the observed velocity from one frame of reference to another frame of reference First we note that as Vc and vc approachzero the equation approaches those which would be derived from the Galilean transformation
2 2
( ) 1 01 1 ( )
Vx v Vx vV
VxV x
Vx v
v cVxx
v
c
v
cc
Another property of this equation is that it is impossible to choose V and v such that Vrsquo the magnitude of the velocity which is seen in the new frame is greater than c
1
2
09 ( 09 ) 18009876543
( 09 ) 1811 (09 )
c c cV c c
cc
c
Example As seen by O particle 1 has velocity 09c in the direction of positive x and particle 2 has velocity 09c in the negative x direction
Einstein equation however predicts To evaluate the velocity of particle 1 with respect to particle 2 we transform from the O frame to the Orsquo frame moving in the negative x direction with velocity v = -09c using equation 1-16 We obtain
JK LOGIC will observe the particles separating at 18c
A confusing INCORRECT and ILLOGICAL answer
left is a dir rightsub of a
1
2
1
09 ( 09 ) 18009876543
181( 09 )1 ( 09 )
The multiplication of dir left time a dir right is ONLY a dir left
09
dir
direction leftdirection rightsubtraction
c c cV c c
cc
c
V
dir
22
22
09 18 18009876543
181( 081)1(09)(09)
1
left rightsub
sub
c c c cc c
ccc
c
Analyzing Einsteinrsquos BS math Equation
Using SM math a logical answer can be obtained
Cannot multiply a subtraction operator times a direction in space
Can only subtract and change its direction
Cannot multiply a direction to the left times a direction to the right
Cannot multiply opposite directions in space
Can only solve for the resultant
rightleft
subtraction
1
subtraction
2
sub of a dir to the left is a dir to the right
1 resultant 0 sub 2
2sub 2
2
09 09
09 amp 091
09 09 18 1
(0)1
(09) amp (09)1
c cV
c cc
c c cV
cc
cc
8
181
cc
SM math predicts a logical answer using logic
21st century scientists state in many articles that Einsteins General Relativity and Quantum Mechanics are incompatible
They are not only incompatible they are both mathematically incorrect
They both use illogical BS math
Any equations in math that use imaginary numbers should be abandoned
Imaginary numbers are only necessary in the current BS math system
This means that Schroumldingers equations should be abandoned
Any equation in quantum mechanics that uses imaginary number should be abandoned
Prof Kanarev has developed logical physics and chemistrytheories without the use of Quantum mechanics or relativity
A partial list of his theories coverbull Mechanics ndash replaces Newtonbull Photon - bull Electrons ndash Protons- Neutronsbull Spectroscopybull Nucleusbull Atoms amp Moleculesbull Electrodynamicsbull Heatbull Curvature of Space ndash replaces Schwarzcchildrsquos equation
Kanarevrsquos Hypothesis
bull Electron align linearly with proton They do NOT spin or form clouds around the nucleus
bull Neutrons are in the center of atom Proton are outside of Neutrons
bull The wavelength of all particles is the experimental radius
bull Double slit experiment explained correctly
bull Heisenbergrsquos inequality explained correctly
bull is the radius of approach of the magnetic lines around the torus of the electron
( 1 137)
bull Why water expands when it freezes
bull Electron clusters form lightning
rbull The meaning of Plancks constant is explained
22 2 kg m
h m f mr f consts
2210254 42m r kgm bull If the mr-constant describes a ring the equations
describe experiments bull If the mr-constant describes a wave the equations do
NOT describe experiments
Kanarevrsquos 5 Equations of Mechanics
Kanarevrsquos 1st Equation Acceleration ampmv maMma R R
Kanarevrsquos 2nd Equation Uniform motion amp mv maMma R R
maM mvR R
Kanarevrsquos 3rd Equation
Kanarevrsquos 4th Equation Equal and opposite reaction
1 21 21 2
12
a m
m a m ama
Kanarevrsquos 5th Equation 1 2 3 Mi M M M Mna R R R R R
The sum of the Actions working on a moving body is never equal to zero
Kanarevrsquos Photon model is a rolling ring
His equations accurately describe known experiments with photonsbull Electromagnetic radiation should be called photon radiationbull Diffraction and interference are the same phenomenabull Correctly explains the double slit experiment without quantum equationsbull Correctly explains the meaning of Heisenbergrsquos inequality
e
2 (2)(3142)(2817 15 ) 10007297352568
R 2426 12 137mag
xp
r m
m
Charge
Magnetic action
Proton
The intensity of the magnetic action near the geometrical center of the proton ring is so great that we theorize that it is the strong nuclear force
(9383 8)(1602 19)8476 14
4 (4)(3142)(1411 26)p
pp
EB T
M
(9395 8)(1602 19)1035 15
4 (4)(3142)(0966 26)N
NN
EB T
M
Neutron
1a Hydrogenavi
Since there can be no negative direction from a squared term in BS math BS math should be modified or abandoned
The BS distributive law should be modified or abandoned
The numbers +1+2+3+4 and ndash1-2-3-4hellip should be abandoned
BS math gives incorrect answers because their Rule-of-Signs allows multiplying an arrow going in one direction by an arrow going in another direction
Clearly this is not logical
Arrows should be used that are specific for directions
W h at d o es it m e an to m u ltip ly ( 4) b y (4 ) A s w e k n o w 1
m u ltip lica tio n is ju s t ad d itio n W h a t w o u ld ( 4) tim e (4 ) b e 2
In BS math the answer is by definition -16 unless you square -4 and then it is +16
In SM you cannot multiply arrows going in opposite directions
The middle two terms are not logical You cannot multiply opposite directions
This is a MAJOR error in BS math They multiply a dash [(-) a direction to the left] by a cross [(+) a direction to the right] And then ldquoby definitionrdquo label the answer dash (-) This is illogical and produces incorrect answers
2
2 2
This is Einsteinrsquos math error
in Special relativity
Magnitude 4 Direction 1 Magnitude 3 Direction 2 Magnitude 5 Direction 3 Relative to an observer at 0 4
Magnitude 4 Direction 1 Magnitude 3 Direction 2 Magnitude 5 Direction 3 Relative to an observer at 0 4
Magnitude and Direction (MD) An arrow has both a magnitude andA direction relative to a 2nd objectobserver
bull Magnitude and velocity (MV) relative to a 2nd objectobserver
bull Magnitude And acceleration (MA) relative to a 2nd objectobserver2
v da
t t
Introducing a new symbol the ldquoamprdquo sign The ldquoamprdquo sign means resultant ADD all the Arrows for each direction and then SUBTRACT to find the final resultant
1 1amp 1 2 2amp 2 4 If =
1 amp1 2 2 amp2 4 If =
amp 5 amp 3 2 If gt
amp 3 amp 5= 2 If gt
amp 0 3 amp 3=0 if =
only the resultant equals zero
the vectors do not - they are still there
Does not exist There is no multiplication of arrows
There are two arrows One has a magnitude of 3 One has a magnitude of 5 The maximum these two arrows can be is 8
Only the resultant of arrows exist
An SM arrow has a direction and a magnitude
Arrows and magnitudes are still there even if the resultant is zero
There are no negative SM Arrows
There are no positive SM Arrows
There are no ZERO SM Arrows A zero would negate the definition of an SM Arrow something with no magnitude and no direction cannot be the definition of an arrow that is defined as having magnitude and direction
bull A resultant is the addition and subtraction of the differences in magnitude and direction arrows
A B and C are the arrows (a) A amp B exist(b) A amp B amp C exist(c) A-B Do Not Exist in Sm there are no negatives or positives(d) A+B-C Do Not Exist in SM there are no negatives or positives
There are only resultants when there are multiple arrows
SM Graphs versus BS Graphs
Numbers Raised to Powers
When numbers are raised to powers the negative sides of all BS math graphs are ILLOGICAL and INCORRECT
Incorrect BS Math for 2x
2x
4 16 correct3 8 correct2 4 correct1 2 correct0 1 correct-1 005 incorrect-2 025 incorrect-3 0125 incorrect-4 00625 incorrect
negative side of graph is incorrectpositive side of graph is correct
16
8
42105025012500625
-4 -3 -2 -1 0 1 2 3 4
po
sit
ive
negative positive
2x BS Math
correctincorrect
In BS math the dash sign (-) used in this example represents three different operations They arebull (-) Subtractionbull (-) A direction to the leftbull (-) As an exponent to mean divide by
All positive numbers produce correct results and graphs
All negative numbers produce incorrect results and graphs
A direction of three units to the left (-3) is placed in the equation as an exponent
A direction in space (3 units toward the left) changes a whole number to a fraction by using the dash (-) as an exponent rather than a direction in space
This is illogical and incorrect BS math should have performed this operation as
Unfortunately the answers to problems can come out correct because data points of a real problem fit the curve
However the answers will not be due to logical reasoning
Because the equation produced a curve that real problem data points follow it has been falsely believed that the equations and graphs are correct
1
2
x
Incorrect BK graph for (12)x
(12)x
4 00625 correct3 0125 correct2 025 correct1 1 correct0 1 correct-1 2 incorrect-2 4 incorrect-3 8 incorrect-4 16 incorrect
negative side of graph is incorrectpositive side of graph is correct
00625012502505124
8
16
-5 -4 -3 -2 -1 0 1 2 3 4 5P
osit
ive
negative positive
BS Math for (12)x
correctincorrect
Incorrect BS math and graph for
All graphs are symmetrical with SM
Two BS Math Equations that Produce Incorrect Answers
2 0x x starting equation 2
2
x x x x
x x
Add lsquoxrsquo to both sides
2(1) 1 let lsquoxrsquo = 1
2 1 This is an incorrect answer 2(0)-(0)=0 The only answer that BK math can give is x=0
1
a b starting equation ( )( ) ( )( )a a a b multiply both sides by ( )a
( )( ) ( )( ) ( )( ) ( )( )a a b b a b b b subtract ( )( )b b from both sides
( )( ) ( )( )a b a b a b b factor ( )a b from both sides
( )a b b divide both sides by ( )a b
( )b b b replace ( )a with ( )b 2b b simplify 2(1) (1) Let ( )b = 1
2 1 Again BS math yields and incorrect answer 1
This incorrect result is a problem with the distributive law SM explains how this incorrect answer is obtained and the problem with the distributive law that allows the error
Page 148 ndash149 Barry Mazur ndash Imagining Numbers
3
2
31
Do the computation carefully on paper using the rules we agreed to Then ponder your answer which should be something of a surprise to you donrsquot stop there Think of what your answer might possibly mean or might imply
Solve this equation using the distributive law
JK definitely a surprise the answer wasincorrect and illogical
Gerhard Gade University Professor doing math at Harvard University
Here is the BS math answer that Barry Mazur calculated
2
31
2
31
2
31
1 3 1 3 1 3 3 3 2 2 3 1 3
2 2 2 2 4 4 4 4 4 4 2 2
dash
1 3 1 3 1 31
2 2 2 2 4 4
3 3
4 4
When the distributive law is used with the BS Rule-of-Signs and one of the signs is a (-) dash sign the middle two terms always cancel each other This leads to the incorrect answer of (-1) in the above equation
Barry Mazur did not go back and check his answer with the original equation If he had he would have known that the answer was incorrect
3
1 31
2
check answer by setting equal
3
331 3
12
1 31
2
Take the Cube root of both sides
1 3 2 multiply both sides by 2
3 3 subtract 1 from both sides
2 2
3 3 square both sides
3 9 This is an incorrect answer A simple equation check could have shown that the distributive law using BS math provides an incorrect answer
1
BS Illogical and Incorrect Numbers Raised to Powers
nC D n DC
642 DDD
In BS math only a positive number can have an answer if ldquonrdquo is an even number (246---)
All negative numbers have no solutions
642 DDD
With SM just add a direction arrow and any number gives the correct answer
or n nC D C D
3 3( 3 8 8 or 8 2 or 2 )n D C
( 2 16 C= 16 or 16 4 or 4 )n D
Since a negative number has no square root answer in BS math BS mathematicians invented imaginary numbers
i2=-1
24414 iiC n
These equations are easily solved without the use of imaginary numbers with SM math
If 4 then 4 2x
If 4 then 4 2x
The 360-degree circle is a 4400 years old extremely outdated system based on a year of 12 months of 30 days each that is 360 days
SM will use a more logical system that will have a circle of 6283185307o
d
( )
(deg )
d r radians
d r rees
the arc distance around
the circumference of a circle
d
radiusr
2 2
2 2
d r
dv rd t t
v dda rd t t t
1
d
r
( )av tv
The average angular velocity of a body is the rate of its angular displacement
( )0
liminstt
vt
The instantaneous angular velocity is the limit of the ratio
t
as t approaches zero
( )d
av
d
tv
The average circumference velocity of the body is the rate of its distance traveled d divided by the time t
dvv
r
1 dd vdrv
t t t r r
The equation of the relationship between the angular velocity and the velocity of a point on that body
( )o
av
v v v
t ta
The average angular acceleration is the change in the angular velocity
( )0
liminstt
v
ta
The instantaneous angular acceleration is the
limit of the ration of v
t
as t approaches
zero
daa
r
1
1
d
d s d
vva
t r tv a
r t r
The equation of the relationship between the angular acceleration and the acceleration of a point on the body
21
2v a t
If a body with an initial angular velocity ov has a constant
angular acceleration a it will turn through an angle in a time t
1
Straight Line Symbol Units Rotational Symbol Units Distance d m Angular
Distance radians or degrees
Velocity dt
v m
s Angular Velocity
V
t
radians or degrees
s
Acceleration vat
2ms
Angular Acceleration
VA
t
2
radians or degrees
s
Mass M Kg Mass distribution
dM 2Kgm
Force actionA Ma
2
Kgm
sNewton
Torque
dT M A 2
2
Kgm
s
Momentum Mv Kgm
s
Angular Momentum
dM V 2Kgm
s
Work (Force)(distance)
Ad 2
Kgmm
sNm Joule
Work T
Using AS arrow math to solve and direction in space problems using the equation 1 2
2 2 2 2 2 2( ) amp and ( ) amp 3
4
2 2 2
2 2
( ) amp
(5 ) 4 3
(5 ) 16 9
(5 ) 25
(5 ) (5 )
2 2(5 ) 4 3
(5 ) 16 9
(5 ) 25
(5 ) (5 )
In SM math Pythagorasrsquo theorem provides both magnitude and direction
In BS math Pythagorasrsquo theorem provides magnitude only
time set equal to zero
23
( ) ( 0 )10 1
x t xx
3( ) ( 0)
10 1W t f
In BS math this is an equation for a traveling wave
must use a square term in order to make the curve symmetric
NO SQUARE TERM IS NECESSARY
In SM math when t is set to ldquo0rdquo
This is a major fault in BS math
x f(x)
-4 0019
-3 0033
-2 0073
-1 0273
0 3000
1 0273
2 0073
3 0033
4 0019 -4 -3 -2 -1 0 1 2 3 400
05
10
15
20
25
30
35
BS wave (must use square term)
snap shot of wave at x = 0
Am
plit
ude
lt to gt lt to gt f(ltgt)
lt -4 0073
lt -3 0097
lt -2 0143
lt -1 0273
origin 0 3000
gt 1 0273
gt 2 0143
gt 3 0097
gt 4 0073lt0073 lt0097 lt0143 lt0273
3000
0273gt 0143gt 0097gt 0073gt
Wave moving
snap shot of wave
Am
plit
ude
lt4 lt3 lt2 lt1 0 1gt 2gt 3gt 4gt
3
2
1
In SM NO squared terms are needed
SPRING
An Action is needed to stretch or compress a spring to a starting point where it can execute cycle motion
tsA (ts to start) is the work required to move a mass attached to a spring a distance (d1) from
its initial rest position RAK 0 the spring constant is the stiffness of the spring
o A is attraction o R is repulsion o
ARRA
00 is for the back and forth motion through the zero initial rest position A
stretched spring goes from ldquoArdquo to ldquo0rdquo to ldquoRrdquo to ldquoArdquo to ldquo0rdquo to ldquoRrdquo d1 is the distance a mass is moved from its initial rest position
0
0 1ts A RA K d
Once ( )mad
is performed to move the mass to position d1 (against the restraint of the spring) a of the ( )mad
is transferred into the spring as potential attractionrepulsion (PAR) There is always
more ( )mad
performed than is stored as PAR When the spring is released from the restraining Ats the PAR starts the oscillation Conservation of ( )mad
requires that the PAR and the motion ( )mad
( )( )madM be a constant
( )amp ( ) constantmadPAR M 02
0 0
1
2A R A RPAR K d
02
( ) 0 0
1( )
2mad A R A RM m v
0 0 02 2 2
0 0 0 1
2 2 21
2 2 21
1 1 1amp
2 2 2
1 1 1amp
2 2 2
amp
A R A R A RK d m v K d
Kd mv Kd
Kd mv Kd
0
1d is frac12 the height from the base to the peak of a cycle
An experiment of moving paper under a pendulum through a small angle filled with sand that can escape through a small hole in the bottom shows that a pendulum cycling back and forth follows a curve that is mathematically analogous to a lsquosinrsquo or lsquocosrsquo
The projection of the curved area onto the straight line exactly duplicate the back and forth motion of CM This allows for the math of lsquosinrsquo or lsquocosrsquo to be used
SM will use lsquosinrdquo math It could use lsquocosrsquo or a combination of both lsquosinrsquo and lsquocosrsquo Using just the lsquosinrsquo math makes the understanding and math easier
0 sin 0 0SM 0 sin157 1SM The diagram at the left shows a spring being stretched from its (zero rest) position-2 and the PAR being released at position-1 There are 4 sections (1 to 2 2 to 3 3 to 4 and 4 to5) that alternate between maximum acceleration and zero velocity to max velocity and zero acceleration
At position-1 there is maximum acceleration and zero velocity o The lsquosinrsquo values from 0o to 157o will be used for velocity (values from 0 to 1) o The lsquosinrsquo values from 157o to 0o will be used for acceleration (values from 1 to 0)
At position-2 there is maximum velocity and zero acceleration Acceleration changes to deceleration
o The lsquosinrsquo values from 157o to 0o will be used for deceleration (values from 1 to 0) o The lsquosinrsquo values from 0o to 157o will be used for acceleration (values from 0 to 1)
At position-3 there is maximum acceleration and zero velocity o The lsquosinrsquo values from 0o to 157o will be used for velocity (values from 0 to 1) o The lsquosinrsquo values from 157o to 0o will be used for acceleration (values from 1 to 0)
At position-4 there is maximum velocity and zero acceleration Acceleration changes to deceleration
o The lsquosinrsquo values from 157o to 0o will be used for deceleration (values from 1 to 0) o The lsquosinrsquo values from 0o to 157o will be used for acceleration (values from 0 to 1)
At position-5 the cycle starts again
2
)539(
cT
MK
spring constant (K) when mass (M) and cycle-time (T) are known
)(
))(89)((
111 md
uKgM
d
MG
d
FK sts
spring constant (K) when mass (M) and amplitude ( d1) are known
2 21( )
Kv d d
m
velocity (v) at any point during the cycle
velocity
dforM
Kdv
max
012 velocity (v2) maximum (when d = 0 at position-2 and 4)
acceleration (a1) maximum when spring constant (K) mass (M) and amplitude (d1) are known
acceleration (a1) maximum when amplitude (d1) and cycle-time (Tc) are known
acceleration (a) at any point during the cycle-time (T)
1
151
286
v
dTc
cycle-time (Tc) when amplitude (d1) and velocity (v1) are known
K
MTc 28651
cycle-time (Tc) when mass (M) and spring constant (K) are known
fTc
1
cycle-time (Tc) when the frequency is known
cTf
1
frequency (f) when the cycle-time (Tc) is known
of cycles
elapsed timef
frequency (f) when the number of cycles in an elapsed time is known
d1 amplitude (distance from rest to start position)
Also frac12 the distance from the base to the peak
3131 oror dM
Ka
251
3131 539
c
oror T
da
2539
cT
da
In December of 2006 I discovered the mathematical error in Einsteinrsquos Special Relativity Einstein used BS math and that math only works in an imaginary world
It cannot describe real symmetrical space
Einsteinrsquos Special Relativity should be abandoned and a theory of real space using SM should be used to replace it
Numerous articles on the Internet show Einsteinrsquos theory to be incorrect but no one in academia is willing to publish the information
Figure 1-10 A moving particle observed from two frames of reference in uniform translation
Einsteinrsquos Math Error in Special Relativity Transformation of Velocity (From Eisenberg- Fundamentals of Modern Physics pages 27-29) Consider the particle shown in figure (1-10) moving with velocity v as
seen in a frame of reference O
dx dy dz
Vx Vy Vzdt dt dt
dx dy dz
V x V y V zdy dt dt
We would like to evaluate the velocity v of the particle as seen in the frame of reference 0 which is itself moving relative to 0 with velocity v Measured in the 0 frame the velocity vector of the particle has com ponents
The same velocity vector as measured in the 0 frame has components
2
2
2
2
2
1
1
x vtx
v
cy y
z z
vxt
ctv
c
Now from equations (1-13) we know that the relation between the primed and the unprimed coordinates and times is
Take the differential of these equations remembering that v is a constant This gives
2
2
2
2
2
1
1
dx vdtdx
v
cdy dy
dz dz
vxdt
cdtv
c
2
2
2 2 2 2 2
2
2
2
11 ( )
1( ) 1
1
1
1
dx vdt
v dxdx vdt v
dx dx vdtc dt dtvdx vdx vdx v dxdt dt dt dt
dt dtc c c c
VxV x
Vx
VxV x
vv
vv
c
x
v
c
cV
( )
coordinate of particle relative to moving primed system
coordinate of particle relative to stationary system
velocity of primed coordina
V x
Vx
te system relative to stationaryv sys
tem
Page 28 Fundamentals of Modern Physics
(1-16)
According to Einstein this equation tells you how to transform the observed velocity from one frame of reference to another frame of reference First we note that as Vc and vc approachzero the equation approaches those which would be derived from the Galilean transformation
2 2
( ) 1 01 1 ( )
Vx v Vx vV
VxV x
Vx v
v cVxx
v
c
v
cc
Another property of this equation is that it is impossible to choose V and v such that Vrsquo the magnitude of the velocity which is seen in the new frame is greater than c
1
2
09 ( 09 ) 18009876543
( 09 ) 1811 (09 )
c c cV c c
cc
c
Example As seen by O particle 1 has velocity 09c in the direction of positive x and particle 2 has velocity 09c in the negative x direction
Einstein equation however predicts To evaluate the velocity of particle 1 with respect to particle 2 we transform from the O frame to the Orsquo frame moving in the negative x direction with velocity v = -09c using equation 1-16 We obtain
JK LOGIC will observe the particles separating at 18c
A confusing INCORRECT and ILLOGICAL answer
left is a dir rightsub of a
1
2
1
09 ( 09 ) 18009876543
181( 09 )1 ( 09 )
The multiplication of dir left time a dir right is ONLY a dir left
09
dir
direction leftdirection rightsubtraction
c c cV c c
cc
c
V
dir
22
22
09 18 18009876543
181( 081)1(09)(09)
1
left rightsub
sub
c c c cc c
ccc
c
Analyzing Einsteinrsquos BS math Equation
Using SM math a logical answer can be obtained
Cannot multiply a subtraction operator times a direction in space
Can only subtract and change its direction
Cannot multiply a direction to the left times a direction to the right
Cannot multiply opposite directions in space
Can only solve for the resultant
rightleft
subtraction
1
subtraction
2
sub of a dir to the left is a dir to the right
1 resultant 0 sub 2
2sub 2
2
09 09
09 amp 091
09 09 18 1
(0)1
(09) amp (09)1
c cV
c cc
c c cV
cc
cc
8
181
cc
SM math predicts a logical answer using logic
21st century scientists state in many articles that Einsteins General Relativity and Quantum Mechanics are incompatible
They are not only incompatible they are both mathematically incorrect
They both use illogical BS math
Any equations in math that use imaginary numbers should be abandoned
Imaginary numbers are only necessary in the current BS math system
This means that Schroumldingers equations should be abandoned
Any equation in quantum mechanics that uses imaginary number should be abandoned
Prof Kanarev has developed logical physics and chemistrytheories without the use of Quantum mechanics or relativity
A partial list of his theories coverbull Mechanics ndash replaces Newtonbull Photon - bull Electrons ndash Protons- Neutronsbull Spectroscopybull Nucleusbull Atoms amp Moleculesbull Electrodynamicsbull Heatbull Curvature of Space ndash replaces Schwarzcchildrsquos equation
Kanarevrsquos Hypothesis
bull Electron align linearly with proton They do NOT spin or form clouds around the nucleus
bull Neutrons are in the center of atom Proton are outside of Neutrons
bull The wavelength of all particles is the experimental radius
bull Double slit experiment explained correctly
bull Heisenbergrsquos inequality explained correctly
bull is the radius of approach of the magnetic lines around the torus of the electron
( 1 137)
bull Why water expands when it freezes
bull Electron clusters form lightning
rbull The meaning of Plancks constant is explained
22 2 kg m
h m f mr f consts
2210254 42m r kgm bull If the mr-constant describes a ring the equations
describe experiments bull If the mr-constant describes a wave the equations do
NOT describe experiments
Kanarevrsquos 5 Equations of Mechanics
Kanarevrsquos 1st Equation Acceleration ampmv maMma R R
Kanarevrsquos 2nd Equation Uniform motion amp mv maMma R R
maM mvR R
Kanarevrsquos 3rd Equation
Kanarevrsquos 4th Equation Equal and opposite reaction
1 21 21 2
12
a m
m a m ama
Kanarevrsquos 5th Equation 1 2 3 Mi M M M Mna R R R R R
The sum of the Actions working on a moving body is never equal to zero
Kanarevrsquos Photon model is a rolling ring
His equations accurately describe known experiments with photonsbull Electromagnetic radiation should be called photon radiationbull Diffraction and interference are the same phenomenabull Correctly explains the double slit experiment without quantum equationsbull Correctly explains the meaning of Heisenbergrsquos inequality
e
2 (2)(3142)(2817 15 ) 10007297352568
R 2426 12 137mag
xp
r m
m
Charge
Magnetic action
Proton
The intensity of the magnetic action near the geometrical center of the proton ring is so great that we theorize that it is the strong nuclear force
(9383 8)(1602 19)8476 14
4 (4)(3142)(1411 26)p
pp
EB T
M
(9395 8)(1602 19)1035 15
4 (4)(3142)(0966 26)N
NN
EB T
M
Neutron
1a Hydrogenavi
W h at d o es it m e an to m u ltip ly ( 4) b y (4 ) A s w e k n o w 1
m u ltip lica tio n is ju s t ad d itio n W h a t w o u ld ( 4) tim e (4 ) b e 2
In BS math the answer is by definition -16 unless you square -4 and then it is +16
In SM you cannot multiply arrows going in opposite directions
The middle two terms are not logical You cannot multiply opposite directions
This is a MAJOR error in BS math They multiply a dash [(-) a direction to the left] by a cross [(+) a direction to the right] And then ldquoby definitionrdquo label the answer dash (-) This is illogical and produces incorrect answers
2
2 2
This is Einsteinrsquos math error
in Special relativity
Magnitude 4 Direction 1 Magnitude 3 Direction 2 Magnitude 5 Direction 3 Relative to an observer at 0 4
Magnitude 4 Direction 1 Magnitude 3 Direction 2 Magnitude 5 Direction 3 Relative to an observer at 0 4
Magnitude and Direction (MD) An arrow has both a magnitude andA direction relative to a 2nd objectobserver
bull Magnitude and velocity (MV) relative to a 2nd objectobserver
bull Magnitude And acceleration (MA) relative to a 2nd objectobserver2
v da
t t
Introducing a new symbol the ldquoamprdquo sign The ldquoamprdquo sign means resultant ADD all the Arrows for each direction and then SUBTRACT to find the final resultant
1 1amp 1 2 2amp 2 4 If =
1 amp1 2 2 amp2 4 If =
amp 5 amp 3 2 If gt
amp 3 amp 5= 2 If gt
amp 0 3 amp 3=0 if =
only the resultant equals zero
the vectors do not - they are still there
Does not exist There is no multiplication of arrows
There are two arrows One has a magnitude of 3 One has a magnitude of 5 The maximum these two arrows can be is 8
Only the resultant of arrows exist
An SM arrow has a direction and a magnitude
Arrows and magnitudes are still there even if the resultant is zero
There are no negative SM Arrows
There are no positive SM Arrows
There are no ZERO SM Arrows A zero would negate the definition of an SM Arrow something with no magnitude and no direction cannot be the definition of an arrow that is defined as having magnitude and direction
bull A resultant is the addition and subtraction of the differences in magnitude and direction arrows
A B and C are the arrows (a) A amp B exist(b) A amp B amp C exist(c) A-B Do Not Exist in Sm there are no negatives or positives(d) A+B-C Do Not Exist in SM there are no negatives or positives
There are only resultants when there are multiple arrows
SM Graphs versus BS Graphs
Numbers Raised to Powers
When numbers are raised to powers the negative sides of all BS math graphs are ILLOGICAL and INCORRECT
Incorrect BS Math for 2x
2x
4 16 correct3 8 correct2 4 correct1 2 correct0 1 correct-1 005 incorrect-2 025 incorrect-3 0125 incorrect-4 00625 incorrect
negative side of graph is incorrectpositive side of graph is correct
16
8
42105025012500625
-4 -3 -2 -1 0 1 2 3 4
po
sit
ive
negative positive
2x BS Math
correctincorrect
In BS math the dash sign (-) used in this example represents three different operations They arebull (-) Subtractionbull (-) A direction to the leftbull (-) As an exponent to mean divide by
All positive numbers produce correct results and graphs
All negative numbers produce incorrect results and graphs
A direction of three units to the left (-3) is placed in the equation as an exponent
A direction in space (3 units toward the left) changes a whole number to a fraction by using the dash (-) as an exponent rather than a direction in space
This is illogical and incorrect BS math should have performed this operation as
Unfortunately the answers to problems can come out correct because data points of a real problem fit the curve
However the answers will not be due to logical reasoning
Because the equation produced a curve that real problem data points follow it has been falsely believed that the equations and graphs are correct
1
2
x
Incorrect BK graph for (12)x
(12)x
4 00625 correct3 0125 correct2 025 correct1 1 correct0 1 correct-1 2 incorrect-2 4 incorrect-3 8 incorrect-4 16 incorrect
negative side of graph is incorrectpositive side of graph is correct
00625012502505124
8
16
-5 -4 -3 -2 -1 0 1 2 3 4 5P
osit
ive
negative positive
BS Math for (12)x
correctincorrect
Incorrect BS math and graph for
All graphs are symmetrical with SM
Two BS Math Equations that Produce Incorrect Answers
2 0x x starting equation 2
2
x x x x
x x
Add lsquoxrsquo to both sides
2(1) 1 let lsquoxrsquo = 1
2 1 This is an incorrect answer 2(0)-(0)=0 The only answer that BK math can give is x=0
1
a b starting equation ( )( ) ( )( )a a a b multiply both sides by ( )a
( )( ) ( )( ) ( )( ) ( )( )a a b b a b b b subtract ( )( )b b from both sides
( )( ) ( )( )a b a b a b b factor ( )a b from both sides
( )a b b divide both sides by ( )a b
( )b b b replace ( )a with ( )b 2b b simplify 2(1) (1) Let ( )b = 1
2 1 Again BS math yields and incorrect answer 1
This incorrect result is a problem with the distributive law SM explains how this incorrect answer is obtained and the problem with the distributive law that allows the error
Page 148 ndash149 Barry Mazur ndash Imagining Numbers
3
2
31
Do the computation carefully on paper using the rules we agreed to Then ponder your answer which should be something of a surprise to you donrsquot stop there Think of what your answer might possibly mean or might imply
Solve this equation using the distributive law
JK definitely a surprise the answer wasincorrect and illogical
Gerhard Gade University Professor doing math at Harvard University
Here is the BS math answer that Barry Mazur calculated
2
31
2
31
2
31
1 3 1 3 1 3 3 3 2 2 3 1 3
2 2 2 2 4 4 4 4 4 4 2 2
dash
1 3 1 3 1 31
2 2 2 2 4 4
3 3
4 4
When the distributive law is used with the BS Rule-of-Signs and one of the signs is a (-) dash sign the middle two terms always cancel each other This leads to the incorrect answer of (-1) in the above equation
Barry Mazur did not go back and check his answer with the original equation If he had he would have known that the answer was incorrect
3
1 31
2
check answer by setting equal
3
331 3
12
1 31
2
Take the Cube root of both sides
1 3 2 multiply both sides by 2
3 3 subtract 1 from both sides
2 2
3 3 square both sides
3 9 This is an incorrect answer A simple equation check could have shown that the distributive law using BS math provides an incorrect answer
1
BS Illogical and Incorrect Numbers Raised to Powers
nC D n DC
642 DDD
In BS math only a positive number can have an answer if ldquonrdquo is an even number (246---)
All negative numbers have no solutions
642 DDD
With SM just add a direction arrow and any number gives the correct answer
or n nC D C D
3 3( 3 8 8 or 8 2 or 2 )n D C
( 2 16 C= 16 or 16 4 or 4 )n D
Since a negative number has no square root answer in BS math BS mathematicians invented imaginary numbers
i2=-1
24414 iiC n
These equations are easily solved without the use of imaginary numbers with SM math
If 4 then 4 2x
If 4 then 4 2x
The 360-degree circle is a 4400 years old extremely outdated system based on a year of 12 months of 30 days each that is 360 days
SM will use a more logical system that will have a circle of 6283185307o
d
( )
(deg )
d r radians
d r rees
the arc distance around
the circumference of a circle
d
radiusr
2 2
2 2
d r
dv rd t t
v dda rd t t t
1
d
r
( )av tv
The average angular velocity of a body is the rate of its angular displacement
( )0
liminstt
vt
The instantaneous angular velocity is the limit of the ratio
t
as t approaches zero
( )d
av
d
tv
The average circumference velocity of the body is the rate of its distance traveled d divided by the time t
dvv
r
1 dd vdrv
t t t r r
The equation of the relationship between the angular velocity and the velocity of a point on that body
( )o
av
v v v
t ta
The average angular acceleration is the change in the angular velocity
( )0
liminstt
v
ta
The instantaneous angular acceleration is the
limit of the ration of v
t
as t approaches
zero
daa
r
1
1
d
d s d
vva
t r tv a
r t r
The equation of the relationship between the angular acceleration and the acceleration of a point on the body
21
2v a t
If a body with an initial angular velocity ov has a constant
angular acceleration a it will turn through an angle in a time t
1
Straight Line Symbol Units Rotational Symbol Units Distance d m Angular
Distance radians or degrees
Velocity dt
v m
s Angular Velocity
V
t
radians or degrees
s
Acceleration vat
2ms
Angular Acceleration
VA
t
2
radians or degrees
s
Mass M Kg Mass distribution
dM 2Kgm
Force actionA Ma
2
Kgm
sNewton
Torque
dT M A 2
2
Kgm
s
Momentum Mv Kgm
s
Angular Momentum
dM V 2Kgm
s
Work (Force)(distance)
Ad 2
Kgmm
sNm Joule
Work T
Using AS arrow math to solve and direction in space problems using the equation 1 2
2 2 2 2 2 2( ) amp and ( ) amp 3
4
2 2 2
2 2
( ) amp
(5 ) 4 3
(5 ) 16 9
(5 ) 25
(5 ) (5 )
2 2(5 ) 4 3
(5 ) 16 9
(5 ) 25
(5 ) (5 )
In SM math Pythagorasrsquo theorem provides both magnitude and direction
In BS math Pythagorasrsquo theorem provides magnitude only
time set equal to zero
23
( ) ( 0 )10 1
x t xx
3( ) ( 0)
10 1W t f
In BS math this is an equation for a traveling wave
must use a square term in order to make the curve symmetric
NO SQUARE TERM IS NECESSARY
In SM math when t is set to ldquo0rdquo
This is a major fault in BS math
x f(x)
-4 0019
-3 0033
-2 0073
-1 0273
0 3000
1 0273
2 0073
3 0033
4 0019 -4 -3 -2 -1 0 1 2 3 400
05
10
15
20
25
30
35
BS wave (must use square term)
snap shot of wave at x = 0
Am
plit
ude
lt to gt lt to gt f(ltgt)
lt -4 0073
lt -3 0097
lt -2 0143
lt -1 0273
origin 0 3000
gt 1 0273
gt 2 0143
gt 3 0097
gt 4 0073lt0073 lt0097 lt0143 lt0273
3000
0273gt 0143gt 0097gt 0073gt
Wave moving
snap shot of wave
Am
plit
ude
lt4 lt3 lt2 lt1 0 1gt 2gt 3gt 4gt
3
2
1
In SM NO squared terms are needed
SPRING
An Action is needed to stretch or compress a spring to a starting point where it can execute cycle motion
tsA (ts to start) is the work required to move a mass attached to a spring a distance (d1) from
its initial rest position RAK 0 the spring constant is the stiffness of the spring
o A is attraction o R is repulsion o
ARRA
00 is for the back and forth motion through the zero initial rest position A
stretched spring goes from ldquoArdquo to ldquo0rdquo to ldquoRrdquo to ldquoArdquo to ldquo0rdquo to ldquoRrdquo d1 is the distance a mass is moved from its initial rest position
0
0 1ts A RA K d
Once ( )mad
is performed to move the mass to position d1 (against the restraint of the spring) a of the ( )mad
is transferred into the spring as potential attractionrepulsion (PAR) There is always
more ( )mad
performed than is stored as PAR When the spring is released from the restraining Ats the PAR starts the oscillation Conservation of ( )mad
requires that the PAR and the motion ( )mad
( )( )madM be a constant
( )amp ( ) constantmadPAR M 02
0 0
1
2A R A RPAR K d
02
( ) 0 0
1( )
2mad A R A RM m v
0 0 02 2 2
0 0 0 1
2 2 21
2 2 21
1 1 1amp
2 2 2
1 1 1amp
2 2 2
amp
A R A R A RK d m v K d
Kd mv Kd
Kd mv Kd
0
1d is frac12 the height from the base to the peak of a cycle
An experiment of moving paper under a pendulum through a small angle filled with sand that can escape through a small hole in the bottom shows that a pendulum cycling back and forth follows a curve that is mathematically analogous to a lsquosinrsquo or lsquocosrsquo
The projection of the curved area onto the straight line exactly duplicate the back and forth motion of CM This allows for the math of lsquosinrsquo or lsquocosrsquo to be used
SM will use lsquosinrdquo math It could use lsquocosrsquo or a combination of both lsquosinrsquo and lsquocosrsquo Using just the lsquosinrsquo math makes the understanding and math easier
0 sin 0 0SM 0 sin157 1SM The diagram at the left shows a spring being stretched from its (zero rest) position-2 and the PAR being released at position-1 There are 4 sections (1 to 2 2 to 3 3 to 4 and 4 to5) that alternate between maximum acceleration and zero velocity to max velocity and zero acceleration
At position-1 there is maximum acceleration and zero velocity o The lsquosinrsquo values from 0o to 157o will be used for velocity (values from 0 to 1) o The lsquosinrsquo values from 157o to 0o will be used for acceleration (values from 1 to 0)
At position-2 there is maximum velocity and zero acceleration Acceleration changes to deceleration
o The lsquosinrsquo values from 157o to 0o will be used for deceleration (values from 1 to 0) o The lsquosinrsquo values from 0o to 157o will be used for acceleration (values from 0 to 1)
At position-3 there is maximum acceleration and zero velocity o The lsquosinrsquo values from 0o to 157o will be used for velocity (values from 0 to 1) o The lsquosinrsquo values from 157o to 0o will be used for acceleration (values from 1 to 0)
At position-4 there is maximum velocity and zero acceleration Acceleration changes to deceleration
o The lsquosinrsquo values from 157o to 0o will be used for deceleration (values from 1 to 0) o The lsquosinrsquo values from 0o to 157o will be used for acceleration (values from 0 to 1)
At position-5 the cycle starts again
2
)539(
cT
MK
spring constant (K) when mass (M) and cycle-time (T) are known
)(
))(89)((
111 md
uKgM
d
MG
d
FK sts
spring constant (K) when mass (M) and amplitude ( d1) are known
2 21( )
Kv d d
m
velocity (v) at any point during the cycle
velocity
dforM
Kdv
max
012 velocity (v2) maximum (when d = 0 at position-2 and 4)
acceleration (a1) maximum when spring constant (K) mass (M) and amplitude (d1) are known
acceleration (a1) maximum when amplitude (d1) and cycle-time (Tc) are known
acceleration (a) at any point during the cycle-time (T)
1
151
286
v
dTc
cycle-time (Tc) when amplitude (d1) and velocity (v1) are known
K
MTc 28651
cycle-time (Tc) when mass (M) and spring constant (K) are known
fTc
1
cycle-time (Tc) when the frequency is known
cTf
1
frequency (f) when the cycle-time (Tc) is known
of cycles
elapsed timef
frequency (f) when the number of cycles in an elapsed time is known
d1 amplitude (distance from rest to start position)
Also frac12 the distance from the base to the peak
3131 oror dM
Ka
251
3131 539
c
oror T
da
2539
cT
da
In December of 2006 I discovered the mathematical error in Einsteinrsquos Special Relativity Einstein used BS math and that math only works in an imaginary world
It cannot describe real symmetrical space
Einsteinrsquos Special Relativity should be abandoned and a theory of real space using SM should be used to replace it
Numerous articles on the Internet show Einsteinrsquos theory to be incorrect but no one in academia is willing to publish the information
Figure 1-10 A moving particle observed from two frames of reference in uniform translation
Einsteinrsquos Math Error in Special Relativity Transformation of Velocity (From Eisenberg- Fundamentals of Modern Physics pages 27-29) Consider the particle shown in figure (1-10) moving with velocity v as
seen in a frame of reference O
dx dy dz
Vx Vy Vzdt dt dt
dx dy dz
V x V y V zdy dt dt
We would like to evaluate the velocity v of the particle as seen in the frame of reference 0 which is itself moving relative to 0 with velocity v Measured in the 0 frame the velocity vector of the particle has com ponents
The same velocity vector as measured in the 0 frame has components
2
2
2
2
2
1
1
x vtx
v
cy y
z z
vxt
ctv
c
Now from equations (1-13) we know that the relation between the primed and the unprimed coordinates and times is
Take the differential of these equations remembering that v is a constant This gives
2
2
2
2
2
1
1
dx vdtdx
v
cdy dy
dz dz
vxdt
cdtv
c
2
2
2 2 2 2 2
2
2
2
11 ( )
1( ) 1
1
1
1
dx vdt
v dxdx vdt v
dx dx vdtc dt dtvdx vdx vdx v dxdt dt dt dt
dt dtc c c c
VxV x
Vx
VxV x
vv
vv
c
x
v
c
cV
( )
coordinate of particle relative to moving primed system
coordinate of particle relative to stationary system
velocity of primed coordina
V x
Vx
te system relative to stationaryv sys
tem
Page 28 Fundamentals of Modern Physics
(1-16)
According to Einstein this equation tells you how to transform the observed velocity from one frame of reference to another frame of reference First we note that as Vc and vc approachzero the equation approaches those which would be derived from the Galilean transformation
2 2
( ) 1 01 1 ( )
Vx v Vx vV
VxV x
Vx v
v cVxx
v
c
v
cc
Another property of this equation is that it is impossible to choose V and v such that Vrsquo the magnitude of the velocity which is seen in the new frame is greater than c
1
2
09 ( 09 ) 18009876543
( 09 ) 1811 (09 )
c c cV c c
cc
c
Example As seen by O particle 1 has velocity 09c in the direction of positive x and particle 2 has velocity 09c in the negative x direction
Einstein equation however predicts To evaluate the velocity of particle 1 with respect to particle 2 we transform from the O frame to the Orsquo frame moving in the negative x direction with velocity v = -09c using equation 1-16 We obtain
JK LOGIC will observe the particles separating at 18c
A confusing INCORRECT and ILLOGICAL answer
left is a dir rightsub of a
1
2
1
09 ( 09 ) 18009876543
181( 09 )1 ( 09 )
The multiplication of dir left time a dir right is ONLY a dir left
09
dir
direction leftdirection rightsubtraction
c c cV c c
cc
c
V
dir
22
22
09 18 18009876543
181( 081)1(09)(09)
1
left rightsub
sub
c c c cc c
ccc
c
Analyzing Einsteinrsquos BS math Equation
Using SM math a logical answer can be obtained
Cannot multiply a subtraction operator times a direction in space
Can only subtract and change its direction
Cannot multiply a direction to the left times a direction to the right
Cannot multiply opposite directions in space
Can only solve for the resultant
rightleft
subtraction
1
subtraction
2
sub of a dir to the left is a dir to the right
1 resultant 0 sub 2
2sub 2
2
09 09
09 amp 091
09 09 18 1
(0)1
(09) amp (09)1
c cV
c cc
c c cV
cc
cc
8
181
cc
SM math predicts a logical answer using logic
21st century scientists state in many articles that Einsteins General Relativity and Quantum Mechanics are incompatible
They are not only incompatible they are both mathematically incorrect
They both use illogical BS math
Any equations in math that use imaginary numbers should be abandoned
Imaginary numbers are only necessary in the current BS math system
This means that Schroumldingers equations should be abandoned
Any equation in quantum mechanics that uses imaginary number should be abandoned
Prof Kanarev has developed logical physics and chemistrytheories without the use of Quantum mechanics or relativity
A partial list of his theories coverbull Mechanics ndash replaces Newtonbull Photon - bull Electrons ndash Protons- Neutronsbull Spectroscopybull Nucleusbull Atoms amp Moleculesbull Electrodynamicsbull Heatbull Curvature of Space ndash replaces Schwarzcchildrsquos equation
Kanarevrsquos Hypothesis
bull Electron align linearly with proton They do NOT spin or form clouds around the nucleus
bull Neutrons are in the center of atom Proton are outside of Neutrons
bull The wavelength of all particles is the experimental radius
bull Double slit experiment explained correctly
bull Heisenbergrsquos inequality explained correctly
bull is the radius of approach of the magnetic lines around the torus of the electron
( 1 137)
bull Why water expands when it freezes
bull Electron clusters form lightning
rbull The meaning of Plancks constant is explained
22 2 kg m
h m f mr f consts
2210254 42m r kgm bull If the mr-constant describes a ring the equations
describe experiments bull If the mr-constant describes a wave the equations do
NOT describe experiments
Kanarevrsquos 5 Equations of Mechanics
Kanarevrsquos 1st Equation Acceleration ampmv maMma R R
Kanarevrsquos 2nd Equation Uniform motion amp mv maMma R R
maM mvR R
Kanarevrsquos 3rd Equation
Kanarevrsquos 4th Equation Equal and opposite reaction
1 21 21 2
12
a m
m a m ama
Kanarevrsquos 5th Equation 1 2 3 Mi M M M Mna R R R R R
The sum of the Actions working on a moving body is never equal to zero
Kanarevrsquos Photon model is a rolling ring
His equations accurately describe known experiments with photonsbull Electromagnetic radiation should be called photon radiationbull Diffraction and interference are the same phenomenabull Correctly explains the double slit experiment without quantum equationsbull Correctly explains the meaning of Heisenbergrsquos inequality
e
2 (2)(3142)(2817 15 ) 10007297352568
R 2426 12 137mag
xp
r m
m
Charge
Magnetic action
Proton
The intensity of the magnetic action near the geometrical center of the proton ring is so great that we theorize that it is the strong nuclear force
(9383 8)(1602 19)8476 14
4 (4)(3142)(1411 26)p
pp
EB T
M
(9395 8)(1602 19)1035 15
4 (4)(3142)(0966 26)N
NN
EB T
M
Neutron
1a Hydrogenavi
In SM you cannot multiply arrows going in opposite directions
The middle two terms are not logical You cannot multiply opposite directions
This is a MAJOR error in BS math They multiply a dash [(-) a direction to the left] by a cross [(+) a direction to the right] And then ldquoby definitionrdquo label the answer dash (-) This is illogical and produces incorrect answers
2
2 2
This is Einsteinrsquos math error
in Special relativity
Magnitude 4 Direction 1 Magnitude 3 Direction 2 Magnitude 5 Direction 3 Relative to an observer at 0 4
Magnitude 4 Direction 1 Magnitude 3 Direction 2 Magnitude 5 Direction 3 Relative to an observer at 0 4
Magnitude and Direction (MD) An arrow has both a magnitude andA direction relative to a 2nd objectobserver
bull Magnitude and velocity (MV) relative to a 2nd objectobserver
bull Magnitude And acceleration (MA) relative to a 2nd objectobserver2
v da
t t
Introducing a new symbol the ldquoamprdquo sign The ldquoamprdquo sign means resultant ADD all the Arrows for each direction and then SUBTRACT to find the final resultant
1 1amp 1 2 2amp 2 4 If =
1 amp1 2 2 amp2 4 If =
amp 5 amp 3 2 If gt
amp 3 amp 5= 2 If gt
amp 0 3 amp 3=0 if =
only the resultant equals zero
the vectors do not - they are still there
Does not exist There is no multiplication of arrows
There are two arrows One has a magnitude of 3 One has a magnitude of 5 The maximum these two arrows can be is 8
Only the resultant of arrows exist
An SM arrow has a direction and a magnitude
Arrows and magnitudes are still there even if the resultant is zero
There are no negative SM Arrows
There are no positive SM Arrows
There are no ZERO SM Arrows A zero would negate the definition of an SM Arrow something with no magnitude and no direction cannot be the definition of an arrow that is defined as having magnitude and direction
bull A resultant is the addition and subtraction of the differences in magnitude and direction arrows
A B and C are the arrows (a) A amp B exist(b) A amp B amp C exist(c) A-B Do Not Exist in Sm there are no negatives or positives(d) A+B-C Do Not Exist in SM there are no negatives or positives
There are only resultants when there are multiple arrows
SM Graphs versus BS Graphs
Numbers Raised to Powers
When numbers are raised to powers the negative sides of all BS math graphs are ILLOGICAL and INCORRECT
Incorrect BS Math for 2x
2x
4 16 correct3 8 correct2 4 correct1 2 correct0 1 correct-1 005 incorrect-2 025 incorrect-3 0125 incorrect-4 00625 incorrect
negative side of graph is incorrectpositive side of graph is correct
16
8
42105025012500625
-4 -3 -2 -1 0 1 2 3 4
po
sit
ive
negative positive
2x BS Math
correctincorrect
In BS math the dash sign (-) used in this example represents three different operations They arebull (-) Subtractionbull (-) A direction to the leftbull (-) As an exponent to mean divide by
All positive numbers produce correct results and graphs
All negative numbers produce incorrect results and graphs
A direction of three units to the left (-3) is placed in the equation as an exponent
A direction in space (3 units toward the left) changes a whole number to a fraction by using the dash (-) as an exponent rather than a direction in space
This is illogical and incorrect BS math should have performed this operation as
Unfortunately the answers to problems can come out correct because data points of a real problem fit the curve
However the answers will not be due to logical reasoning
Because the equation produced a curve that real problem data points follow it has been falsely believed that the equations and graphs are correct
1
2
x
Incorrect BK graph for (12)x
(12)x
4 00625 correct3 0125 correct2 025 correct1 1 correct0 1 correct-1 2 incorrect-2 4 incorrect-3 8 incorrect-4 16 incorrect
negative side of graph is incorrectpositive side of graph is correct
00625012502505124
8
16
-5 -4 -3 -2 -1 0 1 2 3 4 5P
osit
ive
negative positive
BS Math for (12)x
correctincorrect
Incorrect BS math and graph for
All graphs are symmetrical with SM
Two BS Math Equations that Produce Incorrect Answers
2 0x x starting equation 2
2
x x x x
x x
Add lsquoxrsquo to both sides
2(1) 1 let lsquoxrsquo = 1
2 1 This is an incorrect answer 2(0)-(0)=0 The only answer that BK math can give is x=0
1
a b starting equation ( )( ) ( )( )a a a b multiply both sides by ( )a
( )( ) ( )( ) ( )( ) ( )( )a a b b a b b b subtract ( )( )b b from both sides
( )( ) ( )( )a b a b a b b factor ( )a b from both sides
( )a b b divide both sides by ( )a b
( )b b b replace ( )a with ( )b 2b b simplify 2(1) (1) Let ( )b = 1
2 1 Again BS math yields and incorrect answer 1
This incorrect result is a problem with the distributive law SM explains how this incorrect answer is obtained and the problem with the distributive law that allows the error
Page 148 ndash149 Barry Mazur ndash Imagining Numbers
3
2
31
Do the computation carefully on paper using the rules we agreed to Then ponder your answer which should be something of a surprise to you donrsquot stop there Think of what your answer might possibly mean or might imply
Solve this equation using the distributive law
JK definitely a surprise the answer wasincorrect and illogical
Gerhard Gade University Professor doing math at Harvard University
Here is the BS math answer that Barry Mazur calculated
2
31
2
31
2
31
1 3 1 3 1 3 3 3 2 2 3 1 3
2 2 2 2 4 4 4 4 4 4 2 2
dash
1 3 1 3 1 31
2 2 2 2 4 4
3 3
4 4
When the distributive law is used with the BS Rule-of-Signs and one of the signs is a (-) dash sign the middle two terms always cancel each other This leads to the incorrect answer of (-1) in the above equation
Barry Mazur did not go back and check his answer with the original equation If he had he would have known that the answer was incorrect
3
1 31
2
check answer by setting equal
3
331 3
12
1 31
2
Take the Cube root of both sides
1 3 2 multiply both sides by 2
3 3 subtract 1 from both sides
2 2
3 3 square both sides
3 9 This is an incorrect answer A simple equation check could have shown that the distributive law using BS math provides an incorrect answer
1
BS Illogical and Incorrect Numbers Raised to Powers
nC D n DC
642 DDD
In BS math only a positive number can have an answer if ldquonrdquo is an even number (246---)
All negative numbers have no solutions
642 DDD
With SM just add a direction arrow and any number gives the correct answer
or n nC D C D
3 3( 3 8 8 or 8 2 or 2 )n D C
( 2 16 C= 16 or 16 4 or 4 )n D
Since a negative number has no square root answer in BS math BS mathematicians invented imaginary numbers
i2=-1
24414 iiC n
These equations are easily solved without the use of imaginary numbers with SM math
If 4 then 4 2x
If 4 then 4 2x
The 360-degree circle is a 4400 years old extremely outdated system based on a year of 12 months of 30 days each that is 360 days
SM will use a more logical system that will have a circle of 6283185307o
d
( )
(deg )
d r radians
d r rees
the arc distance around
the circumference of a circle
d
radiusr
2 2
2 2
d r
dv rd t t
v dda rd t t t
1
d
r
( )av tv
The average angular velocity of a body is the rate of its angular displacement
( )0
liminstt
vt
The instantaneous angular velocity is the limit of the ratio
t
as t approaches zero
( )d
av
d
tv
The average circumference velocity of the body is the rate of its distance traveled d divided by the time t
dvv
r
1 dd vdrv
t t t r r
The equation of the relationship between the angular velocity and the velocity of a point on that body
( )o
av
v v v
t ta
The average angular acceleration is the change in the angular velocity
( )0
liminstt
v
ta
The instantaneous angular acceleration is the
limit of the ration of v
t
as t approaches
zero
daa
r
1
1
d
d s d
vva
t r tv a
r t r
The equation of the relationship between the angular acceleration and the acceleration of a point on the body
21
2v a t
If a body with an initial angular velocity ov has a constant
angular acceleration a it will turn through an angle in a time t
1
Straight Line Symbol Units Rotational Symbol Units Distance d m Angular
Distance radians or degrees
Velocity dt
v m
s Angular Velocity
V
t
radians or degrees
s
Acceleration vat
2ms
Angular Acceleration
VA
t
2
radians or degrees
s
Mass M Kg Mass distribution
dM 2Kgm
Force actionA Ma
2
Kgm
sNewton
Torque
dT M A 2
2
Kgm
s
Momentum Mv Kgm
s
Angular Momentum
dM V 2Kgm
s
Work (Force)(distance)
Ad 2
Kgmm
sNm Joule
Work T
Using AS arrow math to solve and direction in space problems using the equation 1 2
2 2 2 2 2 2( ) amp and ( ) amp 3
4
2 2 2
2 2
( ) amp
(5 ) 4 3
(5 ) 16 9
(5 ) 25
(5 ) (5 )
2 2(5 ) 4 3
(5 ) 16 9
(5 ) 25
(5 ) (5 )
In SM math Pythagorasrsquo theorem provides both magnitude and direction
In BS math Pythagorasrsquo theorem provides magnitude only
time set equal to zero
23
( ) ( 0 )10 1
x t xx
3( ) ( 0)
10 1W t f
In BS math this is an equation for a traveling wave
must use a square term in order to make the curve symmetric
NO SQUARE TERM IS NECESSARY
In SM math when t is set to ldquo0rdquo
This is a major fault in BS math
x f(x)
-4 0019
-3 0033
-2 0073
-1 0273
0 3000
1 0273
2 0073
3 0033
4 0019 -4 -3 -2 -1 0 1 2 3 400
05
10
15
20
25
30
35
BS wave (must use square term)
snap shot of wave at x = 0
Am
plit
ude
lt to gt lt to gt f(ltgt)
lt -4 0073
lt -3 0097
lt -2 0143
lt -1 0273
origin 0 3000
gt 1 0273
gt 2 0143
gt 3 0097
gt 4 0073lt0073 lt0097 lt0143 lt0273
3000
0273gt 0143gt 0097gt 0073gt
Wave moving
snap shot of wave
Am
plit
ude
lt4 lt3 lt2 lt1 0 1gt 2gt 3gt 4gt
3
2
1
In SM NO squared terms are needed
SPRING
An Action is needed to stretch or compress a spring to a starting point where it can execute cycle motion
tsA (ts to start) is the work required to move a mass attached to a spring a distance (d1) from
its initial rest position RAK 0 the spring constant is the stiffness of the spring
o A is attraction o R is repulsion o
ARRA
00 is for the back and forth motion through the zero initial rest position A
stretched spring goes from ldquoArdquo to ldquo0rdquo to ldquoRrdquo to ldquoArdquo to ldquo0rdquo to ldquoRrdquo d1 is the distance a mass is moved from its initial rest position
0
0 1ts A RA K d
Once ( )mad
is performed to move the mass to position d1 (against the restraint of the spring) a of the ( )mad
is transferred into the spring as potential attractionrepulsion (PAR) There is always
more ( )mad
performed than is stored as PAR When the spring is released from the restraining Ats the PAR starts the oscillation Conservation of ( )mad
requires that the PAR and the motion ( )mad
( )( )madM be a constant
( )amp ( ) constantmadPAR M 02
0 0
1
2A R A RPAR K d
02
( ) 0 0
1( )
2mad A R A RM m v
0 0 02 2 2
0 0 0 1
2 2 21
2 2 21
1 1 1amp
2 2 2
1 1 1amp
2 2 2
amp
A R A R A RK d m v K d
Kd mv Kd
Kd mv Kd
0
1d is frac12 the height from the base to the peak of a cycle
An experiment of moving paper under a pendulum through a small angle filled with sand that can escape through a small hole in the bottom shows that a pendulum cycling back and forth follows a curve that is mathematically analogous to a lsquosinrsquo or lsquocosrsquo
The projection of the curved area onto the straight line exactly duplicate the back and forth motion of CM This allows for the math of lsquosinrsquo or lsquocosrsquo to be used
SM will use lsquosinrdquo math It could use lsquocosrsquo or a combination of both lsquosinrsquo and lsquocosrsquo Using just the lsquosinrsquo math makes the understanding and math easier
0 sin 0 0SM 0 sin157 1SM The diagram at the left shows a spring being stretched from its (zero rest) position-2 and the PAR being released at position-1 There are 4 sections (1 to 2 2 to 3 3 to 4 and 4 to5) that alternate between maximum acceleration and zero velocity to max velocity and zero acceleration
At position-1 there is maximum acceleration and zero velocity o The lsquosinrsquo values from 0o to 157o will be used for velocity (values from 0 to 1) o The lsquosinrsquo values from 157o to 0o will be used for acceleration (values from 1 to 0)
At position-2 there is maximum velocity and zero acceleration Acceleration changes to deceleration
o The lsquosinrsquo values from 157o to 0o will be used for deceleration (values from 1 to 0) o The lsquosinrsquo values from 0o to 157o will be used for acceleration (values from 0 to 1)
At position-3 there is maximum acceleration and zero velocity o The lsquosinrsquo values from 0o to 157o will be used for velocity (values from 0 to 1) o The lsquosinrsquo values from 157o to 0o will be used for acceleration (values from 1 to 0)
At position-4 there is maximum velocity and zero acceleration Acceleration changes to deceleration
o The lsquosinrsquo values from 157o to 0o will be used for deceleration (values from 1 to 0) o The lsquosinrsquo values from 0o to 157o will be used for acceleration (values from 0 to 1)
At position-5 the cycle starts again
2
)539(
cT
MK
spring constant (K) when mass (M) and cycle-time (T) are known
)(
))(89)((
111 md
uKgM
d
MG
d
FK sts
spring constant (K) when mass (M) and amplitude ( d1) are known
2 21( )
Kv d d
m
velocity (v) at any point during the cycle
velocity
dforM
Kdv
max
012 velocity (v2) maximum (when d = 0 at position-2 and 4)
acceleration (a1) maximum when spring constant (K) mass (M) and amplitude (d1) are known
acceleration (a1) maximum when amplitude (d1) and cycle-time (Tc) are known
acceleration (a) at any point during the cycle-time (T)
1
151
286
v
dTc
cycle-time (Tc) when amplitude (d1) and velocity (v1) are known
K
MTc 28651
cycle-time (Tc) when mass (M) and spring constant (K) are known
fTc
1
cycle-time (Tc) when the frequency is known
cTf
1
frequency (f) when the cycle-time (Tc) is known
of cycles
elapsed timef
frequency (f) when the number of cycles in an elapsed time is known
d1 amplitude (distance from rest to start position)
Also frac12 the distance from the base to the peak
3131 oror dM
Ka
251
3131 539
c
oror T
da
2539
cT
da
In December of 2006 I discovered the mathematical error in Einsteinrsquos Special Relativity Einstein used BS math and that math only works in an imaginary world
It cannot describe real symmetrical space
Einsteinrsquos Special Relativity should be abandoned and a theory of real space using SM should be used to replace it
Numerous articles on the Internet show Einsteinrsquos theory to be incorrect but no one in academia is willing to publish the information
Figure 1-10 A moving particle observed from two frames of reference in uniform translation
Einsteinrsquos Math Error in Special Relativity Transformation of Velocity (From Eisenberg- Fundamentals of Modern Physics pages 27-29) Consider the particle shown in figure (1-10) moving with velocity v as
seen in a frame of reference O
dx dy dz
Vx Vy Vzdt dt dt
dx dy dz
V x V y V zdy dt dt
We would like to evaluate the velocity v of the particle as seen in the frame of reference 0 which is itself moving relative to 0 with velocity v Measured in the 0 frame the velocity vector of the particle has com ponents
The same velocity vector as measured in the 0 frame has components
2
2
2
2
2
1
1
x vtx
v
cy y
z z
vxt
ctv
c
Now from equations (1-13) we know that the relation between the primed and the unprimed coordinates and times is
Take the differential of these equations remembering that v is a constant This gives
2
2
2
2
2
1
1
dx vdtdx
v
cdy dy
dz dz
vxdt
cdtv
c
2
2
2 2 2 2 2
2
2
2
11 ( )
1( ) 1
1
1
1
dx vdt
v dxdx vdt v
dx dx vdtc dt dtvdx vdx vdx v dxdt dt dt dt
dt dtc c c c
VxV x
Vx
VxV x
vv
vv
c
x
v
c
cV
( )
coordinate of particle relative to moving primed system
coordinate of particle relative to stationary system
velocity of primed coordina
V x
Vx
te system relative to stationaryv sys
tem
Page 28 Fundamentals of Modern Physics
(1-16)
According to Einstein this equation tells you how to transform the observed velocity from one frame of reference to another frame of reference First we note that as Vc and vc approachzero the equation approaches those which would be derived from the Galilean transformation
2 2
( ) 1 01 1 ( )
Vx v Vx vV
VxV x
Vx v
v cVxx
v
c
v
cc
Another property of this equation is that it is impossible to choose V and v such that Vrsquo the magnitude of the velocity which is seen in the new frame is greater than c
1
2
09 ( 09 ) 18009876543
( 09 ) 1811 (09 )
c c cV c c
cc
c
Example As seen by O particle 1 has velocity 09c in the direction of positive x and particle 2 has velocity 09c in the negative x direction
Einstein equation however predicts To evaluate the velocity of particle 1 with respect to particle 2 we transform from the O frame to the Orsquo frame moving in the negative x direction with velocity v = -09c using equation 1-16 We obtain
JK LOGIC will observe the particles separating at 18c
A confusing INCORRECT and ILLOGICAL answer
left is a dir rightsub of a
1
2
1
09 ( 09 ) 18009876543
181( 09 )1 ( 09 )
The multiplication of dir left time a dir right is ONLY a dir left
09
dir
direction leftdirection rightsubtraction
c c cV c c
cc
c
V
dir
22
22
09 18 18009876543
181( 081)1(09)(09)
1
left rightsub
sub
c c c cc c
ccc
c
Analyzing Einsteinrsquos BS math Equation
Using SM math a logical answer can be obtained
Cannot multiply a subtraction operator times a direction in space
Can only subtract and change its direction
Cannot multiply a direction to the left times a direction to the right
Cannot multiply opposite directions in space
Can only solve for the resultant
rightleft
subtraction
1
subtraction
2
sub of a dir to the left is a dir to the right
1 resultant 0 sub 2
2sub 2
2
09 09
09 amp 091
09 09 18 1
(0)1
(09) amp (09)1
c cV
c cc
c c cV
cc
cc
8
181
cc
SM math predicts a logical answer using logic
21st century scientists state in many articles that Einsteins General Relativity and Quantum Mechanics are incompatible
They are not only incompatible they are both mathematically incorrect
They both use illogical BS math
Any equations in math that use imaginary numbers should be abandoned
Imaginary numbers are only necessary in the current BS math system
This means that Schroumldingers equations should be abandoned
Any equation in quantum mechanics that uses imaginary number should be abandoned
Prof Kanarev has developed logical physics and chemistrytheories without the use of Quantum mechanics or relativity
A partial list of his theories coverbull Mechanics ndash replaces Newtonbull Photon - bull Electrons ndash Protons- Neutronsbull Spectroscopybull Nucleusbull Atoms amp Moleculesbull Electrodynamicsbull Heatbull Curvature of Space ndash replaces Schwarzcchildrsquos equation
Kanarevrsquos Hypothesis
bull Electron align linearly with proton They do NOT spin or form clouds around the nucleus
bull Neutrons are in the center of atom Proton are outside of Neutrons
bull The wavelength of all particles is the experimental radius
bull Double slit experiment explained correctly
bull Heisenbergrsquos inequality explained correctly
bull is the radius of approach of the magnetic lines around the torus of the electron
( 1 137)
bull Why water expands when it freezes
bull Electron clusters form lightning
rbull The meaning of Plancks constant is explained
22 2 kg m
h m f mr f consts
2210254 42m r kgm bull If the mr-constant describes a ring the equations
describe experiments bull If the mr-constant describes a wave the equations do
NOT describe experiments
Kanarevrsquos 5 Equations of Mechanics
Kanarevrsquos 1st Equation Acceleration ampmv maMma R R
Kanarevrsquos 2nd Equation Uniform motion amp mv maMma R R
maM mvR R
Kanarevrsquos 3rd Equation
Kanarevrsquos 4th Equation Equal and opposite reaction
1 21 21 2
12
a m
m a m ama
Kanarevrsquos 5th Equation 1 2 3 Mi M M M Mna R R R R R
The sum of the Actions working on a moving body is never equal to zero
Kanarevrsquos Photon model is a rolling ring
His equations accurately describe known experiments with photonsbull Electromagnetic radiation should be called photon radiationbull Diffraction and interference are the same phenomenabull Correctly explains the double slit experiment without quantum equationsbull Correctly explains the meaning of Heisenbergrsquos inequality
e
2 (2)(3142)(2817 15 ) 10007297352568
R 2426 12 137mag
xp
r m
m
Charge
Magnetic action
Proton
The intensity of the magnetic action near the geometrical center of the proton ring is so great that we theorize that it is the strong nuclear force
(9383 8)(1602 19)8476 14
4 (4)(3142)(1411 26)p
pp
EB T
M
(9395 8)(1602 19)1035 15
4 (4)(3142)(0966 26)N
NN
EB T
M
Neutron
1a Hydrogenavi
This is Einsteinrsquos math error
in Special relativity
Magnitude 4 Direction 1 Magnitude 3 Direction 2 Magnitude 5 Direction 3 Relative to an observer at 0 4
Magnitude 4 Direction 1 Magnitude 3 Direction 2 Magnitude 5 Direction 3 Relative to an observer at 0 4
Magnitude and Direction (MD) An arrow has both a magnitude andA direction relative to a 2nd objectobserver
bull Magnitude and velocity (MV) relative to a 2nd objectobserver
bull Magnitude And acceleration (MA) relative to a 2nd objectobserver2
v da
t t
Introducing a new symbol the ldquoamprdquo sign The ldquoamprdquo sign means resultant ADD all the Arrows for each direction and then SUBTRACT to find the final resultant
1 1amp 1 2 2amp 2 4 If =
1 amp1 2 2 amp2 4 If =
amp 5 amp 3 2 If gt
amp 3 amp 5= 2 If gt
amp 0 3 amp 3=0 if =
only the resultant equals zero
the vectors do not - they are still there
Does not exist There is no multiplication of arrows
There are two arrows One has a magnitude of 3 One has a magnitude of 5 The maximum these two arrows can be is 8
Only the resultant of arrows exist
An SM arrow has a direction and a magnitude
Arrows and magnitudes are still there even if the resultant is zero
There are no negative SM Arrows
There are no positive SM Arrows
There are no ZERO SM Arrows A zero would negate the definition of an SM Arrow something with no magnitude and no direction cannot be the definition of an arrow that is defined as having magnitude and direction
bull A resultant is the addition and subtraction of the differences in magnitude and direction arrows
A B and C are the arrows (a) A amp B exist(b) A amp B amp C exist(c) A-B Do Not Exist in Sm there are no negatives or positives(d) A+B-C Do Not Exist in SM there are no negatives or positives
There are only resultants when there are multiple arrows
SM Graphs versus BS Graphs
Numbers Raised to Powers
When numbers are raised to powers the negative sides of all BS math graphs are ILLOGICAL and INCORRECT
Incorrect BS Math for 2x
2x
4 16 correct3 8 correct2 4 correct1 2 correct0 1 correct-1 005 incorrect-2 025 incorrect-3 0125 incorrect-4 00625 incorrect
negative side of graph is incorrectpositive side of graph is correct
16
8
42105025012500625
-4 -3 -2 -1 0 1 2 3 4
po
sit
ive
negative positive
2x BS Math
correctincorrect
In BS math the dash sign (-) used in this example represents three different operations They arebull (-) Subtractionbull (-) A direction to the leftbull (-) As an exponent to mean divide by
All positive numbers produce correct results and graphs
All negative numbers produce incorrect results and graphs
A direction of three units to the left (-3) is placed in the equation as an exponent
A direction in space (3 units toward the left) changes a whole number to a fraction by using the dash (-) as an exponent rather than a direction in space
This is illogical and incorrect BS math should have performed this operation as
Unfortunately the answers to problems can come out correct because data points of a real problem fit the curve
However the answers will not be due to logical reasoning
Because the equation produced a curve that real problem data points follow it has been falsely believed that the equations and graphs are correct
1
2
x
Incorrect BK graph for (12)x
(12)x
4 00625 correct3 0125 correct2 025 correct1 1 correct0 1 correct-1 2 incorrect-2 4 incorrect-3 8 incorrect-4 16 incorrect
negative side of graph is incorrectpositive side of graph is correct
00625012502505124
8
16
-5 -4 -3 -2 -1 0 1 2 3 4 5P
osit
ive
negative positive
BS Math for (12)x
correctincorrect
Incorrect BS math and graph for
All graphs are symmetrical with SM
Two BS Math Equations that Produce Incorrect Answers
2 0x x starting equation 2
2
x x x x
x x
Add lsquoxrsquo to both sides
2(1) 1 let lsquoxrsquo = 1
2 1 This is an incorrect answer 2(0)-(0)=0 The only answer that BK math can give is x=0
1
a b starting equation ( )( ) ( )( )a a a b multiply both sides by ( )a
( )( ) ( )( ) ( )( ) ( )( )a a b b a b b b subtract ( )( )b b from both sides
( )( ) ( )( )a b a b a b b factor ( )a b from both sides
( )a b b divide both sides by ( )a b
( )b b b replace ( )a with ( )b 2b b simplify 2(1) (1) Let ( )b = 1
2 1 Again BS math yields and incorrect answer 1
This incorrect result is a problem with the distributive law SM explains how this incorrect answer is obtained and the problem with the distributive law that allows the error
Page 148 ndash149 Barry Mazur ndash Imagining Numbers
3
2
31
Do the computation carefully on paper using the rules we agreed to Then ponder your answer which should be something of a surprise to you donrsquot stop there Think of what your answer might possibly mean or might imply
Solve this equation using the distributive law
JK definitely a surprise the answer wasincorrect and illogical
Gerhard Gade University Professor doing math at Harvard University
Here is the BS math answer that Barry Mazur calculated
2
31
2
31
2
31
1 3 1 3 1 3 3 3 2 2 3 1 3
2 2 2 2 4 4 4 4 4 4 2 2
dash
1 3 1 3 1 31
2 2 2 2 4 4
3 3
4 4
When the distributive law is used with the BS Rule-of-Signs and one of the signs is a (-) dash sign the middle two terms always cancel each other This leads to the incorrect answer of (-1) in the above equation
Barry Mazur did not go back and check his answer with the original equation If he had he would have known that the answer was incorrect
3
1 31
2
check answer by setting equal
3
331 3
12
1 31
2
Take the Cube root of both sides
1 3 2 multiply both sides by 2
3 3 subtract 1 from both sides
2 2
3 3 square both sides
3 9 This is an incorrect answer A simple equation check could have shown that the distributive law using BS math provides an incorrect answer
1
BS Illogical and Incorrect Numbers Raised to Powers
nC D n DC
642 DDD
In BS math only a positive number can have an answer if ldquonrdquo is an even number (246---)
All negative numbers have no solutions
642 DDD
With SM just add a direction arrow and any number gives the correct answer
or n nC D C D
3 3( 3 8 8 or 8 2 or 2 )n D C
( 2 16 C= 16 or 16 4 or 4 )n D
Since a negative number has no square root answer in BS math BS mathematicians invented imaginary numbers
i2=-1
24414 iiC n
These equations are easily solved without the use of imaginary numbers with SM math
If 4 then 4 2x
If 4 then 4 2x
The 360-degree circle is a 4400 years old extremely outdated system based on a year of 12 months of 30 days each that is 360 days
SM will use a more logical system that will have a circle of 6283185307o
d
( )
(deg )
d r radians
d r rees
the arc distance around
the circumference of a circle
d
radiusr
2 2
2 2
d r
dv rd t t
v dda rd t t t
1
d
r
( )av tv
The average angular velocity of a body is the rate of its angular displacement
( )0
liminstt
vt
The instantaneous angular velocity is the limit of the ratio
t
as t approaches zero
( )d
av
d
tv
The average circumference velocity of the body is the rate of its distance traveled d divided by the time t
dvv
r
1 dd vdrv
t t t r r
The equation of the relationship between the angular velocity and the velocity of a point on that body
( )o
av
v v v
t ta
The average angular acceleration is the change in the angular velocity
( )0
liminstt
v
ta
The instantaneous angular acceleration is the
limit of the ration of v
t
as t approaches
zero
daa
r
1
1
d
d s d
vva
t r tv a
r t r
The equation of the relationship between the angular acceleration and the acceleration of a point on the body
21
2v a t
If a body with an initial angular velocity ov has a constant
angular acceleration a it will turn through an angle in a time t
1
Straight Line Symbol Units Rotational Symbol Units Distance d m Angular
Distance radians or degrees
Velocity dt
v m
s Angular Velocity
V
t
radians or degrees
s
Acceleration vat
2ms
Angular Acceleration
VA
t
2
radians or degrees
s
Mass M Kg Mass distribution
dM 2Kgm
Force actionA Ma
2
Kgm
sNewton
Torque
dT M A 2
2
Kgm
s
Momentum Mv Kgm
s
Angular Momentum
dM V 2Kgm
s
Work (Force)(distance)
Ad 2
Kgmm
sNm Joule
Work T
Using AS arrow math to solve and direction in space problems using the equation 1 2
2 2 2 2 2 2( ) amp and ( ) amp 3
4
2 2 2
2 2
( ) amp
(5 ) 4 3
(5 ) 16 9
(5 ) 25
(5 ) (5 )
2 2(5 ) 4 3
(5 ) 16 9
(5 ) 25
(5 ) (5 )
In SM math Pythagorasrsquo theorem provides both magnitude and direction
In BS math Pythagorasrsquo theorem provides magnitude only
time set equal to zero
23
( ) ( 0 )10 1
x t xx
3( ) ( 0)
10 1W t f
In BS math this is an equation for a traveling wave
must use a square term in order to make the curve symmetric
NO SQUARE TERM IS NECESSARY
In SM math when t is set to ldquo0rdquo
This is a major fault in BS math
x f(x)
-4 0019
-3 0033
-2 0073
-1 0273
0 3000
1 0273
2 0073
3 0033
4 0019 -4 -3 -2 -1 0 1 2 3 400
05
10
15
20
25
30
35
BS wave (must use square term)
snap shot of wave at x = 0
Am
plit
ude
lt to gt lt to gt f(ltgt)
lt -4 0073
lt -3 0097
lt -2 0143
lt -1 0273
origin 0 3000
gt 1 0273
gt 2 0143
gt 3 0097
gt 4 0073lt0073 lt0097 lt0143 lt0273
3000
0273gt 0143gt 0097gt 0073gt
Wave moving
snap shot of wave
Am
plit
ude
lt4 lt3 lt2 lt1 0 1gt 2gt 3gt 4gt
3
2
1
In SM NO squared terms are needed
SPRING
An Action is needed to stretch or compress a spring to a starting point where it can execute cycle motion
tsA (ts to start) is the work required to move a mass attached to a spring a distance (d1) from
its initial rest position RAK 0 the spring constant is the stiffness of the spring
o A is attraction o R is repulsion o
ARRA
00 is for the back and forth motion through the zero initial rest position A
stretched spring goes from ldquoArdquo to ldquo0rdquo to ldquoRrdquo to ldquoArdquo to ldquo0rdquo to ldquoRrdquo d1 is the distance a mass is moved from its initial rest position
0
0 1ts A RA K d
Once ( )mad
is performed to move the mass to position d1 (against the restraint of the spring) a of the ( )mad
is transferred into the spring as potential attractionrepulsion (PAR) There is always
more ( )mad
performed than is stored as PAR When the spring is released from the restraining Ats the PAR starts the oscillation Conservation of ( )mad
requires that the PAR and the motion ( )mad
( )( )madM be a constant
( )amp ( ) constantmadPAR M 02
0 0
1
2A R A RPAR K d
02
( ) 0 0
1( )
2mad A R A RM m v
0 0 02 2 2
0 0 0 1
2 2 21
2 2 21
1 1 1amp
2 2 2
1 1 1amp
2 2 2
amp
A R A R A RK d m v K d
Kd mv Kd
Kd mv Kd
0
1d is frac12 the height from the base to the peak of a cycle
An experiment of moving paper under a pendulum through a small angle filled with sand that can escape through a small hole in the bottom shows that a pendulum cycling back and forth follows a curve that is mathematically analogous to a lsquosinrsquo or lsquocosrsquo
The projection of the curved area onto the straight line exactly duplicate the back and forth motion of CM This allows for the math of lsquosinrsquo or lsquocosrsquo to be used
SM will use lsquosinrdquo math It could use lsquocosrsquo or a combination of both lsquosinrsquo and lsquocosrsquo Using just the lsquosinrsquo math makes the understanding and math easier
0 sin 0 0SM 0 sin157 1SM The diagram at the left shows a spring being stretched from its (zero rest) position-2 and the PAR being released at position-1 There are 4 sections (1 to 2 2 to 3 3 to 4 and 4 to5) that alternate between maximum acceleration and zero velocity to max velocity and zero acceleration
At position-1 there is maximum acceleration and zero velocity o The lsquosinrsquo values from 0o to 157o will be used for velocity (values from 0 to 1) o The lsquosinrsquo values from 157o to 0o will be used for acceleration (values from 1 to 0)
At position-2 there is maximum velocity and zero acceleration Acceleration changes to deceleration
o The lsquosinrsquo values from 157o to 0o will be used for deceleration (values from 1 to 0) o The lsquosinrsquo values from 0o to 157o will be used for acceleration (values from 0 to 1)
At position-3 there is maximum acceleration and zero velocity o The lsquosinrsquo values from 0o to 157o will be used for velocity (values from 0 to 1) o The lsquosinrsquo values from 157o to 0o will be used for acceleration (values from 1 to 0)
At position-4 there is maximum velocity and zero acceleration Acceleration changes to deceleration
o The lsquosinrsquo values from 157o to 0o will be used for deceleration (values from 1 to 0) o The lsquosinrsquo values from 0o to 157o will be used for acceleration (values from 0 to 1)
At position-5 the cycle starts again
2
)539(
cT
MK
spring constant (K) when mass (M) and cycle-time (T) are known
)(
))(89)((
111 md
uKgM
d
MG
d
FK sts
spring constant (K) when mass (M) and amplitude ( d1) are known
2 21( )
Kv d d
m
velocity (v) at any point during the cycle
velocity
dforM
Kdv
max
012 velocity (v2) maximum (when d = 0 at position-2 and 4)
acceleration (a1) maximum when spring constant (K) mass (M) and amplitude (d1) are known
acceleration (a1) maximum when amplitude (d1) and cycle-time (Tc) are known
acceleration (a) at any point during the cycle-time (T)
1
151
286
v
dTc
cycle-time (Tc) when amplitude (d1) and velocity (v1) are known
K
MTc 28651
cycle-time (Tc) when mass (M) and spring constant (K) are known
fTc
1
cycle-time (Tc) when the frequency is known
cTf
1
frequency (f) when the cycle-time (Tc) is known
of cycles
elapsed timef
frequency (f) when the number of cycles in an elapsed time is known
d1 amplitude (distance from rest to start position)
Also frac12 the distance from the base to the peak
3131 oror dM
Ka
251
3131 539
c
oror T
da
2539
cT
da
In December of 2006 I discovered the mathematical error in Einsteinrsquos Special Relativity Einstein used BS math and that math only works in an imaginary world
It cannot describe real symmetrical space
Einsteinrsquos Special Relativity should be abandoned and a theory of real space using SM should be used to replace it
Numerous articles on the Internet show Einsteinrsquos theory to be incorrect but no one in academia is willing to publish the information
Figure 1-10 A moving particle observed from two frames of reference in uniform translation
Einsteinrsquos Math Error in Special Relativity Transformation of Velocity (From Eisenberg- Fundamentals of Modern Physics pages 27-29) Consider the particle shown in figure (1-10) moving with velocity v as
seen in a frame of reference O
dx dy dz
Vx Vy Vzdt dt dt
dx dy dz
V x V y V zdy dt dt
We would like to evaluate the velocity v of the particle as seen in the frame of reference 0 which is itself moving relative to 0 with velocity v Measured in the 0 frame the velocity vector of the particle has com ponents
The same velocity vector as measured in the 0 frame has components
2
2
2
2
2
1
1
x vtx
v
cy y
z z
vxt
ctv
c
Now from equations (1-13) we know that the relation between the primed and the unprimed coordinates and times is
Take the differential of these equations remembering that v is a constant This gives
2
2
2
2
2
1
1
dx vdtdx
v
cdy dy
dz dz
vxdt
cdtv
c
2
2
2 2 2 2 2
2
2
2
11 ( )
1( ) 1
1
1
1
dx vdt
v dxdx vdt v
dx dx vdtc dt dtvdx vdx vdx v dxdt dt dt dt
dt dtc c c c
VxV x
Vx
VxV x
vv
vv
c
x
v
c
cV
( )
coordinate of particle relative to moving primed system
coordinate of particle relative to stationary system
velocity of primed coordina
V x
Vx
te system relative to stationaryv sys
tem
Page 28 Fundamentals of Modern Physics
(1-16)
According to Einstein this equation tells you how to transform the observed velocity from one frame of reference to another frame of reference First we note that as Vc and vc approachzero the equation approaches those which would be derived from the Galilean transformation
2 2
( ) 1 01 1 ( )
Vx v Vx vV
VxV x
Vx v
v cVxx
v
c
v
cc
Another property of this equation is that it is impossible to choose V and v such that Vrsquo the magnitude of the velocity which is seen in the new frame is greater than c
1
2
09 ( 09 ) 18009876543
( 09 ) 1811 (09 )
c c cV c c
cc
c
Example As seen by O particle 1 has velocity 09c in the direction of positive x and particle 2 has velocity 09c in the negative x direction
Einstein equation however predicts To evaluate the velocity of particle 1 with respect to particle 2 we transform from the O frame to the Orsquo frame moving in the negative x direction with velocity v = -09c using equation 1-16 We obtain
JK LOGIC will observe the particles separating at 18c
A confusing INCORRECT and ILLOGICAL answer
left is a dir rightsub of a
1
2
1
09 ( 09 ) 18009876543
181( 09 )1 ( 09 )
The multiplication of dir left time a dir right is ONLY a dir left
09
dir
direction leftdirection rightsubtraction
c c cV c c
cc
c
V
dir
22
22
09 18 18009876543
181( 081)1(09)(09)
1
left rightsub
sub
c c c cc c
ccc
c
Analyzing Einsteinrsquos BS math Equation
Using SM math a logical answer can be obtained
Cannot multiply a subtraction operator times a direction in space
Can only subtract and change its direction
Cannot multiply a direction to the left times a direction to the right
Cannot multiply opposite directions in space
Can only solve for the resultant
rightleft
subtraction
1
subtraction
2
sub of a dir to the left is a dir to the right
1 resultant 0 sub 2
2sub 2
2
09 09
09 amp 091
09 09 18 1
(0)1
(09) amp (09)1
c cV
c cc
c c cV
cc
cc
8
181
cc
SM math predicts a logical answer using logic
21st century scientists state in many articles that Einsteins General Relativity and Quantum Mechanics are incompatible
They are not only incompatible they are both mathematically incorrect
They both use illogical BS math
Any equations in math that use imaginary numbers should be abandoned
Imaginary numbers are only necessary in the current BS math system
This means that Schroumldingers equations should be abandoned
Any equation in quantum mechanics that uses imaginary number should be abandoned
Prof Kanarev has developed logical physics and chemistrytheories without the use of Quantum mechanics or relativity
A partial list of his theories coverbull Mechanics ndash replaces Newtonbull Photon - bull Electrons ndash Protons- Neutronsbull Spectroscopybull Nucleusbull Atoms amp Moleculesbull Electrodynamicsbull Heatbull Curvature of Space ndash replaces Schwarzcchildrsquos equation
Kanarevrsquos Hypothesis
bull Electron align linearly with proton They do NOT spin or form clouds around the nucleus
bull Neutrons are in the center of atom Proton are outside of Neutrons
bull The wavelength of all particles is the experimental radius
bull Double slit experiment explained correctly
bull Heisenbergrsquos inequality explained correctly
bull is the radius of approach of the magnetic lines around the torus of the electron
( 1 137)
bull Why water expands when it freezes
bull Electron clusters form lightning
rbull The meaning of Plancks constant is explained
22 2 kg m
h m f mr f consts
2210254 42m r kgm bull If the mr-constant describes a ring the equations
describe experiments bull If the mr-constant describes a wave the equations do
NOT describe experiments
Kanarevrsquos 5 Equations of Mechanics
Kanarevrsquos 1st Equation Acceleration ampmv maMma R R
Kanarevrsquos 2nd Equation Uniform motion amp mv maMma R R
maM mvR R
Kanarevrsquos 3rd Equation
Kanarevrsquos 4th Equation Equal and opposite reaction
1 21 21 2
12
a m
m a m ama
Kanarevrsquos 5th Equation 1 2 3 Mi M M M Mna R R R R R
The sum of the Actions working on a moving body is never equal to zero
Kanarevrsquos Photon model is a rolling ring
His equations accurately describe known experiments with photonsbull Electromagnetic radiation should be called photon radiationbull Diffraction and interference are the same phenomenabull Correctly explains the double slit experiment without quantum equationsbull Correctly explains the meaning of Heisenbergrsquos inequality
e
2 (2)(3142)(2817 15 ) 10007297352568
R 2426 12 137mag
xp
r m
m
Charge
Magnetic action
Proton
The intensity of the magnetic action near the geometrical center of the proton ring is so great that we theorize that it is the strong nuclear force
(9383 8)(1602 19)8476 14
4 (4)(3142)(1411 26)p
pp
EB T
M
(9395 8)(1602 19)1035 15
4 (4)(3142)(0966 26)N
NN
EB T
M
Neutron
1a Hydrogenavi
Magnitude 4 Direction 1 Magnitude 3 Direction 2 Magnitude 5 Direction 3 Relative to an observer at 0 4
Magnitude 4 Direction 1 Magnitude 3 Direction 2 Magnitude 5 Direction 3 Relative to an observer at 0 4
Magnitude and Direction (MD) An arrow has both a magnitude andA direction relative to a 2nd objectobserver
bull Magnitude and velocity (MV) relative to a 2nd objectobserver
bull Magnitude And acceleration (MA) relative to a 2nd objectobserver2
v da
t t
Introducing a new symbol the ldquoamprdquo sign The ldquoamprdquo sign means resultant ADD all the Arrows for each direction and then SUBTRACT to find the final resultant
1 1amp 1 2 2amp 2 4 If =
1 amp1 2 2 amp2 4 If =
amp 5 amp 3 2 If gt
amp 3 amp 5= 2 If gt
amp 0 3 amp 3=0 if =
only the resultant equals zero
the vectors do not - they are still there
Does not exist There is no multiplication of arrows
There are two arrows One has a magnitude of 3 One has a magnitude of 5 The maximum these two arrows can be is 8
Only the resultant of arrows exist
An SM arrow has a direction and a magnitude
Arrows and magnitudes are still there even if the resultant is zero
There are no negative SM Arrows
There are no positive SM Arrows
There are no ZERO SM Arrows A zero would negate the definition of an SM Arrow something with no magnitude and no direction cannot be the definition of an arrow that is defined as having magnitude and direction
bull A resultant is the addition and subtraction of the differences in magnitude and direction arrows
A B and C are the arrows (a) A amp B exist(b) A amp B amp C exist(c) A-B Do Not Exist in Sm there are no negatives or positives(d) A+B-C Do Not Exist in SM there are no negatives or positives
There are only resultants when there are multiple arrows
SM Graphs versus BS Graphs
Numbers Raised to Powers
When numbers are raised to powers the negative sides of all BS math graphs are ILLOGICAL and INCORRECT
Incorrect BS Math for 2x
2x
4 16 correct3 8 correct2 4 correct1 2 correct0 1 correct-1 005 incorrect-2 025 incorrect-3 0125 incorrect-4 00625 incorrect
negative side of graph is incorrectpositive side of graph is correct
16
8
42105025012500625
-4 -3 -2 -1 0 1 2 3 4
po
sit
ive
negative positive
2x BS Math
correctincorrect
In BS math the dash sign (-) used in this example represents three different operations They arebull (-) Subtractionbull (-) A direction to the leftbull (-) As an exponent to mean divide by
All positive numbers produce correct results and graphs
All negative numbers produce incorrect results and graphs
A direction of three units to the left (-3) is placed in the equation as an exponent
A direction in space (3 units toward the left) changes a whole number to a fraction by using the dash (-) as an exponent rather than a direction in space
This is illogical and incorrect BS math should have performed this operation as
Unfortunately the answers to problems can come out correct because data points of a real problem fit the curve
However the answers will not be due to logical reasoning
Because the equation produced a curve that real problem data points follow it has been falsely believed that the equations and graphs are correct
1
2
x
Incorrect BK graph for (12)x
(12)x
4 00625 correct3 0125 correct2 025 correct1 1 correct0 1 correct-1 2 incorrect-2 4 incorrect-3 8 incorrect-4 16 incorrect
negative side of graph is incorrectpositive side of graph is correct
00625012502505124
8
16
-5 -4 -3 -2 -1 0 1 2 3 4 5P
osit
ive
negative positive
BS Math for (12)x
correctincorrect
Incorrect BS math and graph for
All graphs are symmetrical with SM
Two BS Math Equations that Produce Incorrect Answers
2 0x x starting equation 2
2
x x x x
x x
Add lsquoxrsquo to both sides
2(1) 1 let lsquoxrsquo = 1
2 1 This is an incorrect answer 2(0)-(0)=0 The only answer that BK math can give is x=0
1
a b starting equation ( )( ) ( )( )a a a b multiply both sides by ( )a
( )( ) ( )( ) ( )( ) ( )( )a a b b a b b b subtract ( )( )b b from both sides
( )( ) ( )( )a b a b a b b factor ( )a b from both sides
( )a b b divide both sides by ( )a b
( )b b b replace ( )a with ( )b 2b b simplify 2(1) (1) Let ( )b = 1
2 1 Again BS math yields and incorrect answer 1
This incorrect result is a problem with the distributive law SM explains how this incorrect answer is obtained and the problem with the distributive law that allows the error
Page 148 ndash149 Barry Mazur ndash Imagining Numbers
3
2
31
Do the computation carefully on paper using the rules we agreed to Then ponder your answer which should be something of a surprise to you donrsquot stop there Think of what your answer might possibly mean or might imply
Solve this equation using the distributive law
JK definitely a surprise the answer wasincorrect and illogical
Gerhard Gade University Professor doing math at Harvard University
Here is the BS math answer that Barry Mazur calculated
2
31
2
31
2
31
1 3 1 3 1 3 3 3 2 2 3 1 3
2 2 2 2 4 4 4 4 4 4 2 2
dash
1 3 1 3 1 31
2 2 2 2 4 4
3 3
4 4
When the distributive law is used with the BS Rule-of-Signs and one of the signs is a (-) dash sign the middle two terms always cancel each other This leads to the incorrect answer of (-1) in the above equation
Barry Mazur did not go back and check his answer with the original equation If he had he would have known that the answer was incorrect
3
1 31
2
check answer by setting equal
3
331 3
12
1 31
2
Take the Cube root of both sides
1 3 2 multiply both sides by 2
3 3 subtract 1 from both sides
2 2
3 3 square both sides
3 9 This is an incorrect answer A simple equation check could have shown that the distributive law using BS math provides an incorrect answer
1
BS Illogical and Incorrect Numbers Raised to Powers
nC D n DC
642 DDD
In BS math only a positive number can have an answer if ldquonrdquo is an even number (246---)
All negative numbers have no solutions
642 DDD
With SM just add a direction arrow and any number gives the correct answer
or n nC D C D
3 3( 3 8 8 or 8 2 or 2 )n D C
( 2 16 C= 16 or 16 4 or 4 )n D
Since a negative number has no square root answer in BS math BS mathematicians invented imaginary numbers
i2=-1
24414 iiC n
These equations are easily solved without the use of imaginary numbers with SM math
If 4 then 4 2x
If 4 then 4 2x
The 360-degree circle is a 4400 years old extremely outdated system based on a year of 12 months of 30 days each that is 360 days
SM will use a more logical system that will have a circle of 6283185307o
d
( )
(deg )
d r radians
d r rees
the arc distance around
the circumference of a circle
d
radiusr
2 2
2 2
d r
dv rd t t
v dda rd t t t
1
d
r
( )av tv
The average angular velocity of a body is the rate of its angular displacement
( )0
liminstt
vt
The instantaneous angular velocity is the limit of the ratio
t
as t approaches zero
( )d
av
d
tv
The average circumference velocity of the body is the rate of its distance traveled d divided by the time t
dvv
r
1 dd vdrv
t t t r r
The equation of the relationship between the angular velocity and the velocity of a point on that body
( )o
av
v v v
t ta
The average angular acceleration is the change in the angular velocity
( )0
liminstt
v
ta
The instantaneous angular acceleration is the
limit of the ration of v
t
as t approaches
zero
daa
r
1
1
d
d s d
vva
t r tv a
r t r
The equation of the relationship between the angular acceleration and the acceleration of a point on the body
21
2v a t
If a body with an initial angular velocity ov has a constant
angular acceleration a it will turn through an angle in a time t
1
Straight Line Symbol Units Rotational Symbol Units Distance d m Angular
Distance radians or degrees
Velocity dt
v m
s Angular Velocity
V
t
radians or degrees
s
Acceleration vat
2ms
Angular Acceleration
VA
t
2
radians or degrees
s
Mass M Kg Mass distribution
dM 2Kgm
Force actionA Ma
2
Kgm
sNewton
Torque
dT M A 2
2
Kgm
s
Momentum Mv Kgm
s
Angular Momentum
dM V 2Kgm
s
Work (Force)(distance)
Ad 2
Kgmm
sNm Joule
Work T
Using AS arrow math to solve and direction in space problems using the equation 1 2
2 2 2 2 2 2( ) amp and ( ) amp 3
4
2 2 2
2 2
( ) amp
(5 ) 4 3
(5 ) 16 9
(5 ) 25
(5 ) (5 )
2 2(5 ) 4 3
(5 ) 16 9
(5 ) 25
(5 ) (5 )
In SM math Pythagorasrsquo theorem provides both magnitude and direction
In BS math Pythagorasrsquo theorem provides magnitude only
time set equal to zero
23
( ) ( 0 )10 1
x t xx
3( ) ( 0)
10 1W t f
In BS math this is an equation for a traveling wave
must use a square term in order to make the curve symmetric
NO SQUARE TERM IS NECESSARY
In SM math when t is set to ldquo0rdquo
This is a major fault in BS math
x f(x)
-4 0019
-3 0033
-2 0073
-1 0273
0 3000
1 0273
2 0073
3 0033
4 0019 -4 -3 -2 -1 0 1 2 3 400
05
10
15
20
25
30
35
BS wave (must use square term)
snap shot of wave at x = 0
Am
plit
ude
lt to gt lt to gt f(ltgt)
lt -4 0073
lt -3 0097
lt -2 0143
lt -1 0273
origin 0 3000
gt 1 0273
gt 2 0143
gt 3 0097
gt 4 0073lt0073 lt0097 lt0143 lt0273
3000
0273gt 0143gt 0097gt 0073gt
Wave moving
snap shot of wave
Am
plit
ude
lt4 lt3 lt2 lt1 0 1gt 2gt 3gt 4gt
3
2
1
In SM NO squared terms are needed
SPRING
An Action is needed to stretch or compress a spring to a starting point where it can execute cycle motion
tsA (ts to start) is the work required to move a mass attached to a spring a distance (d1) from
its initial rest position RAK 0 the spring constant is the stiffness of the spring
o A is attraction o R is repulsion o
ARRA
00 is for the back and forth motion through the zero initial rest position A
stretched spring goes from ldquoArdquo to ldquo0rdquo to ldquoRrdquo to ldquoArdquo to ldquo0rdquo to ldquoRrdquo d1 is the distance a mass is moved from its initial rest position
0
0 1ts A RA K d
Once ( )mad
is performed to move the mass to position d1 (against the restraint of the spring) a of the ( )mad
is transferred into the spring as potential attractionrepulsion (PAR) There is always
more ( )mad
performed than is stored as PAR When the spring is released from the restraining Ats the PAR starts the oscillation Conservation of ( )mad
requires that the PAR and the motion ( )mad
( )( )madM be a constant
( )amp ( ) constantmadPAR M 02
0 0
1
2A R A RPAR K d
02
( ) 0 0
1( )
2mad A R A RM m v
0 0 02 2 2
0 0 0 1
2 2 21
2 2 21
1 1 1amp
2 2 2
1 1 1amp
2 2 2
amp
A R A R A RK d m v K d
Kd mv Kd
Kd mv Kd
0
1d is frac12 the height from the base to the peak of a cycle
An experiment of moving paper under a pendulum through a small angle filled with sand that can escape through a small hole in the bottom shows that a pendulum cycling back and forth follows a curve that is mathematically analogous to a lsquosinrsquo or lsquocosrsquo
The projection of the curved area onto the straight line exactly duplicate the back and forth motion of CM This allows for the math of lsquosinrsquo or lsquocosrsquo to be used
SM will use lsquosinrdquo math It could use lsquocosrsquo or a combination of both lsquosinrsquo and lsquocosrsquo Using just the lsquosinrsquo math makes the understanding and math easier
0 sin 0 0SM 0 sin157 1SM The diagram at the left shows a spring being stretched from its (zero rest) position-2 and the PAR being released at position-1 There are 4 sections (1 to 2 2 to 3 3 to 4 and 4 to5) that alternate between maximum acceleration and zero velocity to max velocity and zero acceleration
At position-1 there is maximum acceleration and zero velocity o The lsquosinrsquo values from 0o to 157o will be used for velocity (values from 0 to 1) o The lsquosinrsquo values from 157o to 0o will be used for acceleration (values from 1 to 0)
At position-2 there is maximum velocity and zero acceleration Acceleration changes to deceleration
o The lsquosinrsquo values from 157o to 0o will be used for deceleration (values from 1 to 0) o The lsquosinrsquo values from 0o to 157o will be used for acceleration (values from 0 to 1)
At position-3 there is maximum acceleration and zero velocity o The lsquosinrsquo values from 0o to 157o will be used for velocity (values from 0 to 1) o The lsquosinrsquo values from 157o to 0o will be used for acceleration (values from 1 to 0)
At position-4 there is maximum velocity and zero acceleration Acceleration changes to deceleration
o The lsquosinrsquo values from 157o to 0o will be used for deceleration (values from 1 to 0) o The lsquosinrsquo values from 0o to 157o will be used for acceleration (values from 0 to 1)
At position-5 the cycle starts again
2
)539(
cT
MK
spring constant (K) when mass (M) and cycle-time (T) are known
)(
))(89)((
111 md
uKgM
d
MG
d
FK sts
spring constant (K) when mass (M) and amplitude ( d1) are known
2 21( )
Kv d d
m
velocity (v) at any point during the cycle
velocity
dforM
Kdv
max
012 velocity (v2) maximum (when d = 0 at position-2 and 4)
acceleration (a1) maximum when spring constant (K) mass (M) and amplitude (d1) are known
acceleration (a1) maximum when amplitude (d1) and cycle-time (Tc) are known
acceleration (a) at any point during the cycle-time (T)
1
151
286
v
dTc
cycle-time (Tc) when amplitude (d1) and velocity (v1) are known
K
MTc 28651
cycle-time (Tc) when mass (M) and spring constant (K) are known
fTc
1
cycle-time (Tc) when the frequency is known
cTf
1
frequency (f) when the cycle-time (Tc) is known
of cycles
elapsed timef
frequency (f) when the number of cycles in an elapsed time is known
d1 amplitude (distance from rest to start position)
Also frac12 the distance from the base to the peak
3131 oror dM
Ka
251
3131 539
c
oror T
da
2539
cT
da
In December of 2006 I discovered the mathematical error in Einsteinrsquos Special Relativity Einstein used BS math and that math only works in an imaginary world
It cannot describe real symmetrical space
Einsteinrsquos Special Relativity should be abandoned and a theory of real space using SM should be used to replace it
Numerous articles on the Internet show Einsteinrsquos theory to be incorrect but no one in academia is willing to publish the information
Figure 1-10 A moving particle observed from two frames of reference in uniform translation
Einsteinrsquos Math Error in Special Relativity Transformation of Velocity (From Eisenberg- Fundamentals of Modern Physics pages 27-29) Consider the particle shown in figure (1-10) moving with velocity v as
seen in a frame of reference O
dx dy dz
Vx Vy Vzdt dt dt
dx dy dz
V x V y V zdy dt dt
We would like to evaluate the velocity v of the particle as seen in the frame of reference 0 which is itself moving relative to 0 with velocity v Measured in the 0 frame the velocity vector of the particle has com ponents
The same velocity vector as measured in the 0 frame has components
2
2
2
2
2
1
1
x vtx
v
cy y
z z
vxt
ctv
c
Now from equations (1-13) we know that the relation between the primed and the unprimed coordinates and times is
Take the differential of these equations remembering that v is a constant This gives
2
2
2
2
2
1
1
dx vdtdx
v
cdy dy
dz dz
vxdt
cdtv
c
2
2
2 2 2 2 2
2
2
2
11 ( )
1( ) 1
1
1
1
dx vdt
v dxdx vdt v
dx dx vdtc dt dtvdx vdx vdx v dxdt dt dt dt
dt dtc c c c
VxV x
Vx
VxV x
vv
vv
c
x
v
c
cV
( )
coordinate of particle relative to moving primed system
coordinate of particle relative to stationary system
velocity of primed coordina
V x
Vx
te system relative to stationaryv sys
tem
Page 28 Fundamentals of Modern Physics
(1-16)
According to Einstein this equation tells you how to transform the observed velocity from one frame of reference to another frame of reference First we note that as Vc and vc approachzero the equation approaches those which would be derived from the Galilean transformation
2 2
( ) 1 01 1 ( )
Vx v Vx vV
VxV x
Vx v
v cVxx
v
c
v
cc
Another property of this equation is that it is impossible to choose V and v such that Vrsquo the magnitude of the velocity which is seen in the new frame is greater than c
1
2
09 ( 09 ) 18009876543
( 09 ) 1811 (09 )
c c cV c c
cc
c
Example As seen by O particle 1 has velocity 09c in the direction of positive x and particle 2 has velocity 09c in the negative x direction
Einstein equation however predicts To evaluate the velocity of particle 1 with respect to particle 2 we transform from the O frame to the Orsquo frame moving in the negative x direction with velocity v = -09c using equation 1-16 We obtain
JK LOGIC will observe the particles separating at 18c
A confusing INCORRECT and ILLOGICAL answer
left is a dir rightsub of a
1
2
1
09 ( 09 ) 18009876543
181( 09 )1 ( 09 )
The multiplication of dir left time a dir right is ONLY a dir left
09
dir
direction leftdirection rightsubtraction
c c cV c c
cc
c
V
dir
22
22
09 18 18009876543
181( 081)1(09)(09)
1
left rightsub
sub
c c c cc c
ccc
c
Analyzing Einsteinrsquos BS math Equation
Using SM math a logical answer can be obtained
Cannot multiply a subtraction operator times a direction in space
Can only subtract and change its direction
Cannot multiply a direction to the left times a direction to the right
Cannot multiply opposite directions in space
Can only solve for the resultant
rightleft
subtraction
1
subtraction
2
sub of a dir to the left is a dir to the right
1 resultant 0 sub 2
2sub 2
2
09 09
09 amp 091
09 09 18 1
(0)1
(09) amp (09)1
c cV
c cc
c c cV
cc
cc
8
181
cc
SM math predicts a logical answer using logic
21st century scientists state in many articles that Einsteins General Relativity and Quantum Mechanics are incompatible
They are not only incompatible they are both mathematically incorrect
They both use illogical BS math
Any equations in math that use imaginary numbers should be abandoned
Imaginary numbers are only necessary in the current BS math system
This means that Schroumldingers equations should be abandoned
Any equation in quantum mechanics that uses imaginary number should be abandoned
Prof Kanarev has developed logical physics and chemistrytheories without the use of Quantum mechanics or relativity
A partial list of his theories coverbull Mechanics ndash replaces Newtonbull Photon - bull Electrons ndash Protons- Neutronsbull Spectroscopybull Nucleusbull Atoms amp Moleculesbull Electrodynamicsbull Heatbull Curvature of Space ndash replaces Schwarzcchildrsquos equation
Kanarevrsquos Hypothesis
bull Electron align linearly with proton They do NOT spin or form clouds around the nucleus
bull Neutrons are in the center of atom Proton are outside of Neutrons
bull The wavelength of all particles is the experimental radius
bull Double slit experiment explained correctly
bull Heisenbergrsquos inequality explained correctly
bull is the radius of approach of the magnetic lines around the torus of the electron
( 1 137)
bull Why water expands when it freezes
bull Electron clusters form lightning
rbull The meaning of Plancks constant is explained
22 2 kg m
h m f mr f consts
2210254 42m r kgm bull If the mr-constant describes a ring the equations
describe experiments bull If the mr-constant describes a wave the equations do
NOT describe experiments
Kanarevrsquos 5 Equations of Mechanics
Kanarevrsquos 1st Equation Acceleration ampmv maMma R R
Kanarevrsquos 2nd Equation Uniform motion amp mv maMma R R
maM mvR R
Kanarevrsquos 3rd Equation
Kanarevrsquos 4th Equation Equal and opposite reaction
1 21 21 2
12
a m
m a m ama
Kanarevrsquos 5th Equation 1 2 3 Mi M M M Mna R R R R R
The sum of the Actions working on a moving body is never equal to zero
Kanarevrsquos Photon model is a rolling ring
His equations accurately describe known experiments with photonsbull Electromagnetic radiation should be called photon radiationbull Diffraction and interference are the same phenomenabull Correctly explains the double slit experiment without quantum equationsbull Correctly explains the meaning of Heisenbergrsquos inequality
e
2 (2)(3142)(2817 15 ) 10007297352568
R 2426 12 137mag
xp
r m
m
Charge
Magnetic action
Proton
The intensity of the magnetic action near the geometrical center of the proton ring is so great that we theorize that it is the strong nuclear force
(9383 8)(1602 19)8476 14
4 (4)(3142)(1411 26)p
pp
EB T
M
(9395 8)(1602 19)1035 15
4 (4)(3142)(0966 26)N
NN
EB T
M
Neutron
1a Hydrogenavi
Introducing a new symbol the ldquoamprdquo sign The ldquoamprdquo sign means resultant ADD all the Arrows for each direction and then SUBTRACT to find the final resultant
1 1amp 1 2 2amp 2 4 If =
1 amp1 2 2 amp2 4 If =
amp 5 amp 3 2 If gt
amp 3 amp 5= 2 If gt
amp 0 3 amp 3=0 if =
only the resultant equals zero
the vectors do not - they are still there
Does not exist There is no multiplication of arrows
There are two arrows One has a magnitude of 3 One has a magnitude of 5 The maximum these two arrows can be is 8
Only the resultant of arrows exist
An SM arrow has a direction and a magnitude
Arrows and magnitudes are still there even if the resultant is zero
There are no negative SM Arrows
There are no positive SM Arrows
There are no ZERO SM Arrows A zero would negate the definition of an SM Arrow something with no magnitude and no direction cannot be the definition of an arrow that is defined as having magnitude and direction
bull A resultant is the addition and subtraction of the differences in magnitude and direction arrows
A B and C are the arrows (a) A amp B exist(b) A amp B amp C exist(c) A-B Do Not Exist in Sm there are no negatives or positives(d) A+B-C Do Not Exist in SM there are no negatives or positives
There are only resultants when there are multiple arrows
SM Graphs versus BS Graphs
Numbers Raised to Powers
When numbers are raised to powers the negative sides of all BS math graphs are ILLOGICAL and INCORRECT
Incorrect BS Math for 2x
2x
4 16 correct3 8 correct2 4 correct1 2 correct0 1 correct-1 005 incorrect-2 025 incorrect-3 0125 incorrect-4 00625 incorrect
negative side of graph is incorrectpositive side of graph is correct
16
8
42105025012500625
-4 -3 -2 -1 0 1 2 3 4
po
sit
ive
negative positive
2x BS Math
correctincorrect
In BS math the dash sign (-) used in this example represents three different operations They arebull (-) Subtractionbull (-) A direction to the leftbull (-) As an exponent to mean divide by
All positive numbers produce correct results and graphs
All negative numbers produce incorrect results and graphs
A direction of three units to the left (-3) is placed in the equation as an exponent
A direction in space (3 units toward the left) changes a whole number to a fraction by using the dash (-) as an exponent rather than a direction in space
This is illogical and incorrect BS math should have performed this operation as
Unfortunately the answers to problems can come out correct because data points of a real problem fit the curve
However the answers will not be due to logical reasoning
Because the equation produced a curve that real problem data points follow it has been falsely believed that the equations and graphs are correct
1
2
x
Incorrect BK graph for (12)x
(12)x
4 00625 correct3 0125 correct2 025 correct1 1 correct0 1 correct-1 2 incorrect-2 4 incorrect-3 8 incorrect-4 16 incorrect
negative side of graph is incorrectpositive side of graph is correct
00625012502505124
8
16
-5 -4 -3 -2 -1 0 1 2 3 4 5P
osit
ive
negative positive
BS Math for (12)x
correctincorrect
Incorrect BS math and graph for
All graphs are symmetrical with SM
Two BS Math Equations that Produce Incorrect Answers
2 0x x starting equation 2
2
x x x x
x x
Add lsquoxrsquo to both sides
2(1) 1 let lsquoxrsquo = 1
2 1 This is an incorrect answer 2(0)-(0)=0 The only answer that BK math can give is x=0
1
a b starting equation ( )( ) ( )( )a a a b multiply both sides by ( )a
( )( ) ( )( ) ( )( ) ( )( )a a b b a b b b subtract ( )( )b b from both sides
( )( ) ( )( )a b a b a b b factor ( )a b from both sides
( )a b b divide both sides by ( )a b
( )b b b replace ( )a with ( )b 2b b simplify 2(1) (1) Let ( )b = 1
2 1 Again BS math yields and incorrect answer 1
This incorrect result is a problem with the distributive law SM explains how this incorrect answer is obtained and the problem with the distributive law that allows the error
Page 148 ndash149 Barry Mazur ndash Imagining Numbers
3
2
31
Do the computation carefully on paper using the rules we agreed to Then ponder your answer which should be something of a surprise to you donrsquot stop there Think of what your answer might possibly mean or might imply
Solve this equation using the distributive law
JK definitely a surprise the answer wasincorrect and illogical
Gerhard Gade University Professor doing math at Harvard University
Here is the BS math answer that Barry Mazur calculated
2
31
2
31
2
31
1 3 1 3 1 3 3 3 2 2 3 1 3
2 2 2 2 4 4 4 4 4 4 2 2
dash
1 3 1 3 1 31
2 2 2 2 4 4
3 3
4 4
When the distributive law is used with the BS Rule-of-Signs and one of the signs is a (-) dash sign the middle two terms always cancel each other This leads to the incorrect answer of (-1) in the above equation
Barry Mazur did not go back and check his answer with the original equation If he had he would have known that the answer was incorrect
3
1 31
2
check answer by setting equal
3
331 3
12
1 31
2
Take the Cube root of both sides
1 3 2 multiply both sides by 2
3 3 subtract 1 from both sides
2 2
3 3 square both sides
3 9 This is an incorrect answer A simple equation check could have shown that the distributive law using BS math provides an incorrect answer
1
BS Illogical and Incorrect Numbers Raised to Powers
nC D n DC
642 DDD
In BS math only a positive number can have an answer if ldquonrdquo is an even number (246---)
All negative numbers have no solutions
642 DDD
With SM just add a direction arrow and any number gives the correct answer
or n nC D C D
3 3( 3 8 8 or 8 2 or 2 )n D C
( 2 16 C= 16 or 16 4 or 4 )n D
Since a negative number has no square root answer in BS math BS mathematicians invented imaginary numbers
i2=-1
24414 iiC n
These equations are easily solved without the use of imaginary numbers with SM math
If 4 then 4 2x
If 4 then 4 2x
The 360-degree circle is a 4400 years old extremely outdated system based on a year of 12 months of 30 days each that is 360 days
SM will use a more logical system that will have a circle of 6283185307o
d
( )
(deg )
d r radians
d r rees
the arc distance around
the circumference of a circle
d
radiusr
2 2
2 2
d r
dv rd t t
v dda rd t t t
1
d
r
( )av tv
The average angular velocity of a body is the rate of its angular displacement
( )0
liminstt
vt
The instantaneous angular velocity is the limit of the ratio
t
as t approaches zero
( )d
av
d
tv
The average circumference velocity of the body is the rate of its distance traveled d divided by the time t
dvv
r
1 dd vdrv
t t t r r
The equation of the relationship between the angular velocity and the velocity of a point on that body
( )o
av
v v v
t ta
The average angular acceleration is the change in the angular velocity
( )0
liminstt
v
ta
The instantaneous angular acceleration is the
limit of the ration of v
t
as t approaches
zero
daa
r
1
1
d
d s d
vva
t r tv a
r t r
The equation of the relationship between the angular acceleration and the acceleration of a point on the body
21
2v a t
If a body with an initial angular velocity ov has a constant
angular acceleration a it will turn through an angle in a time t
1
Straight Line Symbol Units Rotational Symbol Units Distance d m Angular
Distance radians or degrees
Velocity dt
v m
s Angular Velocity
V
t
radians or degrees
s
Acceleration vat
2ms
Angular Acceleration
VA
t
2
radians or degrees
s
Mass M Kg Mass distribution
dM 2Kgm
Force actionA Ma
2
Kgm
sNewton
Torque
dT M A 2
2
Kgm
s
Momentum Mv Kgm
s
Angular Momentum
dM V 2Kgm
s
Work (Force)(distance)
Ad 2
Kgmm
sNm Joule
Work T
Using AS arrow math to solve and direction in space problems using the equation 1 2
2 2 2 2 2 2( ) amp and ( ) amp 3
4
2 2 2
2 2
( ) amp
(5 ) 4 3
(5 ) 16 9
(5 ) 25
(5 ) (5 )
2 2(5 ) 4 3
(5 ) 16 9
(5 ) 25
(5 ) (5 )
In SM math Pythagorasrsquo theorem provides both magnitude and direction
In BS math Pythagorasrsquo theorem provides magnitude only
time set equal to zero
23
( ) ( 0 )10 1
x t xx
3( ) ( 0)
10 1W t f
In BS math this is an equation for a traveling wave
must use a square term in order to make the curve symmetric
NO SQUARE TERM IS NECESSARY
In SM math when t is set to ldquo0rdquo
This is a major fault in BS math
x f(x)
-4 0019
-3 0033
-2 0073
-1 0273
0 3000
1 0273
2 0073
3 0033
4 0019 -4 -3 -2 -1 0 1 2 3 400
05
10
15
20
25
30
35
BS wave (must use square term)
snap shot of wave at x = 0
Am
plit
ude
lt to gt lt to gt f(ltgt)
lt -4 0073
lt -3 0097
lt -2 0143
lt -1 0273
origin 0 3000
gt 1 0273
gt 2 0143
gt 3 0097
gt 4 0073lt0073 lt0097 lt0143 lt0273
3000
0273gt 0143gt 0097gt 0073gt
Wave moving
snap shot of wave
Am
plit
ude
lt4 lt3 lt2 lt1 0 1gt 2gt 3gt 4gt
3
2
1
In SM NO squared terms are needed
SPRING
An Action is needed to stretch or compress a spring to a starting point where it can execute cycle motion
tsA (ts to start) is the work required to move a mass attached to a spring a distance (d1) from
its initial rest position RAK 0 the spring constant is the stiffness of the spring
o A is attraction o R is repulsion o
ARRA
00 is for the back and forth motion through the zero initial rest position A
stretched spring goes from ldquoArdquo to ldquo0rdquo to ldquoRrdquo to ldquoArdquo to ldquo0rdquo to ldquoRrdquo d1 is the distance a mass is moved from its initial rest position
0
0 1ts A RA K d
Once ( )mad
is performed to move the mass to position d1 (against the restraint of the spring) a of the ( )mad
is transferred into the spring as potential attractionrepulsion (PAR) There is always
more ( )mad
performed than is stored as PAR When the spring is released from the restraining Ats the PAR starts the oscillation Conservation of ( )mad
requires that the PAR and the motion ( )mad
( )( )madM be a constant
( )amp ( ) constantmadPAR M 02
0 0
1
2A R A RPAR K d
02
( ) 0 0
1( )
2mad A R A RM m v
0 0 02 2 2
0 0 0 1
2 2 21
2 2 21
1 1 1amp
2 2 2
1 1 1amp
2 2 2
amp
A R A R A RK d m v K d
Kd mv Kd
Kd mv Kd
0
1d is frac12 the height from the base to the peak of a cycle
An experiment of moving paper under a pendulum through a small angle filled with sand that can escape through a small hole in the bottom shows that a pendulum cycling back and forth follows a curve that is mathematically analogous to a lsquosinrsquo or lsquocosrsquo
The projection of the curved area onto the straight line exactly duplicate the back and forth motion of CM This allows for the math of lsquosinrsquo or lsquocosrsquo to be used
SM will use lsquosinrdquo math It could use lsquocosrsquo or a combination of both lsquosinrsquo and lsquocosrsquo Using just the lsquosinrsquo math makes the understanding and math easier
0 sin 0 0SM 0 sin157 1SM The diagram at the left shows a spring being stretched from its (zero rest) position-2 and the PAR being released at position-1 There are 4 sections (1 to 2 2 to 3 3 to 4 and 4 to5) that alternate between maximum acceleration and zero velocity to max velocity and zero acceleration
At position-1 there is maximum acceleration and zero velocity o The lsquosinrsquo values from 0o to 157o will be used for velocity (values from 0 to 1) o The lsquosinrsquo values from 157o to 0o will be used for acceleration (values from 1 to 0)
At position-2 there is maximum velocity and zero acceleration Acceleration changes to deceleration
o The lsquosinrsquo values from 157o to 0o will be used for deceleration (values from 1 to 0) o The lsquosinrsquo values from 0o to 157o will be used for acceleration (values from 0 to 1)
At position-3 there is maximum acceleration and zero velocity o The lsquosinrsquo values from 0o to 157o will be used for velocity (values from 0 to 1) o The lsquosinrsquo values from 157o to 0o will be used for acceleration (values from 1 to 0)
At position-4 there is maximum velocity and zero acceleration Acceleration changes to deceleration
o The lsquosinrsquo values from 157o to 0o will be used for deceleration (values from 1 to 0) o The lsquosinrsquo values from 0o to 157o will be used for acceleration (values from 0 to 1)
At position-5 the cycle starts again
2
)539(
cT
MK
spring constant (K) when mass (M) and cycle-time (T) are known
)(
))(89)((
111 md
uKgM
d
MG
d
FK sts
spring constant (K) when mass (M) and amplitude ( d1) are known
2 21( )
Kv d d
m
velocity (v) at any point during the cycle
velocity
dforM
Kdv
max
012 velocity (v2) maximum (when d = 0 at position-2 and 4)
acceleration (a1) maximum when spring constant (K) mass (M) and amplitude (d1) are known
acceleration (a1) maximum when amplitude (d1) and cycle-time (Tc) are known
acceleration (a) at any point during the cycle-time (T)
1
151
286
v
dTc
cycle-time (Tc) when amplitude (d1) and velocity (v1) are known
K
MTc 28651
cycle-time (Tc) when mass (M) and spring constant (K) are known
fTc
1
cycle-time (Tc) when the frequency is known
cTf
1
frequency (f) when the cycle-time (Tc) is known
of cycles
elapsed timef
frequency (f) when the number of cycles in an elapsed time is known
d1 amplitude (distance from rest to start position)
Also frac12 the distance from the base to the peak
3131 oror dM
Ka
251
3131 539
c
oror T
da
2539
cT
da
In December of 2006 I discovered the mathematical error in Einsteinrsquos Special Relativity Einstein used BS math and that math only works in an imaginary world
It cannot describe real symmetrical space
Einsteinrsquos Special Relativity should be abandoned and a theory of real space using SM should be used to replace it
Numerous articles on the Internet show Einsteinrsquos theory to be incorrect but no one in academia is willing to publish the information
Figure 1-10 A moving particle observed from two frames of reference in uniform translation
Einsteinrsquos Math Error in Special Relativity Transformation of Velocity (From Eisenberg- Fundamentals of Modern Physics pages 27-29) Consider the particle shown in figure (1-10) moving with velocity v as
seen in a frame of reference O
dx dy dz
Vx Vy Vzdt dt dt
dx dy dz
V x V y V zdy dt dt
We would like to evaluate the velocity v of the particle as seen in the frame of reference 0 which is itself moving relative to 0 with velocity v Measured in the 0 frame the velocity vector of the particle has com ponents
The same velocity vector as measured in the 0 frame has components
2
2
2
2
2
1
1
x vtx
v
cy y
z z
vxt
ctv
c
Now from equations (1-13) we know that the relation between the primed and the unprimed coordinates and times is
Take the differential of these equations remembering that v is a constant This gives
2
2
2
2
2
1
1
dx vdtdx
v
cdy dy
dz dz
vxdt
cdtv
c
2
2
2 2 2 2 2
2
2
2
11 ( )
1( ) 1
1
1
1
dx vdt
v dxdx vdt v
dx dx vdtc dt dtvdx vdx vdx v dxdt dt dt dt
dt dtc c c c
VxV x
Vx
VxV x
vv
vv
c
x
v
c
cV
( )
coordinate of particle relative to moving primed system
coordinate of particle relative to stationary system
velocity of primed coordina
V x
Vx
te system relative to stationaryv sys
tem
Page 28 Fundamentals of Modern Physics
(1-16)
According to Einstein this equation tells you how to transform the observed velocity from one frame of reference to another frame of reference First we note that as Vc and vc approachzero the equation approaches those which would be derived from the Galilean transformation
2 2
( ) 1 01 1 ( )
Vx v Vx vV
VxV x
Vx v
v cVxx
v
c
v
cc
Another property of this equation is that it is impossible to choose V and v such that Vrsquo the magnitude of the velocity which is seen in the new frame is greater than c
1
2
09 ( 09 ) 18009876543
( 09 ) 1811 (09 )
c c cV c c
cc
c
Example As seen by O particle 1 has velocity 09c in the direction of positive x and particle 2 has velocity 09c in the negative x direction
Einstein equation however predicts To evaluate the velocity of particle 1 with respect to particle 2 we transform from the O frame to the Orsquo frame moving in the negative x direction with velocity v = -09c using equation 1-16 We obtain
JK LOGIC will observe the particles separating at 18c
A confusing INCORRECT and ILLOGICAL answer
left is a dir rightsub of a
1
2
1
09 ( 09 ) 18009876543
181( 09 )1 ( 09 )
The multiplication of dir left time a dir right is ONLY a dir left
09
dir
direction leftdirection rightsubtraction
c c cV c c
cc
c
V
dir
22
22
09 18 18009876543
181( 081)1(09)(09)
1
left rightsub
sub
c c c cc c
ccc
c
Analyzing Einsteinrsquos BS math Equation
Using SM math a logical answer can be obtained
Cannot multiply a subtraction operator times a direction in space
Can only subtract and change its direction
Cannot multiply a direction to the left times a direction to the right
Cannot multiply opposite directions in space
Can only solve for the resultant
rightleft
subtraction
1
subtraction
2
sub of a dir to the left is a dir to the right
1 resultant 0 sub 2
2sub 2
2
09 09
09 amp 091
09 09 18 1
(0)1
(09) amp (09)1
c cV
c cc
c c cV
cc
cc
8
181
cc
SM math predicts a logical answer using logic
21st century scientists state in many articles that Einsteins General Relativity and Quantum Mechanics are incompatible
They are not only incompatible they are both mathematically incorrect
They both use illogical BS math
Any equations in math that use imaginary numbers should be abandoned
Imaginary numbers are only necessary in the current BS math system
This means that Schroumldingers equations should be abandoned
Any equation in quantum mechanics that uses imaginary number should be abandoned
Prof Kanarev has developed logical physics and chemistrytheories without the use of Quantum mechanics or relativity
A partial list of his theories coverbull Mechanics ndash replaces Newtonbull Photon - bull Electrons ndash Protons- Neutronsbull Spectroscopybull Nucleusbull Atoms amp Moleculesbull Electrodynamicsbull Heatbull Curvature of Space ndash replaces Schwarzcchildrsquos equation
Kanarevrsquos Hypothesis
bull Electron align linearly with proton They do NOT spin or form clouds around the nucleus
bull Neutrons are in the center of atom Proton are outside of Neutrons
bull The wavelength of all particles is the experimental radius
bull Double slit experiment explained correctly
bull Heisenbergrsquos inequality explained correctly
bull is the radius of approach of the magnetic lines around the torus of the electron
( 1 137)
bull Why water expands when it freezes
bull Electron clusters form lightning
rbull The meaning of Plancks constant is explained
22 2 kg m
h m f mr f consts
2210254 42m r kgm bull If the mr-constant describes a ring the equations
describe experiments bull If the mr-constant describes a wave the equations do
NOT describe experiments
Kanarevrsquos 5 Equations of Mechanics
Kanarevrsquos 1st Equation Acceleration ampmv maMma R R
Kanarevrsquos 2nd Equation Uniform motion amp mv maMma R R
maM mvR R
Kanarevrsquos 3rd Equation
Kanarevrsquos 4th Equation Equal and opposite reaction
1 21 21 2
12
a m
m a m ama
Kanarevrsquos 5th Equation 1 2 3 Mi M M M Mna R R R R R
The sum of the Actions working on a moving body is never equal to zero
Kanarevrsquos Photon model is a rolling ring
His equations accurately describe known experiments with photonsbull Electromagnetic radiation should be called photon radiationbull Diffraction and interference are the same phenomenabull Correctly explains the double slit experiment without quantum equationsbull Correctly explains the meaning of Heisenbergrsquos inequality
e
2 (2)(3142)(2817 15 ) 10007297352568
R 2426 12 137mag
xp
r m
m
Charge
Magnetic action
Proton
The intensity of the magnetic action near the geometrical center of the proton ring is so great that we theorize that it is the strong nuclear force
(9383 8)(1602 19)8476 14
4 (4)(3142)(1411 26)p
pp
EB T
M
(9395 8)(1602 19)1035 15
4 (4)(3142)(0966 26)N
NN
EB T
M
Neutron
1a Hydrogenavi
Does not exist There is no multiplication of arrows
There are two arrows One has a magnitude of 3 One has a magnitude of 5 The maximum these two arrows can be is 8
Only the resultant of arrows exist
An SM arrow has a direction and a magnitude
Arrows and magnitudes are still there even if the resultant is zero
There are no negative SM Arrows
There are no positive SM Arrows
There are no ZERO SM Arrows A zero would negate the definition of an SM Arrow something with no magnitude and no direction cannot be the definition of an arrow that is defined as having magnitude and direction
bull A resultant is the addition and subtraction of the differences in magnitude and direction arrows
A B and C are the arrows (a) A amp B exist(b) A amp B amp C exist(c) A-B Do Not Exist in Sm there are no negatives or positives(d) A+B-C Do Not Exist in SM there are no negatives or positives
There are only resultants when there are multiple arrows
SM Graphs versus BS Graphs
Numbers Raised to Powers
When numbers are raised to powers the negative sides of all BS math graphs are ILLOGICAL and INCORRECT
Incorrect BS Math for 2x
2x
4 16 correct3 8 correct2 4 correct1 2 correct0 1 correct-1 005 incorrect-2 025 incorrect-3 0125 incorrect-4 00625 incorrect
negative side of graph is incorrectpositive side of graph is correct
16
8
42105025012500625
-4 -3 -2 -1 0 1 2 3 4
po
sit
ive
negative positive
2x BS Math
correctincorrect
In BS math the dash sign (-) used in this example represents three different operations They arebull (-) Subtractionbull (-) A direction to the leftbull (-) As an exponent to mean divide by
All positive numbers produce correct results and graphs
All negative numbers produce incorrect results and graphs
A direction of three units to the left (-3) is placed in the equation as an exponent
A direction in space (3 units toward the left) changes a whole number to a fraction by using the dash (-) as an exponent rather than a direction in space
This is illogical and incorrect BS math should have performed this operation as
Unfortunately the answers to problems can come out correct because data points of a real problem fit the curve
However the answers will not be due to logical reasoning
Because the equation produced a curve that real problem data points follow it has been falsely believed that the equations and graphs are correct
1
2
x
Incorrect BK graph for (12)x
(12)x
4 00625 correct3 0125 correct2 025 correct1 1 correct0 1 correct-1 2 incorrect-2 4 incorrect-3 8 incorrect-4 16 incorrect
negative side of graph is incorrectpositive side of graph is correct
00625012502505124
8
16
-5 -4 -3 -2 -1 0 1 2 3 4 5P
osit
ive
negative positive
BS Math for (12)x
correctincorrect
Incorrect BS math and graph for
All graphs are symmetrical with SM
Two BS Math Equations that Produce Incorrect Answers
2 0x x starting equation 2
2
x x x x
x x
Add lsquoxrsquo to both sides
2(1) 1 let lsquoxrsquo = 1
2 1 This is an incorrect answer 2(0)-(0)=0 The only answer that BK math can give is x=0
1
a b starting equation ( )( ) ( )( )a a a b multiply both sides by ( )a
( )( ) ( )( ) ( )( ) ( )( )a a b b a b b b subtract ( )( )b b from both sides
( )( ) ( )( )a b a b a b b factor ( )a b from both sides
( )a b b divide both sides by ( )a b
( )b b b replace ( )a with ( )b 2b b simplify 2(1) (1) Let ( )b = 1
2 1 Again BS math yields and incorrect answer 1
This incorrect result is a problem with the distributive law SM explains how this incorrect answer is obtained and the problem with the distributive law that allows the error
Page 148 ndash149 Barry Mazur ndash Imagining Numbers
3
2
31
Do the computation carefully on paper using the rules we agreed to Then ponder your answer which should be something of a surprise to you donrsquot stop there Think of what your answer might possibly mean or might imply
Solve this equation using the distributive law
JK definitely a surprise the answer wasincorrect and illogical
Gerhard Gade University Professor doing math at Harvard University
Here is the BS math answer that Barry Mazur calculated
2
31
2
31
2
31
1 3 1 3 1 3 3 3 2 2 3 1 3
2 2 2 2 4 4 4 4 4 4 2 2
dash
1 3 1 3 1 31
2 2 2 2 4 4
3 3
4 4
When the distributive law is used with the BS Rule-of-Signs and one of the signs is a (-) dash sign the middle two terms always cancel each other This leads to the incorrect answer of (-1) in the above equation
Barry Mazur did not go back and check his answer with the original equation If he had he would have known that the answer was incorrect
3
1 31
2
check answer by setting equal
3
331 3
12
1 31
2
Take the Cube root of both sides
1 3 2 multiply both sides by 2
3 3 subtract 1 from both sides
2 2
3 3 square both sides
3 9 This is an incorrect answer A simple equation check could have shown that the distributive law using BS math provides an incorrect answer
1
BS Illogical and Incorrect Numbers Raised to Powers
nC D n DC
642 DDD
In BS math only a positive number can have an answer if ldquonrdquo is an even number (246---)
All negative numbers have no solutions
642 DDD
With SM just add a direction arrow and any number gives the correct answer
or n nC D C D
3 3( 3 8 8 or 8 2 or 2 )n D C
( 2 16 C= 16 or 16 4 or 4 )n D
Since a negative number has no square root answer in BS math BS mathematicians invented imaginary numbers
i2=-1
24414 iiC n
These equations are easily solved without the use of imaginary numbers with SM math
If 4 then 4 2x
If 4 then 4 2x
The 360-degree circle is a 4400 years old extremely outdated system based on a year of 12 months of 30 days each that is 360 days
SM will use a more logical system that will have a circle of 6283185307o
d
( )
(deg )
d r radians
d r rees
the arc distance around
the circumference of a circle
d
radiusr
2 2
2 2
d r
dv rd t t
v dda rd t t t
1
d
r
( )av tv
The average angular velocity of a body is the rate of its angular displacement
( )0
liminstt
vt
The instantaneous angular velocity is the limit of the ratio
t
as t approaches zero
( )d
av
d
tv
The average circumference velocity of the body is the rate of its distance traveled d divided by the time t
dvv
r
1 dd vdrv
t t t r r
The equation of the relationship between the angular velocity and the velocity of a point on that body
( )o
av
v v v
t ta
The average angular acceleration is the change in the angular velocity
( )0
liminstt
v
ta
The instantaneous angular acceleration is the
limit of the ration of v
t
as t approaches
zero
daa
r
1
1
d
d s d
vva
t r tv a
r t r
The equation of the relationship between the angular acceleration and the acceleration of a point on the body
21
2v a t
If a body with an initial angular velocity ov has a constant
angular acceleration a it will turn through an angle in a time t
1
Straight Line Symbol Units Rotational Symbol Units Distance d m Angular
Distance radians or degrees
Velocity dt
v m
s Angular Velocity
V
t
radians or degrees
s
Acceleration vat
2ms
Angular Acceleration
VA
t
2
radians or degrees
s
Mass M Kg Mass distribution
dM 2Kgm
Force actionA Ma
2
Kgm
sNewton
Torque
dT M A 2
2
Kgm
s
Momentum Mv Kgm
s
Angular Momentum
dM V 2Kgm
s
Work (Force)(distance)
Ad 2
Kgmm
sNm Joule
Work T
Using AS arrow math to solve and direction in space problems using the equation 1 2
2 2 2 2 2 2( ) amp and ( ) amp 3
4
2 2 2
2 2
( ) amp
(5 ) 4 3
(5 ) 16 9
(5 ) 25
(5 ) (5 )
2 2(5 ) 4 3
(5 ) 16 9
(5 ) 25
(5 ) (5 )
In SM math Pythagorasrsquo theorem provides both magnitude and direction
In BS math Pythagorasrsquo theorem provides magnitude only
time set equal to zero
23
( ) ( 0 )10 1
x t xx
3( ) ( 0)
10 1W t f
In BS math this is an equation for a traveling wave
must use a square term in order to make the curve symmetric
NO SQUARE TERM IS NECESSARY
In SM math when t is set to ldquo0rdquo
This is a major fault in BS math
x f(x)
-4 0019
-3 0033
-2 0073
-1 0273
0 3000
1 0273
2 0073
3 0033
4 0019 -4 -3 -2 -1 0 1 2 3 400
05
10
15
20
25
30
35
BS wave (must use square term)
snap shot of wave at x = 0
Am
plit
ude
lt to gt lt to gt f(ltgt)
lt -4 0073
lt -3 0097
lt -2 0143
lt -1 0273
origin 0 3000
gt 1 0273
gt 2 0143
gt 3 0097
gt 4 0073lt0073 lt0097 lt0143 lt0273
3000
0273gt 0143gt 0097gt 0073gt
Wave moving
snap shot of wave
Am
plit
ude
lt4 lt3 lt2 lt1 0 1gt 2gt 3gt 4gt
3
2
1
In SM NO squared terms are needed
SPRING
An Action is needed to stretch or compress a spring to a starting point where it can execute cycle motion
tsA (ts to start) is the work required to move a mass attached to a spring a distance (d1) from
its initial rest position RAK 0 the spring constant is the stiffness of the spring
o A is attraction o R is repulsion o
ARRA
00 is for the back and forth motion through the zero initial rest position A
stretched spring goes from ldquoArdquo to ldquo0rdquo to ldquoRrdquo to ldquoArdquo to ldquo0rdquo to ldquoRrdquo d1 is the distance a mass is moved from its initial rest position
0
0 1ts A RA K d
Once ( )mad
is performed to move the mass to position d1 (against the restraint of the spring) a of the ( )mad
is transferred into the spring as potential attractionrepulsion (PAR) There is always
more ( )mad
performed than is stored as PAR When the spring is released from the restraining Ats the PAR starts the oscillation Conservation of ( )mad
requires that the PAR and the motion ( )mad
( )( )madM be a constant
( )amp ( ) constantmadPAR M 02
0 0
1
2A R A RPAR K d
02
( ) 0 0
1( )
2mad A R A RM m v
0 0 02 2 2
0 0 0 1
2 2 21
2 2 21
1 1 1amp
2 2 2
1 1 1amp
2 2 2
amp
A R A R A RK d m v K d
Kd mv Kd
Kd mv Kd
0
1d is frac12 the height from the base to the peak of a cycle
An experiment of moving paper under a pendulum through a small angle filled with sand that can escape through a small hole in the bottom shows that a pendulum cycling back and forth follows a curve that is mathematically analogous to a lsquosinrsquo or lsquocosrsquo
The projection of the curved area onto the straight line exactly duplicate the back and forth motion of CM This allows for the math of lsquosinrsquo or lsquocosrsquo to be used
SM will use lsquosinrdquo math It could use lsquocosrsquo or a combination of both lsquosinrsquo and lsquocosrsquo Using just the lsquosinrsquo math makes the understanding and math easier
0 sin 0 0SM 0 sin157 1SM The diagram at the left shows a spring being stretched from its (zero rest) position-2 and the PAR being released at position-1 There are 4 sections (1 to 2 2 to 3 3 to 4 and 4 to5) that alternate between maximum acceleration and zero velocity to max velocity and zero acceleration
At position-1 there is maximum acceleration and zero velocity o The lsquosinrsquo values from 0o to 157o will be used for velocity (values from 0 to 1) o The lsquosinrsquo values from 157o to 0o will be used for acceleration (values from 1 to 0)
At position-2 there is maximum velocity and zero acceleration Acceleration changes to deceleration
o The lsquosinrsquo values from 157o to 0o will be used for deceleration (values from 1 to 0) o The lsquosinrsquo values from 0o to 157o will be used for acceleration (values from 0 to 1)
At position-3 there is maximum acceleration and zero velocity o The lsquosinrsquo values from 0o to 157o will be used for velocity (values from 0 to 1) o The lsquosinrsquo values from 157o to 0o will be used for acceleration (values from 1 to 0)
At position-4 there is maximum velocity and zero acceleration Acceleration changes to deceleration
o The lsquosinrsquo values from 157o to 0o will be used for deceleration (values from 1 to 0) o The lsquosinrsquo values from 0o to 157o will be used for acceleration (values from 0 to 1)
At position-5 the cycle starts again
2
)539(
cT
MK
spring constant (K) when mass (M) and cycle-time (T) are known
)(
))(89)((
111 md
uKgM
d
MG
d
FK sts
spring constant (K) when mass (M) and amplitude ( d1) are known
2 21( )
Kv d d
m
velocity (v) at any point during the cycle
velocity
dforM
Kdv
max
012 velocity (v2) maximum (when d = 0 at position-2 and 4)
acceleration (a1) maximum when spring constant (K) mass (M) and amplitude (d1) are known
acceleration (a1) maximum when amplitude (d1) and cycle-time (Tc) are known
acceleration (a) at any point during the cycle-time (T)
1
151
286
v
dTc
cycle-time (Tc) when amplitude (d1) and velocity (v1) are known
K
MTc 28651
cycle-time (Tc) when mass (M) and spring constant (K) are known
fTc
1
cycle-time (Tc) when the frequency is known
cTf
1
frequency (f) when the cycle-time (Tc) is known
of cycles
elapsed timef
frequency (f) when the number of cycles in an elapsed time is known
d1 amplitude (distance from rest to start position)
Also frac12 the distance from the base to the peak
3131 oror dM
Ka
251
3131 539
c
oror T
da
2539
cT
da
In December of 2006 I discovered the mathematical error in Einsteinrsquos Special Relativity Einstein used BS math and that math only works in an imaginary world
It cannot describe real symmetrical space
Einsteinrsquos Special Relativity should be abandoned and a theory of real space using SM should be used to replace it
Numerous articles on the Internet show Einsteinrsquos theory to be incorrect but no one in academia is willing to publish the information
Figure 1-10 A moving particle observed from two frames of reference in uniform translation
Einsteinrsquos Math Error in Special Relativity Transformation of Velocity (From Eisenberg- Fundamentals of Modern Physics pages 27-29) Consider the particle shown in figure (1-10) moving with velocity v as
seen in a frame of reference O
dx dy dz
Vx Vy Vzdt dt dt
dx dy dz
V x V y V zdy dt dt
We would like to evaluate the velocity v of the particle as seen in the frame of reference 0 which is itself moving relative to 0 with velocity v Measured in the 0 frame the velocity vector of the particle has com ponents
The same velocity vector as measured in the 0 frame has components
2
2
2
2
2
1
1
x vtx
v
cy y
z z
vxt
ctv
c
Now from equations (1-13) we know that the relation between the primed and the unprimed coordinates and times is
Take the differential of these equations remembering that v is a constant This gives
2
2
2
2
2
1
1
dx vdtdx
v
cdy dy
dz dz
vxdt
cdtv
c
2
2
2 2 2 2 2
2
2
2
11 ( )
1( ) 1
1
1
1
dx vdt
v dxdx vdt v
dx dx vdtc dt dtvdx vdx vdx v dxdt dt dt dt
dt dtc c c c
VxV x
Vx
VxV x
vv
vv
c
x
v
c
cV
( )
coordinate of particle relative to moving primed system
coordinate of particle relative to stationary system
velocity of primed coordina
V x
Vx
te system relative to stationaryv sys
tem
Page 28 Fundamentals of Modern Physics
(1-16)
According to Einstein this equation tells you how to transform the observed velocity from one frame of reference to another frame of reference First we note that as Vc and vc approachzero the equation approaches those which would be derived from the Galilean transformation
2 2
( ) 1 01 1 ( )
Vx v Vx vV
VxV x
Vx v
v cVxx
v
c
v
cc
Another property of this equation is that it is impossible to choose V and v such that Vrsquo the magnitude of the velocity which is seen in the new frame is greater than c
1
2
09 ( 09 ) 18009876543
( 09 ) 1811 (09 )
c c cV c c
cc
c
Example As seen by O particle 1 has velocity 09c in the direction of positive x and particle 2 has velocity 09c in the negative x direction
Einstein equation however predicts To evaluate the velocity of particle 1 with respect to particle 2 we transform from the O frame to the Orsquo frame moving in the negative x direction with velocity v = -09c using equation 1-16 We obtain
JK LOGIC will observe the particles separating at 18c
A confusing INCORRECT and ILLOGICAL answer
left is a dir rightsub of a
1
2
1
09 ( 09 ) 18009876543
181( 09 )1 ( 09 )
The multiplication of dir left time a dir right is ONLY a dir left
09
dir
direction leftdirection rightsubtraction
c c cV c c
cc
c
V
dir
22
22
09 18 18009876543
181( 081)1(09)(09)
1
left rightsub
sub
c c c cc c
ccc
c
Analyzing Einsteinrsquos BS math Equation
Using SM math a logical answer can be obtained
Cannot multiply a subtraction operator times a direction in space
Can only subtract and change its direction
Cannot multiply a direction to the left times a direction to the right
Cannot multiply opposite directions in space
Can only solve for the resultant
rightleft
subtraction
1
subtraction
2
sub of a dir to the left is a dir to the right
1 resultant 0 sub 2
2sub 2
2
09 09
09 amp 091
09 09 18 1
(0)1
(09) amp (09)1
c cV
c cc
c c cV
cc
cc
8
181
cc
SM math predicts a logical answer using logic
21st century scientists state in many articles that Einsteins General Relativity and Quantum Mechanics are incompatible
They are not only incompatible they are both mathematically incorrect
They both use illogical BS math
Any equations in math that use imaginary numbers should be abandoned
Imaginary numbers are only necessary in the current BS math system
This means that Schroumldingers equations should be abandoned
Any equation in quantum mechanics that uses imaginary number should be abandoned
Prof Kanarev has developed logical physics and chemistrytheories without the use of Quantum mechanics or relativity
A partial list of his theories coverbull Mechanics ndash replaces Newtonbull Photon - bull Electrons ndash Protons- Neutronsbull Spectroscopybull Nucleusbull Atoms amp Moleculesbull Electrodynamicsbull Heatbull Curvature of Space ndash replaces Schwarzcchildrsquos equation
Kanarevrsquos Hypothesis
bull Electron align linearly with proton They do NOT spin or form clouds around the nucleus
bull Neutrons are in the center of atom Proton are outside of Neutrons
bull The wavelength of all particles is the experimental radius
bull Double slit experiment explained correctly
bull Heisenbergrsquos inequality explained correctly
bull is the radius of approach of the magnetic lines around the torus of the electron
( 1 137)
bull Why water expands when it freezes
bull Electron clusters form lightning
rbull The meaning of Plancks constant is explained
22 2 kg m
h m f mr f consts
2210254 42m r kgm bull If the mr-constant describes a ring the equations
describe experiments bull If the mr-constant describes a wave the equations do
NOT describe experiments
Kanarevrsquos 5 Equations of Mechanics
Kanarevrsquos 1st Equation Acceleration ampmv maMma R R
Kanarevrsquos 2nd Equation Uniform motion amp mv maMma R R
maM mvR R
Kanarevrsquos 3rd Equation
Kanarevrsquos 4th Equation Equal and opposite reaction
1 21 21 2
12
a m
m a m ama
Kanarevrsquos 5th Equation 1 2 3 Mi M M M Mna R R R R R
The sum of the Actions working on a moving body is never equal to zero
Kanarevrsquos Photon model is a rolling ring
His equations accurately describe known experiments with photonsbull Electromagnetic radiation should be called photon radiationbull Diffraction and interference are the same phenomenabull Correctly explains the double slit experiment without quantum equationsbull Correctly explains the meaning of Heisenbergrsquos inequality
e
2 (2)(3142)(2817 15 ) 10007297352568
R 2426 12 137mag
xp
r m
m
Charge
Magnetic action
Proton
The intensity of the magnetic action near the geometrical center of the proton ring is so great that we theorize that it is the strong nuclear force
(9383 8)(1602 19)8476 14
4 (4)(3142)(1411 26)p
pp
EB T
M
(9395 8)(1602 19)1035 15
4 (4)(3142)(0966 26)N
NN
EB T
M
Neutron
1a Hydrogenavi
An SM arrow has a direction and a magnitude
Arrows and magnitudes are still there even if the resultant is zero
There are no negative SM Arrows
There are no positive SM Arrows
There are no ZERO SM Arrows A zero would negate the definition of an SM Arrow something with no magnitude and no direction cannot be the definition of an arrow that is defined as having magnitude and direction
bull A resultant is the addition and subtraction of the differences in magnitude and direction arrows
A B and C are the arrows (a) A amp B exist(b) A amp B amp C exist(c) A-B Do Not Exist in Sm there are no negatives or positives(d) A+B-C Do Not Exist in SM there are no negatives or positives
There are only resultants when there are multiple arrows
SM Graphs versus BS Graphs
Numbers Raised to Powers
When numbers are raised to powers the negative sides of all BS math graphs are ILLOGICAL and INCORRECT
Incorrect BS Math for 2x
2x
4 16 correct3 8 correct2 4 correct1 2 correct0 1 correct-1 005 incorrect-2 025 incorrect-3 0125 incorrect-4 00625 incorrect
negative side of graph is incorrectpositive side of graph is correct
16
8
42105025012500625
-4 -3 -2 -1 0 1 2 3 4
po
sit
ive
negative positive
2x BS Math
correctincorrect
In BS math the dash sign (-) used in this example represents three different operations They arebull (-) Subtractionbull (-) A direction to the leftbull (-) As an exponent to mean divide by
All positive numbers produce correct results and graphs
All negative numbers produce incorrect results and graphs
A direction of three units to the left (-3) is placed in the equation as an exponent
A direction in space (3 units toward the left) changes a whole number to a fraction by using the dash (-) as an exponent rather than a direction in space
This is illogical and incorrect BS math should have performed this operation as
Unfortunately the answers to problems can come out correct because data points of a real problem fit the curve
However the answers will not be due to logical reasoning
Because the equation produced a curve that real problem data points follow it has been falsely believed that the equations and graphs are correct
1
2
x
Incorrect BK graph for (12)x
(12)x
4 00625 correct3 0125 correct2 025 correct1 1 correct0 1 correct-1 2 incorrect-2 4 incorrect-3 8 incorrect-4 16 incorrect
negative side of graph is incorrectpositive side of graph is correct
00625012502505124
8
16
-5 -4 -3 -2 -1 0 1 2 3 4 5P
osit
ive
negative positive
BS Math for (12)x
correctincorrect
Incorrect BS math and graph for
All graphs are symmetrical with SM
Two BS Math Equations that Produce Incorrect Answers
2 0x x starting equation 2
2
x x x x
x x
Add lsquoxrsquo to both sides
2(1) 1 let lsquoxrsquo = 1
2 1 This is an incorrect answer 2(0)-(0)=0 The only answer that BK math can give is x=0
1
a b starting equation ( )( ) ( )( )a a a b multiply both sides by ( )a
( )( ) ( )( ) ( )( ) ( )( )a a b b a b b b subtract ( )( )b b from both sides
( )( ) ( )( )a b a b a b b factor ( )a b from both sides
( )a b b divide both sides by ( )a b
( )b b b replace ( )a with ( )b 2b b simplify 2(1) (1) Let ( )b = 1
2 1 Again BS math yields and incorrect answer 1
This incorrect result is a problem with the distributive law SM explains how this incorrect answer is obtained and the problem with the distributive law that allows the error
Page 148 ndash149 Barry Mazur ndash Imagining Numbers
3
2
31
Do the computation carefully on paper using the rules we agreed to Then ponder your answer which should be something of a surprise to you donrsquot stop there Think of what your answer might possibly mean or might imply
Solve this equation using the distributive law
JK definitely a surprise the answer wasincorrect and illogical
Gerhard Gade University Professor doing math at Harvard University
Here is the BS math answer that Barry Mazur calculated
2
31
2
31
2
31
1 3 1 3 1 3 3 3 2 2 3 1 3
2 2 2 2 4 4 4 4 4 4 2 2
dash
1 3 1 3 1 31
2 2 2 2 4 4
3 3
4 4
When the distributive law is used with the BS Rule-of-Signs and one of the signs is a (-) dash sign the middle two terms always cancel each other This leads to the incorrect answer of (-1) in the above equation
Barry Mazur did not go back and check his answer with the original equation If he had he would have known that the answer was incorrect
3
1 31
2
check answer by setting equal
3
331 3
12
1 31
2
Take the Cube root of both sides
1 3 2 multiply both sides by 2
3 3 subtract 1 from both sides
2 2
3 3 square both sides
3 9 This is an incorrect answer A simple equation check could have shown that the distributive law using BS math provides an incorrect answer
1
BS Illogical and Incorrect Numbers Raised to Powers
nC D n DC
642 DDD
In BS math only a positive number can have an answer if ldquonrdquo is an even number (246---)
All negative numbers have no solutions
642 DDD
With SM just add a direction arrow and any number gives the correct answer
or n nC D C D
3 3( 3 8 8 or 8 2 or 2 )n D C
( 2 16 C= 16 or 16 4 or 4 )n D
Since a negative number has no square root answer in BS math BS mathematicians invented imaginary numbers
i2=-1
24414 iiC n
These equations are easily solved without the use of imaginary numbers with SM math
If 4 then 4 2x
If 4 then 4 2x
The 360-degree circle is a 4400 years old extremely outdated system based on a year of 12 months of 30 days each that is 360 days
SM will use a more logical system that will have a circle of 6283185307o
d
( )
(deg )
d r radians
d r rees
the arc distance around
the circumference of a circle
d
radiusr
2 2
2 2
d r
dv rd t t
v dda rd t t t
1
d
r
( )av tv
The average angular velocity of a body is the rate of its angular displacement
( )0
liminstt
vt
The instantaneous angular velocity is the limit of the ratio
t
as t approaches zero
( )d
av
d
tv
The average circumference velocity of the body is the rate of its distance traveled d divided by the time t
dvv
r
1 dd vdrv
t t t r r
The equation of the relationship between the angular velocity and the velocity of a point on that body
( )o
av
v v v
t ta
The average angular acceleration is the change in the angular velocity
( )0
liminstt
v
ta
The instantaneous angular acceleration is the
limit of the ration of v
t
as t approaches
zero
daa
r
1
1
d
d s d
vva
t r tv a
r t r
The equation of the relationship between the angular acceleration and the acceleration of a point on the body
21
2v a t
If a body with an initial angular velocity ov has a constant
angular acceleration a it will turn through an angle in a time t
1
Straight Line Symbol Units Rotational Symbol Units Distance d m Angular
Distance radians or degrees
Velocity dt
v m
s Angular Velocity
V
t
radians or degrees
s
Acceleration vat
2ms
Angular Acceleration
VA
t
2
radians or degrees
s
Mass M Kg Mass distribution
dM 2Kgm
Force actionA Ma
2
Kgm
sNewton
Torque
dT M A 2
2
Kgm
s
Momentum Mv Kgm
s
Angular Momentum
dM V 2Kgm
s
Work (Force)(distance)
Ad 2
Kgmm
sNm Joule
Work T
Using AS arrow math to solve and direction in space problems using the equation 1 2
2 2 2 2 2 2( ) amp and ( ) amp 3
4
2 2 2
2 2
( ) amp
(5 ) 4 3
(5 ) 16 9
(5 ) 25
(5 ) (5 )
2 2(5 ) 4 3
(5 ) 16 9
(5 ) 25
(5 ) (5 )
In SM math Pythagorasrsquo theorem provides both magnitude and direction
In BS math Pythagorasrsquo theorem provides magnitude only
time set equal to zero
23
( ) ( 0 )10 1
x t xx
3( ) ( 0)
10 1W t f
In BS math this is an equation for a traveling wave
must use a square term in order to make the curve symmetric
NO SQUARE TERM IS NECESSARY
In SM math when t is set to ldquo0rdquo
This is a major fault in BS math
x f(x)
-4 0019
-3 0033
-2 0073
-1 0273
0 3000
1 0273
2 0073
3 0033
4 0019 -4 -3 -2 -1 0 1 2 3 400
05
10
15
20
25
30
35
BS wave (must use square term)
snap shot of wave at x = 0
Am
plit
ude
lt to gt lt to gt f(ltgt)
lt -4 0073
lt -3 0097
lt -2 0143
lt -1 0273
origin 0 3000
gt 1 0273
gt 2 0143
gt 3 0097
gt 4 0073lt0073 lt0097 lt0143 lt0273
3000
0273gt 0143gt 0097gt 0073gt
Wave moving
snap shot of wave
Am
plit
ude
lt4 lt3 lt2 lt1 0 1gt 2gt 3gt 4gt
3
2
1
In SM NO squared terms are needed
SPRING
An Action is needed to stretch or compress a spring to a starting point where it can execute cycle motion
tsA (ts to start) is the work required to move a mass attached to a spring a distance (d1) from
its initial rest position RAK 0 the spring constant is the stiffness of the spring
o A is attraction o R is repulsion o
ARRA
00 is for the back and forth motion through the zero initial rest position A
stretched spring goes from ldquoArdquo to ldquo0rdquo to ldquoRrdquo to ldquoArdquo to ldquo0rdquo to ldquoRrdquo d1 is the distance a mass is moved from its initial rest position
0
0 1ts A RA K d
Once ( )mad
is performed to move the mass to position d1 (against the restraint of the spring) a of the ( )mad
is transferred into the spring as potential attractionrepulsion (PAR) There is always
more ( )mad
performed than is stored as PAR When the spring is released from the restraining Ats the PAR starts the oscillation Conservation of ( )mad
requires that the PAR and the motion ( )mad
( )( )madM be a constant
( )amp ( ) constantmadPAR M 02
0 0
1
2A R A RPAR K d
02
( ) 0 0
1( )
2mad A R A RM m v
0 0 02 2 2
0 0 0 1
2 2 21
2 2 21
1 1 1amp
2 2 2
1 1 1amp
2 2 2
amp
A R A R A RK d m v K d
Kd mv Kd
Kd mv Kd
0
1d is frac12 the height from the base to the peak of a cycle
An experiment of moving paper under a pendulum through a small angle filled with sand that can escape through a small hole in the bottom shows that a pendulum cycling back and forth follows a curve that is mathematically analogous to a lsquosinrsquo or lsquocosrsquo
The projection of the curved area onto the straight line exactly duplicate the back and forth motion of CM This allows for the math of lsquosinrsquo or lsquocosrsquo to be used
SM will use lsquosinrdquo math It could use lsquocosrsquo or a combination of both lsquosinrsquo and lsquocosrsquo Using just the lsquosinrsquo math makes the understanding and math easier
0 sin 0 0SM 0 sin157 1SM The diagram at the left shows a spring being stretched from its (zero rest) position-2 and the PAR being released at position-1 There are 4 sections (1 to 2 2 to 3 3 to 4 and 4 to5) that alternate between maximum acceleration and zero velocity to max velocity and zero acceleration
At position-1 there is maximum acceleration and zero velocity o The lsquosinrsquo values from 0o to 157o will be used for velocity (values from 0 to 1) o The lsquosinrsquo values from 157o to 0o will be used for acceleration (values from 1 to 0)
At position-2 there is maximum velocity and zero acceleration Acceleration changes to deceleration
o The lsquosinrsquo values from 157o to 0o will be used for deceleration (values from 1 to 0) o The lsquosinrsquo values from 0o to 157o will be used for acceleration (values from 0 to 1)
At position-3 there is maximum acceleration and zero velocity o The lsquosinrsquo values from 0o to 157o will be used for velocity (values from 0 to 1) o The lsquosinrsquo values from 157o to 0o will be used for acceleration (values from 1 to 0)
At position-4 there is maximum velocity and zero acceleration Acceleration changes to deceleration
o The lsquosinrsquo values from 157o to 0o will be used for deceleration (values from 1 to 0) o The lsquosinrsquo values from 0o to 157o will be used for acceleration (values from 0 to 1)
At position-5 the cycle starts again
2
)539(
cT
MK
spring constant (K) when mass (M) and cycle-time (T) are known
)(
))(89)((
111 md
uKgM
d
MG
d
FK sts
spring constant (K) when mass (M) and amplitude ( d1) are known
2 21( )
Kv d d
m
velocity (v) at any point during the cycle
velocity
dforM
Kdv
max
012 velocity (v2) maximum (when d = 0 at position-2 and 4)
acceleration (a1) maximum when spring constant (K) mass (M) and amplitude (d1) are known
acceleration (a1) maximum when amplitude (d1) and cycle-time (Tc) are known
acceleration (a) at any point during the cycle-time (T)
1
151
286
v
dTc
cycle-time (Tc) when amplitude (d1) and velocity (v1) are known
K
MTc 28651
cycle-time (Tc) when mass (M) and spring constant (K) are known
fTc
1
cycle-time (Tc) when the frequency is known
cTf
1
frequency (f) when the cycle-time (Tc) is known
of cycles
elapsed timef
frequency (f) when the number of cycles in an elapsed time is known
d1 amplitude (distance from rest to start position)
Also frac12 the distance from the base to the peak
3131 oror dM
Ka
251
3131 539
c
oror T
da
2539
cT
da
In December of 2006 I discovered the mathematical error in Einsteinrsquos Special Relativity Einstein used BS math and that math only works in an imaginary world
It cannot describe real symmetrical space
Einsteinrsquos Special Relativity should be abandoned and a theory of real space using SM should be used to replace it
Numerous articles on the Internet show Einsteinrsquos theory to be incorrect but no one in academia is willing to publish the information
Figure 1-10 A moving particle observed from two frames of reference in uniform translation
Einsteinrsquos Math Error in Special Relativity Transformation of Velocity (From Eisenberg- Fundamentals of Modern Physics pages 27-29) Consider the particle shown in figure (1-10) moving with velocity v as
seen in a frame of reference O
dx dy dz
Vx Vy Vzdt dt dt
dx dy dz
V x V y V zdy dt dt
We would like to evaluate the velocity v of the particle as seen in the frame of reference 0 which is itself moving relative to 0 with velocity v Measured in the 0 frame the velocity vector of the particle has com ponents
The same velocity vector as measured in the 0 frame has components
2
2
2
2
2
1
1
x vtx
v
cy y
z z
vxt
ctv
c
Now from equations (1-13) we know that the relation between the primed and the unprimed coordinates and times is
Take the differential of these equations remembering that v is a constant This gives
2
2
2
2
2
1
1
dx vdtdx
v
cdy dy
dz dz
vxdt
cdtv
c
2
2
2 2 2 2 2
2
2
2
11 ( )
1( ) 1
1
1
1
dx vdt
v dxdx vdt v
dx dx vdtc dt dtvdx vdx vdx v dxdt dt dt dt
dt dtc c c c
VxV x
Vx
VxV x
vv
vv
c
x
v
c
cV
( )
coordinate of particle relative to moving primed system
coordinate of particle relative to stationary system
velocity of primed coordina
V x
Vx
te system relative to stationaryv sys
tem
Page 28 Fundamentals of Modern Physics
(1-16)
According to Einstein this equation tells you how to transform the observed velocity from one frame of reference to another frame of reference First we note that as Vc and vc approachzero the equation approaches those which would be derived from the Galilean transformation
2 2
( ) 1 01 1 ( )
Vx v Vx vV
VxV x
Vx v
v cVxx
v
c
v
cc
Another property of this equation is that it is impossible to choose V and v such that Vrsquo the magnitude of the velocity which is seen in the new frame is greater than c
1
2
09 ( 09 ) 18009876543
( 09 ) 1811 (09 )
c c cV c c
cc
c
Example As seen by O particle 1 has velocity 09c in the direction of positive x and particle 2 has velocity 09c in the negative x direction
Einstein equation however predicts To evaluate the velocity of particle 1 with respect to particle 2 we transform from the O frame to the Orsquo frame moving in the negative x direction with velocity v = -09c using equation 1-16 We obtain
JK LOGIC will observe the particles separating at 18c
A confusing INCORRECT and ILLOGICAL answer
left is a dir rightsub of a
1
2
1
09 ( 09 ) 18009876543
181( 09 )1 ( 09 )
The multiplication of dir left time a dir right is ONLY a dir left
09
dir
direction leftdirection rightsubtraction
c c cV c c
cc
c
V
dir
22
22
09 18 18009876543
181( 081)1(09)(09)
1
left rightsub
sub
c c c cc c
ccc
c
Analyzing Einsteinrsquos BS math Equation
Using SM math a logical answer can be obtained
Cannot multiply a subtraction operator times a direction in space
Can only subtract and change its direction
Cannot multiply a direction to the left times a direction to the right
Cannot multiply opposite directions in space
Can only solve for the resultant
rightleft
subtraction
1
subtraction
2
sub of a dir to the left is a dir to the right
1 resultant 0 sub 2
2sub 2
2
09 09
09 amp 091
09 09 18 1
(0)1
(09) amp (09)1
c cV
c cc
c c cV
cc
cc
8
181
cc
SM math predicts a logical answer using logic
21st century scientists state in many articles that Einsteins General Relativity and Quantum Mechanics are incompatible
They are not only incompatible they are both mathematically incorrect
They both use illogical BS math
Any equations in math that use imaginary numbers should be abandoned
Imaginary numbers are only necessary in the current BS math system
This means that Schroumldingers equations should be abandoned
Any equation in quantum mechanics that uses imaginary number should be abandoned
Prof Kanarev has developed logical physics and chemistrytheories without the use of Quantum mechanics or relativity
A partial list of his theories coverbull Mechanics ndash replaces Newtonbull Photon - bull Electrons ndash Protons- Neutronsbull Spectroscopybull Nucleusbull Atoms amp Moleculesbull Electrodynamicsbull Heatbull Curvature of Space ndash replaces Schwarzcchildrsquos equation
Kanarevrsquos Hypothesis
bull Electron align linearly with proton They do NOT spin or form clouds around the nucleus
bull Neutrons are in the center of atom Proton are outside of Neutrons
bull The wavelength of all particles is the experimental radius
bull Double slit experiment explained correctly
bull Heisenbergrsquos inequality explained correctly
bull is the radius of approach of the magnetic lines around the torus of the electron
( 1 137)
bull Why water expands when it freezes
bull Electron clusters form lightning
rbull The meaning of Plancks constant is explained
22 2 kg m
h m f mr f consts
2210254 42m r kgm bull If the mr-constant describes a ring the equations
describe experiments bull If the mr-constant describes a wave the equations do
NOT describe experiments
Kanarevrsquos 5 Equations of Mechanics
Kanarevrsquos 1st Equation Acceleration ampmv maMma R R
Kanarevrsquos 2nd Equation Uniform motion amp mv maMma R R
maM mvR R
Kanarevrsquos 3rd Equation
Kanarevrsquos 4th Equation Equal and opposite reaction
1 21 21 2
12
a m
m a m ama
Kanarevrsquos 5th Equation 1 2 3 Mi M M M Mna R R R R R
The sum of the Actions working on a moving body is never equal to zero
Kanarevrsquos Photon model is a rolling ring
His equations accurately describe known experiments with photonsbull Electromagnetic radiation should be called photon radiationbull Diffraction and interference are the same phenomenabull Correctly explains the double slit experiment without quantum equationsbull Correctly explains the meaning of Heisenbergrsquos inequality
e
2 (2)(3142)(2817 15 ) 10007297352568
R 2426 12 137mag
xp
r m
m
Charge
Magnetic action
Proton
The intensity of the magnetic action near the geometrical center of the proton ring is so great that we theorize that it is the strong nuclear force
(9383 8)(1602 19)8476 14
4 (4)(3142)(1411 26)p
pp
EB T
M
(9395 8)(1602 19)1035 15
4 (4)(3142)(0966 26)N
NN
EB T
M
Neutron
1a Hydrogenavi
A B and C are the arrows (a) A amp B exist(b) A amp B amp C exist(c) A-B Do Not Exist in Sm there are no negatives or positives(d) A+B-C Do Not Exist in SM there are no negatives or positives
There are only resultants when there are multiple arrows
SM Graphs versus BS Graphs
Numbers Raised to Powers
When numbers are raised to powers the negative sides of all BS math graphs are ILLOGICAL and INCORRECT
Incorrect BS Math for 2x
2x
4 16 correct3 8 correct2 4 correct1 2 correct0 1 correct-1 005 incorrect-2 025 incorrect-3 0125 incorrect-4 00625 incorrect
negative side of graph is incorrectpositive side of graph is correct
16
8
42105025012500625
-4 -3 -2 -1 0 1 2 3 4
po
sit
ive
negative positive
2x BS Math
correctincorrect
In BS math the dash sign (-) used in this example represents three different operations They arebull (-) Subtractionbull (-) A direction to the leftbull (-) As an exponent to mean divide by
All positive numbers produce correct results and graphs
All negative numbers produce incorrect results and graphs
A direction of three units to the left (-3) is placed in the equation as an exponent
A direction in space (3 units toward the left) changes a whole number to a fraction by using the dash (-) as an exponent rather than a direction in space
This is illogical and incorrect BS math should have performed this operation as
Unfortunately the answers to problems can come out correct because data points of a real problem fit the curve
However the answers will not be due to logical reasoning
Because the equation produced a curve that real problem data points follow it has been falsely believed that the equations and graphs are correct
1
2
x
Incorrect BK graph for (12)x
(12)x
4 00625 correct3 0125 correct2 025 correct1 1 correct0 1 correct-1 2 incorrect-2 4 incorrect-3 8 incorrect-4 16 incorrect
negative side of graph is incorrectpositive side of graph is correct
00625012502505124
8
16
-5 -4 -3 -2 -1 0 1 2 3 4 5P
osit
ive
negative positive
BS Math for (12)x
correctincorrect
Incorrect BS math and graph for
All graphs are symmetrical with SM
Two BS Math Equations that Produce Incorrect Answers
2 0x x starting equation 2
2
x x x x
x x
Add lsquoxrsquo to both sides
2(1) 1 let lsquoxrsquo = 1
2 1 This is an incorrect answer 2(0)-(0)=0 The only answer that BK math can give is x=0
1
a b starting equation ( )( ) ( )( )a a a b multiply both sides by ( )a
( )( ) ( )( ) ( )( ) ( )( )a a b b a b b b subtract ( )( )b b from both sides
( )( ) ( )( )a b a b a b b factor ( )a b from both sides
( )a b b divide both sides by ( )a b
( )b b b replace ( )a with ( )b 2b b simplify 2(1) (1) Let ( )b = 1
2 1 Again BS math yields and incorrect answer 1
This incorrect result is a problem with the distributive law SM explains how this incorrect answer is obtained and the problem with the distributive law that allows the error
Page 148 ndash149 Barry Mazur ndash Imagining Numbers
3
2
31
Do the computation carefully on paper using the rules we agreed to Then ponder your answer which should be something of a surprise to you donrsquot stop there Think of what your answer might possibly mean or might imply
Solve this equation using the distributive law
JK definitely a surprise the answer wasincorrect and illogical
Gerhard Gade University Professor doing math at Harvard University
Here is the BS math answer that Barry Mazur calculated
2
31
2
31
2
31
1 3 1 3 1 3 3 3 2 2 3 1 3
2 2 2 2 4 4 4 4 4 4 2 2
dash
1 3 1 3 1 31
2 2 2 2 4 4
3 3
4 4
When the distributive law is used with the BS Rule-of-Signs and one of the signs is a (-) dash sign the middle two terms always cancel each other This leads to the incorrect answer of (-1) in the above equation
Barry Mazur did not go back and check his answer with the original equation If he had he would have known that the answer was incorrect
3
1 31
2
check answer by setting equal
3
331 3
12
1 31
2
Take the Cube root of both sides
1 3 2 multiply both sides by 2
3 3 subtract 1 from both sides
2 2
3 3 square both sides
3 9 This is an incorrect answer A simple equation check could have shown that the distributive law using BS math provides an incorrect answer
1
BS Illogical and Incorrect Numbers Raised to Powers
nC D n DC
642 DDD
In BS math only a positive number can have an answer if ldquonrdquo is an even number (246---)
All negative numbers have no solutions
642 DDD
With SM just add a direction arrow and any number gives the correct answer
or n nC D C D
3 3( 3 8 8 or 8 2 or 2 )n D C
( 2 16 C= 16 or 16 4 or 4 )n D
Since a negative number has no square root answer in BS math BS mathematicians invented imaginary numbers
i2=-1
24414 iiC n
These equations are easily solved without the use of imaginary numbers with SM math
If 4 then 4 2x
If 4 then 4 2x
The 360-degree circle is a 4400 years old extremely outdated system based on a year of 12 months of 30 days each that is 360 days
SM will use a more logical system that will have a circle of 6283185307o
d
( )
(deg )
d r radians
d r rees
the arc distance around
the circumference of a circle
d
radiusr
2 2
2 2
d r
dv rd t t
v dda rd t t t
1
d
r
( )av tv
The average angular velocity of a body is the rate of its angular displacement
( )0
liminstt
vt
The instantaneous angular velocity is the limit of the ratio
t
as t approaches zero
( )d
av
d
tv
The average circumference velocity of the body is the rate of its distance traveled d divided by the time t
dvv
r
1 dd vdrv
t t t r r
The equation of the relationship between the angular velocity and the velocity of a point on that body
( )o
av
v v v
t ta
The average angular acceleration is the change in the angular velocity
( )0
liminstt
v
ta
The instantaneous angular acceleration is the
limit of the ration of v
t
as t approaches
zero
daa
r
1
1
d
d s d
vva
t r tv a
r t r
The equation of the relationship between the angular acceleration and the acceleration of a point on the body
21
2v a t
If a body with an initial angular velocity ov has a constant
angular acceleration a it will turn through an angle in a time t
1
Straight Line Symbol Units Rotational Symbol Units Distance d m Angular
Distance radians or degrees
Velocity dt
v m
s Angular Velocity
V
t
radians or degrees
s
Acceleration vat
2ms
Angular Acceleration
VA
t
2
radians or degrees
s
Mass M Kg Mass distribution
dM 2Kgm
Force actionA Ma
2
Kgm
sNewton
Torque
dT M A 2
2
Kgm
s
Momentum Mv Kgm
s
Angular Momentum
dM V 2Kgm
s
Work (Force)(distance)
Ad 2
Kgmm
sNm Joule
Work T
Using AS arrow math to solve and direction in space problems using the equation 1 2
2 2 2 2 2 2( ) amp and ( ) amp 3
4
2 2 2
2 2
( ) amp
(5 ) 4 3
(5 ) 16 9
(5 ) 25
(5 ) (5 )
2 2(5 ) 4 3
(5 ) 16 9
(5 ) 25
(5 ) (5 )
In SM math Pythagorasrsquo theorem provides both magnitude and direction
In BS math Pythagorasrsquo theorem provides magnitude only
time set equal to zero
23
( ) ( 0 )10 1
x t xx
3( ) ( 0)
10 1W t f
In BS math this is an equation for a traveling wave
must use a square term in order to make the curve symmetric
NO SQUARE TERM IS NECESSARY
In SM math when t is set to ldquo0rdquo
This is a major fault in BS math
x f(x)
-4 0019
-3 0033
-2 0073
-1 0273
0 3000
1 0273
2 0073
3 0033
4 0019 -4 -3 -2 -1 0 1 2 3 400
05
10
15
20
25
30
35
BS wave (must use square term)
snap shot of wave at x = 0
Am
plit
ude
lt to gt lt to gt f(ltgt)
lt -4 0073
lt -3 0097
lt -2 0143
lt -1 0273
origin 0 3000
gt 1 0273
gt 2 0143
gt 3 0097
gt 4 0073lt0073 lt0097 lt0143 lt0273
3000
0273gt 0143gt 0097gt 0073gt
Wave moving
snap shot of wave
Am
plit
ude
lt4 lt3 lt2 lt1 0 1gt 2gt 3gt 4gt
3
2
1
In SM NO squared terms are needed
SPRING
An Action is needed to stretch or compress a spring to a starting point where it can execute cycle motion
tsA (ts to start) is the work required to move a mass attached to a spring a distance (d1) from
its initial rest position RAK 0 the spring constant is the stiffness of the spring
o A is attraction o R is repulsion o
ARRA
00 is for the back and forth motion through the zero initial rest position A
stretched spring goes from ldquoArdquo to ldquo0rdquo to ldquoRrdquo to ldquoArdquo to ldquo0rdquo to ldquoRrdquo d1 is the distance a mass is moved from its initial rest position
0
0 1ts A RA K d
Once ( )mad
is performed to move the mass to position d1 (against the restraint of the spring) a of the ( )mad
is transferred into the spring as potential attractionrepulsion (PAR) There is always
more ( )mad
performed than is stored as PAR When the spring is released from the restraining Ats the PAR starts the oscillation Conservation of ( )mad
requires that the PAR and the motion ( )mad
( )( )madM be a constant
( )amp ( ) constantmadPAR M 02
0 0
1
2A R A RPAR K d
02
( ) 0 0
1( )
2mad A R A RM m v
0 0 02 2 2
0 0 0 1
2 2 21
2 2 21
1 1 1amp
2 2 2
1 1 1amp
2 2 2
amp
A R A R A RK d m v K d
Kd mv Kd
Kd mv Kd
0
1d is frac12 the height from the base to the peak of a cycle
An experiment of moving paper under a pendulum through a small angle filled with sand that can escape through a small hole in the bottom shows that a pendulum cycling back and forth follows a curve that is mathematically analogous to a lsquosinrsquo or lsquocosrsquo
The projection of the curved area onto the straight line exactly duplicate the back and forth motion of CM This allows for the math of lsquosinrsquo or lsquocosrsquo to be used
SM will use lsquosinrdquo math It could use lsquocosrsquo or a combination of both lsquosinrsquo and lsquocosrsquo Using just the lsquosinrsquo math makes the understanding and math easier
0 sin 0 0SM 0 sin157 1SM The diagram at the left shows a spring being stretched from its (zero rest) position-2 and the PAR being released at position-1 There are 4 sections (1 to 2 2 to 3 3 to 4 and 4 to5) that alternate between maximum acceleration and zero velocity to max velocity and zero acceleration
At position-1 there is maximum acceleration and zero velocity o The lsquosinrsquo values from 0o to 157o will be used for velocity (values from 0 to 1) o The lsquosinrsquo values from 157o to 0o will be used for acceleration (values from 1 to 0)
At position-2 there is maximum velocity and zero acceleration Acceleration changes to deceleration
o The lsquosinrsquo values from 157o to 0o will be used for deceleration (values from 1 to 0) o The lsquosinrsquo values from 0o to 157o will be used for acceleration (values from 0 to 1)
At position-3 there is maximum acceleration and zero velocity o The lsquosinrsquo values from 0o to 157o will be used for velocity (values from 0 to 1) o The lsquosinrsquo values from 157o to 0o will be used for acceleration (values from 1 to 0)
At position-4 there is maximum velocity and zero acceleration Acceleration changes to deceleration
o The lsquosinrsquo values from 157o to 0o will be used for deceleration (values from 1 to 0) o The lsquosinrsquo values from 0o to 157o will be used for acceleration (values from 0 to 1)
At position-5 the cycle starts again
2
)539(
cT
MK
spring constant (K) when mass (M) and cycle-time (T) are known
)(
))(89)((
111 md
uKgM
d
MG
d
FK sts
spring constant (K) when mass (M) and amplitude ( d1) are known
2 21( )
Kv d d
m
velocity (v) at any point during the cycle
velocity
dforM
Kdv
max
012 velocity (v2) maximum (when d = 0 at position-2 and 4)
acceleration (a1) maximum when spring constant (K) mass (M) and amplitude (d1) are known
acceleration (a1) maximum when amplitude (d1) and cycle-time (Tc) are known
acceleration (a) at any point during the cycle-time (T)
1
151
286
v
dTc
cycle-time (Tc) when amplitude (d1) and velocity (v1) are known
K
MTc 28651
cycle-time (Tc) when mass (M) and spring constant (K) are known
fTc
1
cycle-time (Tc) when the frequency is known
cTf
1
frequency (f) when the cycle-time (Tc) is known
of cycles
elapsed timef
frequency (f) when the number of cycles in an elapsed time is known
d1 amplitude (distance from rest to start position)
Also frac12 the distance from the base to the peak
3131 oror dM
Ka
251
3131 539
c
oror T
da
2539
cT
da
In December of 2006 I discovered the mathematical error in Einsteinrsquos Special Relativity Einstein used BS math and that math only works in an imaginary world
It cannot describe real symmetrical space
Einsteinrsquos Special Relativity should be abandoned and a theory of real space using SM should be used to replace it
Numerous articles on the Internet show Einsteinrsquos theory to be incorrect but no one in academia is willing to publish the information
Figure 1-10 A moving particle observed from two frames of reference in uniform translation
Einsteinrsquos Math Error in Special Relativity Transformation of Velocity (From Eisenberg- Fundamentals of Modern Physics pages 27-29) Consider the particle shown in figure (1-10) moving with velocity v as
seen in a frame of reference O
dx dy dz
Vx Vy Vzdt dt dt
dx dy dz
V x V y V zdy dt dt
We would like to evaluate the velocity v of the particle as seen in the frame of reference 0 which is itself moving relative to 0 with velocity v Measured in the 0 frame the velocity vector of the particle has com ponents
The same velocity vector as measured in the 0 frame has components
2
2
2
2
2
1
1
x vtx
v
cy y
z z
vxt
ctv
c
Now from equations (1-13) we know that the relation between the primed and the unprimed coordinates and times is
Take the differential of these equations remembering that v is a constant This gives
2
2
2
2
2
1
1
dx vdtdx
v
cdy dy
dz dz
vxdt
cdtv
c
2
2
2 2 2 2 2
2
2
2
11 ( )
1( ) 1
1
1
1
dx vdt
v dxdx vdt v
dx dx vdtc dt dtvdx vdx vdx v dxdt dt dt dt
dt dtc c c c
VxV x
Vx
VxV x
vv
vv
c
x
v
c
cV
( )
coordinate of particle relative to moving primed system
coordinate of particle relative to stationary system
velocity of primed coordina
V x
Vx
te system relative to stationaryv sys
tem
Page 28 Fundamentals of Modern Physics
(1-16)
According to Einstein this equation tells you how to transform the observed velocity from one frame of reference to another frame of reference First we note that as Vc and vc approachzero the equation approaches those which would be derived from the Galilean transformation
2 2
( ) 1 01 1 ( )
Vx v Vx vV
VxV x
Vx v
v cVxx
v
c
v
cc
Another property of this equation is that it is impossible to choose V and v such that Vrsquo the magnitude of the velocity which is seen in the new frame is greater than c
1
2
09 ( 09 ) 18009876543
( 09 ) 1811 (09 )
c c cV c c
cc
c
Example As seen by O particle 1 has velocity 09c in the direction of positive x and particle 2 has velocity 09c in the negative x direction
Einstein equation however predicts To evaluate the velocity of particle 1 with respect to particle 2 we transform from the O frame to the Orsquo frame moving in the negative x direction with velocity v = -09c using equation 1-16 We obtain
JK LOGIC will observe the particles separating at 18c
A confusing INCORRECT and ILLOGICAL answer
left is a dir rightsub of a
1
2
1
09 ( 09 ) 18009876543
181( 09 )1 ( 09 )
The multiplication of dir left time a dir right is ONLY a dir left
09
dir
direction leftdirection rightsubtraction
c c cV c c
cc
c
V
dir
22
22
09 18 18009876543
181( 081)1(09)(09)
1
left rightsub
sub
c c c cc c
ccc
c
Analyzing Einsteinrsquos BS math Equation
Using SM math a logical answer can be obtained
Cannot multiply a subtraction operator times a direction in space
Can only subtract and change its direction
Cannot multiply a direction to the left times a direction to the right
Cannot multiply opposite directions in space
Can only solve for the resultant
rightleft
subtraction
1
subtraction
2
sub of a dir to the left is a dir to the right
1 resultant 0 sub 2
2sub 2
2
09 09
09 amp 091
09 09 18 1
(0)1
(09) amp (09)1
c cV
c cc
c c cV
cc
cc
8
181
cc
SM math predicts a logical answer using logic
21st century scientists state in many articles that Einsteins General Relativity and Quantum Mechanics are incompatible
They are not only incompatible they are both mathematically incorrect
They both use illogical BS math
Any equations in math that use imaginary numbers should be abandoned
Imaginary numbers are only necessary in the current BS math system
This means that Schroumldingers equations should be abandoned
Any equation in quantum mechanics that uses imaginary number should be abandoned
Prof Kanarev has developed logical physics and chemistrytheories without the use of Quantum mechanics or relativity
A partial list of his theories coverbull Mechanics ndash replaces Newtonbull Photon - bull Electrons ndash Protons- Neutronsbull Spectroscopybull Nucleusbull Atoms amp Moleculesbull Electrodynamicsbull Heatbull Curvature of Space ndash replaces Schwarzcchildrsquos equation
Kanarevrsquos Hypothesis
bull Electron align linearly with proton They do NOT spin or form clouds around the nucleus
bull Neutrons are in the center of atom Proton are outside of Neutrons
bull The wavelength of all particles is the experimental radius
bull Double slit experiment explained correctly
bull Heisenbergrsquos inequality explained correctly
bull is the radius of approach of the magnetic lines around the torus of the electron
( 1 137)
bull Why water expands when it freezes
bull Electron clusters form lightning
rbull The meaning of Plancks constant is explained
22 2 kg m
h m f mr f consts
2210254 42m r kgm bull If the mr-constant describes a ring the equations
describe experiments bull If the mr-constant describes a wave the equations do
NOT describe experiments
Kanarevrsquos 5 Equations of Mechanics
Kanarevrsquos 1st Equation Acceleration ampmv maMma R R
Kanarevrsquos 2nd Equation Uniform motion amp mv maMma R R
maM mvR R
Kanarevrsquos 3rd Equation
Kanarevrsquos 4th Equation Equal and opposite reaction
1 21 21 2
12
a m
m a m ama
Kanarevrsquos 5th Equation 1 2 3 Mi M M M Mna R R R R R
The sum of the Actions working on a moving body is never equal to zero
Kanarevrsquos Photon model is a rolling ring
His equations accurately describe known experiments with photonsbull Electromagnetic radiation should be called photon radiationbull Diffraction and interference are the same phenomenabull Correctly explains the double slit experiment without quantum equationsbull Correctly explains the meaning of Heisenbergrsquos inequality
e
2 (2)(3142)(2817 15 ) 10007297352568
R 2426 12 137mag
xp
r m
m
Charge
Magnetic action
Proton
The intensity of the magnetic action near the geometrical center of the proton ring is so great that we theorize that it is the strong nuclear force
(9383 8)(1602 19)8476 14
4 (4)(3142)(1411 26)p
pp
EB T
M
(9395 8)(1602 19)1035 15
4 (4)(3142)(0966 26)N
NN
EB T
M
Neutron
1a Hydrogenavi
SM Graphs versus BS Graphs
Numbers Raised to Powers
When numbers are raised to powers the negative sides of all BS math graphs are ILLOGICAL and INCORRECT
Incorrect BS Math for 2x
2x
4 16 correct3 8 correct2 4 correct1 2 correct0 1 correct-1 005 incorrect-2 025 incorrect-3 0125 incorrect-4 00625 incorrect
negative side of graph is incorrectpositive side of graph is correct
16
8
42105025012500625
-4 -3 -2 -1 0 1 2 3 4
po
sit
ive
negative positive
2x BS Math
correctincorrect
In BS math the dash sign (-) used in this example represents three different operations They arebull (-) Subtractionbull (-) A direction to the leftbull (-) As an exponent to mean divide by
All positive numbers produce correct results and graphs
All negative numbers produce incorrect results and graphs
A direction of three units to the left (-3) is placed in the equation as an exponent
A direction in space (3 units toward the left) changes a whole number to a fraction by using the dash (-) as an exponent rather than a direction in space
This is illogical and incorrect BS math should have performed this operation as
Unfortunately the answers to problems can come out correct because data points of a real problem fit the curve
However the answers will not be due to logical reasoning
Because the equation produced a curve that real problem data points follow it has been falsely believed that the equations and graphs are correct
1
2
x
Incorrect BK graph for (12)x
(12)x
4 00625 correct3 0125 correct2 025 correct1 1 correct0 1 correct-1 2 incorrect-2 4 incorrect-3 8 incorrect-4 16 incorrect
negative side of graph is incorrectpositive side of graph is correct
00625012502505124
8
16
-5 -4 -3 -2 -1 0 1 2 3 4 5P
osit
ive
negative positive
BS Math for (12)x
correctincorrect
Incorrect BS math and graph for
All graphs are symmetrical with SM
Two BS Math Equations that Produce Incorrect Answers
2 0x x starting equation 2
2
x x x x
x x
Add lsquoxrsquo to both sides
2(1) 1 let lsquoxrsquo = 1
2 1 This is an incorrect answer 2(0)-(0)=0 The only answer that BK math can give is x=0
1
a b starting equation ( )( ) ( )( )a a a b multiply both sides by ( )a
( )( ) ( )( ) ( )( ) ( )( )a a b b a b b b subtract ( )( )b b from both sides
( )( ) ( )( )a b a b a b b factor ( )a b from both sides
( )a b b divide both sides by ( )a b
( )b b b replace ( )a with ( )b 2b b simplify 2(1) (1) Let ( )b = 1
2 1 Again BS math yields and incorrect answer 1
This incorrect result is a problem with the distributive law SM explains how this incorrect answer is obtained and the problem with the distributive law that allows the error
Page 148 ndash149 Barry Mazur ndash Imagining Numbers
3
2
31
Do the computation carefully on paper using the rules we agreed to Then ponder your answer which should be something of a surprise to you donrsquot stop there Think of what your answer might possibly mean or might imply
Solve this equation using the distributive law
JK definitely a surprise the answer wasincorrect and illogical
Gerhard Gade University Professor doing math at Harvard University
Here is the BS math answer that Barry Mazur calculated
2
31
2
31
2
31
1 3 1 3 1 3 3 3 2 2 3 1 3
2 2 2 2 4 4 4 4 4 4 2 2
dash
1 3 1 3 1 31
2 2 2 2 4 4
3 3
4 4
When the distributive law is used with the BS Rule-of-Signs and one of the signs is a (-) dash sign the middle two terms always cancel each other This leads to the incorrect answer of (-1) in the above equation
Barry Mazur did not go back and check his answer with the original equation If he had he would have known that the answer was incorrect
3
1 31
2
check answer by setting equal
3
331 3
12
1 31
2
Take the Cube root of both sides
1 3 2 multiply both sides by 2
3 3 subtract 1 from both sides
2 2
3 3 square both sides
3 9 This is an incorrect answer A simple equation check could have shown that the distributive law using BS math provides an incorrect answer
1
BS Illogical and Incorrect Numbers Raised to Powers
nC D n DC
642 DDD
In BS math only a positive number can have an answer if ldquonrdquo is an even number (246---)
All negative numbers have no solutions
642 DDD
With SM just add a direction arrow and any number gives the correct answer
or n nC D C D
3 3( 3 8 8 or 8 2 or 2 )n D C
( 2 16 C= 16 or 16 4 or 4 )n D
Since a negative number has no square root answer in BS math BS mathematicians invented imaginary numbers
i2=-1
24414 iiC n
These equations are easily solved without the use of imaginary numbers with SM math
If 4 then 4 2x
If 4 then 4 2x
The 360-degree circle is a 4400 years old extremely outdated system based on a year of 12 months of 30 days each that is 360 days
SM will use a more logical system that will have a circle of 6283185307o
d
( )
(deg )
d r radians
d r rees
the arc distance around
the circumference of a circle
d
radiusr
2 2
2 2
d r
dv rd t t
v dda rd t t t
1
d
r
( )av tv
The average angular velocity of a body is the rate of its angular displacement
( )0
liminstt
vt
The instantaneous angular velocity is the limit of the ratio
t
as t approaches zero
( )d
av
d
tv
The average circumference velocity of the body is the rate of its distance traveled d divided by the time t
dvv
r
1 dd vdrv
t t t r r
The equation of the relationship between the angular velocity and the velocity of a point on that body
( )o
av
v v v
t ta
The average angular acceleration is the change in the angular velocity
( )0
liminstt
v
ta
The instantaneous angular acceleration is the
limit of the ration of v
t
as t approaches
zero
daa
r
1
1
d
d s d
vva
t r tv a
r t r
The equation of the relationship between the angular acceleration and the acceleration of a point on the body
21
2v a t
If a body with an initial angular velocity ov has a constant
angular acceleration a it will turn through an angle in a time t
1
Straight Line Symbol Units Rotational Symbol Units Distance d m Angular
Distance radians or degrees
Velocity dt
v m
s Angular Velocity
V
t
radians or degrees
s
Acceleration vat
2ms
Angular Acceleration
VA
t
2
radians or degrees
s
Mass M Kg Mass distribution
dM 2Kgm
Force actionA Ma
2
Kgm
sNewton
Torque
dT M A 2
2
Kgm
s
Momentum Mv Kgm
s
Angular Momentum
dM V 2Kgm
s
Work (Force)(distance)
Ad 2
Kgmm
sNm Joule
Work T
Using AS arrow math to solve and direction in space problems using the equation 1 2
2 2 2 2 2 2( ) amp and ( ) amp 3
4
2 2 2
2 2
( ) amp
(5 ) 4 3
(5 ) 16 9
(5 ) 25
(5 ) (5 )
2 2(5 ) 4 3
(5 ) 16 9
(5 ) 25
(5 ) (5 )
In SM math Pythagorasrsquo theorem provides both magnitude and direction
In BS math Pythagorasrsquo theorem provides magnitude only
time set equal to zero
23
( ) ( 0 )10 1
x t xx
3( ) ( 0)
10 1W t f
In BS math this is an equation for a traveling wave
must use a square term in order to make the curve symmetric
NO SQUARE TERM IS NECESSARY
In SM math when t is set to ldquo0rdquo
This is a major fault in BS math
x f(x)
-4 0019
-3 0033
-2 0073
-1 0273
0 3000
1 0273
2 0073
3 0033
4 0019 -4 -3 -2 -1 0 1 2 3 400
05
10
15
20
25
30
35
BS wave (must use square term)
snap shot of wave at x = 0
Am
plit
ude
lt to gt lt to gt f(ltgt)
lt -4 0073
lt -3 0097
lt -2 0143
lt -1 0273
origin 0 3000
gt 1 0273
gt 2 0143
gt 3 0097
gt 4 0073lt0073 lt0097 lt0143 lt0273
3000
0273gt 0143gt 0097gt 0073gt
Wave moving
snap shot of wave
Am
plit
ude
lt4 lt3 lt2 lt1 0 1gt 2gt 3gt 4gt
3
2
1
In SM NO squared terms are needed
SPRING
An Action is needed to stretch or compress a spring to a starting point where it can execute cycle motion
tsA (ts to start) is the work required to move a mass attached to a spring a distance (d1) from
its initial rest position RAK 0 the spring constant is the stiffness of the spring
o A is attraction o R is repulsion o
ARRA
00 is for the back and forth motion through the zero initial rest position A
stretched spring goes from ldquoArdquo to ldquo0rdquo to ldquoRrdquo to ldquoArdquo to ldquo0rdquo to ldquoRrdquo d1 is the distance a mass is moved from its initial rest position
0
0 1ts A RA K d
Once ( )mad
is performed to move the mass to position d1 (against the restraint of the spring) a of the ( )mad
is transferred into the spring as potential attractionrepulsion (PAR) There is always
more ( )mad
performed than is stored as PAR When the spring is released from the restraining Ats the PAR starts the oscillation Conservation of ( )mad
requires that the PAR and the motion ( )mad
( )( )madM be a constant
( )amp ( ) constantmadPAR M 02
0 0
1
2A R A RPAR K d
02
( ) 0 0
1( )
2mad A R A RM m v
0 0 02 2 2
0 0 0 1
2 2 21
2 2 21
1 1 1amp
2 2 2
1 1 1amp
2 2 2
amp
A R A R A RK d m v K d
Kd mv Kd
Kd mv Kd
0
1d is frac12 the height from the base to the peak of a cycle
An experiment of moving paper under a pendulum through a small angle filled with sand that can escape through a small hole in the bottom shows that a pendulum cycling back and forth follows a curve that is mathematically analogous to a lsquosinrsquo or lsquocosrsquo
The projection of the curved area onto the straight line exactly duplicate the back and forth motion of CM This allows for the math of lsquosinrsquo or lsquocosrsquo to be used
SM will use lsquosinrdquo math It could use lsquocosrsquo or a combination of both lsquosinrsquo and lsquocosrsquo Using just the lsquosinrsquo math makes the understanding and math easier
0 sin 0 0SM 0 sin157 1SM The diagram at the left shows a spring being stretched from its (zero rest) position-2 and the PAR being released at position-1 There are 4 sections (1 to 2 2 to 3 3 to 4 and 4 to5) that alternate between maximum acceleration and zero velocity to max velocity and zero acceleration
At position-1 there is maximum acceleration and zero velocity o The lsquosinrsquo values from 0o to 157o will be used for velocity (values from 0 to 1) o The lsquosinrsquo values from 157o to 0o will be used for acceleration (values from 1 to 0)
At position-2 there is maximum velocity and zero acceleration Acceleration changes to deceleration
o The lsquosinrsquo values from 157o to 0o will be used for deceleration (values from 1 to 0) o The lsquosinrsquo values from 0o to 157o will be used for acceleration (values from 0 to 1)
At position-3 there is maximum acceleration and zero velocity o The lsquosinrsquo values from 0o to 157o will be used for velocity (values from 0 to 1) o The lsquosinrsquo values from 157o to 0o will be used for acceleration (values from 1 to 0)
At position-4 there is maximum velocity and zero acceleration Acceleration changes to deceleration
o The lsquosinrsquo values from 157o to 0o will be used for deceleration (values from 1 to 0) o The lsquosinrsquo values from 0o to 157o will be used for acceleration (values from 0 to 1)
At position-5 the cycle starts again
2
)539(
cT
MK
spring constant (K) when mass (M) and cycle-time (T) are known
)(
))(89)((
111 md
uKgM
d
MG
d
FK sts
spring constant (K) when mass (M) and amplitude ( d1) are known
2 21( )
Kv d d
m
velocity (v) at any point during the cycle
velocity
dforM
Kdv
max
012 velocity (v2) maximum (when d = 0 at position-2 and 4)
acceleration (a1) maximum when spring constant (K) mass (M) and amplitude (d1) are known
acceleration (a1) maximum when amplitude (d1) and cycle-time (Tc) are known
acceleration (a) at any point during the cycle-time (T)
1
151
286
v
dTc
cycle-time (Tc) when amplitude (d1) and velocity (v1) are known
K
MTc 28651
cycle-time (Tc) when mass (M) and spring constant (K) are known
fTc
1
cycle-time (Tc) when the frequency is known
cTf
1
frequency (f) when the cycle-time (Tc) is known
of cycles
elapsed timef
frequency (f) when the number of cycles in an elapsed time is known
d1 amplitude (distance from rest to start position)
Also frac12 the distance from the base to the peak
3131 oror dM
Ka
251
3131 539
c
oror T
da
2539
cT
da
In December of 2006 I discovered the mathematical error in Einsteinrsquos Special Relativity Einstein used BS math and that math only works in an imaginary world
It cannot describe real symmetrical space
Einsteinrsquos Special Relativity should be abandoned and a theory of real space using SM should be used to replace it
Numerous articles on the Internet show Einsteinrsquos theory to be incorrect but no one in academia is willing to publish the information
Figure 1-10 A moving particle observed from two frames of reference in uniform translation
Einsteinrsquos Math Error in Special Relativity Transformation of Velocity (From Eisenberg- Fundamentals of Modern Physics pages 27-29) Consider the particle shown in figure (1-10) moving with velocity v as
seen in a frame of reference O
dx dy dz
Vx Vy Vzdt dt dt
dx dy dz
V x V y V zdy dt dt
We would like to evaluate the velocity v of the particle as seen in the frame of reference 0 which is itself moving relative to 0 with velocity v Measured in the 0 frame the velocity vector of the particle has com ponents
The same velocity vector as measured in the 0 frame has components
2
2
2
2
2
1
1
x vtx
v
cy y
z z
vxt
ctv
c
Now from equations (1-13) we know that the relation between the primed and the unprimed coordinates and times is
Take the differential of these equations remembering that v is a constant This gives
2
2
2
2
2
1
1
dx vdtdx
v
cdy dy
dz dz
vxdt
cdtv
c
2
2
2 2 2 2 2
2
2
2
11 ( )
1( ) 1
1
1
1
dx vdt
v dxdx vdt v
dx dx vdtc dt dtvdx vdx vdx v dxdt dt dt dt
dt dtc c c c
VxV x
Vx
VxV x
vv
vv
c
x
v
c
cV
( )
coordinate of particle relative to moving primed system
coordinate of particle relative to stationary system
velocity of primed coordina
V x
Vx
te system relative to stationaryv sys
tem
Page 28 Fundamentals of Modern Physics
(1-16)
According to Einstein this equation tells you how to transform the observed velocity from one frame of reference to another frame of reference First we note that as Vc and vc approachzero the equation approaches those which would be derived from the Galilean transformation
2 2
( ) 1 01 1 ( )
Vx v Vx vV
VxV x
Vx v
v cVxx
v
c
v
cc
Another property of this equation is that it is impossible to choose V and v such that Vrsquo the magnitude of the velocity which is seen in the new frame is greater than c
1
2
09 ( 09 ) 18009876543
( 09 ) 1811 (09 )
c c cV c c
cc
c
Example As seen by O particle 1 has velocity 09c in the direction of positive x and particle 2 has velocity 09c in the negative x direction
Einstein equation however predicts To evaluate the velocity of particle 1 with respect to particle 2 we transform from the O frame to the Orsquo frame moving in the negative x direction with velocity v = -09c using equation 1-16 We obtain
JK LOGIC will observe the particles separating at 18c
A confusing INCORRECT and ILLOGICAL answer
left is a dir rightsub of a
1
2
1
09 ( 09 ) 18009876543
181( 09 )1 ( 09 )
The multiplication of dir left time a dir right is ONLY a dir left
09
dir
direction leftdirection rightsubtraction
c c cV c c
cc
c
V
dir
22
22
09 18 18009876543
181( 081)1(09)(09)
1
left rightsub
sub
c c c cc c
ccc
c
Analyzing Einsteinrsquos BS math Equation
Using SM math a logical answer can be obtained
Cannot multiply a subtraction operator times a direction in space
Can only subtract and change its direction
Cannot multiply a direction to the left times a direction to the right
Cannot multiply opposite directions in space
Can only solve for the resultant
rightleft
subtraction
1
subtraction
2
sub of a dir to the left is a dir to the right
1 resultant 0 sub 2
2sub 2
2
09 09
09 amp 091
09 09 18 1
(0)1
(09) amp (09)1
c cV
c cc
c c cV
cc
cc
8
181
cc
SM math predicts a logical answer using logic
21st century scientists state in many articles that Einsteins General Relativity and Quantum Mechanics are incompatible
They are not only incompatible they are both mathematically incorrect
They both use illogical BS math
Any equations in math that use imaginary numbers should be abandoned
Imaginary numbers are only necessary in the current BS math system
This means that Schroumldingers equations should be abandoned
Any equation in quantum mechanics that uses imaginary number should be abandoned
Prof Kanarev has developed logical physics and chemistrytheories without the use of Quantum mechanics or relativity
A partial list of his theories coverbull Mechanics ndash replaces Newtonbull Photon - bull Electrons ndash Protons- Neutronsbull Spectroscopybull Nucleusbull Atoms amp Moleculesbull Electrodynamicsbull Heatbull Curvature of Space ndash replaces Schwarzcchildrsquos equation
Kanarevrsquos Hypothesis
bull Electron align linearly with proton They do NOT spin or form clouds around the nucleus
bull Neutrons are in the center of atom Proton are outside of Neutrons
bull The wavelength of all particles is the experimental radius
bull Double slit experiment explained correctly
bull Heisenbergrsquos inequality explained correctly
bull is the radius of approach of the magnetic lines around the torus of the electron
( 1 137)
bull Why water expands when it freezes
bull Electron clusters form lightning
rbull The meaning of Plancks constant is explained
22 2 kg m
h m f mr f consts
2210254 42m r kgm bull If the mr-constant describes a ring the equations
describe experiments bull If the mr-constant describes a wave the equations do
NOT describe experiments
Kanarevrsquos 5 Equations of Mechanics
Kanarevrsquos 1st Equation Acceleration ampmv maMma R R
Kanarevrsquos 2nd Equation Uniform motion amp mv maMma R R
maM mvR R
Kanarevrsquos 3rd Equation
Kanarevrsquos 4th Equation Equal and opposite reaction
1 21 21 2
12
a m
m a m ama
Kanarevrsquos 5th Equation 1 2 3 Mi M M M Mna R R R R R
The sum of the Actions working on a moving body is never equal to zero
Kanarevrsquos Photon model is a rolling ring
His equations accurately describe known experiments with photonsbull Electromagnetic radiation should be called photon radiationbull Diffraction and interference are the same phenomenabull Correctly explains the double slit experiment without quantum equationsbull Correctly explains the meaning of Heisenbergrsquos inequality
e
2 (2)(3142)(2817 15 ) 10007297352568
R 2426 12 137mag
xp
r m
m
Charge
Magnetic action
Proton
The intensity of the magnetic action near the geometrical center of the proton ring is so great that we theorize that it is the strong nuclear force
(9383 8)(1602 19)8476 14
4 (4)(3142)(1411 26)p
pp
EB T
M
(9395 8)(1602 19)1035 15
4 (4)(3142)(0966 26)N
NN
EB T
M
Neutron
1a Hydrogenavi
In BS math the dash sign (-) used in this example represents three different operations They arebull (-) Subtractionbull (-) A direction to the leftbull (-) As an exponent to mean divide by
All positive numbers produce correct results and graphs
All negative numbers produce incorrect results and graphs
A direction of three units to the left (-3) is placed in the equation as an exponent
A direction in space (3 units toward the left) changes a whole number to a fraction by using the dash (-) as an exponent rather than a direction in space
This is illogical and incorrect BS math should have performed this operation as
Unfortunately the answers to problems can come out correct because data points of a real problem fit the curve
However the answers will not be due to logical reasoning
Because the equation produced a curve that real problem data points follow it has been falsely believed that the equations and graphs are correct
1
2
x
Incorrect BK graph for (12)x
(12)x
4 00625 correct3 0125 correct2 025 correct1 1 correct0 1 correct-1 2 incorrect-2 4 incorrect-3 8 incorrect-4 16 incorrect
negative side of graph is incorrectpositive side of graph is correct
00625012502505124
8
16
-5 -4 -3 -2 -1 0 1 2 3 4 5P
osit
ive
negative positive
BS Math for (12)x
correctincorrect
Incorrect BS math and graph for
All graphs are symmetrical with SM
Two BS Math Equations that Produce Incorrect Answers
2 0x x starting equation 2
2
x x x x
x x
Add lsquoxrsquo to both sides
2(1) 1 let lsquoxrsquo = 1
2 1 This is an incorrect answer 2(0)-(0)=0 The only answer that BK math can give is x=0
1
a b starting equation ( )( ) ( )( )a a a b multiply both sides by ( )a
( )( ) ( )( ) ( )( ) ( )( )a a b b a b b b subtract ( )( )b b from both sides
( )( ) ( )( )a b a b a b b factor ( )a b from both sides
( )a b b divide both sides by ( )a b
( )b b b replace ( )a with ( )b 2b b simplify 2(1) (1) Let ( )b = 1
2 1 Again BS math yields and incorrect answer 1
This incorrect result is a problem with the distributive law SM explains how this incorrect answer is obtained and the problem with the distributive law that allows the error
Page 148 ndash149 Barry Mazur ndash Imagining Numbers
3
2
31
Do the computation carefully on paper using the rules we agreed to Then ponder your answer which should be something of a surprise to you donrsquot stop there Think of what your answer might possibly mean or might imply
Solve this equation using the distributive law
JK definitely a surprise the answer wasincorrect and illogical
Gerhard Gade University Professor doing math at Harvard University
Here is the BS math answer that Barry Mazur calculated
2
31
2
31
2
31
1 3 1 3 1 3 3 3 2 2 3 1 3
2 2 2 2 4 4 4 4 4 4 2 2
dash
1 3 1 3 1 31
2 2 2 2 4 4
3 3
4 4
When the distributive law is used with the BS Rule-of-Signs and one of the signs is a (-) dash sign the middle two terms always cancel each other This leads to the incorrect answer of (-1) in the above equation
Barry Mazur did not go back and check his answer with the original equation If he had he would have known that the answer was incorrect
3
1 31
2
check answer by setting equal
3
331 3
12
1 31
2
Take the Cube root of both sides
1 3 2 multiply both sides by 2
3 3 subtract 1 from both sides
2 2
3 3 square both sides
3 9 This is an incorrect answer A simple equation check could have shown that the distributive law using BS math provides an incorrect answer
1
BS Illogical and Incorrect Numbers Raised to Powers
nC D n DC
642 DDD
In BS math only a positive number can have an answer if ldquonrdquo is an even number (246---)
All negative numbers have no solutions
642 DDD
With SM just add a direction arrow and any number gives the correct answer
or n nC D C D
3 3( 3 8 8 or 8 2 or 2 )n D C
( 2 16 C= 16 or 16 4 or 4 )n D
Since a negative number has no square root answer in BS math BS mathematicians invented imaginary numbers
i2=-1
24414 iiC n
These equations are easily solved without the use of imaginary numbers with SM math
If 4 then 4 2x
If 4 then 4 2x
The 360-degree circle is a 4400 years old extremely outdated system based on a year of 12 months of 30 days each that is 360 days
SM will use a more logical system that will have a circle of 6283185307o
d
( )
(deg )
d r radians
d r rees
the arc distance around
the circumference of a circle
d
radiusr
2 2
2 2
d r
dv rd t t
v dda rd t t t
1
d
r
( )av tv
The average angular velocity of a body is the rate of its angular displacement
( )0
liminstt
vt
The instantaneous angular velocity is the limit of the ratio
t
as t approaches zero
( )d
av
d
tv
The average circumference velocity of the body is the rate of its distance traveled d divided by the time t
dvv
r
1 dd vdrv
t t t r r
The equation of the relationship between the angular velocity and the velocity of a point on that body
( )o
av
v v v
t ta
The average angular acceleration is the change in the angular velocity
( )0
liminstt
v
ta
The instantaneous angular acceleration is the
limit of the ration of v
t
as t approaches
zero
daa
r
1
1
d
d s d
vva
t r tv a
r t r
The equation of the relationship between the angular acceleration and the acceleration of a point on the body
21
2v a t
If a body with an initial angular velocity ov has a constant
angular acceleration a it will turn through an angle in a time t
1
Straight Line Symbol Units Rotational Symbol Units Distance d m Angular
Distance radians or degrees
Velocity dt
v m
s Angular Velocity
V
t
radians or degrees
s
Acceleration vat
2ms
Angular Acceleration
VA
t
2
radians or degrees
s
Mass M Kg Mass distribution
dM 2Kgm
Force actionA Ma
2
Kgm
sNewton
Torque
dT M A 2
2
Kgm
s
Momentum Mv Kgm
s
Angular Momentum
dM V 2Kgm
s
Work (Force)(distance)
Ad 2
Kgmm
sNm Joule
Work T
Using AS arrow math to solve and direction in space problems using the equation 1 2
2 2 2 2 2 2( ) amp and ( ) amp 3
4
2 2 2
2 2
( ) amp
(5 ) 4 3
(5 ) 16 9
(5 ) 25
(5 ) (5 )
2 2(5 ) 4 3
(5 ) 16 9
(5 ) 25
(5 ) (5 )
In SM math Pythagorasrsquo theorem provides both magnitude and direction
In BS math Pythagorasrsquo theorem provides magnitude only
time set equal to zero
23
( ) ( 0 )10 1
x t xx
3( ) ( 0)
10 1W t f
In BS math this is an equation for a traveling wave
must use a square term in order to make the curve symmetric
NO SQUARE TERM IS NECESSARY
In SM math when t is set to ldquo0rdquo
This is a major fault in BS math
x f(x)
-4 0019
-3 0033
-2 0073
-1 0273
0 3000
1 0273
2 0073
3 0033
4 0019 -4 -3 -2 -1 0 1 2 3 400
05
10
15
20
25
30
35
BS wave (must use square term)
snap shot of wave at x = 0
Am
plit
ude
lt to gt lt to gt f(ltgt)
lt -4 0073
lt -3 0097
lt -2 0143
lt -1 0273
origin 0 3000
gt 1 0273
gt 2 0143
gt 3 0097
gt 4 0073lt0073 lt0097 lt0143 lt0273
3000
0273gt 0143gt 0097gt 0073gt
Wave moving
snap shot of wave
Am
plit
ude
lt4 lt3 lt2 lt1 0 1gt 2gt 3gt 4gt
3
2
1
In SM NO squared terms are needed
SPRING
An Action is needed to stretch or compress a spring to a starting point where it can execute cycle motion
tsA (ts to start) is the work required to move a mass attached to a spring a distance (d1) from
its initial rest position RAK 0 the spring constant is the stiffness of the spring
o A is attraction o R is repulsion o
ARRA
00 is for the back and forth motion through the zero initial rest position A
stretched spring goes from ldquoArdquo to ldquo0rdquo to ldquoRrdquo to ldquoArdquo to ldquo0rdquo to ldquoRrdquo d1 is the distance a mass is moved from its initial rest position
0
0 1ts A RA K d
Once ( )mad
is performed to move the mass to position d1 (against the restraint of the spring) a of the ( )mad
is transferred into the spring as potential attractionrepulsion (PAR) There is always
more ( )mad
performed than is stored as PAR When the spring is released from the restraining Ats the PAR starts the oscillation Conservation of ( )mad
requires that the PAR and the motion ( )mad
( )( )madM be a constant
( )amp ( ) constantmadPAR M 02
0 0
1
2A R A RPAR K d
02
( ) 0 0
1( )
2mad A R A RM m v
0 0 02 2 2
0 0 0 1
2 2 21
2 2 21
1 1 1amp
2 2 2
1 1 1amp
2 2 2
amp
A R A R A RK d m v K d
Kd mv Kd
Kd mv Kd
0
1d is frac12 the height from the base to the peak of a cycle
An experiment of moving paper under a pendulum through a small angle filled with sand that can escape through a small hole in the bottom shows that a pendulum cycling back and forth follows a curve that is mathematically analogous to a lsquosinrsquo or lsquocosrsquo
The projection of the curved area onto the straight line exactly duplicate the back and forth motion of CM This allows for the math of lsquosinrsquo or lsquocosrsquo to be used
SM will use lsquosinrdquo math It could use lsquocosrsquo or a combination of both lsquosinrsquo and lsquocosrsquo Using just the lsquosinrsquo math makes the understanding and math easier
0 sin 0 0SM 0 sin157 1SM The diagram at the left shows a spring being stretched from its (zero rest) position-2 and the PAR being released at position-1 There are 4 sections (1 to 2 2 to 3 3 to 4 and 4 to5) that alternate between maximum acceleration and zero velocity to max velocity and zero acceleration
At position-1 there is maximum acceleration and zero velocity o The lsquosinrsquo values from 0o to 157o will be used for velocity (values from 0 to 1) o The lsquosinrsquo values from 157o to 0o will be used for acceleration (values from 1 to 0)
At position-2 there is maximum velocity and zero acceleration Acceleration changes to deceleration
o The lsquosinrsquo values from 157o to 0o will be used for deceleration (values from 1 to 0) o The lsquosinrsquo values from 0o to 157o will be used for acceleration (values from 0 to 1)
At position-3 there is maximum acceleration and zero velocity o The lsquosinrsquo values from 0o to 157o will be used for velocity (values from 0 to 1) o The lsquosinrsquo values from 157o to 0o will be used for acceleration (values from 1 to 0)
At position-4 there is maximum velocity and zero acceleration Acceleration changes to deceleration
o The lsquosinrsquo values from 157o to 0o will be used for deceleration (values from 1 to 0) o The lsquosinrsquo values from 0o to 157o will be used for acceleration (values from 0 to 1)
At position-5 the cycle starts again
2
)539(
cT
MK
spring constant (K) when mass (M) and cycle-time (T) are known
)(
))(89)((
111 md
uKgM
d
MG
d
FK sts
spring constant (K) when mass (M) and amplitude ( d1) are known
2 21( )
Kv d d
m
velocity (v) at any point during the cycle
velocity
dforM
Kdv
max
012 velocity (v2) maximum (when d = 0 at position-2 and 4)
acceleration (a1) maximum when spring constant (K) mass (M) and amplitude (d1) are known
acceleration (a1) maximum when amplitude (d1) and cycle-time (Tc) are known
acceleration (a) at any point during the cycle-time (T)
1
151
286
v
dTc
cycle-time (Tc) when amplitude (d1) and velocity (v1) are known
K
MTc 28651
cycle-time (Tc) when mass (M) and spring constant (K) are known
fTc
1
cycle-time (Tc) when the frequency is known
cTf
1
frequency (f) when the cycle-time (Tc) is known
of cycles
elapsed timef
frequency (f) when the number of cycles in an elapsed time is known
d1 amplitude (distance from rest to start position)
Also frac12 the distance from the base to the peak
3131 oror dM
Ka
251
3131 539
c
oror T
da
2539
cT
da
In December of 2006 I discovered the mathematical error in Einsteinrsquos Special Relativity Einstein used BS math and that math only works in an imaginary world
It cannot describe real symmetrical space
Einsteinrsquos Special Relativity should be abandoned and a theory of real space using SM should be used to replace it
Numerous articles on the Internet show Einsteinrsquos theory to be incorrect but no one in academia is willing to publish the information
Figure 1-10 A moving particle observed from two frames of reference in uniform translation
Einsteinrsquos Math Error in Special Relativity Transformation of Velocity (From Eisenberg- Fundamentals of Modern Physics pages 27-29) Consider the particle shown in figure (1-10) moving with velocity v as
seen in a frame of reference O
dx dy dz
Vx Vy Vzdt dt dt
dx dy dz
V x V y V zdy dt dt
We would like to evaluate the velocity v of the particle as seen in the frame of reference 0 which is itself moving relative to 0 with velocity v Measured in the 0 frame the velocity vector of the particle has com ponents
The same velocity vector as measured in the 0 frame has components
2
2
2
2
2
1
1
x vtx
v
cy y
z z
vxt
ctv
c
Now from equations (1-13) we know that the relation between the primed and the unprimed coordinates and times is
Take the differential of these equations remembering that v is a constant This gives
2
2
2
2
2
1
1
dx vdtdx
v
cdy dy
dz dz
vxdt
cdtv
c
2
2
2 2 2 2 2
2
2
2
11 ( )
1( ) 1
1
1
1
dx vdt
v dxdx vdt v
dx dx vdtc dt dtvdx vdx vdx v dxdt dt dt dt
dt dtc c c c
VxV x
Vx
VxV x
vv
vv
c
x
v
c
cV
( )
coordinate of particle relative to moving primed system
coordinate of particle relative to stationary system
velocity of primed coordina
V x
Vx
te system relative to stationaryv sys
tem
Page 28 Fundamentals of Modern Physics
(1-16)
According to Einstein this equation tells you how to transform the observed velocity from one frame of reference to another frame of reference First we note that as Vc and vc approachzero the equation approaches those which would be derived from the Galilean transformation
2 2
( ) 1 01 1 ( )
Vx v Vx vV
VxV x
Vx v
v cVxx
v
c
v
cc
Another property of this equation is that it is impossible to choose V and v such that Vrsquo the magnitude of the velocity which is seen in the new frame is greater than c
1
2
09 ( 09 ) 18009876543
( 09 ) 1811 (09 )
c c cV c c
cc
c
Example As seen by O particle 1 has velocity 09c in the direction of positive x and particle 2 has velocity 09c in the negative x direction
Einstein equation however predicts To evaluate the velocity of particle 1 with respect to particle 2 we transform from the O frame to the Orsquo frame moving in the negative x direction with velocity v = -09c using equation 1-16 We obtain
JK LOGIC will observe the particles separating at 18c
A confusing INCORRECT and ILLOGICAL answer
left is a dir rightsub of a
1
2
1
09 ( 09 ) 18009876543
181( 09 )1 ( 09 )
The multiplication of dir left time a dir right is ONLY a dir left
09
dir
direction leftdirection rightsubtraction
c c cV c c
cc
c
V
dir
22
22
09 18 18009876543
181( 081)1(09)(09)
1
left rightsub
sub
c c c cc c
ccc
c
Analyzing Einsteinrsquos BS math Equation
Using SM math a logical answer can be obtained
Cannot multiply a subtraction operator times a direction in space
Can only subtract and change its direction
Cannot multiply a direction to the left times a direction to the right
Cannot multiply opposite directions in space
Can only solve for the resultant
rightleft
subtraction
1
subtraction
2
sub of a dir to the left is a dir to the right
1 resultant 0 sub 2
2sub 2
2
09 09
09 amp 091
09 09 18 1
(0)1
(09) amp (09)1
c cV
c cc
c c cV
cc
cc
8
181
cc
SM math predicts a logical answer using logic
21st century scientists state in many articles that Einsteins General Relativity and Quantum Mechanics are incompatible
They are not only incompatible they are both mathematically incorrect
They both use illogical BS math
Any equations in math that use imaginary numbers should be abandoned
Imaginary numbers are only necessary in the current BS math system
This means that Schroumldingers equations should be abandoned
Any equation in quantum mechanics that uses imaginary number should be abandoned
Prof Kanarev has developed logical physics and chemistrytheories without the use of Quantum mechanics or relativity
A partial list of his theories coverbull Mechanics ndash replaces Newtonbull Photon - bull Electrons ndash Protons- Neutronsbull Spectroscopybull Nucleusbull Atoms amp Moleculesbull Electrodynamicsbull Heatbull Curvature of Space ndash replaces Schwarzcchildrsquos equation
Kanarevrsquos Hypothesis
bull Electron align linearly with proton They do NOT spin or form clouds around the nucleus
bull Neutrons are in the center of atom Proton are outside of Neutrons
bull The wavelength of all particles is the experimental radius
bull Double slit experiment explained correctly
bull Heisenbergrsquos inequality explained correctly
bull is the radius of approach of the magnetic lines around the torus of the electron
( 1 137)
bull Why water expands when it freezes
bull Electron clusters form lightning
rbull The meaning of Plancks constant is explained
22 2 kg m
h m f mr f consts
2210254 42m r kgm bull If the mr-constant describes a ring the equations
describe experiments bull If the mr-constant describes a wave the equations do
NOT describe experiments
Kanarevrsquos 5 Equations of Mechanics
Kanarevrsquos 1st Equation Acceleration ampmv maMma R R
Kanarevrsquos 2nd Equation Uniform motion amp mv maMma R R
maM mvR R
Kanarevrsquos 3rd Equation
Kanarevrsquos 4th Equation Equal and opposite reaction
1 21 21 2
12
a m
m a m ama
Kanarevrsquos 5th Equation 1 2 3 Mi M M M Mna R R R R R
The sum of the Actions working on a moving body is never equal to zero
Kanarevrsquos Photon model is a rolling ring
His equations accurately describe known experiments with photonsbull Electromagnetic radiation should be called photon radiationbull Diffraction and interference are the same phenomenabull Correctly explains the double slit experiment without quantum equationsbull Correctly explains the meaning of Heisenbergrsquos inequality
e
2 (2)(3142)(2817 15 ) 10007297352568
R 2426 12 137mag
xp
r m
m
Charge
Magnetic action
Proton
The intensity of the magnetic action near the geometrical center of the proton ring is so great that we theorize that it is the strong nuclear force
(9383 8)(1602 19)8476 14
4 (4)(3142)(1411 26)p
pp
EB T
M
(9395 8)(1602 19)1035 15
4 (4)(3142)(0966 26)N
NN
EB T
M
Neutron
1a Hydrogenavi
A direction in space (3 units toward the left) changes a whole number to a fraction by using the dash (-) as an exponent rather than a direction in space
This is illogical and incorrect BS math should have performed this operation as
Unfortunately the answers to problems can come out correct because data points of a real problem fit the curve
However the answers will not be due to logical reasoning
Because the equation produced a curve that real problem data points follow it has been falsely believed that the equations and graphs are correct
1
2
x
Incorrect BK graph for (12)x
(12)x
4 00625 correct3 0125 correct2 025 correct1 1 correct0 1 correct-1 2 incorrect-2 4 incorrect-3 8 incorrect-4 16 incorrect
negative side of graph is incorrectpositive side of graph is correct
00625012502505124
8
16
-5 -4 -3 -2 -1 0 1 2 3 4 5P
osit
ive
negative positive
BS Math for (12)x
correctincorrect
Incorrect BS math and graph for
All graphs are symmetrical with SM
Two BS Math Equations that Produce Incorrect Answers
2 0x x starting equation 2
2
x x x x
x x
Add lsquoxrsquo to both sides
2(1) 1 let lsquoxrsquo = 1
2 1 This is an incorrect answer 2(0)-(0)=0 The only answer that BK math can give is x=0
1
a b starting equation ( )( ) ( )( )a a a b multiply both sides by ( )a
( )( ) ( )( ) ( )( ) ( )( )a a b b a b b b subtract ( )( )b b from both sides
( )( ) ( )( )a b a b a b b factor ( )a b from both sides
( )a b b divide both sides by ( )a b
( )b b b replace ( )a with ( )b 2b b simplify 2(1) (1) Let ( )b = 1
2 1 Again BS math yields and incorrect answer 1
This incorrect result is a problem with the distributive law SM explains how this incorrect answer is obtained and the problem with the distributive law that allows the error
Page 148 ndash149 Barry Mazur ndash Imagining Numbers
3
2
31
Do the computation carefully on paper using the rules we agreed to Then ponder your answer which should be something of a surprise to you donrsquot stop there Think of what your answer might possibly mean or might imply
Solve this equation using the distributive law
JK definitely a surprise the answer wasincorrect and illogical
Gerhard Gade University Professor doing math at Harvard University
Here is the BS math answer that Barry Mazur calculated
2
31
2
31
2
31
1 3 1 3 1 3 3 3 2 2 3 1 3
2 2 2 2 4 4 4 4 4 4 2 2
dash
1 3 1 3 1 31
2 2 2 2 4 4
3 3
4 4
When the distributive law is used with the BS Rule-of-Signs and one of the signs is a (-) dash sign the middle two terms always cancel each other This leads to the incorrect answer of (-1) in the above equation
Barry Mazur did not go back and check his answer with the original equation If he had he would have known that the answer was incorrect
3
1 31
2
check answer by setting equal
3
331 3
12
1 31
2
Take the Cube root of both sides
1 3 2 multiply both sides by 2
3 3 subtract 1 from both sides
2 2
3 3 square both sides
3 9 This is an incorrect answer A simple equation check could have shown that the distributive law using BS math provides an incorrect answer
1
BS Illogical and Incorrect Numbers Raised to Powers
nC D n DC
642 DDD
In BS math only a positive number can have an answer if ldquonrdquo is an even number (246---)
All negative numbers have no solutions
642 DDD
With SM just add a direction arrow and any number gives the correct answer
or n nC D C D
3 3( 3 8 8 or 8 2 or 2 )n D C
( 2 16 C= 16 or 16 4 or 4 )n D
Since a negative number has no square root answer in BS math BS mathematicians invented imaginary numbers
i2=-1
24414 iiC n
These equations are easily solved without the use of imaginary numbers with SM math
If 4 then 4 2x
If 4 then 4 2x
The 360-degree circle is a 4400 years old extremely outdated system based on a year of 12 months of 30 days each that is 360 days
SM will use a more logical system that will have a circle of 6283185307o
d
( )
(deg )
d r radians
d r rees
the arc distance around
the circumference of a circle
d
radiusr
2 2
2 2
d r
dv rd t t
v dda rd t t t
1
d
r
( )av tv
The average angular velocity of a body is the rate of its angular displacement
( )0
liminstt
vt
The instantaneous angular velocity is the limit of the ratio
t
as t approaches zero
( )d
av
d
tv
The average circumference velocity of the body is the rate of its distance traveled d divided by the time t
dvv
r
1 dd vdrv
t t t r r
The equation of the relationship between the angular velocity and the velocity of a point on that body
( )o
av
v v v
t ta
The average angular acceleration is the change in the angular velocity
( )0
liminstt
v
ta
The instantaneous angular acceleration is the
limit of the ration of v
t
as t approaches
zero
daa
r
1
1
d
d s d
vva
t r tv a
r t r
The equation of the relationship between the angular acceleration and the acceleration of a point on the body
21
2v a t
If a body with an initial angular velocity ov has a constant
angular acceleration a it will turn through an angle in a time t
1
Straight Line Symbol Units Rotational Symbol Units Distance d m Angular
Distance radians or degrees
Velocity dt
v m
s Angular Velocity
V
t
radians or degrees
s
Acceleration vat
2ms
Angular Acceleration
VA
t
2
radians or degrees
s
Mass M Kg Mass distribution
dM 2Kgm
Force actionA Ma
2
Kgm
sNewton
Torque
dT M A 2
2
Kgm
s
Momentum Mv Kgm
s
Angular Momentum
dM V 2Kgm
s
Work (Force)(distance)
Ad 2
Kgmm
sNm Joule
Work T
Using AS arrow math to solve and direction in space problems using the equation 1 2
2 2 2 2 2 2( ) amp and ( ) amp 3
4
2 2 2
2 2
( ) amp
(5 ) 4 3
(5 ) 16 9
(5 ) 25
(5 ) (5 )
2 2(5 ) 4 3
(5 ) 16 9
(5 ) 25
(5 ) (5 )
In SM math Pythagorasrsquo theorem provides both magnitude and direction
In BS math Pythagorasrsquo theorem provides magnitude only
time set equal to zero
23
( ) ( 0 )10 1
x t xx
3( ) ( 0)
10 1W t f
In BS math this is an equation for a traveling wave
must use a square term in order to make the curve symmetric
NO SQUARE TERM IS NECESSARY
In SM math when t is set to ldquo0rdquo
This is a major fault in BS math
x f(x)
-4 0019
-3 0033
-2 0073
-1 0273
0 3000
1 0273
2 0073
3 0033
4 0019 -4 -3 -2 -1 0 1 2 3 400
05
10
15
20
25
30
35
BS wave (must use square term)
snap shot of wave at x = 0
Am
plit
ude
lt to gt lt to gt f(ltgt)
lt -4 0073
lt -3 0097
lt -2 0143
lt -1 0273
origin 0 3000
gt 1 0273
gt 2 0143
gt 3 0097
gt 4 0073lt0073 lt0097 lt0143 lt0273
3000
0273gt 0143gt 0097gt 0073gt
Wave moving
snap shot of wave
Am
plit
ude
lt4 lt3 lt2 lt1 0 1gt 2gt 3gt 4gt
3
2
1
In SM NO squared terms are needed
SPRING
An Action is needed to stretch or compress a spring to a starting point where it can execute cycle motion
tsA (ts to start) is the work required to move a mass attached to a spring a distance (d1) from
its initial rest position RAK 0 the spring constant is the stiffness of the spring
o A is attraction o R is repulsion o
ARRA
00 is for the back and forth motion through the zero initial rest position A
stretched spring goes from ldquoArdquo to ldquo0rdquo to ldquoRrdquo to ldquoArdquo to ldquo0rdquo to ldquoRrdquo d1 is the distance a mass is moved from its initial rest position
0
0 1ts A RA K d
Once ( )mad
is performed to move the mass to position d1 (against the restraint of the spring) a of the ( )mad
is transferred into the spring as potential attractionrepulsion (PAR) There is always
more ( )mad
performed than is stored as PAR When the spring is released from the restraining Ats the PAR starts the oscillation Conservation of ( )mad
requires that the PAR and the motion ( )mad
( )( )madM be a constant
( )amp ( ) constantmadPAR M 02
0 0
1
2A R A RPAR K d
02
( ) 0 0
1( )
2mad A R A RM m v
0 0 02 2 2
0 0 0 1
2 2 21
2 2 21
1 1 1amp
2 2 2
1 1 1amp
2 2 2
amp
A R A R A RK d m v K d
Kd mv Kd
Kd mv Kd
0
1d is frac12 the height from the base to the peak of a cycle
An experiment of moving paper under a pendulum through a small angle filled with sand that can escape through a small hole in the bottom shows that a pendulum cycling back and forth follows a curve that is mathematically analogous to a lsquosinrsquo or lsquocosrsquo
The projection of the curved area onto the straight line exactly duplicate the back and forth motion of CM This allows for the math of lsquosinrsquo or lsquocosrsquo to be used
SM will use lsquosinrdquo math It could use lsquocosrsquo or a combination of both lsquosinrsquo and lsquocosrsquo Using just the lsquosinrsquo math makes the understanding and math easier
0 sin 0 0SM 0 sin157 1SM The diagram at the left shows a spring being stretched from its (zero rest) position-2 and the PAR being released at position-1 There are 4 sections (1 to 2 2 to 3 3 to 4 and 4 to5) that alternate between maximum acceleration and zero velocity to max velocity and zero acceleration
At position-1 there is maximum acceleration and zero velocity o The lsquosinrsquo values from 0o to 157o will be used for velocity (values from 0 to 1) o The lsquosinrsquo values from 157o to 0o will be used for acceleration (values from 1 to 0)
At position-2 there is maximum velocity and zero acceleration Acceleration changes to deceleration
o The lsquosinrsquo values from 157o to 0o will be used for deceleration (values from 1 to 0) o The lsquosinrsquo values from 0o to 157o will be used for acceleration (values from 0 to 1)
At position-3 there is maximum acceleration and zero velocity o The lsquosinrsquo values from 0o to 157o will be used for velocity (values from 0 to 1) o The lsquosinrsquo values from 157o to 0o will be used for acceleration (values from 1 to 0)
At position-4 there is maximum velocity and zero acceleration Acceleration changes to deceleration
o The lsquosinrsquo values from 157o to 0o will be used for deceleration (values from 1 to 0) o The lsquosinrsquo values from 0o to 157o will be used for acceleration (values from 0 to 1)
At position-5 the cycle starts again
2
)539(
cT
MK
spring constant (K) when mass (M) and cycle-time (T) are known
)(
))(89)((
111 md
uKgM
d
MG
d
FK sts
spring constant (K) when mass (M) and amplitude ( d1) are known
2 21( )
Kv d d
m
velocity (v) at any point during the cycle
velocity
dforM
Kdv
max
012 velocity (v2) maximum (when d = 0 at position-2 and 4)
acceleration (a1) maximum when spring constant (K) mass (M) and amplitude (d1) are known
acceleration (a1) maximum when amplitude (d1) and cycle-time (Tc) are known
acceleration (a) at any point during the cycle-time (T)
1
151
286
v
dTc
cycle-time (Tc) when amplitude (d1) and velocity (v1) are known
K
MTc 28651
cycle-time (Tc) when mass (M) and spring constant (K) are known
fTc
1
cycle-time (Tc) when the frequency is known
cTf
1
frequency (f) when the cycle-time (Tc) is known
of cycles
elapsed timef
frequency (f) when the number of cycles in an elapsed time is known
d1 amplitude (distance from rest to start position)
Also frac12 the distance from the base to the peak
3131 oror dM
Ka
251
3131 539
c
oror T
da
2539
cT
da
In December of 2006 I discovered the mathematical error in Einsteinrsquos Special Relativity Einstein used BS math and that math only works in an imaginary world
It cannot describe real symmetrical space
Einsteinrsquos Special Relativity should be abandoned and a theory of real space using SM should be used to replace it
Numerous articles on the Internet show Einsteinrsquos theory to be incorrect but no one in academia is willing to publish the information
Figure 1-10 A moving particle observed from two frames of reference in uniform translation
Einsteinrsquos Math Error in Special Relativity Transformation of Velocity (From Eisenberg- Fundamentals of Modern Physics pages 27-29) Consider the particle shown in figure (1-10) moving with velocity v as
seen in a frame of reference O
dx dy dz
Vx Vy Vzdt dt dt
dx dy dz
V x V y V zdy dt dt
We would like to evaluate the velocity v of the particle as seen in the frame of reference 0 which is itself moving relative to 0 with velocity v Measured in the 0 frame the velocity vector of the particle has com ponents
The same velocity vector as measured in the 0 frame has components
2
2
2
2
2
1
1
x vtx
v
cy y
z z
vxt
ctv
c
Now from equations (1-13) we know that the relation between the primed and the unprimed coordinates and times is
Take the differential of these equations remembering that v is a constant This gives
2
2
2
2
2
1
1
dx vdtdx
v
cdy dy
dz dz
vxdt
cdtv
c
2
2
2 2 2 2 2
2
2
2
11 ( )
1( ) 1
1
1
1
dx vdt
v dxdx vdt v
dx dx vdtc dt dtvdx vdx vdx v dxdt dt dt dt
dt dtc c c c
VxV x
Vx
VxV x
vv
vv
c
x
v
c
cV
( )
coordinate of particle relative to moving primed system
coordinate of particle relative to stationary system
velocity of primed coordina
V x
Vx
te system relative to stationaryv sys
tem
Page 28 Fundamentals of Modern Physics
(1-16)
According to Einstein this equation tells you how to transform the observed velocity from one frame of reference to another frame of reference First we note that as Vc and vc approachzero the equation approaches those which would be derived from the Galilean transformation
2 2
( ) 1 01 1 ( )
Vx v Vx vV
VxV x
Vx v
v cVxx
v
c
v
cc
Another property of this equation is that it is impossible to choose V and v such that Vrsquo the magnitude of the velocity which is seen in the new frame is greater than c
1
2
09 ( 09 ) 18009876543
( 09 ) 1811 (09 )
c c cV c c
cc
c
Example As seen by O particle 1 has velocity 09c in the direction of positive x and particle 2 has velocity 09c in the negative x direction
Einstein equation however predicts To evaluate the velocity of particle 1 with respect to particle 2 we transform from the O frame to the Orsquo frame moving in the negative x direction with velocity v = -09c using equation 1-16 We obtain
JK LOGIC will observe the particles separating at 18c
A confusing INCORRECT and ILLOGICAL answer
left is a dir rightsub of a
1
2
1
09 ( 09 ) 18009876543
181( 09 )1 ( 09 )
The multiplication of dir left time a dir right is ONLY a dir left
09
dir
direction leftdirection rightsubtraction
c c cV c c
cc
c
V
dir
22
22
09 18 18009876543
181( 081)1(09)(09)
1
left rightsub
sub
c c c cc c
ccc
c
Analyzing Einsteinrsquos BS math Equation
Using SM math a logical answer can be obtained
Cannot multiply a subtraction operator times a direction in space
Can only subtract and change its direction
Cannot multiply a direction to the left times a direction to the right
Cannot multiply opposite directions in space
Can only solve for the resultant
rightleft
subtraction
1
subtraction
2
sub of a dir to the left is a dir to the right
1 resultant 0 sub 2
2sub 2
2
09 09
09 amp 091
09 09 18 1
(0)1
(09) amp (09)1
c cV
c cc
c c cV
cc
cc
8
181
cc
SM math predicts a logical answer using logic
21st century scientists state in many articles that Einsteins General Relativity and Quantum Mechanics are incompatible
They are not only incompatible they are both mathematically incorrect
They both use illogical BS math
Any equations in math that use imaginary numbers should be abandoned
Imaginary numbers are only necessary in the current BS math system
This means that Schroumldingers equations should be abandoned
Any equation in quantum mechanics that uses imaginary number should be abandoned
Prof Kanarev has developed logical physics and chemistrytheories without the use of Quantum mechanics or relativity
A partial list of his theories coverbull Mechanics ndash replaces Newtonbull Photon - bull Electrons ndash Protons- Neutronsbull Spectroscopybull Nucleusbull Atoms amp Moleculesbull Electrodynamicsbull Heatbull Curvature of Space ndash replaces Schwarzcchildrsquos equation
Kanarevrsquos Hypothesis
bull Electron align linearly with proton They do NOT spin or form clouds around the nucleus
bull Neutrons are in the center of atom Proton are outside of Neutrons
bull The wavelength of all particles is the experimental radius
bull Double slit experiment explained correctly
bull Heisenbergrsquos inequality explained correctly
bull is the radius of approach of the magnetic lines around the torus of the electron
( 1 137)
bull Why water expands when it freezes
bull Electron clusters form lightning
rbull The meaning of Plancks constant is explained
22 2 kg m
h m f mr f consts
2210254 42m r kgm bull If the mr-constant describes a ring the equations
describe experiments bull If the mr-constant describes a wave the equations do
NOT describe experiments
Kanarevrsquos 5 Equations of Mechanics
Kanarevrsquos 1st Equation Acceleration ampmv maMma R R
Kanarevrsquos 2nd Equation Uniform motion amp mv maMma R R
maM mvR R
Kanarevrsquos 3rd Equation
Kanarevrsquos 4th Equation Equal and opposite reaction
1 21 21 2
12
a m
m a m ama
Kanarevrsquos 5th Equation 1 2 3 Mi M M M Mna R R R R R
The sum of the Actions working on a moving body is never equal to zero
Kanarevrsquos Photon model is a rolling ring
His equations accurately describe known experiments with photonsbull Electromagnetic radiation should be called photon radiationbull Diffraction and interference are the same phenomenabull Correctly explains the double slit experiment without quantum equationsbull Correctly explains the meaning of Heisenbergrsquos inequality
e
2 (2)(3142)(2817 15 ) 10007297352568
R 2426 12 137mag
xp
r m
m
Charge
Magnetic action
Proton
The intensity of the magnetic action near the geometrical center of the proton ring is so great that we theorize that it is the strong nuclear force
(9383 8)(1602 19)8476 14
4 (4)(3142)(1411 26)p
pp
EB T
M
(9395 8)(1602 19)1035 15
4 (4)(3142)(0966 26)N
NN
EB T
M
Neutron
1a Hydrogenavi
1
2
x
Incorrect BK graph for (12)x
(12)x
4 00625 correct3 0125 correct2 025 correct1 1 correct0 1 correct-1 2 incorrect-2 4 incorrect-3 8 incorrect-4 16 incorrect
negative side of graph is incorrectpositive side of graph is correct
00625012502505124
8
16
-5 -4 -3 -2 -1 0 1 2 3 4 5P
osit
ive
negative positive
BS Math for (12)x
correctincorrect
Incorrect BS math and graph for
All graphs are symmetrical with SM
Two BS Math Equations that Produce Incorrect Answers
2 0x x starting equation 2
2
x x x x
x x
Add lsquoxrsquo to both sides
2(1) 1 let lsquoxrsquo = 1
2 1 This is an incorrect answer 2(0)-(0)=0 The only answer that BK math can give is x=0
1
a b starting equation ( )( ) ( )( )a a a b multiply both sides by ( )a
( )( ) ( )( ) ( )( ) ( )( )a a b b a b b b subtract ( )( )b b from both sides
( )( ) ( )( )a b a b a b b factor ( )a b from both sides
( )a b b divide both sides by ( )a b
( )b b b replace ( )a with ( )b 2b b simplify 2(1) (1) Let ( )b = 1
2 1 Again BS math yields and incorrect answer 1
This incorrect result is a problem with the distributive law SM explains how this incorrect answer is obtained and the problem with the distributive law that allows the error
Page 148 ndash149 Barry Mazur ndash Imagining Numbers
3
2
31
Do the computation carefully on paper using the rules we agreed to Then ponder your answer which should be something of a surprise to you donrsquot stop there Think of what your answer might possibly mean or might imply
Solve this equation using the distributive law
JK definitely a surprise the answer wasincorrect and illogical
Gerhard Gade University Professor doing math at Harvard University
Here is the BS math answer that Barry Mazur calculated
2
31
2
31
2
31
1 3 1 3 1 3 3 3 2 2 3 1 3
2 2 2 2 4 4 4 4 4 4 2 2
dash
1 3 1 3 1 31
2 2 2 2 4 4
3 3
4 4
When the distributive law is used with the BS Rule-of-Signs and one of the signs is a (-) dash sign the middle two terms always cancel each other This leads to the incorrect answer of (-1) in the above equation
Barry Mazur did not go back and check his answer with the original equation If he had he would have known that the answer was incorrect
3
1 31
2
check answer by setting equal
3
331 3
12
1 31
2
Take the Cube root of both sides
1 3 2 multiply both sides by 2
3 3 subtract 1 from both sides
2 2
3 3 square both sides
3 9 This is an incorrect answer A simple equation check could have shown that the distributive law using BS math provides an incorrect answer
1
BS Illogical and Incorrect Numbers Raised to Powers
nC D n DC
642 DDD
In BS math only a positive number can have an answer if ldquonrdquo is an even number (246---)
All negative numbers have no solutions
642 DDD
With SM just add a direction arrow and any number gives the correct answer
or n nC D C D
3 3( 3 8 8 or 8 2 or 2 )n D C
( 2 16 C= 16 or 16 4 or 4 )n D
Since a negative number has no square root answer in BS math BS mathematicians invented imaginary numbers
i2=-1
24414 iiC n
These equations are easily solved without the use of imaginary numbers with SM math
If 4 then 4 2x
If 4 then 4 2x
The 360-degree circle is a 4400 years old extremely outdated system based on a year of 12 months of 30 days each that is 360 days
SM will use a more logical system that will have a circle of 6283185307o
d
( )
(deg )
d r radians
d r rees
the arc distance around
the circumference of a circle
d
radiusr
2 2
2 2
d r
dv rd t t
v dda rd t t t
1
d
r
( )av tv
The average angular velocity of a body is the rate of its angular displacement
( )0
liminstt
vt
The instantaneous angular velocity is the limit of the ratio
t
as t approaches zero
( )d
av
d
tv
The average circumference velocity of the body is the rate of its distance traveled d divided by the time t
dvv
r
1 dd vdrv
t t t r r
The equation of the relationship between the angular velocity and the velocity of a point on that body
( )o
av
v v v
t ta
The average angular acceleration is the change in the angular velocity
( )0
liminstt
v
ta
The instantaneous angular acceleration is the
limit of the ration of v
t
as t approaches
zero
daa
r
1
1
d
d s d
vva
t r tv a
r t r
The equation of the relationship between the angular acceleration and the acceleration of a point on the body
21
2v a t
If a body with an initial angular velocity ov has a constant
angular acceleration a it will turn through an angle in a time t
1
Straight Line Symbol Units Rotational Symbol Units Distance d m Angular
Distance radians or degrees
Velocity dt
v m
s Angular Velocity
V
t
radians or degrees
s
Acceleration vat
2ms
Angular Acceleration
VA
t
2
radians or degrees
s
Mass M Kg Mass distribution
dM 2Kgm
Force actionA Ma
2
Kgm
sNewton
Torque
dT M A 2
2
Kgm
s
Momentum Mv Kgm
s
Angular Momentum
dM V 2Kgm
s
Work (Force)(distance)
Ad 2
Kgmm
sNm Joule
Work T
Using AS arrow math to solve and direction in space problems using the equation 1 2
2 2 2 2 2 2( ) amp and ( ) amp 3
4
2 2 2
2 2
( ) amp
(5 ) 4 3
(5 ) 16 9
(5 ) 25
(5 ) (5 )
2 2(5 ) 4 3
(5 ) 16 9
(5 ) 25
(5 ) (5 )
In SM math Pythagorasrsquo theorem provides both magnitude and direction
In BS math Pythagorasrsquo theorem provides magnitude only
time set equal to zero
23
( ) ( 0 )10 1
x t xx
3( ) ( 0)
10 1W t f
In BS math this is an equation for a traveling wave
must use a square term in order to make the curve symmetric
NO SQUARE TERM IS NECESSARY
In SM math when t is set to ldquo0rdquo
This is a major fault in BS math
x f(x)
-4 0019
-3 0033
-2 0073
-1 0273
0 3000
1 0273
2 0073
3 0033
4 0019 -4 -3 -2 -1 0 1 2 3 400
05
10
15
20
25
30
35
BS wave (must use square term)
snap shot of wave at x = 0
Am
plit
ude
lt to gt lt to gt f(ltgt)
lt -4 0073
lt -3 0097
lt -2 0143
lt -1 0273
origin 0 3000
gt 1 0273
gt 2 0143
gt 3 0097
gt 4 0073lt0073 lt0097 lt0143 lt0273
3000
0273gt 0143gt 0097gt 0073gt
Wave moving
snap shot of wave
Am
plit
ude
lt4 lt3 lt2 lt1 0 1gt 2gt 3gt 4gt
3
2
1
In SM NO squared terms are needed
SPRING
An Action is needed to stretch or compress a spring to a starting point where it can execute cycle motion
tsA (ts to start) is the work required to move a mass attached to a spring a distance (d1) from
its initial rest position RAK 0 the spring constant is the stiffness of the spring
o A is attraction o R is repulsion o
ARRA
00 is for the back and forth motion through the zero initial rest position A
stretched spring goes from ldquoArdquo to ldquo0rdquo to ldquoRrdquo to ldquoArdquo to ldquo0rdquo to ldquoRrdquo d1 is the distance a mass is moved from its initial rest position
0
0 1ts A RA K d
Once ( )mad
is performed to move the mass to position d1 (against the restraint of the spring) a of the ( )mad
is transferred into the spring as potential attractionrepulsion (PAR) There is always
more ( )mad
performed than is stored as PAR When the spring is released from the restraining Ats the PAR starts the oscillation Conservation of ( )mad
requires that the PAR and the motion ( )mad
( )( )madM be a constant
( )amp ( ) constantmadPAR M 02
0 0
1
2A R A RPAR K d
02
( ) 0 0
1( )
2mad A R A RM m v
0 0 02 2 2
0 0 0 1
2 2 21
2 2 21
1 1 1amp
2 2 2
1 1 1amp
2 2 2
amp
A R A R A RK d m v K d
Kd mv Kd
Kd mv Kd
0
1d is frac12 the height from the base to the peak of a cycle
An experiment of moving paper under a pendulum through a small angle filled with sand that can escape through a small hole in the bottom shows that a pendulum cycling back and forth follows a curve that is mathematically analogous to a lsquosinrsquo or lsquocosrsquo
The projection of the curved area onto the straight line exactly duplicate the back and forth motion of CM This allows for the math of lsquosinrsquo or lsquocosrsquo to be used
SM will use lsquosinrdquo math It could use lsquocosrsquo or a combination of both lsquosinrsquo and lsquocosrsquo Using just the lsquosinrsquo math makes the understanding and math easier
0 sin 0 0SM 0 sin157 1SM The diagram at the left shows a spring being stretched from its (zero rest) position-2 and the PAR being released at position-1 There are 4 sections (1 to 2 2 to 3 3 to 4 and 4 to5) that alternate between maximum acceleration and zero velocity to max velocity and zero acceleration
At position-1 there is maximum acceleration and zero velocity o The lsquosinrsquo values from 0o to 157o will be used for velocity (values from 0 to 1) o The lsquosinrsquo values from 157o to 0o will be used for acceleration (values from 1 to 0)
At position-2 there is maximum velocity and zero acceleration Acceleration changes to deceleration
o The lsquosinrsquo values from 157o to 0o will be used for deceleration (values from 1 to 0) o The lsquosinrsquo values from 0o to 157o will be used for acceleration (values from 0 to 1)
At position-3 there is maximum acceleration and zero velocity o The lsquosinrsquo values from 0o to 157o will be used for velocity (values from 0 to 1) o The lsquosinrsquo values from 157o to 0o will be used for acceleration (values from 1 to 0)
At position-4 there is maximum velocity and zero acceleration Acceleration changes to deceleration
o The lsquosinrsquo values from 157o to 0o will be used for deceleration (values from 1 to 0) o The lsquosinrsquo values from 0o to 157o will be used for acceleration (values from 0 to 1)
At position-5 the cycle starts again
2
)539(
cT
MK
spring constant (K) when mass (M) and cycle-time (T) are known
)(
))(89)((
111 md
uKgM
d
MG
d
FK sts
spring constant (K) when mass (M) and amplitude ( d1) are known
2 21( )
Kv d d
m
velocity (v) at any point during the cycle
velocity
dforM
Kdv
max
012 velocity (v2) maximum (when d = 0 at position-2 and 4)
acceleration (a1) maximum when spring constant (K) mass (M) and amplitude (d1) are known
acceleration (a1) maximum when amplitude (d1) and cycle-time (Tc) are known
acceleration (a) at any point during the cycle-time (T)
1
151
286
v
dTc
cycle-time (Tc) when amplitude (d1) and velocity (v1) are known
K
MTc 28651
cycle-time (Tc) when mass (M) and spring constant (K) are known
fTc
1
cycle-time (Tc) when the frequency is known
cTf
1
frequency (f) when the cycle-time (Tc) is known
of cycles
elapsed timef
frequency (f) when the number of cycles in an elapsed time is known
d1 amplitude (distance from rest to start position)
Also frac12 the distance from the base to the peak
3131 oror dM
Ka
251
3131 539
c
oror T
da
2539
cT
da
In December of 2006 I discovered the mathematical error in Einsteinrsquos Special Relativity Einstein used BS math and that math only works in an imaginary world
It cannot describe real symmetrical space
Einsteinrsquos Special Relativity should be abandoned and a theory of real space using SM should be used to replace it
Numerous articles on the Internet show Einsteinrsquos theory to be incorrect but no one in academia is willing to publish the information
Figure 1-10 A moving particle observed from two frames of reference in uniform translation
Einsteinrsquos Math Error in Special Relativity Transformation of Velocity (From Eisenberg- Fundamentals of Modern Physics pages 27-29) Consider the particle shown in figure (1-10) moving with velocity v as
seen in a frame of reference O
dx dy dz
Vx Vy Vzdt dt dt
dx dy dz
V x V y V zdy dt dt
We would like to evaluate the velocity v of the particle as seen in the frame of reference 0 which is itself moving relative to 0 with velocity v Measured in the 0 frame the velocity vector of the particle has com ponents
The same velocity vector as measured in the 0 frame has components
2
2
2
2
2
1
1
x vtx
v
cy y
z z
vxt
ctv
c
Now from equations (1-13) we know that the relation between the primed and the unprimed coordinates and times is
Take the differential of these equations remembering that v is a constant This gives
2
2
2
2
2
1
1
dx vdtdx
v
cdy dy
dz dz
vxdt
cdtv
c
2
2
2 2 2 2 2
2
2
2
11 ( )
1( ) 1
1
1
1
dx vdt
v dxdx vdt v
dx dx vdtc dt dtvdx vdx vdx v dxdt dt dt dt
dt dtc c c c
VxV x
Vx
VxV x
vv
vv
c
x
v
c
cV
( )
coordinate of particle relative to moving primed system
coordinate of particle relative to stationary system
velocity of primed coordina
V x
Vx
te system relative to stationaryv sys
tem
Page 28 Fundamentals of Modern Physics
(1-16)
According to Einstein this equation tells you how to transform the observed velocity from one frame of reference to another frame of reference First we note that as Vc and vc approachzero the equation approaches those which would be derived from the Galilean transformation
2 2
( ) 1 01 1 ( )
Vx v Vx vV
VxV x
Vx v
v cVxx
v
c
v
cc
Another property of this equation is that it is impossible to choose V and v such that Vrsquo the magnitude of the velocity which is seen in the new frame is greater than c
1
2
09 ( 09 ) 18009876543
( 09 ) 1811 (09 )
c c cV c c
cc
c
Example As seen by O particle 1 has velocity 09c in the direction of positive x and particle 2 has velocity 09c in the negative x direction
Einstein equation however predicts To evaluate the velocity of particle 1 with respect to particle 2 we transform from the O frame to the Orsquo frame moving in the negative x direction with velocity v = -09c using equation 1-16 We obtain
JK LOGIC will observe the particles separating at 18c
A confusing INCORRECT and ILLOGICAL answer
left is a dir rightsub of a
1
2
1
09 ( 09 ) 18009876543
181( 09 )1 ( 09 )
The multiplication of dir left time a dir right is ONLY a dir left
09
dir
direction leftdirection rightsubtraction
c c cV c c
cc
c
V
dir
22
22
09 18 18009876543
181( 081)1(09)(09)
1
left rightsub
sub
c c c cc c
ccc
c
Analyzing Einsteinrsquos BS math Equation
Using SM math a logical answer can be obtained
Cannot multiply a subtraction operator times a direction in space
Can only subtract and change its direction
Cannot multiply a direction to the left times a direction to the right
Cannot multiply opposite directions in space
Can only solve for the resultant
rightleft
subtraction
1
subtraction
2
sub of a dir to the left is a dir to the right
1 resultant 0 sub 2
2sub 2
2
09 09
09 amp 091
09 09 18 1
(0)1
(09) amp (09)1
c cV
c cc
c c cV
cc
cc
8
181
cc
SM math predicts a logical answer using logic
21st century scientists state in many articles that Einsteins General Relativity and Quantum Mechanics are incompatible
They are not only incompatible they are both mathematically incorrect
They both use illogical BS math
Any equations in math that use imaginary numbers should be abandoned
Imaginary numbers are only necessary in the current BS math system
This means that Schroumldingers equations should be abandoned
Any equation in quantum mechanics that uses imaginary number should be abandoned
Prof Kanarev has developed logical physics and chemistrytheories without the use of Quantum mechanics or relativity
A partial list of his theories coverbull Mechanics ndash replaces Newtonbull Photon - bull Electrons ndash Protons- Neutronsbull Spectroscopybull Nucleusbull Atoms amp Moleculesbull Electrodynamicsbull Heatbull Curvature of Space ndash replaces Schwarzcchildrsquos equation
Kanarevrsquos Hypothesis
bull Electron align linearly with proton They do NOT spin or form clouds around the nucleus
bull Neutrons are in the center of atom Proton are outside of Neutrons
bull The wavelength of all particles is the experimental radius
bull Double slit experiment explained correctly
bull Heisenbergrsquos inequality explained correctly
bull is the radius of approach of the magnetic lines around the torus of the electron
( 1 137)
bull Why water expands when it freezes
bull Electron clusters form lightning
rbull The meaning of Plancks constant is explained
22 2 kg m
h m f mr f consts
2210254 42m r kgm bull If the mr-constant describes a ring the equations
describe experiments bull If the mr-constant describes a wave the equations do
NOT describe experiments
Kanarevrsquos 5 Equations of Mechanics
Kanarevrsquos 1st Equation Acceleration ampmv maMma R R
Kanarevrsquos 2nd Equation Uniform motion amp mv maMma R R
maM mvR R
Kanarevrsquos 3rd Equation
Kanarevrsquos 4th Equation Equal and opposite reaction
1 21 21 2
12
a m
m a m ama
Kanarevrsquos 5th Equation 1 2 3 Mi M M M Mna R R R R R
The sum of the Actions working on a moving body is never equal to zero
Kanarevrsquos Photon model is a rolling ring
His equations accurately describe known experiments with photonsbull Electromagnetic radiation should be called photon radiationbull Diffraction and interference are the same phenomenabull Correctly explains the double slit experiment without quantum equationsbull Correctly explains the meaning of Heisenbergrsquos inequality
e
2 (2)(3142)(2817 15 ) 10007297352568
R 2426 12 137mag
xp
r m
m
Charge
Magnetic action
Proton
The intensity of the magnetic action near the geometrical center of the proton ring is so great that we theorize that it is the strong nuclear force
(9383 8)(1602 19)8476 14
4 (4)(3142)(1411 26)p
pp
EB T
M
(9395 8)(1602 19)1035 15
4 (4)(3142)(0966 26)N
NN
EB T
M
Neutron
1a Hydrogenavi
All graphs are symmetrical with SM
Two BS Math Equations that Produce Incorrect Answers
2 0x x starting equation 2
2
x x x x
x x
Add lsquoxrsquo to both sides
2(1) 1 let lsquoxrsquo = 1
2 1 This is an incorrect answer 2(0)-(0)=0 The only answer that BK math can give is x=0
1
a b starting equation ( )( ) ( )( )a a a b multiply both sides by ( )a
( )( ) ( )( ) ( )( ) ( )( )a a b b a b b b subtract ( )( )b b from both sides
( )( ) ( )( )a b a b a b b factor ( )a b from both sides
( )a b b divide both sides by ( )a b
( )b b b replace ( )a with ( )b 2b b simplify 2(1) (1) Let ( )b = 1
2 1 Again BS math yields and incorrect answer 1
This incorrect result is a problem with the distributive law SM explains how this incorrect answer is obtained and the problem with the distributive law that allows the error
Page 148 ndash149 Barry Mazur ndash Imagining Numbers
3
2
31
Do the computation carefully on paper using the rules we agreed to Then ponder your answer which should be something of a surprise to you donrsquot stop there Think of what your answer might possibly mean or might imply
Solve this equation using the distributive law
JK definitely a surprise the answer wasincorrect and illogical
Gerhard Gade University Professor doing math at Harvard University
Here is the BS math answer that Barry Mazur calculated
2
31
2
31
2
31
1 3 1 3 1 3 3 3 2 2 3 1 3
2 2 2 2 4 4 4 4 4 4 2 2
dash
1 3 1 3 1 31
2 2 2 2 4 4
3 3
4 4
When the distributive law is used with the BS Rule-of-Signs and one of the signs is a (-) dash sign the middle two terms always cancel each other This leads to the incorrect answer of (-1) in the above equation
Barry Mazur did not go back and check his answer with the original equation If he had he would have known that the answer was incorrect
3
1 31
2
check answer by setting equal
3
331 3
12
1 31
2
Take the Cube root of both sides
1 3 2 multiply both sides by 2
3 3 subtract 1 from both sides
2 2
3 3 square both sides
3 9 This is an incorrect answer A simple equation check could have shown that the distributive law using BS math provides an incorrect answer
1
BS Illogical and Incorrect Numbers Raised to Powers
nC D n DC
642 DDD
In BS math only a positive number can have an answer if ldquonrdquo is an even number (246---)
All negative numbers have no solutions
642 DDD
With SM just add a direction arrow and any number gives the correct answer
or n nC D C D
3 3( 3 8 8 or 8 2 or 2 )n D C
( 2 16 C= 16 or 16 4 or 4 )n D
Since a negative number has no square root answer in BS math BS mathematicians invented imaginary numbers
i2=-1
24414 iiC n
These equations are easily solved without the use of imaginary numbers with SM math
If 4 then 4 2x
If 4 then 4 2x
The 360-degree circle is a 4400 years old extremely outdated system based on a year of 12 months of 30 days each that is 360 days
SM will use a more logical system that will have a circle of 6283185307o
d
( )
(deg )
d r radians
d r rees
the arc distance around
the circumference of a circle
d
radiusr
2 2
2 2
d r
dv rd t t
v dda rd t t t
1
d
r
( )av tv
The average angular velocity of a body is the rate of its angular displacement
( )0
liminstt
vt
The instantaneous angular velocity is the limit of the ratio
t
as t approaches zero
( )d
av
d
tv
The average circumference velocity of the body is the rate of its distance traveled d divided by the time t
dvv
r
1 dd vdrv
t t t r r
The equation of the relationship between the angular velocity and the velocity of a point on that body
( )o
av
v v v
t ta
The average angular acceleration is the change in the angular velocity
( )0
liminstt
v
ta
The instantaneous angular acceleration is the
limit of the ration of v
t
as t approaches
zero
daa
r
1
1
d
d s d
vva
t r tv a
r t r
The equation of the relationship between the angular acceleration and the acceleration of a point on the body
21
2v a t
If a body with an initial angular velocity ov has a constant
angular acceleration a it will turn through an angle in a time t
1
Straight Line Symbol Units Rotational Symbol Units Distance d m Angular
Distance radians or degrees
Velocity dt
v m
s Angular Velocity
V
t
radians or degrees
s
Acceleration vat
2ms
Angular Acceleration
VA
t
2
radians or degrees
s
Mass M Kg Mass distribution
dM 2Kgm
Force actionA Ma
2
Kgm
sNewton
Torque
dT M A 2
2
Kgm
s
Momentum Mv Kgm
s
Angular Momentum
dM V 2Kgm
s
Work (Force)(distance)
Ad 2
Kgmm
sNm Joule
Work T
Using AS arrow math to solve and direction in space problems using the equation 1 2
2 2 2 2 2 2( ) amp and ( ) amp 3
4
2 2 2
2 2
( ) amp
(5 ) 4 3
(5 ) 16 9
(5 ) 25
(5 ) (5 )
2 2(5 ) 4 3
(5 ) 16 9
(5 ) 25
(5 ) (5 )
In SM math Pythagorasrsquo theorem provides both magnitude and direction
In BS math Pythagorasrsquo theorem provides magnitude only
time set equal to zero
23
( ) ( 0 )10 1
x t xx
3( ) ( 0)
10 1W t f
In BS math this is an equation for a traveling wave
must use a square term in order to make the curve symmetric
NO SQUARE TERM IS NECESSARY
In SM math when t is set to ldquo0rdquo
This is a major fault in BS math
x f(x)
-4 0019
-3 0033
-2 0073
-1 0273
0 3000
1 0273
2 0073
3 0033
4 0019 -4 -3 -2 -1 0 1 2 3 400
05
10
15
20
25
30
35
BS wave (must use square term)
snap shot of wave at x = 0
Am
plit
ude
lt to gt lt to gt f(ltgt)
lt -4 0073
lt -3 0097
lt -2 0143
lt -1 0273
origin 0 3000
gt 1 0273
gt 2 0143
gt 3 0097
gt 4 0073lt0073 lt0097 lt0143 lt0273
3000
0273gt 0143gt 0097gt 0073gt
Wave moving
snap shot of wave
Am
plit
ude
lt4 lt3 lt2 lt1 0 1gt 2gt 3gt 4gt
3
2
1
In SM NO squared terms are needed
SPRING
An Action is needed to stretch or compress a spring to a starting point where it can execute cycle motion
tsA (ts to start) is the work required to move a mass attached to a spring a distance (d1) from
its initial rest position RAK 0 the spring constant is the stiffness of the spring
o A is attraction o R is repulsion o
ARRA
00 is for the back and forth motion through the zero initial rest position A
stretched spring goes from ldquoArdquo to ldquo0rdquo to ldquoRrdquo to ldquoArdquo to ldquo0rdquo to ldquoRrdquo d1 is the distance a mass is moved from its initial rest position
0
0 1ts A RA K d
Once ( )mad
is performed to move the mass to position d1 (against the restraint of the spring) a of the ( )mad
is transferred into the spring as potential attractionrepulsion (PAR) There is always
more ( )mad
performed than is stored as PAR When the spring is released from the restraining Ats the PAR starts the oscillation Conservation of ( )mad
requires that the PAR and the motion ( )mad
( )( )madM be a constant
( )amp ( ) constantmadPAR M 02
0 0
1
2A R A RPAR K d
02
( ) 0 0
1( )
2mad A R A RM m v
0 0 02 2 2
0 0 0 1
2 2 21
2 2 21
1 1 1amp
2 2 2
1 1 1amp
2 2 2
amp
A R A R A RK d m v K d
Kd mv Kd
Kd mv Kd
0
1d is frac12 the height from the base to the peak of a cycle
An experiment of moving paper under a pendulum through a small angle filled with sand that can escape through a small hole in the bottom shows that a pendulum cycling back and forth follows a curve that is mathematically analogous to a lsquosinrsquo or lsquocosrsquo
The projection of the curved area onto the straight line exactly duplicate the back and forth motion of CM This allows for the math of lsquosinrsquo or lsquocosrsquo to be used
SM will use lsquosinrdquo math It could use lsquocosrsquo or a combination of both lsquosinrsquo and lsquocosrsquo Using just the lsquosinrsquo math makes the understanding and math easier
0 sin 0 0SM 0 sin157 1SM The diagram at the left shows a spring being stretched from its (zero rest) position-2 and the PAR being released at position-1 There are 4 sections (1 to 2 2 to 3 3 to 4 and 4 to5) that alternate between maximum acceleration and zero velocity to max velocity and zero acceleration
At position-1 there is maximum acceleration and zero velocity o The lsquosinrsquo values from 0o to 157o will be used for velocity (values from 0 to 1) o The lsquosinrsquo values from 157o to 0o will be used for acceleration (values from 1 to 0)
At position-2 there is maximum velocity and zero acceleration Acceleration changes to deceleration
o The lsquosinrsquo values from 157o to 0o will be used for deceleration (values from 1 to 0) o The lsquosinrsquo values from 0o to 157o will be used for acceleration (values from 0 to 1)
At position-3 there is maximum acceleration and zero velocity o The lsquosinrsquo values from 0o to 157o will be used for velocity (values from 0 to 1) o The lsquosinrsquo values from 157o to 0o will be used for acceleration (values from 1 to 0)
At position-4 there is maximum velocity and zero acceleration Acceleration changes to deceleration
o The lsquosinrsquo values from 157o to 0o will be used for deceleration (values from 1 to 0) o The lsquosinrsquo values from 0o to 157o will be used for acceleration (values from 0 to 1)
At position-5 the cycle starts again
2
)539(
cT
MK
spring constant (K) when mass (M) and cycle-time (T) are known
)(
))(89)((
111 md
uKgM
d
MG
d
FK sts
spring constant (K) when mass (M) and amplitude ( d1) are known
2 21( )
Kv d d
m
velocity (v) at any point during the cycle
velocity
dforM
Kdv
max
012 velocity (v2) maximum (when d = 0 at position-2 and 4)
acceleration (a1) maximum when spring constant (K) mass (M) and amplitude (d1) are known
acceleration (a1) maximum when amplitude (d1) and cycle-time (Tc) are known
acceleration (a) at any point during the cycle-time (T)
1
151
286
v
dTc
cycle-time (Tc) when amplitude (d1) and velocity (v1) are known
K
MTc 28651
cycle-time (Tc) when mass (M) and spring constant (K) are known
fTc
1
cycle-time (Tc) when the frequency is known
cTf
1
frequency (f) when the cycle-time (Tc) is known
of cycles
elapsed timef
frequency (f) when the number of cycles in an elapsed time is known
d1 amplitude (distance from rest to start position)
Also frac12 the distance from the base to the peak
3131 oror dM
Ka
251
3131 539
c
oror T
da
2539
cT
da
In December of 2006 I discovered the mathematical error in Einsteinrsquos Special Relativity Einstein used BS math and that math only works in an imaginary world
It cannot describe real symmetrical space
Einsteinrsquos Special Relativity should be abandoned and a theory of real space using SM should be used to replace it
Numerous articles on the Internet show Einsteinrsquos theory to be incorrect but no one in academia is willing to publish the information
Figure 1-10 A moving particle observed from two frames of reference in uniform translation
Einsteinrsquos Math Error in Special Relativity Transformation of Velocity (From Eisenberg- Fundamentals of Modern Physics pages 27-29) Consider the particle shown in figure (1-10) moving with velocity v as
seen in a frame of reference O
dx dy dz
Vx Vy Vzdt dt dt
dx dy dz
V x V y V zdy dt dt
We would like to evaluate the velocity v of the particle as seen in the frame of reference 0 which is itself moving relative to 0 with velocity v Measured in the 0 frame the velocity vector of the particle has com ponents
The same velocity vector as measured in the 0 frame has components
2
2
2
2
2
1
1
x vtx
v
cy y
z z
vxt
ctv
c
Now from equations (1-13) we know that the relation between the primed and the unprimed coordinates and times is
Take the differential of these equations remembering that v is a constant This gives
2
2
2
2
2
1
1
dx vdtdx
v
cdy dy
dz dz
vxdt
cdtv
c
2
2
2 2 2 2 2
2
2
2
11 ( )
1( ) 1
1
1
1
dx vdt
v dxdx vdt v
dx dx vdtc dt dtvdx vdx vdx v dxdt dt dt dt
dt dtc c c c
VxV x
Vx
VxV x
vv
vv
c
x
v
c
cV
( )
coordinate of particle relative to moving primed system
coordinate of particle relative to stationary system
velocity of primed coordina
V x
Vx
te system relative to stationaryv sys
tem
Page 28 Fundamentals of Modern Physics
(1-16)
According to Einstein this equation tells you how to transform the observed velocity from one frame of reference to another frame of reference First we note that as Vc and vc approachzero the equation approaches those which would be derived from the Galilean transformation
2 2
( ) 1 01 1 ( )
Vx v Vx vV
VxV x
Vx v
v cVxx
v
c
v
cc
Another property of this equation is that it is impossible to choose V and v such that Vrsquo the magnitude of the velocity which is seen in the new frame is greater than c
1
2
09 ( 09 ) 18009876543
( 09 ) 1811 (09 )
c c cV c c
cc
c
Example As seen by O particle 1 has velocity 09c in the direction of positive x and particle 2 has velocity 09c in the negative x direction
Einstein equation however predicts To evaluate the velocity of particle 1 with respect to particle 2 we transform from the O frame to the Orsquo frame moving in the negative x direction with velocity v = -09c using equation 1-16 We obtain
JK LOGIC will observe the particles separating at 18c
A confusing INCORRECT and ILLOGICAL answer
left is a dir rightsub of a
1
2
1
09 ( 09 ) 18009876543
181( 09 )1 ( 09 )
The multiplication of dir left time a dir right is ONLY a dir left
09
dir
direction leftdirection rightsubtraction
c c cV c c
cc
c
V
dir
22
22
09 18 18009876543
181( 081)1(09)(09)
1
left rightsub
sub
c c c cc c
ccc
c
Analyzing Einsteinrsquos BS math Equation
Using SM math a logical answer can be obtained
Cannot multiply a subtraction operator times a direction in space
Can only subtract and change its direction
Cannot multiply a direction to the left times a direction to the right
Cannot multiply opposite directions in space
Can only solve for the resultant
rightleft
subtraction
1
subtraction
2
sub of a dir to the left is a dir to the right
1 resultant 0 sub 2
2sub 2
2
09 09
09 amp 091
09 09 18 1
(0)1
(09) amp (09)1
c cV
c cc
c c cV
cc
cc
8
181
cc
SM math predicts a logical answer using logic
21st century scientists state in many articles that Einsteins General Relativity and Quantum Mechanics are incompatible
They are not only incompatible they are both mathematically incorrect
They both use illogical BS math
Any equations in math that use imaginary numbers should be abandoned
Imaginary numbers are only necessary in the current BS math system
This means that Schroumldingers equations should be abandoned
Any equation in quantum mechanics that uses imaginary number should be abandoned
Prof Kanarev has developed logical physics and chemistrytheories without the use of Quantum mechanics or relativity
A partial list of his theories coverbull Mechanics ndash replaces Newtonbull Photon - bull Electrons ndash Protons- Neutronsbull Spectroscopybull Nucleusbull Atoms amp Moleculesbull Electrodynamicsbull Heatbull Curvature of Space ndash replaces Schwarzcchildrsquos equation
Kanarevrsquos Hypothesis
bull Electron align linearly with proton They do NOT spin or form clouds around the nucleus
bull Neutrons are in the center of atom Proton are outside of Neutrons
bull The wavelength of all particles is the experimental radius
bull Double slit experiment explained correctly
bull Heisenbergrsquos inequality explained correctly
bull is the radius of approach of the magnetic lines around the torus of the electron
( 1 137)
bull Why water expands when it freezes
bull Electron clusters form lightning
rbull The meaning of Plancks constant is explained
22 2 kg m
h m f mr f consts
2210254 42m r kgm bull If the mr-constant describes a ring the equations
describe experiments bull If the mr-constant describes a wave the equations do
NOT describe experiments
Kanarevrsquos 5 Equations of Mechanics
Kanarevrsquos 1st Equation Acceleration ampmv maMma R R
Kanarevrsquos 2nd Equation Uniform motion amp mv maMma R R
maM mvR R
Kanarevrsquos 3rd Equation
Kanarevrsquos 4th Equation Equal and opposite reaction
1 21 21 2
12
a m
m a m ama
Kanarevrsquos 5th Equation 1 2 3 Mi M M M Mna R R R R R
The sum of the Actions working on a moving body is never equal to zero
Kanarevrsquos Photon model is a rolling ring
His equations accurately describe known experiments with photonsbull Electromagnetic radiation should be called photon radiationbull Diffraction and interference are the same phenomenabull Correctly explains the double slit experiment without quantum equationsbull Correctly explains the meaning of Heisenbergrsquos inequality
e
2 (2)(3142)(2817 15 ) 10007297352568
R 2426 12 137mag
xp
r m
m
Charge
Magnetic action
Proton
The intensity of the magnetic action near the geometrical center of the proton ring is so great that we theorize that it is the strong nuclear force
(9383 8)(1602 19)8476 14
4 (4)(3142)(1411 26)p
pp
EB T
M
(9395 8)(1602 19)1035 15
4 (4)(3142)(0966 26)N
NN
EB T
M
Neutron
1a Hydrogenavi
Two BS Math Equations that Produce Incorrect Answers
2 0x x starting equation 2
2
x x x x
x x
Add lsquoxrsquo to both sides
2(1) 1 let lsquoxrsquo = 1
2 1 This is an incorrect answer 2(0)-(0)=0 The only answer that BK math can give is x=0
1
a b starting equation ( )( ) ( )( )a a a b multiply both sides by ( )a
( )( ) ( )( ) ( )( ) ( )( )a a b b a b b b subtract ( )( )b b from both sides
( )( ) ( )( )a b a b a b b factor ( )a b from both sides
( )a b b divide both sides by ( )a b
( )b b b replace ( )a with ( )b 2b b simplify 2(1) (1) Let ( )b = 1
2 1 Again BS math yields and incorrect answer 1
This incorrect result is a problem with the distributive law SM explains how this incorrect answer is obtained and the problem with the distributive law that allows the error
Page 148 ndash149 Barry Mazur ndash Imagining Numbers
3
2
31
Do the computation carefully on paper using the rules we agreed to Then ponder your answer which should be something of a surprise to you donrsquot stop there Think of what your answer might possibly mean or might imply
Solve this equation using the distributive law
JK definitely a surprise the answer wasincorrect and illogical
Gerhard Gade University Professor doing math at Harvard University
Here is the BS math answer that Barry Mazur calculated
2
31
2
31
2
31
1 3 1 3 1 3 3 3 2 2 3 1 3
2 2 2 2 4 4 4 4 4 4 2 2
dash
1 3 1 3 1 31
2 2 2 2 4 4
3 3
4 4
When the distributive law is used with the BS Rule-of-Signs and one of the signs is a (-) dash sign the middle two terms always cancel each other This leads to the incorrect answer of (-1) in the above equation
Barry Mazur did not go back and check his answer with the original equation If he had he would have known that the answer was incorrect
3
1 31
2
check answer by setting equal
3
331 3
12
1 31
2
Take the Cube root of both sides
1 3 2 multiply both sides by 2
3 3 subtract 1 from both sides
2 2
3 3 square both sides
3 9 This is an incorrect answer A simple equation check could have shown that the distributive law using BS math provides an incorrect answer
1
BS Illogical and Incorrect Numbers Raised to Powers
nC D n DC
642 DDD
In BS math only a positive number can have an answer if ldquonrdquo is an even number (246---)
All negative numbers have no solutions
642 DDD
With SM just add a direction arrow and any number gives the correct answer
or n nC D C D
3 3( 3 8 8 or 8 2 or 2 )n D C
( 2 16 C= 16 or 16 4 or 4 )n D
Since a negative number has no square root answer in BS math BS mathematicians invented imaginary numbers
i2=-1
24414 iiC n
These equations are easily solved without the use of imaginary numbers with SM math
If 4 then 4 2x
If 4 then 4 2x
The 360-degree circle is a 4400 years old extremely outdated system based on a year of 12 months of 30 days each that is 360 days
SM will use a more logical system that will have a circle of 6283185307o
d
( )
(deg )
d r radians
d r rees
the arc distance around
the circumference of a circle
d
radiusr
2 2
2 2
d r
dv rd t t
v dda rd t t t
1
d
r
( )av tv
The average angular velocity of a body is the rate of its angular displacement
( )0
liminstt
vt
The instantaneous angular velocity is the limit of the ratio
t
as t approaches zero
( )d
av
d
tv
The average circumference velocity of the body is the rate of its distance traveled d divided by the time t
dvv
r
1 dd vdrv
t t t r r
The equation of the relationship between the angular velocity and the velocity of a point on that body
( )o
av
v v v
t ta
The average angular acceleration is the change in the angular velocity
( )0
liminstt
v
ta
The instantaneous angular acceleration is the
limit of the ration of v
t
as t approaches
zero
daa
r
1
1
d
d s d
vva
t r tv a
r t r
The equation of the relationship between the angular acceleration and the acceleration of a point on the body
21
2v a t
If a body with an initial angular velocity ov has a constant
angular acceleration a it will turn through an angle in a time t
1
Straight Line Symbol Units Rotational Symbol Units Distance d m Angular
Distance radians or degrees
Velocity dt
v m
s Angular Velocity
V
t
radians or degrees
s
Acceleration vat
2ms
Angular Acceleration
VA
t
2
radians or degrees
s
Mass M Kg Mass distribution
dM 2Kgm
Force actionA Ma
2
Kgm
sNewton
Torque
dT M A 2
2
Kgm
s
Momentum Mv Kgm
s
Angular Momentum
dM V 2Kgm
s
Work (Force)(distance)
Ad 2
Kgmm
sNm Joule
Work T
Using AS arrow math to solve and direction in space problems using the equation 1 2
2 2 2 2 2 2( ) amp and ( ) amp 3
4
2 2 2
2 2
( ) amp
(5 ) 4 3
(5 ) 16 9
(5 ) 25
(5 ) (5 )
2 2(5 ) 4 3
(5 ) 16 9
(5 ) 25
(5 ) (5 )
In SM math Pythagorasrsquo theorem provides both magnitude and direction
In BS math Pythagorasrsquo theorem provides magnitude only
time set equal to zero
23
( ) ( 0 )10 1
x t xx
3( ) ( 0)
10 1W t f
In BS math this is an equation for a traveling wave
must use a square term in order to make the curve symmetric
NO SQUARE TERM IS NECESSARY
In SM math when t is set to ldquo0rdquo
This is a major fault in BS math
x f(x)
-4 0019
-3 0033
-2 0073
-1 0273
0 3000
1 0273
2 0073
3 0033
4 0019 -4 -3 -2 -1 0 1 2 3 400
05
10
15
20
25
30
35
BS wave (must use square term)
snap shot of wave at x = 0
Am
plit
ude
lt to gt lt to gt f(ltgt)
lt -4 0073
lt -3 0097
lt -2 0143
lt -1 0273
origin 0 3000
gt 1 0273
gt 2 0143
gt 3 0097
gt 4 0073lt0073 lt0097 lt0143 lt0273
3000
0273gt 0143gt 0097gt 0073gt
Wave moving
snap shot of wave
Am
plit
ude
lt4 lt3 lt2 lt1 0 1gt 2gt 3gt 4gt
3
2
1
In SM NO squared terms are needed
SPRING
An Action is needed to stretch or compress a spring to a starting point where it can execute cycle motion
tsA (ts to start) is the work required to move a mass attached to a spring a distance (d1) from
its initial rest position RAK 0 the spring constant is the stiffness of the spring
o A is attraction o R is repulsion o
ARRA
00 is for the back and forth motion through the zero initial rest position A
stretched spring goes from ldquoArdquo to ldquo0rdquo to ldquoRrdquo to ldquoArdquo to ldquo0rdquo to ldquoRrdquo d1 is the distance a mass is moved from its initial rest position
0
0 1ts A RA K d
Once ( )mad
is performed to move the mass to position d1 (against the restraint of the spring) a of the ( )mad
is transferred into the spring as potential attractionrepulsion (PAR) There is always
more ( )mad
performed than is stored as PAR When the spring is released from the restraining Ats the PAR starts the oscillation Conservation of ( )mad
requires that the PAR and the motion ( )mad
( )( )madM be a constant
( )amp ( ) constantmadPAR M 02
0 0
1
2A R A RPAR K d
02
( ) 0 0
1( )
2mad A R A RM m v
0 0 02 2 2
0 0 0 1
2 2 21
2 2 21
1 1 1amp
2 2 2
1 1 1amp
2 2 2
amp
A R A R A RK d m v K d
Kd mv Kd
Kd mv Kd
0
1d is frac12 the height from the base to the peak of a cycle
An experiment of moving paper under a pendulum through a small angle filled with sand that can escape through a small hole in the bottom shows that a pendulum cycling back and forth follows a curve that is mathematically analogous to a lsquosinrsquo or lsquocosrsquo
The projection of the curved area onto the straight line exactly duplicate the back and forth motion of CM This allows for the math of lsquosinrsquo or lsquocosrsquo to be used
SM will use lsquosinrdquo math It could use lsquocosrsquo or a combination of both lsquosinrsquo and lsquocosrsquo Using just the lsquosinrsquo math makes the understanding and math easier
0 sin 0 0SM 0 sin157 1SM The diagram at the left shows a spring being stretched from its (zero rest) position-2 and the PAR being released at position-1 There are 4 sections (1 to 2 2 to 3 3 to 4 and 4 to5) that alternate between maximum acceleration and zero velocity to max velocity and zero acceleration
At position-1 there is maximum acceleration and zero velocity o The lsquosinrsquo values from 0o to 157o will be used for velocity (values from 0 to 1) o The lsquosinrsquo values from 157o to 0o will be used for acceleration (values from 1 to 0)
At position-2 there is maximum velocity and zero acceleration Acceleration changes to deceleration
o The lsquosinrsquo values from 157o to 0o will be used for deceleration (values from 1 to 0) o The lsquosinrsquo values from 0o to 157o will be used for acceleration (values from 0 to 1)
At position-3 there is maximum acceleration and zero velocity o The lsquosinrsquo values from 0o to 157o will be used for velocity (values from 0 to 1) o The lsquosinrsquo values from 157o to 0o will be used for acceleration (values from 1 to 0)
At position-4 there is maximum velocity and zero acceleration Acceleration changes to deceleration
o The lsquosinrsquo values from 157o to 0o will be used for deceleration (values from 1 to 0) o The lsquosinrsquo values from 0o to 157o will be used for acceleration (values from 0 to 1)
At position-5 the cycle starts again
2
)539(
cT
MK
spring constant (K) when mass (M) and cycle-time (T) are known
)(
))(89)((
111 md
uKgM
d
MG
d
FK sts
spring constant (K) when mass (M) and amplitude ( d1) are known
2 21( )
Kv d d
m
velocity (v) at any point during the cycle
velocity
dforM
Kdv
max
012 velocity (v2) maximum (when d = 0 at position-2 and 4)
acceleration (a1) maximum when spring constant (K) mass (M) and amplitude (d1) are known
acceleration (a1) maximum when amplitude (d1) and cycle-time (Tc) are known
acceleration (a) at any point during the cycle-time (T)
1
151
286
v
dTc
cycle-time (Tc) when amplitude (d1) and velocity (v1) are known
K
MTc 28651
cycle-time (Tc) when mass (M) and spring constant (K) are known
fTc
1
cycle-time (Tc) when the frequency is known
cTf
1
frequency (f) when the cycle-time (Tc) is known
of cycles
elapsed timef
frequency (f) when the number of cycles in an elapsed time is known
d1 amplitude (distance from rest to start position)
Also frac12 the distance from the base to the peak
3131 oror dM
Ka
251
3131 539
c
oror T
da
2539
cT
da
In December of 2006 I discovered the mathematical error in Einsteinrsquos Special Relativity Einstein used BS math and that math only works in an imaginary world
It cannot describe real symmetrical space
Einsteinrsquos Special Relativity should be abandoned and a theory of real space using SM should be used to replace it
Numerous articles on the Internet show Einsteinrsquos theory to be incorrect but no one in academia is willing to publish the information
Figure 1-10 A moving particle observed from two frames of reference in uniform translation
Einsteinrsquos Math Error in Special Relativity Transformation of Velocity (From Eisenberg- Fundamentals of Modern Physics pages 27-29) Consider the particle shown in figure (1-10) moving with velocity v as
seen in a frame of reference O
dx dy dz
Vx Vy Vzdt dt dt
dx dy dz
V x V y V zdy dt dt
We would like to evaluate the velocity v of the particle as seen in the frame of reference 0 which is itself moving relative to 0 with velocity v Measured in the 0 frame the velocity vector of the particle has com ponents
The same velocity vector as measured in the 0 frame has components
2
2
2
2
2
1
1
x vtx
v
cy y
z z
vxt
ctv
c
Now from equations (1-13) we know that the relation between the primed and the unprimed coordinates and times is
Take the differential of these equations remembering that v is a constant This gives
2
2
2
2
2
1
1
dx vdtdx
v
cdy dy
dz dz
vxdt
cdtv
c
2
2
2 2 2 2 2
2
2
2
11 ( )
1( ) 1
1
1
1
dx vdt
v dxdx vdt v
dx dx vdtc dt dtvdx vdx vdx v dxdt dt dt dt
dt dtc c c c
VxV x
Vx
VxV x
vv
vv
c
x
v
c
cV
( )
coordinate of particle relative to moving primed system
coordinate of particle relative to stationary system
velocity of primed coordina
V x
Vx
te system relative to stationaryv sys
tem
Page 28 Fundamentals of Modern Physics
(1-16)
According to Einstein this equation tells you how to transform the observed velocity from one frame of reference to another frame of reference First we note that as Vc and vc approachzero the equation approaches those which would be derived from the Galilean transformation
2 2
( ) 1 01 1 ( )
Vx v Vx vV
VxV x
Vx v
v cVxx
v
c
v
cc
Another property of this equation is that it is impossible to choose V and v such that Vrsquo the magnitude of the velocity which is seen in the new frame is greater than c
1
2
09 ( 09 ) 18009876543
( 09 ) 1811 (09 )
c c cV c c
cc
c
Example As seen by O particle 1 has velocity 09c in the direction of positive x and particle 2 has velocity 09c in the negative x direction
Einstein equation however predicts To evaluate the velocity of particle 1 with respect to particle 2 we transform from the O frame to the Orsquo frame moving in the negative x direction with velocity v = -09c using equation 1-16 We obtain
JK LOGIC will observe the particles separating at 18c
A confusing INCORRECT and ILLOGICAL answer
left is a dir rightsub of a
1
2
1
09 ( 09 ) 18009876543
181( 09 )1 ( 09 )
The multiplication of dir left time a dir right is ONLY a dir left
09
dir
direction leftdirection rightsubtraction
c c cV c c
cc
c
V
dir
22
22
09 18 18009876543
181( 081)1(09)(09)
1
left rightsub
sub
c c c cc c
ccc
c
Analyzing Einsteinrsquos BS math Equation
Using SM math a logical answer can be obtained
Cannot multiply a subtraction operator times a direction in space
Can only subtract and change its direction
Cannot multiply a direction to the left times a direction to the right
Cannot multiply opposite directions in space
Can only solve for the resultant
rightleft
subtraction
1
subtraction
2
sub of a dir to the left is a dir to the right
1 resultant 0 sub 2
2sub 2
2
09 09
09 amp 091
09 09 18 1
(0)1
(09) amp (09)1
c cV
c cc
c c cV
cc
cc
8
181
cc
SM math predicts a logical answer using logic
21st century scientists state in many articles that Einsteins General Relativity and Quantum Mechanics are incompatible
They are not only incompatible they are both mathematically incorrect
They both use illogical BS math
Any equations in math that use imaginary numbers should be abandoned
Imaginary numbers are only necessary in the current BS math system
This means that Schroumldingers equations should be abandoned
Any equation in quantum mechanics that uses imaginary number should be abandoned
Prof Kanarev has developed logical physics and chemistrytheories without the use of Quantum mechanics or relativity
A partial list of his theories coverbull Mechanics ndash replaces Newtonbull Photon - bull Electrons ndash Protons- Neutronsbull Spectroscopybull Nucleusbull Atoms amp Moleculesbull Electrodynamicsbull Heatbull Curvature of Space ndash replaces Schwarzcchildrsquos equation
Kanarevrsquos Hypothesis
bull Electron align linearly with proton They do NOT spin or form clouds around the nucleus
bull Neutrons are in the center of atom Proton are outside of Neutrons
bull The wavelength of all particles is the experimental radius
bull Double slit experiment explained correctly
bull Heisenbergrsquos inequality explained correctly
bull is the radius of approach of the magnetic lines around the torus of the electron
( 1 137)
bull Why water expands when it freezes
bull Electron clusters form lightning
rbull The meaning of Plancks constant is explained
22 2 kg m
h m f mr f consts
2210254 42m r kgm bull If the mr-constant describes a ring the equations
describe experiments bull If the mr-constant describes a wave the equations do
NOT describe experiments
Kanarevrsquos 5 Equations of Mechanics
Kanarevrsquos 1st Equation Acceleration ampmv maMma R R
Kanarevrsquos 2nd Equation Uniform motion amp mv maMma R R
maM mvR R
Kanarevrsquos 3rd Equation
Kanarevrsquos 4th Equation Equal and opposite reaction
1 21 21 2
12
a m
m a m ama
Kanarevrsquos 5th Equation 1 2 3 Mi M M M Mna R R R R R
The sum of the Actions working on a moving body is never equal to zero
Kanarevrsquos Photon model is a rolling ring
His equations accurately describe known experiments with photonsbull Electromagnetic radiation should be called photon radiationbull Diffraction and interference are the same phenomenabull Correctly explains the double slit experiment without quantum equationsbull Correctly explains the meaning of Heisenbergrsquos inequality
e
2 (2)(3142)(2817 15 ) 10007297352568
R 2426 12 137mag
xp
r m
m
Charge
Magnetic action
Proton
The intensity of the magnetic action near the geometrical center of the proton ring is so great that we theorize that it is the strong nuclear force
(9383 8)(1602 19)8476 14
4 (4)(3142)(1411 26)p
pp
EB T
M
(9395 8)(1602 19)1035 15
4 (4)(3142)(0966 26)N
NN
EB T
M
Neutron
1a Hydrogenavi
a b starting equation ( )( ) ( )( )a a a b multiply both sides by ( )a
( )( ) ( )( ) ( )( ) ( )( )a a b b a b b b subtract ( )( )b b from both sides
( )( ) ( )( )a b a b a b b factor ( )a b from both sides
( )a b b divide both sides by ( )a b
( )b b b replace ( )a with ( )b 2b b simplify 2(1) (1) Let ( )b = 1
2 1 Again BS math yields and incorrect answer 1
This incorrect result is a problem with the distributive law SM explains how this incorrect answer is obtained and the problem with the distributive law that allows the error
Page 148 ndash149 Barry Mazur ndash Imagining Numbers
3
2
31
Do the computation carefully on paper using the rules we agreed to Then ponder your answer which should be something of a surprise to you donrsquot stop there Think of what your answer might possibly mean or might imply
Solve this equation using the distributive law
JK definitely a surprise the answer wasincorrect and illogical
Gerhard Gade University Professor doing math at Harvard University
Here is the BS math answer that Barry Mazur calculated
2
31
2
31
2
31
1 3 1 3 1 3 3 3 2 2 3 1 3
2 2 2 2 4 4 4 4 4 4 2 2
dash
1 3 1 3 1 31
2 2 2 2 4 4
3 3
4 4
When the distributive law is used with the BS Rule-of-Signs and one of the signs is a (-) dash sign the middle two terms always cancel each other This leads to the incorrect answer of (-1) in the above equation
Barry Mazur did not go back and check his answer with the original equation If he had he would have known that the answer was incorrect
3
1 31
2
check answer by setting equal
3
331 3
12
1 31
2
Take the Cube root of both sides
1 3 2 multiply both sides by 2
3 3 subtract 1 from both sides
2 2
3 3 square both sides
3 9 This is an incorrect answer A simple equation check could have shown that the distributive law using BS math provides an incorrect answer
1
BS Illogical and Incorrect Numbers Raised to Powers
nC D n DC
642 DDD
In BS math only a positive number can have an answer if ldquonrdquo is an even number (246---)
All negative numbers have no solutions
642 DDD
With SM just add a direction arrow and any number gives the correct answer
or n nC D C D
3 3( 3 8 8 or 8 2 or 2 )n D C
( 2 16 C= 16 or 16 4 or 4 )n D
Since a negative number has no square root answer in BS math BS mathematicians invented imaginary numbers
i2=-1
24414 iiC n
These equations are easily solved without the use of imaginary numbers with SM math
If 4 then 4 2x
If 4 then 4 2x
The 360-degree circle is a 4400 years old extremely outdated system based on a year of 12 months of 30 days each that is 360 days
SM will use a more logical system that will have a circle of 6283185307o
d
( )
(deg )
d r radians
d r rees
the arc distance around
the circumference of a circle
d
radiusr
2 2
2 2
d r
dv rd t t
v dda rd t t t
1
d
r
( )av tv
The average angular velocity of a body is the rate of its angular displacement
( )0
liminstt
vt
The instantaneous angular velocity is the limit of the ratio
t
as t approaches zero
( )d
av
d
tv
The average circumference velocity of the body is the rate of its distance traveled d divided by the time t
dvv
r
1 dd vdrv
t t t r r
The equation of the relationship between the angular velocity and the velocity of a point on that body
( )o
av
v v v
t ta
The average angular acceleration is the change in the angular velocity
( )0
liminstt
v
ta
The instantaneous angular acceleration is the
limit of the ration of v
t
as t approaches
zero
daa
r
1
1
d
d s d
vva
t r tv a
r t r
The equation of the relationship between the angular acceleration and the acceleration of a point on the body
21
2v a t
If a body with an initial angular velocity ov has a constant
angular acceleration a it will turn through an angle in a time t
1
Straight Line Symbol Units Rotational Symbol Units Distance d m Angular
Distance radians or degrees
Velocity dt
v m
s Angular Velocity
V
t
radians or degrees
s
Acceleration vat
2ms
Angular Acceleration
VA
t
2
radians or degrees
s
Mass M Kg Mass distribution
dM 2Kgm
Force actionA Ma
2
Kgm
sNewton
Torque
dT M A 2
2
Kgm
s
Momentum Mv Kgm
s
Angular Momentum
dM V 2Kgm
s
Work (Force)(distance)
Ad 2
Kgmm
sNm Joule
Work T
Using AS arrow math to solve and direction in space problems using the equation 1 2
2 2 2 2 2 2( ) amp and ( ) amp 3
4
2 2 2
2 2
( ) amp
(5 ) 4 3
(5 ) 16 9
(5 ) 25
(5 ) (5 )
2 2(5 ) 4 3
(5 ) 16 9
(5 ) 25
(5 ) (5 )
In SM math Pythagorasrsquo theorem provides both magnitude and direction
In BS math Pythagorasrsquo theorem provides magnitude only
time set equal to zero
23
( ) ( 0 )10 1
x t xx
3( ) ( 0)
10 1W t f
In BS math this is an equation for a traveling wave
must use a square term in order to make the curve symmetric
NO SQUARE TERM IS NECESSARY
In SM math when t is set to ldquo0rdquo
This is a major fault in BS math
x f(x)
-4 0019
-3 0033
-2 0073
-1 0273
0 3000
1 0273
2 0073
3 0033
4 0019 -4 -3 -2 -1 0 1 2 3 400
05
10
15
20
25
30
35
BS wave (must use square term)
snap shot of wave at x = 0
Am
plit
ude
lt to gt lt to gt f(ltgt)
lt -4 0073
lt -3 0097
lt -2 0143
lt -1 0273
origin 0 3000
gt 1 0273
gt 2 0143
gt 3 0097
gt 4 0073lt0073 lt0097 lt0143 lt0273
3000
0273gt 0143gt 0097gt 0073gt
Wave moving
snap shot of wave
Am
plit
ude
lt4 lt3 lt2 lt1 0 1gt 2gt 3gt 4gt
3
2
1
In SM NO squared terms are needed
SPRING
An Action is needed to stretch or compress a spring to a starting point where it can execute cycle motion
tsA (ts to start) is the work required to move a mass attached to a spring a distance (d1) from
its initial rest position RAK 0 the spring constant is the stiffness of the spring
o A is attraction o R is repulsion o
ARRA
00 is for the back and forth motion through the zero initial rest position A
stretched spring goes from ldquoArdquo to ldquo0rdquo to ldquoRrdquo to ldquoArdquo to ldquo0rdquo to ldquoRrdquo d1 is the distance a mass is moved from its initial rest position
0
0 1ts A RA K d
Once ( )mad
is performed to move the mass to position d1 (against the restraint of the spring) a of the ( )mad
is transferred into the spring as potential attractionrepulsion (PAR) There is always
more ( )mad
performed than is stored as PAR When the spring is released from the restraining Ats the PAR starts the oscillation Conservation of ( )mad
requires that the PAR and the motion ( )mad
( )( )madM be a constant
( )amp ( ) constantmadPAR M 02
0 0
1
2A R A RPAR K d
02
( ) 0 0
1( )
2mad A R A RM m v
0 0 02 2 2
0 0 0 1
2 2 21
2 2 21
1 1 1amp
2 2 2
1 1 1amp
2 2 2
amp
A R A R A RK d m v K d
Kd mv Kd
Kd mv Kd
0
1d is frac12 the height from the base to the peak of a cycle
An experiment of moving paper under a pendulum through a small angle filled with sand that can escape through a small hole in the bottom shows that a pendulum cycling back and forth follows a curve that is mathematically analogous to a lsquosinrsquo or lsquocosrsquo
The projection of the curved area onto the straight line exactly duplicate the back and forth motion of CM This allows for the math of lsquosinrsquo or lsquocosrsquo to be used
SM will use lsquosinrdquo math It could use lsquocosrsquo or a combination of both lsquosinrsquo and lsquocosrsquo Using just the lsquosinrsquo math makes the understanding and math easier
0 sin 0 0SM 0 sin157 1SM The diagram at the left shows a spring being stretched from its (zero rest) position-2 and the PAR being released at position-1 There are 4 sections (1 to 2 2 to 3 3 to 4 and 4 to5) that alternate between maximum acceleration and zero velocity to max velocity and zero acceleration
At position-1 there is maximum acceleration and zero velocity o The lsquosinrsquo values from 0o to 157o will be used for velocity (values from 0 to 1) o The lsquosinrsquo values from 157o to 0o will be used for acceleration (values from 1 to 0)
At position-2 there is maximum velocity and zero acceleration Acceleration changes to deceleration
o The lsquosinrsquo values from 157o to 0o will be used for deceleration (values from 1 to 0) o The lsquosinrsquo values from 0o to 157o will be used for acceleration (values from 0 to 1)
At position-3 there is maximum acceleration and zero velocity o The lsquosinrsquo values from 0o to 157o will be used for velocity (values from 0 to 1) o The lsquosinrsquo values from 157o to 0o will be used for acceleration (values from 1 to 0)
At position-4 there is maximum velocity and zero acceleration Acceleration changes to deceleration
o The lsquosinrsquo values from 157o to 0o will be used for deceleration (values from 1 to 0) o The lsquosinrsquo values from 0o to 157o will be used for acceleration (values from 0 to 1)
At position-5 the cycle starts again
2
)539(
cT
MK
spring constant (K) when mass (M) and cycle-time (T) are known
)(
))(89)((
111 md
uKgM
d
MG
d
FK sts
spring constant (K) when mass (M) and amplitude ( d1) are known
2 21( )
Kv d d
m
velocity (v) at any point during the cycle
velocity
dforM
Kdv
max
012 velocity (v2) maximum (when d = 0 at position-2 and 4)
acceleration (a1) maximum when spring constant (K) mass (M) and amplitude (d1) are known
acceleration (a1) maximum when amplitude (d1) and cycle-time (Tc) are known
acceleration (a) at any point during the cycle-time (T)
1
151
286
v
dTc
cycle-time (Tc) when amplitude (d1) and velocity (v1) are known
K
MTc 28651
cycle-time (Tc) when mass (M) and spring constant (K) are known
fTc
1
cycle-time (Tc) when the frequency is known
cTf
1
frequency (f) when the cycle-time (Tc) is known
of cycles
elapsed timef
frequency (f) when the number of cycles in an elapsed time is known
d1 amplitude (distance from rest to start position)
Also frac12 the distance from the base to the peak
3131 oror dM
Ka
251
3131 539
c
oror T
da
2539
cT
da
In December of 2006 I discovered the mathematical error in Einsteinrsquos Special Relativity Einstein used BS math and that math only works in an imaginary world
It cannot describe real symmetrical space
Einsteinrsquos Special Relativity should be abandoned and a theory of real space using SM should be used to replace it
Numerous articles on the Internet show Einsteinrsquos theory to be incorrect but no one in academia is willing to publish the information
Figure 1-10 A moving particle observed from two frames of reference in uniform translation
Einsteinrsquos Math Error in Special Relativity Transformation of Velocity (From Eisenberg- Fundamentals of Modern Physics pages 27-29) Consider the particle shown in figure (1-10) moving with velocity v as
seen in a frame of reference O
dx dy dz
Vx Vy Vzdt dt dt
dx dy dz
V x V y V zdy dt dt
We would like to evaluate the velocity v of the particle as seen in the frame of reference 0 which is itself moving relative to 0 with velocity v Measured in the 0 frame the velocity vector of the particle has com ponents
The same velocity vector as measured in the 0 frame has components
2
2
2
2
2
1
1
x vtx
v
cy y
z z
vxt
ctv
c
Now from equations (1-13) we know that the relation between the primed and the unprimed coordinates and times is
Take the differential of these equations remembering that v is a constant This gives
2
2
2
2
2
1
1
dx vdtdx
v
cdy dy
dz dz
vxdt
cdtv
c
2
2
2 2 2 2 2
2
2
2
11 ( )
1( ) 1
1
1
1
dx vdt
v dxdx vdt v
dx dx vdtc dt dtvdx vdx vdx v dxdt dt dt dt
dt dtc c c c
VxV x
Vx
VxV x
vv
vv
c
x
v
c
cV
( )
coordinate of particle relative to moving primed system
coordinate of particle relative to stationary system
velocity of primed coordina
V x
Vx
te system relative to stationaryv sys
tem
Page 28 Fundamentals of Modern Physics
(1-16)
According to Einstein this equation tells you how to transform the observed velocity from one frame of reference to another frame of reference First we note that as Vc and vc approachzero the equation approaches those which would be derived from the Galilean transformation
2 2
( ) 1 01 1 ( )
Vx v Vx vV
VxV x
Vx v
v cVxx
v
c
v
cc
Another property of this equation is that it is impossible to choose V and v such that Vrsquo the magnitude of the velocity which is seen in the new frame is greater than c
1
2
09 ( 09 ) 18009876543
( 09 ) 1811 (09 )
c c cV c c
cc
c
Example As seen by O particle 1 has velocity 09c in the direction of positive x and particle 2 has velocity 09c in the negative x direction
Einstein equation however predicts To evaluate the velocity of particle 1 with respect to particle 2 we transform from the O frame to the Orsquo frame moving in the negative x direction with velocity v = -09c using equation 1-16 We obtain
JK LOGIC will observe the particles separating at 18c
A confusing INCORRECT and ILLOGICAL answer
left is a dir rightsub of a
1
2
1
09 ( 09 ) 18009876543
181( 09 )1 ( 09 )
The multiplication of dir left time a dir right is ONLY a dir left
09
dir
direction leftdirection rightsubtraction
c c cV c c
cc
c
V
dir
22
22
09 18 18009876543
181( 081)1(09)(09)
1
left rightsub
sub
c c c cc c
ccc
c
Analyzing Einsteinrsquos BS math Equation
Using SM math a logical answer can be obtained
Cannot multiply a subtraction operator times a direction in space
Can only subtract and change its direction
Cannot multiply a direction to the left times a direction to the right
Cannot multiply opposite directions in space
Can only solve for the resultant
rightleft
subtraction
1
subtraction
2
sub of a dir to the left is a dir to the right
1 resultant 0 sub 2
2sub 2
2
09 09
09 amp 091
09 09 18 1
(0)1
(09) amp (09)1
c cV
c cc
c c cV
cc
cc
8
181
cc
SM math predicts a logical answer using logic
21st century scientists state in many articles that Einsteins General Relativity and Quantum Mechanics are incompatible
They are not only incompatible they are both mathematically incorrect
They both use illogical BS math
Any equations in math that use imaginary numbers should be abandoned
Imaginary numbers are only necessary in the current BS math system
This means that Schroumldingers equations should be abandoned
Any equation in quantum mechanics that uses imaginary number should be abandoned
Prof Kanarev has developed logical physics and chemistrytheories without the use of Quantum mechanics or relativity
A partial list of his theories coverbull Mechanics ndash replaces Newtonbull Photon - bull Electrons ndash Protons- Neutronsbull Spectroscopybull Nucleusbull Atoms amp Moleculesbull Electrodynamicsbull Heatbull Curvature of Space ndash replaces Schwarzcchildrsquos equation
Kanarevrsquos Hypothesis
bull Electron align linearly with proton They do NOT spin or form clouds around the nucleus
bull Neutrons are in the center of atom Proton are outside of Neutrons
bull The wavelength of all particles is the experimental radius
bull Double slit experiment explained correctly
bull Heisenbergrsquos inequality explained correctly
bull is the radius of approach of the magnetic lines around the torus of the electron
( 1 137)
bull Why water expands when it freezes
bull Electron clusters form lightning
rbull The meaning of Plancks constant is explained
22 2 kg m
h m f mr f consts
2210254 42m r kgm bull If the mr-constant describes a ring the equations
describe experiments bull If the mr-constant describes a wave the equations do
NOT describe experiments
Kanarevrsquos 5 Equations of Mechanics
Kanarevrsquos 1st Equation Acceleration ampmv maMma R R
Kanarevrsquos 2nd Equation Uniform motion amp mv maMma R R
maM mvR R
Kanarevrsquos 3rd Equation
Kanarevrsquos 4th Equation Equal and opposite reaction
1 21 21 2
12
a m
m a m ama
Kanarevrsquos 5th Equation 1 2 3 Mi M M M Mna R R R R R
The sum of the Actions working on a moving body is never equal to zero
Kanarevrsquos Photon model is a rolling ring
His equations accurately describe known experiments with photonsbull Electromagnetic radiation should be called photon radiationbull Diffraction and interference are the same phenomenabull Correctly explains the double slit experiment without quantum equationsbull Correctly explains the meaning of Heisenbergrsquos inequality
e
2 (2)(3142)(2817 15 ) 10007297352568
R 2426 12 137mag
xp
r m
m
Charge
Magnetic action
Proton
The intensity of the magnetic action near the geometrical center of the proton ring is so great that we theorize that it is the strong nuclear force
(9383 8)(1602 19)8476 14
4 (4)(3142)(1411 26)p
pp
EB T
M
(9395 8)(1602 19)1035 15
4 (4)(3142)(0966 26)N
NN
EB T
M
Neutron
1a Hydrogenavi
Page 148 ndash149 Barry Mazur ndash Imagining Numbers
3
2
31
Do the computation carefully on paper using the rules we agreed to Then ponder your answer which should be something of a surprise to you donrsquot stop there Think of what your answer might possibly mean or might imply
Solve this equation using the distributive law
JK definitely a surprise the answer wasincorrect and illogical
Gerhard Gade University Professor doing math at Harvard University
Here is the BS math answer that Barry Mazur calculated
2
31
2
31
2
31
1 3 1 3 1 3 3 3 2 2 3 1 3
2 2 2 2 4 4 4 4 4 4 2 2
dash
1 3 1 3 1 31
2 2 2 2 4 4
3 3
4 4
When the distributive law is used with the BS Rule-of-Signs and one of the signs is a (-) dash sign the middle two terms always cancel each other This leads to the incorrect answer of (-1) in the above equation
Barry Mazur did not go back and check his answer with the original equation If he had he would have known that the answer was incorrect
3
1 31
2
check answer by setting equal
3
331 3
12
1 31
2
Take the Cube root of both sides
1 3 2 multiply both sides by 2
3 3 subtract 1 from both sides
2 2
3 3 square both sides
3 9 This is an incorrect answer A simple equation check could have shown that the distributive law using BS math provides an incorrect answer
1
BS Illogical and Incorrect Numbers Raised to Powers
nC D n DC
642 DDD
In BS math only a positive number can have an answer if ldquonrdquo is an even number (246---)
All negative numbers have no solutions
642 DDD
With SM just add a direction arrow and any number gives the correct answer
or n nC D C D
3 3( 3 8 8 or 8 2 or 2 )n D C
( 2 16 C= 16 or 16 4 or 4 )n D
Since a negative number has no square root answer in BS math BS mathematicians invented imaginary numbers
i2=-1
24414 iiC n
These equations are easily solved without the use of imaginary numbers with SM math
If 4 then 4 2x
If 4 then 4 2x
The 360-degree circle is a 4400 years old extremely outdated system based on a year of 12 months of 30 days each that is 360 days
SM will use a more logical system that will have a circle of 6283185307o
d
( )
(deg )
d r radians
d r rees
the arc distance around
the circumference of a circle
d
radiusr
2 2
2 2
d r
dv rd t t
v dda rd t t t
1
d
r
( )av tv
The average angular velocity of a body is the rate of its angular displacement
( )0
liminstt
vt
The instantaneous angular velocity is the limit of the ratio
t
as t approaches zero
( )d
av
d
tv
The average circumference velocity of the body is the rate of its distance traveled d divided by the time t
dvv
r
1 dd vdrv
t t t r r
The equation of the relationship between the angular velocity and the velocity of a point on that body
( )o
av
v v v
t ta
The average angular acceleration is the change in the angular velocity
( )0
liminstt
v
ta
The instantaneous angular acceleration is the
limit of the ration of v
t
as t approaches
zero
daa
r
1
1
d
d s d
vva
t r tv a
r t r
The equation of the relationship between the angular acceleration and the acceleration of a point on the body
21
2v a t
If a body with an initial angular velocity ov has a constant
angular acceleration a it will turn through an angle in a time t
1
Straight Line Symbol Units Rotational Symbol Units Distance d m Angular
Distance radians or degrees
Velocity dt
v m
s Angular Velocity
V
t
radians or degrees
s
Acceleration vat
2ms
Angular Acceleration
VA
t
2
radians or degrees
s
Mass M Kg Mass distribution
dM 2Kgm
Force actionA Ma
2
Kgm
sNewton
Torque
dT M A 2
2
Kgm
s
Momentum Mv Kgm
s
Angular Momentum
dM V 2Kgm
s
Work (Force)(distance)
Ad 2
Kgmm
sNm Joule
Work T
Using AS arrow math to solve and direction in space problems using the equation 1 2
2 2 2 2 2 2( ) amp and ( ) amp 3
4
2 2 2
2 2
( ) amp
(5 ) 4 3
(5 ) 16 9
(5 ) 25
(5 ) (5 )
2 2(5 ) 4 3
(5 ) 16 9
(5 ) 25
(5 ) (5 )
In SM math Pythagorasrsquo theorem provides both magnitude and direction
In BS math Pythagorasrsquo theorem provides magnitude only
time set equal to zero
23
( ) ( 0 )10 1
x t xx
3( ) ( 0)
10 1W t f
In BS math this is an equation for a traveling wave
must use a square term in order to make the curve symmetric
NO SQUARE TERM IS NECESSARY
In SM math when t is set to ldquo0rdquo
This is a major fault in BS math
x f(x)
-4 0019
-3 0033
-2 0073
-1 0273
0 3000
1 0273
2 0073
3 0033
4 0019 -4 -3 -2 -1 0 1 2 3 400
05
10
15
20
25
30
35
BS wave (must use square term)
snap shot of wave at x = 0
Am
plit
ude
lt to gt lt to gt f(ltgt)
lt -4 0073
lt -3 0097
lt -2 0143
lt -1 0273
origin 0 3000
gt 1 0273
gt 2 0143
gt 3 0097
gt 4 0073lt0073 lt0097 lt0143 lt0273
3000
0273gt 0143gt 0097gt 0073gt
Wave moving
snap shot of wave
Am
plit
ude
lt4 lt3 lt2 lt1 0 1gt 2gt 3gt 4gt
3
2
1
In SM NO squared terms are needed
SPRING
An Action is needed to stretch or compress a spring to a starting point where it can execute cycle motion
tsA (ts to start) is the work required to move a mass attached to a spring a distance (d1) from
its initial rest position RAK 0 the spring constant is the stiffness of the spring
o A is attraction o R is repulsion o
ARRA
00 is for the back and forth motion through the zero initial rest position A
stretched spring goes from ldquoArdquo to ldquo0rdquo to ldquoRrdquo to ldquoArdquo to ldquo0rdquo to ldquoRrdquo d1 is the distance a mass is moved from its initial rest position
0
0 1ts A RA K d
Once ( )mad
is performed to move the mass to position d1 (against the restraint of the spring) a of the ( )mad
is transferred into the spring as potential attractionrepulsion (PAR) There is always
more ( )mad
performed than is stored as PAR When the spring is released from the restraining Ats the PAR starts the oscillation Conservation of ( )mad
requires that the PAR and the motion ( )mad
( )( )madM be a constant
( )amp ( ) constantmadPAR M 02
0 0
1
2A R A RPAR K d
02
( ) 0 0
1( )
2mad A R A RM m v
0 0 02 2 2
0 0 0 1
2 2 21
2 2 21
1 1 1amp
2 2 2
1 1 1amp
2 2 2
amp
A R A R A RK d m v K d
Kd mv Kd
Kd mv Kd
0
1d is frac12 the height from the base to the peak of a cycle
An experiment of moving paper under a pendulum through a small angle filled with sand that can escape through a small hole in the bottom shows that a pendulum cycling back and forth follows a curve that is mathematically analogous to a lsquosinrsquo or lsquocosrsquo
The projection of the curved area onto the straight line exactly duplicate the back and forth motion of CM This allows for the math of lsquosinrsquo or lsquocosrsquo to be used
SM will use lsquosinrdquo math It could use lsquocosrsquo or a combination of both lsquosinrsquo and lsquocosrsquo Using just the lsquosinrsquo math makes the understanding and math easier
0 sin 0 0SM 0 sin157 1SM The diagram at the left shows a spring being stretched from its (zero rest) position-2 and the PAR being released at position-1 There are 4 sections (1 to 2 2 to 3 3 to 4 and 4 to5) that alternate between maximum acceleration and zero velocity to max velocity and zero acceleration
At position-1 there is maximum acceleration and zero velocity o The lsquosinrsquo values from 0o to 157o will be used for velocity (values from 0 to 1) o The lsquosinrsquo values from 157o to 0o will be used for acceleration (values from 1 to 0)
At position-2 there is maximum velocity and zero acceleration Acceleration changes to deceleration
o The lsquosinrsquo values from 157o to 0o will be used for deceleration (values from 1 to 0) o The lsquosinrsquo values from 0o to 157o will be used for acceleration (values from 0 to 1)
At position-3 there is maximum acceleration and zero velocity o The lsquosinrsquo values from 0o to 157o will be used for velocity (values from 0 to 1) o The lsquosinrsquo values from 157o to 0o will be used for acceleration (values from 1 to 0)
At position-4 there is maximum velocity and zero acceleration Acceleration changes to deceleration
o The lsquosinrsquo values from 157o to 0o will be used for deceleration (values from 1 to 0) o The lsquosinrsquo values from 0o to 157o will be used for acceleration (values from 0 to 1)
At position-5 the cycle starts again
2
)539(
cT
MK
spring constant (K) when mass (M) and cycle-time (T) are known
)(
))(89)((
111 md
uKgM
d
MG
d
FK sts
spring constant (K) when mass (M) and amplitude ( d1) are known
2 21( )
Kv d d
m
velocity (v) at any point during the cycle
velocity
dforM
Kdv
max
012 velocity (v2) maximum (when d = 0 at position-2 and 4)
acceleration (a1) maximum when spring constant (K) mass (M) and amplitude (d1) are known
acceleration (a1) maximum when amplitude (d1) and cycle-time (Tc) are known
acceleration (a) at any point during the cycle-time (T)
1
151
286
v
dTc
cycle-time (Tc) when amplitude (d1) and velocity (v1) are known
K
MTc 28651
cycle-time (Tc) when mass (M) and spring constant (K) are known
fTc
1
cycle-time (Tc) when the frequency is known
cTf
1
frequency (f) when the cycle-time (Tc) is known
of cycles
elapsed timef
frequency (f) when the number of cycles in an elapsed time is known
d1 amplitude (distance from rest to start position)
Also frac12 the distance from the base to the peak
3131 oror dM
Ka
251
3131 539
c
oror T
da
2539
cT
da
In December of 2006 I discovered the mathematical error in Einsteinrsquos Special Relativity Einstein used BS math and that math only works in an imaginary world
It cannot describe real symmetrical space
Einsteinrsquos Special Relativity should be abandoned and a theory of real space using SM should be used to replace it
Numerous articles on the Internet show Einsteinrsquos theory to be incorrect but no one in academia is willing to publish the information
Figure 1-10 A moving particle observed from two frames of reference in uniform translation
Einsteinrsquos Math Error in Special Relativity Transformation of Velocity (From Eisenberg- Fundamentals of Modern Physics pages 27-29) Consider the particle shown in figure (1-10) moving with velocity v as
seen in a frame of reference O
dx dy dz
Vx Vy Vzdt dt dt
dx dy dz
V x V y V zdy dt dt
We would like to evaluate the velocity v of the particle as seen in the frame of reference 0 which is itself moving relative to 0 with velocity v Measured in the 0 frame the velocity vector of the particle has com ponents
The same velocity vector as measured in the 0 frame has components
2
2
2
2
2
1
1
x vtx
v
cy y
z z
vxt
ctv
c
Now from equations (1-13) we know that the relation between the primed and the unprimed coordinates and times is
Take the differential of these equations remembering that v is a constant This gives
2
2
2
2
2
1
1
dx vdtdx
v
cdy dy
dz dz
vxdt
cdtv
c
2
2
2 2 2 2 2
2
2
2
11 ( )
1( ) 1
1
1
1
dx vdt
v dxdx vdt v
dx dx vdtc dt dtvdx vdx vdx v dxdt dt dt dt
dt dtc c c c
VxV x
Vx
VxV x
vv
vv
c
x
v
c
cV
( )
coordinate of particle relative to moving primed system
coordinate of particle relative to stationary system
velocity of primed coordina
V x
Vx
te system relative to stationaryv sys
tem
Page 28 Fundamentals of Modern Physics
(1-16)
According to Einstein this equation tells you how to transform the observed velocity from one frame of reference to another frame of reference First we note that as Vc and vc approachzero the equation approaches those which would be derived from the Galilean transformation
2 2
( ) 1 01 1 ( )
Vx v Vx vV
VxV x
Vx v
v cVxx
v
c
v
cc
Another property of this equation is that it is impossible to choose V and v such that Vrsquo the magnitude of the velocity which is seen in the new frame is greater than c
1
2
09 ( 09 ) 18009876543
( 09 ) 1811 (09 )
c c cV c c
cc
c
Example As seen by O particle 1 has velocity 09c in the direction of positive x and particle 2 has velocity 09c in the negative x direction
Einstein equation however predicts To evaluate the velocity of particle 1 with respect to particle 2 we transform from the O frame to the Orsquo frame moving in the negative x direction with velocity v = -09c using equation 1-16 We obtain
JK LOGIC will observe the particles separating at 18c
A confusing INCORRECT and ILLOGICAL answer
left is a dir rightsub of a
1
2
1
09 ( 09 ) 18009876543
181( 09 )1 ( 09 )
The multiplication of dir left time a dir right is ONLY a dir left
09
dir
direction leftdirection rightsubtraction
c c cV c c
cc
c
V
dir
22
22
09 18 18009876543
181( 081)1(09)(09)
1
left rightsub
sub
c c c cc c
ccc
c
Analyzing Einsteinrsquos BS math Equation
Using SM math a logical answer can be obtained
Cannot multiply a subtraction operator times a direction in space
Can only subtract and change its direction
Cannot multiply a direction to the left times a direction to the right
Cannot multiply opposite directions in space
Can only solve for the resultant
rightleft
subtraction
1
subtraction
2
sub of a dir to the left is a dir to the right
1 resultant 0 sub 2
2sub 2
2
09 09
09 amp 091
09 09 18 1
(0)1
(09) amp (09)1
c cV
c cc
c c cV
cc
cc
8
181
cc
SM math predicts a logical answer using logic
21st century scientists state in many articles that Einsteins General Relativity and Quantum Mechanics are incompatible
They are not only incompatible they are both mathematically incorrect
They both use illogical BS math
Any equations in math that use imaginary numbers should be abandoned
Imaginary numbers are only necessary in the current BS math system
This means that Schroumldingers equations should be abandoned
Any equation in quantum mechanics that uses imaginary number should be abandoned
Prof Kanarev has developed logical physics and chemistrytheories without the use of Quantum mechanics or relativity
A partial list of his theories coverbull Mechanics ndash replaces Newtonbull Photon - bull Electrons ndash Protons- Neutronsbull Spectroscopybull Nucleusbull Atoms amp Moleculesbull Electrodynamicsbull Heatbull Curvature of Space ndash replaces Schwarzcchildrsquos equation
Kanarevrsquos Hypothesis
bull Electron align linearly with proton They do NOT spin or form clouds around the nucleus
bull Neutrons are in the center of atom Proton are outside of Neutrons
bull The wavelength of all particles is the experimental radius
bull Double slit experiment explained correctly
bull Heisenbergrsquos inequality explained correctly
bull is the radius of approach of the magnetic lines around the torus of the electron
( 1 137)
bull Why water expands when it freezes
bull Electron clusters form lightning
rbull The meaning of Plancks constant is explained
22 2 kg m
h m f mr f consts
2210254 42m r kgm bull If the mr-constant describes a ring the equations
describe experiments bull If the mr-constant describes a wave the equations do
NOT describe experiments
Kanarevrsquos 5 Equations of Mechanics
Kanarevrsquos 1st Equation Acceleration ampmv maMma R R
Kanarevrsquos 2nd Equation Uniform motion amp mv maMma R R
maM mvR R
Kanarevrsquos 3rd Equation
Kanarevrsquos 4th Equation Equal and opposite reaction
1 21 21 2
12
a m
m a m ama
Kanarevrsquos 5th Equation 1 2 3 Mi M M M Mna R R R R R
The sum of the Actions working on a moving body is never equal to zero
Kanarevrsquos Photon model is a rolling ring
His equations accurately describe known experiments with photonsbull Electromagnetic radiation should be called photon radiationbull Diffraction and interference are the same phenomenabull Correctly explains the double slit experiment without quantum equationsbull Correctly explains the meaning of Heisenbergrsquos inequality
e
2 (2)(3142)(2817 15 ) 10007297352568
R 2426 12 137mag
xp
r m
m
Charge
Magnetic action
Proton
The intensity of the magnetic action near the geometrical center of the proton ring is so great that we theorize that it is the strong nuclear force
(9383 8)(1602 19)8476 14
4 (4)(3142)(1411 26)p
pp
EB T
M
(9395 8)(1602 19)1035 15
4 (4)(3142)(0966 26)N
NN
EB T
M
Neutron
1a Hydrogenavi
Here is the BS math answer that Barry Mazur calculated
2
31
2
31
2
31
1 3 1 3 1 3 3 3 2 2 3 1 3
2 2 2 2 4 4 4 4 4 4 2 2
dash
1 3 1 3 1 31
2 2 2 2 4 4
3 3
4 4
When the distributive law is used with the BS Rule-of-Signs and one of the signs is a (-) dash sign the middle two terms always cancel each other This leads to the incorrect answer of (-1) in the above equation
Barry Mazur did not go back and check his answer with the original equation If he had he would have known that the answer was incorrect
3
1 31
2
check answer by setting equal
3
331 3
12
1 31
2
Take the Cube root of both sides
1 3 2 multiply both sides by 2
3 3 subtract 1 from both sides
2 2
3 3 square both sides
3 9 This is an incorrect answer A simple equation check could have shown that the distributive law using BS math provides an incorrect answer
1
BS Illogical and Incorrect Numbers Raised to Powers
nC D n DC
642 DDD
In BS math only a positive number can have an answer if ldquonrdquo is an even number (246---)
All negative numbers have no solutions
642 DDD
With SM just add a direction arrow and any number gives the correct answer
or n nC D C D
3 3( 3 8 8 or 8 2 or 2 )n D C
( 2 16 C= 16 or 16 4 or 4 )n D
Since a negative number has no square root answer in BS math BS mathematicians invented imaginary numbers
i2=-1
24414 iiC n
These equations are easily solved without the use of imaginary numbers with SM math
If 4 then 4 2x
If 4 then 4 2x
The 360-degree circle is a 4400 years old extremely outdated system based on a year of 12 months of 30 days each that is 360 days
SM will use a more logical system that will have a circle of 6283185307o
d
( )
(deg )
d r radians
d r rees
the arc distance around
the circumference of a circle
d
radiusr
2 2
2 2
d r
dv rd t t
v dda rd t t t
1
d
r
( )av tv
The average angular velocity of a body is the rate of its angular displacement
( )0
liminstt
vt
The instantaneous angular velocity is the limit of the ratio
t
as t approaches zero
( )d
av
d
tv
The average circumference velocity of the body is the rate of its distance traveled d divided by the time t
dvv
r
1 dd vdrv
t t t r r
The equation of the relationship between the angular velocity and the velocity of a point on that body
( )o
av
v v v
t ta
The average angular acceleration is the change in the angular velocity
( )0
liminstt
v
ta
The instantaneous angular acceleration is the
limit of the ration of v
t
as t approaches
zero
daa
r
1
1
d
d s d
vva
t r tv a
r t r
The equation of the relationship between the angular acceleration and the acceleration of a point on the body
21
2v a t
If a body with an initial angular velocity ov has a constant
angular acceleration a it will turn through an angle in a time t
1
Straight Line Symbol Units Rotational Symbol Units Distance d m Angular
Distance radians or degrees
Velocity dt
v m
s Angular Velocity
V
t
radians or degrees
s
Acceleration vat
2ms
Angular Acceleration
VA
t
2
radians or degrees
s
Mass M Kg Mass distribution
dM 2Kgm
Force actionA Ma
2
Kgm
sNewton
Torque
dT M A 2
2
Kgm
s
Momentum Mv Kgm
s
Angular Momentum
dM V 2Kgm
s
Work (Force)(distance)
Ad 2
Kgmm
sNm Joule
Work T
Using AS arrow math to solve and direction in space problems using the equation 1 2
2 2 2 2 2 2( ) amp and ( ) amp 3
4
2 2 2
2 2
( ) amp
(5 ) 4 3
(5 ) 16 9
(5 ) 25
(5 ) (5 )
2 2(5 ) 4 3
(5 ) 16 9
(5 ) 25
(5 ) (5 )
In SM math Pythagorasrsquo theorem provides both magnitude and direction
In BS math Pythagorasrsquo theorem provides magnitude only
time set equal to zero
23
( ) ( 0 )10 1
x t xx
3( ) ( 0)
10 1W t f
In BS math this is an equation for a traveling wave
must use a square term in order to make the curve symmetric
NO SQUARE TERM IS NECESSARY
In SM math when t is set to ldquo0rdquo
This is a major fault in BS math
x f(x)
-4 0019
-3 0033
-2 0073
-1 0273
0 3000
1 0273
2 0073
3 0033
4 0019 -4 -3 -2 -1 0 1 2 3 400
05
10
15
20
25
30
35
BS wave (must use square term)
snap shot of wave at x = 0
Am
plit
ude
lt to gt lt to gt f(ltgt)
lt -4 0073
lt -3 0097
lt -2 0143
lt -1 0273
origin 0 3000
gt 1 0273
gt 2 0143
gt 3 0097
gt 4 0073lt0073 lt0097 lt0143 lt0273
3000
0273gt 0143gt 0097gt 0073gt
Wave moving
snap shot of wave
Am
plit
ude
lt4 lt3 lt2 lt1 0 1gt 2gt 3gt 4gt
3
2
1
In SM NO squared terms are needed
SPRING
An Action is needed to stretch or compress a spring to a starting point where it can execute cycle motion
tsA (ts to start) is the work required to move a mass attached to a spring a distance (d1) from
its initial rest position RAK 0 the spring constant is the stiffness of the spring
o A is attraction o R is repulsion o
ARRA
00 is for the back and forth motion through the zero initial rest position A
stretched spring goes from ldquoArdquo to ldquo0rdquo to ldquoRrdquo to ldquoArdquo to ldquo0rdquo to ldquoRrdquo d1 is the distance a mass is moved from its initial rest position
0
0 1ts A RA K d
Once ( )mad
is performed to move the mass to position d1 (against the restraint of the spring) a of the ( )mad
is transferred into the spring as potential attractionrepulsion (PAR) There is always
more ( )mad
performed than is stored as PAR When the spring is released from the restraining Ats the PAR starts the oscillation Conservation of ( )mad
requires that the PAR and the motion ( )mad
( )( )madM be a constant
( )amp ( ) constantmadPAR M 02
0 0
1
2A R A RPAR K d
02
( ) 0 0
1( )
2mad A R A RM m v
0 0 02 2 2
0 0 0 1
2 2 21
2 2 21
1 1 1amp
2 2 2
1 1 1amp
2 2 2
amp
A R A R A RK d m v K d
Kd mv Kd
Kd mv Kd
0
1d is frac12 the height from the base to the peak of a cycle
An experiment of moving paper under a pendulum through a small angle filled with sand that can escape through a small hole in the bottom shows that a pendulum cycling back and forth follows a curve that is mathematically analogous to a lsquosinrsquo or lsquocosrsquo
The projection of the curved area onto the straight line exactly duplicate the back and forth motion of CM This allows for the math of lsquosinrsquo or lsquocosrsquo to be used
SM will use lsquosinrdquo math It could use lsquocosrsquo or a combination of both lsquosinrsquo and lsquocosrsquo Using just the lsquosinrsquo math makes the understanding and math easier
0 sin 0 0SM 0 sin157 1SM The diagram at the left shows a spring being stretched from its (zero rest) position-2 and the PAR being released at position-1 There are 4 sections (1 to 2 2 to 3 3 to 4 and 4 to5) that alternate between maximum acceleration and zero velocity to max velocity and zero acceleration
At position-1 there is maximum acceleration and zero velocity o The lsquosinrsquo values from 0o to 157o will be used for velocity (values from 0 to 1) o The lsquosinrsquo values from 157o to 0o will be used for acceleration (values from 1 to 0)
At position-2 there is maximum velocity and zero acceleration Acceleration changes to deceleration
o The lsquosinrsquo values from 157o to 0o will be used for deceleration (values from 1 to 0) o The lsquosinrsquo values from 0o to 157o will be used for acceleration (values from 0 to 1)
At position-3 there is maximum acceleration and zero velocity o The lsquosinrsquo values from 0o to 157o will be used for velocity (values from 0 to 1) o The lsquosinrsquo values from 157o to 0o will be used for acceleration (values from 1 to 0)
At position-4 there is maximum velocity and zero acceleration Acceleration changes to deceleration
o The lsquosinrsquo values from 157o to 0o will be used for deceleration (values from 1 to 0) o The lsquosinrsquo values from 0o to 157o will be used for acceleration (values from 0 to 1)
At position-5 the cycle starts again
2
)539(
cT
MK
spring constant (K) when mass (M) and cycle-time (T) are known
)(
))(89)((
111 md
uKgM
d
MG
d
FK sts
spring constant (K) when mass (M) and amplitude ( d1) are known
2 21( )
Kv d d
m
velocity (v) at any point during the cycle
velocity
dforM
Kdv
max
012 velocity (v2) maximum (when d = 0 at position-2 and 4)
acceleration (a1) maximum when spring constant (K) mass (M) and amplitude (d1) are known
acceleration (a1) maximum when amplitude (d1) and cycle-time (Tc) are known
acceleration (a) at any point during the cycle-time (T)
1
151
286
v
dTc
cycle-time (Tc) when amplitude (d1) and velocity (v1) are known
K
MTc 28651
cycle-time (Tc) when mass (M) and spring constant (K) are known
fTc
1
cycle-time (Tc) when the frequency is known
cTf
1
frequency (f) when the cycle-time (Tc) is known
of cycles
elapsed timef
frequency (f) when the number of cycles in an elapsed time is known
d1 amplitude (distance from rest to start position)
Also frac12 the distance from the base to the peak
3131 oror dM
Ka
251
3131 539
c
oror T
da
2539
cT
da
In December of 2006 I discovered the mathematical error in Einsteinrsquos Special Relativity Einstein used BS math and that math only works in an imaginary world
It cannot describe real symmetrical space
Einsteinrsquos Special Relativity should be abandoned and a theory of real space using SM should be used to replace it
Numerous articles on the Internet show Einsteinrsquos theory to be incorrect but no one in academia is willing to publish the information
Figure 1-10 A moving particle observed from two frames of reference in uniform translation
Einsteinrsquos Math Error in Special Relativity Transformation of Velocity (From Eisenberg- Fundamentals of Modern Physics pages 27-29) Consider the particle shown in figure (1-10) moving with velocity v as
seen in a frame of reference O
dx dy dz
Vx Vy Vzdt dt dt
dx dy dz
V x V y V zdy dt dt
We would like to evaluate the velocity v of the particle as seen in the frame of reference 0 which is itself moving relative to 0 with velocity v Measured in the 0 frame the velocity vector of the particle has com ponents
The same velocity vector as measured in the 0 frame has components
2
2
2
2
2
1
1
x vtx
v
cy y
z z
vxt
ctv
c
Now from equations (1-13) we know that the relation between the primed and the unprimed coordinates and times is
Take the differential of these equations remembering that v is a constant This gives
2
2
2
2
2
1
1
dx vdtdx
v
cdy dy
dz dz
vxdt
cdtv
c
2
2
2 2 2 2 2
2
2
2
11 ( )
1( ) 1
1
1
1
dx vdt
v dxdx vdt v
dx dx vdtc dt dtvdx vdx vdx v dxdt dt dt dt
dt dtc c c c
VxV x
Vx
VxV x
vv
vv
c
x
v
c
cV
( )
coordinate of particle relative to moving primed system
coordinate of particle relative to stationary system
velocity of primed coordina
V x
Vx
te system relative to stationaryv sys
tem
Page 28 Fundamentals of Modern Physics
(1-16)
According to Einstein this equation tells you how to transform the observed velocity from one frame of reference to another frame of reference First we note that as Vc and vc approachzero the equation approaches those which would be derived from the Galilean transformation
2 2
( ) 1 01 1 ( )
Vx v Vx vV
VxV x
Vx v
v cVxx
v
c
v
cc
Another property of this equation is that it is impossible to choose V and v such that Vrsquo the magnitude of the velocity which is seen in the new frame is greater than c
1
2
09 ( 09 ) 18009876543
( 09 ) 1811 (09 )
c c cV c c
cc
c
Example As seen by O particle 1 has velocity 09c in the direction of positive x and particle 2 has velocity 09c in the negative x direction
Einstein equation however predicts To evaluate the velocity of particle 1 with respect to particle 2 we transform from the O frame to the Orsquo frame moving in the negative x direction with velocity v = -09c using equation 1-16 We obtain
JK LOGIC will observe the particles separating at 18c
A confusing INCORRECT and ILLOGICAL answer
left is a dir rightsub of a
1
2
1
09 ( 09 ) 18009876543
181( 09 )1 ( 09 )
The multiplication of dir left time a dir right is ONLY a dir left
09
dir
direction leftdirection rightsubtraction
c c cV c c
cc
c
V
dir
22
22
09 18 18009876543
181( 081)1(09)(09)
1
left rightsub
sub
c c c cc c
ccc
c
Analyzing Einsteinrsquos BS math Equation
Using SM math a logical answer can be obtained
Cannot multiply a subtraction operator times a direction in space
Can only subtract and change its direction
Cannot multiply a direction to the left times a direction to the right
Cannot multiply opposite directions in space
Can only solve for the resultant
rightleft
subtraction
1
subtraction
2
sub of a dir to the left is a dir to the right
1 resultant 0 sub 2
2sub 2
2
09 09
09 amp 091
09 09 18 1
(0)1
(09) amp (09)1
c cV
c cc
c c cV
cc
cc
8
181
cc
SM math predicts a logical answer using logic
21st century scientists state in many articles that Einsteins General Relativity and Quantum Mechanics are incompatible
They are not only incompatible they are both mathematically incorrect
They both use illogical BS math
Any equations in math that use imaginary numbers should be abandoned
Imaginary numbers are only necessary in the current BS math system
This means that Schroumldingers equations should be abandoned
Any equation in quantum mechanics that uses imaginary number should be abandoned
Prof Kanarev has developed logical physics and chemistrytheories without the use of Quantum mechanics or relativity
A partial list of his theories coverbull Mechanics ndash replaces Newtonbull Photon - bull Electrons ndash Protons- Neutronsbull Spectroscopybull Nucleusbull Atoms amp Moleculesbull Electrodynamicsbull Heatbull Curvature of Space ndash replaces Schwarzcchildrsquos equation
Kanarevrsquos Hypothesis
bull Electron align linearly with proton They do NOT spin or form clouds around the nucleus
bull Neutrons are in the center of atom Proton are outside of Neutrons
bull The wavelength of all particles is the experimental radius
bull Double slit experiment explained correctly
bull Heisenbergrsquos inequality explained correctly
bull is the radius of approach of the magnetic lines around the torus of the electron
( 1 137)
bull Why water expands when it freezes
bull Electron clusters form lightning
rbull The meaning of Plancks constant is explained
22 2 kg m
h m f mr f consts
2210254 42m r kgm bull If the mr-constant describes a ring the equations
describe experiments bull If the mr-constant describes a wave the equations do
NOT describe experiments
Kanarevrsquos 5 Equations of Mechanics
Kanarevrsquos 1st Equation Acceleration ampmv maMma R R
Kanarevrsquos 2nd Equation Uniform motion amp mv maMma R R
maM mvR R
Kanarevrsquos 3rd Equation
Kanarevrsquos 4th Equation Equal and opposite reaction
1 21 21 2
12
a m
m a m ama
Kanarevrsquos 5th Equation 1 2 3 Mi M M M Mna R R R R R
The sum of the Actions working on a moving body is never equal to zero
Kanarevrsquos Photon model is a rolling ring
His equations accurately describe known experiments with photonsbull Electromagnetic radiation should be called photon radiationbull Diffraction and interference are the same phenomenabull Correctly explains the double slit experiment without quantum equationsbull Correctly explains the meaning of Heisenbergrsquos inequality
e
2 (2)(3142)(2817 15 ) 10007297352568
R 2426 12 137mag
xp
r m
m
Charge
Magnetic action
Proton
The intensity of the magnetic action near the geometrical center of the proton ring is so great that we theorize that it is the strong nuclear force
(9383 8)(1602 19)8476 14
4 (4)(3142)(1411 26)p
pp
EB T
M
(9395 8)(1602 19)1035 15
4 (4)(3142)(0966 26)N
NN
EB T
M
Neutron
1a Hydrogenavi
Barry Mazur did not go back and check his answer with the original equation If he had he would have known that the answer was incorrect
3
1 31
2
check answer by setting equal
3
331 3
12
1 31
2
Take the Cube root of both sides
1 3 2 multiply both sides by 2
3 3 subtract 1 from both sides
2 2
3 3 square both sides
3 9 This is an incorrect answer A simple equation check could have shown that the distributive law using BS math provides an incorrect answer
1
BS Illogical and Incorrect Numbers Raised to Powers
nC D n DC
642 DDD
In BS math only a positive number can have an answer if ldquonrdquo is an even number (246---)
All negative numbers have no solutions
642 DDD
With SM just add a direction arrow and any number gives the correct answer
or n nC D C D
3 3( 3 8 8 or 8 2 or 2 )n D C
( 2 16 C= 16 or 16 4 or 4 )n D
Since a negative number has no square root answer in BS math BS mathematicians invented imaginary numbers
i2=-1
24414 iiC n
These equations are easily solved without the use of imaginary numbers with SM math
If 4 then 4 2x
If 4 then 4 2x
The 360-degree circle is a 4400 years old extremely outdated system based on a year of 12 months of 30 days each that is 360 days
SM will use a more logical system that will have a circle of 6283185307o
d
( )
(deg )
d r radians
d r rees
the arc distance around
the circumference of a circle
d
radiusr
2 2
2 2
d r
dv rd t t
v dda rd t t t
1
d
r
( )av tv
The average angular velocity of a body is the rate of its angular displacement
( )0
liminstt
vt
The instantaneous angular velocity is the limit of the ratio
t
as t approaches zero
( )d
av
d
tv
The average circumference velocity of the body is the rate of its distance traveled d divided by the time t
dvv
r
1 dd vdrv
t t t r r
The equation of the relationship between the angular velocity and the velocity of a point on that body
( )o
av
v v v
t ta
The average angular acceleration is the change in the angular velocity
( )0
liminstt
v
ta
The instantaneous angular acceleration is the
limit of the ration of v
t
as t approaches
zero
daa
r
1
1
d
d s d
vva
t r tv a
r t r
The equation of the relationship between the angular acceleration and the acceleration of a point on the body
21
2v a t
If a body with an initial angular velocity ov has a constant
angular acceleration a it will turn through an angle in a time t
1
Straight Line Symbol Units Rotational Symbol Units Distance d m Angular
Distance radians or degrees
Velocity dt
v m
s Angular Velocity
V
t
radians or degrees
s
Acceleration vat
2ms
Angular Acceleration
VA
t
2
radians or degrees
s
Mass M Kg Mass distribution
dM 2Kgm
Force actionA Ma
2
Kgm
sNewton
Torque
dT M A 2
2
Kgm
s
Momentum Mv Kgm
s
Angular Momentum
dM V 2Kgm
s
Work (Force)(distance)
Ad 2
Kgmm
sNm Joule
Work T
Using AS arrow math to solve and direction in space problems using the equation 1 2
2 2 2 2 2 2( ) amp and ( ) amp 3
4
2 2 2
2 2
( ) amp
(5 ) 4 3
(5 ) 16 9
(5 ) 25
(5 ) (5 )
2 2(5 ) 4 3
(5 ) 16 9
(5 ) 25
(5 ) (5 )
In SM math Pythagorasrsquo theorem provides both magnitude and direction
In BS math Pythagorasrsquo theorem provides magnitude only
time set equal to zero
23
( ) ( 0 )10 1
x t xx
3( ) ( 0)
10 1W t f
In BS math this is an equation for a traveling wave
must use a square term in order to make the curve symmetric
NO SQUARE TERM IS NECESSARY
In SM math when t is set to ldquo0rdquo
This is a major fault in BS math
x f(x)
-4 0019
-3 0033
-2 0073
-1 0273
0 3000
1 0273
2 0073
3 0033
4 0019 -4 -3 -2 -1 0 1 2 3 400
05
10
15
20
25
30
35
BS wave (must use square term)
snap shot of wave at x = 0
Am
plit
ude
lt to gt lt to gt f(ltgt)
lt -4 0073
lt -3 0097
lt -2 0143
lt -1 0273
origin 0 3000
gt 1 0273
gt 2 0143
gt 3 0097
gt 4 0073lt0073 lt0097 lt0143 lt0273
3000
0273gt 0143gt 0097gt 0073gt
Wave moving
snap shot of wave
Am
plit
ude
lt4 lt3 lt2 lt1 0 1gt 2gt 3gt 4gt
3
2
1
In SM NO squared terms are needed
SPRING
An Action is needed to stretch or compress a spring to a starting point where it can execute cycle motion
tsA (ts to start) is the work required to move a mass attached to a spring a distance (d1) from
its initial rest position RAK 0 the spring constant is the stiffness of the spring
o A is attraction o R is repulsion o
ARRA
00 is for the back and forth motion through the zero initial rest position A
stretched spring goes from ldquoArdquo to ldquo0rdquo to ldquoRrdquo to ldquoArdquo to ldquo0rdquo to ldquoRrdquo d1 is the distance a mass is moved from its initial rest position
0
0 1ts A RA K d
Once ( )mad
is performed to move the mass to position d1 (against the restraint of the spring) a of the ( )mad
is transferred into the spring as potential attractionrepulsion (PAR) There is always
more ( )mad
performed than is stored as PAR When the spring is released from the restraining Ats the PAR starts the oscillation Conservation of ( )mad
requires that the PAR and the motion ( )mad
( )( )madM be a constant
( )amp ( ) constantmadPAR M 02
0 0
1
2A R A RPAR K d
02
( ) 0 0
1( )
2mad A R A RM m v
0 0 02 2 2
0 0 0 1
2 2 21
2 2 21
1 1 1amp
2 2 2
1 1 1amp
2 2 2
amp
A R A R A RK d m v K d
Kd mv Kd
Kd mv Kd
0
1d is frac12 the height from the base to the peak of a cycle
An experiment of moving paper under a pendulum through a small angle filled with sand that can escape through a small hole in the bottom shows that a pendulum cycling back and forth follows a curve that is mathematically analogous to a lsquosinrsquo or lsquocosrsquo
The projection of the curved area onto the straight line exactly duplicate the back and forth motion of CM This allows for the math of lsquosinrsquo or lsquocosrsquo to be used
SM will use lsquosinrdquo math It could use lsquocosrsquo or a combination of both lsquosinrsquo and lsquocosrsquo Using just the lsquosinrsquo math makes the understanding and math easier
0 sin 0 0SM 0 sin157 1SM The diagram at the left shows a spring being stretched from its (zero rest) position-2 and the PAR being released at position-1 There are 4 sections (1 to 2 2 to 3 3 to 4 and 4 to5) that alternate between maximum acceleration and zero velocity to max velocity and zero acceleration
At position-1 there is maximum acceleration and zero velocity o The lsquosinrsquo values from 0o to 157o will be used for velocity (values from 0 to 1) o The lsquosinrsquo values from 157o to 0o will be used for acceleration (values from 1 to 0)
At position-2 there is maximum velocity and zero acceleration Acceleration changes to deceleration
o The lsquosinrsquo values from 157o to 0o will be used for deceleration (values from 1 to 0) o The lsquosinrsquo values from 0o to 157o will be used for acceleration (values from 0 to 1)
At position-3 there is maximum acceleration and zero velocity o The lsquosinrsquo values from 0o to 157o will be used for velocity (values from 0 to 1) o The lsquosinrsquo values from 157o to 0o will be used for acceleration (values from 1 to 0)
At position-4 there is maximum velocity and zero acceleration Acceleration changes to deceleration
o The lsquosinrsquo values from 157o to 0o will be used for deceleration (values from 1 to 0) o The lsquosinrsquo values from 0o to 157o will be used for acceleration (values from 0 to 1)
At position-5 the cycle starts again
2
)539(
cT
MK
spring constant (K) when mass (M) and cycle-time (T) are known
)(
))(89)((
111 md
uKgM
d
MG
d
FK sts
spring constant (K) when mass (M) and amplitude ( d1) are known
2 21( )
Kv d d
m
velocity (v) at any point during the cycle
velocity
dforM
Kdv
max
012 velocity (v2) maximum (when d = 0 at position-2 and 4)
acceleration (a1) maximum when spring constant (K) mass (M) and amplitude (d1) are known
acceleration (a1) maximum when amplitude (d1) and cycle-time (Tc) are known
acceleration (a) at any point during the cycle-time (T)
1
151
286
v
dTc
cycle-time (Tc) when amplitude (d1) and velocity (v1) are known
K
MTc 28651
cycle-time (Tc) when mass (M) and spring constant (K) are known
fTc
1
cycle-time (Tc) when the frequency is known
cTf
1
frequency (f) when the cycle-time (Tc) is known
of cycles
elapsed timef
frequency (f) when the number of cycles in an elapsed time is known
d1 amplitude (distance from rest to start position)
Also frac12 the distance from the base to the peak
3131 oror dM
Ka
251
3131 539
c
oror T
da
2539
cT
da
In December of 2006 I discovered the mathematical error in Einsteinrsquos Special Relativity Einstein used BS math and that math only works in an imaginary world
It cannot describe real symmetrical space
Einsteinrsquos Special Relativity should be abandoned and a theory of real space using SM should be used to replace it
Numerous articles on the Internet show Einsteinrsquos theory to be incorrect but no one in academia is willing to publish the information
Figure 1-10 A moving particle observed from two frames of reference in uniform translation
Einsteinrsquos Math Error in Special Relativity Transformation of Velocity (From Eisenberg- Fundamentals of Modern Physics pages 27-29) Consider the particle shown in figure (1-10) moving with velocity v as
seen in a frame of reference O
dx dy dz
Vx Vy Vzdt dt dt
dx dy dz
V x V y V zdy dt dt
We would like to evaluate the velocity v of the particle as seen in the frame of reference 0 which is itself moving relative to 0 with velocity v Measured in the 0 frame the velocity vector of the particle has com ponents
The same velocity vector as measured in the 0 frame has components
2
2
2
2
2
1
1
x vtx
v
cy y
z z
vxt
ctv
c
Now from equations (1-13) we know that the relation between the primed and the unprimed coordinates and times is
Take the differential of these equations remembering that v is a constant This gives
2
2
2
2
2
1
1
dx vdtdx
v
cdy dy
dz dz
vxdt
cdtv
c
2
2
2 2 2 2 2
2
2
2
11 ( )
1( ) 1
1
1
1
dx vdt
v dxdx vdt v
dx dx vdtc dt dtvdx vdx vdx v dxdt dt dt dt
dt dtc c c c
VxV x
Vx
VxV x
vv
vv
c
x
v
c
cV
( )
coordinate of particle relative to moving primed system
coordinate of particle relative to stationary system
velocity of primed coordina
V x
Vx
te system relative to stationaryv sys
tem
Page 28 Fundamentals of Modern Physics
(1-16)
According to Einstein this equation tells you how to transform the observed velocity from one frame of reference to another frame of reference First we note that as Vc and vc approachzero the equation approaches those which would be derived from the Galilean transformation
2 2
( ) 1 01 1 ( )
Vx v Vx vV
VxV x
Vx v
v cVxx
v
c
v
cc
Another property of this equation is that it is impossible to choose V and v such that Vrsquo the magnitude of the velocity which is seen in the new frame is greater than c
1
2
09 ( 09 ) 18009876543
( 09 ) 1811 (09 )
c c cV c c
cc
c
Example As seen by O particle 1 has velocity 09c in the direction of positive x and particle 2 has velocity 09c in the negative x direction
Einstein equation however predicts To evaluate the velocity of particle 1 with respect to particle 2 we transform from the O frame to the Orsquo frame moving in the negative x direction with velocity v = -09c using equation 1-16 We obtain
JK LOGIC will observe the particles separating at 18c
A confusing INCORRECT and ILLOGICAL answer
left is a dir rightsub of a
1
2
1
09 ( 09 ) 18009876543
181( 09 )1 ( 09 )
The multiplication of dir left time a dir right is ONLY a dir left
09
dir
direction leftdirection rightsubtraction
c c cV c c
cc
c
V
dir
22
22
09 18 18009876543
181( 081)1(09)(09)
1
left rightsub
sub
c c c cc c
ccc
c
Analyzing Einsteinrsquos BS math Equation
Using SM math a logical answer can be obtained
Cannot multiply a subtraction operator times a direction in space
Can only subtract and change its direction
Cannot multiply a direction to the left times a direction to the right
Cannot multiply opposite directions in space
Can only solve for the resultant
rightleft
subtraction
1
subtraction
2
sub of a dir to the left is a dir to the right
1 resultant 0 sub 2
2sub 2
2
09 09
09 amp 091
09 09 18 1
(0)1
(09) amp (09)1
c cV
c cc
c c cV
cc
cc
8
181
cc
SM math predicts a logical answer using logic
21st century scientists state in many articles that Einsteins General Relativity and Quantum Mechanics are incompatible
They are not only incompatible they are both mathematically incorrect
They both use illogical BS math
Any equations in math that use imaginary numbers should be abandoned
Imaginary numbers are only necessary in the current BS math system
This means that Schroumldingers equations should be abandoned
Any equation in quantum mechanics that uses imaginary number should be abandoned
Prof Kanarev has developed logical physics and chemistrytheories without the use of Quantum mechanics or relativity
A partial list of his theories coverbull Mechanics ndash replaces Newtonbull Photon - bull Electrons ndash Protons- Neutronsbull Spectroscopybull Nucleusbull Atoms amp Moleculesbull Electrodynamicsbull Heatbull Curvature of Space ndash replaces Schwarzcchildrsquos equation
Kanarevrsquos Hypothesis
bull Electron align linearly with proton They do NOT spin or form clouds around the nucleus
bull Neutrons are in the center of atom Proton are outside of Neutrons
bull The wavelength of all particles is the experimental radius
bull Double slit experiment explained correctly
bull Heisenbergrsquos inequality explained correctly
bull is the radius of approach of the magnetic lines around the torus of the electron
( 1 137)
bull Why water expands when it freezes
bull Electron clusters form lightning
rbull The meaning of Plancks constant is explained
22 2 kg m
h m f mr f consts
2210254 42m r kgm bull If the mr-constant describes a ring the equations
describe experiments bull If the mr-constant describes a wave the equations do
NOT describe experiments
Kanarevrsquos 5 Equations of Mechanics
Kanarevrsquos 1st Equation Acceleration ampmv maMma R R
Kanarevrsquos 2nd Equation Uniform motion amp mv maMma R R
maM mvR R
Kanarevrsquos 3rd Equation
Kanarevrsquos 4th Equation Equal and opposite reaction
1 21 21 2
12
a m
m a m ama
Kanarevrsquos 5th Equation 1 2 3 Mi M M M Mna R R R R R
The sum of the Actions working on a moving body is never equal to zero
Kanarevrsquos Photon model is a rolling ring
His equations accurately describe known experiments with photonsbull Electromagnetic radiation should be called photon radiationbull Diffraction and interference are the same phenomenabull Correctly explains the double slit experiment without quantum equationsbull Correctly explains the meaning of Heisenbergrsquos inequality
e
2 (2)(3142)(2817 15 ) 10007297352568
R 2426 12 137mag
xp
r m
m
Charge
Magnetic action
Proton
The intensity of the magnetic action near the geometrical center of the proton ring is so great that we theorize that it is the strong nuclear force
(9383 8)(1602 19)8476 14
4 (4)(3142)(1411 26)p
pp
EB T
M
(9395 8)(1602 19)1035 15
4 (4)(3142)(0966 26)N
NN
EB T
M
Neutron
1a Hydrogenavi
BS Illogical and Incorrect Numbers Raised to Powers
nC D n DC
642 DDD
In BS math only a positive number can have an answer if ldquonrdquo is an even number (246---)
All negative numbers have no solutions
642 DDD
With SM just add a direction arrow and any number gives the correct answer
or n nC D C D
3 3( 3 8 8 or 8 2 or 2 )n D C
( 2 16 C= 16 or 16 4 or 4 )n D
Since a negative number has no square root answer in BS math BS mathematicians invented imaginary numbers
i2=-1
24414 iiC n
These equations are easily solved without the use of imaginary numbers with SM math
If 4 then 4 2x
If 4 then 4 2x
The 360-degree circle is a 4400 years old extremely outdated system based on a year of 12 months of 30 days each that is 360 days
SM will use a more logical system that will have a circle of 6283185307o
d
( )
(deg )
d r radians
d r rees
the arc distance around
the circumference of a circle
d
radiusr
2 2
2 2
d r
dv rd t t
v dda rd t t t
1
d
r
( )av tv
The average angular velocity of a body is the rate of its angular displacement
( )0
liminstt
vt
The instantaneous angular velocity is the limit of the ratio
t
as t approaches zero
( )d
av
d
tv
The average circumference velocity of the body is the rate of its distance traveled d divided by the time t
dvv
r
1 dd vdrv
t t t r r
The equation of the relationship between the angular velocity and the velocity of a point on that body
( )o
av
v v v
t ta
The average angular acceleration is the change in the angular velocity
( )0
liminstt
v
ta
The instantaneous angular acceleration is the
limit of the ration of v
t
as t approaches
zero
daa
r
1
1
d
d s d
vva
t r tv a
r t r
The equation of the relationship between the angular acceleration and the acceleration of a point on the body
21
2v a t
If a body with an initial angular velocity ov has a constant
angular acceleration a it will turn through an angle in a time t
1
Straight Line Symbol Units Rotational Symbol Units Distance d m Angular
Distance radians or degrees
Velocity dt
v m
s Angular Velocity
V
t
radians or degrees
s
Acceleration vat
2ms
Angular Acceleration
VA
t
2
radians or degrees
s
Mass M Kg Mass distribution
dM 2Kgm
Force actionA Ma
2
Kgm
sNewton
Torque
dT M A 2
2
Kgm
s
Momentum Mv Kgm
s
Angular Momentum
dM V 2Kgm
s
Work (Force)(distance)
Ad 2
Kgmm
sNm Joule
Work T
Using AS arrow math to solve and direction in space problems using the equation 1 2
2 2 2 2 2 2( ) amp and ( ) amp 3
4
2 2 2
2 2
( ) amp
(5 ) 4 3
(5 ) 16 9
(5 ) 25
(5 ) (5 )
2 2(5 ) 4 3
(5 ) 16 9
(5 ) 25
(5 ) (5 )
In SM math Pythagorasrsquo theorem provides both magnitude and direction
In BS math Pythagorasrsquo theorem provides magnitude only
time set equal to zero
23
( ) ( 0 )10 1
x t xx
3( ) ( 0)
10 1W t f
In BS math this is an equation for a traveling wave
must use a square term in order to make the curve symmetric
NO SQUARE TERM IS NECESSARY
In SM math when t is set to ldquo0rdquo
This is a major fault in BS math
x f(x)
-4 0019
-3 0033
-2 0073
-1 0273
0 3000
1 0273
2 0073
3 0033
4 0019 -4 -3 -2 -1 0 1 2 3 400
05
10
15
20
25
30
35
BS wave (must use square term)
snap shot of wave at x = 0
Am
plit
ude
lt to gt lt to gt f(ltgt)
lt -4 0073
lt -3 0097
lt -2 0143
lt -1 0273
origin 0 3000
gt 1 0273
gt 2 0143
gt 3 0097
gt 4 0073lt0073 lt0097 lt0143 lt0273
3000
0273gt 0143gt 0097gt 0073gt
Wave moving
snap shot of wave
Am
plit
ude
lt4 lt3 lt2 lt1 0 1gt 2gt 3gt 4gt
3
2
1
In SM NO squared terms are needed
SPRING
An Action is needed to stretch or compress a spring to a starting point where it can execute cycle motion
tsA (ts to start) is the work required to move a mass attached to a spring a distance (d1) from
its initial rest position RAK 0 the spring constant is the stiffness of the spring
o A is attraction o R is repulsion o
ARRA
00 is for the back and forth motion through the zero initial rest position A
stretched spring goes from ldquoArdquo to ldquo0rdquo to ldquoRrdquo to ldquoArdquo to ldquo0rdquo to ldquoRrdquo d1 is the distance a mass is moved from its initial rest position
0
0 1ts A RA K d
Once ( )mad
is performed to move the mass to position d1 (against the restraint of the spring) a of the ( )mad
is transferred into the spring as potential attractionrepulsion (PAR) There is always
more ( )mad
performed than is stored as PAR When the spring is released from the restraining Ats the PAR starts the oscillation Conservation of ( )mad
requires that the PAR and the motion ( )mad
( )( )madM be a constant
( )amp ( ) constantmadPAR M 02
0 0
1
2A R A RPAR K d
02
( ) 0 0
1( )
2mad A R A RM m v
0 0 02 2 2
0 0 0 1
2 2 21
2 2 21
1 1 1amp
2 2 2
1 1 1amp
2 2 2
amp
A R A R A RK d m v K d
Kd mv Kd
Kd mv Kd
0
1d is frac12 the height from the base to the peak of a cycle
An experiment of moving paper under a pendulum through a small angle filled with sand that can escape through a small hole in the bottom shows that a pendulum cycling back and forth follows a curve that is mathematically analogous to a lsquosinrsquo or lsquocosrsquo
The projection of the curved area onto the straight line exactly duplicate the back and forth motion of CM This allows for the math of lsquosinrsquo or lsquocosrsquo to be used
SM will use lsquosinrdquo math It could use lsquocosrsquo or a combination of both lsquosinrsquo and lsquocosrsquo Using just the lsquosinrsquo math makes the understanding and math easier
0 sin 0 0SM 0 sin157 1SM The diagram at the left shows a spring being stretched from its (zero rest) position-2 and the PAR being released at position-1 There are 4 sections (1 to 2 2 to 3 3 to 4 and 4 to5) that alternate between maximum acceleration and zero velocity to max velocity and zero acceleration
At position-1 there is maximum acceleration and zero velocity o The lsquosinrsquo values from 0o to 157o will be used for velocity (values from 0 to 1) o The lsquosinrsquo values from 157o to 0o will be used for acceleration (values from 1 to 0)
At position-2 there is maximum velocity and zero acceleration Acceleration changes to deceleration
o The lsquosinrsquo values from 157o to 0o will be used for deceleration (values from 1 to 0) o The lsquosinrsquo values from 0o to 157o will be used for acceleration (values from 0 to 1)
At position-3 there is maximum acceleration and zero velocity o The lsquosinrsquo values from 0o to 157o will be used for velocity (values from 0 to 1) o The lsquosinrsquo values from 157o to 0o will be used for acceleration (values from 1 to 0)
At position-4 there is maximum velocity and zero acceleration Acceleration changes to deceleration
o The lsquosinrsquo values from 157o to 0o will be used for deceleration (values from 1 to 0) o The lsquosinrsquo values from 0o to 157o will be used for acceleration (values from 0 to 1)
At position-5 the cycle starts again
2
)539(
cT
MK
spring constant (K) when mass (M) and cycle-time (T) are known
)(
))(89)((
111 md
uKgM
d
MG
d
FK sts
spring constant (K) when mass (M) and amplitude ( d1) are known
2 21( )
Kv d d
m
velocity (v) at any point during the cycle
velocity
dforM
Kdv
max
012 velocity (v2) maximum (when d = 0 at position-2 and 4)
acceleration (a1) maximum when spring constant (K) mass (M) and amplitude (d1) are known
acceleration (a1) maximum when amplitude (d1) and cycle-time (Tc) are known
acceleration (a) at any point during the cycle-time (T)
1
151
286
v
dTc
cycle-time (Tc) when amplitude (d1) and velocity (v1) are known
K
MTc 28651
cycle-time (Tc) when mass (M) and spring constant (K) are known
fTc
1
cycle-time (Tc) when the frequency is known
cTf
1
frequency (f) when the cycle-time (Tc) is known
of cycles
elapsed timef
frequency (f) when the number of cycles in an elapsed time is known
d1 amplitude (distance from rest to start position)
Also frac12 the distance from the base to the peak
3131 oror dM
Ka
251
3131 539
c
oror T
da
2539
cT
da
In December of 2006 I discovered the mathematical error in Einsteinrsquos Special Relativity Einstein used BS math and that math only works in an imaginary world
It cannot describe real symmetrical space
Einsteinrsquos Special Relativity should be abandoned and a theory of real space using SM should be used to replace it
Numerous articles on the Internet show Einsteinrsquos theory to be incorrect but no one in academia is willing to publish the information
Figure 1-10 A moving particle observed from two frames of reference in uniform translation
Einsteinrsquos Math Error in Special Relativity Transformation of Velocity (From Eisenberg- Fundamentals of Modern Physics pages 27-29) Consider the particle shown in figure (1-10) moving with velocity v as
seen in a frame of reference O
dx dy dz
Vx Vy Vzdt dt dt
dx dy dz
V x V y V zdy dt dt
We would like to evaluate the velocity v of the particle as seen in the frame of reference 0 which is itself moving relative to 0 with velocity v Measured in the 0 frame the velocity vector of the particle has com ponents
The same velocity vector as measured in the 0 frame has components
2
2
2
2
2
1
1
x vtx
v
cy y
z z
vxt
ctv
c
Now from equations (1-13) we know that the relation between the primed and the unprimed coordinates and times is
Take the differential of these equations remembering that v is a constant This gives
2
2
2
2
2
1
1
dx vdtdx
v
cdy dy
dz dz
vxdt
cdtv
c
2
2
2 2 2 2 2
2
2
2
11 ( )
1( ) 1
1
1
1
dx vdt
v dxdx vdt v
dx dx vdtc dt dtvdx vdx vdx v dxdt dt dt dt
dt dtc c c c
VxV x
Vx
VxV x
vv
vv
c
x
v
c
cV
( )
coordinate of particle relative to moving primed system
coordinate of particle relative to stationary system
velocity of primed coordina
V x
Vx
te system relative to stationaryv sys
tem
Page 28 Fundamentals of Modern Physics
(1-16)
According to Einstein this equation tells you how to transform the observed velocity from one frame of reference to another frame of reference First we note that as Vc and vc approachzero the equation approaches those which would be derived from the Galilean transformation
2 2
( ) 1 01 1 ( )
Vx v Vx vV
VxV x
Vx v
v cVxx
v
c
v
cc
Another property of this equation is that it is impossible to choose V and v such that Vrsquo the magnitude of the velocity which is seen in the new frame is greater than c
1
2
09 ( 09 ) 18009876543
( 09 ) 1811 (09 )
c c cV c c
cc
c
Example As seen by O particle 1 has velocity 09c in the direction of positive x and particle 2 has velocity 09c in the negative x direction
Einstein equation however predicts To evaluate the velocity of particle 1 with respect to particle 2 we transform from the O frame to the Orsquo frame moving in the negative x direction with velocity v = -09c using equation 1-16 We obtain
JK LOGIC will observe the particles separating at 18c
A confusing INCORRECT and ILLOGICAL answer
left is a dir rightsub of a
1
2
1
09 ( 09 ) 18009876543
181( 09 )1 ( 09 )
The multiplication of dir left time a dir right is ONLY a dir left
09
dir
direction leftdirection rightsubtraction
c c cV c c
cc
c
V
dir
22
22
09 18 18009876543
181( 081)1(09)(09)
1
left rightsub
sub
c c c cc c
ccc
c
Analyzing Einsteinrsquos BS math Equation
Using SM math a logical answer can be obtained
Cannot multiply a subtraction operator times a direction in space
Can only subtract and change its direction
Cannot multiply a direction to the left times a direction to the right
Cannot multiply opposite directions in space
Can only solve for the resultant
rightleft
subtraction
1
subtraction
2
sub of a dir to the left is a dir to the right
1 resultant 0 sub 2
2sub 2
2
09 09
09 amp 091
09 09 18 1
(0)1
(09) amp (09)1
c cV
c cc
c c cV
cc
cc
8
181
cc
SM math predicts a logical answer using logic
21st century scientists state in many articles that Einsteins General Relativity and Quantum Mechanics are incompatible
They are not only incompatible they are both mathematically incorrect
They both use illogical BS math
Any equations in math that use imaginary numbers should be abandoned
Imaginary numbers are only necessary in the current BS math system
This means that Schroumldingers equations should be abandoned
Any equation in quantum mechanics that uses imaginary number should be abandoned
Prof Kanarev has developed logical physics and chemistrytheories without the use of Quantum mechanics or relativity
A partial list of his theories coverbull Mechanics ndash replaces Newtonbull Photon - bull Electrons ndash Protons- Neutronsbull Spectroscopybull Nucleusbull Atoms amp Moleculesbull Electrodynamicsbull Heatbull Curvature of Space ndash replaces Schwarzcchildrsquos equation
Kanarevrsquos Hypothesis
bull Electron align linearly with proton They do NOT spin or form clouds around the nucleus
bull Neutrons are in the center of atom Proton are outside of Neutrons
bull The wavelength of all particles is the experimental radius
bull Double slit experiment explained correctly
bull Heisenbergrsquos inequality explained correctly
bull is the radius of approach of the magnetic lines around the torus of the electron
( 1 137)
bull Why water expands when it freezes
bull Electron clusters form lightning
rbull The meaning of Plancks constant is explained
22 2 kg m
h m f mr f consts
2210254 42m r kgm bull If the mr-constant describes a ring the equations
describe experiments bull If the mr-constant describes a wave the equations do
NOT describe experiments
Kanarevrsquos 5 Equations of Mechanics
Kanarevrsquos 1st Equation Acceleration ampmv maMma R R
Kanarevrsquos 2nd Equation Uniform motion amp mv maMma R R
maM mvR R
Kanarevrsquos 3rd Equation
Kanarevrsquos 4th Equation Equal and opposite reaction
1 21 21 2
12
a m
m a m ama
Kanarevrsquos 5th Equation 1 2 3 Mi M M M Mna R R R R R
The sum of the Actions working on a moving body is never equal to zero
Kanarevrsquos Photon model is a rolling ring
His equations accurately describe known experiments with photonsbull Electromagnetic radiation should be called photon radiationbull Diffraction and interference are the same phenomenabull Correctly explains the double slit experiment without quantum equationsbull Correctly explains the meaning of Heisenbergrsquos inequality
e
2 (2)(3142)(2817 15 ) 10007297352568
R 2426 12 137mag
xp
r m
m
Charge
Magnetic action
Proton
The intensity of the magnetic action near the geometrical center of the proton ring is so great that we theorize that it is the strong nuclear force
(9383 8)(1602 19)8476 14
4 (4)(3142)(1411 26)p
pp
EB T
M
(9395 8)(1602 19)1035 15
4 (4)(3142)(0966 26)N
NN
EB T
M
Neutron
1a Hydrogenavi
With SM just add a direction arrow and any number gives the correct answer
or n nC D C D
3 3( 3 8 8 or 8 2 or 2 )n D C
( 2 16 C= 16 or 16 4 or 4 )n D
Since a negative number has no square root answer in BS math BS mathematicians invented imaginary numbers
i2=-1
24414 iiC n
These equations are easily solved without the use of imaginary numbers with SM math
If 4 then 4 2x
If 4 then 4 2x
The 360-degree circle is a 4400 years old extremely outdated system based on a year of 12 months of 30 days each that is 360 days
SM will use a more logical system that will have a circle of 6283185307o
d
( )
(deg )
d r radians
d r rees
the arc distance around
the circumference of a circle
d
radiusr
2 2
2 2
d r
dv rd t t
v dda rd t t t
1
d
r
( )av tv
The average angular velocity of a body is the rate of its angular displacement
( )0
liminstt
vt
The instantaneous angular velocity is the limit of the ratio
t
as t approaches zero
( )d
av
d
tv
The average circumference velocity of the body is the rate of its distance traveled d divided by the time t
dvv
r
1 dd vdrv
t t t r r
The equation of the relationship between the angular velocity and the velocity of a point on that body
( )o
av
v v v
t ta
The average angular acceleration is the change in the angular velocity
( )0
liminstt
v
ta
The instantaneous angular acceleration is the
limit of the ration of v
t
as t approaches
zero
daa
r
1
1
d
d s d
vva
t r tv a
r t r
The equation of the relationship between the angular acceleration and the acceleration of a point on the body
21
2v a t
If a body with an initial angular velocity ov has a constant
angular acceleration a it will turn through an angle in a time t
1
Straight Line Symbol Units Rotational Symbol Units Distance d m Angular
Distance radians or degrees
Velocity dt
v m
s Angular Velocity
V
t
radians or degrees
s
Acceleration vat
2ms
Angular Acceleration
VA
t
2
radians or degrees
s
Mass M Kg Mass distribution
dM 2Kgm
Force actionA Ma
2
Kgm
sNewton
Torque
dT M A 2
2
Kgm
s
Momentum Mv Kgm
s
Angular Momentum
dM V 2Kgm
s
Work (Force)(distance)
Ad 2
Kgmm
sNm Joule
Work T
Using AS arrow math to solve and direction in space problems using the equation 1 2
2 2 2 2 2 2( ) amp and ( ) amp 3
4
2 2 2
2 2
( ) amp
(5 ) 4 3
(5 ) 16 9
(5 ) 25
(5 ) (5 )
2 2(5 ) 4 3
(5 ) 16 9
(5 ) 25
(5 ) (5 )
In SM math Pythagorasrsquo theorem provides both magnitude and direction
In BS math Pythagorasrsquo theorem provides magnitude only
time set equal to zero
23
( ) ( 0 )10 1
x t xx
3( ) ( 0)
10 1W t f
In BS math this is an equation for a traveling wave
must use a square term in order to make the curve symmetric
NO SQUARE TERM IS NECESSARY
In SM math when t is set to ldquo0rdquo
This is a major fault in BS math
x f(x)
-4 0019
-3 0033
-2 0073
-1 0273
0 3000
1 0273
2 0073
3 0033
4 0019 -4 -3 -2 -1 0 1 2 3 400
05
10
15
20
25
30
35
BS wave (must use square term)
snap shot of wave at x = 0
Am
plit
ude
lt to gt lt to gt f(ltgt)
lt -4 0073
lt -3 0097
lt -2 0143
lt -1 0273
origin 0 3000
gt 1 0273
gt 2 0143
gt 3 0097
gt 4 0073lt0073 lt0097 lt0143 lt0273
3000
0273gt 0143gt 0097gt 0073gt
Wave moving
snap shot of wave
Am
plit
ude
lt4 lt3 lt2 lt1 0 1gt 2gt 3gt 4gt
3
2
1
In SM NO squared terms are needed
SPRING
An Action is needed to stretch or compress a spring to a starting point where it can execute cycle motion
tsA (ts to start) is the work required to move a mass attached to a spring a distance (d1) from
its initial rest position RAK 0 the spring constant is the stiffness of the spring
o A is attraction o R is repulsion o
ARRA
00 is for the back and forth motion through the zero initial rest position A
stretched spring goes from ldquoArdquo to ldquo0rdquo to ldquoRrdquo to ldquoArdquo to ldquo0rdquo to ldquoRrdquo d1 is the distance a mass is moved from its initial rest position
0
0 1ts A RA K d
Once ( )mad
is performed to move the mass to position d1 (against the restraint of the spring) a of the ( )mad
is transferred into the spring as potential attractionrepulsion (PAR) There is always
more ( )mad
performed than is stored as PAR When the spring is released from the restraining Ats the PAR starts the oscillation Conservation of ( )mad
requires that the PAR and the motion ( )mad
( )( )madM be a constant
( )amp ( ) constantmadPAR M 02
0 0
1
2A R A RPAR K d
02
( ) 0 0
1( )
2mad A R A RM m v
0 0 02 2 2
0 0 0 1
2 2 21
2 2 21
1 1 1amp
2 2 2
1 1 1amp
2 2 2
amp
A R A R A RK d m v K d
Kd mv Kd
Kd mv Kd
0
1d is frac12 the height from the base to the peak of a cycle
An experiment of moving paper under a pendulum through a small angle filled with sand that can escape through a small hole in the bottom shows that a pendulum cycling back and forth follows a curve that is mathematically analogous to a lsquosinrsquo or lsquocosrsquo
The projection of the curved area onto the straight line exactly duplicate the back and forth motion of CM This allows for the math of lsquosinrsquo or lsquocosrsquo to be used
SM will use lsquosinrdquo math It could use lsquocosrsquo or a combination of both lsquosinrsquo and lsquocosrsquo Using just the lsquosinrsquo math makes the understanding and math easier
0 sin 0 0SM 0 sin157 1SM The diagram at the left shows a spring being stretched from its (zero rest) position-2 and the PAR being released at position-1 There are 4 sections (1 to 2 2 to 3 3 to 4 and 4 to5) that alternate between maximum acceleration and zero velocity to max velocity and zero acceleration
At position-1 there is maximum acceleration and zero velocity o The lsquosinrsquo values from 0o to 157o will be used for velocity (values from 0 to 1) o The lsquosinrsquo values from 157o to 0o will be used for acceleration (values from 1 to 0)
At position-2 there is maximum velocity and zero acceleration Acceleration changes to deceleration
o The lsquosinrsquo values from 157o to 0o will be used for deceleration (values from 1 to 0) o The lsquosinrsquo values from 0o to 157o will be used for acceleration (values from 0 to 1)
At position-3 there is maximum acceleration and zero velocity o The lsquosinrsquo values from 0o to 157o will be used for velocity (values from 0 to 1) o The lsquosinrsquo values from 157o to 0o will be used for acceleration (values from 1 to 0)
At position-4 there is maximum velocity and zero acceleration Acceleration changes to deceleration
o The lsquosinrsquo values from 157o to 0o will be used for deceleration (values from 1 to 0) o The lsquosinrsquo values from 0o to 157o will be used for acceleration (values from 0 to 1)
At position-5 the cycle starts again
2
)539(
cT
MK
spring constant (K) when mass (M) and cycle-time (T) are known
)(
))(89)((
111 md
uKgM
d
MG
d
FK sts
spring constant (K) when mass (M) and amplitude ( d1) are known
2 21( )
Kv d d
m
velocity (v) at any point during the cycle
velocity
dforM
Kdv
max
012 velocity (v2) maximum (when d = 0 at position-2 and 4)
acceleration (a1) maximum when spring constant (K) mass (M) and amplitude (d1) are known
acceleration (a1) maximum when amplitude (d1) and cycle-time (Tc) are known
acceleration (a) at any point during the cycle-time (T)
1
151
286
v
dTc
cycle-time (Tc) when amplitude (d1) and velocity (v1) are known
K
MTc 28651
cycle-time (Tc) when mass (M) and spring constant (K) are known
fTc
1
cycle-time (Tc) when the frequency is known
cTf
1
frequency (f) when the cycle-time (Tc) is known
of cycles
elapsed timef
frequency (f) when the number of cycles in an elapsed time is known
d1 amplitude (distance from rest to start position)
Also frac12 the distance from the base to the peak
3131 oror dM
Ka
251
3131 539
c
oror T
da
2539
cT
da
In December of 2006 I discovered the mathematical error in Einsteinrsquos Special Relativity Einstein used BS math and that math only works in an imaginary world
It cannot describe real symmetrical space
Einsteinrsquos Special Relativity should be abandoned and a theory of real space using SM should be used to replace it
Numerous articles on the Internet show Einsteinrsquos theory to be incorrect but no one in academia is willing to publish the information
Figure 1-10 A moving particle observed from two frames of reference in uniform translation
Einsteinrsquos Math Error in Special Relativity Transformation of Velocity (From Eisenberg- Fundamentals of Modern Physics pages 27-29) Consider the particle shown in figure (1-10) moving with velocity v as
seen in a frame of reference O
dx dy dz
Vx Vy Vzdt dt dt
dx dy dz
V x V y V zdy dt dt
We would like to evaluate the velocity v of the particle as seen in the frame of reference 0 which is itself moving relative to 0 with velocity v Measured in the 0 frame the velocity vector of the particle has com ponents
The same velocity vector as measured in the 0 frame has components
2
2
2
2
2
1
1
x vtx
v
cy y
z z
vxt
ctv
c
Now from equations (1-13) we know that the relation between the primed and the unprimed coordinates and times is
Take the differential of these equations remembering that v is a constant This gives
2
2
2
2
2
1
1
dx vdtdx
v
cdy dy
dz dz
vxdt
cdtv
c
2
2
2 2 2 2 2
2
2
2
11 ( )
1( ) 1
1
1
1
dx vdt
v dxdx vdt v
dx dx vdtc dt dtvdx vdx vdx v dxdt dt dt dt
dt dtc c c c
VxV x
Vx
VxV x
vv
vv
c
x
v
c
cV
( )
coordinate of particle relative to moving primed system
coordinate of particle relative to stationary system
velocity of primed coordina
V x
Vx
te system relative to stationaryv sys
tem
Page 28 Fundamentals of Modern Physics
(1-16)
According to Einstein this equation tells you how to transform the observed velocity from one frame of reference to another frame of reference First we note that as Vc and vc approachzero the equation approaches those which would be derived from the Galilean transformation
2 2
( ) 1 01 1 ( )
Vx v Vx vV
VxV x
Vx v
v cVxx
v
c
v
cc
Another property of this equation is that it is impossible to choose V and v such that Vrsquo the magnitude of the velocity which is seen in the new frame is greater than c
1
2
09 ( 09 ) 18009876543
( 09 ) 1811 (09 )
c c cV c c
cc
c
Example As seen by O particle 1 has velocity 09c in the direction of positive x and particle 2 has velocity 09c in the negative x direction
Einstein equation however predicts To evaluate the velocity of particle 1 with respect to particle 2 we transform from the O frame to the Orsquo frame moving in the negative x direction with velocity v = -09c using equation 1-16 We obtain
JK LOGIC will observe the particles separating at 18c
A confusing INCORRECT and ILLOGICAL answer
left is a dir rightsub of a
1
2
1
09 ( 09 ) 18009876543
181( 09 )1 ( 09 )
The multiplication of dir left time a dir right is ONLY a dir left
09
dir
direction leftdirection rightsubtraction
c c cV c c
cc
c
V
dir
22
22
09 18 18009876543
181( 081)1(09)(09)
1
left rightsub
sub
c c c cc c
ccc
c
Analyzing Einsteinrsquos BS math Equation
Using SM math a logical answer can be obtained
Cannot multiply a subtraction operator times a direction in space
Can only subtract and change its direction
Cannot multiply a direction to the left times a direction to the right
Cannot multiply opposite directions in space
Can only solve for the resultant
rightleft
subtraction
1
subtraction
2
sub of a dir to the left is a dir to the right
1 resultant 0 sub 2
2sub 2
2
09 09
09 amp 091
09 09 18 1
(0)1
(09) amp (09)1
c cV
c cc
c c cV
cc
cc
8
181
cc
SM math predicts a logical answer using logic
21st century scientists state in many articles that Einsteins General Relativity and Quantum Mechanics are incompatible
They are not only incompatible they are both mathematically incorrect
They both use illogical BS math
Any equations in math that use imaginary numbers should be abandoned
Imaginary numbers are only necessary in the current BS math system
This means that Schroumldingers equations should be abandoned
Any equation in quantum mechanics that uses imaginary number should be abandoned
Prof Kanarev has developed logical physics and chemistrytheories without the use of Quantum mechanics or relativity
A partial list of his theories coverbull Mechanics ndash replaces Newtonbull Photon - bull Electrons ndash Protons- Neutronsbull Spectroscopybull Nucleusbull Atoms amp Moleculesbull Electrodynamicsbull Heatbull Curvature of Space ndash replaces Schwarzcchildrsquos equation
Kanarevrsquos Hypothesis
bull Electron align linearly with proton They do NOT spin or form clouds around the nucleus
bull Neutrons are in the center of atom Proton are outside of Neutrons
bull The wavelength of all particles is the experimental radius
bull Double slit experiment explained correctly
bull Heisenbergrsquos inequality explained correctly
bull is the radius of approach of the magnetic lines around the torus of the electron
( 1 137)
bull Why water expands when it freezes
bull Electron clusters form lightning
rbull The meaning of Plancks constant is explained
22 2 kg m
h m f mr f consts
2210254 42m r kgm bull If the mr-constant describes a ring the equations
describe experiments bull If the mr-constant describes a wave the equations do
NOT describe experiments
Kanarevrsquos 5 Equations of Mechanics
Kanarevrsquos 1st Equation Acceleration ampmv maMma R R
Kanarevrsquos 2nd Equation Uniform motion amp mv maMma R R
maM mvR R
Kanarevrsquos 3rd Equation
Kanarevrsquos 4th Equation Equal and opposite reaction
1 21 21 2
12
a m
m a m ama
Kanarevrsquos 5th Equation 1 2 3 Mi M M M Mna R R R R R
The sum of the Actions working on a moving body is never equal to zero
Kanarevrsquos Photon model is a rolling ring
His equations accurately describe known experiments with photonsbull Electromagnetic radiation should be called photon radiationbull Diffraction and interference are the same phenomenabull Correctly explains the double slit experiment without quantum equationsbull Correctly explains the meaning of Heisenbergrsquos inequality
e
2 (2)(3142)(2817 15 ) 10007297352568
R 2426 12 137mag
xp
r m
m
Charge
Magnetic action
Proton
The intensity of the magnetic action near the geometrical center of the proton ring is so great that we theorize that it is the strong nuclear force
(9383 8)(1602 19)8476 14
4 (4)(3142)(1411 26)p
pp
EB T
M
(9395 8)(1602 19)1035 15
4 (4)(3142)(0966 26)N
NN
EB T
M
Neutron
1a Hydrogenavi
These equations are easily solved without the use of imaginary numbers with SM math
If 4 then 4 2x
If 4 then 4 2x
The 360-degree circle is a 4400 years old extremely outdated system based on a year of 12 months of 30 days each that is 360 days
SM will use a more logical system that will have a circle of 6283185307o
d
( )
(deg )
d r radians
d r rees
the arc distance around
the circumference of a circle
d
radiusr
2 2
2 2
d r
dv rd t t
v dda rd t t t
1
d
r
( )av tv
The average angular velocity of a body is the rate of its angular displacement
( )0
liminstt
vt
The instantaneous angular velocity is the limit of the ratio
t
as t approaches zero
( )d
av
d
tv
The average circumference velocity of the body is the rate of its distance traveled d divided by the time t
dvv
r
1 dd vdrv
t t t r r
The equation of the relationship between the angular velocity and the velocity of a point on that body
( )o
av
v v v
t ta
The average angular acceleration is the change in the angular velocity
( )0
liminstt
v
ta
The instantaneous angular acceleration is the
limit of the ration of v
t
as t approaches
zero
daa
r
1
1
d
d s d
vva
t r tv a
r t r
The equation of the relationship between the angular acceleration and the acceleration of a point on the body
21
2v a t
If a body with an initial angular velocity ov has a constant
angular acceleration a it will turn through an angle in a time t
1
Straight Line Symbol Units Rotational Symbol Units Distance d m Angular
Distance radians or degrees
Velocity dt
v m
s Angular Velocity
V
t
radians or degrees
s
Acceleration vat
2ms
Angular Acceleration
VA
t
2
radians or degrees
s
Mass M Kg Mass distribution
dM 2Kgm
Force actionA Ma
2
Kgm
sNewton
Torque
dT M A 2
2
Kgm
s
Momentum Mv Kgm
s
Angular Momentum
dM V 2Kgm
s
Work (Force)(distance)
Ad 2
Kgmm
sNm Joule
Work T
Using AS arrow math to solve and direction in space problems using the equation 1 2
2 2 2 2 2 2( ) amp and ( ) amp 3
4
2 2 2
2 2
( ) amp
(5 ) 4 3
(5 ) 16 9
(5 ) 25
(5 ) (5 )
2 2(5 ) 4 3
(5 ) 16 9
(5 ) 25
(5 ) (5 )
In SM math Pythagorasrsquo theorem provides both magnitude and direction
In BS math Pythagorasrsquo theorem provides magnitude only
time set equal to zero
23
( ) ( 0 )10 1
x t xx
3( ) ( 0)
10 1W t f
In BS math this is an equation for a traveling wave
must use a square term in order to make the curve symmetric
NO SQUARE TERM IS NECESSARY
In SM math when t is set to ldquo0rdquo
This is a major fault in BS math
x f(x)
-4 0019
-3 0033
-2 0073
-1 0273
0 3000
1 0273
2 0073
3 0033
4 0019 -4 -3 -2 -1 0 1 2 3 400
05
10
15
20
25
30
35
BS wave (must use square term)
snap shot of wave at x = 0
Am
plit
ude
lt to gt lt to gt f(ltgt)
lt -4 0073
lt -3 0097
lt -2 0143
lt -1 0273
origin 0 3000
gt 1 0273
gt 2 0143
gt 3 0097
gt 4 0073lt0073 lt0097 lt0143 lt0273
3000
0273gt 0143gt 0097gt 0073gt
Wave moving
snap shot of wave
Am
plit
ude
lt4 lt3 lt2 lt1 0 1gt 2gt 3gt 4gt
3
2
1
In SM NO squared terms are needed
SPRING
An Action is needed to stretch or compress a spring to a starting point where it can execute cycle motion
tsA (ts to start) is the work required to move a mass attached to a spring a distance (d1) from
its initial rest position RAK 0 the spring constant is the stiffness of the spring
o A is attraction o R is repulsion o
ARRA
00 is for the back and forth motion through the zero initial rest position A
stretched spring goes from ldquoArdquo to ldquo0rdquo to ldquoRrdquo to ldquoArdquo to ldquo0rdquo to ldquoRrdquo d1 is the distance a mass is moved from its initial rest position
0
0 1ts A RA K d
Once ( )mad
is performed to move the mass to position d1 (against the restraint of the spring) a of the ( )mad
is transferred into the spring as potential attractionrepulsion (PAR) There is always
more ( )mad
performed than is stored as PAR When the spring is released from the restraining Ats the PAR starts the oscillation Conservation of ( )mad
requires that the PAR and the motion ( )mad
( )( )madM be a constant
( )amp ( ) constantmadPAR M 02
0 0
1
2A R A RPAR K d
02
( ) 0 0
1( )
2mad A R A RM m v
0 0 02 2 2
0 0 0 1
2 2 21
2 2 21
1 1 1amp
2 2 2
1 1 1amp
2 2 2
amp
A R A R A RK d m v K d
Kd mv Kd
Kd mv Kd
0
1d is frac12 the height from the base to the peak of a cycle
An experiment of moving paper under a pendulum through a small angle filled with sand that can escape through a small hole in the bottom shows that a pendulum cycling back and forth follows a curve that is mathematically analogous to a lsquosinrsquo or lsquocosrsquo
The projection of the curved area onto the straight line exactly duplicate the back and forth motion of CM This allows for the math of lsquosinrsquo or lsquocosrsquo to be used
SM will use lsquosinrdquo math It could use lsquocosrsquo or a combination of both lsquosinrsquo and lsquocosrsquo Using just the lsquosinrsquo math makes the understanding and math easier
0 sin 0 0SM 0 sin157 1SM The diagram at the left shows a spring being stretched from its (zero rest) position-2 and the PAR being released at position-1 There are 4 sections (1 to 2 2 to 3 3 to 4 and 4 to5) that alternate between maximum acceleration and zero velocity to max velocity and zero acceleration
At position-1 there is maximum acceleration and zero velocity o The lsquosinrsquo values from 0o to 157o will be used for velocity (values from 0 to 1) o The lsquosinrsquo values from 157o to 0o will be used for acceleration (values from 1 to 0)
At position-2 there is maximum velocity and zero acceleration Acceleration changes to deceleration
o The lsquosinrsquo values from 157o to 0o will be used for deceleration (values from 1 to 0) o The lsquosinrsquo values from 0o to 157o will be used for acceleration (values from 0 to 1)
At position-3 there is maximum acceleration and zero velocity o The lsquosinrsquo values from 0o to 157o will be used for velocity (values from 0 to 1) o The lsquosinrsquo values from 157o to 0o will be used for acceleration (values from 1 to 0)
At position-4 there is maximum velocity and zero acceleration Acceleration changes to deceleration
o The lsquosinrsquo values from 157o to 0o will be used for deceleration (values from 1 to 0) o The lsquosinrsquo values from 0o to 157o will be used for acceleration (values from 0 to 1)
At position-5 the cycle starts again
2
)539(
cT
MK
spring constant (K) when mass (M) and cycle-time (T) are known
)(
))(89)((
111 md
uKgM
d
MG
d
FK sts
spring constant (K) when mass (M) and amplitude ( d1) are known
2 21( )
Kv d d
m
velocity (v) at any point during the cycle
velocity
dforM
Kdv
max
012 velocity (v2) maximum (when d = 0 at position-2 and 4)
acceleration (a1) maximum when spring constant (K) mass (M) and amplitude (d1) are known
acceleration (a1) maximum when amplitude (d1) and cycle-time (Tc) are known
acceleration (a) at any point during the cycle-time (T)
1
151
286
v
dTc
cycle-time (Tc) when amplitude (d1) and velocity (v1) are known
K
MTc 28651
cycle-time (Tc) when mass (M) and spring constant (K) are known
fTc
1
cycle-time (Tc) when the frequency is known
cTf
1
frequency (f) when the cycle-time (Tc) is known
of cycles
elapsed timef
frequency (f) when the number of cycles in an elapsed time is known
d1 amplitude (distance from rest to start position)
Also frac12 the distance from the base to the peak
3131 oror dM
Ka
251
3131 539
c
oror T
da
2539
cT
da
In December of 2006 I discovered the mathematical error in Einsteinrsquos Special Relativity Einstein used BS math and that math only works in an imaginary world
It cannot describe real symmetrical space
Einsteinrsquos Special Relativity should be abandoned and a theory of real space using SM should be used to replace it
Numerous articles on the Internet show Einsteinrsquos theory to be incorrect but no one in academia is willing to publish the information
Figure 1-10 A moving particle observed from two frames of reference in uniform translation
Einsteinrsquos Math Error in Special Relativity Transformation of Velocity (From Eisenberg- Fundamentals of Modern Physics pages 27-29) Consider the particle shown in figure (1-10) moving with velocity v as
seen in a frame of reference O
dx dy dz
Vx Vy Vzdt dt dt
dx dy dz
V x V y V zdy dt dt
We would like to evaluate the velocity v of the particle as seen in the frame of reference 0 which is itself moving relative to 0 with velocity v Measured in the 0 frame the velocity vector of the particle has com ponents
The same velocity vector as measured in the 0 frame has components
2
2
2
2
2
1
1
x vtx
v
cy y
z z
vxt
ctv
c
Now from equations (1-13) we know that the relation between the primed and the unprimed coordinates and times is
Take the differential of these equations remembering that v is a constant This gives
2
2
2
2
2
1
1
dx vdtdx
v
cdy dy
dz dz
vxdt
cdtv
c
2
2
2 2 2 2 2
2
2
2
11 ( )
1( ) 1
1
1
1
dx vdt
v dxdx vdt v
dx dx vdtc dt dtvdx vdx vdx v dxdt dt dt dt
dt dtc c c c
VxV x
Vx
VxV x
vv
vv
c
x
v
c
cV
( )
coordinate of particle relative to moving primed system
coordinate of particle relative to stationary system
velocity of primed coordina
V x
Vx
te system relative to stationaryv sys
tem
Page 28 Fundamentals of Modern Physics
(1-16)
According to Einstein this equation tells you how to transform the observed velocity from one frame of reference to another frame of reference First we note that as Vc and vc approachzero the equation approaches those which would be derived from the Galilean transformation
2 2
( ) 1 01 1 ( )
Vx v Vx vV
VxV x
Vx v
v cVxx
v
c
v
cc
Another property of this equation is that it is impossible to choose V and v such that Vrsquo the magnitude of the velocity which is seen in the new frame is greater than c
1
2
09 ( 09 ) 18009876543
( 09 ) 1811 (09 )
c c cV c c
cc
c
Example As seen by O particle 1 has velocity 09c in the direction of positive x and particle 2 has velocity 09c in the negative x direction
Einstein equation however predicts To evaluate the velocity of particle 1 with respect to particle 2 we transform from the O frame to the Orsquo frame moving in the negative x direction with velocity v = -09c using equation 1-16 We obtain
JK LOGIC will observe the particles separating at 18c
A confusing INCORRECT and ILLOGICAL answer
left is a dir rightsub of a
1
2
1
09 ( 09 ) 18009876543
181( 09 )1 ( 09 )
The multiplication of dir left time a dir right is ONLY a dir left
09
dir
direction leftdirection rightsubtraction
c c cV c c
cc
c
V
dir
22
22
09 18 18009876543
181( 081)1(09)(09)
1
left rightsub
sub
c c c cc c
ccc
c
Analyzing Einsteinrsquos BS math Equation
Using SM math a logical answer can be obtained
Cannot multiply a subtraction operator times a direction in space
Can only subtract and change its direction
Cannot multiply a direction to the left times a direction to the right
Cannot multiply opposite directions in space
Can only solve for the resultant
rightleft
subtraction
1
subtraction
2
sub of a dir to the left is a dir to the right
1 resultant 0 sub 2
2sub 2
2
09 09
09 amp 091
09 09 18 1
(0)1
(09) amp (09)1
c cV
c cc
c c cV
cc
cc
8
181
cc
SM math predicts a logical answer using logic
21st century scientists state in many articles that Einsteins General Relativity and Quantum Mechanics are incompatible
They are not only incompatible they are both mathematically incorrect
They both use illogical BS math
Any equations in math that use imaginary numbers should be abandoned
Imaginary numbers are only necessary in the current BS math system
This means that Schroumldingers equations should be abandoned
Any equation in quantum mechanics that uses imaginary number should be abandoned
Prof Kanarev has developed logical physics and chemistrytheories without the use of Quantum mechanics or relativity
A partial list of his theories coverbull Mechanics ndash replaces Newtonbull Photon - bull Electrons ndash Protons- Neutronsbull Spectroscopybull Nucleusbull Atoms amp Moleculesbull Electrodynamicsbull Heatbull Curvature of Space ndash replaces Schwarzcchildrsquos equation
Kanarevrsquos Hypothesis
bull Electron align linearly with proton They do NOT spin or form clouds around the nucleus
bull Neutrons are in the center of atom Proton are outside of Neutrons
bull The wavelength of all particles is the experimental radius
bull Double slit experiment explained correctly
bull Heisenbergrsquos inequality explained correctly
bull is the radius of approach of the magnetic lines around the torus of the electron
( 1 137)
bull Why water expands when it freezes
bull Electron clusters form lightning
rbull The meaning of Plancks constant is explained
22 2 kg m
h m f mr f consts
2210254 42m r kgm bull If the mr-constant describes a ring the equations
describe experiments bull If the mr-constant describes a wave the equations do
NOT describe experiments
Kanarevrsquos 5 Equations of Mechanics
Kanarevrsquos 1st Equation Acceleration ampmv maMma R R
Kanarevrsquos 2nd Equation Uniform motion amp mv maMma R R
maM mvR R
Kanarevrsquos 3rd Equation
Kanarevrsquos 4th Equation Equal and opposite reaction
1 21 21 2
12
a m
m a m ama
Kanarevrsquos 5th Equation 1 2 3 Mi M M M Mna R R R R R
The sum of the Actions working on a moving body is never equal to zero
Kanarevrsquos Photon model is a rolling ring
His equations accurately describe known experiments with photonsbull Electromagnetic radiation should be called photon radiationbull Diffraction and interference are the same phenomenabull Correctly explains the double slit experiment without quantum equationsbull Correctly explains the meaning of Heisenbergrsquos inequality
e
2 (2)(3142)(2817 15 ) 10007297352568
R 2426 12 137mag
xp
r m
m
Charge
Magnetic action
Proton
The intensity of the magnetic action near the geometrical center of the proton ring is so great that we theorize that it is the strong nuclear force
(9383 8)(1602 19)8476 14
4 (4)(3142)(1411 26)p
pp
EB T
M
(9395 8)(1602 19)1035 15
4 (4)(3142)(0966 26)N
NN
EB T
M
Neutron
1a Hydrogenavi
The 360-degree circle is a 4400 years old extremely outdated system based on a year of 12 months of 30 days each that is 360 days
SM will use a more logical system that will have a circle of 6283185307o
d
( )
(deg )
d r radians
d r rees
the arc distance around
the circumference of a circle
d
radiusr
2 2
2 2
d r
dv rd t t
v dda rd t t t
1
d
r
( )av tv
The average angular velocity of a body is the rate of its angular displacement
( )0
liminstt
vt
The instantaneous angular velocity is the limit of the ratio
t
as t approaches zero
( )d
av
d
tv
The average circumference velocity of the body is the rate of its distance traveled d divided by the time t
dvv
r
1 dd vdrv
t t t r r
The equation of the relationship between the angular velocity and the velocity of a point on that body
( )o
av
v v v
t ta
The average angular acceleration is the change in the angular velocity
( )0
liminstt
v
ta
The instantaneous angular acceleration is the
limit of the ration of v
t
as t approaches
zero
daa
r
1
1
d
d s d
vva
t r tv a
r t r
The equation of the relationship between the angular acceleration and the acceleration of a point on the body
21
2v a t
If a body with an initial angular velocity ov has a constant
angular acceleration a it will turn through an angle in a time t
1
Straight Line Symbol Units Rotational Symbol Units Distance d m Angular
Distance radians or degrees
Velocity dt
v m
s Angular Velocity
V
t
radians or degrees
s
Acceleration vat
2ms
Angular Acceleration
VA
t
2
radians or degrees
s
Mass M Kg Mass distribution
dM 2Kgm
Force actionA Ma
2
Kgm
sNewton
Torque
dT M A 2
2
Kgm
s
Momentum Mv Kgm
s
Angular Momentum
dM V 2Kgm
s
Work (Force)(distance)
Ad 2
Kgmm
sNm Joule
Work T
Using AS arrow math to solve and direction in space problems using the equation 1 2
2 2 2 2 2 2( ) amp and ( ) amp 3
4
2 2 2
2 2
( ) amp
(5 ) 4 3
(5 ) 16 9
(5 ) 25
(5 ) (5 )
2 2(5 ) 4 3
(5 ) 16 9
(5 ) 25
(5 ) (5 )
In SM math Pythagorasrsquo theorem provides both magnitude and direction
In BS math Pythagorasrsquo theorem provides magnitude only
time set equal to zero
23
( ) ( 0 )10 1
x t xx
3( ) ( 0)
10 1W t f
In BS math this is an equation for a traveling wave
must use a square term in order to make the curve symmetric
NO SQUARE TERM IS NECESSARY
In SM math when t is set to ldquo0rdquo
This is a major fault in BS math
x f(x)
-4 0019
-3 0033
-2 0073
-1 0273
0 3000
1 0273
2 0073
3 0033
4 0019 -4 -3 -2 -1 0 1 2 3 400
05
10
15
20
25
30
35
BS wave (must use square term)
snap shot of wave at x = 0
Am
plit
ude
lt to gt lt to gt f(ltgt)
lt -4 0073
lt -3 0097
lt -2 0143
lt -1 0273
origin 0 3000
gt 1 0273
gt 2 0143
gt 3 0097
gt 4 0073lt0073 lt0097 lt0143 lt0273
3000
0273gt 0143gt 0097gt 0073gt
Wave moving
snap shot of wave
Am
plit
ude
lt4 lt3 lt2 lt1 0 1gt 2gt 3gt 4gt
3
2
1
In SM NO squared terms are needed
SPRING
An Action is needed to stretch or compress a spring to a starting point where it can execute cycle motion
tsA (ts to start) is the work required to move a mass attached to a spring a distance (d1) from
its initial rest position RAK 0 the spring constant is the stiffness of the spring
o A is attraction o R is repulsion o
ARRA
00 is for the back and forth motion through the zero initial rest position A
stretched spring goes from ldquoArdquo to ldquo0rdquo to ldquoRrdquo to ldquoArdquo to ldquo0rdquo to ldquoRrdquo d1 is the distance a mass is moved from its initial rest position
0
0 1ts A RA K d
Once ( )mad
is performed to move the mass to position d1 (against the restraint of the spring) a of the ( )mad
is transferred into the spring as potential attractionrepulsion (PAR) There is always
more ( )mad
performed than is stored as PAR When the spring is released from the restraining Ats the PAR starts the oscillation Conservation of ( )mad
requires that the PAR and the motion ( )mad
( )( )madM be a constant
( )amp ( ) constantmadPAR M 02
0 0
1
2A R A RPAR K d
02
( ) 0 0
1( )
2mad A R A RM m v
0 0 02 2 2
0 0 0 1
2 2 21
2 2 21
1 1 1amp
2 2 2
1 1 1amp
2 2 2
amp
A R A R A RK d m v K d
Kd mv Kd
Kd mv Kd
0
1d is frac12 the height from the base to the peak of a cycle
An experiment of moving paper under a pendulum through a small angle filled with sand that can escape through a small hole in the bottom shows that a pendulum cycling back and forth follows a curve that is mathematically analogous to a lsquosinrsquo or lsquocosrsquo
The projection of the curved area onto the straight line exactly duplicate the back and forth motion of CM This allows for the math of lsquosinrsquo or lsquocosrsquo to be used
SM will use lsquosinrdquo math It could use lsquocosrsquo or a combination of both lsquosinrsquo and lsquocosrsquo Using just the lsquosinrsquo math makes the understanding and math easier
0 sin 0 0SM 0 sin157 1SM The diagram at the left shows a spring being stretched from its (zero rest) position-2 and the PAR being released at position-1 There are 4 sections (1 to 2 2 to 3 3 to 4 and 4 to5) that alternate between maximum acceleration and zero velocity to max velocity and zero acceleration
At position-1 there is maximum acceleration and zero velocity o The lsquosinrsquo values from 0o to 157o will be used for velocity (values from 0 to 1) o The lsquosinrsquo values from 157o to 0o will be used for acceleration (values from 1 to 0)
At position-2 there is maximum velocity and zero acceleration Acceleration changes to deceleration
o The lsquosinrsquo values from 157o to 0o will be used for deceleration (values from 1 to 0) o The lsquosinrsquo values from 0o to 157o will be used for acceleration (values from 0 to 1)
At position-3 there is maximum acceleration and zero velocity o The lsquosinrsquo values from 0o to 157o will be used for velocity (values from 0 to 1) o The lsquosinrsquo values from 157o to 0o will be used for acceleration (values from 1 to 0)
At position-4 there is maximum velocity and zero acceleration Acceleration changes to deceleration
o The lsquosinrsquo values from 157o to 0o will be used for deceleration (values from 1 to 0) o The lsquosinrsquo values from 0o to 157o will be used for acceleration (values from 0 to 1)
At position-5 the cycle starts again
2
)539(
cT
MK
spring constant (K) when mass (M) and cycle-time (T) are known
)(
))(89)((
111 md
uKgM
d
MG
d
FK sts
spring constant (K) when mass (M) and amplitude ( d1) are known
2 21( )
Kv d d
m
velocity (v) at any point during the cycle
velocity
dforM
Kdv
max
012 velocity (v2) maximum (when d = 0 at position-2 and 4)
acceleration (a1) maximum when spring constant (K) mass (M) and amplitude (d1) are known
acceleration (a1) maximum when amplitude (d1) and cycle-time (Tc) are known
acceleration (a) at any point during the cycle-time (T)
1
151
286
v
dTc
cycle-time (Tc) when amplitude (d1) and velocity (v1) are known
K
MTc 28651
cycle-time (Tc) when mass (M) and spring constant (K) are known
fTc
1
cycle-time (Tc) when the frequency is known
cTf
1
frequency (f) when the cycle-time (Tc) is known
of cycles
elapsed timef
frequency (f) when the number of cycles in an elapsed time is known
d1 amplitude (distance from rest to start position)
Also frac12 the distance from the base to the peak
3131 oror dM
Ka
251
3131 539
c
oror T
da
2539
cT
da
In December of 2006 I discovered the mathematical error in Einsteinrsquos Special Relativity Einstein used BS math and that math only works in an imaginary world
It cannot describe real symmetrical space
Einsteinrsquos Special Relativity should be abandoned and a theory of real space using SM should be used to replace it
Numerous articles on the Internet show Einsteinrsquos theory to be incorrect but no one in academia is willing to publish the information
Figure 1-10 A moving particle observed from two frames of reference in uniform translation
Einsteinrsquos Math Error in Special Relativity Transformation of Velocity (From Eisenberg- Fundamentals of Modern Physics pages 27-29) Consider the particle shown in figure (1-10) moving with velocity v as
seen in a frame of reference O
dx dy dz
Vx Vy Vzdt dt dt
dx dy dz
V x V y V zdy dt dt
We would like to evaluate the velocity v of the particle as seen in the frame of reference 0 which is itself moving relative to 0 with velocity v Measured in the 0 frame the velocity vector of the particle has com ponents
The same velocity vector as measured in the 0 frame has components
2
2
2
2
2
1
1
x vtx
v
cy y
z z
vxt
ctv
c
Now from equations (1-13) we know that the relation between the primed and the unprimed coordinates and times is
Take the differential of these equations remembering that v is a constant This gives
2
2
2
2
2
1
1
dx vdtdx
v
cdy dy
dz dz
vxdt
cdtv
c
2
2
2 2 2 2 2
2
2
2
11 ( )
1( ) 1
1
1
1
dx vdt
v dxdx vdt v
dx dx vdtc dt dtvdx vdx vdx v dxdt dt dt dt
dt dtc c c c
VxV x
Vx
VxV x
vv
vv
c
x
v
c
cV
( )
coordinate of particle relative to moving primed system
coordinate of particle relative to stationary system
velocity of primed coordina
V x
Vx
te system relative to stationaryv sys
tem
Page 28 Fundamentals of Modern Physics
(1-16)
According to Einstein this equation tells you how to transform the observed velocity from one frame of reference to another frame of reference First we note that as Vc and vc approachzero the equation approaches those which would be derived from the Galilean transformation
2 2
( ) 1 01 1 ( )
Vx v Vx vV
VxV x
Vx v
v cVxx
v
c
v
cc
Another property of this equation is that it is impossible to choose V and v such that Vrsquo the magnitude of the velocity which is seen in the new frame is greater than c
1
2
09 ( 09 ) 18009876543
( 09 ) 1811 (09 )
c c cV c c
cc
c
Example As seen by O particle 1 has velocity 09c in the direction of positive x and particle 2 has velocity 09c in the negative x direction
Einstein equation however predicts To evaluate the velocity of particle 1 with respect to particle 2 we transform from the O frame to the Orsquo frame moving in the negative x direction with velocity v = -09c using equation 1-16 We obtain
JK LOGIC will observe the particles separating at 18c
A confusing INCORRECT and ILLOGICAL answer
left is a dir rightsub of a
1
2
1
09 ( 09 ) 18009876543
181( 09 )1 ( 09 )
The multiplication of dir left time a dir right is ONLY a dir left
09
dir
direction leftdirection rightsubtraction
c c cV c c
cc
c
V
dir
22
22
09 18 18009876543
181( 081)1(09)(09)
1
left rightsub
sub
c c c cc c
ccc
c
Analyzing Einsteinrsquos BS math Equation
Using SM math a logical answer can be obtained
Cannot multiply a subtraction operator times a direction in space
Can only subtract and change its direction
Cannot multiply a direction to the left times a direction to the right
Cannot multiply opposite directions in space
Can only solve for the resultant
rightleft
subtraction
1
subtraction
2
sub of a dir to the left is a dir to the right
1 resultant 0 sub 2
2sub 2
2
09 09
09 amp 091
09 09 18 1
(0)1
(09) amp (09)1
c cV
c cc
c c cV
cc
cc
8
181
cc
SM math predicts a logical answer using logic
21st century scientists state in many articles that Einsteins General Relativity and Quantum Mechanics are incompatible
They are not only incompatible they are both mathematically incorrect
They both use illogical BS math
Any equations in math that use imaginary numbers should be abandoned
Imaginary numbers are only necessary in the current BS math system
This means that Schroumldingers equations should be abandoned
Any equation in quantum mechanics that uses imaginary number should be abandoned
Prof Kanarev has developed logical physics and chemistrytheories without the use of Quantum mechanics or relativity
A partial list of his theories coverbull Mechanics ndash replaces Newtonbull Photon - bull Electrons ndash Protons- Neutronsbull Spectroscopybull Nucleusbull Atoms amp Moleculesbull Electrodynamicsbull Heatbull Curvature of Space ndash replaces Schwarzcchildrsquos equation
Kanarevrsquos Hypothesis
bull Electron align linearly with proton They do NOT spin or form clouds around the nucleus
bull Neutrons are in the center of atom Proton are outside of Neutrons
bull The wavelength of all particles is the experimental radius
bull Double slit experiment explained correctly
bull Heisenbergrsquos inequality explained correctly
bull is the radius of approach of the magnetic lines around the torus of the electron
( 1 137)
bull Why water expands when it freezes
bull Electron clusters form lightning
rbull The meaning of Plancks constant is explained
22 2 kg m
h m f mr f consts
2210254 42m r kgm bull If the mr-constant describes a ring the equations
describe experiments bull If the mr-constant describes a wave the equations do
NOT describe experiments
Kanarevrsquos 5 Equations of Mechanics
Kanarevrsquos 1st Equation Acceleration ampmv maMma R R
Kanarevrsquos 2nd Equation Uniform motion amp mv maMma R R
maM mvR R
Kanarevrsquos 3rd Equation
Kanarevrsquos 4th Equation Equal and opposite reaction
1 21 21 2
12
a m
m a m ama
Kanarevrsquos 5th Equation 1 2 3 Mi M M M Mna R R R R R
The sum of the Actions working on a moving body is never equal to zero
Kanarevrsquos Photon model is a rolling ring
His equations accurately describe known experiments with photonsbull Electromagnetic radiation should be called photon radiationbull Diffraction and interference are the same phenomenabull Correctly explains the double slit experiment without quantum equationsbull Correctly explains the meaning of Heisenbergrsquos inequality
e
2 (2)(3142)(2817 15 ) 10007297352568
R 2426 12 137mag
xp
r m
m
Charge
Magnetic action
Proton
The intensity of the magnetic action near the geometrical center of the proton ring is so great that we theorize that it is the strong nuclear force
(9383 8)(1602 19)8476 14
4 (4)(3142)(1411 26)p
pp
EB T
M
(9395 8)(1602 19)1035 15
4 (4)(3142)(0966 26)N
NN
EB T
M
Neutron
1a Hydrogenavi
d
( )
(deg )
d r radians
d r rees
the arc distance around
the circumference of a circle
d
radiusr
2 2
2 2
d r
dv rd t t
v dda rd t t t
1
d
r
( )av tv
The average angular velocity of a body is the rate of its angular displacement
( )0
liminstt
vt
The instantaneous angular velocity is the limit of the ratio
t
as t approaches zero
( )d
av
d
tv
The average circumference velocity of the body is the rate of its distance traveled d divided by the time t
dvv
r
1 dd vdrv
t t t r r
The equation of the relationship between the angular velocity and the velocity of a point on that body
( )o
av
v v v
t ta
The average angular acceleration is the change in the angular velocity
( )0
liminstt
v
ta
The instantaneous angular acceleration is the
limit of the ration of v
t
as t approaches
zero
daa
r
1
1
d
d s d
vva
t r tv a
r t r
The equation of the relationship between the angular acceleration and the acceleration of a point on the body
21
2v a t
If a body with an initial angular velocity ov has a constant
angular acceleration a it will turn through an angle in a time t
1
Straight Line Symbol Units Rotational Symbol Units Distance d m Angular
Distance radians or degrees
Velocity dt
v m
s Angular Velocity
V
t
radians or degrees
s
Acceleration vat
2ms
Angular Acceleration
VA
t
2
radians or degrees
s
Mass M Kg Mass distribution
dM 2Kgm
Force actionA Ma
2
Kgm
sNewton
Torque
dT M A 2
2
Kgm
s
Momentum Mv Kgm
s
Angular Momentum
dM V 2Kgm
s
Work (Force)(distance)
Ad 2
Kgmm
sNm Joule
Work T
Using AS arrow math to solve and direction in space problems using the equation 1 2
2 2 2 2 2 2( ) amp and ( ) amp 3
4
2 2 2
2 2
( ) amp
(5 ) 4 3
(5 ) 16 9
(5 ) 25
(5 ) (5 )
2 2(5 ) 4 3
(5 ) 16 9
(5 ) 25
(5 ) (5 )
In SM math Pythagorasrsquo theorem provides both magnitude and direction
In BS math Pythagorasrsquo theorem provides magnitude only
time set equal to zero
23
( ) ( 0 )10 1
x t xx
3( ) ( 0)
10 1W t f
In BS math this is an equation for a traveling wave
must use a square term in order to make the curve symmetric
NO SQUARE TERM IS NECESSARY
In SM math when t is set to ldquo0rdquo
This is a major fault in BS math
x f(x)
-4 0019
-3 0033
-2 0073
-1 0273
0 3000
1 0273
2 0073
3 0033
4 0019 -4 -3 -2 -1 0 1 2 3 400
05
10
15
20
25
30
35
BS wave (must use square term)
snap shot of wave at x = 0
Am
plit
ude
lt to gt lt to gt f(ltgt)
lt -4 0073
lt -3 0097
lt -2 0143
lt -1 0273
origin 0 3000
gt 1 0273
gt 2 0143
gt 3 0097
gt 4 0073lt0073 lt0097 lt0143 lt0273
3000
0273gt 0143gt 0097gt 0073gt
Wave moving
snap shot of wave
Am
plit
ude
lt4 lt3 lt2 lt1 0 1gt 2gt 3gt 4gt
3
2
1
In SM NO squared terms are needed
SPRING
An Action is needed to stretch or compress a spring to a starting point where it can execute cycle motion
tsA (ts to start) is the work required to move a mass attached to a spring a distance (d1) from
its initial rest position RAK 0 the spring constant is the stiffness of the spring
o A is attraction o R is repulsion o
ARRA
00 is for the back and forth motion through the zero initial rest position A
stretched spring goes from ldquoArdquo to ldquo0rdquo to ldquoRrdquo to ldquoArdquo to ldquo0rdquo to ldquoRrdquo d1 is the distance a mass is moved from its initial rest position
0
0 1ts A RA K d
Once ( )mad
is performed to move the mass to position d1 (against the restraint of the spring) a of the ( )mad
is transferred into the spring as potential attractionrepulsion (PAR) There is always
more ( )mad
performed than is stored as PAR When the spring is released from the restraining Ats the PAR starts the oscillation Conservation of ( )mad
requires that the PAR and the motion ( )mad
( )( )madM be a constant
( )amp ( ) constantmadPAR M 02
0 0
1
2A R A RPAR K d
02
( ) 0 0
1( )
2mad A R A RM m v
0 0 02 2 2
0 0 0 1
2 2 21
2 2 21
1 1 1amp
2 2 2
1 1 1amp
2 2 2
amp
A R A R A RK d m v K d
Kd mv Kd
Kd mv Kd
0
1d is frac12 the height from the base to the peak of a cycle
An experiment of moving paper under a pendulum through a small angle filled with sand that can escape through a small hole in the bottom shows that a pendulum cycling back and forth follows a curve that is mathematically analogous to a lsquosinrsquo or lsquocosrsquo
The projection of the curved area onto the straight line exactly duplicate the back and forth motion of CM This allows for the math of lsquosinrsquo or lsquocosrsquo to be used
SM will use lsquosinrdquo math It could use lsquocosrsquo or a combination of both lsquosinrsquo and lsquocosrsquo Using just the lsquosinrsquo math makes the understanding and math easier
0 sin 0 0SM 0 sin157 1SM The diagram at the left shows a spring being stretched from its (zero rest) position-2 and the PAR being released at position-1 There are 4 sections (1 to 2 2 to 3 3 to 4 and 4 to5) that alternate between maximum acceleration and zero velocity to max velocity and zero acceleration
At position-1 there is maximum acceleration and zero velocity o The lsquosinrsquo values from 0o to 157o will be used for velocity (values from 0 to 1) o The lsquosinrsquo values from 157o to 0o will be used for acceleration (values from 1 to 0)
At position-2 there is maximum velocity and zero acceleration Acceleration changes to deceleration
o The lsquosinrsquo values from 157o to 0o will be used for deceleration (values from 1 to 0) o The lsquosinrsquo values from 0o to 157o will be used for acceleration (values from 0 to 1)
At position-3 there is maximum acceleration and zero velocity o The lsquosinrsquo values from 0o to 157o will be used for velocity (values from 0 to 1) o The lsquosinrsquo values from 157o to 0o will be used for acceleration (values from 1 to 0)
At position-4 there is maximum velocity and zero acceleration Acceleration changes to deceleration
o The lsquosinrsquo values from 157o to 0o will be used for deceleration (values from 1 to 0) o The lsquosinrsquo values from 0o to 157o will be used for acceleration (values from 0 to 1)
At position-5 the cycle starts again
2
)539(
cT
MK
spring constant (K) when mass (M) and cycle-time (T) are known
)(
))(89)((
111 md
uKgM
d
MG
d
FK sts
spring constant (K) when mass (M) and amplitude ( d1) are known
2 21( )
Kv d d
m
velocity (v) at any point during the cycle
velocity
dforM
Kdv
max
012 velocity (v2) maximum (when d = 0 at position-2 and 4)
acceleration (a1) maximum when spring constant (K) mass (M) and amplitude (d1) are known
acceleration (a1) maximum when amplitude (d1) and cycle-time (Tc) are known
acceleration (a) at any point during the cycle-time (T)
1
151
286
v
dTc
cycle-time (Tc) when amplitude (d1) and velocity (v1) are known
K
MTc 28651
cycle-time (Tc) when mass (M) and spring constant (K) are known
fTc
1
cycle-time (Tc) when the frequency is known
cTf
1
frequency (f) when the cycle-time (Tc) is known
of cycles
elapsed timef
frequency (f) when the number of cycles in an elapsed time is known
d1 amplitude (distance from rest to start position)
Also frac12 the distance from the base to the peak
3131 oror dM
Ka
251
3131 539
c
oror T
da
2539
cT
da
In December of 2006 I discovered the mathematical error in Einsteinrsquos Special Relativity Einstein used BS math and that math only works in an imaginary world
It cannot describe real symmetrical space
Einsteinrsquos Special Relativity should be abandoned and a theory of real space using SM should be used to replace it
Numerous articles on the Internet show Einsteinrsquos theory to be incorrect but no one in academia is willing to publish the information
Figure 1-10 A moving particle observed from two frames of reference in uniform translation
Einsteinrsquos Math Error in Special Relativity Transformation of Velocity (From Eisenberg- Fundamentals of Modern Physics pages 27-29) Consider the particle shown in figure (1-10) moving with velocity v as
seen in a frame of reference O
dx dy dz
Vx Vy Vzdt dt dt
dx dy dz
V x V y V zdy dt dt
We would like to evaluate the velocity v of the particle as seen in the frame of reference 0 which is itself moving relative to 0 with velocity v Measured in the 0 frame the velocity vector of the particle has com ponents
The same velocity vector as measured in the 0 frame has components
2
2
2
2
2
1
1
x vtx
v
cy y
z z
vxt
ctv
c
Now from equations (1-13) we know that the relation between the primed and the unprimed coordinates and times is
Take the differential of these equations remembering that v is a constant This gives
2
2
2
2
2
1
1
dx vdtdx
v
cdy dy
dz dz
vxdt
cdtv
c
2
2
2 2 2 2 2
2
2
2
11 ( )
1( ) 1
1
1
1
dx vdt
v dxdx vdt v
dx dx vdtc dt dtvdx vdx vdx v dxdt dt dt dt
dt dtc c c c
VxV x
Vx
VxV x
vv
vv
c
x
v
c
cV
( )
coordinate of particle relative to moving primed system
coordinate of particle relative to stationary system
velocity of primed coordina
V x
Vx
te system relative to stationaryv sys
tem
Page 28 Fundamentals of Modern Physics
(1-16)
According to Einstein this equation tells you how to transform the observed velocity from one frame of reference to another frame of reference First we note that as Vc and vc approachzero the equation approaches those which would be derived from the Galilean transformation
2 2
( ) 1 01 1 ( )
Vx v Vx vV
VxV x
Vx v
v cVxx
v
c
v
cc
Another property of this equation is that it is impossible to choose V and v such that Vrsquo the magnitude of the velocity which is seen in the new frame is greater than c
1
2
09 ( 09 ) 18009876543
( 09 ) 1811 (09 )
c c cV c c
cc
c
Example As seen by O particle 1 has velocity 09c in the direction of positive x and particle 2 has velocity 09c in the negative x direction
Einstein equation however predicts To evaluate the velocity of particle 1 with respect to particle 2 we transform from the O frame to the Orsquo frame moving in the negative x direction with velocity v = -09c using equation 1-16 We obtain
JK LOGIC will observe the particles separating at 18c
A confusing INCORRECT and ILLOGICAL answer
left is a dir rightsub of a
1
2
1
09 ( 09 ) 18009876543
181( 09 )1 ( 09 )
The multiplication of dir left time a dir right is ONLY a dir left
09
dir
direction leftdirection rightsubtraction
c c cV c c
cc
c
V
dir
22
22
09 18 18009876543
181( 081)1(09)(09)
1
left rightsub
sub
c c c cc c
ccc
c
Analyzing Einsteinrsquos BS math Equation
Using SM math a logical answer can be obtained
Cannot multiply a subtraction operator times a direction in space
Can only subtract and change its direction
Cannot multiply a direction to the left times a direction to the right
Cannot multiply opposite directions in space
Can only solve for the resultant
rightleft
subtraction
1
subtraction
2
sub of a dir to the left is a dir to the right
1 resultant 0 sub 2
2sub 2
2
09 09
09 amp 091
09 09 18 1
(0)1
(09) amp (09)1
c cV
c cc
c c cV
cc
cc
8
181
cc
SM math predicts a logical answer using logic
21st century scientists state in many articles that Einsteins General Relativity and Quantum Mechanics are incompatible
They are not only incompatible they are both mathematically incorrect
They both use illogical BS math
Any equations in math that use imaginary numbers should be abandoned
Imaginary numbers are only necessary in the current BS math system
This means that Schroumldingers equations should be abandoned
Any equation in quantum mechanics that uses imaginary number should be abandoned
Prof Kanarev has developed logical physics and chemistrytheories without the use of Quantum mechanics or relativity
A partial list of his theories coverbull Mechanics ndash replaces Newtonbull Photon - bull Electrons ndash Protons- Neutronsbull Spectroscopybull Nucleusbull Atoms amp Moleculesbull Electrodynamicsbull Heatbull Curvature of Space ndash replaces Schwarzcchildrsquos equation
Kanarevrsquos Hypothesis
bull Electron align linearly with proton They do NOT spin or form clouds around the nucleus
bull Neutrons are in the center of atom Proton are outside of Neutrons
bull The wavelength of all particles is the experimental radius
bull Double slit experiment explained correctly
bull Heisenbergrsquos inequality explained correctly
bull is the radius of approach of the magnetic lines around the torus of the electron
( 1 137)
bull Why water expands when it freezes
bull Electron clusters form lightning
rbull The meaning of Plancks constant is explained
22 2 kg m
h m f mr f consts
2210254 42m r kgm bull If the mr-constant describes a ring the equations
describe experiments bull If the mr-constant describes a wave the equations do
NOT describe experiments
Kanarevrsquos 5 Equations of Mechanics
Kanarevrsquos 1st Equation Acceleration ampmv maMma R R
Kanarevrsquos 2nd Equation Uniform motion amp mv maMma R R
maM mvR R
Kanarevrsquos 3rd Equation
Kanarevrsquos 4th Equation Equal and opposite reaction
1 21 21 2
12
a m
m a m ama
Kanarevrsquos 5th Equation 1 2 3 Mi M M M Mna R R R R R
The sum of the Actions working on a moving body is never equal to zero
Kanarevrsquos Photon model is a rolling ring
His equations accurately describe known experiments with photonsbull Electromagnetic radiation should be called photon radiationbull Diffraction and interference are the same phenomenabull Correctly explains the double slit experiment without quantum equationsbull Correctly explains the meaning of Heisenbergrsquos inequality
e
2 (2)(3142)(2817 15 ) 10007297352568
R 2426 12 137mag
xp
r m
m
Charge
Magnetic action
Proton
The intensity of the magnetic action near the geometrical center of the proton ring is so great that we theorize that it is the strong nuclear force
(9383 8)(1602 19)8476 14
4 (4)(3142)(1411 26)p
pp
EB T
M
(9395 8)(1602 19)1035 15
4 (4)(3142)(0966 26)N
NN
EB T
M
Neutron
1a Hydrogenavi
( )
(deg )
d r radians
d r rees
the arc distance around
the circumference of a circle
d
radiusr
2 2
2 2
d r
dv rd t t
v dda rd t t t
1
d
r
( )av tv
The average angular velocity of a body is the rate of its angular displacement
( )0
liminstt
vt
The instantaneous angular velocity is the limit of the ratio
t
as t approaches zero
( )d
av
d
tv
The average circumference velocity of the body is the rate of its distance traveled d divided by the time t
dvv
r
1 dd vdrv
t t t r r
The equation of the relationship between the angular velocity and the velocity of a point on that body
( )o
av
v v v
t ta
The average angular acceleration is the change in the angular velocity
( )0
liminstt
v
ta
The instantaneous angular acceleration is the
limit of the ration of v
t
as t approaches
zero
daa
r
1
1
d
d s d
vva
t r tv a
r t r
The equation of the relationship between the angular acceleration and the acceleration of a point on the body
21
2v a t
If a body with an initial angular velocity ov has a constant
angular acceleration a it will turn through an angle in a time t
1
Straight Line Symbol Units Rotational Symbol Units Distance d m Angular
Distance radians or degrees
Velocity dt
v m
s Angular Velocity
V
t
radians or degrees
s
Acceleration vat
2ms
Angular Acceleration
VA
t
2
radians or degrees
s
Mass M Kg Mass distribution
dM 2Kgm
Force actionA Ma
2
Kgm
sNewton
Torque
dT M A 2
2
Kgm
s
Momentum Mv Kgm
s
Angular Momentum
dM V 2Kgm
s
Work (Force)(distance)
Ad 2
Kgmm
sNm Joule
Work T
Using AS arrow math to solve and direction in space problems using the equation 1 2
2 2 2 2 2 2( ) amp and ( ) amp 3
4
2 2 2
2 2
( ) amp
(5 ) 4 3
(5 ) 16 9
(5 ) 25
(5 ) (5 )
2 2(5 ) 4 3
(5 ) 16 9
(5 ) 25
(5 ) (5 )
In SM math Pythagorasrsquo theorem provides both magnitude and direction
In BS math Pythagorasrsquo theorem provides magnitude only
time set equal to zero
23
( ) ( 0 )10 1
x t xx
3( ) ( 0)
10 1W t f
In BS math this is an equation for a traveling wave
must use a square term in order to make the curve symmetric
NO SQUARE TERM IS NECESSARY
In SM math when t is set to ldquo0rdquo
This is a major fault in BS math
x f(x)
-4 0019
-3 0033
-2 0073
-1 0273
0 3000
1 0273
2 0073
3 0033
4 0019 -4 -3 -2 -1 0 1 2 3 400
05
10
15
20
25
30
35
BS wave (must use square term)
snap shot of wave at x = 0
Am
plit
ude
lt to gt lt to gt f(ltgt)
lt -4 0073
lt -3 0097
lt -2 0143
lt -1 0273
origin 0 3000
gt 1 0273
gt 2 0143
gt 3 0097
gt 4 0073lt0073 lt0097 lt0143 lt0273
3000
0273gt 0143gt 0097gt 0073gt
Wave moving
snap shot of wave
Am
plit
ude
lt4 lt3 lt2 lt1 0 1gt 2gt 3gt 4gt
3
2
1
In SM NO squared terms are needed
SPRING
An Action is needed to stretch or compress a spring to a starting point where it can execute cycle motion
tsA (ts to start) is the work required to move a mass attached to a spring a distance (d1) from
its initial rest position RAK 0 the spring constant is the stiffness of the spring
o A is attraction o R is repulsion o
ARRA
00 is for the back and forth motion through the zero initial rest position A
stretched spring goes from ldquoArdquo to ldquo0rdquo to ldquoRrdquo to ldquoArdquo to ldquo0rdquo to ldquoRrdquo d1 is the distance a mass is moved from its initial rest position
0
0 1ts A RA K d
Once ( )mad
is performed to move the mass to position d1 (against the restraint of the spring) a of the ( )mad
is transferred into the spring as potential attractionrepulsion (PAR) There is always
more ( )mad
performed than is stored as PAR When the spring is released from the restraining Ats the PAR starts the oscillation Conservation of ( )mad
requires that the PAR and the motion ( )mad
( )( )madM be a constant
( )amp ( ) constantmadPAR M 02
0 0
1
2A R A RPAR K d
02
( ) 0 0
1( )
2mad A R A RM m v
0 0 02 2 2
0 0 0 1
2 2 21
2 2 21
1 1 1amp
2 2 2
1 1 1amp
2 2 2
amp
A R A R A RK d m v K d
Kd mv Kd
Kd mv Kd
0
1d is frac12 the height from the base to the peak of a cycle
An experiment of moving paper under a pendulum through a small angle filled with sand that can escape through a small hole in the bottom shows that a pendulum cycling back and forth follows a curve that is mathematically analogous to a lsquosinrsquo or lsquocosrsquo
The projection of the curved area onto the straight line exactly duplicate the back and forth motion of CM This allows for the math of lsquosinrsquo or lsquocosrsquo to be used
SM will use lsquosinrdquo math It could use lsquocosrsquo or a combination of both lsquosinrsquo and lsquocosrsquo Using just the lsquosinrsquo math makes the understanding and math easier
0 sin 0 0SM 0 sin157 1SM The diagram at the left shows a spring being stretched from its (zero rest) position-2 and the PAR being released at position-1 There are 4 sections (1 to 2 2 to 3 3 to 4 and 4 to5) that alternate between maximum acceleration and zero velocity to max velocity and zero acceleration
At position-1 there is maximum acceleration and zero velocity o The lsquosinrsquo values from 0o to 157o will be used for velocity (values from 0 to 1) o The lsquosinrsquo values from 157o to 0o will be used for acceleration (values from 1 to 0)
At position-2 there is maximum velocity and zero acceleration Acceleration changes to deceleration
o The lsquosinrsquo values from 157o to 0o will be used for deceleration (values from 1 to 0) o The lsquosinrsquo values from 0o to 157o will be used for acceleration (values from 0 to 1)
At position-3 there is maximum acceleration and zero velocity o The lsquosinrsquo values from 0o to 157o will be used for velocity (values from 0 to 1) o The lsquosinrsquo values from 157o to 0o will be used for acceleration (values from 1 to 0)
At position-4 there is maximum velocity and zero acceleration Acceleration changes to deceleration
o The lsquosinrsquo values from 157o to 0o will be used for deceleration (values from 1 to 0) o The lsquosinrsquo values from 0o to 157o will be used for acceleration (values from 0 to 1)
At position-5 the cycle starts again
2
)539(
cT
MK
spring constant (K) when mass (M) and cycle-time (T) are known
)(
))(89)((
111 md
uKgM
d
MG
d
FK sts
spring constant (K) when mass (M) and amplitude ( d1) are known
2 21( )
Kv d d
m
velocity (v) at any point during the cycle
velocity
dforM
Kdv
max
012 velocity (v2) maximum (when d = 0 at position-2 and 4)
acceleration (a1) maximum when spring constant (K) mass (M) and amplitude (d1) are known
acceleration (a1) maximum when amplitude (d1) and cycle-time (Tc) are known
acceleration (a) at any point during the cycle-time (T)
1
151
286
v
dTc
cycle-time (Tc) when amplitude (d1) and velocity (v1) are known
K
MTc 28651
cycle-time (Tc) when mass (M) and spring constant (K) are known
fTc
1
cycle-time (Tc) when the frequency is known
cTf
1
frequency (f) when the cycle-time (Tc) is known
of cycles
elapsed timef
frequency (f) when the number of cycles in an elapsed time is known
d1 amplitude (distance from rest to start position)
Also frac12 the distance from the base to the peak
3131 oror dM
Ka
251
3131 539
c
oror T
da
2539
cT
da
In December of 2006 I discovered the mathematical error in Einsteinrsquos Special Relativity Einstein used BS math and that math only works in an imaginary world
It cannot describe real symmetrical space
Einsteinrsquos Special Relativity should be abandoned and a theory of real space using SM should be used to replace it
Numerous articles on the Internet show Einsteinrsquos theory to be incorrect but no one in academia is willing to publish the information
Figure 1-10 A moving particle observed from two frames of reference in uniform translation
Einsteinrsquos Math Error in Special Relativity Transformation of Velocity (From Eisenberg- Fundamentals of Modern Physics pages 27-29) Consider the particle shown in figure (1-10) moving with velocity v as
seen in a frame of reference O
dx dy dz
Vx Vy Vzdt dt dt
dx dy dz
V x V y V zdy dt dt
We would like to evaluate the velocity v of the particle as seen in the frame of reference 0 which is itself moving relative to 0 with velocity v Measured in the 0 frame the velocity vector of the particle has com ponents
The same velocity vector as measured in the 0 frame has components
2
2
2
2
2
1
1
x vtx
v
cy y
z z
vxt
ctv
c
Now from equations (1-13) we know that the relation between the primed and the unprimed coordinates and times is
Take the differential of these equations remembering that v is a constant This gives
2
2
2
2
2
1
1
dx vdtdx
v
cdy dy
dz dz
vxdt
cdtv
c
2
2
2 2 2 2 2
2
2
2
11 ( )
1( ) 1
1
1
1
dx vdt
v dxdx vdt v
dx dx vdtc dt dtvdx vdx vdx v dxdt dt dt dt
dt dtc c c c
VxV x
Vx
VxV x
vv
vv
c
x
v
c
cV
( )
coordinate of particle relative to moving primed system
coordinate of particle relative to stationary system
velocity of primed coordina
V x
Vx
te system relative to stationaryv sys
tem
Page 28 Fundamentals of Modern Physics
(1-16)
According to Einstein this equation tells you how to transform the observed velocity from one frame of reference to another frame of reference First we note that as Vc and vc approachzero the equation approaches those which would be derived from the Galilean transformation
2 2
( ) 1 01 1 ( )
Vx v Vx vV
VxV x
Vx v
v cVxx
v
c
v
cc
Another property of this equation is that it is impossible to choose V and v such that Vrsquo the magnitude of the velocity which is seen in the new frame is greater than c
1
2
09 ( 09 ) 18009876543
( 09 ) 1811 (09 )
c c cV c c
cc
c
Example As seen by O particle 1 has velocity 09c in the direction of positive x and particle 2 has velocity 09c in the negative x direction
Einstein equation however predicts To evaluate the velocity of particle 1 with respect to particle 2 we transform from the O frame to the Orsquo frame moving in the negative x direction with velocity v = -09c using equation 1-16 We obtain
JK LOGIC will observe the particles separating at 18c
A confusing INCORRECT and ILLOGICAL answer
left is a dir rightsub of a
1
2
1
09 ( 09 ) 18009876543
181( 09 )1 ( 09 )
The multiplication of dir left time a dir right is ONLY a dir left
09
dir
direction leftdirection rightsubtraction
c c cV c c
cc
c
V
dir
22
22
09 18 18009876543
181( 081)1(09)(09)
1
left rightsub
sub
c c c cc c
ccc
c
Analyzing Einsteinrsquos BS math Equation
Using SM math a logical answer can be obtained
Cannot multiply a subtraction operator times a direction in space
Can only subtract and change its direction
Cannot multiply a direction to the left times a direction to the right
Cannot multiply opposite directions in space
Can only solve for the resultant
rightleft
subtraction
1
subtraction
2
sub of a dir to the left is a dir to the right
1 resultant 0 sub 2
2sub 2
2
09 09
09 amp 091
09 09 18 1
(0)1
(09) amp (09)1
c cV
c cc
c c cV
cc
cc
8
181
cc
SM math predicts a logical answer using logic
21st century scientists state in many articles that Einsteins General Relativity and Quantum Mechanics are incompatible
They are not only incompatible they are both mathematically incorrect
They both use illogical BS math
Any equations in math that use imaginary numbers should be abandoned
Imaginary numbers are only necessary in the current BS math system
This means that Schroumldingers equations should be abandoned
Any equation in quantum mechanics that uses imaginary number should be abandoned
Prof Kanarev has developed logical physics and chemistrytheories without the use of Quantum mechanics or relativity
A partial list of his theories coverbull Mechanics ndash replaces Newtonbull Photon - bull Electrons ndash Protons- Neutronsbull Spectroscopybull Nucleusbull Atoms amp Moleculesbull Electrodynamicsbull Heatbull Curvature of Space ndash replaces Schwarzcchildrsquos equation
Kanarevrsquos Hypothesis
bull Electron align linearly with proton They do NOT spin or form clouds around the nucleus
bull Neutrons are in the center of atom Proton are outside of Neutrons
bull The wavelength of all particles is the experimental radius
bull Double slit experiment explained correctly
bull Heisenbergrsquos inequality explained correctly
bull is the radius of approach of the magnetic lines around the torus of the electron
( 1 137)
bull Why water expands when it freezes
bull Electron clusters form lightning
rbull The meaning of Plancks constant is explained
22 2 kg m
h m f mr f consts
2210254 42m r kgm bull If the mr-constant describes a ring the equations
describe experiments bull If the mr-constant describes a wave the equations do
NOT describe experiments
Kanarevrsquos 5 Equations of Mechanics
Kanarevrsquos 1st Equation Acceleration ampmv maMma R R
Kanarevrsquos 2nd Equation Uniform motion amp mv maMma R R
maM mvR R
Kanarevrsquos 3rd Equation
Kanarevrsquos 4th Equation Equal and opposite reaction
1 21 21 2
12
a m
m a m ama
Kanarevrsquos 5th Equation 1 2 3 Mi M M M Mna R R R R R
The sum of the Actions working on a moving body is never equal to zero
Kanarevrsquos Photon model is a rolling ring
His equations accurately describe known experiments with photonsbull Electromagnetic radiation should be called photon radiationbull Diffraction and interference are the same phenomenabull Correctly explains the double slit experiment without quantum equationsbull Correctly explains the meaning of Heisenbergrsquos inequality
e
2 (2)(3142)(2817 15 ) 10007297352568
R 2426 12 137mag
xp
r m
m
Charge
Magnetic action
Proton
The intensity of the magnetic action near the geometrical center of the proton ring is so great that we theorize that it is the strong nuclear force
(9383 8)(1602 19)8476 14
4 (4)(3142)(1411 26)p
pp
EB T
M
(9395 8)(1602 19)1035 15
4 (4)(3142)(0966 26)N
NN
EB T
M
Neutron
1a Hydrogenavi
( )av tv
The average angular velocity of a body is the rate of its angular displacement
( )0
liminstt
vt
The instantaneous angular velocity is the limit of the ratio
t
as t approaches zero
( )d
av
d
tv
The average circumference velocity of the body is the rate of its distance traveled d divided by the time t
dvv
r
1 dd vdrv
t t t r r
The equation of the relationship between the angular velocity and the velocity of a point on that body
( )o
av
v v v
t ta
The average angular acceleration is the change in the angular velocity
( )0
liminstt
v
ta
The instantaneous angular acceleration is the
limit of the ration of v
t
as t approaches
zero
daa
r
1
1
d
d s d
vva
t r tv a
r t r
The equation of the relationship between the angular acceleration and the acceleration of a point on the body
21
2v a t
If a body with an initial angular velocity ov has a constant
angular acceleration a it will turn through an angle in a time t
1
Straight Line Symbol Units Rotational Symbol Units Distance d m Angular
Distance radians or degrees
Velocity dt
v m
s Angular Velocity
V
t
radians or degrees
s
Acceleration vat
2ms
Angular Acceleration
VA
t
2
radians or degrees
s
Mass M Kg Mass distribution
dM 2Kgm
Force actionA Ma
2
Kgm
sNewton
Torque
dT M A 2
2
Kgm
s
Momentum Mv Kgm
s
Angular Momentum
dM V 2Kgm
s
Work (Force)(distance)
Ad 2
Kgmm
sNm Joule
Work T
Using AS arrow math to solve and direction in space problems using the equation 1 2
2 2 2 2 2 2( ) amp and ( ) amp 3
4
2 2 2
2 2
( ) amp
(5 ) 4 3
(5 ) 16 9
(5 ) 25
(5 ) (5 )
2 2(5 ) 4 3
(5 ) 16 9
(5 ) 25
(5 ) (5 )
In SM math Pythagorasrsquo theorem provides both magnitude and direction
In BS math Pythagorasrsquo theorem provides magnitude only
time set equal to zero
23
( ) ( 0 )10 1
x t xx
3( ) ( 0)
10 1W t f
In BS math this is an equation for a traveling wave
must use a square term in order to make the curve symmetric
NO SQUARE TERM IS NECESSARY
In SM math when t is set to ldquo0rdquo
This is a major fault in BS math
x f(x)
-4 0019
-3 0033
-2 0073
-1 0273
0 3000
1 0273
2 0073
3 0033
4 0019 -4 -3 -2 -1 0 1 2 3 400
05
10
15
20
25
30
35
BS wave (must use square term)
snap shot of wave at x = 0
Am
plit
ude
lt to gt lt to gt f(ltgt)
lt -4 0073
lt -3 0097
lt -2 0143
lt -1 0273
origin 0 3000
gt 1 0273
gt 2 0143
gt 3 0097
gt 4 0073lt0073 lt0097 lt0143 lt0273
3000
0273gt 0143gt 0097gt 0073gt
Wave moving
snap shot of wave
Am
plit
ude
lt4 lt3 lt2 lt1 0 1gt 2gt 3gt 4gt
3
2
1
In SM NO squared terms are needed
SPRING
An Action is needed to stretch or compress a spring to a starting point where it can execute cycle motion
tsA (ts to start) is the work required to move a mass attached to a spring a distance (d1) from
its initial rest position RAK 0 the spring constant is the stiffness of the spring
o A is attraction o R is repulsion o
ARRA
00 is for the back and forth motion through the zero initial rest position A
stretched spring goes from ldquoArdquo to ldquo0rdquo to ldquoRrdquo to ldquoArdquo to ldquo0rdquo to ldquoRrdquo d1 is the distance a mass is moved from its initial rest position
0
0 1ts A RA K d
Once ( )mad
is performed to move the mass to position d1 (against the restraint of the spring) a of the ( )mad
is transferred into the spring as potential attractionrepulsion (PAR) There is always
more ( )mad
performed than is stored as PAR When the spring is released from the restraining Ats the PAR starts the oscillation Conservation of ( )mad
requires that the PAR and the motion ( )mad
( )( )madM be a constant
( )amp ( ) constantmadPAR M 02
0 0
1
2A R A RPAR K d
02
( ) 0 0
1( )
2mad A R A RM m v
0 0 02 2 2
0 0 0 1
2 2 21
2 2 21
1 1 1amp
2 2 2
1 1 1amp
2 2 2
amp
A R A R A RK d m v K d
Kd mv Kd
Kd mv Kd
0
1d is frac12 the height from the base to the peak of a cycle
An experiment of moving paper under a pendulum through a small angle filled with sand that can escape through a small hole in the bottom shows that a pendulum cycling back and forth follows a curve that is mathematically analogous to a lsquosinrsquo or lsquocosrsquo
The projection of the curved area onto the straight line exactly duplicate the back and forth motion of CM This allows for the math of lsquosinrsquo or lsquocosrsquo to be used
SM will use lsquosinrdquo math It could use lsquocosrsquo or a combination of both lsquosinrsquo and lsquocosrsquo Using just the lsquosinrsquo math makes the understanding and math easier
0 sin 0 0SM 0 sin157 1SM The diagram at the left shows a spring being stretched from its (zero rest) position-2 and the PAR being released at position-1 There are 4 sections (1 to 2 2 to 3 3 to 4 and 4 to5) that alternate between maximum acceleration and zero velocity to max velocity and zero acceleration
At position-1 there is maximum acceleration and zero velocity o The lsquosinrsquo values from 0o to 157o will be used for velocity (values from 0 to 1) o The lsquosinrsquo values from 157o to 0o will be used for acceleration (values from 1 to 0)
At position-2 there is maximum velocity and zero acceleration Acceleration changes to deceleration
o The lsquosinrsquo values from 157o to 0o will be used for deceleration (values from 1 to 0) o The lsquosinrsquo values from 0o to 157o will be used for acceleration (values from 0 to 1)
At position-3 there is maximum acceleration and zero velocity o The lsquosinrsquo values from 0o to 157o will be used for velocity (values from 0 to 1) o The lsquosinrsquo values from 157o to 0o will be used for acceleration (values from 1 to 0)
At position-4 there is maximum velocity and zero acceleration Acceleration changes to deceleration
o The lsquosinrsquo values from 157o to 0o will be used for deceleration (values from 1 to 0) o The lsquosinrsquo values from 0o to 157o will be used for acceleration (values from 0 to 1)
At position-5 the cycle starts again
2
)539(
cT
MK
spring constant (K) when mass (M) and cycle-time (T) are known
)(
))(89)((
111 md
uKgM
d
MG
d
FK sts
spring constant (K) when mass (M) and amplitude ( d1) are known
2 21( )
Kv d d
m
velocity (v) at any point during the cycle
velocity
dforM
Kdv
max
012 velocity (v2) maximum (when d = 0 at position-2 and 4)
acceleration (a1) maximum when spring constant (K) mass (M) and amplitude (d1) are known
acceleration (a1) maximum when amplitude (d1) and cycle-time (Tc) are known
acceleration (a) at any point during the cycle-time (T)
1
151
286
v
dTc
cycle-time (Tc) when amplitude (d1) and velocity (v1) are known
K
MTc 28651
cycle-time (Tc) when mass (M) and spring constant (K) are known
fTc
1
cycle-time (Tc) when the frequency is known
cTf
1
frequency (f) when the cycle-time (Tc) is known
of cycles
elapsed timef
frequency (f) when the number of cycles in an elapsed time is known
d1 amplitude (distance from rest to start position)
Also frac12 the distance from the base to the peak
3131 oror dM
Ka
251
3131 539
c
oror T
da
2539
cT
da
In December of 2006 I discovered the mathematical error in Einsteinrsquos Special Relativity Einstein used BS math and that math only works in an imaginary world
It cannot describe real symmetrical space
Einsteinrsquos Special Relativity should be abandoned and a theory of real space using SM should be used to replace it
Numerous articles on the Internet show Einsteinrsquos theory to be incorrect but no one in academia is willing to publish the information
Figure 1-10 A moving particle observed from two frames of reference in uniform translation
Einsteinrsquos Math Error in Special Relativity Transformation of Velocity (From Eisenberg- Fundamentals of Modern Physics pages 27-29) Consider the particle shown in figure (1-10) moving with velocity v as
seen in a frame of reference O
dx dy dz
Vx Vy Vzdt dt dt
dx dy dz
V x V y V zdy dt dt
We would like to evaluate the velocity v of the particle as seen in the frame of reference 0 which is itself moving relative to 0 with velocity v Measured in the 0 frame the velocity vector of the particle has com ponents
The same velocity vector as measured in the 0 frame has components
2
2
2
2
2
1
1
x vtx
v
cy y
z z
vxt
ctv
c
Now from equations (1-13) we know that the relation between the primed and the unprimed coordinates and times is
Take the differential of these equations remembering that v is a constant This gives
2
2
2
2
2
1
1
dx vdtdx
v
cdy dy
dz dz
vxdt
cdtv
c
2
2
2 2 2 2 2
2
2
2
11 ( )
1( ) 1
1
1
1
dx vdt
v dxdx vdt v
dx dx vdtc dt dtvdx vdx vdx v dxdt dt dt dt
dt dtc c c c
VxV x
Vx
VxV x
vv
vv
c
x
v
c
cV
( )
coordinate of particle relative to moving primed system
coordinate of particle relative to stationary system
velocity of primed coordina
V x
Vx
te system relative to stationaryv sys
tem
Page 28 Fundamentals of Modern Physics
(1-16)
According to Einstein this equation tells you how to transform the observed velocity from one frame of reference to another frame of reference First we note that as Vc and vc approachzero the equation approaches those which would be derived from the Galilean transformation
2 2
( ) 1 01 1 ( )
Vx v Vx vV
VxV x
Vx v
v cVxx
v
c
v
cc
Another property of this equation is that it is impossible to choose V and v such that Vrsquo the magnitude of the velocity which is seen in the new frame is greater than c
1
2
09 ( 09 ) 18009876543
( 09 ) 1811 (09 )
c c cV c c
cc
c
Example As seen by O particle 1 has velocity 09c in the direction of positive x and particle 2 has velocity 09c in the negative x direction
Einstein equation however predicts To evaluate the velocity of particle 1 with respect to particle 2 we transform from the O frame to the Orsquo frame moving in the negative x direction with velocity v = -09c using equation 1-16 We obtain
JK LOGIC will observe the particles separating at 18c
A confusing INCORRECT and ILLOGICAL answer
left is a dir rightsub of a
1
2
1
09 ( 09 ) 18009876543
181( 09 )1 ( 09 )
The multiplication of dir left time a dir right is ONLY a dir left
09
dir
direction leftdirection rightsubtraction
c c cV c c
cc
c
V
dir
22
22
09 18 18009876543
181( 081)1(09)(09)
1
left rightsub
sub
c c c cc c
ccc
c
Analyzing Einsteinrsquos BS math Equation
Using SM math a logical answer can be obtained
Cannot multiply a subtraction operator times a direction in space
Can only subtract and change its direction
Cannot multiply a direction to the left times a direction to the right
Cannot multiply opposite directions in space
Can only solve for the resultant
rightleft
subtraction
1
subtraction
2
sub of a dir to the left is a dir to the right
1 resultant 0 sub 2
2sub 2
2
09 09
09 amp 091
09 09 18 1
(0)1
(09) amp (09)1
c cV
c cc
c c cV
cc
cc
8
181
cc
SM math predicts a logical answer using logic
21st century scientists state in many articles that Einsteins General Relativity and Quantum Mechanics are incompatible
They are not only incompatible they are both mathematically incorrect
They both use illogical BS math
Any equations in math that use imaginary numbers should be abandoned
Imaginary numbers are only necessary in the current BS math system
This means that Schroumldingers equations should be abandoned
Any equation in quantum mechanics that uses imaginary number should be abandoned
Prof Kanarev has developed logical physics and chemistrytheories without the use of Quantum mechanics or relativity
A partial list of his theories coverbull Mechanics ndash replaces Newtonbull Photon - bull Electrons ndash Protons- Neutronsbull Spectroscopybull Nucleusbull Atoms amp Moleculesbull Electrodynamicsbull Heatbull Curvature of Space ndash replaces Schwarzcchildrsquos equation
Kanarevrsquos Hypothesis
bull Electron align linearly with proton They do NOT spin or form clouds around the nucleus
bull Neutrons are in the center of atom Proton are outside of Neutrons
bull The wavelength of all particles is the experimental radius
bull Double slit experiment explained correctly
bull Heisenbergrsquos inequality explained correctly
bull is the radius of approach of the magnetic lines around the torus of the electron
( 1 137)
bull Why water expands when it freezes
bull Electron clusters form lightning
rbull The meaning of Plancks constant is explained
22 2 kg m
h m f mr f consts
2210254 42m r kgm bull If the mr-constant describes a ring the equations
describe experiments bull If the mr-constant describes a wave the equations do
NOT describe experiments
Kanarevrsquos 5 Equations of Mechanics
Kanarevrsquos 1st Equation Acceleration ampmv maMma R R
Kanarevrsquos 2nd Equation Uniform motion amp mv maMma R R
maM mvR R
Kanarevrsquos 3rd Equation
Kanarevrsquos 4th Equation Equal and opposite reaction
1 21 21 2
12
a m
m a m ama
Kanarevrsquos 5th Equation 1 2 3 Mi M M M Mna R R R R R
The sum of the Actions working on a moving body is never equal to zero
Kanarevrsquos Photon model is a rolling ring
His equations accurately describe known experiments with photonsbull Electromagnetic radiation should be called photon radiationbull Diffraction and interference are the same phenomenabull Correctly explains the double slit experiment without quantum equationsbull Correctly explains the meaning of Heisenbergrsquos inequality
e
2 (2)(3142)(2817 15 ) 10007297352568
R 2426 12 137mag
xp
r m
m
Charge
Magnetic action
Proton
The intensity of the magnetic action near the geometrical center of the proton ring is so great that we theorize that it is the strong nuclear force
(9383 8)(1602 19)8476 14
4 (4)(3142)(1411 26)p
pp
EB T
M
(9395 8)(1602 19)1035 15
4 (4)(3142)(0966 26)N
NN
EB T
M
Neutron
1a Hydrogenavi
Straight Line Symbol Units Rotational Symbol Units Distance d m Angular
Distance radians or degrees
Velocity dt
v m
s Angular Velocity
V
t
radians or degrees
s
Acceleration vat
2ms
Angular Acceleration
VA
t
2
radians or degrees
s
Mass M Kg Mass distribution
dM 2Kgm
Force actionA Ma
2
Kgm
sNewton
Torque
dT M A 2
2
Kgm
s
Momentum Mv Kgm
s
Angular Momentum
dM V 2Kgm
s
Work (Force)(distance)
Ad 2
Kgmm
sNm Joule
Work T
Using AS arrow math to solve and direction in space problems using the equation 1 2
2 2 2 2 2 2( ) amp and ( ) amp 3
4
2 2 2
2 2
( ) amp
(5 ) 4 3
(5 ) 16 9
(5 ) 25
(5 ) (5 )
2 2(5 ) 4 3
(5 ) 16 9
(5 ) 25
(5 ) (5 )
In SM math Pythagorasrsquo theorem provides both magnitude and direction
In BS math Pythagorasrsquo theorem provides magnitude only
time set equal to zero
23
( ) ( 0 )10 1
x t xx
3( ) ( 0)
10 1W t f
In BS math this is an equation for a traveling wave
must use a square term in order to make the curve symmetric
NO SQUARE TERM IS NECESSARY
In SM math when t is set to ldquo0rdquo
This is a major fault in BS math
x f(x)
-4 0019
-3 0033
-2 0073
-1 0273
0 3000
1 0273
2 0073
3 0033
4 0019 -4 -3 -2 -1 0 1 2 3 400
05
10
15
20
25
30
35
BS wave (must use square term)
snap shot of wave at x = 0
Am
plit
ude
lt to gt lt to gt f(ltgt)
lt -4 0073
lt -3 0097
lt -2 0143
lt -1 0273
origin 0 3000
gt 1 0273
gt 2 0143
gt 3 0097
gt 4 0073lt0073 lt0097 lt0143 lt0273
3000
0273gt 0143gt 0097gt 0073gt
Wave moving
snap shot of wave
Am
plit
ude
lt4 lt3 lt2 lt1 0 1gt 2gt 3gt 4gt
3
2
1
In SM NO squared terms are needed
SPRING
An Action is needed to stretch or compress a spring to a starting point where it can execute cycle motion
tsA (ts to start) is the work required to move a mass attached to a spring a distance (d1) from
its initial rest position RAK 0 the spring constant is the stiffness of the spring
o A is attraction o R is repulsion o
ARRA
00 is for the back and forth motion through the zero initial rest position A
stretched spring goes from ldquoArdquo to ldquo0rdquo to ldquoRrdquo to ldquoArdquo to ldquo0rdquo to ldquoRrdquo d1 is the distance a mass is moved from its initial rest position
0
0 1ts A RA K d
Once ( )mad
is performed to move the mass to position d1 (against the restraint of the spring) a of the ( )mad
is transferred into the spring as potential attractionrepulsion (PAR) There is always
more ( )mad
performed than is stored as PAR When the spring is released from the restraining Ats the PAR starts the oscillation Conservation of ( )mad
requires that the PAR and the motion ( )mad
( )( )madM be a constant
( )amp ( ) constantmadPAR M 02
0 0
1
2A R A RPAR K d
02
( ) 0 0
1( )
2mad A R A RM m v
0 0 02 2 2
0 0 0 1
2 2 21
2 2 21
1 1 1amp
2 2 2
1 1 1amp
2 2 2
amp
A R A R A RK d m v K d
Kd mv Kd
Kd mv Kd
0
1d is frac12 the height from the base to the peak of a cycle
An experiment of moving paper under a pendulum through a small angle filled with sand that can escape through a small hole in the bottom shows that a pendulum cycling back and forth follows a curve that is mathematically analogous to a lsquosinrsquo or lsquocosrsquo
The projection of the curved area onto the straight line exactly duplicate the back and forth motion of CM This allows for the math of lsquosinrsquo or lsquocosrsquo to be used
SM will use lsquosinrdquo math It could use lsquocosrsquo or a combination of both lsquosinrsquo and lsquocosrsquo Using just the lsquosinrsquo math makes the understanding and math easier
0 sin 0 0SM 0 sin157 1SM The diagram at the left shows a spring being stretched from its (zero rest) position-2 and the PAR being released at position-1 There are 4 sections (1 to 2 2 to 3 3 to 4 and 4 to5) that alternate between maximum acceleration and zero velocity to max velocity and zero acceleration
At position-1 there is maximum acceleration and zero velocity o The lsquosinrsquo values from 0o to 157o will be used for velocity (values from 0 to 1) o The lsquosinrsquo values from 157o to 0o will be used for acceleration (values from 1 to 0)
At position-2 there is maximum velocity and zero acceleration Acceleration changes to deceleration
o The lsquosinrsquo values from 157o to 0o will be used for deceleration (values from 1 to 0) o The lsquosinrsquo values from 0o to 157o will be used for acceleration (values from 0 to 1)
At position-3 there is maximum acceleration and zero velocity o The lsquosinrsquo values from 0o to 157o will be used for velocity (values from 0 to 1) o The lsquosinrsquo values from 157o to 0o will be used for acceleration (values from 1 to 0)
At position-4 there is maximum velocity and zero acceleration Acceleration changes to deceleration
o The lsquosinrsquo values from 157o to 0o will be used for deceleration (values from 1 to 0) o The lsquosinrsquo values from 0o to 157o will be used for acceleration (values from 0 to 1)
At position-5 the cycle starts again
2
)539(
cT
MK
spring constant (K) when mass (M) and cycle-time (T) are known
)(
))(89)((
111 md
uKgM
d
MG
d
FK sts
spring constant (K) when mass (M) and amplitude ( d1) are known
2 21( )
Kv d d
m
velocity (v) at any point during the cycle
velocity
dforM
Kdv
max
012 velocity (v2) maximum (when d = 0 at position-2 and 4)
acceleration (a1) maximum when spring constant (K) mass (M) and amplitude (d1) are known
acceleration (a1) maximum when amplitude (d1) and cycle-time (Tc) are known
acceleration (a) at any point during the cycle-time (T)
1
151
286
v
dTc
cycle-time (Tc) when amplitude (d1) and velocity (v1) are known
K
MTc 28651
cycle-time (Tc) when mass (M) and spring constant (K) are known
fTc
1
cycle-time (Tc) when the frequency is known
cTf
1
frequency (f) when the cycle-time (Tc) is known
of cycles
elapsed timef
frequency (f) when the number of cycles in an elapsed time is known
d1 amplitude (distance from rest to start position)
Also frac12 the distance from the base to the peak
3131 oror dM
Ka
251
3131 539
c
oror T
da
2539
cT
da
In December of 2006 I discovered the mathematical error in Einsteinrsquos Special Relativity Einstein used BS math and that math only works in an imaginary world
It cannot describe real symmetrical space
Einsteinrsquos Special Relativity should be abandoned and a theory of real space using SM should be used to replace it
Numerous articles on the Internet show Einsteinrsquos theory to be incorrect but no one in academia is willing to publish the information
Figure 1-10 A moving particle observed from two frames of reference in uniform translation
Einsteinrsquos Math Error in Special Relativity Transformation of Velocity (From Eisenberg- Fundamentals of Modern Physics pages 27-29) Consider the particle shown in figure (1-10) moving with velocity v as
seen in a frame of reference O
dx dy dz
Vx Vy Vzdt dt dt
dx dy dz
V x V y V zdy dt dt
We would like to evaluate the velocity v of the particle as seen in the frame of reference 0 which is itself moving relative to 0 with velocity v Measured in the 0 frame the velocity vector of the particle has com ponents
The same velocity vector as measured in the 0 frame has components
2
2
2
2
2
1
1
x vtx
v
cy y
z z
vxt
ctv
c
Now from equations (1-13) we know that the relation between the primed and the unprimed coordinates and times is
Take the differential of these equations remembering that v is a constant This gives
2
2
2
2
2
1
1
dx vdtdx
v
cdy dy
dz dz
vxdt
cdtv
c
2
2
2 2 2 2 2
2
2
2
11 ( )
1( ) 1
1
1
1
dx vdt
v dxdx vdt v
dx dx vdtc dt dtvdx vdx vdx v dxdt dt dt dt
dt dtc c c c
VxV x
Vx
VxV x
vv
vv
c
x
v
c
cV
( )
coordinate of particle relative to moving primed system
coordinate of particle relative to stationary system
velocity of primed coordina
V x
Vx
te system relative to stationaryv sys
tem
Page 28 Fundamentals of Modern Physics
(1-16)
According to Einstein this equation tells you how to transform the observed velocity from one frame of reference to another frame of reference First we note that as Vc and vc approachzero the equation approaches those which would be derived from the Galilean transformation
2 2
( ) 1 01 1 ( )
Vx v Vx vV
VxV x
Vx v
v cVxx
v
c
v
cc
Another property of this equation is that it is impossible to choose V and v such that Vrsquo the magnitude of the velocity which is seen in the new frame is greater than c
1
2
09 ( 09 ) 18009876543
( 09 ) 1811 (09 )
c c cV c c
cc
c
Example As seen by O particle 1 has velocity 09c in the direction of positive x and particle 2 has velocity 09c in the negative x direction
Einstein equation however predicts To evaluate the velocity of particle 1 with respect to particle 2 we transform from the O frame to the Orsquo frame moving in the negative x direction with velocity v = -09c using equation 1-16 We obtain
JK LOGIC will observe the particles separating at 18c
A confusing INCORRECT and ILLOGICAL answer
left is a dir rightsub of a
1
2
1
09 ( 09 ) 18009876543
181( 09 )1 ( 09 )
The multiplication of dir left time a dir right is ONLY a dir left
09
dir
direction leftdirection rightsubtraction
c c cV c c
cc
c
V
dir
22
22
09 18 18009876543
181( 081)1(09)(09)
1
left rightsub
sub
c c c cc c
ccc
c
Analyzing Einsteinrsquos BS math Equation
Using SM math a logical answer can be obtained
Cannot multiply a subtraction operator times a direction in space
Can only subtract and change its direction
Cannot multiply a direction to the left times a direction to the right
Cannot multiply opposite directions in space
Can only solve for the resultant
rightleft
subtraction
1
subtraction
2
sub of a dir to the left is a dir to the right
1 resultant 0 sub 2
2sub 2
2
09 09
09 amp 091
09 09 18 1
(0)1
(09) amp (09)1
c cV
c cc
c c cV
cc
cc
8
181
cc
SM math predicts a logical answer using logic
21st century scientists state in many articles that Einsteins General Relativity and Quantum Mechanics are incompatible
They are not only incompatible they are both mathematically incorrect
They both use illogical BS math
Any equations in math that use imaginary numbers should be abandoned
Imaginary numbers are only necessary in the current BS math system
This means that Schroumldingers equations should be abandoned
Any equation in quantum mechanics that uses imaginary number should be abandoned
Prof Kanarev has developed logical physics and chemistrytheories without the use of Quantum mechanics or relativity
A partial list of his theories coverbull Mechanics ndash replaces Newtonbull Photon - bull Electrons ndash Protons- Neutronsbull Spectroscopybull Nucleusbull Atoms amp Moleculesbull Electrodynamicsbull Heatbull Curvature of Space ndash replaces Schwarzcchildrsquos equation
Kanarevrsquos Hypothesis
bull Electron align linearly with proton They do NOT spin or form clouds around the nucleus
bull Neutrons are in the center of atom Proton are outside of Neutrons
bull The wavelength of all particles is the experimental radius
bull Double slit experiment explained correctly
bull Heisenbergrsquos inequality explained correctly
bull is the radius of approach of the magnetic lines around the torus of the electron
( 1 137)
bull Why water expands when it freezes
bull Electron clusters form lightning
rbull The meaning of Plancks constant is explained
22 2 kg m
h m f mr f consts
2210254 42m r kgm bull If the mr-constant describes a ring the equations
describe experiments bull If the mr-constant describes a wave the equations do
NOT describe experiments
Kanarevrsquos 5 Equations of Mechanics
Kanarevrsquos 1st Equation Acceleration ampmv maMma R R
Kanarevrsquos 2nd Equation Uniform motion amp mv maMma R R
maM mvR R
Kanarevrsquos 3rd Equation
Kanarevrsquos 4th Equation Equal and opposite reaction
1 21 21 2
12
a m
m a m ama
Kanarevrsquos 5th Equation 1 2 3 Mi M M M Mna R R R R R
The sum of the Actions working on a moving body is never equal to zero
Kanarevrsquos Photon model is a rolling ring
His equations accurately describe known experiments with photonsbull Electromagnetic radiation should be called photon radiationbull Diffraction and interference are the same phenomenabull Correctly explains the double slit experiment without quantum equationsbull Correctly explains the meaning of Heisenbergrsquos inequality
e
2 (2)(3142)(2817 15 ) 10007297352568
R 2426 12 137mag
xp
r m
m
Charge
Magnetic action
Proton
The intensity of the magnetic action near the geometrical center of the proton ring is so great that we theorize that it is the strong nuclear force
(9383 8)(1602 19)8476 14
4 (4)(3142)(1411 26)p
pp
EB T
M
(9395 8)(1602 19)1035 15
4 (4)(3142)(0966 26)N
NN
EB T
M
Neutron
1a Hydrogenavi
Using AS arrow math to solve and direction in space problems using the equation 1 2
2 2 2 2 2 2( ) amp and ( ) amp 3
4
2 2 2
2 2
( ) amp
(5 ) 4 3
(5 ) 16 9
(5 ) 25
(5 ) (5 )
2 2(5 ) 4 3
(5 ) 16 9
(5 ) 25
(5 ) (5 )
In SM math Pythagorasrsquo theorem provides both magnitude and direction
In BS math Pythagorasrsquo theorem provides magnitude only
time set equal to zero
23
( ) ( 0 )10 1
x t xx
3( ) ( 0)
10 1W t f
In BS math this is an equation for a traveling wave
must use a square term in order to make the curve symmetric
NO SQUARE TERM IS NECESSARY
In SM math when t is set to ldquo0rdquo
This is a major fault in BS math
x f(x)
-4 0019
-3 0033
-2 0073
-1 0273
0 3000
1 0273
2 0073
3 0033
4 0019 -4 -3 -2 -1 0 1 2 3 400
05
10
15
20
25
30
35
BS wave (must use square term)
snap shot of wave at x = 0
Am
plit
ude
lt to gt lt to gt f(ltgt)
lt -4 0073
lt -3 0097
lt -2 0143
lt -1 0273
origin 0 3000
gt 1 0273
gt 2 0143
gt 3 0097
gt 4 0073lt0073 lt0097 lt0143 lt0273
3000
0273gt 0143gt 0097gt 0073gt
Wave moving
snap shot of wave
Am
plit
ude
lt4 lt3 lt2 lt1 0 1gt 2gt 3gt 4gt
3
2
1
In SM NO squared terms are needed
SPRING
An Action is needed to stretch or compress a spring to a starting point where it can execute cycle motion
tsA (ts to start) is the work required to move a mass attached to a spring a distance (d1) from
its initial rest position RAK 0 the spring constant is the stiffness of the spring
o A is attraction o R is repulsion o
ARRA
00 is for the back and forth motion through the zero initial rest position A
stretched spring goes from ldquoArdquo to ldquo0rdquo to ldquoRrdquo to ldquoArdquo to ldquo0rdquo to ldquoRrdquo d1 is the distance a mass is moved from its initial rest position
0
0 1ts A RA K d
Once ( )mad
is performed to move the mass to position d1 (against the restraint of the spring) a of the ( )mad
is transferred into the spring as potential attractionrepulsion (PAR) There is always
more ( )mad
performed than is stored as PAR When the spring is released from the restraining Ats the PAR starts the oscillation Conservation of ( )mad
requires that the PAR and the motion ( )mad
( )( )madM be a constant
( )amp ( ) constantmadPAR M 02
0 0
1
2A R A RPAR K d
02
( ) 0 0
1( )
2mad A R A RM m v
0 0 02 2 2
0 0 0 1
2 2 21
2 2 21
1 1 1amp
2 2 2
1 1 1amp
2 2 2
amp
A R A R A RK d m v K d
Kd mv Kd
Kd mv Kd
0
1d is frac12 the height from the base to the peak of a cycle
An experiment of moving paper under a pendulum through a small angle filled with sand that can escape through a small hole in the bottom shows that a pendulum cycling back and forth follows a curve that is mathematically analogous to a lsquosinrsquo or lsquocosrsquo
The projection of the curved area onto the straight line exactly duplicate the back and forth motion of CM This allows for the math of lsquosinrsquo or lsquocosrsquo to be used
SM will use lsquosinrdquo math It could use lsquocosrsquo or a combination of both lsquosinrsquo and lsquocosrsquo Using just the lsquosinrsquo math makes the understanding and math easier
0 sin 0 0SM 0 sin157 1SM The diagram at the left shows a spring being stretched from its (zero rest) position-2 and the PAR being released at position-1 There are 4 sections (1 to 2 2 to 3 3 to 4 and 4 to5) that alternate between maximum acceleration and zero velocity to max velocity and zero acceleration
At position-1 there is maximum acceleration and zero velocity o The lsquosinrsquo values from 0o to 157o will be used for velocity (values from 0 to 1) o The lsquosinrsquo values from 157o to 0o will be used for acceleration (values from 1 to 0)
At position-2 there is maximum velocity and zero acceleration Acceleration changes to deceleration
o The lsquosinrsquo values from 157o to 0o will be used for deceleration (values from 1 to 0) o The lsquosinrsquo values from 0o to 157o will be used for acceleration (values from 0 to 1)
At position-3 there is maximum acceleration and zero velocity o The lsquosinrsquo values from 0o to 157o will be used for velocity (values from 0 to 1) o The lsquosinrsquo values from 157o to 0o will be used for acceleration (values from 1 to 0)
At position-4 there is maximum velocity and zero acceleration Acceleration changes to deceleration
o The lsquosinrsquo values from 157o to 0o will be used for deceleration (values from 1 to 0) o The lsquosinrsquo values from 0o to 157o will be used for acceleration (values from 0 to 1)
At position-5 the cycle starts again
2
)539(
cT
MK
spring constant (K) when mass (M) and cycle-time (T) are known
)(
))(89)((
111 md
uKgM
d
MG
d
FK sts
spring constant (K) when mass (M) and amplitude ( d1) are known
2 21( )
Kv d d
m
velocity (v) at any point during the cycle
velocity
dforM
Kdv
max
012 velocity (v2) maximum (when d = 0 at position-2 and 4)
acceleration (a1) maximum when spring constant (K) mass (M) and amplitude (d1) are known
acceleration (a1) maximum when amplitude (d1) and cycle-time (Tc) are known
acceleration (a) at any point during the cycle-time (T)
1
151
286
v
dTc
cycle-time (Tc) when amplitude (d1) and velocity (v1) are known
K
MTc 28651
cycle-time (Tc) when mass (M) and spring constant (K) are known
fTc
1
cycle-time (Tc) when the frequency is known
cTf
1
frequency (f) when the cycle-time (Tc) is known
of cycles
elapsed timef
frequency (f) when the number of cycles in an elapsed time is known
d1 amplitude (distance from rest to start position)
Also frac12 the distance from the base to the peak
3131 oror dM
Ka
251
3131 539
c
oror T
da
2539
cT
da
In December of 2006 I discovered the mathematical error in Einsteinrsquos Special Relativity Einstein used BS math and that math only works in an imaginary world
It cannot describe real symmetrical space
Einsteinrsquos Special Relativity should be abandoned and a theory of real space using SM should be used to replace it
Numerous articles on the Internet show Einsteinrsquos theory to be incorrect but no one in academia is willing to publish the information
Figure 1-10 A moving particle observed from two frames of reference in uniform translation
Einsteinrsquos Math Error in Special Relativity Transformation of Velocity (From Eisenberg- Fundamentals of Modern Physics pages 27-29) Consider the particle shown in figure (1-10) moving with velocity v as
seen in a frame of reference O
dx dy dz
Vx Vy Vzdt dt dt
dx dy dz
V x V y V zdy dt dt
We would like to evaluate the velocity v of the particle as seen in the frame of reference 0 which is itself moving relative to 0 with velocity v Measured in the 0 frame the velocity vector of the particle has com ponents
The same velocity vector as measured in the 0 frame has components
2
2
2
2
2
1
1
x vtx
v
cy y
z z
vxt
ctv
c
Now from equations (1-13) we know that the relation between the primed and the unprimed coordinates and times is
Take the differential of these equations remembering that v is a constant This gives
2
2
2
2
2
1
1
dx vdtdx
v
cdy dy
dz dz
vxdt
cdtv
c
2
2
2 2 2 2 2
2
2
2
11 ( )
1( ) 1
1
1
1
dx vdt
v dxdx vdt v
dx dx vdtc dt dtvdx vdx vdx v dxdt dt dt dt
dt dtc c c c
VxV x
Vx
VxV x
vv
vv
c
x
v
c
cV
( )
coordinate of particle relative to moving primed system
coordinate of particle relative to stationary system
velocity of primed coordina
V x
Vx
te system relative to stationaryv sys
tem
Page 28 Fundamentals of Modern Physics
(1-16)
According to Einstein this equation tells you how to transform the observed velocity from one frame of reference to another frame of reference First we note that as Vc and vc approachzero the equation approaches those which would be derived from the Galilean transformation
2 2
( ) 1 01 1 ( )
Vx v Vx vV
VxV x
Vx v
v cVxx
v
c
v
cc
Another property of this equation is that it is impossible to choose V and v such that Vrsquo the magnitude of the velocity which is seen in the new frame is greater than c
1
2
09 ( 09 ) 18009876543
( 09 ) 1811 (09 )
c c cV c c
cc
c
Example As seen by O particle 1 has velocity 09c in the direction of positive x and particle 2 has velocity 09c in the negative x direction
Einstein equation however predicts To evaluate the velocity of particle 1 with respect to particle 2 we transform from the O frame to the Orsquo frame moving in the negative x direction with velocity v = -09c using equation 1-16 We obtain
JK LOGIC will observe the particles separating at 18c
A confusing INCORRECT and ILLOGICAL answer
left is a dir rightsub of a
1
2
1
09 ( 09 ) 18009876543
181( 09 )1 ( 09 )
The multiplication of dir left time a dir right is ONLY a dir left
09
dir
direction leftdirection rightsubtraction
c c cV c c
cc
c
V
dir
22
22
09 18 18009876543
181( 081)1(09)(09)
1
left rightsub
sub
c c c cc c
ccc
c
Analyzing Einsteinrsquos BS math Equation
Using SM math a logical answer can be obtained
Cannot multiply a subtraction operator times a direction in space
Can only subtract and change its direction
Cannot multiply a direction to the left times a direction to the right
Cannot multiply opposite directions in space
Can only solve for the resultant
rightleft
subtraction
1
subtraction
2
sub of a dir to the left is a dir to the right
1 resultant 0 sub 2
2sub 2
2
09 09
09 amp 091
09 09 18 1
(0)1
(09) amp (09)1
c cV
c cc
c c cV
cc
cc
8
181
cc
SM math predicts a logical answer using logic
21st century scientists state in many articles that Einsteins General Relativity and Quantum Mechanics are incompatible
They are not only incompatible they are both mathematically incorrect
They both use illogical BS math
Any equations in math that use imaginary numbers should be abandoned
Imaginary numbers are only necessary in the current BS math system
This means that Schroumldingers equations should be abandoned
Any equation in quantum mechanics that uses imaginary number should be abandoned
Prof Kanarev has developed logical physics and chemistrytheories without the use of Quantum mechanics or relativity
A partial list of his theories coverbull Mechanics ndash replaces Newtonbull Photon - bull Electrons ndash Protons- Neutronsbull Spectroscopybull Nucleusbull Atoms amp Moleculesbull Electrodynamicsbull Heatbull Curvature of Space ndash replaces Schwarzcchildrsquos equation
Kanarevrsquos Hypothesis
bull Electron align linearly with proton They do NOT spin or form clouds around the nucleus
bull Neutrons are in the center of atom Proton are outside of Neutrons
bull The wavelength of all particles is the experimental radius
bull Double slit experiment explained correctly
bull Heisenbergrsquos inequality explained correctly
bull is the radius of approach of the magnetic lines around the torus of the electron
( 1 137)
bull Why water expands when it freezes
bull Electron clusters form lightning
rbull The meaning of Plancks constant is explained
22 2 kg m
h m f mr f consts
2210254 42m r kgm bull If the mr-constant describes a ring the equations
describe experiments bull If the mr-constant describes a wave the equations do
NOT describe experiments
Kanarevrsquos 5 Equations of Mechanics
Kanarevrsquos 1st Equation Acceleration ampmv maMma R R
Kanarevrsquos 2nd Equation Uniform motion amp mv maMma R R
maM mvR R
Kanarevrsquos 3rd Equation
Kanarevrsquos 4th Equation Equal and opposite reaction
1 21 21 2
12
a m
m a m ama
Kanarevrsquos 5th Equation 1 2 3 Mi M M M Mna R R R R R
The sum of the Actions working on a moving body is never equal to zero
Kanarevrsquos Photon model is a rolling ring
His equations accurately describe known experiments with photonsbull Electromagnetic radiation should be called photon radiationbull Diffraction and interference are the same phenomenabull Correctly explains the double slit experiment without quantum equationsbull Correctly explains the meaning of Heisenbergrsquos inequality
e
2 (2)(3142)(2817 15 ) 10007297352568
R 2426 12 137mag
xp
r m
m
Charge
Magnetic action
Proton
The intensity of the magnetic action near the geometrical center of the proton ring is so great that we theorize that it is the strong nuclear force
(9383 8)(1602 19)8476 14
4 (4)(3142)(1411 26)p
pp
EB T
M
(9395 8)(1602 19)1035 15
4 (4)(3142)(0966 26)N
NN
EB T
M
Neutron
1a Hydrogenavi
time set equal to zero
23
( ) ( 0 )10 1
x t xx
3( ) ( 0)
10 1W t f
In BS math this is an equation for a traveling wave
must use a square term in order to make the curve symmetric
NO SQUARE TERM IS NECESSARY
In SM math when t is set to ldquo0rdquo
This is a major fault in BS math
x f(x)
-4 0019
-3 0033
-2 0073
-1 0273
0 3000
1 0273
2 0073
3 0033
4 0019 -4 -3 -2 -1 0 1 2 3 400
05
10
15
20
25
30
35
BS wave (must use square term)
snap shot of wave at x = 0
Am
plit
ude
lt to gt lt to gt f(ltgt)
lt -4 0073
lt -3 0097
lt -2 0143
lt -1 0273
origin 0 3000
gt 1 0273
gt 2 0143
gt 3 0097
gt 4 0073lt0073 lt0097 lt0143 lt0273
3000
0273gt 0143gt 0097gt 0073gt
Wave moving
snap shot of wave
Am
plit
ude
lt4 lt3 lt2 lt1 0 1gt 2gt 3gt 4gt
3
2
1
In SM NO squared terms are needed
SPRING
An Action is needed to stretch or compress a spring to a starting point where it can execute cycle motion
tsA (ts to start) is the work required to move a mass attached to a spring a distance (d1) from
its initial rest position RAK 0 the spring constant is the stiffness of the spring
o A is attraction o R is repulsion o
ARRA
00 is for the back and forth motion through the zero initial rest position A
stretched spring goes from ldquoArdquo to ldquo0rdquo to ldquoRrdquo to ldquoArdquo to ldquo0rdquo to ldquoRrdquo d1 is the distance a mass is moved from its initial rest position
0
0 1ts A RA K d
Once ( )mad
is performed to move the mass to position d1 (against the restraint of the spring) a of the ( )mad
is transferred into the spring as potential attractionrepulsion (PAR) There is always
more ( )mad
performed than is stored as PAR When the spring is released from the restraining Ats the PAR starts the oscillation Conservation of ( )mad
requires that the PAR and the motion ( )mad
( )( )madM be a constant
( )amp ( ) constantmadPAR M 02
0 0
1
2A R A RPAR K d
02
( ) 0 0
1( )
2mad A R A RM m v
0 0 02 2 2
0 0 0 1
2 2 21
2 2 21
1 1 1amp
2 2 2
1 1 1amp
2 2 2
amp
A R A R A RK d m v K d
Kd mv Kd
Kd mv Kd
0
1d is frac12 the height from the base to the peak of a cycle
An experiment of moving paper under a pendulum through a small angle filled with sand that can escape through a small hole in the bottom shows that a pendulum cycling back and forth follows a curve that is mathematically analogous to a lsquosinrsquo or lsquocosrsquo
The projection of the curved area onto the straight line exactly duplicate the back and forth motion of CM This allows for the math of lsquosinrsquo or lsquocosrsquo to be used
SM will use lsquosinrdquo math It could use lsquocosrsquo or a combination of both lsquosinrsquo and lsquocosrsquo Using just the lsquosinrsquo math makes the understanding and math easier
0 sin 0 0SM 0 sin157 1SM The diagram at the left shows a spring being stretched from its (zero rest) position-2 and the PAR being released at position-1 There are 4 sections (1 to 2 2 to 3 3 to 4 and 4 to5) that alternate between maximum acceleration and zero velocity to max velocity and zero acceleration
At position-1 there is maximum acceleration and zero velocity o The lsquosinrsquo values from 0o to 157o will be used for velocity (values from 0 to 1) o The lsquosinrsquo values from 157o to 0o will be used for acceleration (values from 1 to 0)
At position-2 there is maximum velocity and zero acceleration Acceleration changes to deceleration
o The lsquosinrsquo values from 157o to 0o will be used for deceleration (values from 1 to 0) o The lsquosinrsquo values from 0o to 157o will be used for acceleration (values from 0 to 1)
At position-3 there is maximum acceleration and zero velocity o The lsquosinrsquo values from 0o to 157o will be used for velocity (values from 0 to 1) o The lsquosinrsquo values from 157o to 0o will be used for acceleration (values from 1 to 0)
At position-4 there is maximum velocity and zero acceleration Acceleration changes to deceleration
o The lsquosinrsquo values from 157o to 0o will be used for deceleration (values from 1 to 0) o The lsquosinrsquo values from 0o to 157o will be used for acceleration (values from 0 to 1)
At position-5 the cycle starts again
2
)539(
cT
MK
spring constant (K) when mass (M) and cycle-time (T) are known
)(
))(89)((
111 md
uKgM
d
MG
d
FK sts
spring constant (K) when mass (M) and amplitude ( d1) are known
2 21( )
Kv d d
m
velocity (v) at any point during the cycle
velocity
dforM
Kdv
max
012 velocity (v2) maximum (when d = 0 at position-2 and 4)
acceleration (a1) maximum when spring constant (K) mass (M) and amplitude (d1) are known
acceleration (a1) maximum when amplitude (d1) and cycle-time (Tc) are known
acceleration (a) at any point during the cycle-time (T)
1
151
286
v
dTc
cycle-time (Tc) when amplitude (d1) and velocity (v1) are known
K
MTc 28651
cycle-time (Tc) when mass (M) and spring constant (K) are known
fTc
1
cycle-time (Tc) when the frequency is known
cTf
1
frequency (f) when the cycle-time (Tc) is known
of cycles
elapsed timef
frequency (f) when the number of cycles in an elapsed time is known
d1 amplitude (distance from rest to start position)
Also frac12 the distance from the base to the peak
3131 oror dM
Ka
251
3131 539
c
oror T
da
2539
cT
da
In December of 2006 I discovered the mathematical error in Einsteinrsquos Special Relativity Einstein used BS math and that math only works in an imaginary world
It cannot describe real symmetrical space
Einsteinrsquos Special Relativity should be abandoned and a theory of real space using SM should be used to replace it
Numerous articles on the Internet show Einsteinrsquos theory to be incorrect but no one in academia is willing to publish the information
Figure 1-10 A moving particle observed from two frames of reference in uniform translation
Einsteinrsquos Math Error in Special Relativity Transformation of Velocity (From Eisenberg- Fundamentals of Modern Physics pages 27-29) Consider the particle shown in figure (1-10) moving with velocity v as
seen in a frame of reference O
dx dy dz
Vx Vy Vzdt dt dt
dx dy dz
V x V y V zdy dt dt
We would like to evaluate the velocity v of the particle as seen in the frame of reference 0 which is itself moving relative to 0 with velocity v Measured in the 0 frame the velocity vector of the particle has com ponents
The same velocity vector as measured in the 0 frame has components
2
2
2
2
2
1
1
x vtx
v
cy y
z z
vxt
ctv
c
Now from equations (1-13) we know that the relation between the primed and the unprimed coordinates and times is
Take the differential of these equations remembering that v is a constant This gives
2
2
2
2
2
1
1
dx vdtdx
v
cdy dy
dz dz
vxdt
cdtv
c
2
2
2 2 2 2 2
2
2
2
11 ( )
1( ) 1
1
1
1
dx vdt
v dxdx vdt v
dx dx vdtc dt dtvdx vdx vdx v dxdt dt dt dt
dt dtc c c c
VxV x
Vx
VxV x
vv
vv
c
x
v
c
cV
( )
coordinate of particle relative to moving primed system
coordinate of particle relative to stationary system
velocity of primed coordina
V x
Vx
te system relative to stationaryv sys
tem
Page 28 Fundamentals of Modern Physics
(1-16)
According to Einstein this equation tells you how to transform the observed velocity from one frame of reference to another frame of reference First we note that as Vc and vc approachzero the equation approaches those which would be derived from the Galilean transformation
2 2
( ) 1 01 1 ( )
Vx v Vx vV
VxV x
Vx v
v cVxx
v
c
v
cc
Another property of this equation is that it is impossible to choose V and v such that Vrsquo the magnitude of the velocity which is seen in the new frame is greater than c
1
2
09 ( 09 ) 18009876543
( 09 ) 1811 (09 )
c c cV c c
cc
c
Example As seen by O particle 1 has velocity 09c in the direction of positive x and particle 2 has velocity 09c in the negative x direction
Einstein equation however predicts To evaluate the velocity of particle 1 with respect to particle 2 we transform from the O frame to the Orsquo frame moving in the negative x direction with velocity v = -09c using equation 1-16 We obtain
JK LOGIC will observe the particles separating at 18c
A confusing INCORRECT and ILLOGICAL answer
left is a dir rightsub of a
1
2
1
09 ( 09 ) 18009876543
181( 09 )1 ( 09 )
The multiplication of dir left time a dir right is ONLY a dir left
09
dir
direction leftdirection rightsubtraction
c c cV c c
cc
c
V
dir
22
22
09 18 18009876543
181( 081)1(09)(09)
1
left rightsub
sub
c c c cc c
ccc
c
Analyzing Einsteinrsquos BS math Equation
Using SM math a logical answer can be obtained
Cannot multiply a subtraction operator times a direction in space
Can only subtract and change its direction
Cannot multiply a direction to the left times a direction to the right
Cannot multiply opposite directions in space
Can only solve for the resultant
rightleft
subtraction
1
subtraction
2
sub of a dir to the left is a dir to the right
1 resultant 0 sub 2
2sub 2
2
09 09
09 amp 091
09 09 18 1
(0)1
(09) amp (09)1
c cV
c cc
c c cV
cc
cc
8
181
cc
SM math predicts a logical answer using logic
21st century scientists state in many articles that Einsteins General Relativity and Quantum Mechanics are incompatible
They are not only incompatible they are both mathematically incorrect
They both use illogical BS math
Any equations in math that use imaginary numbers should be abandoned
Imaginary numbers are only necessary in the current BS math system
This means that Schroumldingers equations should be abandoned
Any equation in quantum mechanics that uses imaginary number should be abandoned
Prof Kanarev has developed logical physics and chemistrytheories without the use of Quantum mechanics or relativity
A partial list of his theories coverbull Mechanics ndash replaces Newtonbull Photon - bull Electrons ndash Protons- Neutronsbull Spectroscopybull Nucleusbull Atoms amp Moleculesbull Electrodynamicsbull Heatbull Curvature of Space ndash replaces Schwarzcchildrsquos equation
Kanarevrsquos Hypothesis
bull Electron align linearly with proton They do NOT spin or form clouds around the nucleus
bull Neutrons are in the center of atom Proton are outside of Neutrons
bull The wavelength of all particles is the experimental radius
bull Double slit experiment explained correctly
bull Heisenbergrsquos inequality explained correctly
bull is the radius of approach of the magnetic lines around the torus of the electron
( 1 137)
bull Why water expands when it freezes
bull Electron clusters form lightning
rbull The meaning of Plancks constant is explained
22 2 kg m
h m f mr f consts
2210254 42m r kgm bull If the mr-constant describes a ring the equations
describe experiments bull If the mr-constant describes a wave the equations do
NOT describe experiments
Kanarevrsquos 5 Equations of Mechanics
Kanarevrsquos 1st Equation Acceleration ampmv maMma R R
Kanarevrsquos 2nd Equation Uniform motion amp mv maMma R R
maM mvR R
Kanarevrsquos 3rd Equation
Kanarevrsquos 4th Equation Equal and opposite reaction
1 21 21 2
12
a m
m a m ama
Kanarevrsquos 5th Equation 1 2 3 Mi M M M Mna R R R R R
The sum of the Actions working on a moving body is never equal to zero
Kanarevrsquos Photon model is a rolling ring
His equations accurately describe known experiments with photonsbull Electromagnetic radiation should be called photon radiationbull Diffraction and interference are the same phenomenabull Correctly explains the double slit experiment without quantum equationsbull Correctly explains the meaning of Heisenbergrsquos inequality
e
2 (2)(3142)(2817 15 ) 10007297352568
R 2426 12 137mag
xp
r m
m
Charge
Magnetic action
Proton
The intensity of the magnetic action near the geometrical center of the proton ring is so great that we theorize that it is the strong nuclear force
(9383 8)(1602 19)8476 14
4 (4)(3142)(1411 26)p
pp
EB T
M
(9395 8)(1602 19)1035 15
4 (4)(3142)(0966 26)N
NN
EB T
M
Neutron
1a Hydrogenavi
x f(x)
-4 0019
-3 0033
-2 0073
-1 0273
0 3000
1 0273
2 0073
3 0033
4 0019 -4 -3 -2 -1 0 1 2 3 400
05
10
15
20
25
30
35
BS wave (must use square term)
snap shot of wave at x = 0
Am
plit
ude
lt to gt lt to gt f(ltgt)
lt -4 0073
lt -3 0097
lt -2 0143
lt -1 0273
origin 0 3000
gt 1 0273
gt 2 0143
gt 3 0097
gt 4 0073lt0073 lt0097 lt0143 lt0273
3000
0273gt 0143gt 0097gt 0073gt
Wave moving
snap shot of wave
Am
plit
ude
lt4 lt3 lt2 lt1 0 1gt 2gt 3gt 4gt
3
2
1
In SM NO squared terms are needed
SPRING
An Action is needed to stretch or compress a spring to a starting point where it can execute cycle motion
tsA (ts to start) is the work required to move a mass attached to a spring a distance (d1) from
its initial rest position RAK 0 the spring constant is the stiffness of the spring
o A is attraction o R is repulsion o
ARRA
00 is for the back and forth motion through the zero initial rest position A
stretched spring goes from ldquoArdquo to ldquo0rdquo to ldquoRrdquo to ldquoArdquo to ldquo0rdquo to ldquoRrdquo d1 is the distance a mass is moved from its initial rest position
0
0 1ts A RA K d
Once ( )mad
is performed to move the mass to position d1 (against the restraint of the spring) a of the ( )mad
is transferred into the spring as potential attractionrepulsion (PAR) There is always
more ( )mad
performed than is stored as PAR When the spring is released from the restraining Ats the PAR starts the oscillation Conservation of ( )mad
requires that the PAR and the motion ( )mad
( )( )madM be a constant
( )amp ( ) constantmadPAR M 02
0 0
1
2A R A RPAR K d
02
( ) 0 0
1( )
2mad A R A RM m v
0 0 02 2 2
0 0 0 1
2 2 21
2 2 21
1 1 1amp
2 2 2
1 1 1amp
2 2 2
amp
A R A R A RK d m v K d
Kd mv Kd
Kd mv Kd
0
1d is frac12 the height from the base to the peak of a cycle
An experiment of moving paper under a pendulum through a small angle filled with sand that can escape through a small hole in the bottom shows that a pendulum cycling back and forth follows a curve that is mathematically analogous to a lsquosinrsquo or lsquocosrsquo
The projection of the curved area onto the straight line exactly duplicate the back and forth motion of CM This allows for the math of lsquosinrsquo or lsquocosrsquo to be used
SM will use lsquosinrdquo math It could use lsquocosrsquo or a combination of both lsquosinrsquo and lsquocosrsquo Using just the lsquosinrsquo math makes the understanding and math easier
0 sin 0 0SM 0 sin157 1SM The diagram at the left shows a spring being stretched from its (zero rest) position-2 and the PAR being released at position-1 There are 4 sections (1 to 2 2 to 3 3 to 4 and 4 to5) that alternate between maximum acceleration and zero velocity to max velocity and zero acceleration
At position-1 there is maximum acceleration and zero velocity o The lsquosinrsquo values from 0o to 157o will be used for velocity (values from 0 to 1) o The lsquosinrsquo values from 157o to 0o will be used for acceleration (values from 1 to 0)
At position-2 there is maximum velocity and zero acceleration Acceleration changes to deceleration
o The lsquosinrsquo values from 157o to 0o will be used for deceleration (values from 1 to 0) o The lsquosinrsquo values from 0o to 157o will be used for acceleration (values from 0 to 1)
At position-3 there is maximum acceleration and zero velocity o The lsquosinrsquo values from 0o to 157o will be used for velocity (values from 0 to 1) o The lsquosinrsquo values from 157o to 0o will be used for acceleration (values from 1 to 0)
At position-4 there is maximum velocity and zero acceleration Acceleration changes to deceleration
o The lsquosinrsquo values from 157o to 0o will be used for deceleration (values from 1 to 0) o The lsquosinrsquo values from 0o to 157o will be used for acceleration (values from 0 to 1)
At position-5 the cycle starts again
2
)539(
cT
MK
spring constant (K) when mass (M) and cycle-time (T) are known
)(
))(89)((
111 md
uKgM
d
MG
d
FK sts
spring constant (K) when mass (M) and amplitude ( d1) are known
2 21( )
Kv d d
m
velocity (v) at any point during the cycle
velocity
dforM
Kdv
max
012 velocity (v2) maximum (when d = 0 at position-2 and 4)
acceleration (a1) maximum when spring constant (K) mass (M) and amplitude (d1) are known
acceleration (a1) maximum when amplitude (d1) and cycle-time (Tc) are known
acceleration (a) at any point during the cycle-time (T)
1
151
286
v
dTc
cycle-time (Tc) when amplitude (d1) and velocity (v1) are known
K
MTc 28651
cycle-time (Tc) when mass (M) and spring constant (K) are known
fTc
1
cycle-time (Tc) when the frequency is known
cTf
1
frequency (f) when the cycle-time (Tc) is known
of cycles
elapsed timef
frequency (f) when the number of cycles in an elapsed time is known
d1 amplitude (distance from rest to start position)
Also frac12 the distance from the base to the peak
3131 oror dM
Ka
251
3131 539
c
oror T
da
2539
cT
da
In December of 2006 I discovered the mathematical error in Einsteinrsquos Special Relativity Einstein used BS math and that math only works in an imaginary world
It cannot describe real symmetrical space
Einsteinrsquos Special Relativity should be abandoned and a theory of real space using SM should be used to replace it
Numerous articles on the Internet show Einsteinrsquos theory to be incorrect but no one in academia is willing to publish the information
Figure 1-10 A moving particle observed from two frames of reference in uniform translation
Einsteinrsquos Math Error in Special Relativity Transformation of Velocity (From Eisenberg- Fundamentals of Modern Physics pages 27-29) Consider the particle shown in figure (1-10) moving with velocity v as
seen in a frame of reference O
dx dy dz
Vx Vy Vzdt dt dt
dx dy dz
V x V y V zdy dt dt
We would like to evaluate the velocity v of the particle as seen in the frame of reference 0 which is itself moving relative to 0 with velocity v Measured in the 0 frame the velocity vector of the particle has com ponents
The same velocity vector as measured in the 0 frame has components
2
2
2
2
2
1
1
x vtx
v
cy y
z z
vxt
ctv
c
Now from equations (1-13) we know that the relation between the primed and the unprimed coordinates and times is
Take the differential of these equations remembering that v is a constant This gives
2
2
2
2
2
1
1
dx vdtdx
v
cdy dy
dz dz
vxdt
cdtv
c
2
2
2 2 2 2 2
2
2
2
11 ( )
1( ) 1
1
1
1
dx vdt
v dxdx vdt v
dx dx vdtc dt dtvdx vdx vdx v dxdt dt dt dt
dt dtc c c c
VxV x
Vx
VxV x
vv
vv
c
x
v
c
cV
( )
coordinate of particle relative to moving primed system
coordinate of particle relative to stationary system
velocity of primed coordina
V x
Vx
te system relative to stationaryv sys
tem
Page 28 Fundamentals of Modern Physics
(1-16)
According to Einstein this equation tells you how to transform the observed velocity from one frame of reference to another frame of reference First we note that as Vc and vc approachzero the equation approaches those which would be derived from the Galilean transformation
2 2
( ) 1 01 1 ( )
Vx v Vx vV
VxV x
Vx v
v cVxx
v
c
v
cc
Another property of this equation is that it is impossible to choose V and v such that Vrsquo the magnitude of the velocity which is seen in the new frame is greater than c
1
2
09 ( 09 ) 18009876543
( 09 ) 1811 (09 )
c c cV c c
cc
c
Example As seen by O particle 1 has velocity 09c in the direction of positive x and particle 2 has velocity 09c in the negative x direction
Einstein equation however predicts To evaluate the velocity of particle 1 with respect to particle 2 we transform from the O frame to the Orsquo frame moving in the negative x direction with velocity v = -09c using equation 1-16 We obtain
JK LOGIC will observe the particles separating at 18c
A confusing INCORRECT and ILLOGICAL answer
left is a dir rightsub of a
1
2
1
09 ( 09 ) 18009876543
181( 09 )1 ( 09 )
The multiplication of dir left time a dir right is ONLY a dir left
09
dir
direction leftdirection rightsubtraction
c c cV c c
cc
c
V
dir
22
22
09 18 18009876543
181( 081)1(09)(09)
1
left rightsub
sub
c c c cc c
ccc
c
Analyzing Einsteinrsquos BS math Equation
Using SM math a logical answer can be obtained
Cannot multiply a subtraction operator times a direction in space
Can only subtract and change its direction
Cannot multiply a direction to the left times a direction to the right
Cannot multiply opposite directions in space
Can only solve for the resultant
rightleft
subtraction
1
subtraction
2
sub of a dir to the left is a dir to the right
1 resultant 0 sub 2
2sub 2
2
09 09
09 amp 091
09 09 18 1
(0)1
(09) amp (09)1
c cV
c cc
c c cV
cc
cc
8
181
cc
SM math predicts a logical answer using logic
21st century scientists state in many articles that Einsteins General Relativity and Quantum Mechanics are incompatible
They are not only incompatible they are both mathematically incorrect
They both use illogical BS math
Any equations in math that use imaginary numbers should be abandoned
Imaginary numbers are only necessary in the current BS math system
This means that Schroumldingers equations should be abandoned
Any equation in quantum mechanics that uses imaginary number should be abandoned
Prof Kanarev has developed logical physics and chemistrytheories without the use of Quantum mechanics or relativity
A partial list of his theories coverbull Mechanics ndash replaces Newtonbull Photon - bull Electrons ndash Protons- Neutronsbull Spectroscopybull Nucleusbull Atoms amp Moleculesbull Electrodynamicsbull Heatbull Curvature of Space ndash replaces Schwarzcchildrsquos equation
Kanarevrsquos Hypothesis
bull Electron align linearly with proton They do NOT spin or form clouds around the nucleus
bull Neutrons are in the center of atom Proton are outside of Neutrons
bull The wavelength of all particles is the experimental radius
bull Double slit experiment explained correctly
bull Heisenbergrsquos inequality explained correctly
bull is the radius of approach of the magnetic lines around the torus of the electron
( 1 137)
bull Why water expands when it freezes
bull Electron clusters form lightning
rbull The meaning of Plancks constant is explained
22 2 kg m
h m f mr f consts
2210254 42m r kgm bull If the mr-constant describes a ring the equations
describe experiments bull If the mr-constant describes a wave the equations do
NOT describe experiments
Kanarevrsquos 5 Equations of Mechanics
Kanarevrsquos 1st Equation Acceleration ampmv maMma R R
Kanarevrsquos 2nd Equation Uniform motion amp mv maMma R R
maM mvR R
Kanarevrsquos 3rd Equation
Kanarevrsquos 4th Equation Equal and opposite reaction
1 21 21 2
12
a m
m a m ama
Kanarevrsquos 5th Equation 1 2 3 Mi M M M Mna R R R R R
The sum of the Actions working on a moving body is never equal to zero
Kanarevrsquos Photon model is a rolling ring
His equations accurately describe known experiments with photonsbull Electromagnetic radiation should be called photon radiationbull Diffraction and interference are the same phenomenabull Correctly explains the double slit experiment without quantum equationsbull Correctly explains the meaning of Heisenbergrsquos inequality
e
2 (2)(3142)(2817 15 ) 10007297352568
R 2426 12 137mag
xp
r m
m
Charge
Magnetic action
Proton
The intensity of the magnetic action near the geometrical center of the proton ring is so great that we theorize that it is the strong nuclear force
(9383 8)(1602 19)8476 14
4 (4)(3142)(1411 26)p
pp
EB T
M
(9395 8)(1602 19)1035 15
4 (4)(3142)(0966 26)N
NN
EB T
M
Neutron
1a Hydrogenavi
lt to gt lt to gt f(ltgt)
lt -4 0073
lt -3 0097
lt -2 0143
lt -1 0273
origin 0 3000
gt 1 0273
gt 2 0143
gt 3 0097
gt 4 0073lt0073 lt0097 lt0143 lt0273
3000
0273gt 0143gt 0097gt 0073gt
Wave moving
snap shot of wave
Am
plit
ude
lt4 lt3 lt2 lt1 0 1gt 2gt 3gt 4gt
3
2
1
In SM NO squared terms are needed
SPRING
An Action is needed to stretch or compress a spring to a starting point where it can execute cycle motion
tsA (ts to start) is the work required to move a mass attached to a spring a distance (d1) from
its initial rest position RAK 0 the spring constant is the stiffness of the spring
o A is attraction o R is repulsion o
ARRA
00 is for the back and forth motion through the zero initial rest position A
stretched spring goes from ldquoArdquo to ldquo0rdquo to ldquoRrdquo to ldquoArdquo to ldquo0rdquo to ldquoRrdquo d1 is the distance a mass is moved from its initial rest position
0
0 1ts A RA K d
Once ( )mad
is performed to move the mass to position d1 (against the restraint of the spring) a of the ( )mad
is transferred into the spring as potential attractionrepulsion (PAR) There is always
more ( )mad
performed than is stored as PAR When the spring is released from the restraining Ats the PAR starts the oscillation Conservation of ( )mad
requires that the PAR and the motion ( )mad
( )( )madM be a constant
( )amp ( ) constantmadPAR M 02
0 0
1
2A R A RPAR K d
02
( ) 0 0
1( )
2mad A R A RM m v
0 0 02 2 2
0 0 0 1
2 2 21
2 2 21
1 1 1amp
2 2 2
1 1 1amp
2 2 2
amp
A R A R A RK d m v K d
Kd mv Kd
Kd mv Kd
0
1d is frac12 the height from the base to the peak of a cycle
An experiment of moving paper under a pendulum through a small angle filled with sand that can escape through a small hole in the bottom shows that a pendulum cycling back and forth follows a curve that is mathematically analogous to a lsquosinrsquo or lsquocosrsquo
The projection of the curved area onto the straight line exactly duplicate the back and forth motion of CM This allows for the math of lsquosinrsquo or lsquocosrsquo to be used
SM will use lsquosinrdquo math It could use lsquocosrsquo or a combination of both lsquosinrsquo and lsquocosrsquo Using just the lsquosinrsquo math makes the understanding and math easier
0 sin 0 0SM 0 sin157 1SM The diagram at the left shows a spring being stretched from its (zero rest) position-2 and the PAR being released at position-1 There are 4 sections (1 to 2 2 to 3 3 to 4 and 4 to5) that alternate between maximum acceleration and zero velocity to max velocity and zero acceleration
At position-1 there is maximum acceleration and zero velocity o The lsquosinrsquo values from 0o to 157o will be used for velocity (values from 0 to 1) o The lsquosinrsquo values from 157o to 0o will be used for acceleration (values from 1 to 0)
At position-2 there is maximum velocity and zero acceleration Acceleration changes to deceleration
o The lsquosinrsquo values from 157o to 0o will be used for deceleration (values from 1 to 0) o The lsquosinrsquo values from 0o to 157o will be used for acceleration (values from 0 to 1)
At position-3 there is maximum acceleration and zero velocity o The lsquosinrsquo values from 0o to 157o will be used for velocity (values from 0 to 1) o The lsquosinrsquo values from 157o to 0o will be used for acceleration (values from 1 to 0)
At position-4 there is maximum velocity and zero acceleration Acceleration changes to deceleration
o The lsquosinrsquo values from 157o to 0o will be used for deceleration (values from 1 to 0) o The lsquosinrsquo values from 0o to 157o will be used for acceleration (values from 0 to 1)
At position-5 the cycle starts again
2
)539(
cT
MK
spring constant (K) when mass (M) and cycle-time (T) are known
)(
))(89)((
111 md
uKgM
d
MG
d
FK sts
spring constant (K) when mass (M) and amplitude ( d1) are known
2 21( )
Kv d d
m
velocity (v) at any point during the cycle
velocity
dforM
Kdv
max
012 velocity (v2) maximum (when d = 0 at position-2 and 4)
acceleration (a1) maximum when spring constant (K) mass (M) and amplitude (d1) are known
acceleration (a1) maximum when amplitude (d1) and cycle-time (Tc) are known
acceleration (a) at any point during the cycle-time (T)
1
151
286
v
dTc
cycle-time (Tc) when amplitude (d1) and velocity (v1) are known
K
MTc 28651
cycle-time (Tc) when mass (M) and spring constant (K) are known
fTc
1
cycle-time (Tc) when the frequency is known
cTf
1
frequency (f) when the cycle-time (Tc) is known
of cycles
elapsed timef
frequency (f) when the number of cycles in an elapsed time is known
d1 amplitude (distance from rest to start position)
Also frac12 the distance from the base to the peak
3131 oror dM
Ka
251
3131 539
c
oror T
da
2539
cT
da
In December of 2006 I discovered the mathematical error in Einsteinrsquos Special Relativity Einstein used BS math and that math only works in an imaginary world
It cannot describe real symmetrical space
Einsteinrsquos Special Relativity should be abandoned and a theory of real space using SM should be used to replace it
Numerous articles on the Internet show Einsteinrsquos theory to be incorrect but no one in academia is willing to publish the information
Figure 1-10 A moving particle observed from two frames of reference in uniform translation
Einsteinrsquos Math Error in Special Relativity Transformation of Velocity (From Eisenberg- Fundamentals of Modern Physics pages 27-29) Consider the particle shown in figure (1-10) moving with velocity v as
seen in a frame of reference O
dx dy dz
Vx Vy Vzdt dt dt
dx dy dz
V x V y V zdy dt dt
We would like to evaluate the velocity v of the particle as seen in the frame of reference 0 which is itself moving relative to 0 with velocity v Measured in the 0 frame the velocity vector of the particle has com ponents
The same velocity vector as measured in the 0 frame has components
2
2
2
2
2
1
1
x vtx
v
cy y
z z
vxt
ctv
c
Now from equations (1-13) we know that the relation between the primed and the unprimed coordinates and times is
Take the differential of these equations remembering that v is a constant This gives
2
2
2
2
2
1
1
dx vdtdx
v
cdy dy
dz dz
vxdt
cdtv
c
2
2
2 2 2 2 2
2
2
2
11 ( )
1( ) 1
1
1
1
dx vdt
v dxdx vdt v
dx dx vdtc dt dtvdx vdx vdx v dxdt dt dt dt
dt dtc c c c
VxV x
Vx
VxV x
vv
vv
c
x
v
c
cV
( )
coordinate of particle relative to moving primed system
coordinate of particle relative to stationary system
velocity of primed coordina
V x
Vx
te system relative to stationaryv sys
tem
Page 28 Fundamentals of Modern Physics
(1-16)
According to Einstein this equation tells you how to transform the observed velocity from one frame of reference to another frame of reference First we note that as Vc and vc approachzero the equation approaches those which would be derived from the Galilean transformation
2 2
( ) 1 01 1 ( )
Vx v Vx vV
VxV x
Vx v
v cVxx
v
c
v
cc
Another property of this equation is that it is impossible to choose V and v such that Vrsquo the magnitude of the velocity which is seen in the new frame is greater than c
1
2
09 ( 09 ) 18009876543
( 09 ) 1811 (09 )
c c cV c c
cc
c
Example As seen by O particle 1 has velocity 09c in the direction of positive x and particle 2 has velocity 09c in the negative x direction
Einstein equation however predicts To evaluate the velocity of particle 1 with respect to particle 2 we transform from the O frame to the Orsquo frame moving in the negative x direction with velocity v = -09c using equation 1-16 We obtain
JK LOGIC will observe the particles separating at 18c
A confusing INCORRECT and ILLOGICAL answer
left is a dir rightsub of a
1
2
1
09 ( 09 ) 18009876543
181( 09 )1 ( 09 )
The multiplication of dir left time a dir right is ONLY a dir left
09
dir
direction leftdirection rightsubtraction
c c cV c c
cc
c
V
dir
22
22
09 18 18009876543
181( 081)1(09)(09)
1
left rightsub
sub
c c c cc c
ccc
c
Analyzing Einsteinrsquos BS math Equation
Using SM math a logical answer can be obtained
Cannot multiply a subtraction operator times a direction in space
Can only subtract and change its direction
Cannot multiply a direction to the left times a direction to the right
Cannot multiply opposite directions in space
Can only solve for the resultant
rightleft
subtraction
1
subtraction
2
sub of a dir to the left is a dir to the right
1 resultant 0 sub 2
2sub 2
2
09 09
09 amp 091
09 09 18 1
(0)1
(09) amp (09)1
c cV
c cc
c c cV
cc
cc
8
181
cc
SM math predicts a logical answer using logic
21st century scientists state in many articles that Einsteins General Relativity and Quantum Mechanics are incompatible
They are not only incompatible they are both mathematically incorrect
They both use illogical BS math
Any equations in math that use imaginary numbers should be abandoned
Imaginary numbers are only necessary in the current BS math system
This means that Schroumldingers equations should be abandoned
Any equation in quantum mechanics that uses imaginary number should be abandoned
Prof Kanarev has developed logical physics and chemistrytheories without the use of Quantum mechanics or relativity
A partial list of his theories coverbull Mechanics ndash replaces Newtonbull Photon - bull Electrons ndash Protons- Neutronsbull Spectroscopybull Nucleusbull Atoms amp Moleculesbull Electrodynamicsbull Heatbull Curvature of Space ndash replaces Schwarzcchildrsquos equation
Kanarevrsquos Hypothesis
bull Electron align linearly with proton They do NOT spin or form clouds around the nucleus
bull Neutrons are in the center of atom Proton are outside of Neutrons
bull The wavelength of all particles is the experimental radius
bull Double slit experiment explained correctly
bull Heisenbergrsquos inequality explained correctly
bull is the radius of approach of the magnetic lines around the torus of the electron
( 1 137)
bull Why water expands when it freezes
bull Electron clusters form lightning
rbull The meaning of Plancks constant is explained
22 2 kg m
h m f mr f consts
2210254 42m r kgm bull If the mr-constant describes a ring the equations
describe experiments bull If the mr-constant describes a wave the equations do
NOT describe experiments
Kanarevrsquos 5 Equations of Mechanics
Kanarevrsquos 1st Equation Acceleration ampmv maMma R R
Kanarevrsquos 2nd Equation Uniform motion amp mv maMma R R
maM mvR R
Kanarevrsquos 3rd Equation
Kanarevrsquos 4th Equation Equal and opposite reaction
1 21 21 2
12
a m
m a m ama
Kanarevrsquos 5th Equation 1 2 3 Mi M M M Mna R R R R R
The sum of the Actions working on a moving body is never equal to zero
Kanarevrsquos Photon model is a rolling ring
His equations accurately describe known experiments with photonsbull Electromagnetic radiation should be called photon radiationbull Diffraction and interference are the same phenomenabull Correctly explains the double slit experiment without quantum equationsbull Correctly explains the meaning of Heisenbergrsquos inequality
e
2 (2)(3142)(2817 15 ) 10007297352568
R 2426 12 137mag
xp
r m
m
Charge
Magnetic action
Proton
The intensity of the magnetic action near the geometrical center of the proton ring is so great that we theorize that it is the strong nuclear force
(9383 8)(1602 19)8476 14
4 (4)(3142)(1411 26)p
pp
EB T
M
(9395 8)(1602 19)1035 15
4 (4)(3142)(0966 26)N
NN
EB T
M
Neutron
1a Hydrogenavi
SPRING
An Action is needed to stretch or compress a spring to a starting point where it can execute cycle motion
tsA (ts to start) is the work required to move a mass attached to a spring a distance (d1) from
its initial rest position RAK 0 the spring constant is the stiffness of the spring
o A is attraction o R is repulsion o
ARRA
00 is for the back and forth motion through the zero initial rest position A
stretched spring goes from ldquoArdquo to ldquo0rdquo to ldquoRrdquo to ldquoArdquo to ldquo0rdquo to ldquoRrdquo d1 is the distance a mass is moved from its initial rest position
0
0 1ts A RA K d
Once ( )mad
is performed to move the mass to position d1 (against the restraint of the spring) a of the ( )mad
is transferred into the spring as potential attractionrepulsion (PAR) There is always
more ( )mad
performed than is stored as PAR When the spring is released from the restraining Ats the PAR starts the oscillation Conservation of ( )mad
requires that the PAR and the motion ( )mad
( )( )madM be a constant
( )amp ( ) constantmadPAR M 02
0 0
1
2A R A RPAR K d
02
( ) 0 0
1( )
2mad A R A RM m v
0 0 02 2 2
0 0 0 1
2 2 21
2 2 21
1 1 1amp
2 2 2
1 1 1amp
2 2 2
amp
A R A R A RK d m v K d
Kd mv Kd
Kd mv Kd
0
1d is frac12 the height from the base to the peak of a cycle
An experiment of moving paper under a pendulum through a small angle filled with sand that can escape through a small hole in the bottom shows that a pendulum cycling back and forth follows a curve that is mathematically analogous to a lsquosinrsquo or lsquocosrsquo
The projection of the curved area onto the straight line exactly duplicate the back and forth motion of CM This allows for the math of lsquosinrsquo or lsquocosrsquo to be used
SM will use lsquosinrdquo math It could use lsquocosrsquo or a combination of both lsquosinrsquo and lsquocosrsquo Using just the lsquosinrsquo math makes the understanding and math easier
0 sin 0 0SM 0 sin157 1SM The diagram at the left shows a spring being stretched from its (zero rest) position-2 and the PAR being released at position-1 There are 4 sections (1 to 2 2 to 3 3 to 4 and 4 to5) that alternate between maximum acceleration and zero velocity to max velocity and zero acceleration
At position-1 there is maximum acceleration and zero velocity o The lsquosinrsquo values from 0o to 157o will be used for velocity (values from 0 to 1) o The lsquosinrsquo values from 157o to 0o will be used for acceleration (values from 1 to 0)
At position-2 there is maximum velocity and zero acceleration Acceleration changes to deceleration
o The lsquosinrsquo values from 157o to 0o will be used for deceleration (values from 1 to 0) o The lsquosinrsquo values from 0o to 157o will be used for acceleration (values from 0 to 1)
At position-3 there is maximum acceleration and zero velocity o The lsquosinrsquo values from 0o to 157o will be used for velocity (values from 0 to 1) o The lsquosinrsquo values from 157o to 0o will be used for acceleration (values from 1 to 0)
At position-4 there is maximum velocity and zero acceleration Acceleration changes to deceleration
o The lsquosinrsquo values from 157o to 0o will be used for deceleration (values from 1 to 0) o The lsquosinrsquo values from 0o to 157o will be used for acceleration (values from 0 to 1)
At position-5 the cycle starts again
2
)539(
cT
MK
spring constant (K) when mass (M) and cycle-time (T) are known
)(
))(89)((
111 md
uKgM
d
MG
d
FK sts
spring constant (K) when mass (M) and amplitude ( d1) are known
2 21( )
Kv d d
m
velocity (v) at any point during the cycle
velocity
dforM
Kdv
max
012 velocity (v2) maximum (when d = 0 at position-2 and 4)
acceleration (a1) maximum when spring constant (K) mass (M) and amplitude (d1) are known
acceleration (a1) maximum when amplitude (d1) and cycle-time (Tc) are known
acceleration (a) at any point during the cycle-time (T)
1
151
286
v
dTc
cycle-time (Tc) when amplitude (d1) and velocity (v1) are known
K
MTc 28651
cycle-time (Tc) when mass (M) and spring constant (K) are known
fTc
1
cycle-time (Tc) when the frequency is known
cTf
1
frequency (f) when the cycle-time (Tc) is known
of cycles
elapsed timef
frequency (f) when the number of cycles in an elapsed time is known
d1 amplitude (distance from rest to start position)
Also frac12 the distance from the base to the peak
3131 oror dM
Ka
251
3131 539
c
oror T
da
2539
cT
da
In December of 2006 I discovered the mathematical error in Einsteinrsquos Special Relativity Einstein used BS math and that math only works in an imaginary world
It cannot describe real symmetrical space
Einsteinrsquos Special Relativity should be abandoned and a theory of real space using SM should be used to replace it
Numerous articles on the Internet show Einsteinrsquos theory to be incorrect but no one in academia is willing to publish the information
Figure 1-10 A moving particle observed from two frames of reference in uniform translation
Einsteinrsquos Math Error in Special Relativity Transformation of Velocity (From Eisenberg- Fundamentals of Modern Physics pages 27-29) Consider the particle shown in figure (1-10) moving with velocity v as
seen in a frame of reference O
dx dy dz
Vx Vy Vzdt dt dt
dx dy dz
V x V y V zdy dt dt
We would like to evaluate the velocity v of the particle as seen in the frame of reference 0 which is itself moving relative to 0 with velocity v Measured in the 0 frame the velocity vector of the particle has com ponents
The same velocity vector as measured in the 0 frame has components
2
2
2
2
2
1
1
x vtx
v
cy y
z z
vxt
ctv
c
Now from equations (1-13) we know that the relation between the primed and the unprimed coordinates and times is
Take the differential of these equations remembering that v is a constant This gives
2
2
2
2
2
1
1
dx vdtdx
v
cdy dy
dz dz
vxdt
cdtv
c
2
2
2 2 2 2 2
2
2
2
11 ( )
1( ) 1
1
1
1
dx vdt
v dxdx vdt v
dx dx vdtc dt dtvdx vdx vdx v dxdt dt dt dt
dt dtc c c c
VxV x
Vx
VxV x
vv
vv
c
x
v
c
cV
( )
coordinate of particle relative to moving primed system
coordinate of particle relative to stationary system
velocity of primed coordina
V x
Vx
te system relative to stationaryv sys
tem
Page 28 Fundamentals of Modern Physics
(1-16)
According to Einstein this equation tells you how to transform the observed velocity from one frame of reference to another frame of reference First we note that as Vc and vc approachzero the equation approaches those which would be derived from the Galilean transformation
2 2
( ) 1 01 1 ( )
Vx v Vx vV
VxV x
Vx v
v cVxx
v
c
v
cc
Another property of this equation is that it is impossible to choose V and v such that Vrsquo the magnitude of the velocity which is seen in the new frame is greater than c
1
2
09 ( 09 ) 18009876543
( 09 ) 1811 (09 )
c c cV c c
cc
c
Example As seen by O particle 1 has velocity 09c in the direction of positive x and particle 2 has velocity 09c in the negative x direction
Einstein equation however predicts To evaluate the velocity of particle 1 with respect to particle 2 we transform from the O frame to the Orsquo frame moving in the negative x direction with velocity v = -09c using equation 1-16 We obtain
JK LOGIC will observe the particles separating at 18c
A confusing INCORRECT and ILLOGICAL answer
left is a dir rightsub of a
1
2
1
09 ( 09 ) 18009876543
181( 09 )1 ( 09 )
The multiplication of dir left time a dir right is ONLY a dir left
09
dir
direction leftdirection rightsubtraction
c c cV c c
cc
c
V
dir
22
22
09 18 18009876543
181( 081)1(09)(09)
1
left rightsub
sub
c c c cc c
ccc
c
Analyzing Einsteinrsquos BS math Equation
Using SM math a logical answer can be obtained
Cannot multiply a subtraction operator times a direction in space
Can only subtract and change its direction
Cannot multiply a direction to the left times a direction to the right
Cannot multiply opposite directions in space
Can only solve for the resultant
rightleft
subtraction
1
subtraction
2
sub of a dir to the left is a dir to the right
1 resultant 0 sub 2
2sub 2
2
09 09
09 amp 091
09 09 18 1
(0)1
(09) amp (09)1
c cV
c cc
c c cV
cc
cc
8
181
cc
SM math predicts a logical answer using logic
21st century scientists state in many articles that Einsteins General Relativity and Quantum Mechanics are incompatible
They are not only incompatible they are both mathematically incorrect
They both use illogical BS math
Any equations in math that use imaginary numbers should be abandoned
Imaginary numbers are only necessary in the current BS math system
This means that Schroumldingers equations should be abandoned
Any equation in quantum mechanics that uses imaginary number should be abandoned
Prof Kanarev has developed logical physics and chemistrytheories without the use of Quantum mechanics or relativity
A partial list of his theories coverbull Mechanics ndash replaces Newtonbull Photon - bull Electrons ndash Protons- Neutronsbull Spectroscopybull Nucleusbull Atoms amp Moleculesbull Electrodynamicsbull Heatbull Curvature of Space ndash replaces Schwarzcchildrsquos equation
Kanarevrsquos Hypothesis
bull Electron align linearly with proton They do NOT spin or form clouds around the nucleus
bull Neutrons are in the center of atom Proton are outside of Neutrons
bull The wavelength of all particles is the experimental radius
bull Double slit experiment explained correctly
bull Heisenbergrsquos inequality explained correctly
bull is the radius of approach of the magnetic lines around the torus of the electron
( 1 137)
bull Why water expands when it freezes
bull Electron clusters form lightning
rbull The meaning of Plancks constant is explained
22 2 kg m
h m f mr f consts
2210254 42m r kgm bull If the mr-constant describes a ring the equations
describe experiments bull If the mr-constant describes a wave the equations do
NOT describe experiments
Kanarevrsquos 5 Equations of Mechanics
Kanarevrsquos 1st Equation Acceleration ampmv maMma R R
Kanarevrsquos 2nd Equation Uniform motion amp mv maMma R R
maM mvR R
Kanarevrsquos 3rd Equation
Kanarevrsquos 4th Equation Equal and opposite reaction
1 21 21 2
12
a m
m a m ama
Kanarevrsquos 5th Equation 1 2 3 Mi M M M Mna R R R R R
The sum of the Actions working on a moving body is never equal to zero
Kanarevrsquos Photon model is a rolling ring
His equations accurately describe known experiments with photonsbull Electromagnetic radiation should be called photon radiationbull Diffraction and interference are the same phenomenabull Correctly explains the double slit experiment without quantum equationsbull Correctly explains the meaning of Heisenbergrsquos inequality
e
2 (2)(3142)(2817 15 ) 10007297352568
R 2426 12 137mag
xp
r m
m
Charge
Magnetic action
Proton
The intensity of the magnetic action near the geometrical center of the proton ring is so great that we theorize that it is the strong nuclear force
(9383 8)(1602 19)8476 14
4 (4)(3142)(1411 26)p
pp
EB T
M
(9395 8)(1602 19)1035 15
4 (4)(3142)(0966 26)N
NN
EB T
M
Neutron
1a Hydrogenavi
Once ( )mad
is performed to move the mass to position d1 (against the restraint of the spring) a of the ( )mad
is transferred into the spring as potential attractionrepulsion (PAR) There is always
more ( )mad
performed than is stored as PAR When the spring is released from the restraining Ats the PAR starts the oscillation Conservation of ( )mad
requires that the PAR and the motion ( )mad
( )( )madM be a constant
( )amp ( ) constantmadPAR M 02
0 0
1
2A R A RPAR K d
02
( ) 0 0
1( )
2mad A R A RM m v
0 0 02 2 2
0 0 0 1
2 2 21
2 2 21
1 1 1amp
2 2 2
1 1 1amp
2 2 2
amp
A R A R A RK d m v K d
Kd mv Kd
Kd mv Kd
0
1d is frac12 the height from the base to the peak of a cycle
An experiment of moving paper under a pendulum through a small angle filled with sand that can escape through a small hole in the bottom shows that a pendulum cycling back and forth follows a curve that is mathematically analogous to a lsquosinrsquo or lsquocosrsquo
The projection of the curved area onto the straight line exactly duplicate the back and forth motion of CM This allows for the math of lsquosinrsquo or lsquocosrsquo to be used
SM will use lsquosinrdquo math It could use lsquocosrsquo or a combination of both lsquosinrsquo and lsquocosrsquo Using just the lsquosinrsquo math makes the understanding and math easier
0 sin 0 0SM 0 sin157 1SM The diagram at the left shows a spring being stretched from its (zero rest) position-2 and the PAR being released at position-1 There are 4 sections (1 to 2 2 to 3 3 to 4 and 4 to5) that alternate between maximum acceleration and zero velocity to max velocity and zero acceleration
At position-1 there is maximum acceleration and zero velocity o The lsquosinrsquo values from 0o to 157o will be used for velocity (values from 0 to 1) o The lsquosinrsquo values from 157o to 0o will be used for acceleration (values from 1 to 0)
At position-2 there is maximum velocity and zero acceleration Acceleration changes to deceleration
o The lsquosinrsquo values from 157o to 0o will be used for deceleration (values from 1 to 0) o The lsquosinrsquo values from 0o to 157o will be used for acceleration (values from 0 to 1)
At position-3 there is maximum acceleration and zero velocity o The lsquosinrsquo values from 0o to 157o will be used for velocity (values from 0 to 1) o The lsquosinrsquo values from 157o to 0o will be used for acceleration (values from 1 to 0)
At position-4 there is maximum velocity and zero acceleration Acceleration changes to deceleration
o The lsquosinrsquo values from 157o to 0o will be used for deceleration (values from 1 to 0) o The lsquosinrsquo values from 0o to 157o will be used for acceleration (values from 0 to 1)
At position-5 the cycle starts again
2
)539(
cT
MK
spring constant (K) when mass (M) and cycle-time (T) are known
)(
))(89)((
111 md
uKgM
d
MG
d
FK sts
spring constant (K) when mass (M) and amplitude ( d1) are known
2 21( )
Kv d d
m
velocity (v) at any point during the cycle
velocity
dforM
Kdv
max
012 velocity (v2) maximum (when d = 0 at position-2 and 4)
acceleration (a1) maximum when spring constant (K) mass (M) and amplitude (d1) are known
acceleration (a1) maximum when amplitude (d1) and cycle-time (Tc) are known
acceleration (a) at any point during the cycle-time (T)
1
151
286
v
dTc
cycle-time (Tc) when amplitude (d1) and velocity (v1) are known
K
MTc 28651
cycle-time (Tc) when mass (M) and spring constant (K) are known
fTc
1
cycle-time (Tc) when the frequency is known
cTf
1
frequency (f) when the cycle-time (Tc) is known
of cycles
elapsed timef
frequency (f) when the number of cycles in an elapsed time is known
d1 amplitude (distance from rest to start position)
Also frac12 the distance from the base to the peak
3131 oror dM
Ka
251
3131 539
c
oror T
da
2539
cT
da
In December of 2006 I discovered the mathematical error in Einsteinrsquos Special Relativity Einstein used BS math and that math only works in an imaginary world
It cannot describe real symmetrical space
Einsteinrsquos Special Relativity should be abandoned and a theory of real space using SM should be used to replace it
Numerous articles on the Internet show Einsteinrsquos theory to be incorrect but no one in academia is willing to publish the information
Figure 1-10 A moving particle observed from two frames of reference in uniform translation
Einsteinrsquos Math Error in Special Relativity Transformation of Velocity (From Eisenberg- Fundamentals of Modern Physics pages 27-29) Consider the particle shown in figure (1-10) moving with velocity v as
seen in a frame of reference O
dx dy dz
Vx Vy Vzdt dt dt
dx dy dz
V x V y V zdy dt dt
We would like to evaluate the velocity v of the particle as seen in the frame of reference 0 which is itself moving relative to 0 with velocity v Measured in the 0 frame the velocity vector of the particle has com ponents
The same velocity vector as measured in the 0 frame has components
2
2
2
2
2
1
1
x vtx
v
cy y
z z
vxt
ctv
c
Now from equations (1-13) we know that the relation between the primed and the unprimed coordinates and times is
Take the differential of these equations remembering that v is a constant This gives
2
2
2
2
2
1
1
dx vdtdx
v
cdy dy
dz dz
vxdt
cdtv
c
2
2
2 2 2 2 2
2
2
2
11 ( )
1( ) 1
1
1
1
dx vdt
v dxdx vdt v
dx dx vdtc dt dtvdx vdx vdx v dxdt dt dt dt
dt dtc c c c
VxV x
Vx
VxV x
vv
vv
c
x
v
c
cV
( )
coordinate of particle relative to moving primed system
coordinate of particle relative to stationary system
velocity of primed coordina
V x
Vx
te system relative to stationaryv sys
tem
Page 28 Fundamentals of Modern Physics
(1-16)
According to Einstein this equation tells you how to transform the observed velocity from one frame of reference to another frame of reference First we note that as Vc and vc approachzero the equation approaches those which would be derived from the Galilean transformation
2 2
( ) 1 01 1 ( )
Vx v Vx vV
VxV x
Vx v
v cVxx
v
c
v
cc
Another property of this equation is that it is impossible to choose V and v such that Vrsquo the magnitude of the velocity which is seen in the new frame is greater than c
1
2
09 ( 09 ) 18009876543
( 09 ) 1811 (09 )
c c cV c c
cc
c
Example As seen by O particle 1 has velocity 09c in the direction of positive x and particle 2 has velocity 09c in the negative x direction
Einstein equation however predicts To evaluate the velocity of particle 1 with respect to particle 2 we transform from the O frame to the Orsquo frame moving in the negative x direction with velocity v = -09c using equation 1-16 We obtain
JK LOGIC will observe the particles separating at 18c
A confusing INCORRECT and ILLOGICAL answer
left is a dir rightsub of a
1
2
1
09 ( 09 ) 18009876543
181( 09 )1 ( 09 )
The multiplication of dir left time a dir right is ONLY a dir left
09
dir
direction leftdirection rightsubtraction
c c cV c c
cc
c
V
dir
22
22
09 18 18009876543
181( 081)1(09)(09)
1
left rightsub
sub
c c c cc c
ccc
c
Analyzing Einsteinrsquos BS math Equation
Using SM math a logical answer can be obtained
Cannot multiply a subtraction operator times a direction in space
Can only subtract and change its direction
Cannot multiply a direction to the left times a direction to the right
Cannot multiply opposite directions in space
Can only solve for the resultant
rightleft
subtraction
1
subtraction
2
sub of a dir to the left is a dir to the right
1 resultant 0 sub 2
2sub 2
2
09 09
09 amp 091
09 09 18 1
(0)1
(09) amp (09)1
c cV
c cc
c c cV
cc
cc
8
181
cc
SM math predicts a logical answer using logic
21st century scientists state in many articles that Einsteins General Relativity and Quantum Mechanics are incompatible
They are not only incompatible they are both mathematically incorrect
They both use illogical BS math
Any equations in math that use imaginary numbers should be abandoned
Imaginary numbers are only necessary in the current BS math system
This means that Schroumldingers equations should be abandoned
Any equation in quantum mechanics that uses imaginary number should be abandoned
Prof Kanarev has developed logical physics and chemistrytheories without the use of Quantum mechanics or relativity
A partial list of his theories coverbull Mechanics ndash replaces Newtonbull Photon - bull Electrons ndash Protons- Neutronsbull Spectroscopybull Nucleusbull Atoms amp Moleculesbull Electrodynamicsbull Heatbull Curvature of Space ndash replaces Schwarzcchildrsquos equation
Kanarevrsquos Hypothesis
bull Electron align linearly with proton They do NOT spin or form clouds around the nucleus
bull Neutrons are in the center of atom Proton are outside of Neutrons
bull The wavelength of all particles is the experimental radius
bull Double slit experiment explained correctly
bull Heisenbergrsquos inequality explained correctly
bull is the radius of approach of the magnetic lines around the torus of the electron
( 1 137)
bull Why water expands when it freezes
bull Electron clusters form lightning
rbull The meaning of Plancks constant is explained
22 2 kg m
h m f mr f consts
2210254 42m r kgm bull If the mr-constant describes a ring the equations
describe experiments bull If the mr-constant describes a wave the equations do
NOT describe experiments
Kanarevrsquos 5 Equations of Mechanics
Kanarevrsquos 1st Equation Acceleration ampmv maMma R R
Kanarevrsquos 2nd Equation Uniform motion amp mv maMma R R
maM mvR R
Kanarevrsquos 3rd Equation
Kanarevrsquos 4th Equation Equal and opposite reaction
1 21 21 2
12
a m
m a m ama
Kanarevrsquos 5th Equation 1 2 3 Mi M M M Mna R R R R R
The sum of the Actions working on a moving body is never equal to zero
Kanarevrsquos Photon model is a rolling ring
His equations accurately describe known experiments with photonsbull Electromagnetic radiation should be called photon radiationbull Diffraction and interference are the same phenomenabull Correctly explains the double slit experiment without quantum equationsbull Correctly explains the meaning of Heisenbergrsquos inequality
e
2 (2)(3142)(2817 15 ) 10007297352568
R 2426 12 137mag
xp
r m
m
Charge
Magnetic action
Proton
The intensity of the magnetic action near the geometrical center of the proton ring is so great that we theorize that it is the strong nuclear force
(9383 8)(1602 19)8476 14
4 (4)(3142)(1411 26)p
pp
EB T
M
(9395 8)(1602 19)1035 15
4 (4)(3142)(0966 26)N
NN
EB T
M
Neutron
1a Hydrogenavi
An experiment of moving paper under a pendulum through a small angle filled with sand that can escape through a small hole in the bottom shows that a pendulum cycling back and forth follows a curve that is mathematically analogous to a lsquosinrsquo or lsquocosrsquo
The projection of the curved area onto the straight line exactly duplicate the back and forth motion of CM This allows for the math of lsquosinrsquo or lsquocosrsquo to be used
SM will use lsquosinrdquo math It could use lsquocosrsquo or a combination of both lsquosinrsquo and lsquocosrsquo Using just the lsquosinrsquo math makes the understanding and math easier
0 sin 0 0SM 0 sin157 1SM The diagram at the left shows a spring being stretched from its (zero rest) position-2 and the PAR being released at position-1 There are 4 sections (1 to 2 2 to 3 3 to 4 and 4 to5) that alternate between maximum acceleration and zero velocity to max velocity and zero acceleration
At position-1 there is maximum acceleration and zero velocity o The lsquosinrsquo values from 0o to 157o will be used for velocity (values from 0 to 1) o The lsquosinrsquo values from 157o to 0o will be used for acceleration (values from 1 to 0)
At position-2 there is maximum velocity and zero acceleration Acceleration changes to deceleration
o The lsquosinrsquo values from 157o to 0o will be used for deceleration (values from 1 to 0) o The lsquosinrsquo values from 0o to 157o will be used for acceleration (values from 0 to 1)
At position-3 there is maximum acceleration and zero velocity o The lsquosinrsquo values from 0o to 157o will be used for velocity (values from 0 to 1) o The lsquosinrsquo values from 157o to 0o will be used for acceleration (values from 1 to 0)
At position-4 there is maximum velocity and zero acceleration Acceleration changes to deceleration
o The lsquosinrsquo values from 157o to 0o will be used for deceleration (values from 1 to 0) o The lsquosinrsquo values from 0o to 157o will be used for acceleration (values from 0 to 1)
At position-5 the cycle starts again
2
)539(
cT
MK
spring constant (K) when mass (M) and cycle-time (T) are known
)(
))(89)((
111 md
uKgM
d
MG
d
FK sts
spring constant (K) when mass (M) and amplitude ( d1) are known
2 21( )
Kv d d
m
velocity (v) at any point during the cycle
velocity
dforM
Kdv
max
012 velocity (v2) maximum (when d = 0 at position-2 and 4)
acceleration (a1) maximum when spring constant (K) mass (M) and amplitude (d1) are known
acceleration (a1) maximum when amplitude (d1) and cycle-time (Tc) are known
acceleration (a) at any point during the cycle-time (T)
1
151
286
v
dTc
cycle-time (Tc) when amplitude (d1) and velocity (v1) are known
K
MTc 28651
cycle-time (Tc) when mass (M) and spring constant (K) are known
fTc
1
cycle-time (Tc) when the frequency is known
cTf
1
frequency (f) when the cycle-time (Tc) is known
of cycles
elapsed timef
frequency (f) when the number of cycles in an elapsed time is known
d1 amplitude (distance from rest to start position)
Also frac12 the distance from the base to the peak
3131 oror dM
Ka
251
3131 539
c
oror T
da
2539
cT
da
In December of 2006 I discovered the mathematical error in Einsteinrsquos Special Relativity Einstein used BS math and that math only works in an imaginary world
It cannot describe real symmetrical space
Einsteinrsquos Special Relativity should be abandoned and a theory of real space using SM should be used to replace it
Numerous articles on the Internet show Einsteinrsquos theory to be incorrect but no one in academia is willing to publish the information
Figure 1-10 A moving particle observed from two frames of reference in uniform translation
Einsteinrsquos Math Error in Special Relativity Transformation of Velocity (From Eisenberg- Fundamentals of Modern Physics pages 27-29) Consider the particle shown in figure (1-10) moving with velocity v as
seen in a frame of reference O
dx dy dz
Vx Vy Vzdt dt dt
dx dy dz
V x V y V zdy dt dt
We would like to evaluate the velocity v of the particle as seen in the frame of reference 0 which is itself moving relative to 0 with velocity v Measured in the 0 frame the velocity vector of the particle has com ponents
The same velocity vector as measured in the 0 frame has components
2
2
2
2
2
1
1
x vtx
v
cy y
z z
vxt
ctv
c
Now from equations (1-13) we know that the relation between the primed and the unprimed coordinates and times is
Take the differential of these equations remembering that v is a constant This gives
2
2
2
2
2
1
1
dx vdtdx
v
cdy dy
dz dz
vxdt
cdtv
c
2
2
2 2 2 2 2
2
2
2
11 ( )
1( ) 1
1
1
1
dx vdt
v dxdx vdt v
dx dx vdtc dt dtvdx vdx vdx v dxdt dt dt dt
dt dtc c c c
VxV x
Vx
VxV x
vv
vv
c
x
v
c
cV
( )
coordinate of particle relative to moving primed system
coordinate of particle relative to stationary system
velocity of primed coordina
V x
Vx
te system relative to stationaryv sys
tem
Page 28 Fundamentals of Modern Physics
(1-16)
According to Einstein this equation tells you how to transform the observed velocity from one frame of reference to another frame of reference First we note that as Vc and vc approachzero the equation approaches those which would be derived from the Galilean transformation
2 2
( ) 1 01 1 ( )
Vx v Vx vV
VxV x
Vx v
v cVxx
v
c
v
cc
Another property of this equation is that it is impossible to choose V and v such that Vrsquo the magnitude of the velocity which is seen in the new frame is greater than c
1
2
09 ( 09 ) 18009876543
( 09 ) 1811 (09 )
c c cV c c
cc
c
Example As seen by O particle 1 has velocity 09c in the direction of positive x and particle 2 has velocity 09c in the negative x direction
Einstein equation however predicts To evaluate the velocity of particle 1 with respect to particle 2 we transform from the O frame to the Orsquo frame moving in the negative x direction with velocity v = -09c using equation 1-16 We obtain
JK LOGIC will observe the particles separating at 18c
A confusing INCORRECT and ILLOGICAL answer
left is a dir rightsub of a
1
2
1
09 ( 09 ) 18009876543
181( 09 )1 ( 09 )
The multiplication of dir left time a dir right is ONLY a dir left
09
dir
direction leftdirection rightsubtraction
c c cV c c
cc
c
V
dir
22
22
09 18 18009876543
181( 081)1(09)(09)
1
left rightsub
sub
c c c cc c
ccc
c
Analyzing Einsteinrsquos BS math Equation
Using SM math a logical answer can be obtained
Cannot multiply a subtraction operator times a direction in space
Can only subtract and change its direction
Cannot multiply a direction to the left times a direction to the right
Cannot multiply opposite directions in space
Can only solve for the resultant
rightleft
subtraction
1
subtraction
2
sub of a dir to the left is a dir to the right
1 resultant 0 sub 2
2sub 2
2
09 09
09 amp 091
09 09 18 1
(0)1
(09) amp (09)1
c cV
c cc
c c cV
cc
cc
8
181
cc
SM math predicts a logical answer using logic
21st century scientists state in many articles that Einsteins General Relativity and Quantum Mechanics are incompatible
They are not only incompatible they are both mathematically incorrect
They both use illogical BS math
Any equations in math that use imaginary numbers should be abandoned
Imaginary numbers are only necessary in the current BS math system
This means that Schroumldingers equations should be abandoned
Any equation in quantum mechanics that uses imaginary number should be abandoned
Prof Kanarev has developed logical physics and chemistrytheories without the use of Quantum mechanics or relativity
A partial list of his theories coverbull Mechanics ndash replaces Newtonbull Photon - bull Electrons ndash Protons- Neutronsbull Spectroscopybull Nucleusbull Atoms amp Moleculesbull Electrodynamicsbull Heatbull Curvature of Space ndash replaces Schwarzcchildrsquos equation
Kanarevrsquos Hypothesis
bull Electron align linearly with proton They do NOT spin or form clouds around the nucleus
bull Neutrons are in the center of atom Proton are outside of Neutrons
bull The wavelength of all particles is the experimental radius
bull Double slit experiment explained correctly
bull Heisenbergrsquos inequality explained correctly
bull is the radius of approach of the magnetic lines around the torus of the electron
( 1 137)
bull Why water expands when it freezes
bull Electron clusters form lightning
rbull The meaning of Plancks constant is explained
22 2 kg m
h m f mr f consts
2210254 42m r kgm bull If the mr-constant describes a ring the equations
describe experiments bull If the mr-constant describes a wave the equations do
NOT describe experiments
Kanarevrsquos 5 Equations of Mechanics
Kanarevrsquos 1st Equation Acceleration ampmv maMma R R
Kanarevrsquos 2nd Equation Uniform motion amp mv maMma R R
maM mvR R
Kanarevrsquos 3rd Equation
Kanarevrsquos 4th Equation Equal and opposite reaction
1 21 21 2
12
a m
m a m ama
Kanarevrsquos 5th Equation 1 2 3 Mi M M M Mna R R R R R
The sum of the Actions working on a moving body is never equal to zero
Kanarevrsquos Photon model is a rolling ring
His equations accurately describe known experiments with photonsbull Electromagnetic radiation should be called photon radiationbull Diffraction and interference are the same phenomenabull Correctly explains the double slit experiment without quantum equationsbull Correctly explains the meaning of Heisenbergrsquos inequality
e
2 (2)(3142)(2817 15 ) 10007297352568
R 2426 12 137mag
xp
r m
m
Charge
Magnetic action
Proton
The intensity of the magnetic action near the geometrical center of the proton ring is so great that we theorize that it is the strong nuclear force
(9383 8)(1602 19)8476 14
4 (4)(3142)(1411 26)p
pp
EB T
M
(9395 8)(1602 19)1035 15
4 (4)(3142)(0966 26)N
NN
EB T
M
Neutron
1a Hydrogenavi
SM will use lsquosinrdquo math It could use lsquocosrsquo or a combination of both lsquosinrsquo and lsquocosrsquo Using just the lsquosinrsquo math makes the understanding and math easier
0 sin 0 0SM 0 sin157 1SM The diagram at the left shows a spring being stretched from its (zero rest) position-2 and the PAR being released at position-1 There are 4 sections (1 to 2 2 to 3 3 to 4 and 4 to5) that alternate between maximum acceleration and zero velocity to max velocity and zero acceleration
At position-1 there is maximum acceleration and zero velocity o The lsquosinrsquo values from 0o to 157o will be used for velocity (values from 0 to 1) o The lsquosinrsquo values from 157o to 0o will be used for acceleration (values from 1 to 0)
At position-2 there is maximum velocity and zero acceleration Acceleration changes to deceleration
o The lsquosinrsquo values from 157o to 0o will be used for deceleration (values from 1 to 0) o The lsquosinrsquo values from 0o to 157o will be used for acceleration (values from 0 to 1)
At position-3 there is maximum acceleration and zero velocity o The lsquosinrsquo values from 0o to 157o will be used for velocity (values from 0 to 1) o The lsquosinrsquo values from 157o to 0o will be used for acceleration (values from 1 to 0)
At position-4 there is maximum velocity and zero acceleration Acceleration changes to deceleration
o The lsquosinrsquo values from 157o to 0o will be used for deceleration (values from 1 to 0) o The lsquosinrsquo values from 0o to 157o will be used for acceleration (values from 0 to 1)
At position-5 the cycle starts again
2
)539(
cT
MK
spring constant (K) when mass (M) and cycle-time (T) are known
)(
))(89)((
111 md
uKgM
d
MG
d
FK sts
spring constant (K) when mass (M) and amplitude ( d1) are known
2 21( )
Kv d d
m
velocity (v) at any point during the cycle
velocity
dforM
Kdv
max
012 velocity (v2) maximum (when d = 0 at position-2 and 4)
acceleration (a1) maximum when spring constant (K) mass (M) and amplitude (d1) are known
acceleration (a1) maximum when amplitude (d1) and cycle-time (Tc) are known
acceleration (a) at any point during the cycle-time (T)
1
151
286
v
dTc
cycle-time (Tc) when amplitude (d1) and velocity (v1) are known
K
MTc 28651
cycle-time (Tc) when mass (M) and spring constant (K) are known
fTc
1
cycle-time (Tc) when the frequency is known
cTf
1
frequency (f) when the cycle-time (Tc) is known
of cycles
elapsed timef
frequency (f) when the number of cycles in an elapsed time is known
d1 amplitude (distance from rest to start position)
Also frac12 the distance from the base to the peak
3131 oror dM
Ka
251
3131 539
c
oror T
da
2539
cT
da
In December of 2006 I discovered the mathematical error in Einsteinrsquos Special Relativity Einstein used BS math and that math only works in an imaginary world
It cannot describe real symmetrical space
Einsteinrsquos Special Relativity should be abandoned and a theory of real space using SM should be used to replace it
Numerous articles on the Internet show Einsteinrsquos theory to be incorrect but no one in academia is willing to publish the information
Figure 1-10 A moving particle observed from two frames of reference in uniform translation
Einsteinrsquos Math Error in Special Relativity Transformation of Velocity (From Eisenberg- Fundamentals of Modern Physics pages 27-29) Consider the particle shown in figure (1-10) moving with velocity v as
seen in a frame of reference O
dx dy dz
Vx Vy Vzdt dt dt
dx dy dz
V x V y V zdy dt dt
We would like to evaluate the velocity v of the particle as seen in the frame of reference 0 which is itself moving relative to 0 with velocity v Measured in the 0 frame the velocity vector of the particle has com ponents
The same velocity vector as measured in the 0 frame has components
2
2
2
2
2
1
1
x vtx
v
cy y
z z
vxt
ctv
c
Now from equations (1-13) we know that the relation between the primed and the unprimed coordinates and times is
Take the differential of these equations remembering that v is a constant This gives
2
2
2
2
2
1
1
dx vdtdx
v
cdy dy
dz dz
vxdt
cdtv
c
2
2
2 2 2 2 2
2
2
2
11 ( )
1( ) 1
1
1
1
dx vdt
v dxdx vdt v
dx dx vdtc dt dtvdx vdx vdx v dxdt dt dt dt
dt dtc c c c
VxV x
Vx
VxV x
vv
vv
c
x
v
c
cV
( )
coordinate of particle relative to moving primed system
coordinate of particle relative to stationary system
velocity of primed coordina
V x
Vx
te system relative to stationaryv sys
tem
Page 28 Fundamentals of Modern Physics
(1-16)
According to Einstein this equation tells you how to transform the observed velocity from one frame of reference to another frame of reference First we note that as Vc and vc approachzero the equation approaches those which would be derived from the Galilean transformation
2 2
( ) 1 01 1 ( )
Vx v Vx vV
VxV x
Vx v
v cVxx
v
c
v
cc
Another property of this equation is that it is impossible to choose V and v such that Vrsquo the magnitude of the velocity which is seen in the new frame is greater than c
1
2
09 ( 09 ) 18009876543
( 09 ) 1811 (09 )
c c cV c c
cc
c
Example As seen by O particle 1 has velocity 09c in the direction of positive x and particle 2 has velocity 09c in the negative x direction
Einstein equation however predicts To evaluate the velocity of particle 1 with respect to particle 2 we transform from the O frame to the Orsquo frame moving in the negative x direction with velocity v = -09c using equation 1-16 We obtain
JK LOGIC will observe the particles separating at 18c
A confusing INCORRECT and ILLOGICAL answer
left is a dir rightsub of a
1
2
1
09 ( 09 ) 18009876543
181( 09 )1 ( 09 )
The multiplication of dir left time a dir right is ONLY a dir left
09
dir
direction leftdirection rightsubtraction
c c cV c c
cc
c
V
dir
22
22
09 18 18009876543
181( 081)1(09)(09)
1
left rightsub
sub
c c c cc c
ccc
c
Analyzing Einsteinrsquos BS math Equation
Using SM math a logical answer can be obtained
Cannot multiply a subtraction operator times a direction in space
Can only subtract and change its direction
Cannot multiply a direction to the left times a direction to the right
Cannot multiply opposite directions in space
Can only solve for the resultant
rightleft
subtraction
1
subtraction
2
sub of a dir to the left is a dir to the right
1 resultant 0 sub 2
2sub 2
2
09 09
09 amp 091
09 09 18 1
(0)1
(09) amp (09)1
c cV
c cc
c c cV
cc
cc
8
181
cc
SM math predicts a logical answer using logic
21st century scientists state in many articles that Einsteins General Relativity and Quantum Mechanics are incompatible
They are not only incompatible they are both mathematically incorrect
They both use illogical BS math
Any equations in math that use imaginary numbers should be abandoned
Imaginary numbers are only necessary in the current BS math system
This means that Schroumldingers equations should be abandoned
Any equation in quantum mechanics that uses imaginary number should be abandoned
Prof Kanarev has developed logical physics and chemistrytheories without the use of Quantum mechanics or relativity
A partial list of his theories coverbull Mechanics ndash replaces Newtonbull Photon - bull Electrons ndash Protons- Neutronsbull Spectroscopybull Nucleusbull Atoms amp Moleculesbull Electrodynamicsbull Heatbull Curvature of Space ndash replaces Schwarzcchildrsquos equation
Kanarevrsquos Hypothesis
bull Electron align linearly with proton They do NOT spin or form clouds around the nucleus
bull Neutrons are in the center of atom Proton are outside of Neutrons
bull The wavelength of all particles is the experimental radius
bull Double slit experiment explained correctly
bull Heisenbergrsquos inequality explained correctly
bull is the radius of approach of the magnetic lines around the torus of the electron
( 1 137)
bull Why water expands when it freezes
bull Electron clusters form lightning
rbull The meaning of Plancks constant is explained
22 2 kg m
h m f mr f consts
2210254 42m r kgm bull If the mr-constant describes a ring the equations
describe experiments bull If the mr-constant describes a wave the equations do
NOT describe experiments
Kanarevrsquos 5 Equations of Mechanics
Kanarevrsquos 1st Equation Acceleration ampmv maMma R R
Kanarevrsquos 2nd Equation Uniform motion amp mv maMma R R
maM mvR R
Kanarevrsquos 3rd Equation
Kanarevrsquos 4th Equation Equal and opposite reaction
1 21 21 2
12
a m
m a m ama
Kanarevrsquos 5th Equation 1 2 3 Mi M M M Mna R R R R R
The sum of the Actions working on a moving body is never equal to zero
Kanarevrsquos Photon model is a rolling ring
His equations accurately describe known experiments with photonsbull Electromagnetic radiation should be called photon radiationbull Diffraction and interference are the same phenomenabull Correctly explains the double slit experiment without quantum equationsbull Correctly explains the meaning of Heisenbergrsquos inequality
e
2 (2)(3142)(2817 15 ) 10007297352568
R 2426 12 137mag
xp
r m
m
Charge
Magnetic action
Proton
The intensity of the magnetic action near the geometrical center of the proton ring is so great that we theorize that it is the strong nuclear force
(9383 8)(1602 19)8476 14
4 (4)(3142)(1411 26)p
pp
EB T
M
(9395 8)(1602 19)1035 15
4 (4)(3142)(0966 26)N
NN
EB T
M
Neutron
1a Hydrogenavi
At position-1 there is maximum acceleration and zero velocity o The lsquosinrsquo values from 0o to 157o will be used for velocity (values from 0 to 1) o The lsquosinrsquo values from 157o to 0o will be used for acceleration (values from 1 to 0)
At position-2 there is maximum velocity and zero acceleration Acceleration changes to deceleration
o The lsquosinrsquo values from 157o to 0o will be used for deceleration (values from 1 to 0) o The lsquosinrsquo values from 0o to 157o will be used for acceleration (values from 0 to 1)
At position-3 there is maximum acceleration and zero velocity o The lsquosinrsquo values from 0o to 157o will be used for velocity (values from 0 to 1) o The lsquosinrsquo values from 157o to 0o will be used for acceleration (values from 1 to 0)
At position-4 there is maximum velocity and zero acceleration Acceleration changes to deceleration
o The lsquosinrsquo values from 157o to 0o will be used for deceleration (values from 1 to 0) o The lsquosinrsquo values from 0o to 157o will be used for acceleration (values from 0 to 1)
At position-5 the cycle starts again
2
)539(
cT
MK
spring constant (K) when mass (M) and cycle-time (T) are known
)(
))(89)((
111 md
uKgM
d
MG
d
FK sts
spring constant (K) when mass (M) and amplitude ( d1) are known
2 21( )
Kv d d
m
velocity (v) at any point during the cycle
velocity
dforM
Kdv
max
012 velocity (v2) maximum (when d = 0 at position-2 and 4)
acceleration (a1) maximum when spring constant (K) mass (M) and amplitude (d1) are known
acceleration (a1) maximum when amplitude (d1) and cycle-time (Tc) are known
acceleration (a) at any point during the cycle-time (T)
1
151
286
v
dTc
cycle-time (Tc) when amplitude (d1) and velocity (v1) are known
K
MTc 28651
cycle-time (Tc) when mass (M) and spring constant (K) are known
fTc
1
cycle-time (Tc) when the frequency is known
cTf
1
frequency (f) when the cycle-time (Tc) is known
of cycles
elapsed timef
frequency (f) when the number of cycles in an elapsed time is known
d1 amplitude (distance from rest to start position)
Also frac12 the distance from the base to the peak
3131 oror dM
Ka
251
3131 539
c
oror T
da
2539
cT
da
In December of 2006 I discovered the mathematical error in Einsteinrsquos Special Relativity Einstein used BS math and that math only works in an imaginary world
It cannot describe real symmetrical space
Einsteinrsquos Special Relativity should be abandoned and a theory of real space using SM should be used to replace it
Numerous articles on the Internet show Einsteinrsquos theory to be incorrect but no one in academia is willing to publish the information
Figure 1-10 A moving particle observed from two frames of reference in uniform translation
Einsteinrsquos Math Error in Special Relativity Transformation of Velocity (From Eisenberg- Fundamentals of Modern Physics pages 27-29) Consider the particle shown in figure (1-10) moving with velocity v as
seen in a frame of reference O
dx dy dz
Vx Vy Vzdt dt dt
dx dy dz
V x V y V zdy dt dt
We would like to evaluate the velocity v of the particle as seen in the frame of reference 0 which is itself moving relative to 0 with velocity v Measured in the 0 frame the velocity vector of the particle has com ponents
The same velocity vector as measured in the 0 frame has components
2
2
2
2
2
1
1
x vtx
v
cy y
z z
vxt
ctv
c
Now from equations (1-13) we know that the relation between the primed and the unprimed coordinates and times is
Take the differential of these equations remembering that v is a constant This gives
2
2
2
2
2
1
1
dx vdtdx
v
cdy dy
dz dz
vxdt
cdtv
c
2
2
2 2 2 2 2
2
2
2
11 ( )
1( ) 1
1
1
1
dx vdt
v dxdx vdt v
dx dx vdtc dt dtvdx vdx vdx v dxdt dt dt dt
dt dtc c c c
VxV x
Vx
VxV x
vv
vv
c
x
v
c
cV
( )
coordinate of particle relative to moving primed system
coordinate of particle relative to stationary system
velocity of primed coordina
V x
Vx
te system relative to stationaryv sys
tem
Page 28 Fundamentals of Modern Physics
(1-16)
According to Einstein this equation tells you how to transform the observed velocity from one frame of reference to another frame of reference First we note that as Vc and vc approachzero the equation approaches those which would be derived from the Galilean transformation
2 2
( ) 1 01 1 ( )
Vx v Vx vV
VxV x
Vx v
v cVxx
v
c
v
cc
Another property of this equation is that it is impossible to choose V and v such that Vrsquo the magnitude of the velocity which is seen in the new frame is greater than c
1
2
09 ( 09 ) 18009876543
( 09 ) 1811 (09 )
c c cV c c
cc
c
Example As seen by O particle 1 has velocity 09c in the direction of positive x and particle 2 has velocity 09c in the negative x direction
Einstein equation however predicts To evaluate the velocity of particle 1 with respect to particle 2 we transform from the O frame to the Orsquo frame moving in the negative x direction with velocity v = -09c using equation 1-16 We obtain
JK LOGIC will observe the particles separating at 18c
A confusing INCORRECT and ILLOGICAL answer
left is a dir rightsub of a
1
2
1
09 ( 09 ) 18009876543
181( 09 )1 ( 09 )
The multiplication of dir left time a dir right is ONLY a dir left
09
dir
direction leftdirection rightsubtraction
c c cV c c
cc
c
V
dir
22
22
09 18 18009876543
181( 081)1(09)(09)
1
left rightsub
sub
c c c cc c
ccc
c
Analyzing Einsteinrsquos BS math Equation
Using SM math a logical answer can be obtained
Cannot multiply a subtraction operator times a direction in space
Can only subtract and change its direction
Cannot multiply a direction to the left times a direction to the right
Cannot multiply opposite directions in space
Can only solve for the resultant
rightleft
subtraction
1
subtraction
2
sub of a dir to the left is a dir to the right
1 resultant 0 sub 2
2sub 2
2
09 09
09 amp 091
09 09 18 1
(0)1
(09) amp (09)1
c cV
c cc
c c cV
cc
cc
8
181
cc
SM math predicts a logical answer using logic
21st century scientists state in many articles that Einsteins General Relativity and Quantum Mechanics are incompatible
They are not only incompatible they are both mathematically incorrect
They both use illogical BS math
Any equations in math that use imaginary numbers should be abandoned
Imaginary numbers are only necessary in the current BS math system
This means that Schroumldingers equations should be abandoned
Any equation in quantum mechanics that uses imaginary number should be abandoned
Prof Kanarev has developed logical physics and chemistrytheories without the use of Quantum mechanics or relativity
A partial list of his theories coverbull Mechanics ndash replaces Newtonbull Photon - bull Electrons ndash Protons- Neutronsbull Spectroscopybull Nucleusbull Atoms amp Moleculesbull Electrodynamicsbull Heatbull Curvature of Space ndash replaces Schwarzcchildrsquos equation
Kanarevrsquos Hypothesis
bull Electron align linearly with proton They do NOT spin or form clouds around the nucleus
bull Neutrons are in the center of atom Proton are outside of Neutrons
bull The wavelength of all particles is the experimental radius
bull Double slit experiment explained correctly
bull Heisenbergrsquos inequality explained correctly
bull is the radius of approach of the magnetic lines around the torus of the electron
( 1 137)
bull Why water expands when it freezes
bull Electron clusters form lightning
rbull The meaning of Plancks constant is explained
22 2 kg m
h m f mr f consts
2210254 42m r kgm bull If the mr-constant describes a ring the equations
describe experiments bull If the mr-constant describes a wave the equations do
NOT describe experiments
Kanarevrsquos 5 Equations of Mechanics
Kanarevrsquos 1st Equation Acceleration ampmv maMma R R
Kanarevrsquos 2nd Equation Uniform motion amp mv maMma R R
maM mvR R
Kanarevrsquos 3rd Equation
Kanarevrsquos 4th Equation Equal and opposite reaction
1 21 21 2
12
a m
m a m ama
Kanarevrsquos 5th Equation 1 2 3 Mi M M M Mna R R R R R
The sum of the Actions working on a moving body is never equal to zero
Kanarevrsquos Photon model is a rolling ring
His equations accurately describe known experiments with photonsbull Electromagnetic radiation should be called photon radiationbull Diffraction and interference are the same phenomenabull Correctly explains the double slit experiment without quantum equationsbull Correctly explains the meaning of Heisenbergrsquos inequality
e
2 (2)(3142)(2817 15 ) 10007297352568
R 2426 12 137mag
xp
r m
m
Charge
Magnetic action
Proton
The intensity of the magnetic action near the geometrical center of the proton ring is so great that we theorize that it is the strong nuclear force
(9383 8)(1602 19)8476 14
4 (4)(3142)(1411 26)p
pp
EB T
M
(9395 8)(1602 19)1035 15
4 (4)(3142)(0966 26)N
NN
EB T
M
Neutron
1a Hydrogenavi
2
)539(
cT
MK
spring constant (K) when mass (M) and cycle-time (T) are known
)(
))(89)((
111 md
uKgM
d
MG
d
FK sts
spring constant (K) when mass (M) and amplitude ( d1) are known
2 21( )
Kv d d
m
velocity (v) at any point during the cycle
velocity
dforM
Kdv
max
012 velocity (v2) maximum (when d = 0 at position-2 and 4)
acceleration (a1) maximum when spring constant (K) mass (M) and amplitude (d1) are known
acceleration (a1) maximum when amplitude (d1) and cycle-time (Tc) are known
acceleration (a) at any point during the cycle-time (T)
1
151
286
v
dTc
cycle-time (Tc) when amplitude (d1) and velocity (v1) are known
K
MTc 28651
cycle-time (Tc) when mass (M) and spring constant (K) are known
fTc
1
cycle-time (Tc) when the frequency is known
cTf
1
frequency (f) when the cycle-time (Tc) is known
of cycles
elapsed timef
frequency (f) when the number of cycles in an elapsed time is known
d1 amplitude (distance from rest to start position)
Also frac12 the distance from the base to the peak
3131 oror dM
Ka
251
3131 539
c
oror T
da
2539
cT
da
In December of 2006 I discovered the mathematical error in Einsteinrsquos Special Relativity Einstein used BS math and that math only works in an imaginary world
It cannot describe real symmetrical space
Einsteinrsquos Special Relativity should be abandoned and a theory of real space using SM should be used to replace it
Numerous articles on the Internet show Einsteinrsquos theory to be incorrect but no one in academia is willing to publish the information
Figure 1-10 A moving particle observed from two frames of reference in uniform translation
Einsteinrsquos Math Error in Special Relativity Transformation of Velocity (From Eisenberg- Fundamentals of Modern Physics pages 27-29) Consider the particle shown in figure (1-10) moving with velocity v as
seen in a frame of reference O
dx dy dz
Vx Vy Vzdt dt dt
dx dy dz
V x V y V zdy dt dt
We would like to evaluate the velocity v of the particle as seen in the frame of reference 0 which is itself moving relative to 0 with velocity v Measured in the 0 frame the velocity vector of the particle has com ponents
The same velocity vector as measured in the 0 frame has components
2
2
2
2
2
1
1
x vtx
v
cy y
z z
vxt
ctv
c
Now from equations (1-13) we know that the relation between the primed and the unprimed coordinates and times is
Take the differential of these equations remembering that v is a constant This gives
2
2
2
2
2
1
1
dx vdtdx
v
cdy dy
dz dz
vxdt
cdtv
c
2
2
2 2 2 2 2
2
2
2
11 ( )
1( ) 1
1
1
1
dx vdt
v dxdx vdt v
dx dx vdtc dt dtvdx vdx vdx v dxdt dt dt dt
dt dtc c c c
VxV x
Vx
VxV x
vv
vv
c
x
v
c
cV
( )
coordinate of particle relative to moving primed system
coordinate of particle relative to stationary system
velocity of primed coordina
V x
Vx
te system relative to stationaryv sys
tem
Page 28 Fundamentals of Modern Physics
(1-16)
According to Einstein this equation tells you how to transform the observed velocity from one frame of reference to another frame of reference First we note that as Vc and vc approachzero the equation approaches those which would be derived from the Galilean transformation
2 2
( ) 1 01 1 ( )
Vx v Vx vV
VxV x
Vx v
v cVxx
v
c
v
cc
Another property of this equation is that it is impossible to choose V and v such that Vrsquo the magnitude of the velocity which is seen in the new frame is greater than c
1
2
09 ( 09 ) 18009876543
( 09 ) 1811 (09 )
c c cV c c
cc
c
Example As seen by O particle 1 has velocity 09c in the direction of positive x and particle 2 has velocity 09c in the negative x direction
Einstein equation however predicts To evaluate the velocity of particle 1 with respect to particle 2 we transform from the O frame to the Orsquo frame moving in the negative x direction with velocity v = -09c using equation 1-16 We obtain
JK LOGIC will observe the particles separating at 18c
A confusing INCORRECT and ILLOGICAL answer
left is a dir rightsub of a
1
2
1
09 ( 09 ) 18009876543
181( 09 )1 ( 09 )
The multiplication of dir left time a dir right is ONLY a dir left
09
dir
direction leftdirection rightsubtraction
c c cV c c
cc
c
V
dir
22
22
09 18 18009876543
181( 081)1(09)(09)
1
left rightsub
sub
c c c cc c
ccc
c
Analyzing Einsteinrsquos BS math Equation
Using SM math a logical answer can be obtained
Cannot multiply a subtraction operator times a direction in space
Can only subtract and change its direction
Cannot multiply a direction to the left times a direction to the right
Cannot multiply opposite directions in space
Can only solve for the resultant
rightleft
subtraction
1
subtraction
2
sub of a dir to the left is a dir to the right
1 resultant 0 sub 2
2sub 2
2
09 09
09 amp 091
09 09 18 1
(0)1
(09) amp (09)1
c cV
c cc
c c cV
cc
cc
8
181
cc
SM math predicts a logical answer using logic
21st century scientists state in many articles that Einsteins General Relativity and Quantum Mechanics are incompatible
They are not only incompatible they are both mathematically incorrect
They both use illogical BS math
Any equations in math that use imaginary numbers should be abandoned
Imaginary numbers are only necessary in the current BS math system
This means that Schroumldingers equations should be abandoned
Any equation in quantum mechanics that uses imaginary number should be abandoned
Prof Kanarev has developed logical physics and chemistrytheories without the use of Quantum mechanics or relativity
A partial list of his theories coverbull Mechanics ndash replaces Newtonbull Photon - bull Electrons ndash Protons- Neutronsbull Spectroscopybull Nucleusbull Atoms amp Moleculesbull Electrodynamicsbull Heatbull Curvature of Space ndash replaces Schwarzcchildrsquos equation
Kanarevrsquos Hypothesis
bull Electron align linearly with proton They do NOT spin or form clouds around the nucleus
bull Neutrons are in the center of atom Proton are outside of Neutrons
bull The wavelength of all particles is the experimental radius
bull Double slit experiment explained correctly
bull Heisenbergrsquos inequality explained correctly
bull is the radius of approach of the magnetic lines around the torus of the electron
( 1 137)
bull Why water expands when it freezes
bull Electron clusters form lightning
rbull The meaning of Plancks constant is explained
22 2 kg m
h m f mr f consts
2210254 42m r kgm bull If the mr-constant describes a ring the equations
describe experiments bull If the mr-constant describes a wave the equations do
NOT describe experiments
Kanarevrsquos 5 Equations of Mechanics
Kanarevrsquos 1st Equation Acceleration ampmv maMma R R
Kanarevrsquos 2nd Equation Uniform motion amp mv maMma R R
maM mvR R
Kanarevrsquos 3rd Equation
Kanarevrsquos 4th Equation Equal and opposite reaction
1 21 21 2
12
a m
m a m ama
Kanarevrsquos 5th Equation 1 2 3 Mi M M M Mna R R R R R
The sum of the Actions working on a moving body is never equal to zero
Kanarevrsquos Photon model is a rolling ring
His equations accurately describe known experiments with photonsbull Electromagnetic radiation should be called photon radiationbull Diffraction and interference are the same phenomenabull Correctly explains the double slit experiment without quantum equationsbull Correctly explains the meaning of Heisenbergrsquos inequality
e
2 (2)(3142)(2817 15 ) 10007297352568
R 2426 12 137mag
xp
r m
m
Charge
Magnetic action
Proton
The intensity of the magnetic action near the geometrical center of the proton ring is so great that we theorize that it is the strong nuclear force
(9383 8)(1602 19)8476 14
4 (4)(3142)(1411 26)p
pp
EB T
M
(9395 8)(1602 19)1035 15
4 (4)(3142)(0966 26)N
NN
EB T
M
Neutron
1a Hydrogenavi
In December of 2006 I discovered the mathematical error in Einsteinrsquos Special Relativity Einstein used BS math and that math only works in an imaginary world
It cannot describe real symmetrical space
Einsteinrsquos Special Relativity should be abandoned and a theory of real space using SM should be used to replace it
Numerous articles on the Internet show Einsteinrsquos theory to be incorrect but no one in academia is willing to publish the information
Figure 1-10 A moving particle observed from two frames of reference in uniform translation
Einsteinrsquos Math Error in Special Relativity Transformation of Velocity (From Eisenberg- Fundamentals of Modern Physics pages 27-29) Consider the particle shown in figure (1-10) moving with velocity v as
seen in a frame of reference O
dx dy dz
Vx Vy Vzdt dt dt
dx dy dz
V x V y V zdy dt dt
We would like to evaluate the velocity v of the particle as seen in the frame of reference 0 which is itself moving relative to 0 with velocity v Measured in the 0 frame the velocity vector of the particle has com ponents
The same velocity vector as measured in the 0 frame has components
2
2
2
2
2
1
1
x vtx
v
cy y
z z
vxt
ctv
c
Now from equations (1-13) we know that the relation between the primed and the unprimed coordinates and times is
Take the differential of these equations remembering that v is a constant This gives
2
2
2
2
2
1
1
dx vdtdx
v
cdy dy
dz dz
vxdt
cdtv
c
2
2
2 2 2 2 2
2
2
2
11 ( )
1( ) 1
1
1
1
dx vdt
v dxdx vdt v
dx dx vdtc dt dtvdx vdx vdx v dxdt dt dt dt
dt dtc c c c
VxV x
Vx
VxV x
vv
vv
c
x
v
c
cV
( )
coordinate of particle relative to moving primed system
coordinate of particle relative to stationary system
velocity of primed coordina
V x
Vx
te system relative to stationaryv sys
tem
Page 28 Fundamentals of Modern Physics
(1-16)
According to Einstein this equation tells you how to transform the observed velocity from one frame of reference to another frame of reference First we note that as Vc and vc approachzero the equation approaches those which would be derived from the Galilean transformation
2 2
( ) 1 01 1 ( )
Vx v Vx vV
VxV x
Vx v
v cVxx
v
c
v
cc
Another property of this equation is that it is impossible to choose V and v such that Vrsquo the magnitude of the velocity which is seen in the new frame is greater than c
1
2
09 ( 09 ) 18009876543
( 09 ) 1811 (09 )
c c cV c c
cc
c
Example As seen by O particle 1 has velocity 09c in the direction of positive x and particle 2 has velocity 09c in the negative x direction
Einstein equation however predicts To evaluate the velocity of particle 1 with respect to particle 2 we transform from the O frame to the Orsquo frame moving in the negative x direction with velocity v = -09c using equation 1-16 We obtain
JK LOGIC will observe the particles separating at 18c
A confusing INCORRECT and ILLOGICAL answer
left is a dir rightsub of a
1
2
1
09 ( 09 ) 18009876543
181( 09 )1 ( 09 )
The multiplication of dir left time a dir right is ONLY a dir left
09
dir
direction leftdirection rightsubtraction
c c cV c c
cc
c
V
dir
22
22
09 18 18009876543
181( 081)1(09)(09)
1
left rightsub
sub
c c c cc c
ccc
c
Analyzing Einsteinrsquos BS math Equation
Using SM math a logical answer can be obtained
Cannot multiply a subtraction operator times a direction in space
Can only subtract and change its direction
Cannot multiply a direction to the left times a direction to the right
Cannot multiply opposite directions in space
Can only solve for the resultant
rightleft
subtraction
1
subtraction
2
sub of a dir to the left is a dir to the right
1 resultant 0 sub 2
2sub 2
2
09 09
09 amp 091
09 09 18 1
(0)1
(09) amp (09)1
c cV
c cc
c c cV
cc
cc
8
181
cc
SM math predicts a logical answer using logic
21st century scientists state in many articles that Einsteins General Relativity and Quantum Mechanics are incompatible
They are not only incompatible they are both mathematically incorrect
They both use illogical BS math
Any equations in math that use imaginary numbers should be abandoned
Imaginary numbers are only necessary in the current BS math system
This means that Schroumldingers equations should be abandoned
Any equation in quantum mechanics that uses imaginary number should be abandoned
Prof Kanarev has developed logical physics and chemistrytheories without the use of Quantum mechanics or relativity
A partial list of his theories coverbull Mechanics ndash replaces Newtonbull Photon - bull Electrons ndash Protons- Neutronsbull Spectroscopybull Nucleusbull Atoms amp Moleculesbull Electrodynamicsbull Heatbull Curvature of Space ndash replaces Schwarzcchildrsquos equation
Kanarevrsquos Hypothesis
bull Electron align linearly with proton They do NOT spin or form clouds around the nucleus
bull Neutrons are in the center of atom Proton are outside of Neutrons
bull The wavelength of all particles is the experimental radius
bull Double slit experiment explained correctly
bull Heisenbergrsquos inequality explained correctly
bull is the radius of approach of the magnetic lines around the torus of the electron
( 1 137)
bull Why water expands when it freezes
bull Electron clusters form lightning
rbull The meaning of Plancks constant is explained
22 2 kg m
h m f mr f consts
2210254 42m r kgm bull If the mr-constant describes a ring the equations
describe experiments bull If the mr-constant describes a wave the equations do
NOT describe experiments
Kanarevrsquos 5 Equations of Mechanics
Kanarevrsquos 1st Equation Acceleration ampmv maMma R R
Kanarevrsquos 2nd Equation Uniform motion amp mv maMma R R
maM mvR R
Kanarevrsquos 3rd Equation
Kanarevrsquos 4th Equation Equal and opposite reaction
1 21 21 2
12
a m
m a m ama
Kanarevrsquos 5th Equation 1 2 3 Mi M M M Mna R R R R R
The sum of the Actions working on a moving body is never equal to zero
Kanarevrsquos Photon model is a rolling ring
His equations accurately describe known experiments with photonsbull Electromagnetic radiation should be called photon radiationbull Diffraction and interference are the same phenomenabull Correctly explains the double slit experiment without quantum equationsbull Correctly explains the meaning of Heisenbergrsquos inequality
e
2 (2)(3142)(2817 15 ) 10007297352568
R 2426 12 137mag
xp
r m
m
Charge
Magnetic action
Proton
The intensity of the magnetic action near the geometrical center of the proton ring is so great that we theorize that it is the strong nuclear force
(9383 8)(1602 19)8476 14
4 (4)(3142)(1411 26)p
pp
EB T
M
(9395 8)(1602 19)1035 15
4 (4)(3142)(0966 26)N
NN
EB T
M
Neutron
1a Hydrogenavi
Figure 1-10 A moving particle observed from two frames of reference in uniform translation
Einsteinrsquos Math Error in Special Relativity Transformation of Velocity (From Eisenberg- Fundamentals of Modern Physics pages 27-29) Consider the particle shown in figure (1-10) moving with velocity v as
seen in a frame of reference O
dx dy dz
Vx Vy Vzdt dt dt
dx dy dz
V x V y V zdy dt dt
We would like to evaluate the velocity v of the particle as seen in the frame of reference 0 which is itself moving relative to 0 with velocity v Measured in the 0 frame the velocity vector of the particle has com ponents
The same velocity vector as measured in the 0 frame has components
2
2
2
2
2
1
1
x vtx
v
cy y
z z
vxt
ctv
c
Now from equations (1-13) we know that the relation between the primed and the unprimed coordinates and times is
Take the differential of these equations remembering that v is a constant This gives
2
2
2
2
2
1
1
dx vdtdx
v
cdy dy
dz dz
vxdt
cdtv
c
2
2
2 2 2 2 2
2
2
2
11 ( )
1( ) 1
1
1
1
dx vdt
v dxdx vdt v
dx dx vdtc dt dtvdx vdx vdx v dxdt dt dt dt
dt dtc c c c
VxV x
Vx
VxV x
vv
vv
c
x
v
c
cV
( )
coordinate of particle relative to moving primed system
coordinate of particle relative to stationary system
velocity of primed coordina
V x
Vx
te system relative to stationaryv sys
tem
Page 28 Fundamentals of Modern Physics
(1-16)
According to Einstein this equation tells you how to transform the observed velocity from one frame of reference to another frame of reference First we note that as Vc and vc approachzero the equation approaches those which would be derived from the Galilean transformation
2 2
( ) 1 01 1 ( )
Vx v Vx vV
VxV x
Vx v
v cVxx
v
c
v
cc
Another property of this equation is that it is impossible to choose V and v such that Vrsquo the magnitude of the velocity which is seen in the new frame is greater than c
1
2
09 ( 09 ) 18009876543
( 09 ) 1811 (09 )
c c cV c c
cc
c
Example As seen by O particle 1 has velocity 09c in the direction of positive x and particle 2 has velocity 09c in the negative x direction
Einstein equation however predicts To evaluate the velocity of particle 1 with respect to particle 2 we transform from the O frame to the Orsquo frame moving in the negative x direction with velocity v = -09c using equation 1-16 We obtain
JK LOGIC will observe the particles separating at 18c
A confusing INCORRECT and ILLOGICAL answer
left is a dir rightsub of a
1
2
1
09 ( 09 ) 18009876543
181( 09 )1 ( 09 )
The multiplication of dir left time a dir right is ONLY a dir left
09
dir
direction leftdirection rightsubtraction
c c cV c c
cc
c
V
dir
22
22
09 18 18009876543
181( 081)1(09)(09)
1
left rightsub
sub
c c c cc c
ccc
c
Analyzing Einsteinrsquos BS math Equation
Using SM math a logical answer can be obtained
Cannot multiply a subtraction operator times a direction in space
Can only subtract and change its direction
Cannot multiply a direction to the left times a direction to the right
Cannot multiply opposite directions in space
Can only solve for the resultant
rightleft
subtraction
1
subtraction
2
sub of a dir to the left is a dir to the right
1 resultant 0 sub 2
2sub 2
2
09 09
09 amp 091
09 09 18 1
(0)1
(09) amp (09)1
c cV
c cc
c c cV
cc
cc
8
181
cc
SM math predicts a logical answer using logic
21st century scientists state in many articles that Einsteins General Relativity and Quantum Mechanics are incompatible
They are not only incompatible they are both mathematically incorrect
They both use illogical BS math
Any equations in math that use imaginary numbers should be abandoned
Imaginary numbers are only necessary in the current BS math system
This means that Schroumldingers equations should be abandoned
Any equation in quantum mechanics that uses imaginary number should be abandoned
Prof Kanarev has developed logical physics and chemistrytheories without the use of Quantum mechanics or relativity
A partial list of his theories coverbull Mechanics ndash replaces Newtonbull Photon - bull Electrons ndash Protons- Neutronsbull Spectroscopybull Nucleusbull Atoms amp Moleculesbull Electrodynamicsbull Heatbull Curvature of Space ndash replaces Schwarzcchildrsquos equation
Kanarevrsquos Hypothesis
bull Electron align linearly with proton They do NOT spin or form clouds around the nucleus
bull Neutrons are in the center of atom Proton are outside of Neutrons
bull The wavelength of all particles is the experimental radius
bull Double slit experiment explained correctly
bull Heisenbergrsquos inequality explained correctly
bull is the radius of approach of the magnetic lines around the torus of the electron
( 1 137)
bull Why water expands when it freezes
bull Electron clusters form lightning
rbull The meaning of Plancks constant is explained
22 2 kg m
h m f mr f consts
2210254 42m r kgm bull If the mr-constant describes a ring the equations
describe experiments bull If the mr-constant describes a wave the equations do
NOT describe experiments
Kanarevrsquos 5 Equations of Mechanics
Kanarevrsquos 1st Equation Acceleration ampmv maMma R R
Kanarevrsquos 2nd Equation Uniform motion amp mv maMma R R
maM mvR R
Kanarevrsquos 3rd Equation
Kanarevrsquos 4th Equation Equal and opposite reaction
1 21 21 2
12
a m
m a m ama
Kanarevrsquos 5th Equation 1 2 3 Mi M M M Mna R R R R R
The sum of the Actions working on a moving body is never equal to zero
Kanarevrsquos Photon model is a rolling ring
His equations accurately describe known experiments with photonsbull Electromagnetic radiation should be called photon radiationbull Diffraction and interference are the same phenomenabull Correctly explains the double slit experiment without quantum equationsbull Correctly explains the meaning of Heisenbergrsquos inequality
e
2 (2)(3142)(2817 15 ) 10007297352568
R 2426 12 137mag
xp
r m
m
Charge
Magnetic action
Proton
The intensity of the magnetic action near the geometrical center of the proton ring is so great that we theorize that it is the strong nuclear force
(9383 8)(1602 19)8476 14
4 (4)(3142)(1411 26)p
pp
EB T
M
(9395 8)(1602 19)1035 15
4 (4)(3142)(0966 26)N
NN
EB T
M
Neutron
1a Hydrogenavi
dx dy dz
Vx Vy Vzdt dt dt
dx dy dz
V x V y V zdy dt dt
We would like to evaluate the velocity v of the particle as seen in the frame of reference 0 which is itself moving relative to 0 with velocity v Measured in the 0 frame the velocity vector of the particle has com ponents
The same velocity vector as measured in the 0 frame has components
2
2
2
2
2
1
1
x vtx
v
cy y
z z
vxt
ctv
c
Now from equations (1-13) we know that the relation between the primed and the unprimed coordinates and times is
Take the differential of these equations remembering that v is a constant This gives
2
2
2
2
2
1
1
dx vdtdx
v
cdy dy
dz dz
vxdt
cdtv
c
2
2
2 2 2 2 2
2
2
2
11 ( )
1( ) 1
1
1
1
dx vdt
v dxdx vdt v
dx dx vdtc dt dtvdx vdx vdx v dxdt dt dt dt
dt dtc c c c
VxV x
Vx
VxV x
vv
vv
c
x
v
c
cV
( )
coordinate of particle relative to moving primed system
coordinate of particle relative to stationary system
velocity of primed coordina
V x
Vx
te system relative to stationaryv sys
tem
Page 28 Fundamentals of Modern Physics
(1-16)
According to Einstein this equation tells you how to transform the observed velocity from one frame of reference to another frame of reference First we note that as Vc and vc approachzero the equation approaches those which would be derived from the Galilean transformation
2 2
( ) 1 01 1 ( )
Vx v Vx vV
VxV x
Vx v
v cVxx
v
c
v
cc
Another property of this equation is that it is impossible to choose V and v such that Vrsquo the magnitude of the velocity which is seen in the new frame is greater than c
1
2
09 ( 09 ) 18009876543
( 09 ) 1811 (09 )
c c cV c c
cc
c
Example As seen by O particle 1 has velocity 09c in the direction of positive x and particle 2 has velocity 09c in the negative x direction
Einstein equation however predicts To evaluate the velocity of particle 1 with respect to particle 2 we transform from the O frame to the Orsquo frame moving in the negative x direction with velocity v = -09c using equation 1-16 We obtain
JK LOGIC will observe the particles separating at 18c
A confusing INCORRECT and ILLOGICAL answer
left is a dir rightsub of a
1
2
1
09 ( 09 ) 18009876543
181( 09 )1 ( 09 )
The multiplication of dir left time a dir right is ONLY a dir left
09
dir
direction leftdirection rightsubtraction
c c cV c c
cc
c
V
dir
22
22
09 18 18009876543
181( 081)1(09)(09)
1
left rightsub
sub
c c c cc c
ccc
c
Analyzing Einsteinrsquos BS math Equation
Using SM math a logical answer can be obtained
Cannot multiply a subtraction operator times a direction in space
Can only subtract and change its direction
Cannot multiply a direction to the left times a direction to the right
Cannot multiply opposite directions in space
Can only solve for the resultant
rightleft
subtraction
1
subtraction
2
sub of a dir to the left is a dir to the right
1 resultant 0 sub 2
2sub 2
2
09 09
09 amp 091
09 09 18 1
(0)1
(09) amp (09)1
c cV
c cc
c c cV
cc
cc
8
181
cc
SM math predicts a logical answer using logic
21st century scientists state in many articles that Einsteins General Relativity and Quantum Mechanics are incompatible
They are not only incompatible they are both mathematically incorrect
They both use illogical BS math
Any equations in math that use imaginary numbers should be abandoned
Imaginary numbers are only necessary in the current BS math system
This means that Schroumldingers equations should be abandoned
Any equation in quantum mechanics that uses imaginary number should be abandoned
Prof Kanarev has developed logical physics and chemistrytheories without the use of Quantum mechanics or relativity
A partial list of his theories coverbull Mechanics ndash replaces Newtonbull Photon - bull Electrons ndash Protons- Neutronsbull Spectroscopybull Nucleusbull Atoms amp Moleculesbull Electrodynamicsbull Heatbull Curvature of Space ndash replaces Schwarzcchildrsquos equation
Kanarevrsquos Hypothesis
bull Electron align linearly with proton They do NOT spin or form clouds around the nucleus
bull Neutrons are in the center of atom Proton are outside of Neutrons
bull The wavelength of all particles is the experimental radius
bull Double slit experiment explained correctly
bull Heisenbergrsquos inequality explained correctly
bull is the radius of approach of the magnetic lines around the torus of the electron
( 1 137)
bull Why water expands when it freezes
bull Electron clusters form lightning
rbull The meaning of Plancks constant is explained
22 2 kg m
h m f mr f consts
2210254 42m r kgm bull If the mr-constant describes a ring the equations
describe experiments bull If the mr-constant describes a wave the equations do
NOT describe experiments
Kanarevrsquos 5 Equations of Mechanics
Kanarevrsquos 1st Equation Acceleration ampmv maMma R R
Kanarevrsquos 2nd Equation Uniform motion amp mv maMma R R
maM mvR R
Kanarevrsquos 3rd Equation
Kanarevrsquos 4th Equation Equal and opposite reaction
1 21 21 2
12
a m
m a m ama
Kanarevrsquos 5th Equation 1 2 3 Mi M M M Mna R R R R R
The sum of the Actions working on a moving body is never equal to zero
Kanarevrsquos Photon model is a rolling ring
His equations accurately describe known experiments with photonsbull Electromagnetic radiation should be called photon radiationbull Diffraction and interference are the same phenomenabull Correctly explains the double slit experiment without quantum equationsbull Correctly explains the meaning of Heisenbergrsquos inequality
e
2 (2)(3142)(2817 15 ) 10007297352568
R 2426 12 137mag
xp
r m
m
Charge
Magnetic action
Proton
The intensity of the magnetic action near the geometrical center of the proton ring is so great that we theorize that it is the strong nuclear force
(9383 8)(1602 19)8476 14
4 (4)(3142)(1411 26)p
pp
EB T
M
(9395 8)(1602 19)1035 15
4 (4)(3142)(0966 26)N
NN
EB T
M
Neutron
1a Hydrogenavi
2
2
2
2
2
1
1
x vtx
v
cy y
z z
vxt
ctv
c
Now from equations (1-13) we know that the relation between the primed and the unprimed coordinates and times is
Take the differential of these equations remembering that v is a constant This gives
2
2
2
2
2
1
1
dx vdtdx
v
cdy dy
dz dz
vxdt
cdtv
c
2
2
2 2 2 2 2
2
2
2
11 ( )
1( ) 1
1
1
1
dx vdt
v dxdx vdt v
dx dx vdtc dt dtvdx vdx vdx v dxdt dt dt dt
dt dtc c c c
VxV x
Vx
VxV x
vv
vv
c
x
v
c
cV
( )
coordinate of particle relative to moving primed system
coordinate of particle relative to stationary system
velocity of primed coordina
V x
Vx
te system relative to stationaryv sys
tem
Page 28 Fundamentals of Modern Physics
(1-16)
According to Einstein this equation tells you how to transform the observed velocity from one frame of reference to another frame of reference First we note that as Vc and vc approachzero the equation approaches those which would be derived from the Galilean transformation
2 2
( ) 1 01 1 ( )
Vx v Vx vV
VxV x
Vx v
v cVxx
v
c
v
cc
Another property of this equation is that it is impossible to choose V and v such that Vrsquo the magnitude of the velocity which is seen in the new frame is greater than c
1
2
09 ( 09 ) 18009876543
( 09 ) 1811 (09 )
c c cV c c
cc
c
Example As seen by O particle 1 has velocity 09c in the direction of positive x and particle 2 has velocity 09c in the negative x direction
Einstein equation however predicts To evaluate the velocity of particle 1 with respect to particle 2 we transform from the O frame to the Orsquo frame moving in the negative x direction with velocity v = -09c using equation 1-16 We obtain
JK LOGIC will observe the particles separating at 18c
A confusing INCORRECT and ILLOGICAL answer
left is a dir rightsub of a
1
2
1
09 ( 09 ) 18009876543
181( 09 )1 ( 09 )
The multiplication of dir left time a dir right is ONLY a dir left
09
dir
direction leftdirection rightsubtraction
c c cV c c
cc
c
V
dir
22
22
09 18 18009876543
181( 081)1(09)(09)
1
left rightsub
sub
c c c cc c
ccc
c
Analyzing Einsteinrsquos BS math Equation
Using SM math a logical answer can be obtained
Cannot multiply a subtraction operator times a direction in space
Can only subtract and change its direction
Cannot multiply a direction to the left times a direction to the right
Cannot multiply opposite directions in space
Can only solve for the resultant
rightleft
subtraction
1
subtraction
2
sub of a dir to the left is a dir to the right
1 resultant 0 sub 2
2sub 2
2
09 09
09 amp 091
09 09 18 1
(0)1
(09) amp (09)1
c cV
c cc
c c cV
cc
cc
8
181
cc
SM math predicts a logical answer using logic
21st century scientists state in many articles that Einsteins General Relativity and Quantum Mechanics are incompatible
They are not only incompatible they are both mathematically incorrect
They both use illogical BS math
Any equations in math that use imaginary numbers should be abandoned
Imaginary numbers are only necessary in the current BS math system
This means that Schroumldingers equations should be abandoned
Any equation in quantum mechanics that uses imaginary number should be abandoned
Prof Kanarev has developed logical physics and chemistrytheories without the use of Quantum mechanics or relativity
A partial list of his theories coverbull Mechanics ndash replaces Newtonbull Photon - bull Electrons ndash Protons- Neutronsbull Spectroscopybull Nucleusbull Atoms amp Moleculesbull Electrodynamicsbull Heatbull Curvature of Space ndash replaces Schwarzcchildrsquos equation
Kanarevrsquos Hypothesis
bull Electron align linearly with proton They do NOT spin or form clouds around the nucleus
bull Neutrons are in the center of atom Proton are outside of Neutrons
bull The wavelength of all particles is the experimental radius
bull Double slit experiment explained correctly
bull Heisenbergrsquos inequality explained correctly
bull is the radius of approach of the magnetic lines around the torus of the electron
( 1 137)
bull Why water expands when it freezes
bull Electron clusters form lightning
rbull The meaning of Plancks constant is explained
22 2 kg m
h m f mr f consts
2210254 42m r kgm bull If the mr-constant describes a ring the equations
describe experiments bull If the mr-constant describes a wave the equations do
NOT describe experiments
Kanarevrsquos 5 Equations of Mechanics
Kanarevrsquos 1st Equation Acceleration ampmv maMma R R
Kanarevrsquos 2nd Equation Uniform motion amp mv maMma R R
maM mvR R
Kanarevrsquos 3rd Equation
Kanarevrsquos 4th Equation Equal and opposite reaction
1 21 21 2
12
a m
m a m ama
Kanarevrsquos 5th Equation 1 2 3 Mi M M M Mna R R R R R
The sum of the Actions working on a moving body is never equal to zero
Kanarevrsquos Photon model is a rolling ring
His equations accurately describe known experiments with photonsbull Electromagnetic radiation should be called photon radiationbull Diffraction and interference are the same phenomenabull Correctly explains the double slit experiment without quantum equationsbull Correctly explains the meaning of Heisenbergrsquos inequality
e
2 (2)(3142)(2817 15 ) 10007297352568
R 2426 12 137mag
xp
r m
m
Charge
Magnetic action
Proton
The intensity of the magnetic action near the geometrical center of the proton ring is so great that we theorize that it is the strong nuclear force
(9383 8)(1602 19)8476 14
4 (4)(3142)(1411 26)p
pp
EB T
M
(9395 8)(1602 19)1035 15
4 (4)(3142)(0966 26)N
NN
EB T
M
Neutron
1a Hydrogenavi
Take the differential of these equations remembering that v is a constant This gives
2
2
2
2
2
1
1
dx vdtdx
v
cdy dy
dz dz
vxdt
cdtv
c
2
2
2 2 2 2 2
2
2
2
11 ( )
1( ) 1
1
1
1
dx vdt
v dxdx vdt v
dx dx vdtc dt dtvdx vdx vdx v dxdt dt dt dt
dt dtc c c c
VxV x
Vx
VxV x
vv
vv
c
x
v
c
cV
( )
coordinate of particle relative to moving primed system
coordinate of particle relative to stationary system
velocity of primed coordina
V x
Vx
te system relative to stationaryv sys
tem
Page 28 Fundamentals of Modern Physics
(1-16)
According to Einstein this equation tells you how to transform the observed velocity from one frame of reference to another frame of reference First we note that as Vc and vc approachzero the equation approaches those which would be derived from the Galilean transformation
2 2
( ) 1 01 1 ( )
Vx v Vx vV
VxV x
Vx v
v cVxx
v
c
v
cc
Another property of this equation is that it is impossible to choose V and v such that Vrsquo the magnitude of the velocity which is seen in the new frame is greater than c
1
2
09 ( 09 ) 18009876543
( 09 ) 1811 (09 )
c c cV c c
cc
c
Example As seen by O particle 1 has velocity 09c in the direction of positive x and particle 2 has velocity 09c in the negative x direction
Einstein equation however predicts To evaluate the velocity of particle 1 with respect to particle 2 we transform from the O frame to the Orsquo frame moving in the negative x direction with velocity v = -09c using equation 1-16 We obtain
JK LOGIC will observe the particles separating at 18c
A confusing INCORRECT and ILLOGICAL answer
left is a dir rightsub of a
1
2
1
09 ( 09 ) 18009876543
181( 09 )1 ( 09 )
The multiplication of dir left time a dir right is ONLY a dir left
09
dir
direction leftdirection rightsubtraction
c c cV c c
cc
c
V
dir
22
22
09 18 18009876543
181( 081)1(09)(09)
1
left rightsub
sub
c c c cc c
ccc
c
Analyzing Einsteinrsquos BS math Equation
Using SM math a logical answer can be obtained
Cannot multiply a subtraction operator times a direction in space
Can only subtract and change its direction
Cannot multiply a direction to the left times a direction to the right
Cannot multiply opposite directions in space
Can only solve for the resultant
rightleft
subtraction
1
subtraction
2
sub of a dir to the left is a dir to the right
1 resultant 0 sub 2
2sub 2
2
09 09
09 amp 091
09 09 18 1
(0)1
(09) amp (09)1
c cV
c cc
c c cV
cc
cc
8
181
cc
SM math predicts a logical answer using logic
21st century scientists state in many articles that Einsteins General Relativity and Quantum Mechanics are incompatible
They are not only incompatible they are both mathematically incorrect
They both use illogical BS math
Any equations in math that use imaginary numbers should be abandoned
Imaginary numbers are only necessary in the current BS math system
This means that Schroumldingers equations should be abandoned
Any equation in quantum mechanics that uses imaginary number should be abandoned
Prof Kanarev has developed logical physics and chemistrytheories without the use of Quantum mechanics or relativity
A partial list of his theories coverbull Mechanics ndash replaces Newtonbull Photon - bull Electrons ndash Protons- Neutronsbull Spectroscopybull Nucleusbull Atoms amp Moleculesbull Electrodynamicsbull Heatbull Curvature of Space ndash replaces Schwarzcchildrsquos equation
Kanarevrsquos Hypothesis
bull Electron align linearly with proton They do NOT spin or form clouds around the nucleus
bull Neutrons are in the center of atom Proton are outside of Neutrons
bull The wavelength of all particles is the experimental radius
bull Double slit experiment explained correctly
bull Heisenbergrsquos inequality explained correctly
bull is the radius of approach of the magnetic lines around the torus of the electron
( 1 137)
bull Why water expands when it freezes
bull Electron clusters form lightning
rbull The meaning of Plancks constant is explained
22 2 kg m
h m f mr f consts
2210254 42m r kgm bull If the mr-constant describes a ring the equations
describe experiments bull If the mr-constant describes a wave the equations do
NOT describe experiments
Kanarevrsquos 5 Equations of Mechanics
Kanarevrsquos 1st Equation Acceleration ampmv maMma R R
Kanarevrsquos 2nd Equation Uniform motion amp mv maMma R R
maM mvR R
Kanarevrsquos 3rd Equation
Kanarevrsquos 4th Equation Equal and opposite reaction
1 21 21 2
12
a m
m a m ama
Kanarevrsquos 5th Equation 1 2 3 Mi M M M Mna R R R R R
The sum of the Actions working on a moving body is never equal to zero
Kanarevrsquos Photon model is a rolling ring
His equations accurately describe known experiments with photonsbull Electromagnetic radiation should be called photon radiationbull Diffraction and interference are the same phenomenabull Correctly explains the double slit experiment without quantum equationsbull Correctly explains the meaning of Heisenbergrsquos inequality
e
2 (2)(3142)(2817 15 ) 10007297352568
R 2426 12 137mag
xp
r m
m
Charge
Magnetic action
Proton
The intensity of the magnetic action near the geometrical center of the proton ring is so great that we theorize that it is the strong nuclear force
(9383 8)(1602 19)8476 14
4 (4)(3142)(1411 26)p
pp
EB T
M
(9395 8)(1602 19)1035 15
4 (4)(3142)(0966 26)N
NN
EB T
M
Neutron
1a Hydrogenavi
2
2
2 2 2 2 2
2
2
2
11 ( )
1( ) 1
1
1
1
dx vdt
v dxdx vdt v
dx dx vdtc dt dtvdx vdx vdx v dxdt dt dt dt
dt dtc c c c
VxV x
Vx
VxV x
vv
vv
c
x
v
c
cV
( )
coordinate of particle relative to moving primed system
coordinate of particle relative to stationary system
velocity of primed coordina
V x
Vx
te system relative to stationaryv sys
tem
Page 28 Fundamentals of Modern Physics
(1-16)
According to Einstein this equation tells you how to transform the observed velocity from one frame of reference to another frame of reference First we note that as Vc and vc approachzero the equation approaches those which would be derived from the Galilean transformation
2 2
( ) 1 01 1 ( )
Vx v Vx vV
VxV x
Vx v
v cVxx
v
c
v
cc
Another property of this equation is that it is impossible to choose V and v such that Vrsquo the magnitude of the velocity which is seen in the new frame is greater than c
1
2
09 ( 09 ) 18009876543
( 09 ) 1811 (09 )
c c cV c c
cc
c
Example As seen by O particle 1 has velocity 09c in the direction of positive x and particle 2 has velocity 09c in the negative x direction
Einstein equation however predicts To evaluate the velocity of particle 1 with respect to particle 2 we transform from the O frame to the Orsquo frame moving in the negative x direction with velocity v = -09c using equation 1-16 We obtain
JK LOGIC will observe the particles separating at 18c
A confusing INCORRECT and ILLOGICAL answer
left is a dir rightsub of a
1
2
1
09 ( 09 ) 18009876543
181( 09 )1 ( 09 )
The multiplication of dir left time a dir right is ONLY a dir left
09
dir
direction leftdirection rightsubtraction
c c cV c c
cc
c
V
dir
22
22
09 18 18009876543
181( 081)1(09)(09)
1
left rightsub
sub
c c c cc c
ccc
c
Analyzing Einsteinrsquos BS math Equation
Using SM math a logical answer can be obtained
Cannot multiply a subtraction operator times a direction in space
Can only subtract and change its direction
Cannot multiply a direction to the left times a direction to the right
Cannot multiply opposite directions in space
Can only solve for the resultant
rightleft
subtraction
1
subtraction
2
sub of a dir to the left is a dir to the right
1 resultant 0 sub 2
2sub 2
2
09 09
09 amp 091
09 09 18 1
(0)1
(09) amp (09)1
c cV
c cc
c c cV
cc
cc
8
181
cc
SM math predicts a logical answer using logic
21st century scientists state in many articles that Einsteins General Relativity and Quantum Mechanics are incompatible
They are not only incompatible they are both mathematically incorrect
They both use illogical BS math
Any equations in math that use imaginary numbers should be abandoned
Imaginary numbers are only necessary in the current BS math system
This means that Schroumldingers equations should be abandoned
Any equation in quantum mechanics that uses imaginary number should be abandoned
Prof Kanarev has developed logical physics and chemistrytheories without the use of Quantum mechanics or relativity
A partial list of his theories coverbull Mechanics ndash replaces Newtonbull Photon - bull Electrons ndash Protons- Neutronsbull Spectroscopybull Nucleusbull Atoms amp Moleculesbull Electrodynamicsbull Heatbull Curvature of Space ndash replaces Schwarzcchildrsquos equation
Kanarevrsquos Hypothesis
bull Electron align linearly with proton They do NOT spin or form clouds around the nucleus
bull Neutrons are in the center of atom Proton are outside of Neutrons
bull The wavelength of all particles is the experimental radius
bull Double slit experiment explained correctly
bull Heisenbergrsquos inequality explained correctly
bull is the radius of approach of the magnetic lines around the torus of the electron
( 1 137)
bull Why water expands when it freezes
bull Electron clusters form lightning
rbull The meaning of Plancks constant is explained
22 2 kg m
h m f mr f consts
2210254 42m r kgm bull If the mr-constant describes a ring the equations
describe experiments bull If the mr-constant describes a wave the equations do
NOT describe experiments
Kanarevrsquos 5 Equations of Mechanics
Kanarevrsquos 1st Equation Acceleration ampmv maMma R R
Kanarevrsquos 2nd Equation Uniform motion amp mv maMma R R
maM mvR R
Kanarevrsquos 3rd Equation
Kanarevrsquos 4th Equation Equal and opposite reaction
1 21 21 2
12
a m
m a m ama
Kanarevrsquos 5th Equation 1 2 3 Mi M M M Mna R R R R R
The sum of the Actions working on a moving body is never equal to zero
Kanarevrsquos Photon model is a rolling ring
His equations accurately describe known experiments with photonsbull Electromagnetic radiation should be called photon radiationbull Diffraction and interference are the same phenomenabull Correctly explains the double slit experiment without quantum equationsbull Correctly explains the meaning of Heisenbergrsquos inequality
e
2 (2)(3142)(2817 15 ) 10007297352568
R 2426 12 137mag
xp
r m
m
Charge
Magnetic action
Proton
The intensity of the magnetic action near the geometrical center of the proton ring is so great that we theorize that it is the strong nuclear force
(9383 8)(1602 19)8476 14
4 (4)(3142)(1411 26)p
pp
EB T
M
(9395 8)(1602 19)1035 15
4 (4)(3142)(0966 26)N
NN
EB T
M
Neutron
1a Hydrogenavi
According to Einstein this equation tells you how to transform the observed velocity from one frame of reference to another frame of reference First we note that as Vc and vc approachzero the equation approaches those which would be derived from the Galilean transformation
2 2
( ) 1 01 1 ( )
Vx v Vx vV
VxV x
Vx v
v cVxx
v
c
v
cc
Another property of this equation is that it is impossible to choose V and v such that Vrsquo the magnitude of the velocity which is seen in the new frame is greater than c
1
2
09 ( 09 ) 18009876543
( 09 ) 1811 (09 )
c c cV c c
cc
c
Example As seen by O particle 1 has velocity 09c in the direction of positive x and particle 2 has velocity 09c in the negative x direction
Einstein equation however predicts To evaluate the velocity of particle 1 with respect to particle 2 we transform from the O frame to the Orsquo frame moving in the negative x direction with velocity v = -09c using equation 1-16 We obtain
JK LOGIC will observe the particles separating at 18c
A confusing INCORRECT and ILLOGICAL answer
left is a dir rightsub of a
1
2
1
09 ( 09 ) 18009876543
181( 09 )1 ( 09 )
The multiplication of dir left time a dir right is ONLY a dir left
09
dir
direction leftdirection rightsubtraction
c c cV c c
cc
c
V
dir
22
22
09 18 18009876543
181( 081)1(09)(09)
1
left rightsub
sub
c c c cc c
ccc
c
Analyzing Einsteinrsquos BS math Equation
Using SM math a logical answer can be obtained
Cannot multiply a subtraction operator times a direction in space
Can only subtract and change its direction
Cannot multiply a direction to the left times a direction to the right
Cannot multiply opposite directions in space
Can only solve for the resultant
rightleft
subtraction
1
subtraction
2
sub of a dir to the left is a dir to the right
1 resultant 0 sub 2
2sub 2
2
09 09
09 amp 091
09 09 18 1
(0)1
(09) amp (09)1
c cV
c cc
c c cV
cc
cc
8
181
cc
SM math predicts a logical answer using logic
21st century scientists state in many articles that Einsteins General Relativity and Quantum Mechanics are incompatible
They are not only incompatible they are both mathematically incorrect
They both use illogical BS math
Any equations in math that use imaginary numbers should be abandoned
Imaginary numbers are only necessary in the current BS math system
This means that Schroumldingers equations should be abandoned
Any equation in quantum mechanics that uses imaginary number should be abandoned
Prof Kanarev has developed logical physics and chemistrytheories without the use of Quantum mechanics or relativity
A partial list of his theories coverbull Mechanics ndash replaces Newtonbull Photon - bull Electrons ndash Protons- Neutronsbull Spectroscopybull Nucleusbull Atoms amp Moleculesbull Electrodynamicsbull Heatbull Curvature of Space ndash replaces Schwarzcchildrsquos equation
Kanarevrsquos Hypothesis
bull Electron align linearly with proton They do NOT spin or form clouds around the nucleus
bull Neutrons are in the center of atom Proton are outside of Neutrons
bull The wavelength of all particles is the experimental radius
bull Double slit experiment explained correctly
bull Heisenbergrsquos inequality explained correctly
bull is the radius of approach of the magnetic lines around the torus of the electron
( 1 137)
bull Why water expands when it freezes
bull Electron clusters form lightning
rbull The meaning of Plancks constant is explained
22 2 kg m
h m f mr f consts
2210254 42m r kgm bull If the mr-constant describes a ring the equations
describe experiments bull If the mr-constant describes a wave the equations do
NOT describe experiments
Kanarevrsquos 5 Equations of Mechanics
Kanarevrsquos 1st Equation Acceleration ampmv maMma R R
Kanarevrsquos 2nd Equation Uniform motion amp mv maMma R R
maM mvR R
Kanarevrsquos 3rd Equation
Kanarevrsquos 4th Equation Equal and opposite reaction
1 21 21 2
12
a m
m a m ama
Kanarevrsquos 5th Equation 1 2 3 Mi M M M Mna R R R R R
The sum of the Actions working on a moving body is never equal to zero
Kanarevrsquos Photon model is a rolling ring
His equations accurately describe known experiments with photonsbull Electromagnetic radiation should be called photon radiationbull Diffraction and interference are the same phenomenabull Correctly explains the double slit experiment without quantum equationsbull Correctly explains the meaning of Heisenbergrsquos inequality
e
2 (2)(3142)(2817 15 ) 10007297352568
R 2426 12 137mag
xp
r m
m
Charge
Magnetic action
Proton
The intensity of the magnetic action near the geometrical center of the proton ring is so great that we theorize that it is the strong nuclear force
(9383 8)(1602 19)8476 14
4 (4)(3142)(1411 26)p
pp
EB T
M
(9395 8)(1602 19)1035 15
4 (4)(3142)(0966 26)N
NN
EB T
M
Neutron
1a Hydrogenavi
1
2
09 ( 09 ) 18009876543
( 09 ) 1811 (09 )
c c cV c c
cc
c
Example As seen by O particle 1 has velocity 09c in the direction of positive x and particle 2 has velocity 09c in the negative x direction
Einstein equation however predicts To evaluate the velocity of particle 1 with respect to particle 2 we transform from the O frame to the Orsquo frame moving in the negative x direction with velocity v = -09c using equation 1-16 We obtain
JK LOGIC will observe the particles separating at 18c
A confusing INCORRECT and ILLOGICAL answer
left is a dir rightsub of a
1
2
1
09 ( 09 ) 18009876543
181( 09 )1 ( 09 )
The multiplication of dir left time a dir right is ONLY a dir left
09
dir
direction leftdirection rightsubtraction
c c cV c c
cc
c
V
dir
22
22
09 18 18009876543
181( 081)1(09)(09)
1
left rightsub
sub
c c c cc c
ccc
c
Analyzing Einsteinrsquos BS math Equation
Using SM math a logical answer can be obtained
Cannot multiply a subtraction operator times a direction in space
Can only subtract and change its direction
Cannot multiply a direction to the left times a direction to the right
Cannot multiply opposite directions in space
Can only solve for the resultant
rightleft
subtraction
1
subtraction
2
sub of a dir to the left is a dir to the right
1 resultant 0 sub 2
2sub 2
2
09 09
09 amp 091
09 09 18 1
(0)1
(09) amp (09)1
c cV
c cc
c c cV
cc
cc
8
181
cc
SM math predicts a logical answer using logic
21st century scientists state in many articles that Einsteins General Relativity and Quantum Mechanics are incompatible
They are not only incompatible they are both mathematically incorrect
They both use illogical BS math
Any equations in math that use imaginary numbers should be abandoned
Imaginary numbers are only necessary in the current BS math system
This means that Schroumldingers equations should be abandoned
Any equation in quantum mechanics that uses imaginary number should be abandoned
Prof Kanarev has developed logical physics and chemistrytheories without the use of Quantum mechanics or relativity
A partial list of his theories coverbull Mechanics ndash replaces Newtonbull Photon - bull Electrons ndash Protons- Neutronsbull Spectroscopybull Nucleusbull Atoms amp Moleculesbull Electrodynamicsbull Heatbull Curvature of Space ndash replaces Schwarzcchildrsquos equation
Kanarevrsquos Hypothesis
bull Electron align linearly with proton They do NOT spin or form clouds around the nucleus
bull Neutrons are in the center of atom Proton are outside of Neutrons
bull The wavelength of all particles is the experimental radius
bull Double slit experiment explained correctly
bull Heisenbergrsquos inequality explained correctly
bull is the radius of approach of the magnetic lines around the torus of the electron
( 1 137)
bull Why water expands when it freezes
bull Electron clusters form lightning
rbull The meaning of Plancks constant is explained
22 2 kg m
h m f mr f consts
2210254 42m r kgm bull If the mr-constant describes a ring the equations
describe experiments bull If the mr-constant describes a wave the equations do
NOT describe experiments
Kanarevrsquos 5 Equations of Mechanics
Kanarevrsquos 1st Equation Acceleration ampmv maMma R R
Kanarevrsquos 2nd Equation Uniform motion amp mv maMma R R
maM mvR R
Kanarevrsquos 3rd Equation
Kanarevrsquos 4th Equation Equal and opposite reaction
1 21 21 2
12
a m
m a m ama
Kanarevrsquos 5th Equation 1 2 3 Mi M M M Mna R R R R R
The sum of the Actions working on a moving body is never equal to zero
Kanarevrsquos Photon model is a rolling ring
His equations accurately describe known experiments with photonsbull Electromagnetic radiation should be called photon radiationbull Diffraction and interference are the same phenomenabull Correctly explains the double slit experiment without quantum equationsbull Correctly explains the meaning of Heisenbergrsquos inequality
e
2 (2)(3142)(2817 15 ) 10007297352568
R 2426 12 137mag
xp
r m
m
Charge
Magnetic action
Proton
The intensity of the magnetic action near the geometrical center of the proton ring is so great that we theorize that it is the strong nuclear force
(9383 8)(1602 19)8476 14
4 (4)(3142)(1411 26)p
pp
EB T
M
(9395 8)(1602 19)1035 15
4 (4)(3142)(0966 26)N
NN
EB T
M
Neutron
1a Hydrogenavi
left is a dir rightsub of a
1
2
1
09 ( 09 ) 18009876543
181( 09 )1 ( 09 )
The multiplication of dir left time a dir right is ONLY a dir left
09
dir
direction leftdirection rightsubtraction
c c cV c c
cc
c
V
dir
22
22
09 18 18009876543
181( 081)1(09)(09)
1
left rightsub
sub
c c c cc c
ccc
c
Analyzing Einsteinrsquos BS math Equation
Using SM math a logical answer can be obtained
Cannot multiply a subtraction operator times a direction in space
Can only subtract and change its direction
Cannot multiply a direction to the left times a direction to the right
Cannot multiply opposite directions in space
Can only solve for the resultant
rightleft
subtraction
1
subtraction
2
sub of a dir to the left is a dir to the right
1 resultant 0 sub 2
2sub 2
2
09 09
09 amp 091
09 09 18 1
(0)1
(09) amp (09)1
c cV
c cc
c c cV
cc
cc
8
181
cc
SM math predicts a logical answer using logic
21st century scientists state in many articles that Einsteins General Relativity and Quantum Mechanics are incompatible
They are not only incompatible they are both mathematically incorrect
They both use illogical BS math
Any equations in math that use imaginary numbers should be abandoned
Imaginary numbers are only necessary in the current BS math system
This means that Schroumldingers equations should be abandoned
Any equation in quantum mechanics that uses imaginary number should be abandoned
Prof Kanarev has developed logical physics and chemistrytheories without the use of Quantum mechanics or relativity
A partial list of his theories coverbull Mechanics ndash replaces Newtonbull Photon - bull Electrons ndash Protons- Neutronsbull Spectroscopybull Nucleusbull Atoms amp Moleculesbull Electrodynamicsbull Heatbull Curvature of Space ndash replaces Schwarzcchildrsquos equation
Kanarevrsquos Hypothesis
bull Electron align linearly with proton They do NOT spin or form clouds around the nucleus
bull Neutrons are in the center of atom Proton are outside of Neutrons
bull The wavelength of all particles is the experimental radius
bull Double slit experiment explained correctly
bull Heisenbergrsquos inequality explained correctly
bull is the radius of approach of the magnetic lines around the torus of the electron
( 1 137)
bull Why water expands when it freezes
bull Electron clusters form lightning
rbull The meaning of Plancks constant is explained
22 2 kg m
h m f mr f consts
2210254 42m r kgm bull If the mr-constant describes a ring the equations
describe experiments bull If the mr-constant describes a wave the equations do
NOT describe experiments
Kanarevrsquos 5 Equations of Mechanics
Kanarevrsquos 1st Equation Acceleration ampmv maMma R R
Kanarevrsquos 2nd Equation Uniform motion amp mv maMma R R
maM mvR R
Kanarevrsquos 3rd Equation
Kanarevrsquos 4th Equation Equal and opposite reaction
1 21 21 2
12
a m
m a m ama
Kanarevrsquos 5th Equation 1 2 3 Mi M M M Mna R R R R R
The sum of the Actions working on a moving body is never equal to zero
Kanarevrsquos Photon model is a rolling ring
His equations accurately describe known experiments with photonsbull Electromagnetic radiation should be called photon radiationbull Diffraction and interference are the same phenomenabull Correctly explains the double slit experiment without quantum equationsbull Correctly explains the meaning of Heisenbergrsquos inequality
e
2 (2)(3142)(2817 15 ) 10007297352568
R 2426 12 137mag
xp
r m
m
Charge
Magnetic action
Proton
The intensity of the magnetic action near the geometrical center of the proton ring is so great that we theorize that it is the strong nuclear force
(9383 8)(1602 19)8476 14
4 (4)(3142)(1411 26)p
pp
EB T
M
(9395 8)(1602 19)1035 15
4 (4)(3142)(0966 26)N
NN
EB T
M
Neutron
1a Hydrogenavi
Using SM math a logical answer can be obtained
Cannot multiply a subtraction operator times a direction in space
Can only subtract and change its direction
Cannot multiply a direction to the left times a direction to the right
Cannot multiply opposite directions in space
Can only solve for the resultant
rightleft
subtraction
1
subtraction
2
sub of a dir to the left is a dir to the right
1 resultant 0 sub 2
2sub 2
2
09 09
09 amp 091
09 09 18 1
(0)1
(09) amp (09)1
c cV
c cc
c c cV
cc
cc
8
181
cc
SM math predicts a logical answer using logic
21st century scientists state in many articles that Einsteins General Relativity and Quantum Mechanics are incompatible
They are not only incompatible they are both mathematically incorrect
They both use illogical BS math
Any equations in math that use imaginary numbers should be abandoned
Imaginary numbers are only necessary in the current BS math system
This means that Schroumldingers equations should be abandoned
Any equation in quantum mechanics that uses imaginary number should be abandoned
Prof Kanarev has developed logical physics and chemistrytheories without the use of Quantum mechanics or relativity
A partial list of his theories coverbull Mechanics ndash replaces Newtonbull Photon - bull Electrons ndash Protons- Neutronsbull Spectroscopybull Nucleusbull Atoms amp Moleculesbull Electrodynamicsbull Heatbull Curvature of Space ndash replaces Schwarzcchildrsquos equation
Kanarevrsquos Hypothesis
bull Electron align linearly with proton They do NOT spin or form clouds around the nucleus
bull Neutrons are in the center of atom Proton are outside of Neutrons
bull The wavelength of all particles is the experimental radius
bull Double slit experiment explained correctly
bull Heisenbergrsquos inequality explained correctly
bull is the radius of approach of the magnetic lines around the torus of the electron
( 1 137)
bull Why water expands when it freezes
bull Electron clusters form lightning
rbull The meaning of Plancks constant is explained
22 2 kg m
h m f mr f consts
2210254 42m r kgm bull If the mr-constant describes a ring the equations
describe experiments bull If the mr-constant describes a wave the equations do
NOT describe experiments
Kanarevrsquos 5 Equations of Mechanics
Kanarevrsquos 1st Equation Acceleration ampmv maMma R R
Kanarevrsquos 2nd Equation Uniform motion amp mv maMma R R
maM mvR R
Kanarevrsquos 3rd Equation
Kanarevrsquos 4th Equation Equal and opposite reaction
1 21 21 2
12
a m
m a m ama
Kanarevrsquos 5th Equation 1 2 3 Mi M M M Mna R R R R R
The sum of the Actions working on a moving body is never equal to zero
Kanarevrsquos Photon model is a rolling ring
His equations accurately describe known experiments with photonsbull Electromagnetic radiation should be called photon radiationbull Diffraction and interference are the same phenomenabull Correctly explains the double slit experiment without quantum equationsbull Correctly explains the meaning of Heisenbergrsquos inequality
e
2 (2)(3142)(2817 15 ) 10007297352568
R 2426 12 137mag
xp
r m
m
Charge
Magnetic action
Proton
The intensity of the magnetic action near the geometrical center of the proton ring is so great that we theorize that it is the strong nuclear force
(9383 8)(1602 19)8476 14
4 (4)(3142)(1411 26)p
pp
EB T
M
(9395 8)(1602 19)1035 15
4 (4)(3142)(0966 26)N
NN
EB T
M
Neutron
1a Hydrogenavi
rightleft
subtraction
1
subtraction
2
sub of a dir to the left is a dir to the right
1 resultant 0 sub 2
2sub 2
2
09 09
09 amp 091
09 09 18 1
(0)1
(09) amp (09)1
c cV
c cc
c c cV
cc
cc
8
181
cc
SM math predicts a logical answer using logic
21st century scientists state in many articles that Einsteins General Relativity and Quantum Mechanics are incompatible
They are not only incompatible they are both mathematically incorrect
They both use illogical BS math
Any equations in math that use imaginary numbers should be abandoned
Imaginary numbers are only necessary in the current BS math system
This means that Schroumldingers equations should be abandoned
Any equation in quantum mechanics that uses imaginary number should be abandoned
Prof Kanarev has developed logical physics and chemistrytheories without the use of Quantum mechanics or relativity
A partial list of his theories coverbull Mechanics ndash replaces Newtonbull Photon - bull Electrons ndash Protons- Neutronsbull Spectroscopybull Nucleusbull Atoms amp Moleculesbull Electrodynamicsbull Heatbull Curvature of Space ndash replaces Schwarzcchildrsquos equation
Kanarevrsquos Hypothesis
bull Electron align linearly with proton They do NOT spin or form clouds around the nucleus
bull Neutrons are in the center of atom Proton are outside of Neutrons
bull The wavelength of all particles is the experimental radius
bull Double slit experiment explained correctly
bull Heisenbergrsquos inequality explained correctly
bull is the radius of approach of the magnetic lines around the torus of the electron
( 1 137)
bull Why water expands when it freezes
bull Electron clusters form lightning
rbull The meaning of Plancks constant is explained
22 2 kg m
h m f mr f consts
2210254 42m r kgm bull If the mr-constant describes a ring the equations
describe experiments bull If the mr-constant describes a wave the equations do
NOT describe experiments
Kanarevrsquos 5 Equations of Mechanics
Kanarevrsquos 1st Equation Acceleration ampmv maMma R R
Kanarevrsquos 2nd Equation Uniform motion amp mv maMma R R
maM mvR R
Kanarevrsquos 3rd Equation
Kanarevrsquos 4th Equation Equal and opposite reaction
1 21 21 2
12
a m
m a m ama
Kanarevrsquos 5th Equation 1 2 3 Mi M M M Mna R R R R R
The sum of the Actions working on a moving body is never equal to zero
Kanarevrsquos Photon model is a rolling ring
His equations accurately describe known experiments with photonsbull Electromagnetic radiation should be called photon radiationbull Diffraction and interference are the same phenomenabull Correctly explains the double slit experiment without quantum equationsbull Correctly explains the meaning of Heisenbergrsquos inequality
e
2 (2)(3142)(2817 15 ) 10007297352568
R 2426 12 137mag
xp
r m
m
Charge
Magnetic action
Proton
The intensity of the magnetic action near the geometrical center of the proton ring is so great that we theorize that it is the strong nuclear force
(9383 8)(1602 19)8476 14
4 (4)(3142)(1411 26)p
pp
EB T
M
(9395 8)(1602 19)1035 15
4 (4)(3142)(0966 26)N
NN
EB T
M
Neutron
1a Hydrogenavi
21st century scientists state in many articles that Einsteins General Relativity and Quantum Mechanics are incompatible
They are not only incompatible they are both mathematically incorrect
They both use illogical BS math
Any equations in math that use imaginary numbers should be abandoned
Imaginary numbers are only necessary in the current BS math system
This means that Schroumldingers equations should be abandoned
Any equation in quantum mechanics that uses imaginary number should be abandoned
Prof Kanarev has developed logical physics and chemistrytheories without the use of Quantum mechanics or relativity
A partial list of his theories coverbull Mechanics ndash replaces Newtonbull Photon - bull Electrons ndash Protons- Neutronsbull Spectroscopybull Nucleusbull Atoms amp Moleculesbull Electrodynamicsbull Heatbull Curvature of Space ndash replaces Schwarzcchildrsquos equation
Kanarevrsquos Hypothesis
bull Electron align linearly with proton They do NOT spin or form clouds around the nucleus
bull Neutrons are in the center of atom Proton are outside of Neutrons
bull The wavelength of all particles is the experimental radius
bull Double slit experiment explained correctly
bull Heisenbergrsquos inequality explained correctly
bull is the radius of approach of the magnetic lines around the torus of the electron
( 1 137)
bull Why water expands when it freezes
bull Electron clusters form lightning
rbull The meaning of Plancks constant is explained
22 2 kg m
h m f mr f consts
2210254 42m r kgm bull If the mr-constant describes a ring the equations
describe experiments bull If the mr-constant describes a wave the equations do
NOT describe experiments
Kanarevrsquos 5 Equations of Mechanics
Kanarevrsquos 1st Equation Acceleration ampmv maMma R R
Kanarevrsquos 2nd Equation Uniform motion amp mv maMma R R
maM mvR R
Kanarevrsquos 3rd Equation
Kanarevrsquos 4th Equation Equal and opposite reaction
1 21 21 2
12
a m
m a m ama
Kanarevrsquos 5th Equation 1 2 3 Mi M M M Mna R R R R R
The sum of the Actions working on a moving body is never equal to zero
Kanarevrsquos Photon model is a rolling ring
His equations accurately describe known experiments with photonsbull Electromagnetic radiation should be called photon radiationbull Diffraction and interference are the same phenomenabull Correctly explains the double slit experiment without quantum equationsbull Correctly explains the meaning of Heisenbergrsquos inequality
e
2 (2)(3142)(2817 15 ) 10007297352568
R 2426 12 137mag
xp
r m
m
Charge
Magnetic action
Proton
The intensity of the magnetic action near the geometrical center of the proton ring is so great that we theorize that it is the strong nuclear force
(9383 8)(1602 19)8476 14
4 (4)(3142)(1411 26)p
pp
EB T
M
(9395 8)(1602 19)1035 15
4 (4)(3142)(0966 26)N
NN
EB T
M
Neutron
1a Hydrogenavi
Any equations in math that use imaginary numbers should be abandoned
Imaginary numbers are only necessary in the current BS math system
This means that Schroumldingers equations should be abandoned
Any equation in quantum mechanics that uses imaginary number should be abandoned
Prof Kanarev has developed logical physics and chemistrytheories without the use of Quantum mechanics or relativity
A partial list of his theories coverbull Mechanics ndash replaces Newtonbull Photon - bull Electrons ndash Protons- Neutronsbull Spectroscopybull Nucleusbull Atoms amp Moleculesbull Electrodynamicsbull Heatbull Curvature of Space ndash replaces Schwarzcchildrsquos equation
Kanarevrsquos Hypothesis
bull Electron align linearly with proton They do NOT spin or form clouds around the nucleus
bull Neutrons are in the center of atom Proton are outside of Neutrons
bull The wavelength of all particles is the experimental radius
bull Double slit experiment explained correctly
bull Heisenbergrsquos inequality explained correctly
bull is the radius of approach of the magnetic lines around the torus of the electron
( 1 137)
bull Why water expands when it freezes
bull Electron clusters form lightning
rbull The meaning of Plancks constant is explained
22 2 kg m
h m f mr f consts
2210254 42m r kgm bull If the mr-constant describes a ring the equations
describe experiments bull If the mr-constant describes a wave the equations do
NOT describe experiments
Kanarevrsquos 5 Equations of Mechanics
Kanarevrsquos 1st Equation Acceleration ampmv maMma R R
Kanarevrsquos 2nd Equation Uniform motion amp mv maMma R R
maM mvR R
Kanarevrsquos 3rd Equation
Kanarevrsquos 4th Equation Equal and opposite reaction
1 21 21 2
12
a m
m a m ama
Kanarevrsquos 5th Equation 1 2 3 Mi M M M Mna R R R R R
The sum of the Actions working on a moving body is never equal to zero
Kanarevrsquos Photon model is a rolling ring
His equations accurately describe known experiments with photonsbull Electromagnetic radiation should be called photon radiationbull Diffraction and interference are the same phenomenabull Correctly explains the double slit experiment without quantum equationsbull Correctly explains the meaning of Heisenbergrsquos inequality
e
2 (2)(3142)(2817 15 ) 10007297352568
R 2426 12 137mag
xp
r m
m
Charge
Magnetic action
Proton
The intensity of the magnetic action near the geometrical center of the proton ring is so great that we theorize that it is the strong nuclear force
(9383 8)(1602 19)8476 14
4 (4)(3142)(1411 26)p
pp
EB T
M
(9395 8)(1602 19)1035 15
4 (4)(3142)(0966 26)N
NN
EB T
M
Neutron
1a Hydrogenavi
Prof Kanarev has developed logical physics and chemistrytheories without the use of Quantum mechanics or relativity
A partial list of his theories coverbull Mechanics ndash replaces Newtonbull Photon - bull Electrons ndash Protons- Neutronsbull Spectroscopybull Nucleusbull Atoms amp Moleculesbull Electrodynamicsbull Heatbull Curvature of Space ndash replaces Schwarzcchildrsquos equation
Kanarevrsquos Hypothesis
bull Electron align linearly with proton They do NOT spin or form clouds around the nucleus
bull Neutrons are in the center of atom Proton are outside of Neutrons
bull The wavelength of all particles is the experimental radius
bull Double slit experiment explained correctly
bull Heisenbergrsquos inequality explained correctly
bull is the radius of approach of the magnetic lines around the torus of the electron
( 1 137)
bull Why water expands when it freezes
bull Electron clusters form lightning
rbull The meaning of Plancks constant is explained
22 2 kg m
h m f mr f consts
2210254 42m r kgm bull If the mr-constant describes a ring the equations
describe experiments bull If the mr-constant describes a wave the equations do
NOT describe experiments
Kanarevrsquos 5 Equations of Mechanics
Kanarevrsquos 1st Equation Acceleration ampmv maMma R R
Kanarevrsquos 2nd Equation Uniform motion amp mv maMma R R
maM mvR R
Kanarevrsquos 3rd Equation
Kanarevrsquos 4th Equation Equal and opposite reaction
1 21 21 2
12
a m
m a m ama
Kanarevrsquos 5th Equation 1 2 3 Mi M M M Mna R R R R R
The sum of the Actions working on a moving body is never equal to zero
Kanarevrsquos Photon model is a rolling ring
His equations accurately describe known experiments with photonsbull Electromagnetic radiation should be called photon radiationbull Diffraction and interference are the same phenomenabull Correctly explains the double slit experiment without quantum equationsbull Correctly explains the meaning of Heisenbergrsquos inequality
e
2 (2)(3142)(2817 15 ) 10007297352568
R 2426 12 137mag
xp
r m
m
Charge
Magnetic action
Proton
The intensity of the magnetic action near the geometrical center of the proton ring is so great that we theorize that it is the strong nuclear force
(9383 8)(1602 19)8476 14
4 (4)(3142)(1411 26)p
pp
EB T
M
(9395 8)(1602 19)1035 15
4 (4)(3142)(0966 26)N
NN
EB T
M
Neutron
1a Hydrogenavi
Kanarevrsquos Hypothesis
bull Electron align linearly with proton They do NOT spin or form clouds around the nucleus
bull Neutrons are in the center of atom Proton are outside of Neutrons
bull The wavelength of all particles is the experimental radius
bull Double slit experiment explained correctly
bull Heisenbergrsquos inequality explained correctly
bull is the radius of approach of the magnetic lines around the torus of the electron
( 1 137)
bull Why water expands when it freezes
bull Electron clusters form lightning
rbull The meaning of Plancks constant is explained
22 2 kg m
h m f mr f consts
2210254 42m r kgm bull If the mr-constant describes a ring the equations
describe experiments bull If the mr-constant describes a wave the equations do
NOT describe experiments
Kanarevrsquos 5 Equations of Mechanics
Kanarevrsquos 1st Equation Acceleration ampmv maMma R R
Kanarevrsquos 2nd Equation Uniform motion amp mv maMma R R
maM mvR R
Kanarevrsquos 3rd Equation
Kanarevrsquos 4th Equation Equal and opposite reaction
1 21 21 2
12
a m
m a m ama
Kanarevrsquos 5th Equation 1 2 3 Mi M M M Mna R R R R R
The sum of the Actions working on a moving body is never equal to zero
Kanarevrsquos Photon model is a rolling ring
His equations accurately describe known experiments with photonsbull Electromagnetic radiation should be called photon radiationbull Diffraction and interference are the same phenomenabull Correctly explains the double slit experiment without quantum equationsbull Correctly explains the meaning of Heisenbergrsquos inequality
e
2 (2)(3142)(2817 15 ) 10007297352568
R 2426 12 137mag
xp
r m
m
Charge
Magnetic action
Proton
The intensity of the magnetic action near the geometrical center of the proton ring is so great that we theorize that it is the strong nuclear force
(9383 8)(1602 19)8476 14
4 (4)(3142)(1411 26)p
pp
EB T
M
(9395 8)(1602 19)1035 15
4 (4)(3142)(0966 26)N
NN
EB T
M
Neutron
1a Hydrogenavi
bull Why water expands when it freezes
bull Electron clusters form lightning
rbull The meaning of Plancks constant is explained
22 2 kg m
h m f mr f consts
2210254 42m r kgm bull If the mr-constant describes a ring the equations
describe experiments bull If the mr-constant describes a wave the equations do
NOT describe experiments
Kanarevrsquos 5 Equations of Mechanics
Kanarevrsquos 1st Equation Acceleration ampmv maMma R R
Kanarevrsquos 2nd Equation Uniform motion amp mv maMma R R
maM mvR R
Kanarevrsquos 3rd Equation
Kanarevrsquos 4th Equation Equal and opposite reaction
1 21 21 2
12
a m
m a m ama
Kanarevrsquos 5th Equation 1 2 3 Mi M M M Mna R R R R R
The sum of the Actions working on a moving body is never equal to zero
Kanarevrsquos Photon model is a rolling ring
His equations accurately describe known experiments with photonsbull Electromagnetic radiation should be called photon radiationbull Diffraction and interference are the same phenomenabull Correctly explains the double slit experiment without quantum equationsbull Correctly explains the meaning of Heisenbergrsquos inequality
e
2 (2)(3142)(2817 15 ) 10007297352568
R 2426 12 137mag
xp
r m
m
Charge
Magnetic action
Proton
The intensity of the magnetic action near the geometrical center of the proton ring is so great that we theorize that it is the strong nuclear force
(9383 8)(1602 19)8476 14
4 (4)(3142)(1411 26)p
pp
EB T
M
(9395 8)(1602 19)1035 15
4 (4)(3142)(0966 26)N
NN
EB T
M
Neutron
1a Hydrogenavi
Kanarevrsquos 5 Equations of Mechanics
Kanarevrsquos 1st Equation Acceleration ampmv maMma R R
Kanarevrsquos 2nd Equation Uniform motion amp mv maMma R R
maM mvR R
Kanarevrsquos 3rd Equation
Kanarevrsquos 4th Equation Equal and opposite reaction
1 21 21 2
12
a m
m a m ama
Kanarevrsquos 5th Equation 1 2 3 Mi M M M Mna R R R R R
The sum of the Actions working on a moving body is never equal to zero
Kanarevrsquos Photon model is a rolling ring
His equations accurately describe known experiments with photonsbull Electromagnetic radiation should be called photon radiationbull Diffraction and interference are the same phenomenabull Correctly explains the double slit experiment without quantum equationsbull Correctly explains the meaning of Heisenbergrsquos inequality
e
2 (2)(3142)(2817 15 ) 10007297352568
R 2426 12 137mag
xp
r m
m
Charge
Magnetic action
Proton
The intensity of the magnetic action near the geometrical center of the proton ring is so great that we theorize that it is the strong nuclear force
(9383 8)(1602 19)8476 14
4 (4)(3142)(1411 26)p
pp
EB T
M
(9395 8)(1602 19)1035 15
4 (4)(3142)(0966 26)N
NN
EB T
M
Neutron
1a Hydrogenavi
Kanarevrsquos Photon model is a rolling ring
His equations accurately describe known experiments with photonsbull Electromagnetic radiation should be called photon radiationbull Diffraction and interference are the same phenomenabull Correctly explains the double slit experiment without quantum equationsbull Correctly explains the meaning of Heisenbergrsquos inequality
e
2 (2)(3142)(2817 15 ) 10007297352568
R 2426 12 137mag
xp
r m
m
Charge
Magnetic action
Proton
The intensity of the magnetic action near the geometrical center of the proton ring is so great that we theorize that it is the strong nuclear force
(9383 8)(1602 19)8476 14
4 (4)(3142)(1411 26)p
pp
EB T
M
(9395 8)(1602 19)1035 15
4 (4)(3142)(0966 26)N
NN
EB T
M
Neutron
1a Hydrogenavi
e
2 (2)(3142)(2817 15 ) 10007297352568
R 2426 12 137mag
xp
r m
m
Charge
Magnetic action
Proton
The intensity of the magnetic action near the geometrical center of the proton ring is so great that we theorize that it is the strong nuclear force
(9383 8)(1602 19)8476 14
4 (4)(3142)(1411 26)p
pp
EB T
M
(9395 8)(1602 19)1035 15
4 (4)(3142)(0966 26)N
NN
EB T
M
Neutron
1a Hydrogenavi
Proton
The intensity of the magnetic action near the geometrical center of the proton ring is so great that we theorize that it is the strong nuclear force
(9383 8)(1602 19)8476 14
4 (4)(3142)(1411 26)p
pp
EB T
M
(9395 8)(1602 19)1035 15
4 (4)(3142)(0966 26)N
NN
EB T
M
Neutron
1a Hydrogenavi
1a Hydrogenavi
Lithium
Beryllium
Carbon
Diamond
Graphene
H2O H2O Cluster
There are 82gt22 (82000000000000000000000) free electrons in each cubic centimeter of a copper wire
Since no protons are free there cannot be an Attractive charge action proton potential in a wire
There are no protons present at one plate and electrons present at the other plate
Temperature Radius of Photons Equation 0 0(0 ) (27315 )AC T
0
2898 310609555 6
27315o
AoC
mTr m
316
0 0(1 ) (27415 )AС T 1
2898 310570855 6
27415A
o oC
mTr m
317
0 020 (68 ) 29315o
AC F T 20
2898 39885 6
29315o
AoC
mTr m
321
0 030 (86 ) 30315oAC F T
30
2898 39560 6
30315o
AoC
mTr m
322
0 0 0[100 (212 ) 37315 ]AC F T 100
2898 37766 6
37315o C
mKr m
K
323
0 0 0[(1000 )(1832 ) 127315 ]AC F T 1000
2898 32276 6
127315o
AoC
mTr m
324
0 0 0[(1500 )(2732 ) 177315 ]AC F T 1500
2898 31634 6
177315A
o oC
mTr m
325
0 0 0[(2000 )(3632 ) 227315 ]AC F T 2000
2898 31275 6
227315A
o oC
mTr m
326
1
0 0 0(0 2000 )(27315 227315 )oAto C to T
Temperatures between are formed with photons from the infra-red range As the temperature increase the radiuses of the photons decrease
Temperature (felt as HEAT by humans) is formed by photon emission from electrons
What is the exact radius of the photons that form the (highest) temperature felt as heat Modern science has no exact answer to this question We can only theorize that lsquoHEATrsquo is formed by photons which are emitted by electrons during the synthesis of atoms and molecules The border for the smallest radius of these photons is not established yet We theorize that it is in an interval of the ultra-violet range Since the shortest know radiuses (X and gamma photons) are emitted by nuclear synthesis they do not participate in heat formation Temperature (felt as HEAT by humans) is formed by photon emission from electrons If gamma photons participated in formation of heat the greatest possible temperature would be equal
maxmin
2898 3 2898 31 15 1000000000000000
3 18oA A
A AmT mT
T T Tr m
(348)
If such a temperature existed it would destroy molecules atoms and the nucleus of all atoms
Schwarzschildrsquos (incorrect) radius equation
2
2 g
GMR
C
Kanarevrsquos correct equation
1(1087148751 14)
2p
g photon
GMrR Mr
C
Schwarzschild 3 33
3 (3)(2 30 )
4 (4)(314)(297 3182 19
)og
M kgD
R m
kg
m
(403) 1
Kanarev
3 33
3 (3)(2 30 )
4 (4)(314)(439 )565 27red
g
M kgD
R m
kg
m
(404) 2
Kanarev
3 33 3
3 (3)(2 30 )276 35 (
4 (4)(314)(553 32 Kanrev error
0012)
)lightg
M kg kgD
R m m
kg
m
(405) 3
4
Kanarevrsquos 3 33
3 (3)(2 30)
4 (4)(314)(154 8)013 54gamma
g
MD
R
kg
m
(406) 5
6
Fernandesrsquo (18 36 9) 3 3
3 (3)(2 30)265 81
4 (4)(314)(18 17)rg
kg
m
MD
R
(407)
7
Nuclear density is around3
(12 24) 17kg
m 8
If the Sun were compressed to ( )( 439 )g redR m (equation 400) its gravitational attraction would
detain only radiation of the far infra-red spectrum Shorter wavelengths will not be detained If photons of gamma radiation were detained the gravitational radius of the Sun would have to
be ( )( 154 8 )g gammaR m (equation 402) This would not be possible because the density of
the Sun (equation 406) would be 37 orders more dense than the nucleus of atoms
Only one type of Black-hole can be all black That is one with a gravitational attraction strong enough to detain the highest energy gamma photons All other Black-holes should have colors which vary with the change of wavelengths of the photons which cannot be detained The order should be infra-red light ultra-violet x-ray and finally gamma photons
Mass is Frequency of an Elementary Particle All known experiments have produced answers that agree with the following equations ( ) ( )( )(distance)mad Action mass acceleration
( )mad
( ) ( ) (constant)(frequency)mad hf
(1) 2( ) ( )(constant)mad mc mass
(2)
c f rf (3)
Equating (1) and (2) 2( )m h c f Mass = (constant)(frequency of one particle)
Plankrsquos constant 6626069 34 HBCP 662603(11) 57(29)69 34 codatah Js h Js
2
222
2
6626069 34( )( ) ( )6626069 347372496 51
(299792458 8 ) 898755179 16
Joule
Kgmm sh Js s kgs
m mcs s
Equating (1)(2) and (3)
2 2
constant
expexp exp
1 6626069 34 22102 42
299792458 8
h h c h h JsM f mr kgm
mr c r cc cs
All mass is built from one elementary particle a ( 737 51 )kg particle
All mass is in relative motion with other mass Time is a description invented to measure the relative movement between different particles of
mass Energy is a term invented to describe what happens when masses interact Energy is a
description Mass does not convert to energy and energy does not convert to mass The word ldquoEnergyrdquo will be replaced with an Action involving mass acceleration and distance ( )mad
When the motion of masses interacts they change to different motions of masses These different mass forms can be measured and their changes in momentum can be measured When the change produces photon particles they have been incorrectly labeled as ldquoEnergyrdquo
Mass Frequency Frequency is a rotation per second
If ( )f is set to one rotation per second (737249638 51 )(1kgs rot s the equation predicts that the
7372496 51kg particle is the elementary mass out of which all mass is made
27372496 51 07372496 50
hKgs kg
c
2 07372496 50 second
m m rotationsf
h kgsc
22 392984064 11
0737249638 50 s
kg rotf
kgs
Kanarev Smallest measurable photon
24241 35328803 15
07372496 50 s
kg rotf
kgs
13598443ev photon 2424143lt35kg
91093897 31123559 20
07372496 50 s
kg rotf
kgs
05109906 6eV electron
9109235lt31kg
16726230 27226873 23
07372496 50 s
kg rotf
kgs
93827241gt6eV proton 1672623lt27kg
16749286 272271858 23
07372496 50 s
kg rotf
kgs
9395656gt6eV neutron 1674928lt27kg
334357629 27226750117 23
(07372496 50 ) )2( s
kg rotf
kgs
(Deuterium nucleus) Proton+Neutron
22102 42mr kgm
334357629 274535202583 23
(07372496 50 ( ) s1)
kg rotf
kgs
Proton+Neutron
22102 42mr kgm
223368 253029747 25
07372496 50 s
kg rotf
kgs
1243gt9eV Higgs Boson 2215849lt25 kg
256482lt25347890 25
07372496 50 s
kg rotf
kgs
144gt11eV Fermi Lab particle 2567033lt25kg
5703889152 1677367 34
07372496 50 s
kg rotf
kgs
320gt20eV highest eV particle measured 5704518lt16kg
1859222909 925218 41
07372496 50 s
kg rotf
kgs
10429476 27eV Fernandes 1859223lt9kg
2176450474 8295212 42
07372496 50 s
kg rotf
kgs
Planck limit
78434 28106387 23
07372496 50 s
kg rotf
kgs
classical radius of an electron
This frequency and radius are not a particle radius
This mass does not exist
308 Is there experimental evidences that an electron has the so-called classic radius of an electron A No experiments do not exist From Questions and Answers of Kanarev Page 95 Electrons-Protons-Neutrons JK ET and Rewrite The handbook of Chemistry and Physics shows the classical radius of the electron as (2817 15 )m Kanarevs theory shows this to be the radius of approach of the magnetic force lines around the torus ring of the electron ( )magr This is confirmed by calculating the dimensionless fine structure constant
( ) (which is equal to the circumference of the inner circle (2 )magr divided by the experimentally
measured radius of the electron exp( )R
exp
2 (2)(314)(2817 15 ) 10007297352568
2426 12 137036magr m
R m
(191)
Kg Rotationsec meters experimental classical rmag mass frequency radius r=cf mr mr(11615lt3) 427E-41 5791781E+09 517617E-02 221022E-42 256717E-45 smallest photon
242414E-35 328809E+15 911753E-08 221022E-42 256717E-45 136eV photon 91094E-31 123559E+20 242631E-12 221022E-42 256717E-45 electron 0511gt6eV 891331E-30 120899E+21 247968E-13 221022E-42 256717E-45 up quark
178266E-29 241799E+21 123984E-13 221022E-42 256717E-45 down quark 188356E-28 255485E+22 117343E-14 221022E-42 256717E-45 muon
240606E-28 326356E+22 918606E-15 221022E-42 256717E-45 meson 248806E-28 337479E+22 88833E-15 221022E-42 256717E-45 meson 356532E-28 483598E+22 619921E-15 221022E-42 256717E-45 strange quark
880118E-28 119379E+23 251128E-15 221022E-42 256717E-45 meson 887249E-28 120346E+23 249109E-15 221022E-42 256717E-45 meson 887177E-28 120336E+23 249129E-15 221022E-42 256717E-45 meson 978325E-28 132699E+23 225919E-15 221022E-42 256717E-45 meson 16726E-27 22687E+23 132143E-15 221022E-42 256717E-45 Proton 938gt6eV
167491E-27 227183E+23 131961E-15 221022E-42 256717E-45 Neutron 939gt6eV 198872E-27 269748E+23 111138E-15 221022E-42 256717E-45 hyperon
212033E-27 2876E+23 104239E-15 221022E-42 256717E-45 hyperon 212579E-27 28834E+23 103972E-15 221022E-42 256717E-45 21347E-27 289549E+23 103538E-15 221022E-42 256717E-45
231746E-27 314339E+23 953725E-16 221022E-42 256717E-45 charm quark 234367E-27 317893E+23 943061E-16 221022E-42 256717E-45
235543E-27 319489E+23 93835E-16 221022E-42 256717E-45 298061E-27 404288E+23 741532E-16 221022E-42 256717E-45 hyperon
802198E-27 10881E+24 27552E-16 221022E-42 256717E-45 bottom quark 143201E-25 194237E+25 154344E-17 221022E-42 256717E-45 W
162556E-25 220489E+25 135967E-17 221022E-42 256717E-45 Z 223368E-25 302974E+25 989499E-18 221022E-42 256717E-45 Higgs125gt9eV
256482E-25 347890E+25 861745E-18 221022E-42 256717E-45 Fermi labs 144gt9eV 311966E-25 423148E+25 708481E-18 221022E-42 256717E-45 top quark
570389E-16 77367E+34 387494E-27 221022E-42 256717E-45 Gamma 320gt20eV 185922E-09 25218E+41 118879E-33 221022E-42 256717E-45 1043gt27eV 186lt9kg
217645E-08 29521E+42 101552E-34 221022E-42 256717E-45 Planck limits
mass frequency radius r=cf mr mr(11615lt3)
mass frequency radius r=cf mr2f 222E-39 298E+11 100E-03 6626069E-34 smallest photon
242414E-35 328809E+15 911753E-08 6626069E-34 Photon 136eV 91094E-31 123559E+20 242631E-12 6626069E-34 electron 0511gt6eV 891331E-30 120899E+21 247968E-13 6626069E-34 up quark
178266E-29 241799E+21 123984E-13 6626069E-34 down quark 188356E-28 255485E+22 117343E-14 6626069E-34 muon
240606E-28 326356E+22 918606E-15 6626069E-34 meson 248806E-28 337479E+22 88833E-15 6626069E-34 meson 356532E-28 483598E+22 619921E-15 6626069E-34 strange quark
880118E-28 119379E+23 251128E-15 6626069E-34 meson 887249E-28 120346E+23 249109E-15 6626069E-34 meson 887177E-28 120336E+23 249129E-15 6626069E-34 meson 978325E-28 132699E+23 225919E-15 6626069E-34 meson 16726E-27 22687E+23 132143E-15 6626069E-34 Proton 938gt6eV
167491E-27 227183E+23 131961E-15 6626069E-34 Neutron 939gt6eV 198872E-27 269748E+23 111138E-15 6626069E-34 hyperon
212033E-27 2876E+23 104239E-15 6626069E-34 hyperon 212579E-27 28834E+23 103972E-15 6626069E-34 21347E-27 289549E+23 103538E-15 6626069E-34
231746E-27 314339E+23 953725E-16 6626069E-34 charm quark 234367E-27 317893E+23 943061E-16 6626069E-34
235543E-27 319489E+23 93835E-16 6626069E-34 298061E-27 404288E+23 741532E-16 6626069E-34 hyperon
802198E-27 10881E+24 27552E-16 6626069E-34 bottom quark 143201E-25 194237E+25 154344E-17 6626069E-34 W
162556E-25 220489E+25 135967E-17 6626069E-34 Z 223368E-25 302974E+25 989499E-18 6626069E-34 Higgs 125eV
256482E-25 347890E+25 861745E-18 6626069E-34 Fermi labs 144gt9eV 311966E-25 423148E+25 708481E-18 6626069E-34 top quark
570389E-16 77367E+34 387493E-27 6626069E-34 Gamma 320gt20eV 185922E-09 25218E+41 118879E-33 6626069E-34 186ltkg 1043gt27eV 217645E-08 29521E+42 101552E-34 6626071E-34 Planck limits
mass frequency radius r=cf mr2f
frequency radius r=cf rf=C 299792458 8c m s 5791781E+09 517617E-02 2997925E+08 smallest photon
328809E+15 911753E-08 2997925E+08 136eV photon 123559E+20 242631E-12 2997925E+08 electron 0511gt6eV 120899E+21 247968E-13 2997925E+08 up quark
241799E+21 123984E-13 2997925E+08 down quark 255485E+22 117343E-14 2997925E+08 muon
326356E+22 918606E-15 2997925E+08 meson 337479E+22 88833E-15 2997925E+08 meson 483598E+22 619921E-15 2997925E+08 strange quark
119379E+23 251128E-15 2997925E+08 meson 120346E+23 249109E-15 2997925E+08 meson 120336E+23 249129E-15 2997925E+08 meson 132699E+23 225919E-15 2997925E+08 meson 22687E+23 132143E-15 2997925E+08 Proton 938gt6eV
227183E+23 131961E-15 2997925E+08 Neutron 939gt6eV 269748E+23 111138E-15 2997925E+08 hyperon
2876E+23 104239E-15 2997925E+08 hyperon 28834E+23 103972E-15 2997925E+08 289549E+23 103538E-15 2997925E+08
314339E+23 953725E-16 2997925E+08 charm quark 317893E+23 943061E-16 2997925E+08
319489E+23 93835E-16 2997925E+08 404288E+23 741532E-16 2997925E+08 hyperon
10881E+24 27552E-16 2997925E+08 bottom quark 194237E+25 154344E-17 2997925E+08 W
220489E+25 135967E-17 2997925E+08 Z 302974E+25 989499E-18 2997925E+08 Higgs 125eV
347890E+25 861745E-18 2997925E+08 Fermi labs 144gt9eV 423148E+25 708481E-18 2997925E+08 top quark
77367E+34 387493E-27 2997925E+08 Gamma 320gt20eV 25218E+41 118879E-33 2997925E+08 1043gt27eV 186lt9kg
29521E+42 101552E-34 2997925E+08 Planck limits
frequency radius r=cf rf=C
Savior Borg
Two helix makes a mass particle
Fernandes
Kanarevrsquos Photon Model
My website Symmetrymathcomis my interpretation of the Authorrsquos I have referenced
I do not claim that any of these authors will agree withmy interpretations
I leave it to anyone who visits my site to decide the logicof my interpretation
- SYMMETRY MATH (SM) By Jack Kuykendall
- Why is Symmetry Math Needed
- From ldquoWhat Is Mathematicsrdquo by Courant and Robbins ndash page 55 Th
- Slide 4
- Slide 5
- Slide 6
- Slide 7
- Slide 8
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- Slide 122
- Slide 123
-
Beryllium
Carbon
Diamond
Graphene
H2O H2O Cluster
There are 82gt22 (82000000000000000000000) free electrons in each cubic centimeter of a copper wire
Since no protons are free there cannot be an Attractive charge action proton potential in a wire
There are no protons present at one plate and electrons present at the other plate
Temperature Radius of Photons Equation 0 0(0 ) (27315 )AC T
0
2898 310609555 6
27315o
AoC
mTr m
316
0 0(1 ) (27415 )AС T 1
2898 310570855 6
27415A
o oC
mTr m
317
0 020 (68 ) 29315o
AC F T 20
2898 39885 6
29315o
AoC
mTr m
321
0 030 (86 ) 30315oAC F T
30
2898 39560 6
30315o
AoC
mTr m
322
0 0 0[100 (212 ) 37315 ]AC F T 100
2898 37766 6
37315o C
mKr m
K
323
0 0 0[(1000 )(1832 ) 127315 ]AC F T 1000
2898 32276 6
127315o
AoC
mTr m
324
0 0 0[(1500 )(2732 ) 177315 ]AC F T 1500
2898 31634 6
177315A
o oC
mTr m
325
0 0 0[(2000 )(3632 ) 227315 ]AC F T 2000
2898 31275 6
227315A
o oC
mTr m
326
1
0 0 0(0 2000 )(27315 227315 )oAto C to T
Temperatures between are formed with photons from the infra-red range As the temperature increase the radiuses of the photons decrease
Temperature (felt as HEAT by humans) is formed by photon emission from electrons
What is the exact radius of the photons that form the (highest) temperature felt as heat Modern science has no exact answer to this question We can only theorize that lsquoHEATrsquo is formed by photons which are emitted by electrons during the synthesis of atoms and molecules The border for the smallest radius of these photons is not established yet We theorize that it is in an interval of the ultra-violet range Since the shortest know radiuses (X and gamma photons) are emitted by nuclear synthesis they do not participate in heat formation Temperature (felt as HEAT by humans) is formed by photon emission from electrons If gamma photons participated in formation of heat the greatest possible temperature would be equal
maxmin
2898 3 2898 31 15 1000000000000000
3 18oA A
A AmT mT
T T Tr m
(348)
If such a temperature existed it would destroy molecules atoms and the nucleus of all atoms
Schwarzschildrsquos (incorrect) radius equation
2
2 g
GMR
C
Kanarevrsquos correct equation
1(1087148751 14)
2p
g photon
GMrR Mr
C
Schwarzschild 3 33
3 (3)(2 30 )
4 (4)(314)(297 3182 19
)og
M kgD
R m
kg
m
(403) 1
Kanarev
3 33
3 (3)(2 30 )
4 (4)(314)(439 )565 27red
g
M kgD
R m
kg
m
(404) 2
Kanarev
3 33 3
3 (3)(2 30 )276 35 (
4 (4)(314)(553 32 Kanrev error
0012)
)lightg
M kg kgD
R m m
kg
m
(405) 3
4
Kanarevrsquos 3 33
3 (3)(2 30)
4 (4)(314)(154 8)013 54gamma
g
MD
R
kg
m
(406) 5
6
Fernandesrsquo (18 36 9) 3 3
3 (3)(2 30)265 81
4 (4)(314)(18 17)rg
kg
m
MD
R
(407)
7
Nuclear density is around3
(12 24) 17kg
m 8
If the Sun were compressed to ( )( 439 )g redR m (equation 400) its gravitational attraction would
detain only radiation of the far infra-red spectrum Shorter wavelengths will not be detained If photons of gamma radiation were detained the gravitational radius of the Sun would have to
be ( )( 154 8 )g gammaR m (equation 402) This would not be possible because the density of
the Sun (equation 406) would be 37 orders more dense than the nucleus of atoms
Only one type of Black-hole can be all black That is one with a gravitational attraction strong enough to detain the highest energy gamma photons All other Black-holes should have colors which vary with the change of wavelengths of the photons which cannot be detained The order should be infra-red light ultra-violet x-ray and finally gamma photons
Mass is Frequency of an Elementary Particle All known experiments have produced answers that agree with the following equations ( ) ( )( )(distance)mad Action mass acceleration
( )mad
( ) ( ) (constant)(frequency)mad hf
(1) 2( ) ( )(constant)mad mc mass
(2)
c f rf (3)
Equating (1) and (2) 2( )m h c f Mass = (constant)(frequency of one particle)
Plankrsquos constant 6626069 34 HBCP 662603(11) 57(29)69 34 codatah Js h Js
2
222
2
6626069 34( )( ) ( )6626069 347372496 51
(299792458 8 ) 898755179 16
Joule
Kgmm sh Js s kgs
m mcs s
Equating (1)(2) and (3)
2 2
constant
expexp exp
1 6626069 34 22102 42
299792458 8
h h c h h JsM f mr kgm
mr c r cc cs
All mass is built from one elementary particle a ( 737 51 )kg particle
All mass is in relative motion with other mass Time is a description invented to measure the relative movement between different particles of
mass Energy is a term invented to describe what happens when masses interact Energy is a
description Mass does not convert to energy and energy does not convert to mass The word ldquoEnergyrdquo will be replaced with an Action involving mass acceleration and distance ( )mad
When the motion of masses interacts they change to different motions of masses These different mass forms can be measured and their changes in momentum can be measured When the change produces photon particles they have been incorrectly labeled as ldquoEnergyrdquo
Mass Frequency Frequency is a rotation per second
If ( )f is set to one rotation per second (737249638 51 )(1kgs rot s the equation predicts that the
7372496 51kg particle is the elementary mass out of which all mass is made
27372496 51 07372496 50
hKgs kg
c
2 07372496 50 second
m m rotationsf
h kgsc
22 392984064 11
0737249638 50 s
kg rotf
kgs
Kanarev Smallest measurable photon
24241 35328803 15
07372496 50 s
kg rotf
kgs
13598443ev photon 2424143lt35kg
91093897 31123559 20
07372496 50 s
kg rotf
kgs
05109906 6eV electron
9109235lt31kg
16726230 27226873 23
07372496 50 s
kg rotf
kgs
93827241gt6eV proton 1672623lt27kg
16749286 272271858 23
07372496 50 s
kg rotf
kgs
9395656gt6eV neutron 1674928lt27kg
334357629 27226750117 23
(07372496 50 ) )2( s
kg rotf
kgs
(Deuterium nucleus) Proton+Neutron
22102 42mr kgm
334357629 274535202583 23
(07372496 50 ( ) s1)
kg rotf
kgs
Proton+Neutron
22102 42mr kgm
223368 253029747 25
07372496 50 s
kg rotf
kgs
1243gt9eV Higgs Boson 2215849lt25 kg
256482lt25347890 25
07372496 50 s
kg rotf
kgs
144gt11eV Fermi Lab particle 2567033lt25kg
5703889152 1677367 34
07372496 50 s
kg rotf
kgs
320gt20eV highest eV particle measured 5704518lt16kg
1859222909 925218 41
07372496 50 s
kg rotf
kgs
10429476 27eV Fernandes 1859223lt9kg
2176450474 8295212 42
07372496 50 s
kg rotf
kgs
Planck limit
78434 28106387 23
07372496 50 s
kg rotf
kgs
classical radius of an electron
This frequency and radius are not a particle radius
This mass does not exist
308 Is there experimental evidences that an electron has the so-called classic radius of an electron A No experiments do not exist From Questions and Answers of Kanarev Page 95 Electrons-Protons-Neutrons JK ET and Rewrite The handbook of Chemistry and Physics shows the classical radius of the electron as (2817 15 )m Kanarevs theory shows this to be the radius of approach of the magnetic force lines around the torus ring of the electron ( )magr This is confirmed by calculating the dimensionless fine structure constant
( ) (which is equal to the circumference of the inner circle (2 )magr divided by the experimentally
measured radius of the electron exp( )R
exp
2 (2)(314)(2817 15 ) 10007297352568
2426 12 137036magr m
R m
(191)
Kg Rotationsec meters experimental classical rmag mass frequency radius r=cf mr mr(11615lt3) 427E-41 5791781E+09 517617E-02 221022E-42 256717E-45 smallest photon
242414E-35 328809E+15 911753E-08 221022E-42 256717E-45 136eV photon 91094E-31 123559E+20 242631E-12 221022E-42 256717E-45 electron 0511gt6eV 891331E-30 120899E+21 247968E-13 221022E-42 256717E-45 up quark
178266E-29 241799E+21 123984E-13 221022E-42 256717E-45 down quark 188356E-28 255485E+22 117343E-14 221022E-42 256717E-45 muon
240606E-28 326356E+22 918606E-15 221022E-42 256717E-45 meson 248806E-28 337479E+22 88833E-15 221022E-42 256717E-45 meson 356532E-28 483598E+22 619921E-15 221022E-42 256717E-45 strange quark
880118E-28 119379E+23 251128E-15 221022E-42 256717E-45 meson 887249E-28 120346E+23 249109E-15 221022E-42 256717E-45 meson 887177E-28 120336E+23 249129E-15 221022E-42 256717E-45 meson 978325E-28 132699E+23 225919E-15 221022E-42 256717E-45 meson 16726E-27 22687E+23 132143E-15 221022E-42 256717E-45 Proton 938gt6eV
167491E-27 227183E+23 131961E-15 221022E-42 256717E-45 Neutron 939gt6eV 198872E-27 269748E+23 111138E-15 221022E-42 256717E-45 hyperon
212033E-27 2876E+23 104239E-15 221022E-42 256717E-45 hyperon 212579E-27 28834E+23 103972E-15 221022E-42 256717E-45 21347E-27 289549E+23 103538E-15 221022E-42 256717E-45
231746E-27 314339E+23 953725E-16 221022E-42 256717E-45 charm quark 234367E-27 317893E+23 943061E-16 221022E-42 256717E-45
235543E-27 319489E+23 93835E-16 221022E-42 256717E-45 298061E-27 404288E+23 741532E-16 221022E-42 256717E-45 hyperon
802198E-27 10881E+24 27552E-16 221022E-42 256717E-45 bottom quark 143201E-25 194237E+25 154344E-17 221022E-42 256717E-45 W
162556E-25 220489E+25 135967E-17 221022E-42 256717E-45 Z 223368E-25 302974E+25 989499E-18 221022E-42 256717E-45 Higgs125gt9eV
256482E-25 347890E+25 861745E-18 221022E-42 256717E-45 Fermi labs 144gt9eV 311966E-25 423148E+25 708481E-18 221022E-42 256717E-45 top quark
570389E-16 77367E+34 387494E-27 221022E-42 256717E-45 Gamma 320gt20eV 185922E-09 25218E+41 118879E-33 221022E-42 256717E-45 1043gt27eV 186lt9kg
217645E-08 29521E+42 101552E-34 221022E-42 256717E-45 Planck limits
mass frequency radius r=cf mr mr(11615lt3)
mass frequency radius r=cf mr2f 222E-39 298E+11 100E-03 6626069E-34 smallest photon
242414E-35 328809E+15 911753E-08 6626069E-34 Photon 136eV 91094E-31 123559E+20 242631E-12 6626069E-34 electron 0511gt6eV 891331E-30 120899E+21 247968E-13 6626069E-34 up quark
178266E-29 241799E+21 123984E-13 6626069E-34 down quark 188356E-28 255485E+22 117343E-14 6626069E-34 muon
240606E-28 326356E+22 918606E-15 6626069E-34 meson 248806E-28 337479E+22 88833E-15 6626069E-34 meson 356532E-28 483598E+22 619921E-15 6626069E-34 strange quark
880118E-28 119379E+23 251128E-15 6626069E-34 meson 887249E-28 120346E+23 249109E-15 6626069E-34 meson 887177E-28 120336E+23 249129E-15 6626069E-34 meson 978325E-28 132699E+23 225919E-15 6626069E-34 meson 16726E-27 22687E+23 132143E-15 6626069E-34 Proton 938gt6eV
167491E-27 227183E+23 131961E-15 6626069E-34 Neutron 939gt6eV 198872E-27 269748E+23 111138E-15 6626069E-34 hyperon
212033E-27 2876E+23 104239E-15 6626069E-34 hyperon 212579E-27 28834E+23 103972E-15 6626069E-34 21347E-27 289549E+23 103538E-15 6626069E-34
231746E-27 314339E+23 953725E-16 6626069E-34 charm quark 234367E-27 317893E+23 943061E-16 6626069E-34
235543E-27 319489E+23 93835E-16 6626069E-34 298061E-27 404288E+23 741532E-16 6626069E-34 hyperon
802198E-27 10881E+24 27552E-16 6626069E-34 bottom quark 143201E-25 194237E+25 154344E-17 6626069E-34 W
162556E-25 220489E+25 135967E-17 6626069E-34 Z 223368E-25 302974E+25 989499E-18 6626069E-34 Higgs 125eV
256482E-25 347890E+25 861745E-18 6626069E-34 Fermi labs 144gt9eV 311966E-25 423148E+25 708481E-18 6626069E-34 top quark
570389E-16 77367E+34 387493E-27 6626069E-34 Gamma 320gt20eV 185922E-09 25218E+41 118879E-33 6626069E-34 186ltkg 1043gt27eV 217645E-08 29521E+42 101552E-34 6626071E-34 Planck limits
mass frequency radius r=cf mr2f
frequency radius r=cf rf=C 299792458 8c m s 5791781E+09 517617E-02 2997925E+08 smallest photon
328809E+15 911753E-08 2997925E+08 136eV photon 123559E+20 242631E-12 2997925E+08 electron 0511gt6eV 120899E+21 247968E-13 2997925E+08 up quark
241799E+21 123984E-13 2997925E+08 down quark 255485E+22 117343E-14 2997925E+08 muon
326356E+22 918606E-15 2997925E+08 meson 337479E+22 88833E-15 2997925E+08 meson 483598E+22 619921E-15 2997925E+08 strange quark
119379E+23 251128E-15 2997925E+08 meson 120346E+23 249109E-15 2997925E+08 meson 120336E+23 249129E-15 2997925E+08 meson 132699E+23 225919E-15 2997925E+08 meson 22687E+23 132143E-15 2997925E+08 Proton 938gt6eV
227183E+23 131961E-15 2997925E+08 Neutron 939gt6eV 269748E+23 111138E-15 2997925E+08 hyperon
2876E+23 104239E-15 2997925E+08 hyperon 28834E+23 103972E-15 2997925E+08 289549E+23 103538E-15 2997925E+08
314339E+23 953725E-16 2997925E+08 charm quark 317893E+23 943061E-16 2997925E+08
319489E+23 93835E-16 2997925E+08 404288E+23 741532E-16 2997925E+08 hyperon
10881E+24 27552E-16 2997925E+08 bottom quark 194237E+25 154344E-17 2997925E+08 W
220489E+25 135967E-17 2997925E+08 Z 302974E+25 989499E-18 2997925E+08 Higgs 125eV
347890E+25 861745E-18 2997925E+08 Fermi labs 144gt9eV 423148E+25 708481E-18 2997925E+08 top quark
77367E+34 387493E-27 2997925E+08 Gamma 320gt20eV 25218E+41 118879E-33 2997925E+08 1043gt27eV 186lt9kg
29521E+42 101552E-34 2997925E+08 Planck limits
frequency radius r=cf rf=C
Savior Borg
Two helix makes a mass particle
Fernandes
Kanarevrsquos Photon Model
My website Symmetrymathcomis my interpretation of the Authorrsquos I have referenced
I do not claim that any of these authors will agree withmy interpretations
I leave it to anyone who visits my site to decide the logicof my interpretation
- SYMMETRY MATH (SM) By Jack Kuykendall
- Why is Symmetry Math Needed
- From ldquoWhat Is Mathematicsrdquo by Courant and Robbins ndash page 55 Th
- Slide 4
- Slide 5
- Slide 6
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-
Carbon
Diamond
Graphene
H2O H2O Cluster
There are 82gt22 (82000000000000000000000) free electrons in each cubic centimeter of a copper wire
Since no protons are free there cannot be an Attractive charge action proton potential in a wire
There are no protons present at one plate and electrons present at the other plate
Temperature Radius of Photons Equation 0 0(0 ) (27315 )AC T
0
2898 310609555 6
27315o
AoC
mTr m
316
0 0(1 ) (27415 )AС T 1
2898 310570855 6
27415A
o oC
mTr m
317
0 020 (68 ) 29315o
AC F T 20
2898 39885 6
29315o
AoC
mTr m
321
0 030 (86 ) 30315oAC F T
30
2898 39560 6
30315o
AoC
mTr m
322
0 0 0[100 (212 ) 37315 ]AC F T 100
2898 37766 6
37315o C
mKr m
K
323
0 0 0[(1000 )(1832 ) 127315 ]AC F T 1000
2898 32276 6
127315o
AoC
mTr m
324
0 0 0[(1500 )(2732 ) 177315 ]AC F T 1500
2898 31634 6
177315A
o oC
mTr m
325
0 0 0[(2000 )(3632 ) 227315 ]AC F T 2000
2898 31275 6
227315A
o oC
mTr m
326
1
0 0 0(0 2000 )(27315 227315 )oAto C to T
Temperatures between are formed with photons from the infra-red range As the temperature increase the radiuses of the photons decrease
Temperature (felt as HEAT by humans) is formed by photon emission from electrons
What is the exact radius of the photons that form the (highest) temperature felt as heat Modern science has no exact answer to this question We can only theorize that lsquoHEATrsquo is formed by photons which are emitted by electrons during the synthesis of atoms and molecules The border for the smallest radius of these photons is not established yet We theorize that it is in an interval of the ultra-violet range Since the shortest know radiuses (X and gamma photons) are emitted by nuclear synthesis they do not participate in heat formation Temperature (felt as HEAT by humans) is formed by photon emission from electrons If gamma photons participated in formation of heat the greatest possible temperature would be equal
maxmin
2898 3 2898 31 15 1000000000000000
3 18oA A
A AmT mT
T T Tr m
(348)
If such a temperature existed it would destroy molecules atoms and the nucleus of all atoms
Schwarzschildrsquos (incorrect) radius equation
2
2 g
GMR
C
Kanarevrsquos correct equation
1(1087148751 14)
2p
g photon
GMrR Mr
C
Schwarzschild 3 33
3 (3)(2 30 )
4 (4)(314)(297 3182 19
)og
M kgD
R m
kg
m
(403) 1
Kanarev
3 33
3 (3)(2 30 )
4 (4)(314)(439 )565 27red
g
M kgD
R m
kg
m
(404) 2
Kanarev
3 33 3
3 (3)(2 30 )276 35 (
4 (4)(314)(553 32 Kanrev error
0012)
)lightg
M kg kgD
R m m
kg
m
(405) 3
4
Kanarevrsquos 3 33
3 (3)(2 30)
4 (4)(314)(154 8)013 54gamma
g
MD
R
kg
m
(406) 5
6
Fernandesrsquo (18 36 9) 3 3
3 (3)(2 30)265 81
4 (4)(314)(18 17)rg
kg
m
MD
R
(407)
7
Nuclear density is around3
(12 24) 17kg
m 8
If the Sun were compressed to ( )( 439 )g redR m (equation 400) its gravitational attraction would
detain only radiation of the far infra-red spectrum Shorter wavelengths will not be detained If photons of gamma radiation were detained the gravitational radius of the Sun would have to
be ( )( 154 8 )g gammaR m (equation 402) This would not be possible because the density of
the Sun (equation 406) would be 37 orders more dense than the nucleus of atoms
Only one type of Black-hole can be all black That is one with a gravitational attraction strong enough to detain the highest energy gamma photons All other Black-holes should have colors which vary with the change of wavelengths of the photons which cannot be detained The order should be infra-red light ultra-violet x-ray and finally gamma photons
Mass is Frequency of an Elementary Particle All known experiments have produced answers that agree with the following equations ( ) ( )( )(distance)mad Action mass acceleration
( )mad
( ) ( ) (constant)(frequency)mad hf
(1) 2( ) ( )(constant)mad mc mass
(2)
c f rf (3)
Equating (1) and (2) 2( )m h c f Mass = (constant)(frequency of one particle)
Plankrsquos constant 6626069 34 HBCP 662603(11) 57(29)69 34 codatah Js h Js
2
222
2
6626069 34( )( ) ( )6626069 347372496 51
(299792458 8 ) 898755179 16
Joule
Kgmm sh Js s kgs
m mcs s
Equating (1)(2) and (3)
2 2
constant
expexp exp
1 6626069 34 22102 42
299792458 8
h h c h h JsM f mr kgm
mr c r cc cs
All mass is built from one elementary particle a ( 737 51 )kg particle
All mass is in relative motion with other mass Time is a description invented to measure the relative movement between different particles of
mass Energy is a term invented to describe what happens when masses interact Energy is a
description Mass does not convert to energy and energy does not convert to mass The word ldquoEnergyrdquo will be replaced with an Action involving mass acceleration and distance ( )mad
When the motion of masses interacts they change to different motions of masses These different mass forms can be measured and their changes in momentum can be measured When the change produces photon particles they have been incorrectly labeled as ldquoEnergyrdquo
Mass Frequency Frequency is a rotation per second
If ( )f is set to one rotation per second (737249638 51 )(1kgs rot s the equation predicts that the
7372496 51kg particle is the elementary mass out of which all mass is made
27372496 51 07372496 50
hKgs kg
c
2 07372496 50 second
m m rotationsf
h kgsc
22 392984064 11
0737249638 50 s
kg rotf
kgs
Kanarev Smallest measurable photon
24241 35328803 15
07372496 50 s
kg rotf
kgs
13598443ev photon 2424143lt35kg
91093897 31123559 20
07372496 50 s
kg rotf
kgs
05109906 6eV electron
9109235lt31kg
16726230 27226873 23
07372496 50 s
kg rotf
kgs
93827241gt6eV proton 1672623lt27kg
16749286 272271858 23
07372496 50 s
kg rotf
kgs
9395656gt6eV neutron 1674928lt27kg
334357629 27226750117 23
(07372496 50 ) )2( s
kg rotf
kgs
(Deuterium nucleus) Proton+Neutron
22102 42mr kgm
334357629 274535202583 23
(07372496 50 ( ) s1)
kg rotf
kgs
Proton+Neutron
22102 42mr kgm
223368 253029747 25
07372496 50 s
kg rotf
kgs
1243gt9eV Higgs Boson 2215849lt25 kg
256482lt25347890 25
07372496 50 s
kg rotf
kgs
144gt11eV Fermi Lab particle 2567033lt25kg
5703889152 1677367 34
07372496 50 s
kg rotf
kgs
320gt20eV highest eV particle measured 5704518lt16kg
1859222909 925218 41
07372496 50 s
kg rotf
kgs
10429476 27eV Fernandes 1859223lt9kg
2176450474 8295212 42
07372496 50 s
kg rotf
kgs
Planck limit
78434 28106387 23
07372496 50 s
kg rotf
kgs
classical radius of an electron
This frequency and radius are not a particle radius
This mass does not exist
308 Is there experimental evidences that an electron has the so-called classic radius of an electron A No experiments do not exist From Questions and Answers of Kanarev Page 95 Electrons-Protons-Neutrons JK ET and Rewrite The handbook of Chemistry and Physics shows the classical radius of the electron as (2817 15 )m Kanarevs theory shows this to be the radius of approach of the magnetic force lines around the torus ring of the electron ( )magr This is confirmed by calculating the dimensionless fine structure constant
( ) (which is equal to the circumference of the inner circle (2 )magr divided by the experimentally
measured radius of the electron exp( )R
exp
2 (2)(314)(2817 15 ) 10007297352568
2426 12 137036magr m
R m
(191)
Kg Rotationsec meters experimental classical rmag mass frequency radius r=cf mr mr(11615lt3) 427E-41 5791781E+09 517617E-02 221022E-42 256717E-45 smallest photon
242414E-35 328809E+15 911753E-08 221022E-42 256717E-45 136eV photon 91094E-31 123559E+20 242631E-12 221022E-42 256717E-45 electron 0511gt6eV 891331E-30 120899E+21 247968E-13 221022E-42 256717E-45 up quark
178266E-29 241799E+21 123984E-13 221022E-42 256717E-45 down quark 188356E-28 255485E+22 117343E-14 221022E-42 256717E-45 muon
240606E-28 326356E+22 918606E-15 221022E-42 256717E-45 meson 248806E-28 337479E+22 88833E-15 221022E-42 256717E-45 meson 356532E-28 483598E+22 619921E-15 221022E-42 256717E-45 strange quark
880118E-28 119379E+23 251128E-15 221022E-42 256717E-45 meson 887249E-28 120346E+23 249109E-15 221022E-42 256717E-45 meson 887177E-28 120336E+23 249129E-15 221022E-42 256717E-45 meson 978325E-28 132699E+23 225919E-15 221022E-42 256717E-45 meson 16726E-27 22687E+23 132143E-15 221022E-42 256717E-45 Proton 938gt6eV
167491E-27 227183E+23 131961E-15 221022E-42 256717E-45 Neutron 939gt6eV 198872E-27 269748E+23 111138E-15 221022E-42 256717E-45 hyperon
212033E-27 2876E+23 104239E-15 221022E-42 256717E-45 hyperon 212579E-27 28834E+23 103972E-15 221022E-42 256717E-45 21347E-27 289549E+23 103538E-15 221022E-42 256717E-45
231746E-27 314339E+23 953725E-16 221022E-42 256717E-45 charm quark 234367E-27 317893E+23 943061E-16 221022E-42 256717E-45
235543E-27 319489E+23 93835E-16 221022E-42 256717E-45 298061E-27 404288E+23 741532E-16 221022E-42 256717E-45 hyperon
802198E-27 10881E+24 27552E-16 221022E-42 256717E-45 bottom quark 143201E-25 194237E+25 154344E-17 221022E-42 256717E-45 W
162556E-25 220489E+25 135967E-17 221022E-42 256717E-45 Z 223368E-25 302974E+25 989499E-18 221022E-42 256717E-45 Higgs125gt9eV
256482E-25 347890E+25 861745E-18 221022E-42 256717E-45 Fermi labs 144gt9eV 311966E-25 423148E+25 708481E-18 221022E-42 256717E-45 top quark
570389E-16 77367E+34 387494E-27 221022E-42 256717E-45 Gamma 320gt20eV 185922E-09 25218E+41 118879E-33 221022E-42 256717E-45 1043gt27eV 186lt9kg
217645E-08 29521E+42 101552E-34 221022E-42 256717E-45 Planck limits
mass frequency radius r=cf mr mr(11615lt3)
mass frequency radius r=cf mr2f 222E-39 298E+11 100E-03 6626069E-34 smallest photon
242414E-35 328809E+15 911753E-08 6626069E-34 Photon 136eV 91094E-31 123559E+20 242631E-12 6626069E-34 electron 0511gt6eV 891331E-30 120899E+21 247968E-13 6626069E-34 up quark
178266E-29 241799E+21 123984E-13 6626069E-34 down quark 188356E-28 255485E+22 117343E-14 6626069E-34 muon
240606E-28 326356E+22 918606E-15 6626069E-34 meson 248806E-28 337479E+22 88833E-15 6626069E-34 meson 356532E-28 483598E+22 619921E-15 6626069E-34 strange quark
880118E-28 119379E+23 251128E-15 6626069E-34 meson 887249E-28 120346E+23 249109E-15 6626069E-34 meson 887177E-28 120336E+23 249129E-15 6626069E-34 meson 978325E-28 132699E+23 225919E-15 6626069E-34 meson 16726E-27 22687E+23 132143E-15 6626069E-34 Proton 938gt6eV
167491E-27 227183E+23 131961E-15 6626069E-34 Neutron 939gt6eV 198872E-27 269748E+23 111138E-15 6626069E-34 hyperon
212033E-27 2876E+23 104239E-15 6626069E-34 hyperon 212579E-27 28834E+23 103972E-15 6626069E-34 21347E-27 289549E+23 103538E-15 6626069E-34
231746E-27 314339E+23 953725E-16 6626069E-34 charm quark 234367E-27 317893E+23 943061E-16 6626069E-34
235543E-27 319489E+23 93835E-16 6626069E-34 298061E-27 404288E+23 741532E-16 6626069E-34 hyperon
802198E-27 10881E+24 27552E-16 6626069E-34 bottom quark 143201E-25 194237E+25 154344E-17 6626069E-34 W
162556E-25 220489E+25 135967E-17 6626069E-34 Z 223368E-25 302974E+25 989499E-18 6626069E-34 Higgs 125eV
256482E-25 347890E+25 861745E-18 6626069E-34 Fermi labs 144gt9eV 311966E-25 423148E+25 708481E-18 6626069E-34 top quark
570389E-16 77367E+34 387493E-27 6626069E-34 Gamma 320gt20eV 185922E-09 25218E+41 118879E-33 6626069E-34 186ltkg 1043gt27eV 217645E-08 29521E+42 101552E-34 6626071E-34 Planck limits
mass frequency radius r=cf mr2f
frequency radius r=cf rf=C 299792458 8c m s 5791781E+09 517617E-02 2997925E+08 smallest photon
328809E+15 911753E-08 2997925E+08 136eV photon 123559E+20 242631E-12 2997925E+08 electron 0511gt6eV 120899E+21 247968E-13 2997925E+08 up quark
241799E+21 123984E-13 2997925E+08 down quark 255485E+22 117343E-14 2997925E+08 muon
326356E+22 918606E-15 2997925E+08 meson 337479E+22 88833E-15 2997925E+08 meson 483598E+22 619921E-15 2997925E+08 strange quark
119379E+23 251128E-15 2997925E+08 meson 120346E+23 249109E-15 2997925E+08 meson 120336E+23 249129E-15 2997925E+08 meson 132699E+23 225919E-15 2997925E+08 meson 22687E+23 132143E-15 2997925E+08 Proton 938gt6eV
227183E+23 131961E-15 2997925E+08 Neutron 939gt6eV 269748E+23 111138E-15 2997925E+08 hyperon
2876E+23 104239E-15 2997925E+08 hyperon 28834E+23 103972E-15 2997925E+08 289549E+23 103538E-15 2997925E+08
314339E+23 953725E-16 2997925E+08 charm quark 317893E+23 943061E-16 2997925E+08
319489E+23 93835E-16 2997925E+08 404288E+23 741532E-16 2997925E+08 hyperon
10881E+24 27552E-16 2997925E+08 bottom quark 194237E+25 154344E-17 2997925E+08 W
220489E+25 135967E-17 2997925E+08 Z 302974E+25 989499E-18 2997925E+08 Higgs 125eV
347890E+25 861745E-18 2997925E+08 Fermi labs 144gt9eV 423148E+25 708481E-18 2997925E+08 top quark
77367E+34 387493E-27 2997925E+08 Gamma 320gt20eV 25218E+41 118879E-33 2997925E+08 1043gt27eV 186lt9kg
29521E+42 101552E-34 2997925E+08 Planck limits
frequency radius r=cf rf=C
Savior Borg
Two helix makes a mass particle
Fernandes
Kanarevrsquos Photon Model
My website Symmetrymathcomis my interpretation of the Authorrsquos I have referenced
I do not claim that any of these authors will agree withmy interpretations
I leave it to anyone who visits my site to decide the logicof my interpretation
- SYMMETRY MATH (SM) By Jack Kuykendall
- Why is Symmetry Math Needed
- From ldquoWhat Is Mathematicsrdquo by Courant and Robbins ndash page 55 Th
- Slide 4
- Slide 5
- Slide 6
- Slide 7
- Slide 8
- Slide 9
- Slide 10
- Slide 11
- Slide 12
- Slide 13
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- Slide 26
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- Slide 35
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- Slide 39
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- Slide 41
- Slide 42
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- Slide 45
- Slide 46
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- Slide 49
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- Slide 54
- Slide 55
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- Slide 57
- Slide 58
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- Slide 61
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- Slide 64
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- Slide 68
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- Slide 77
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- Slide 83
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- Slide 90
- Slide 91
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- Slide 97
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- Slide 103
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- Slide 106
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- Slide 110
- Slide 111
- Slide 112
- Slide 113
- Slide 114
- Slide 115
- Slide 116
- Slide 117
- Slide 118
- Slide 119
- Slide 120
- Slide 121
- Slide 122
- Slide 123
-
Diamond
Graphene
H2O H2O Cluster
There are 82gt22 (82000000000000000000000) free electrons in each cubic centimeter of a copper wire
Since no protons are free there cannot be an Attractive charge action proton potential in a wire
There are no protons present at one plate and electrons present at the other plate
Temperature Radius of Photons Equation 0 0(0 ) (27315 )AC T
0
2898 310609555 6
27315o
AoC
mTr m
316
0 0(1 ) (27415 )AС T 1
2898 310570855 6
27415A
o oC
mTr m
317
0 020 (68 ) 29315o
AC F T 20
2898 39885 6
29315o
AoC
mTr m
321
0 030 (86 ) 30315oAC F T
30
2898 39560 6
30315o
AoC
mTr m
322
0 0 0[100 (212 ) 37315 ]AC F T 100
2898 37766 6
37315o C
mKr m
K
323
0 0 0[(1000 )(1832 ) 127315 ]AC F T 1000
2898 32276 6
127315o
AoC
mTr m
324
0 0 0[(1500 )(2732 ) 177315 ]AC F T 1500
2898 31634 6
177315A
o oC
mTr m
325
0 0 0[(2000 )(3632 ) 227315 ]AC F T 2000
2898 31275 6
227315A
o oC
mTr m
326
1
0 0 0(0 2000 )(27315 227315 )oAto C to T
Temperatures between are formed with photons from the infra-red range As the temperature increase the radiuses of the photons decrease
Temperature (felt as HEAT by humans) is formed by photon emission from electrons
What is the exact radius of the photons that form the (highest) temperature felt as heat Modern science has no exact answer to this question We can only theorize that lsquoHEATrsquo is formed by photons which are emitted by electrons during the synthesis of atoms and molecules The border for the smallest radius of these photons is not established yet We theorize that it is in an interval of the ultra-violet range Since the shortest know radiuses (X and gamma photons) are emitted by nuclear synthesis they do not participate in heat formation Temperature (felt as HEAT by humans) is formed by photon emission from electrons If gamma photons participated in formation of heat the greatest possible temperature would be equal
maxmin
2898 3 2898 31 15 1000000000000000
3 18oA A
A AmT mT
T T Tr m
(348)
If such a temperature existed it would destroy molecules atoms and the nucleus of all atoms
Schwarzschildrsquos (incorrect) radius equation
2
2 g
GMR
C
Kanarevrsquos correct equation
1(1087148751 14)
2p
g photon
GMrR Mr
C
Schwarzschild 3 33
3 (3)(2 30 )
4 (4)(314)(297 3182 19
)og
M kgD
R m
kg
m
(403) 1
Kanarev
3 33
3 (3)(2 30 )
4 (4)(314)(439 )565 27red
g
M kgD
R m
kg
m
(404) 2
Kanarev
3 33 3
3 (3)(2 30 )276 35 (
4 (4)(314)(553 32 Kanrev error
0012)
)lightg
M kg kgD
R m m
kg
m
(405) 3
4
Kanarevrsquos 3 33
3 (3)(2 30)
4 (4)(314)(154 8)013 54gamma
g
MD
R
kg
m
(406) 5
6
Fernandesrsquo (18 36 9) 3 3
3 (3)(2 30)265 81
4 (4)(314)(18 17)rg
kg
m
MD
R
(407)
7
Nuclear density is around3
(12 24) 17kg
m 8
If the Sun were compressed to ( )( 439 )g redR m (equation 400) its gravitational attraction would
detain only radiation of the far infra-red spectrum Shorter wavelengths will not be detained If photons of gamma radiation were detained the gravitational radius of the Sun would have to
be ( )( 154 8 )g gammaR m (equation 402) This would not be possible because the density of
the Sun (equation 406) would be 37 orders more dense than the nucleus of atoms
Only one type of Black-hole can be all black That is one with a gravitational attraction strong enough to detain the highest energy gamma photons All other Black-holes should have colors which vary with the change of wavelengths of the photons which cannot be detained The order should be infra-red light ultra-violet x-ray and finally gamma photons
Mass is Frequency of an Elementary Particle All known experiments have produced answers that agree with the following equations ( ) ( )( )(distance)mad Action mass acceleration
( )mad
( ) ( ) (constant)(frequency)mad hf
(1) 2( ) ( )(constant)mad mc mass
(2)
c f rf (3)
Equating (1) and (2) 2( )m h c f Mass = (constant)(frequency of one particle)
Plankrsquos constant 6626069 34 HBCP 662603(11) 57(29)69 34 codatah Js h Js
2
222
2
6626069 34( )( ) ( )6626069 347372496 51
(299792458 8 ) 898755179 16
Joule
Kgmm sh Js s kgs
m mcs s
Equating (1)(2) and (3)
2 2
constant
expexp exp
1 6626069 34 22102 42
299792458 8
h h c h h JsM f mr kgm
mr c r cc cs
All mass is built from one elementary particle a ( 737 51 )kg particle
All mass is in relative motion with other mass Time is a description invented to measure the relative movement between different particles of
mass Energy is a term invented to describe what happens when masses interact Energy is a
description Mass does not convert to energy and energy does not convert to mass The word ldquoEnergyrdquo will be replaced with an Action involving mass acceleration and distance ( )mad
When the motion of masses interacts they change to different motions of masses These different mass forms can be measured and their changes in momentum can be measured When the change produces photon particles they have been incorrectly labeled as ldquoEnergyrdquo
Mass Frequency Frequency is a rotation per second
If ( )f is set to one rotation per second (737249638 51 )(1kgs rot s the equation predicts that the
7372496 51kg particle is the elementary mass out of which all mass is made
27372496 51 07372496 50
hKgs kg
c
2 07372496 50 second
m m rotationsf
h kgsc
22 392984064 11
0737249638 50 s
kg rotf
kgs
Kanarev Smallest measurable photon
24241 35328803 15
07372496 50 s
kg rotf
kgs
13598443ev photon 2424143lt35kg
91093897 31123559 20
07372496 50 s
kg rotf
kgs
05109906 6eV electron
9109235lt31kg
16726230 27226873 23
07372496 50 s
kg rotf
kgs
93827241gt6eV proton 1672623lt27kg
16749286 272271858 23
07372496 50 s
kg rotf
kgs
9395656gt6eV neutron 1674928lt27kg
334357629 27226750117 23
(07372496 50 ) )2( s
kg rotf
kgs
(Deuterium nucleus) Proton+Neutron
22102 42mr kgm
334357629 274535202583 23
(07372496 50 ( ) s1)
kg rotf
kgs
Proton+Neutron
22102 42mr kgm
223368 253029747 25
07372496 50 s
kg rotf
kgs
1243gt9eV Higgs Boson 2215849lt25 kg
256482lt25347890 25
07372496 50 s
kg rotf
kgs
144gt11eV Fermi Lab particle 2567033lt25kg
5703889152 1677367 34
07372496 50 s
kg rotf
kgs
320gt20eV highest eV particle measured 5704518lt16kg
1859222909 925218 41
07372496 50 s
kg rotf
kgs
10429476 27eV Fernandes 1859223lt9kg
2176450474 8295212 42
07372496 50 s
kg rotf
kgs
Planck limit
78434 28106387 23
07372496 50 s
kg rotf
kgs
classical radius of an electron
This frequency and radius are not a particle radius
This mass does not exist
308 Is there experimental evidences that an electron has the so-called classic radius of an electron A No experiments do not exist From Questions and Answers of Kanarev Page 95 Electrons-Protons-Neutrons JK ET and Rewrite The handbook of Chemistry and Physics shows the classical radius of the electron as (2817 15 )m Kanarevs theory shows this to be the radius of approach of the magnetic force lines around the torus ring of the electron ( )magr This is confirmed by calculating the dimensionless fine structure constant
( ) (which is equal to the circumference of the inner circle (2 )magr divided by the experimentally
measured radius of the electron exp( )R
exp
2 (2)(314)(2817 15 ) 10007297352568
2426 12 137036magr m
R m
(191)
Kg Rotationsec meters experimental classical rmag mass frequency radius r=cf mr mr(11615lt3) 427E-41 5791781E+09 517617E-02 221022E-42 256717E-45 smallest photon
242414E-35 328809E+15 911753E-08 221022E-42 256717E-45 136eV photon 91094E-31 123559E+20 242631E-12 221022E-42 256717E-45 electron 0511gt6eV 891331E-30 120899E+21 247968E-13 221022E-42 256717E-45 up quark
178266E-29 241799E+21 123984E-13 221022E-42 256717E-45 down quark 188356E-28 255485E+22 117343E-14 221022E-42 256717E-45 muon
240606E-28 326356E+22 918606E-15 221022E-42 256717E-45 meson 248806E-28 337479E+22 88833E-15 221022E-42 256717E-45 meson 356532E-28 483598E+22 619921E-15 221022E-42 256717E-45 strange quark
880118E-28 119379E+23 251128E-15 221022E-42 256717E-45 meson 887249E-28 120346E+23 249109E-15 221022E-42 256717E-45 meson 887177E-28 120336E+23 249129E-15 221022E-42 256717E-45 meson 978325E-28 132699E+23 225919E-15 221022E-42 256717E-45 meson 16726E-27 22687E+23 132143E-15 221022E-42 256717E-45 Proton 938gt6eV
167491E-27 227183E+23 131961E-15 221022E-42 256717E-45 Neutron 939gt6eV 198872E-27 269748E+23 111138E-15 221022E-42 256717E-45 hyperon
212033E-27 2876E+23 104239E-15 221022E-42 256717E-45 hyperon 212579E-27 28834E+23 103972E-15 221022E-42 256717E-45 21347E-27 289549E+23 103538E-15 221022E-42 256717E-45
231746E-27 314339E+23 953725E-16 221022E-42 256717E-45 charm quark 234367E-27 317893E+23 943061E-16 221022E-42 256717E-45
235543E-27 319489E+23 93835E-16 221022E-42 256717E-45 298061E-27 404288E+23 741532E-16 221022E-42 256717E-45 hyperon
802198E-27 10881E+24 27552E-16 221022E-42 256717E-45 bottom quark 143201E-25 194237E+25 154344E-17 221022E-42 256717E-45 W
162556E-25 220489E+25 135967E-17 221022E-42 256717E-45 Z 223368E-25 302974E+25 989499E-18 221022E-42 256717E-45 Higgs125gt9eV
256482E-25 347890E+25 861745E-18 221022E-42 256717E-45 Fermi labs 144gt9eV 311966E-25 423148E+25 708481E-18 221022E-42 256717E-45 top quark
570389E-16 77367E+34 387494E-27 221022E-42 256717E-45 Gamma 320gt20eV 185922E-09 25218E+41 118879E-33 221022E-42 256717E-45 1043gt27eV 186lt9kg
217645E-08 29521E+42 101552E-34 221022E-42 256717E-45 Planck limits
mass frequency radius r=cf mr mr(11615lt3)
mass frequency radius r=cf mr2f 222E-39 298E+11 100E-03 6626069E-34 smallest photon
242414E-35 328809E+15 911753E-08 6626069E-34 Photon 136eV 91094E-31 123559E+20 242631E-12 6626069E-34 electron 0511gt6eV 891331E-30 120899E+21 247968E-13 6626069E-34 up quark
178266E-29 241799E+21 123984E-13 6626069E-34 down quark 188356E-28 255485E+22 117343E-14 6626069E-34 muon
240606E-28 326356E+22 918606E-15 6626069E-34 meson 248806E-28 337479E+22 88833E-15 6626069E-34 meson 356532E-28 483598E+22 619921E-15 6626069E-34 strange quark
880118E-28 119379E+23 251128E-15 6626069E-34 meson 887249E-28 120346E+23 249109E-15 6626069E-34 meson 887177E-28 120336E+23 249129E-15 6626069E-34 meson 978325E-28 132699E+23 225919E-15 6626069E-34 meson 16726E-27 22687E+23 132143E-15 6626069E-34 Proton 938gt6eV
167491E-27 227183E+23 131961E-15 6626069E-34 Neutron 939gt6eV 198872E-27 269748E+23 111138E-15 6626069E-34 hyperon
212033E-27 2876E+23 104239E-15 6626069E-34 hyperon 212579E-27 28834E+23 103972E-15 6626069E-34 21347E-27 289549E+23 103538E-15 6626069E-34
231746E-27 314339E+23 953725E-16 6626069E-34 charm quark 234367E-27 317893E+23 943061E-16 6626069E-34
235543E-27 319489E+23 93835E-16 6626069E-34 298061E-27 404288E+23 741532E-16 6626069E-34 hyperon
802198E-27 10881E+24 27552E-16 6626069E-34 bottom quark 143201E-25 194237E+25 154344E-17 6626069E-34 W
162556E-25 220489E+25 135967E-17 6626069E-34 Z 223368E-25 302974E+25 989499E-18 6626069E-34 Higgs 125eV
256482E-25 347890E+25 861745E-18 6626069E-34 Fermi labs 144gt9eV 311966E-25 423148E+25 708481E-18 6626069E-34 top quark
570389E-16 77367E+34 387493E-27 6626069E-34 Gamma 320gt20eV 185922E-09 25218E+41 118879E-33 6626069E-34 186ltkg 1043gt27eV 217645E-08 29521E+42 101552E-34 6626071E-34 Planck limits
mass frequency radius r=cf mr2f
frequency radius r=cf rf=C 299792458 8c m s 5791781E+09 517617E-02 2997925E+08 smallest photon
328809E+15 911753E-08 2997925E+08 136eV photon 123559E+20 242631E-12 2997925E+08 electron 0511gt6eV 120899E+21 247968E-13 2997925E+08 up quark
241799E+21 123984E-13 2997925E+08 down quark 255485E+22 117343E-14 2997925E+08 muon
326356E+22 918606E-15 2997925E+08 meson 337479E+22 88833E-15 2997925E+08 meson 483598E+22 619921E-15 2997925E+08 strange quark
119379E+23 251128E-15 2997925E+08 meson 120346E+23 249109E-15 2997925E+08 meson 120336E+23 249129E-15 2997925E+08 meson 132699E+23 225919E-15 2997925E+08 meson 22687E+23 132143E-15 2997925E+08 Proton 938gt6eV
227183E+23 131961E-15 2997925E+08 Neutron 939gt6eV 269748E+23 111138E-15 2997925E+08 hyperon
2876E+23 104239E-15 2997925E+08 hyperon 28834E+23 103972E-15 2997925E+08 289549E+23 103538E-15 2997925E+08
314339E+23 953725E-16 2997925E+08 charm quark 317893E+23 943061E-16 2997925E+08
319489E+23 93835E-16 2997925E+08 404288E+23 741532E-16 2997925E+08 hyperon
10881E+24 27552E-16 2997925E+08 bottom quark 194237E+25 154344E-17 2997925E+08 W
220489E+25 135967E-17 2997925E+08 Z 302974E+25 989499E-18 2997925E+08 Higgs 125eV
347890E+25 861745E-18 2997925E+08 Fermi labs 144gt9eV 423148E+25 708481E-18 2997925E+08 top quark
77367E+34 387493E-27 2997925E+08 Gamma 320gt20eV 25218E+41 118879E-33 2997925E+08 1043gt27eV 186lt9kg
29521E+42 101552E-34 2997925E+08 Planck limits
frequency radius r=cf rf=C
Savior Borg
Two helix makes a mass particle
Fernandes
Kanarevrsquos Photon Model
My website Symmetrymathcomis my interpretation of the Authorrsquos I have referenced
I do not claim that any of these authors will agree withmy interpretations
I leave it to anyone who visits my site to decide the logicof my interpretation
- SYMMETRY MATH (SM) By Jack Kuykendall
- Why is Symmetry Math Needed
- From ldquoWhat Is Mathematicsrdquo by Courant and Robbins ndash page 55 Th
- Slide 4
- Slide 5
- Slide 6
- Slide 7
- Slide 8
- Slide 9
- Slide 10
- Slide 11
- Slide 12
- Slide 13
- Slide 14
- Slide 15
- Slide 16
- Slide 17
- Slide 18
- Slide 19
- Slide 20
- Slide 21
- Slide 22
- Slide 23
- Slide 24
- Slide 25
- Slide 26
- Slide 27
- Slide 28
- Slide 29
- Slide 30
- Slide 31
- Slide 32
- Slide 33
- Slide 34
- Slide 35
- Slide 36
- Slide 37
- Slide 38
- Slide 39
- Slide 40
- Slide 41
- Slide 42
- Slide 43
- Slide 44
- Slide 45
- Slide 46
- Slide 47
- Slide 48
- Slide 49
- Slide 50
- Slide 51
- Slide 52
- Slide 53
- Slide 54
- Slide 55
- Slide 56
- Slide 57
- Slide 58
- Slide 59
- Slide 60
- Slide 61
- Slide 62
- Slide 63
- Slide 64
- Slide 65
- Slide 66
- Slide 67
- Slide 68
- Slide 69
- Slide 70
- Slide 71
- Slide 72
- Slide 73
- Slide 74
- Slide 75
- Slide 76
- Slide 77
- Slide 78
- Slide 79
- Slide 80
- Slide 81
- Slide 82
- Slide 83
- Slide 84
- Slide 85
- Slide 86
- Slide 87
- Slide 88
- Slide 89
- Slide 90
- Slide 91
- Slide 92
- Slide 93
- Slide 94
- Slide 95
- Slide 96
- Slide 97
- Slide 98
- Slide 99
- Slide 100
- Slide 101
- Slide 102
- Slide 103
- Slide 104
- Slide 105
- Slide 106
- Slide 107
- Slide 108
- Slide 109
- Slide 110
- Slide 111
- Slide 112
- Slide 113
- Slide 114
- Slide 115
- Slide 116
- Slide 117
- Slide 118
- Slide 119
- Slide 120
- Slide 121
- Slide 122
- Slide 123
-
Graphene
H2O H2O Cluster
There are 82gt22 (82000000000000000000000) free electrons in each cubic centimeter of a copper wire
Since no protons are free there cannot be an Attractive charge action proton potential in a wire
There are no protons present at one plate and electrons present at the other plate
Temperature Radius of Photons Equation 0 0(0 ) (27315 )AC T
0
2898 310609555 6
27315o
AoC
mTr m
316
0 0(1 ) (27415 )AС T 1
2898 310570855 6
27415A
o oC
mTr m
317
0 020 (68 ) 29315o
AC F T 20
2898 39885 6
29315o
AoC
mTr m
321
0 030 (86 ) 30315oAC F T
30
2898 39560 6
30315o
AoC
mTr m
322
0 0 0[100 (212 ) 37315 ]AC F T 100
2898 37766 6
37315o C
mKr m
K
323
0 0 0[(1000 )(1832 ) 127315 ]AC F T 1000
2898 32276 6
127315o
AoC
mTr m
324
0 0 0[(1500 )(2732 ) 177315 ]AC F T 1500
2898 31634 6
177315A
o oC
mTr m
325
0 0 0[(2000 )(3632 ) 227315 ]AC F T 2000
2898 31275 6
227315A
o oC
mTr m
326
1
0 0 0(0 2000 )(27315 227315 )oAto C to T
Temperatures between are formed with photons from the infra-red range As the temperature increase the radiuses of the photons decrease
Temperature (felt as HEAT by humans) is formed by photon emission from electrons
What is the exact radius of the photons that form the (highest) temperature felt as heat Modern science has no exact answer to this question We can only theorize that lsquoHEATrsquo is formed by photons which are emitted by electrons during the synthesis of atoms and molecules The border for the smallest radius of these photons is not established yet We theorize that it is in an interval of the ultra-violet range Since the shortest know radiuses (X and gamma photons) are emitted by nuclear synthesis they do not participate in heat formation Temperature (felt as HEAT by humans) is formed by photon emission from electrons If gamma photons participated in formation of heat the greatest possible temperature would be equal
maxmin
2898 3 2898 31 15 1000000000000000
3 18oA A
A AmT mT
T T Tr m
(348)
If such a temperature existed it would destroy molecules atoms and the nucleus of all atoms
Schwarzschildrsquos (incorrect) radius equation
2
2 g
GMR
C
Kanarevrsquos correct equation
1(1087148751 14)
2p
g photon
GMrR Mr
C
Schwarzschild 3 33
3 (3)(2 30 )
4 (4)(314)(297 3182 19
)og
M kgD
R m
kg
m
(403) 1
Kanarev
3 33
3 (3)(2 30 )
4 (4)(314)(439 )565 27red
g
M kgD
R m
kg
m
(404) 2
Kanarev
3 33 3
3 (3)(2 30 )276 35 (
4 (4)(314)(553 32 Kanrev error
0012)
)lightg
M kg kgD
R m m
kg
m
(405) 3
4
Kanarevrsquos 3 33
3 (3)(2 30)
4 (4)(314)(154 8)013 54gamma
g
MD
R
kg
m
(406) 5
6
Fernandesrsquo (18 36 9) 3 3
3 (3)(2 30)265 81
4 (4)(314)(18 17)rg
kg
m
MD
R
(407)
7
Nuclear density is around3
(12 24) 17kg
m 8
If the Sun were compressed to ( )( 439 )g redR m (equation 400) its gravitational attraction would
detain only radiation of the far infra-red spectrum Shorter wavelengths will not be detained If photons of gamma radiation were detained the gravitational radius of the Sun would have to
be ( )( 154 8 )g gammaR m (equation 402) This would not be possible because the density of
the Sun (equation 406) would be 37 orders more dense than the nucleus of atoms
Only one type of Black-hole can be all black That is one with a gravitational attraction strong enough to detain the highest energy gamma photons All other Black-holes should have colors which vary with the change of wavelengths of the photons which cannot be detained The order should be infra-red light ultra-violet x-ray and finally gamma photons
Mass is Frequency of an Elementary Particle All known experiments have produced answers that agree with the following equations ( ) ( )( )(distance)mad Action mass acceleration
( )mad
( ) ( ) (constant)(frequency)mad hf
(1) 2( ) ( )(constant)mad mc mass
(2)
c f rf (3)
Equating (1) and (2) 2( )m h c f Mass = (constant)(frequency of one particle)
Plankrsquos constant 6626069 34 HBCP 662603(11) 57(29)69 34 codatah Js h Js
2
222
2
6626069 34( )( ) ( )6626069 347372496 51
(299792458 8 ) 898755179 16
Joule
Kgmm sh Js s kgs
m mcs s
Equating (1)(2) and (3)
2 2
constant
expexp exp
1 6626069 34 22102 42
299792458 8
h h c h h JsM f mr kgm
mr c r cc cs
All mass is built from one elementary particle a ( 737 51 )kg particle
All mass is in relative motion with other mass Time is a description invented to measure the relative movement between different particles of
mass Energy is a term invented to describe what happens when masses interact Energy is a
description Mass does not convert to energy and energy does not convert to mass The word ldquoEnergyrdquo will be replaced with an Action involving mass acceleration and distance ( )mad
When the motion of masses interacts they change to different motions of masses These different mass forms can be measured and their changes in momentum can be measured When the change produces photon particles they have been incorrectly labeled as ldquoEnergyrdquo
Mass Frequency Frequency is a rotation per second
If ( )f is set to one rotation per second (737249638 51 )(1kgs rot s the equation predicts that the
7372496 51kg particle is the elementary mass out of which all mass is made
27372496 51 07372496 50
hKgs kg
c
2 07372496 50 second
m m rotationsf
h kgsc
22 392984064 11
0737249638 50 s
kg rotf
kgs
Kanarev Smallest measurable photon
24241 35328803 15
07372496 50 s
kg rotf
kgs
13598443ev photon 2424143lt35kg
91093897 31123559 20
07372496 50 s
kg rotf
kgs
05109906 6eV electron
9109235lt31kg
16726230 27226873 23
07372496 50 s
kg rotf
kgs
93827241gt6eV proton 1672623lt27kg
16749286 272271858 23
07372496 50 s
kg rotf
kgs
9395656gt6eV neutron 1674928lt27kg
334357629 27226750117 23
(07372496 50 ) )2( s
kg rotf
kgs
(Deuterium nucleus) Proton+Neutron
22102 42mr kgm
334357629 274535202583 23
(07372496 50 ( ) s1)
kg rotf
kgs
Proton+Neutron
22102 42mr kgm
223368 253029747 25
07372496 50 s
kg rotf
kgs
1243gt9eV Higgs Boson 2215849lt25 kg
256482lt25347890 25
07372496 50 s
kg rotf
kgs
144gt11eV Fermi Lab particle 2567033lt25kg
5703889152 1677367 34
07372496 50 s
kg rotf
kgs
320gt20eV highest eV particle measured 5704518lt16kg
1859222909 925218 41
07372496 50 s
kg rotf
kgs
10429476 27eV Fernandes 1859223lt9kg
2176450474 8295212 42
07372496 50 s
kg rotf
kgs
Planck limit
78434 28106387 23
07372496 50 s
kg rotf
kgs
classical radius of an electron
This frequency and radius are not a particle radius
This mass does not exist
308 Is there experimental evidences that an electron has the so-called classic radius of an electron A No experiments do not exist From Questions and Answers of Kanarev Page 95 Electrons-Protons-Neutrons JK ET and Rewrite The handbook of Chemistry and Physics shows the classical radius of the electron as (2817 15 )m Kanarevs theory shows this to be the radius of approach of the magnetic force lines around the torus ring of the electron ( )magr This is confirmed by calculating the dimensionless fine structure constant
( ) (which is equal to the circumference of the inner circle (2 )magr divided by the experimentally
measured radius of the electron exp( )R
exp
2 (2)(314)(2817 15 ) 10007297352568
2426 12 137036magr m
R m
(191)
Kg Rotationsec meters experimental classical rmag mass frequency radius r=cf mr mr(11615lt3) 427E-41 5791781E+09 517617E-02 221022E-42 256717E-45 smallest photon
242414E-35 328809E+15 911753E-08 221022E-42 256717E-45 136eV photon 91094E-31 123559E+20 242631E-12 221022E-42 256717E-45 electron 0511gt6eV 891331E-30 120899E+21 247968E-13 221022E-42 256717E-45 up quark
178266E-29 241799E+21 123984E-13 221022E-42 256717E-45 down quark 188356E-28 255485E+22 117343E-14 221022E-42 256717E-45 muon
240606E-28 326356E+22 918606E-15 221022E-42 256717E-45 meson 248806E-28 337479E+22 88833E-15 221022E-42 256717E-45 meson 356532E-28 483598E+22 619921E-15 221022E-42 256717E-45 strange quark
880118E-28 119379E+23 251128E-15 221022E-42 256717E-45 meson 887249E-28 120346E+23 249109E-15 221022E-42 256717E-45 meson 887177E-28 120336E+23 249129E-15 221022E-42 256717E-45 meson 978325E-28 132699E+23 225919E-15 221022E-42 256717E-45 meson 16726E-27 22687E+23 132143E-15 221022E-42 256717E-45 Proton 938gt6eV
167491E-27 227183E+23 131961E-15 221022E-42 256717E-45 Neutron 939gt6eV 198872E-27 269748E+23 111138E-15 221022E-42 256717E-45 hyperon
212033E-27 2876E+23 104239E-15 221022E-42 256717E-45 hyperon 212579E-27 28834E+23 103972E-15 221022E-42 256717E-45 21347E-27 289549E+23 103538E-15 221022E-42 256717E-45
231746E-27 314339E+23 953725E-16 221022E-42 256717E-45 charm quark 234367E-27 317893E+23 943061E-16 221022E-42 256717E-45
235543E-27 319489E+23 93835E-16 221022E-42 256717E-45 298061E-27 404288E+23 741532E-16 221022E-42 256717E-45 hyperon
802198E-27 10881E+24 27552E-16 221022E-42 256717E-45 bottom quark 143201E-25 194237E+25 154344E-17 221022E-42 256717E-45 W
162556E-25 220489E+25 135967E-17 221022E-42 256717E-45 Z 223368E-25 302974E+25 989499E-18 221022E-42 256717E-45 Higgs125gt9eV
256482E-25 347890E+25 861745E-18 221022E-42 256717E-45 Fermi labs 144gt9eV 311966E-25 423148E+25 708481E-18 221022E-42 256717E-45 top quark
570389E-16 77367E+34 387494E-27 221022E-42 256717E-45 Gamma 320gt20eV 185922E-09 25218E+41 118879E-33 221022E-42 256717E-45 1043gt27eV 186lt9kg
217645E-08 29521E+42 101552E-34 221022E-42 256717E-45 Planck limits
mass frequency radius r=cf mr mr(11615lt3)
mass frequency radius r=cf mr2f 222E-39 298E+11 100E-03 6626069E-34 smallest photon
242414E-35 328809E+15 911753E-08 6626069E-34 Photon 136eV 91094E-31 123559E+20 242631E-12 6626069E-34 electron 0511gt6eV 891331E-30 120899E+21 247968E-13 6626069E-34 up quark
178266E-29 241799E+21 123984E-13 6626069E-34 down quark 188356E-28 255485E+22 117343E-14 6626069E-34 muon
240606E-28 326356E+22 918606E-15 6626069E-34 meson 248806E-28 337479E+22 88833E-15 6626069E-34 meson 356532E-28 483598E+22 619921E-15 6626069E-34 strange quark
880118E-28 119379E+23 251128E-15 6626069E-34 meson 887249E-28 120346E+23 249109E-15 6626069E-34 meson 887177E-28 120336E+23 249129E-15 6626069E-34 meson 978325E-28 132699E+23 225919E-15 6626069E-34 meson 16726E-27 22687E+23 132143E-15 6626069E-34 Proton 938gt6eV
167491E-27 227183E+23 131961E-15 6626069E-34 Neutron 939gt6eV 198872E-27 269748E+23 111138E-15 6626069E-34 hyperon
212033E-27 2876E+23 104239E-15 6626069E-34 hyperon 212579E-27 28834E+23 103972E-15 6626069E-34 21347E-27 289549E+23 103538E-15 6626069E-34
231746E-27 314339E+23 953725E-16 6626069E-34 charm quark 234367E-27 317893E+23 943061E-16 6626069E-34
235543E-27 319489E+23 93835E-16 6626069E-34 298061E-27 404288E+23 741532E-16 6626069E-34 hyperon
802198E-27 10881E+24 27552E-16 6626069E-34 bottom quark 143201E-25 194237E+25 154344E-17 6626069E-34 W
162556E-25 220489E+25 135967E-17 6626069E-34 Z 223368E-25 302974E+25 989499E-18 6626069E-34 Higgs 125eV
256482E-25 347890E+25 861745E-18 6626069E-34 Fermi labs 144gt9eV 311966E-25 423148E+25 708481E-18 6626069E-34 top quark
570389E-16 77367E+34 387493E-27 6626069E-34 Gamma 320gt20eV 185922E-09 25218E+41 118879E-33 6626069E-34 186ltkg 1043gt27eV 217645E-08 29521E+42 101552E-34 6626071E-34 Planck limits
mass frequency radius r=cf mr2f
frequency radius r=cf rf=C 299792458 8c m s 5791781E+09 517617E-02 2997925E+08 smallest photon
328809E+15 911753E-08 2997925E+08 136eV photon 123559E+20 242631E-12 2997925E+08 electron 0511gt6eV 120899E+21 247968E-13 2997925E+08 up quark
241799E+21 123984E-13 2997925E+08 down quark 255485E+22 117343E-14 2997925E+08 muon
326356E+22 918606E-15 2997925E+08 meson 337479E+22 88833E-15 2997925E+08 meson 483598E+22 619921E-15 2997925E+08 strange quark
119379E+23 251128E-15 2997925E+08 meson 120346E+23 249109E-15 2997925E+08 meson 120336E+23 249129E-15 2997925E+08 meson 132699E+23 225919E-15 2997925E+08 meson 22687E+23 132143E-15 2997925E+08 Proton 938gt6eV
227183E+23 131961E-15 2997925E+08 Neutron 939gt6eV 269748E+23 111138E-15 2997925E+08 hyperon
2876E+23 104239E-15 2997925E+08 hyperon 28834E+23 103972E-15 2997925E+08 289549E+23 103538E-15 2997925E+08
314339E+23 953725E-16 2997925E+08 charm quark 317893E+23 943061E-16 2997925E+08
319489E+23 93835E-16 2997925E+08 404288E+23 741532E-16 2997925E+08 hyperon
10881E+24 27552E-16 2997925E+08 bottom quark 194237E+25 154344E-17 2997925E+08 W
220489E+25 135967E-17 2997925E+08 Z 302974E+25 989499E-18 2997925E+08 Higgs 125eV
347890E+25 861745E-18 2997925E+08 Fermi labs 144gt9eV 423148E+25 708481E-18 2997925E+08 top quark
77367E+34 387493E-27 2997925E+08 Gamma 320gt20eV 25218E+41 118879E-33 2997925E+08 1043gt27eV 186lt9kg
29521E+42 101552E-34 2997925E+08 Planck limits
frequency radius r=cf rf=C
Savior Borg
Two helix makes a mass particle
Fernandes
Kanarevrsquos Photon Model
My website Symmetrymathcomis my interpretation of the Authorrsquos I have referenced
I do not claim that any of these authors will agree withmy interpretations
I leave it to anyone who visits my site to decide the logicof my interpretation
- SYMMETRY MATH (SM) By Jack Kuykendall
- Why is Symmetry Math Needed
- From ldquoWhat Is Mathematicsrdquo by Courant and Robbins ndash page 55 Th
- Slide 4
- Slide 5
- Slide 6
- Slide 7
- Slide 8
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- Slide 122
- Slide 123
-
H2O H2O Cluster
There are 82gt22 (82000000000000000000000) free electrons in each cubic centimeter of a copper wire
Since no protons are free there cannot be an Attractive charge action proton potential in a wire
There are no protons present at one plate and electrons present at the other plate
Temperature Radius of Photons Equation 0 0(0 ) (27315 )AC T
0
2898 310609555 6
27315o
AoC
mTr m
316
0 0(1 ) (27415 )AС T 1
2898 310570855 6
27415A
o oC
mTr m
317
0 020 (68 ) 29315o
AC F T 20
2898 39885 6
29315o
AoC
mTr m
321
0 030 (86 ) 30315oAC F T
30
2898 39560 6
30315o
AoC
mTr m
322
0 0 0[100 (212 ) 37315 ]AC F T 100
2898 37766 6
37315o C
mKr m
K
323
0 0 0[(1000 )(1832 ) 127315 ]AC F T 1000
2898 32276 6
127315o
AoC
mTr m
324
0 0 0[(1500 )(2732 ) 177315 ]AC F T 1500
2898 31634 6
177315A
o oC
mTr m
325
0 0 0[(2000 )(3632 ) 227315 ]AC F T 2000
2898 31275 6
227315A
o oC
mTr m
326
1
0 0 0(0 2000 )(27315 227315 )oAto C to T
Temperatures between are formed with photons from the infra-red range As the temperature increase the radiuses of the photons decrease
Temperature (felt as HEAT by humans) is formed by photon emission from electrons
What is the exact radius of the photons that form the (highest) temperature felt as heat Modern science has no exact answer to this question We can only theorize that lsquoHEATrsquo is formed by photons which are emitted by electrons during the synthesis of atoms and molecules The border for the smallest radius of these photons is not established yet We theorize that it is in an interval of the ultra-violet range Since the shortest know radiuses (X and gamma photons) are emitted by nuclear synthesis they do not participate in heat formation Temperature (felt as HEAT by humans) is formed by photon emission from electrons If gamma photons participated in formation of heat the greatest possible temperature would be equal
maxmin
2898 3 2898 31 15 1000000000000000
3 18oA A
A AmT mT
T T Tr m
(348)
If such a temperature existed it would destroy molecules atoms and the nucleus of all atoms
Schwarzschildrsquos (incorrect) radius equation
2
2 g
GMR
C
Kanarevrsquos correct equation
1(1087148751 14)
2p
g photon
GMrR Mr
C
Schwarzschild 3 33
3 (3)(2 30 )
4 (4)(314)(297 3182 19
)og
M kgD
R m
kg
m
(403) 1
Kanarev
3 33
3 (3)(2 30 )
4 (4)(314)(439 )565 27red
g
M kgD
R m
kg
m
(404) 2
Kanarev
3 33 3
3 (3)(2 30 )276 35 (
4 (4)(314)(553 32 Kanrev error
0012)
)lightg
M kg kgD
R m m
kg
m
(405) 3
4
Kanarevrsquos 3 33
3 (3)(2 30)
4 (4)(314)(154 8)013 54gamma
g
MD
R
kg
m
(406) 5
6
Fernandesrsquo (18 36 9) 3 3
3 (3)(2 30)265 81
4 (4)(314)(18 17)rg
kg
m
MD
R
(407)
7
Nuclear density is around3
(12 24) 17kg
m 8
If the Sun were compressed to ( )( 439 )g redR m (equation 400) its gravitational attraction would
detain only radiation of the far infra-red spectrum Shorter wavelengths will not be detained If photons of gamma radiation were detained the gravitational radius of the Sun would have to
be ( )( 154 8 )g gammaR m (equation 402) This would not be possible because the density of
the Sun (equation 406) would be 37 orders more dense than the nucleus of atoms
Only one type of Black-hole can be all black That is one with a gravitational attraction strong enough to detain the highest energy gamma photons All other Black-holes should have colors which vary with the change of wavelengths of the photons which cannot be detained The order should be infra-red light ultra-violet x-ray and finally gamma photons
Mass is Frequency of an Elementary Particle All known experiments have produced answers that agree with the following equations ( ) ( )( )(distance)mad Action mass acceleration
( )mad
( ) ( ) (constant)(frequency)mad hf
(1) 2( ) ( )(constant)mad mc mass
(2)
c f rf (3)
Equating (1) and (2) 2( )m h c f Mass = (constant)(frequency of one particle)
Plankrsquos constant 6626069 34 HBCP 662603(11) 57(29)69 34 codatah Js h Js
2
222
2
6626069 34( )( ) ( )6626069 347372496 51
(299792458 8 ) 898755179 16
Joule
Kgmm sh Js s kgs
m mcs s
Equating (1)(2) and (3)
2 2
constant
expexp exp
1 6626069 34 22102 42
299792458 8
h h c h h JsM f mr kgm
mr c r cc cs
All mass is built from one elementary particle a ( 737 51 )kg particle
All mass is in relative motion with other mass Time is a description invented to measure the relative movement between different particles of
mass Energy is a term invented to describe what happens when masses interact Energy is a
description Mass does not convert to energy and energy does not convert to mass The word ldquoEnergyrdquo will be replaced with an Action involving mass acceleration and distance ( )mad
When the motion of masses interacts they change to different motions of masses These different mass forms can be measured and their changes in momentum can be measured When the change produces photon particles they have been incorrectly labeled as ldquoEnergyrdquo
Mass Frequency Frequency is a rotation per second
If ( )f is set to one rotation per second (737249638 51 )(1kgs rot s the equation predicts that the
7372496 51kg particle is the elementary mass out of which all mass is made
27372496 51 07372496 50
hKgs kg
c
2 07372496 50 second
m m rotationsf
h kgsc
22 392984064 11
0737249638 50 s
kg rotf
kgs
Kanarev Smallest measurable photon
24241 35328803 15
07372496 50 s
kg rotf
kgs
13598443ev photon 2424143lt35kg
91093897 31123559 20
07372496 50 s
kg rotf
kgs
05109906 6eV electron
9109235lt31kg
16726230 27226873 23
07372496 50 s
kg rotf
kgs
93827241gt6eV proton 1672623lt27kg
16749286 272271858 23
07372496 50 s
kg rotf
kgs
9395656gt6eV neutron 1674928lt27kg
334357629 27226750117 23
(07372496 50 ) )2( s
kg rotf
kgs
(Deuterium nucleus) Proton+Neutron
22102 42mr kgm
334357629 274535202583 23
(07372496 50 ( ) s1)
kg rotf
kgs
Proton+Neutron
22102 42mr kgm
223368 253029747 25
07372496 50 s
kg rotf
kgs
1243gt9eV Higgs Boson 2215849lt25 kg
256482lt25347890 25
07372496 50 s
kg rotf
kgs
144gt11eV Fermi Lab particle 2567033lt25kg
5703889152 1677367 34
07372496 50 s
kg rotf
kgs
320gt20eV highest eV particle measured 5704518lt16kg
1859222909 925218 41
07372496 50 s
kg rotf
kgs
10429476 27eV Fernandes 1859223lt9kg
2176450474 8295212 42
07372496 50 s
kg rotf
kgs
Planck limit
78434 28106387 23
07372496 50 s
kg rotf
kgs
classical radius of an electron
This frequency and radius are not a particle radius
This mass does not exist
308 Is there experimental evidences that an electron has the so-called classic radius of an electron A No experiments do not exist From Questions and Answers of Kanarev Page 95 Electrons-Protons-Neutrons JK ET and Rewrite The handbook of Chemistry and Physics shows the classical radius of the electron as (2817 15 )m Kanarevs theory shows this to be the radius of approach of the magnetic force lines around the torus ring of the electron ( )magr This is confirmed by calculating the dimensionless fine structure constant
( ) (which is equal to the circumference of the inner circle (2 )magr divided by the experimentally
measured radius of the electron exp( )R
exp
2 (2)(314)(2817 15 ) 10007297352568
2426 12 137036magr m
R m
(191)
Kg Rotationsec meters experimental classical rmag mass frequency radius r=cf mr mr(11615lt3) 427E-41 5791781E+09 517617E-02 221022E-42 256717E-45 smallest photon
242414E-35 328809E+15 911753E-08 221022E-42 256717E-45 136eV photon 91094E-31 123559E+20 242631E-12 221022E-42 256717E-45 electron 0511gt6eV 891331E-30 120899E+21 247968E-13 221022E-42 256717E-45 up quark
178266E-29 241799E+21 123984E-13 221022E-42 256717E-45 down quark 188356E-28 255485E+22 117343E-14 221022E-42 256717E-45 muon
240606E-28 326356E+22 918606E-15 221022E-42 256717E-45 meson 248806E-28 337479E+22 88833E-15 221022E-42 256717E-45 meson 356532E-28 483598E+22 619921E-15 221022E-42 256717E-45 strange quark
880118E-28 119379E+23 251128E-15 221022E-42 256717E-45 meson 887249E-28 120346E+23 249109E-15 221022E-42 256717E-45 meson 887177E-28 120336E+23 249129E-15 221022E-42 256717E-45 meson 978325E-28 132699E+23 225919E-15 221022E-42 256717E-45 meson 16726E-27 22687E+23 132143E-15 221022E-42 256717E-45 Proton 938gt6eV
167491E-27 227183E+23 131961E-15 221022E-42 256717E-45 Neutron 939gt6eV 198872E-27 269748E+23 111138E-15 221022E-42 256717E-45 hyperon
212033E-27 2876E+23 104239E-15 221022E-42 256717E-45 hyperon 212579E-27 28834E+23 103972E-15 221022E-42 256717E-45 21347E-27 289549E+23 103538E-15 221022E-42 256717E-45
231746E-27 314339E+23 953725E-16 221022E-42 256717E-45 charm quark 234367E-27 317893E+23 943061E-16 221022E-42 256717E-45
235543E-27 319489E+23 93835E-16 221022E-42 256717E-45 298061E-27 404288E+23 741532E-16 221022E-42 256717E-45 hyperon
802198E-27 10881E+24 27552E-16 221022E-42 256717E-45 bottom quark 143201E-25 194237E+25 154344E-17 221022E-42 256717E-45 W
162556E-25 220489E+25 135967E-17 221022E-42 256717E-45 Z 223368E-25 302974E+25 989499E-18 221022E-42 256717E-45 Higgs125gt9eV
256482E-25 347890E+25 861745E-18 221022E-42 256717E-45 Fermi labs 144gt9eV 311966E-25 423148E+25 708481E-18 221022E-42 256717E-45 top quark
570389E-16 77367E+34 387494E-27 221022E-42 256717E-45 Gamma 320gt20eV 185922E-09 25218E+41 118879E-33 221022E-42 256717E-45 1043gt27eV 186lt9kg
217645E-08 29521E+42 101552E-34 221022E-42 256717E-45 Planck limits
mass frequency radius r=cf mr mr(11615lt3)
mass frequency radius r=cf mr2f 222E-39 298E+11 100E-03 6626069E-34 smallest photon
242414E-35 328809E+15 911753E-08 6626069E-34 Photon 136eV 91094E-31 123559E+20 242631E-12 6626069E-34 electron 0511gt6eV 891331E-30 120899E+21 247968E-13 6626069E-34 up quark
178266E-29 241799E+21 123984E-13 6626069E-34 down quark 188356E-28 255485E+22 117343E-14 6626069E-34 muon
240606E-28 326356E+22 918606E-15 6626069E-34 meson 248806E-28 337479E+22 88833E-15 6626069E-34 meson 356532E-28 483598E+22 619921E-15 6626069E-34 strange quark
880118E-28 119379E+23 251128E-15 6626069E-34 meson 887249E-28 120346E+23 249109E-15 6626069E-34 meson 887177E-28 120336E+23 249129E-15 6626069E-34 meson 978325E-28 132699E+23 225919E-15 6626069E-34 meson 16726E-27 22687E+23 132143E-15 6626069E-34 Proton 938gt6eV
167491E-27 227183E+23 131961E-15 6626069E-34 Neutron 939gt6eV 198872E-27 269748E+23 111138E-15 6626069E-34 hyperon
212033E-27 2876E+23 104239E-15 6626069E-34 hyperon 212579E-27 28834E+23 103972E-15 6626069E-34 21347E-27 289549E+23 103538E-15 6626069E-34
231746E-27 314339E+23 953725E-16 6626069E-34 charm quark 234367E-27 317893E+23 943061E-16 6626069E-34
235543E-27 319489E+23 93835E-16 6626069E-34 298061E-27 404288E+23 741532E-16 6626069E-34 hyperon
802198E-27 10881E+24 27552E-16 6626069E-34 bottom quark 143201E-25 194237E+25 154344E-17 6626069E-34 W
162556E-25 220489E+25 135967E-17 6626069E-34 Z 223368E-25 302974E+25 989499E-18 6626069E-34 Higgs 125eV
256482E-25 347890E+25 861745E-18 6626069E-34 Fermi labs 144gt9eV 311966E-25 423148E+25 708481E-18 6626069E-34 top quark
570389E-16 77367E+34 387493E-27 6626069E-34 Gamma 320gt20eV 185922E-09 25218E+41 118879E-33 6626069E-34 186ltkg 1043gt27eV 217645E-08 29521E+42 101552E-34 6626071E-34 Planck limits
mass frequency radius r=cf mr2f
frequency radius r=cf rf=C 299792458 8c m s 5791781E+09 517617E-02 2997925E+08 smallest photon
328809E+15 911753E-08 2997925E+08 136eV photon 123559E+20 242631E-12 2997925E+08 electron 0511gt6eV 120899E+21 247968E-13 2997925E+08 up quark
241799E+21 123984E-13 2997925E+08 down quark 255485E+22 117343E-14 2997925E+08 muon
326356E+22 918606E-15 2997925E+08 meson 337479E+22 88833E-15 2997925E+08 meson 483598E+22 619921E-15 2997925E+08 strange quark
119379E+23 251128E-15 2997925E+08 meson 120346E+23 249109E-15 2997925E+08 meson 120336E+23 249129E-15 2997925E+08 meson 132699E+23 225919E-15 2997925E+08 meson 22687E+23 132143E-15 2997925E+08 Proton 938gt6eV
227183E+23 131961E-15 2997925E+08 Neutron 939gt6eV 269748E+23 111138E-15 2997925E+08 hyperon
2876E+23 104239E-15 2997925E+08 hyperon 28834E+23 103972E-15 2997925E+08 289549E+23 103538E-15 2997925E+08
314339E+23 953725E-16 2997925E+08 charm quark 317893E+23 943061E-16 2997925E+08
319489E+23 93835E-16 2997925E+08 404288E+23 741532E-16 2997925E+08 hyperon
10881E+24 27552E-16 2997925E+08 bottom quark 194237E+25 154344E-17 2997925E+08 W
220489E+25 135967E-17 2997925E+08 Z 302974E+25 989499E-18 2997925E+08 Higgs 125eV
347890E+25 861745E-18 2997925E+08 Fermi labs 144gt9eV 423148E+25 708481E-18 2997925E+08 top quark
77367E+34 387493E-27 2997925E+08 Gamma 320gt20eV 25218E+41 118879E-33 2997925E+08 1043gt27eV 186lt9kg
29521E+42 101552E-34 2997925E+08 Planck limits
frequency radius r=cf rf=C
Savior Borg
Two helix makes a mass particle
Fernandes
Kanarevrsquos Photon Model
My website Symmetrymathcomis my interpretation of the Authorrsquos I have referenced
I do not claim that any of these authors will agree withmy interpretations
I leave it to anyone who visits my site to decide the logicof my interpretation
- SYMMETRY MATH (SM) By Jack Kuykendall
- Why is Symmetry Math Needed
- From ldquoWhat Is Mathematicsrdquo by Courant and Robbins ndash page 55 Th
- Slide 4
- Slide 5
- Slide 6
- Slide 7
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-
There are 82gt22 (82000000000000000000000) free electrons in each cubic centimeter of a copper wire
Since no protons are free there cannot be an Attractive charge action proton potential in a wire
There are no protons present at one plate and electrons present at the other plate
Temperature Radius of Photons Equation 0 0(0 ) (27315 )AC T
0
2898 310609555 6
27315o
AoC
mTr m
316
0 0(1 ) (27415 )AС T 1
2898 310570855 6
27415A
o oC
mTr m
317
0 020 (68 ) 29315o
AC F T 20
2898 39885 6
29315o
AoC
mTr m
321
0 030 (86 ) 30315oAC F T
30
2898 39560 6
30315o
AoC
mTr m
322
0 0 0[100 (212 ) 37315 ]AC F T 100
2898 37766 6
37315o C
mKr m
K
323
0 0 0[(1000 )(1832 ) 127315 ]AC F T 1000
2898 32276 6
127315o
AoC
mTr m
324
0 0 0[(1500 )(2732 ) 177315 ]AC F T 1500
2898 31634 6
177315A
o oC
mTr m
325
0 0 0[(2000 )(3632 ) 227315 ]AC F T 2000
2898 31275 6
227315A
o oC
mTr m
326
1
0 0 0(0 2000 )(27315 227315 )oAto C to T
Temperatures between are formed with photons from the infra-red range As the temperature increase the radiuses of the photons decrease
Temperature (felt as HEAT by humans) is formed by photon emission from electrons
What is the exact radius of the photons that form the (highest) temperature felt as heat Modern science has no exact answer to this question We can only theorize that lsquoHEATrsquo is formed by photons which are emitted by electrons during the synthesis of atoms and molecules The border for the smallest radius of these photons is not established yet We theorize that it is in an interval of the ultra-violet range Since the shortest know radiuses (X and gamma photons) are emitted by nuclear synthesis they do not participate in heat formation Temperature (felt as HEAT by humans) is formed by photon emission from electrons If gamma photons participated in formation of heat the greatest possible temperature would be equal
maxmin
2898 3 2898 31 15 1000000000000000
3 18oA A
A AmT mT
T T Tr m
(348)
If such a temperature existed it would destroy molecules atoms and the nucleus of all atoms
Schwarzschildrsquos (incorrect) radius equation
2
2 g
GMR
C
Kanarevrsquos correct equation
1(1087148751 14)
2p
g photon
GMrR Mr
C
Schwarzschild 3 33
3 (3)(2 30 )
4 (4)(314)(297 3182 19
)og
M kgD
R m
kg
m
(403) 1
Kanarev
3 33
3 (3)(2 30 )
4 (4)(314)(439 )565 27red
g
M kgD
R m
kg
m
(404) 2
Kanarev
3 33 3
3 (3)(2 30 )276 35 (
4 (4)(314)(553 32 Kanrev error
0012)
)lightg
M kg kgD
R m m
kg
m
(405) 3
4
Kanarevrsquos 3 33
3 (3)(2 30)
4 (4)(314)(154 8)013 54gamma
g
MD
R
kg
m
(406) 5
6
Fernandesrsquo (18 36 9) 3 3
3 (3)(2 30)265 81
4 (4)(314)(18 17)rg
kg
m
MD
R
(407)
7
Nuclear density is around3
(12 24) 17kg
m 8
If the Sun were compressed to ( )( 439 )g redR m (equation 400) its gravitational attraction would
detain only radiation of the far infra-red spectrum Shorter wavelengths will not be detained If photons of gamma radiation were detained the gravitational radius of the Sun would have to
be ( )( 154 8 )g gammaR m (equation 402) This would not be possible because the density of
the Sun (equation 406) would be 37 orders more dense than the nucleus of atoms
Only one type of Black-hole can be all black That is one with a gravitational attraction strong enough to detain the highest energy gamma photons All other Black-holes should have colors which vary with the change of wavelengths of the photons which cannot be detained The order should be infra-red light ultra-violet x-ray and finally gamma photons
Mass is Frequency of an Elementary Particle All known experiments have produced answers that agree with the following equations ( ) ( )( )(distance)mad Action mass acceleration
( )mad
( ) ( ) (constant)(frequency)mad hf
(1) 2( ) ( )(constant)mad mc mass
(2)
c f rf (3)
Equating (1) and (2) 2( )m h c f Mass = (constant)(frequency of one particle)
Plankrsquos constant 6626069 34 HBCP 662603(11) 57(29)69 34 codatah Js h Js
2
222
2
6626069 34( )( ) ( )6626069 347372496 51
(299792458 8 ) 898755179 16
Joule
Kgmm sh Js s kgs
m mcs s
Equating (1)(2) and (3)
2 2
constant
expexp exp
1 6626069 34 22102 42
299792458 8
h h c h h JsM f mr kgm
mr c r cc cs
All mass is built from one elementary particle a ( 737 51 )kg particle
All mass is in relative motion with other mass Time is a description invented to measure the relative movement between different particles of
mass Energy is a term invented to describe what happens when masses interact Energy is a
description Mass does not convert to energy and energy does not convert to mass The word ldquoEnergyrdquo will be replaced with an Action involving mass acceleration and distance ( )mad
When the motion of masses interacts they change to different motions of masses These different mass forms can be measured and their changes in momentum can be measured When the change produces photon particles they have been incorrectly labeled as ldquoEnergyrdquo
Mass Frequency Frequency is a rotation per second
If ( )f is set to one rotation per second (737249638 51 )(1kgs rot s the equation predicts that the
7372496 51kg particle is the elementary mass out of which all mass is made
27372496 51 07372496 50
hKgs kg
c
2 07372496 50 second
m m rotationsf
h kgsc
22 392984064 11
0737249638 50 s
kg rotf
kgs
Kanarev Smallest measurable photon
24241 35328803 15
07372496 50 s
kg rotf
kgs
13598443ev photon 2424143lt35kg
91093897 31123559 20
07372496 50 s
kg rotf
kgs
05109906 6eV electron
9109235lt31kg
16726230 27226873 23
07372496 50 s
kg rotf
kgs
93827241gt6eV proton 1672623lt27kg
16749286 272271858 23
07372496 50 s
kg rotf
kgs
9395656gt6eV neutron 1674928lt27kg
334357629 27226750117 23
(07372496 50 ) )2( s
kg rotf
kgs
(Deuterium nucleus) Proton+Neutron
22102 42mr kgm
334357629 274535202583 23
(07372496 50 ( ) s1)
kg rotf
kgs
Proton+Neutron
22102 42mr kgm
223368 253029747 25
07372496 50 s
kg rotf
kgs
1243gt9eV Higgs Boson 2215849lt25 kg
256482lt25347890 25
07372496 50 s
kg rotf
kgs
144gt11eV Fermi Lab particle 2567033lt25kg
5703889152 1677367 34
07372496 50 s
kg rotf
kgs
320gt20eV highest eV particle measured 5704518lt16kg
1859222909 925218 41
07372496 50 s
kg rotf
kgs
10429476 27eV Fernandes 1859223lt9kg
2176450474 8295212 42
07372496 50 s
kg rotf
kgs
Planck limit
78434 28106387 23
07372496 50 s
kg rotf
kgs
classical radius of an electron
This frequency and radius are not a particle radius
This mass does not exist
308 Is there experimental evidences that an electron has the so-called classic radius of an electron A No experiments do not exist From Questions and Answers of Kanarev Page 95 Electrons-Protons-Neutrons JK ET and Rewrite The handbook of Chemistry and Physics shows the classical radius of the electron as (2817 15 )m Kanarevs theory shows this to be the radius of approach of the magnetic force lines around the torus ring of the electron ( )magr This is confirmed by calculating the dimensionless fine structure constant
( ) (which is equal to the circumference of the inner circle (2 )magr divided by the experimentally
measured radius of the electron exp( )R
exp
2 (2)(314)(2817 15 ) 10007297352568
2426 12 137036magr m
R m
(191)
Kg Rotationsec meters experimental classical rmag mass frequency radius r=cf mr mr(11615lt3) 427E-41 5791781E+09 517617E-02 221022E-42 256717E-45 smallest photon
242414E-35 328809E+15 911753E-08 221022E-42 256717E-45 136eV photon 91094E-31 123559E+20 242631E-12 221022E-42 256717E-45 electron 0511gt6eV 891331E-30 120899E+21 247968E-13 221022E-42 256717E-45 up quark
178266E-29 241799E+21 123984E-13 221022E-42 256717E-45 down quark 188356E-28 255485E+22 117343E-14 221022E-42 256717E-45 muon
240606E-28 326356E+22 918606E-15 221022E-42 256717E-45 meson 248806E-28 337479E+22 88833E-15 221022E-42 256717E-45 meson 356532E-28 483598E+22 619921E-15 221022E-42 256717E-45 strange quark
880118E-28 119379E+23 251128E-15 221022E-42 256717E-45 meson 887249E-28 120346E+23 249109E-15 221022E-42 256717E-45 meson 887177E-28 120336E+23 249129E-15 221022E-42 256717E-45 meson 978325E-28 132699E+23 225919E-15 221022E-42 256717E-45 meson 16726E-27 22687E+23 132143E-15 221022E-42 256717E-45 Proton 938gt6eV
167491E-27 227183E+23 131961E-15 221022E-42 256717E-45 Neutron 939gt6eV 198872E-27 269748E+23 111138E-15 221022E-42 256717E-45 hyperon
212033E-27 2876E+23 104239E-15 221022E-42 256717E-45 hyperon 212579E-27 28834E+23 103972E-15 221022E-42 256717E-45 21347E-27 289549E+23 103538E-15 221022E-42 256717E-45
231746E-27 314339E+23 953725E-16 221022E-42 256717E-45 charm quark 234367E-27 317893E+23 943061E-16 221022E-42 256717E-45
235543E-27 319489E+23 93835E-16 221022E-42 256717E-45 298061E-27 404288E+23 741532E-16 221022E-42 256717E-45 hyperon
802198E-27 10881E+24 27552E-16 221022E-42 256717E-45 bottom quark 143201E-25 194237E+25 154344E-17 221022E-42 256717E-45 W
162556E-25 220489E+25 135967E-17 221022E-42 256717E-45 Z 223368E-25 302974E+25 989499E-18 221022E-42 256717E-45 Higgs125gt9eV
256482E-25 347890E+25 861745E-18 221022E-42 256717E-45 Fermi labs 144gt9eV 311966E-25 423148E+25 708481E-18 221022E-42 256717E-45 top quark
570389E-16 77367E+34 387494E-27 221022E-42 256717E-45 Gamma 320gt20eV 185922E-09 25218E+41 118879E-33 221022E-42 256717E-45 1043gt27eV 186lt9kg
217645E-08 29521E+42 101552E-34 221022E-42 256717E-45 Planck limits
mass frequency radius r=cf mr mr(11615lt3)
mass frequency radius r=cf mr2f 222E-39 298E+11 100E-03 6626069E-34 smallest photon
242414E-35 328809E+15 911753E-08 6626069E-34 Photon 136eV 91094E-31 123559E+20 242631E-12 6626069E-34 electron 0511gt6eV 891331E-30 120899E+21 247968E-13 6626069E-34 up quark
178266E-29 241799E+21 123984E-13 6626069E-34 down quark 188356E-28 255485E+22 117343E-14 6626069E-34 muon
240606E-28 326356E+22 918606E-15 6626069E-34 meson 248806E-28 337479E+22 88833E-15 6626069E-34 meson 356532E-28 483598E+22 619921E-15 6626069E-34 strange quark
880118E-28 119379E+23 251128E-15 6626069E-34 meson 887249E-28 120346E+23 249109E-15 6626069E-34 meson 887177E-28 120336E+23 249129E-15 6626069E-34 meson 978325E-28 132699E+23 225919E-15 6626069E-34 meson 16726E-27 22687E+23 132143E-15 6626069E-34 Proton 938gt6eV
167491E-27 227183E+23 131961E-15 6626069E-34 Neutron 939gt6eV 198872E-27 269748E+23 111138E-15 6626069E-34 hyperon
212033E-27 2876E+23 104239E-15 6626069E-34 hyperon 212579E-27 28834E+23 103972E-15 6626069E-34 21347E-27 289549E+23 103538E-15 6626069E-34
231746E-27 314339E+23 953725E-16 6626069E-34 charm quark 234367E-27 317893E+23 943061E-16 6626069E-34
235543E-27 319489E+23 93835E-16 6626069E-34 298061E-27 404288E+23 741532E-16 6626069E-34 hyperon
802198E-27 10881E+24 27552E-16 6626069E-34 bottom quark 143201E-25 194237E+25 154344E-17 6626069E-34 W
162556E-25 220489E+25 135967E-17 6626069E-34 Z 223368E-25 302974E+25 989499E-18 6626069E-34 Higgs 125eV
256482E-25 347890E+25 861745E-18 6626069E-34 Fermi labs 144gt9eV 311966E-25 423148E+25 708481E-18 6626069E-34 top quark
570389E-16 77367E+34 387493E-27 6626069E-34 Gamma 320gt20eV 185922E-09 25218E+41 118879E-33 6626069E-34 186ltkg 1043gt27eV 217645E-08 29521E+42 101552E-34 6626071E-34 Planck limits
mass frequency radius r=cf mr2f
frequency radius r=cf rf=C 299792458 8c m s 5791781E+09 517617E-02 2997925E+08 smallest photon
328809E+15 911753E-08 2997925E+08 136eV photon 123559E+20 242631E-12 2997925E+08 electron 0511gt6eV 120899E+21 247968E-13 2997925E+08 up quark
241799E+21 123984E-13 2997925E+08 down quark 255485E+22 117343E-14 2997925E+08 muon
326356E+22 918606E-15 2997925E+08 meson 337479E+22 88833E-15 2997925E+08 meson 483598E+22 619921E-15 2997925E+08 strange quark
119379E+23 251128E-15 2997925E+08 meson 120346E+23 249109E-15 2997925E+08 meson 120336E+23 249129E-15 2997925E+08 meson 132699E+23 225919E-15 2997925E+08 meson 22687E+23 132143E-15 2997925E+08 Proton 938gt6eV
227183E+23 131961E-15 2997925E+08 Neutron 939gt6eV 269748E+23 111138E-15 2997925E+08 hyperon
2876E+23 104239E-15 2997925E+08 hyperon 28834E+23 103972E-15 2997925E+08 289549E+23 103538E-15 2997925E+08
314339E+23 953725E-16 2997925E+08 charm quark 317893E+23 943061E-16 2997925E+08
319489E+23 93835E-16 2997925E+08 404288E+23 741532E-16 2997925E+08 hyperon
10881E+24 27552E-16 2997925E+08 bottom quark 194237E+25 154344E-17 2997925E+08 W
220489E+25 135967E-17 2997925E+08 Z 302974E+25 989499E-18 2997925E+08 Higgs 125eV
347890E+25 861745E-18 2997925E+08 Fermi labs 144gt9eV 423148E+25 708481E-18 2997925E+08 top quark
77367E+34 387493E-27 2997925E+08 Gamma 320gt20eV 25218E+41 118879E-33 2997925E+08 1043gt27eV 186lt9kg
29521E+42 101552E-34 2997925E+08 Planck limits
frequency radius r=cf rf=C
Savior Borg
Two helix makes a mass particle
Fernandes
Kanarevrsquos Photon Model
My website Symmetrymathcomis my interpretation of the Authorrsquos I have referenced
I do not claim that any of these authors will agree withmy interpretations
I leave it to anyone who visits my site to decide the logicof my interpretation
- SYMMETRY MATH (SM) By Jack Kuykendall
- Why is Symmetry Math Needed
- From ldquoWhat Is Mathematicsrdquo by Courant and Robbins ndash page 55 Th
- Slide 4
- Slide 5
- Slide 6
- Slide 7
- Slide 8
- Slide 9
- Slide 10
- Slide 11
- Slide 12
- Slide 13
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- Slide 41
- Slide 42
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- Slide 46
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- Slide 115
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- Slide 117
- Slide 118
- Slide 119
- Slide 120
- Slide 121
- Slide 122
- Slide 123
-
There are no protons present at one plate and electrons present at the other plate
Temperature Radius of Photons Equation 0 0(0 ) (27315 )AC T
0
2898 310609555 6
27315o
AoC
mTr m
316
0 0(1 ) (27415 )AС T 1
2898 310570855 6
27415A
o oC
mTr m
317
0 020 (68 ) 29315o
AC F T 20
2898 39885 6
29315o
AoC
mTr m
321
0 030 (86 ) 30315oAC F T
30
2898 39560 6
30315o
AoC
mTr m
322
0 0 0[100 (212 ) 37315 ]AC F T 100
2898 37766 6
37315o C
mKr m
K
323
0 0 0[(1000 )(1832 ) 127315 ]AC F T 1000
2898 32276 6
127315o
AoC
mTr m
324
0 0 0[(1500 )(2732 ) 177315 ]AC F T 1500
2898 31634 6
177315A
o oC
mTr m
325
0 0 0[(2000 )(3632 ) 227315 ]AC F T 2000
2898 31275 6
227315A
o oC
mTr m
326
1
0 0 0(0 2000 )(27315 227315 )oAto C to T
Temperatures between are formed with photons from the infra-red range As the temperature increase the radiuses of the photons decrease
Temperature (felt as HEAT by humans) is formed by photon emission from electrons
What is the exact radius of the photons that form the (highest) temperature felt as heat Modern science has no exact answer to this question We can only theorize that lsquoHEATrsquo is formed by photons which are emitted by electrons during the synthesis of atoms and molecules The border for the smallest radius of these photons is not established yet We theorize that it is in an interval of the ultra-violet range Since the shortest know radiuses (X and gamma photons) are emitted by nuclear synthesis they do not participate in heat formation Temperature (felt as HEAT by humans) is formed by photon emission from electrons If gamma photons participated in formation of heat the greatest possible temperature would be equal
maxmin
2898 3 2898 31 15 1000000000000000
3 18oA A
A AmT mT
T T Tr m
(348)
If such a temperature existed it would destroy molecules atoms and the nucleus of all atoms
Schwarzschildrsquos (incorrect) radius equation
2
2 g
GMR
C
Kanarevrsquos correct equation
1(1087148751 14)
2p
g photon
GMrR Mr
C
Schwarzschild 3 33
3 (3)(2 30 )
4 (4)(314)(297 3182 19
)og
M kgD
R m
kg
m
(403) 1
Kanarev
3 33
3 (3)(2 30 )
4 (4)(314)(439 )565 27red
g
M kgD
R m
kg
m
(404) 2
Kanarev
3 33 3
3 (3)(2 30 )276 35 (
4 (4)(314)(553 32 Kanrev error
0012)
)lightg
M kg kgD
R m m
kg
m
(405) 3
4
Kanarevrsquos 3 33
3 (3)(2 30)
4 (4)(314)(154 8)013 54gamma
g
MD
R
kg
m
(406) 5
6
Fernandesrsquo (18 36 9) 3 3
3 (3)(2 30)265 81
4 (4)(314)(18 17)rg
kg
m
MD
R
(407)
7
Nuclear density is around3
(12 24) 17kg
m 8
If the Sun were compressed to ( )( 439 )g redR m (equation 400) its gravitational attraction would
detain only radiation of the far infra-red spectrum Shorter wavelengths will not be detained If photons of gamma radiation were detained the gravitational radius of the Sun would have to
be ( )( 154 8 )g gammaR m (equation 402) This would not be possible because the density of
the Sun (equation 406) would be 37 orders more dense than the nucleus of atoms
Only one type of Black-hole can be all black That is one with a gravitational attraction strong enough to detain the highest energy gamma photons All other Black-holes should have colors which vary with the change of wavelengths of the photons which cannot be detained The order should be infra-red light ultra-violet x-ray and finally gamma photons
Mass is Frequency of an Elementary Particle All known experiments have produced answers that agree with the following equations ( ) ( )( )(distance)mad Action mass acceleration
( )mad
( ) ( ) (constant)(frequency)mad hf
(1) 2( ) ( )(constant)mad mc mass
(2)
c f rf (3)
Equating (1) and (2) 2( )m h c f Mass = (constant)(frequency of one particle)
Plankrsquos constant 6626069 34 HBCP 662603(11) 57(29)69 34 codatah Js h Js
2
222
2
6626069 34( )( ) ( )6626069 347372496 51
(299792458 8 ) 898755179 16
Joule
Kgmm sh Js s kgs
m mcs s
Equating (1)(2) and (3)
2 2
constant
expexp exp
1 6626069 34 22102 42
299792458 8
h h c h h JsM f mr kgm
mr c r cc cs
All mass is built from one elementary particle a ( 737 51 )kg particle
All mass is in relative motion with other mass Time is a description invented to measure the relative movement between different particles of
mass Energy is a term invented to describe what happens when masses interact Energy is a
description Mass does not convert to energy and energy does not convert to mass The word ldquoEnergyrdquo will be replaced with an Action involving mass acceleration and distance ( )mad
When the motion of masses interacts they change to different motions of masses These different mass forms can be measured and their changes in momentum can be measured When the change produces photon particles they have been incorrectly labeled as ldquoEnergyrdquo
Mass Frequency Frequency is a rotation per second
If ( )f is set to one rotation per second (737249638 51 )(1kgs rot s the equation predicts that the
7372496 51kg particle is the elementary mass out of which all mass is made
27372496 51 07372496 50
hKgs kg
c
2 07372496 50 second
m m rotationsf
h kgsc
22 392984064 11
0737249638 50 s
kg rotf
kgs
Kanarev Smallest measurable photon
24241 35328803 15
07372496 50 s
kg rotf
kgs
13598443ev photon 2424143lt35kg
91093897 31123559 20
07372496 50 s
kg rotf
kgs
05109906 6eV electron
9109235lt31kg
16726230 27226873 23
07372496 50 s
kg rotf
kgs
93827241gt6eV proton 1672623lt27kg
16749286 272271858 23
07372496 50 s
kg rotf
kgs
9395656gt6eV neutron 1674928lt27kg
334357629 27226750117 23
(07372496 50 ) )2( s
kg rotf
kgs
(Deuterium nucleus) Proton+Neutron
22102 42mr kgm
334357629 274535202583 23
(07372496 50 ( ) s1)
kg rotf
kgs
Proton+Neutron
22102 42mr kgm
223368 253029747 25
07372496 50 s
kg rotf
kgs
1243gt9eV Higgs Boson 2215849lt25 kg
256482lt25347890 25
07372496 50 s
kg rotf
kgs
144gt11eV Fermi Lab particle 2567033lt25kg
5703889152 1677367 34
07372496 50 s
kg rotf
kgs
320gt20eV highest eV particle measured 5704518lt16kg
1859222909 925218 41
07372496 50 s
kg rotf
kgs
10429476 27eV Fernandes 1859223lt9kg
2176450474 8295212 42
07372496 50 s
kg rotf
kgs
Planck limit
78434 28106387 23
07372496 50 s
kg rotf
kgs
classical radius of an electron
This frequency and radius are not a particle radius
This mass does not exist
308 Is there experimental evidences that an electron has the so-called classic radius of an electron A No experiments do not exist From Questions and Answers of Kanarev Page 95 Electrons-Protons-Neutrons JK ET and Rewrite The handbook of Chemistry and Physics shows the classical radius of the electron as (2817 15 )m Kanarevs theory shows this to be the radius of approach of the magnetic force lines around the torus ring of the electron ( )magr This is confirmed by calculating the dimensionless fine structure constant
( ) (which is equal to the circumference of the inner circle (2 )magr divided by the experimentally
measured radius of the electron exp( )R
exp
2 (2)(314)(2817 15 ) 10007297352568
2426 12 137036magr m
R m
(191)
Kg Rotationsec meters experimental classical rmag mass frequency radius r=cf mr mr(11615lt3) 427E-41 5791781E+09 517617E-02 221022E-42 256717E-45 smallest photon
242414E-35 328809E+15 911753E-08 221022E-42 256717E-45 136eV photon 91094E-31 123559E+20 242631E-12 221022E-42 256717E-45 electron 0511gt6eV 891331E-30 120899E+21 247968E-13 221022E-42 256717E-45 up quark
178266E-29 241799E+21 123984E-13 221022E-42 256717E-45 down quark 188356E-28 255485E+22 117343E-14 221022E-42 256717E-45 muon
240606E-28 326356E+22 918606E-15 221022E-42 256717E-45 meson 248806E-28 337479E+22 88833E-15 221022E-42 256717E-45 meson 356532E-28 483598E+22 619921E-15 221022E-42 256717E-45 strange quark
880118E-28 119379E+23 251128E-15 221022E-42 256717E-45 meson 887249E-28 120346E+23 249109E-15 221022E-42 256717E-45 meson 887177E-28 120336E+23 249129E-15 221022E-42 256717E-45 meson 978325E-28 132699E+23 225919E-15 221022E-42 256717E-45 meson 16726E-27 22687E+23 132143E-15 221022E-42 256717E-45 Proton 938gt6eV
167491E-27 227183E+23 131961E-15 221022E-42 256717E-45 Neutron 939gt6eV 198872E-27 269748E+23 111138E-15 221022E-42 256717E-45 hyperon
212033E-27 2876E+23 104239E-15 221022E-42 256717E-45 hyperon 212579E-27 28834E+23 103972E-15 221022E-42 256717E-45 21347E-27 289549E+23 103538E-15 221022E-42 256717E-45
231746E-27 314339E+23 953725E-16 221022E-42 256717E-45 charm quark 234367E-27 317893E+23 943061E-16 221022E-42 256717E-45
235543E-27 319489E+23 93835E-16 221022E-42 256717E-45 298061E-27 404288E+23 741532E-16 221022E-42 256717E-45 hyperon
802198E-27 10881E+24 27552E-16 221022E-42 256717E-45 bottom quark 143201E-25 194237E+25 154344E-17 221022E-42 256717E-45 W
162556E-25 220489E+25 135967E-17 221022E-42 256717E-45 Z 223368E-25 302974E+25 989499E-18 221022E-42 256717E-45 Higgs125gt9eV
256482E-25 347890E+25 861745E-18 221022E-42 256717E-45 Fermi labs 144gt9eV 311966E-25 423148E+25 708481E-18 221022E-42 256717E-45 top quark
570389E-16 77367E+34 387494E-27 221022E-42 256717E-45 Gamma 320gt20eV 185922E-09 25218E+41 118879E-33 221022E-42 256717E-45 1043gt27eV 186lt9kg
217645E-08 29521E+42 101552E-34 221022E-42 256717E-45 Planck limits
mass frequency radius r=cf mr mr(11615lt3)
mass frequency radius r=cf mr2f 222E-39 298E+11 100E-03 6626069E-34 smallest photon
242414E-35 328809E+15 911753E-08 6626069E-34 Photon 136eV 91094E-31 123559E+20 242631E-12 6626069E-34 electron 0511gt6eV 891331E-30 120899E+21 247968E-13 6626069E-34 up quark
178266E-29 241799E+21 123984E-13 6626069E-34 down quark 188356E-28 255485E+22 117343E-14 6626069E-34 muon
240606E-28 326356E+22 918606E-15 6626069E-34 meson 248806E-28 337479E+22 88833E-15 6626069E-34 meson 356532E-28 483598E+22 619921E-15 6626069E-34 strange quark
880118E-28 119379E+23 251128E-15 6626069E-34 meson 887249E-28 120346E+23 249109E-15 6626069E-34 meson 887177E-28 120336E+23 249129E-15 6626069E-34 meson 978325E-28 132699E+23 225919E-15 6626069E-34 meson 16726E-27 22687E+23 132143E-15 6626069E-34 Proton 938gt6eV
167491E-27 227183E+23 131961E-15 6626069E-34 Neutron 939gt6eV 198872E-27 269748E+23 111138E-15 6626069E-34 hyperon
212033E-27 2876E+23 104239E-15 6626069E-34 hyperon 212579E-27 28834E+23 103972E-15 6626069E-34 21347E-27 289549E+23 103538E-15 6626069E-34
231746E-27 314339E+23 953725E-16 6626069E-34 charm quark 234367E-27 317893E+23 943061E-16 6626069E-34
235543E-27 319489E+23 93835E-16 6626069E-34 298061E-27 404288E+23 741532E-16 6626069E-34 hyperon
802198E-27 10881E+24 27552E-16 6626069E-34 bottom quark 143201E-25 194237E+25 154344E-17 6626069E-34 W
162556E-25 220489E+25 135967E-17 6626069E-34 Z 223368E-25 302974E+25 989499E-18 6626069E-34 Higgs 125eV
256482E-25 347890E+25 861745E-18 6626069E-34 Fermi labs 144gt9eV 311966E-25 423148E+25 708481E-18 6626069E-34 top quark
570389E-16 77367E+34 387493E-27 6626069E-34 Gamma 320gt20eV 185922E-09 25218E+41 118879E-33 6626069E-34 186ltkg 1043gt27eV 217645E-08 29521E+42 101552E-34 6626071E-34 Planck limits
mass frequency radius r=cf mr2f
frequency radius r=cf rf=C 299792458 8c m s 5791781E+09 517617E-02 2997925E+08 smallest photon
328809E+15 911753E-08 2997925E+08 136eV photon 123559E+20 242631E-12 2997925E+08 electron 0511gt6eV 120899E+21 247968E-13 2997925E+08 up quark
241799E+21 123984E-13 2997925E+08 down quark 255485E+22 117343E-14 2997925E+08 muon
326356E+22 918606E-15 2997925E+08 meson 337479E+22 88833E-15 2997925E+08 meson 483598E+22 619921E-15 2997925E+08 strange quark
119379E+23 251128E-15 2997925E+08 meson 120346E+23 249109E-15 2997925E+08 meson 120336E+23 249129E-15 2997925E+08 meson 132699E+23 225919E-15 2997925E+08 meson 22687E+23 132143E-15 2997925E+08 Proton 938gt6eV
227183E+23 131961E-15 2997925E+08 Neutron 939gt6eV 269748E+23 111138E-15 2997925E+08 hyperon
2876E+23 104239E-15 2997925E+08 hyperon 28834E+23 103972E-15 2997925E+08 289549E+23 103538E-15 2997925E+08
314339E+23 953725E-16 2997925E+08 charm quark 317893E+23 943061E-16 2997925E+08
319489E+23 93835E-16 2997925E+08 404288E+23 741532E-16 2997925E+08 hyperon
10881E+24 27552E-16 2997925E+08 bottom quark 194237E+25 154344E-17 2997925E+08 W
220489E+25 135967E-17 2997925E+08 Z 302974E+25 989499E-18 2997925E+08 Higgs 125eV
347890E+25 861745E-18 2997925E+08 Fermi labs 144gt9eV 423148E+25 708481E-18 2997925E+08 top quark
77367E+34 387493E-27 2997925E+08 Gamma 320gt20eV 25218E+41 118879E-33 2997925E+08 1043gt27eV 186lt9kg
29521E+42 101552E-34 2997925E+08 Planck limits
frequency radius r=cf rf=C
Savior Borg
Two helix makes a mass particle
Fernandes
Kanarevrsquos Photon Model
My website Symmetrymathcomis my interpretation of the Authorrsquos I have referenced
I do not claim that any of these authors will agree withmy interpretations
I leave it to anyone who visits my site to decide the logicof my interpretation
- SYMMETRY MATH (SM) By Jack Kuykendall
- Why is Symmetry Math Needed
- From ldquoWhat Is Mathematicsrdquo by Courant and Robbins ndash page 55 Th
- Slide 4
- Slide 5
- Slide 6
- Slide 7
- Slide 8
- Slide 9
- Slide 10
- Slide 11
- Slide 12
- Slide 13
- Slide 14
- Slide 15
- Slide 16
- Slide 17
- Slide 18
- Slide 19
- Slide 20
- Slide 21
- Slide 22
- Slide 23
- Slide 24
- Slide 25
- Slide 26
- Slide 27
- Slide 28
- Slide 29
- Slide 30
- Slide 31
- Slide 32
- Slide 33
- Slide 34
- Slide 35
- Slide 36
- Slide 37
- Slide 38
- Slide 39
- Slide 40
- Slide 41
- Slide 42
- Slide 43
- Slide 44
- Slide 45
- Slide 46
- Slide 47
- Slide 48
- Slide 49
- Slide 50
- Slide 51
- Slide 52
- Slide 53
- Slide 54
- Slide 55
- Slide 56
- Slide 57
- Slide 58
- Slide 59
- Slide 60
- Slide 61
- Slide 62
- Slide 63
- Slide 64
- Slide 65
- Slide 66
- Slide 67
- Slide 68
- Slide 69
- Slide 70
- Slide 71
- Slide 72
- Slide 73
- Slide 74
- Slide 75
- Slide 76
- Slide 77
- Slide 78
- Slide 79
- Slide 80
- Slide 81
- Slide 82
- Slide 83
- Slide 84
- Slide 85
- Slide 86
- Slide 87
- Slide 88
- Slide 89
- Slide 90
- Slide 91
- Slide 92
- Slide 93
- Slide 94
- Slide 95
- Slide 96
- Slide 97
- Slide 98
- Slide 99
- Slide 100
- Slide 101
- Slide 102
- Slide 103
- Slide 104
- Slide 105
- Slide 106
- Slide 107
- Slide 108
- Slide 109
- Slide 110
- Slide 111
- Slide 112
- Slide 113
- Slide 114
- Slide 115
- Slide 116
- Slide 117
- Slide 118
- Slide 119
- Slide 120
- Slide 121
- Slide 122
- Slide 123
-
Temperature Radius of Photons Equation 0 0(0 ) (27315 )AC T
0
2898 310609555 6
27315o
AoC
mTr m
316
0 0(1 ) (27415 )AС T 1
2898 310570855 6
27415A
o oC
mTr m
317
0 020 (68 ) 29315o
AC F T 20
2898 39885 6
29315o
AoC
mTr m
321
0 030 (86 ) 30315oAC F T
30
2898 39560 6
30315o
AoC
mTr m
322
0 0 0[100 (212 ) 37315 ]AC F T 100
2898 37766 6
37315o C
mKr m
K
323
0 0 0[(1000 )(1832 ) 127315 ]AC F T 1000
2898 32276 6
127315o
AoC
mTr m
324
0 0 0[(1500 )(2732 ) 177315 ]AC F T 1500
2898 31634 6
177315A
o oC
mTr m
325
0 0 0[(2000 )(3632 ) 227315 ]AC F T 2000
2898 31275 6
227315A
o oC
mTr m
326
1
0 0 0(0 2000 )(27315 227315 )oAto C to T
Temperatures between are formed with photons from the infra-red range As the temperature increase the radiuses of the photons decrease
Temperature (felt as HEAT by humans) is formed by photon emission from electrons
What is the exact radius of the photons that form the (highest) temperature felt as heat Modern science has no exact answer to this question We can only theorize that lsquoHEATrsquo is formed by photons which are emitted by electrons during the synthesis of atoms and molecules The border for the smallest radius of these photons is not established yet We theorize that it is in an interval of the ultra-violet range Since the shortest know radiuses (X and gamma photons) are emitted by nuclear synthesis they do not participate in heat formation Temperature (felt as HEAT by humans) is formed by photon emission from electrons If gamma photons participated in formation of heat the greatest possible temperature would be equal
maxmin
2898 3 2898 31 15 1000000000000000
3 18oA A
A AmT mT
T T Tr m
(348)
If such a temperature existed it would destroy molecules atoms and the nucleus of all atoms
Schwarzschildrsquos (incorrect) radius equation
2
2 g
GMR
C
Kanarevrsquos correct equation
1(1087148751 14)
2p
g photon
GMrR Mr
C
Schwarzschild 3 33
3 (3)(2 30 )
4 (4)(314)(297 3182 19
)og
M kgD
R m
kg
m
(403) 1
Kanarev
3 33
3 (3)(2 30 )
4 (4)(314)(439 )565 27red
g
M kgD
R m
kg
m
(404) 2
Kanarev
3 33 3
3 (3)(2 30 )276 35 (
4 (4)(314)(553 32 Kanrev error
0012)
)lightg
M kg kgD
R m m
kg
m
(405) 3
4
Kanarevrsquos 3 33
3 (3)(2 30)
4 (4)(314)(154 8)013 54gamma
g
MD
R
kg
m
(406) 5
6
Fernandesrsquo (18 36 9) 3 3
3 (3)(2 30)265 81
4 (4)(314)(18 17)rg
kg
m
MD
R
(407)
7
Nuclear density is around3
(12 24) 17kg
m 8
If the Sun were compressed to ( )( 439 )g redR m (equation 400) its gravitational attraction would
detain only radiation of the far infra-red spectrum Shorter wavelengths will not be detained If photons of gamma radiation were detained the gravitational radius of the Sun would have to
be ( )( 154 8 )g gammaR m (equation 402) This would not be possible because the density of
the Sun (equation 406) would be 37 orders more dense than the nucleus of atoms
Only one type of Black-hole can be all black That is one with a gravitational attraction strong enough to detain the highest energy gamma photons All other Black-holes should have colors which vary with the change of wavelengths of the photons which cannot be detained The order should be infra-red light ultra-violet x-ray and finally gamma photons
Mass is Frequency of an Elementary Particle All known experiments have produced answers that agree with the following equations ( ) ( )( )(distance)mad Action mass acceleration
( )mad
( ) ( ) (constant)(frequency)mad hf
(1) 2( ) ( )(constant)mad mc mass
(2)
c f rf (3)
Equating (1) and (2) 2( )m h c f Mass = (constant)(frequency of one particle)
Plankrsquos constant 6626069 34 HBCP 662603(11) 57(29)69 34 codatah Js h Js
2
222
2
6626069 34( )( ) ( )6626069 347372496 51
(299792458 8 ) 898755179 16
Joule
Kgmm sh Js s kgs
m mcs s
Equating (1)(2) and (3)
2 2
constant
expexp exp
1 6626069 34 22102 42
299792458 8
h h c h h JsM f mr kgm
mr c r cc cs
All mass is built from one elementary particle a ( 737 51 )kg particle
All mass is in relative motion with other mass Time is a description invented to measure the relative movement between different particles of
mass Energy is a term invented to describe what happens when masses interact Energy is a
description Mass does not convert to energy and energy does not convert to mass The word ldquoEnergyrdquo will be replaced with an Action involving mass acceleration and distance ( )mad
When the motion of masses interacts they change to different motions of masses These different mass forms can be measured and their changes in momentum can be measured When the change produces photon particles they have been incorrectly labeled as ldquoEnergyrdquo
Mass Frequency Frequency is a rotation per second
If ( )f is set to one rotation per second (737249638 51 )(1kgs rot s the equation predicts that the
7372496 51kg particle is the elementary mass out of which all mass is made
27372496 51 07372496 50
hKgs kg
c
2 07372496 50 second
m m rotationsf
h kgsc
22 392984064 11
0737249638 50 s
kg rotf
kgs
Kanarev Smallest measurable photon
24241 35328803 15
07372496 50 s
kg rotf
kgs
13598443ev photon 2424143lt35kg
91093897 31123559 20
07372496 50 s
kg rotf
kgs
05109906 6eV electron
9109235lt31kg
16726230 27226873 23
07372496 50 s
kg rotf
kgs
93827241gt6eV proton 1672623lt27kg
16749286 272271858 23
07372496 50 s
kg rotf
kgs
9395656gt6eV neutron 1674928lt27kg
334357629 27226750117 23
(07372496 50 ) )2( s
kg rotf
kgs
(Deuterium nucleus) Proton+Neutron
22102 42mr kgm
334357629 274535202583 23
(07372496 50 ( ) s1)
kg rotf
kgs
Proton+Neutron
22102 42mr kgm
223368 253029747 25
07372496 50 s
kg rotf
kgs
1243gt9eV Higgs Boson 2215849lt25 kg
256482lt25347890 25
07372496 50 s
kg rotf
kgs
144gt11eV Fermi Lab particle 2567033lt25kg
5703889152 1677367 34
07372496 50 s
kg rotf
kgs
320gt20eV highest eV particle measured 5704518lt16kg
1859222909 925218 41
07372496 50 s
kg rotf
kgs
10429476 27eV Fernandes 1859223lt9kg
2176450474 8295212 42
07372496 50 s
kg rotf
kgs
Planck limit
78434 28106387 23
07372496 50 s
kg rotf
kgs
classical radius of an electron
This frequency and radius are not a particle radius
This mass does not exist
308 Is there experimental evidences that an electron has the so-called classic radius of an electron A No experiments do not exist From Questions and Answers of Kanarev Page 95 Electrons-Protons-Neutrons JK ET and Rewrite The handbook of Chemistry and Physics shows the classical radius of the electron as (2817 15 )m Kanarevs theory shows this to be the radius of approach of the magnetic force lines around the torus ring of the electron ( )magr This is confirmed by calculating the dimensionless fine structure constant
( ) (which is equal to the circumference of the inner circle (2 )magr divided by the experimentally
measured radius of the electron exp( )R
exp
2 (2)(314)(2817 15 ) 10007297352568
2426 12 137036magr m
R m
(191)
Kg Rotationsec meters experimental classical rmag mass frequency radius r=cf mr mr(11615lt3) 427E-41 5791781E+09 517617E-02 221022E-42 256717E-45 smallest photon
242414E-35 328809E+15 911753E-08 221022E-42 256717E-45 136eV photon 91094E-31 123559E+20 242631E-12 221022E-42 256717E-45 electron 0511gt6eV 891331E-30 120899E+21 247968E-13 221022E-42 256717E-45 up quark
178266E-29 241799E+21 123984E-13 221022E-42 256717E-45 down quark 188356E-28 255485E+22 117343E-14 221022E-42 256717E-45 muon
240606E-28 326356E+22 918606E-15 221022E-42 256717E-45 meson 248806E-28 337479E+22 88833E-15 221022E-42 256717E-45 meson 356532E-28 483598E+22 619921E-15 221022E-42 256717E-45 strange quark
880118E-28 119379E+23 251128E-15 221022E-42 256717E-45 meson 887249E-28 120346E+23 249109E-15 221022E-42 256717E-45 meson 887177E-28 120336E+23 249129E-15 221022E-42 256717E-45 meson 978325E-28 132699E+23 225919E-15 221022E-42 256717E-45 meson 16726E-27 22687E+23 132143E-15 221022E-42 256717E-45 Proton 938gt6eV
167491E-27 227183E+23 131961E-15 221022E-42 256717E-45 Neutron 939gt6eV 198872E-27 269748E+23 111138E-15 221022E-42 256717E-45 hyperon
212033E-27 2876E+23 104239E-15 221022E-42 256717E-45 hyperon 212579E-27 28834E+23 103972E-15 221022E-42 256717E-45 21347E-27 289549E+23 103538E-15 221022E-42 256717E-45
231746E-27 314339E+23 953725E-16 221022E-42 256717E-45 charm quark 234367E-27 317893E+23 943061E-16 221022E-42 256717E-45
235543E-27 319489E+23 93835E-16 221022E-42 256717E-45 298061E-27 404288E+23 741532E-16 221022E-42 256717E-45 hyperon
802198E-27 10881E+24 27552E-16 221022E-42 256717E-45 bottom quark 143201E-25 194237E+25 154344E-17 221022E-42 256717E-45 W
162556E-25 220489E+25 135967E-17 221022E-42 256717E-45 Z 223368E-25 302974E+25 989499E-18 221022E-42 256717E-45 Higgs125gt9eV
256482E-25 347890E+25 861745E-18 221022E-42 256717E-45 Fermi labs 144gt9eV 311966E-25 423148E+25 708481E-18 221022E-42 256717E-45 top quark
570389E-16 77367E+34 387494E-27 221022E-42 256717E-45 Gamma 320gt20eV 185922E-09 25218E+41 118879E-33 221022E-42 256717E-45 1043gt27eV 186lt9kg
217645E-08 29521E+42 101552E-34 221022E-42 256717E-45 Planck limits
mass frequency radius r=cf mr mr(11615lt3)
mass frequency radius r=cf mr2f 222E-39 298E+11 100E-03 6626069E-34 smallest photon
242414E-35 328809E+15 911753E-08 6626069E-34 Photon 136eV 91094E-31 123559E+20 242631E-12 6626069E-34 electron 0511gt6eV 891331E-30 120899E+21 247968E-13 6626069E-34 up quark
178266E-29 241799E+21 123984E-13 6626069E-34 down quark 188356E-28 255485E+22 117343E-14 6626069E-34 muon
240606E-28 326356E+22 918606E-15 6626069E-34 meson 248806E-28 337479E+22 88833E-15 6626069E-34 meson 356532E-28 483598E+22 619921E-15 6626069E-34 strange quark
880118E-28 119379E+23 251128E-15 6626069E-34 meson 887249E-28 120346E+23 249109E-15 6626069E-34 meson 887177E-28 120336E+23 249129E-15 6626069E-34 meson 978325E-28 132699E+23 225919E-15 6626069E-34 meson 16726E-27 22687E+23 132143E-15 6626069E-34 Proton 938gt6eV
167491E-27 227183E+23 131961E-15 6626069E-34 Neutron 939gt6eV 198872E-27 269748E+23 111138E-15 6626069E-34 hyperon
212033E-27 2876E+23 104239E-15 6626069E-34 hyperon 212579E-27 28834E+23 103972E-15 6626069E-34 21347E-27 289549E+23 103538E-15 6626069E-34
231746E-27 314339E+23 953725E-16 6626069E-34 charm quark 234367E-27 317893E+23 943061E-16 6626069E-34
235543E-27 319489E+23 93835E-16 6626069E-34 298061E-27 404288E+23 741532E-16 6626069E-34 hyperon
802198E-27 10881E+24 27552E-16 6626069E-34 bottom quark 143201E-25 194237E+25 154344E-17 6626069E-34 W
162556E-25 220489E+25 135967E-17 6626069E-34 Z 223368E-25 302974E+25 989499E-18 6626069E-34 Higgs 125eV
256482E-25 347890E+25 861745E-18 6626069E-34 Fermi labs 144gt9eV 311966E-25 423148E+25 708481E-18 6626069E-34 top quark
570389E-16 77367E+34 387493E-27 6626069E-34 Gamma 320gt20eV 185922E-09 25218E+41 118879E-33 6626069E-34 186ltkg 1043gt27eV 217645E-08 29521E+42 101552E-34 6626071E-34 Planck limits
mass frequency radius r=cf mr2f
frequency radius r=cf rf=C 299792458 8c m s 5791781E+09 517617E-02 2997925E+08 smallest photon
328809E+15 911753E-08 2997925E+08 136eV photon 123559E+20 242631E-12 2997925E+08 electron 0511gt6eV 120899E+21 247968E-13 2997925E+08 up quark
241799E+21 123984E-13 2997925E+08 down quark 255485E+22 117343E-14 2997925E+08 muon
326356E+22 918606E-15 2997925E+08 meson 337479E+22 88833E-15 2997925E+08 meson 483598E+22 619921E-15 2997925E+08 strange quark
119379E+23 251128E-15 2997925E+08 meson 120346E+23 249109E-15 2997925E+08 meson 120336E+23 249129E-15 2997925E+08 meson 132699E+23 225919E-15 2997925E+08 meson 22687E+23 132143E-15 2997925E+08 Proton 938gt6eV
227183E+23 131961E-15 2997925E+08 Neutron 939gt6eV 269748E+23 111138E-15 2997925E+08 hyperon
2876E+23 104239E-15 2997925E+08 hyperon 28834E+23 103972E-15 2997925E+08 289549E+23 103538E-15 2997925E+08
314339E+23 953725E-16 2997925E+08 charm quark 317893E+23 943061E-16 2997925E+08
319489E+23 93835E-16 2997925E+08 404288E+23 741532E-16 2997925E+08 hyperon
10881E+24 27552E-16 2997925E+08 bottom quark 194237E+25 154344E-17 2997925E+08 W
220489E+25 135967E-17 2997925E+08 Z 302974E+25 989499E-18 2997925E+08 Higgs 125eV
347890E+25 861745E-18 2997925E+08 Fermi labs 144gt9eV 423148E+25 708481E-18 2997925E+08 top quark
77367E+34 387493E-27 2997925E+08 Gamma 320gt20eV 25218E+41 118879E-33 2997925E+08 1043gt27eV 186lt9kg
29521E+42 101552E-34 2997925E+08 Planck limits
frequency radius r=cf rf=C
Savior Borg
Two helix makes a mass particle
Fernandes
Kanarevrsquos Photon Model
My website Symmetrymathcomis my interpretation of the Authorrsquos I have referenced
I do not claim that any of these authors will agree withmy interpretations
I leave it to anyone who visits my site to decide the logicof my interpretation
- SYMMETRY MATH (SM) By Jack Kuykendall
- Why is Symmetry Math Needed
- From ldquoWhat Is Mathematicsrdquo by Courant and Robbins ndash page 55 Th
- Slide 4
- Slide 5
- Slide 6
- Slide 7
- Slide 8
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- Slide 123
-
Temperature (felt as HEAT by humans) is formed by photon emission from electrons
What is the exact radius of the photons that form the (highest) temperature felt as heat Modern science has no exact answer to this question We can only theorize that lsquoHEATrsquo is formed by photons which are emitted by electrons during the synthesis of atoms and molecules The border for the smallest radius of these photons is not established yet We theorize that it is in an interval of the ultra-violet range Since the shortest know radiuses (X and gamma photons) are emitted by nuclear synthesis they do not participate in heat formation Temperature (felt as HEAT by humans) is formed by photon emission from electrons If gamma photons participated in formation of heat the greatest possible temperature would be equal
maxmin
2898 3 2898 31 15 1000000000000000
3 18oA A
A AmT mT
T T Tr m
(348)
If such a temperature existed it would destroy molecules atoms and the nucleus of all atoms
Schwarzschildrsquos (incorrect) radius equation
2
2 g
GMR
C
Kanarevrsquos correct equation
1(1087148751 14)
2p
g photon
GMrR Mr
C
Schwarzschild 3 33
3 (3)(2 30 )
4 (4)(314)(297 3182 19
)og
M kgD
R m
kg
m
(403) 1
Kanarev
3 33
3 (3)(2 30 )
4 (4)(314)(439 )565 27red
g
M kgD
R m
kg
m
(404) 2
Kanarev
3 33 3
3 (3)(2 30 )276 35 (
4 (4)(314)(553 32 Kanrev error
0012)
)lightg
M kg kgD
R m m
kg
m
(405) 3
4
Kanarevrsquos 3 33
3 (3)(2 30)
4 (4)(314)(154 8)013 54gamma
g
MD
R
kg
m
(406) 5
6
Fernandesrsquo (18 36 9) 3 3
3 (3)(2 30)265 81
4 (4)(314)(18 17)rg
kg
m
MD
R
(407)
7
Nuclear density is around3
(12 24) 17kg
m 8
If the Sun were compressed to ( )( 439 )g redR m (equation 400) its gravitational attraction would
detain only radiation of the far infra-red spectrum Shorter wavelengths will not be detained If photons of gamma radiation were detained the gravitational radius of the Sun would have to
be ( )( 154 8 )g gammaR m (equation 402) This would not be possible because the density of
the Sun (equation 406) would be 37 orders more dense than the nucleus of atoms
Only one type of Black-hole can be all black That is one with a gravitational attraction strong enough to detain the highest energy gamma photons All other Black-holes should have colors which vary with the change of wavelengths of the photons which cannot be detained The order should be infra-red light ultra-violet x-ray and finally gamma photons
Mass is Frequency of an Elementary Particle All known experiments have produced answers that agree with the following equations ( ) ( )( )(distance)mad Action mass acceleration
( )mad
( ) ( ) (constant)(frequency)mad hf
(1) 2( ) ( )(constant)mad mc mass
(2)
c f rf (3)
Equating (1) and (2) 2( )m h c f Mass = (constant)(frequency of one particle)
Plankrsquos constant 6626069 34 HBCP 662603(11) 57(29)69 34 codatah Js h Js
2
222
2
6626069 34( )( ) ( )6626069 347372496 51
(299792458 8 ) 898755179 16
Joule
Kgmm sh Js s kgs
m mcs s
Equating (1)(2) and (3)
2 2
constant
expexp exp
1 6626069 34 22102 42
299792458 8
h h c h h JsM f mr kgm
mr c r cc cs
All mass is built from one elementary particle a ( 737 51 )kg particle
All mass is in relative motion with other mass Time is a description invented to measure the relative movement between different particles of
mass Energy is a term invented to describe what happens when masses interact Energy is a
description Mass does not convert to energy and energy does not convert to mass The word ldquoEnergyrdquo will be replaced with an Action involving mass acceleration and distance ( )mad
When the motion of masses interacts they change to different motions of masses These different mass forms can be measured and their changes in momentum can be measured When the change produces photon particles they have been incorrectly labeled as ldquoEnergyrdquo
Mass Frequency Frequency is a rotation per second
If ( )f is set to one rotation per second (737249638 51 )(1kgs rot s the equation predicts that the
7372496 51kg particle is the elementary mass out of which all mass is made
27372496 51 07372496 50
hKgs kg
c
2 07372496 50 second
m m rotationsf
h kgsc
22 392984064 11
0737249638 50 s
kg rotf
kgs
Kanarev Smallest measurable photon
24241 35328803 15
07372496 50 s
kg rotf
kgs
13598443ev photon 2424143lt35kg
91093897 31123559 20
07372496 50 s
kg rotf
kgs
05109906 6eV electron
9109235lt31kg
16726230 27226873 23
07372496 50 s
kg rotf
kgs
93827241gt6eV proton 1672623lt27kg
16749286 272271858 23
07372496 50 s
kg rotf
kgs
9395656gt6eV neutron 1674928lt27kg
334357629 27226750117 23
(07372496 50 ) )2( s
kg rotf
kgs
(Deuterium nucleus) Proton+Neutron
22102 42mr kgm
334357629 274535202583 23
(07372496 50 ( ) s1)
kg rotf
kgs
Proton+Neutron
22102 42mr kgm
223368 253029747 25
07372496 50 s
kg rotf
kgs
1243gt9eV Higgs Boson 2215849lt25 kg
256482lt25347890 25
07372496 50 s
kg rotf
kgs
144gt11eV Fermi Lab particle 2567033lt25kg
5703889152 1677367 34
07372496 50 s
kg rotf
kgs
320gt20eV highest eV particle measured 5704518lt16kg
1859222909 925218 41
07372496 50 s
kg rotf
kgs
10429476 27eV Fernandes 1859223lt9kg
2176450474 8295212 42
07372496 50 s
kg rotf
kgs
Planck limit
78434 28106387 23
07372496 50 s
kg rotf
kgs
classical radius of an electron
This frequency and radius are not a particle radius
This mass does not exist
308 Is there experimental evidences that an electron has the so-called classic radius of an electron A No experiments do not exist From Questions and Answers of Kanarev Page 95 Electrons-Protons-Neutrons JK ET and Rewrite The handbook of Chemistry and Physics shows the classical radius of the electron as (2817 15 )m Kanarevs theory shows this to be the radius of approach of the magnetic force lines around the torus ring of the electron ( )magr This is confirmed by calculating the dimensionless fine structure constant
( ) (which is equal to the circumference of the inner circle (2 )magr divided by the experimentally
measured radius of the electron exp( )R
exp
2 (2)(314)(2817 15 ) 10007297352568
2426 12 137036magr m
R m
(191)
Kg Rotationsec meters experimental classical rmag mass frequency radius r=cf mr mr(11615lt3) 427E-41 5791781E+09 517617E-02 221022E-42 256717E-45 smallest photon
242414E-35 328809E+15 911753E-08 221022E-42 256717E-45 136eV photon 91094E-31 123559E+20 242631E-12 221022E-42 256717E-45 electron 0511gt6eV 891331E-30 120899E+21 247968E-13 221022E-42 256717E-45 up quark
178266E-29 241799E+21 123984E-13 221022E-42 256717E-45 down quark 188356E-28 255485E+22 117343E-14 221022E-42 256717E-45 muon
240606E-28 326356E+22 918606E-15 221022E-42 256717E-45 meson 248806E-28 337479E+22 88833E-15 221022E-42 256717E-45 meson 356532E-28 483598E+22 619921E-15 221022E-42 256717E-45 strange quark
880118E-28 119379E+23 251128E-15 221022E-42 256717E-45 meson 887249E-28 120346E+23 249109E-15 221022E-42 256717E-45 meson 887177E-28 120336E+23 249129E-15 221022E-42 256717E-45 meson 978325E-28 132699E+23 225919E-15 221022E-42 256717E-45 meson 16726E-27 22687E+23 132143E-15 221022E-42 256717E-45 Proton 938gt6eV
167491E-27 227183E+23 131961E-15 221022E-42 256717E-45 Neutron 939gt6eV 198872E-27 269748E+23 111138E-15 221022E-42 256717E-45 hyperon
212033E-27 2876E+23 104239E-15 221022E-42 256717E-45 hyperon 212579E-27 28834E+23 103972E-15 221022E-42 256717E-45 21347E-27 289549E+23 103538E-15 221022E-42 256717E-45
231746E-27 314339E+23 953725E-16 221022E-42 256717E-45 charm quark 234367E-27 317893E+23 943061E-16 221022E-42 256717E-45
235543E-27 319489E+23 93835E-16 221022E-42 256717E-45 298061E-27 404288E+23 741532E-16 221022E-42 256717E-45 hyperon
802198E-27 10881E+24 27552E-16 221022E-42 256717E-45 bottom quark 143201E-25 194237E+25 154344E-17 221022E-42 256717E-45 W
162556E-25 220489E+25 135967E-17 221022E-42 256717E-45 Z 223368E-25 302974E+25 989499E-18 221022E-42 256717E-45 Higgs125gt9eV
256482E-25 347890E+25 861745E-18 221022E-42 256717E-45 Fermi labs 144gt9eV 311966E-25 423148E+25 708481E-18 221022E-42 256717E-45 top quark
570389E-16 77367E+34 387494E-27 221022E-42 256717E-45 Gamma 320gt20eV 185922E-09 25218E+41 118879E-33 221022E-42 256717E-45 1043gt27eV 186lt9kg
217645E-08 29521E+42 101552E-34 221022E-42 256717E-45 Planck limits
mass frequency radius r=cf mr mr(11615lt3)
mass frequency radius r=cf mr2f 222E-39 298E+11 100E-03 6626069E-34 smallest photon
242414E-35 328809E+15 911753E-08 6626069E-34 Photon 136eV 91094E-31 123559E+20 242631E-12 6626069E-34 electron 0511gt6eV 891331E-30 120899E+21 247968E-13 6626069E-34 up quark
178266E-29 241799E+21 123984E-13 6626069E-34 down quark 188356E-28 255485E+22 117343E-14 6626069E-34 muon
240606E-28 326356E+22 918606E-15 6626069E-34 meson 248806E-28 337479E+22 88833E-15 6626069E-34 meson 356532E-28 483598E+22 619921E-15 6626069E-34 strange quark
880118E-28 119379E+23 251128E-15 6626069E-34 meson 887249E-28 120346E+23 249109E-15 6626069E-34 meson 887177E-28 120336E+23 249129E-15 6626069E-34 meson 978325E-28 132699E+23 225919E-15 6626069E-34 meson 16726E-27 22687E+23 132143E-15 6626069E-34 Proton 938gt6eV
167491E-27 227183E+23 131961E-15 6626069E-34 Neutron 939gt6eV 198872E-27 269748E+23 111138E-15 6626069E-34 hyperon
212033E-27 2876E+23 104239E-15 6626069E-34 hyperon 212579E-27 28834E+23 103972E-15 6626069E-34 21347E-27 289549E+23 103538E-15 6626069E-34
231746E-27 314339E+23 953725E-16 6626069E-34 charm quark 234367E-27 317893E+23 943061E-16 6626069E-34
235543E-27 319489E+23 93835E-16 6626069E-34 298061E-27 404288E+23 741532E-16 6626069E-34 hyperon
802198E-27 10881E+24 27552E-16 6626069E-34 bottom quark 143201E-25 194237E+25 154344E-17 6626069E-34 W
162556E-25 220489E+25 135967E-17 6626069E-34 Z 223368E-25 302974E+25 989499E-18 6626069E-34 Higgs 125eV
256482E-25 347890E+25 861745E-18 6626069E-34 Fermi labs 144gt9eV 311966E-25 423148E+25 708481E-18 6626069E-34 top quark
570389E-16 77367E+34 387493E-27 6626069E-34 Gamma 320gt20eV 185922E-09 25218E+41 118879E-33 6626069E-34 186ltkg 1043gt27eV 217645E-08 29521E+42 101552E-34 6626071E-34 Planck limits
mass frequency radius r=cf mr2f
frequency radius r=cf rf=C 299792458 8c m s 5791781E+09 517617E-02 2997925E+08 smallest photon
328809E+15 911753E-08 2997925E+08 136eV photon 123559E+20 242631E-12 2997925E+08 electron 0511gt6eV 120899E+21 247968E-13 2997925E+08 up quark
241799E+21 123984E-13 2997925E+08 down quark 255485E+22 117343E-14 2997925E+08 muon
326356E+22 918606E-15 2997925E+08 meson 337479E+22 88833E-15 2997925E+08 meson 483598E+22 619921E-15 2997925E+08 strange quark
119379E+23 251128E-15 2997925E+08 meson 120346E+23 249109E-15 2997925E+08 meson 120336E+23 249129E-15 2997925E+08 meson 132699E+23 225919E-15 2997925E+08 meson 22687E+23 132143E-15 2997925E+08 Proton 938gt6eV
227183E+23 131961E-15 2997925E+08 Neutron 939gt6eV 269748E+23 111138E-15 2997925E+08 hyperon
2876E+23 104239E-15 2997925E+08 hyperon 28834E+23 103972E-15 2997925E+08 289549E+23 103538E-15 2997925E+08
314339E+23 953725E-16 2997925E+08 charm quark 317893E+23 943061E-16 2997925E+08
319489E+23 93835E-16 2997925E+08 404288E+23 741532E-16 2997925E+08 hyperon
10881E+24 27552E-16 2997925E+08 bottom quark 194237E+25 154344E-17 2997925E+08 W
220489E+25 135967E-17 2997925E+08 Z 302974E+25 989499E-18 2997925E+08 Higgs 125eV
347890E+25 861745E-18 2997925E+08 Fermi labs 144gt9eV 423148E+25 708481E-18 2997925E+08 top quark
77367E+34 387493E-27 2997925E+08 Gamma 320gt20eV 25218E+41 118879E-33 2997925E+08 1043gt27eV 186lt9kg
29521E+42 101552E-34 2997925E+08 Planck limits
frequency radius r=cf rf=C
Savior Borg
Two helix makes a mass particle
Fernandes
Kanarevrsquos Photon Model
My website Symmetrymathcomis my interpretation of the Authorrsquos I have referenced
I do not claim that any of these authors will agree withmy interpretations
I leave it to anyone who visits my site to decide the logicof my interpretation
- SYMMETRY MATH (SM) By Jack Kuykendall
- Why is Symmetry Math Needed
- From ldquoWhat Is Mathematicsrdquo by Courant and Robbins ndash page 55 Th
- Slide 4
- Slide 5
- Slide 6
- Slide 7
- Slide 8
- Slide 9
- Slide 10
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- Slide 12
- Slide 13
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- Slide 68
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- Slide 123
-
Schwarzschildrsquos (incorrect) radius equation
2
2 g
GMR
C
Kanarevrsquos correct equation
1(1087148751 14)
2p
g photon
GMrR Mr
C
Schwarzschild 3 33
3 (3)(2 30 )
4 (4)(314)(297 3182 19
)og
M kgD
R m
kg
m
(403) 1
Kanarev
3 33
3 (3)(2 30 )
4 (4)(314)(439 )565 27red
g
M kgD
R m
kg
m
(404) 2
Kanarev
3 33 3
3 (3)(2 30 )276 35 (
4 (4)(314)(553 32 Kanrev error
0012)
)lightg
M kg kgD
R m m
kg
m
(405) 3
4
Kanarevrsquos 3 33
3 (3)(2 30)
4 (4)(314)(154 8)013 54gamma
g
MD
R
kg
m
(406) 5
6
Fernandesrsquo (18 36 9) 3 3
3 (3)(2 30)265 81
4 (4)(314)(18 17)rg
kg
m
MD
R
(407)
7
Nuclear density is around3
(12 24) 17kg
m 8
If the Sun were compressed to ( )( 439 )g redR m (equation 400) its gravitational attraction would
detain only radiation of the far infra-red spectrum Shorter wavelengths will not be detained If photons of gamma radiation were detained the gravitational radius of the Sun would have to
be ( )( 154 8 )g gammaR m (equation 402) This would not be possible because the density of
the Sun (equation 406) would be 37 orders more dense than the nucleus of atoms
Only one type of Black-hole can be all black That is one with a gravitational attraction strong enough to detain the highest energy gamma photons All other Black-holes should have colors which vary with the change of wavelengths of the photons which cannot be detained The order should be infra-red light ultra-violet x-ray and finally gamma photons
Mass is Frequency of an Elementary Particle All known experiments have produced answers that agree with the following equations ( ) ( )( )(distance)mad Action mass acceleration
( )mad
( ) ( ) (constant)(frequency)mad hf
(1) 2( ) ( )(constant)mad mc mass
(2)
c f rf (3)
Equating (1) and (2) 2( )m h c f Mass = (constant)(frequency of one particle)
Plankrsquos constant 6626069 34 HBCP 662603(11) 57(29)69 34 codatah Js h Js
2
222
2
6626069 34( )( ) ( )6626069 347372496 51
(299792458 8 ) 898755179 16
Joule
Kgmm sh Js s kgs
m mcs s
Equating (1)(2) and (3)
2 2
constant
expexp exp
1 6626069 34 22102 42
299792458 8
h h c h h JsM f mr kgm
mr c r cc cs
All mass is built from one elementary particle a ( 737 51 )kg particle
All mass is in relative motion with other mass Time is a description invented to measure the relative movement between different particles of
mass Energy is a term invented to describe what happens when masses interact Energy is a
description Mass does not convert to energy and energy does not convert to mass The word ldquoEnergyrdquo will be replaced with an Action involving mass acceleration and distance ( )mad
When the motion of masses interacts they change to different motions of masses These different mass forms can be measured and their changes in momentum can be measured When the change produces photon particles they have been incorrectly labeled as ldquoEnergyrdquo
Mass Frequency Frequency is a rotation per second
If ( )f is set to one rotation per second (737249638 51 )(1kgs rot s the equation predicts that the
7372496 51kg particle is the elementary mass out of which all mass is made
27372496 51 07372496 50
hKgs kg
c
2 07372496 50 second
m m rotationsf
h kgsc
22 392984064 11
0737249638 50 s
kg rotf
kgs
Kanarev Smallest measurable photon
24241 35328803 15
07372496 50 s
kg rotf
kgs
13598443ev photon 2424143lt35kg
91093897 31123559 20
07372496 50 s
kg rotf
kgs
05109906 6eV electron
9109235lt31kg
16726230 27226873 23
07372496 50 s
kg rotf
kgs
93827241gt6eV proton 1672623lt27kg
16749286 272271858 23
07372496 50 s
kg rotf
kgs
9395656gt6eV neutron 1674928lt27kg
334357629 27226750117 23
(07372496 50 ) )2( s
kg rotf
kgs
(Deuterium nucleus) Proton+Neutron
22102 42mr kgm
334357629 274535202583 23
(07372496 50 ( ) s1)
kg rotf
kgs
Proton+Neutron
22102 42mr kgm
223368 253029747 25
07372496 50 s
kg rotf
kgs
1243gt9eV Higgs Boson 2215849lt25 kg
256482lt25347890 25
07372496 50 s
kg rotf
kgs
144gt11eV Fermi Lab particle 2567033lt25kg
5703889152 1677367 34
07372496 50 s
kg rotf
kgs
320gt20eV highest eV particle measured 5704518lt16kg
1859222909 925218 41
07372496 50 s
kg rotf
kgs
10429476 27eV Fernandes 1859223lt9kg
2176450474 8295212 42
07372496 50 s
kg rotf
kgs
Planck limit
78434 28106387 23
07372496 50 s
kg rotf
kgs
classical radius of an electron
This frequency and radius are not a particle radius
This mass does not exist
308 Is there experimental evidences that an electron has the so-called classic radius of an electron A No experiments do not exist From Questions and Answers of Kanarev Page 95 Electrons-Protons-Neutrons JK ET and Rewrite The handbook of Chemistry and Physics shows the classical radius of the electron as (2817 15 )m Kanarevs theory shows this to be the radius of approach of the magnetic force lines around the torus ring of the electron ( )magr This is confirmed by calculating the dimensionless fine structure constant
( ) (which is equal to the circumference of the inner circle (2 )magr divided by the experimentally
measured radius of the electron exp( )R
exp
2 (2)(314)(2817 15 ) 10007297352568
2426 12 137036magr m
R m
(191)
Kg Rotationsec meters experimental classical rmag mass frequency radius r=cf mr mr(11615lt3) 427E-41 5791781E+09 517617E-02 221022E-42 256717E-45 smallest photon
242414E-35 328809E+15 911753E-08 221022E-42 256717E-45 136eV photon 91094E-31 123559E+20 242631E-12 221022E-42 256717E-45 electron 0511gt6eV 891331E-30 120899E+21 247968E-13 221022E-42 256717E-45 up quark
178266E-29 241799E+21 123984E-13 221022E-42 256717E-45 down quark 188356E-28 255485E+22 117343E-14 221022E-42 256717E-45 muon
240606E-28 326356E+22 918606E-15 221022E-42 256717E-45 meson 248806E-28 337479E+22 88833E-15 221022E-42 256717E-45 meson 356532E-28 483598E+22 619921E-15 221022E-42 256717E-45 strange quark
880118E-28 119379E+23 251128E-15 221022E-42 256717E-45 meson 887249E-28 120346E+23 249109E-15 221022E-42 256717E-45 meson 887177E-28 120336E+23 249129E-15 221022E-42 256717E-45 meson 978325E-28 132699E+23 225919E-15 221022E-42 256717E-45 meson 16726E-27 22687E+23 132143E-15 221022E-42 256717E-45 Proton 938gt6eV
167491E-27 227183E+23 131961E-15 221022E-42 256717E-45 Neutron 939gt6eV 198872E-27 269748E+23 111138E-15 221022E-42 256717E-45 hyperon
212033E-27 2876E+23 104239E-15 221022E-42 256717E-45 hyperon 212579E-27 28834E+23 103972E-15 221022E-42 256717E-45 21347E-27 289549E+23 103538E-15 221022E-42 256717E-45
231746E-27 314339E+23 953725E-16 221022E-42 256717E-45 charm quark 234367E-27 317893E+23 943061E-16 221022E-42 256717E-45
235543E-27 319489E+23 93835E-16 221022E-42 256717E-45 298061E-27 404288E+23 741532E-16 221022E-42 256717E-45 hyperon
802198E-27 10881E+24 27552E-16 221022E-42 256717E-45 bottom quark 143201E-25 194237E+25 154344E-17 221022E-42 256717E-45 W
162556E-25 220489E+25 135967E-17 221022E-42 256717E-45 Z 223368E-25 302974E+25 989499E-18 221022E-42 256717E-45 Higgs125gt9eV
256482E-25 347890E+25 861745E-18 221022E-42 256717E-45 Fermi labs 144gt9eV 311966E-25 423148E+25 708481E-18 221022E-42 256717E-45 top quark
570389E-16 77367E+34 387494E-27 221022E-42 256717E-45 Gamma 320gt20eV 185922E-09 25218E+41 118879E-33 221022E-42 256717E-45 1043gt27eV 186lt9kg
217645E-08 29521E+42 101552E-34 221022E-42 256717E-45 Planck limits
mass frequency radius r=cf mr mr(11615lt3)
mass frequency radius r=cf mr2f 222E-39 298E+11 100E-03 6626069E-34 smallest photon
242414E-35 328809E+15 911753E-08 6626069E-34 Photon 136eV 91094E-31 123559E+20 242631E-12 6626069E-34 electron 0511gt6eV 891331E-30 120899E+21 247968E-13 6626069E-34 up quark
178266E-29 241799E+21 123984E-13 6626069E-34 down quark 188356E-28 255485E+22 117343E-14 6626069E-34 muon
240606E-28 326356E+22 918606E-15 6626069E-34 meson 248806E-28 337479E+22 88833E-15 6626069E-34 meson 356532E-28 483598E+22 619921E-15 6626069E-34 strange quark
880118E-28 119379E+23 251128E-15 6626069E-34 meson 887249E-28 120346E+23 249109E-15 6626069E-34 meson 887177E-28 120336E+23 249129E-15 6626069E-34 meson 978325E-28 132699E+23 225919E-15 6626069E-34 meson 16726E-27 22687E+23 132143E-15 6626069E-34 Proton 938gt6eV
167491E-27 227183E+23 131961E-15 6626069E-34 Neutron 939gt6eV 198872E-27 269748E+23 111138E-15 6626069E-34 hyperon
212033E-27 2876E+23 104239E-15 6626069E-34 hyperon 212579E-27 28834E+23 103972E-15 6626069E-34 21347E-27 289549E+23 103538E-15 6626069E-34
231746E-27 314339E+23 953725E-16 6626069E-34 charm quark 234367E-27 317893E+23 943061E-16 6626069E-34
235543E-27 319489E+23 93835E-16 6626069E-34 298061E-27 404288E+23 741532E-16 6626069E-34 hyperon
802198E-27 10881E+24 27552E-16 6626069E-34 bottom quark 143201E-25 194237E+25 154344E-17 6626069E-34 W
162556E-25 220489E+25 135967E-17 6626069E-34 Z 223368E-25 302974E+25 989499E-18 6626069E-34 Higgs 125eV
256482E-25 347890E+25 861745E-18 6626069E-34 Fermi labs 144gt9eV 311966E-25 423148E+25 708481E-18 6626069E-34 top quark
570389E-16 77367E+34 387493E-27 6626069E-34 Gamma 320gt20eV 185922E-09 25218E+41 118879E-33 6626069E-34 186ltkg 1043gt27eV 217645E-08 29521E+42 101552E-34 6626071E-34 Planck limits
mass frequency radius r=cf mr2f
frequency radius r=cf rf=C 299792458 8c m s 5791781E+09 517617E-02 2997925E+08 smallest photon
328809E+15 911753E-08 2997925E+08 136eV photon 123559E+20 242631E-12 2997925E+08 electron 0511gt6eV 120899E+21 247968E-13 2997925E+08 up quark
241799E+21 123984E-13 2997925E+08 down quark 255485E+22 117343E-14 2997925E+08 muon
326356E+22 918606E-15 2997925E+08 meson 337479E+22 88833E-15 2997925E+08 meson 483598E+22 619921E-15 2997925E+08 strange quark
119379E+23 251128E-15 2997925E+08 meson 120346E+23 249109E-15 2997925E+08 meson 120336E+23 249129E-15 2997925E+08 meson 132699E+23 225919E-15 2997925E+08 meson 22687E+23 132143E-15 2997925E+08 Proton 938gt6eV
227183E+23 131961E-15 2997925E+08 Neutron 939gt6eV 269748E+23 111138E-15 2997925E+08 hyperon
2876E+23 104239E-15 2997925E+08 hyperon 28834E+23 103972E-15 2997925E+08 289549E+23 103538E-15 2997925E+08
314339E+23 953725E-16 2997925E+08 charm quark 317893E+23 943061E-16 2997925E+08
319489E+23 93835E-16 2997925E+08 404288E+23 741532E-16 2997925E+08 hyperon
10881E+24 27552E-16 2997925E+08 bottom quark 194237E+25 154344E-17 2997925E+08 W
220489E+25 135967E-17 2997925E+08 Z 302974E+25 989499E-18 2997925E+08 Higgs 125eV
347890E+25 861745E-18 2997925E+08 Fermi labs 144gt9eV 423148E+25 708481E-18 2997925E+08 top quark
77367E+34 387493E-27 2997925E+08 Gamma 320gt20eV 25218E+41 118879E-33 2997925E+08 1043gt27eV 186lt9kg
29521E+42 101552E-34 2997925E+08 Planck limits
frequency radius r=cf rf=C
Savior Borg
Two helix makes a mass particle
Fernandes
Kanarevrsquos Photon Model
My website Symmetrymathcomis my interpretation of the Authorrsquos I have referenced
I do not claim that any of these authors will agree withmy interpretations
I leave it to anyone who visits my site to decide the logicof my interpretation
- SYMMETRY MATH (SM) By Jack Kuykendall
- Why is Symmetry Math Needed
- From ldquoWhat Is Mathematicsrdquo by Courant and Robbins ndash page 55 Th
- Slide 4
- Slide 5
- Slide 6
- Slide 7
- Slide 8
- Slide 9
- Slide 10
- Slide 11
- Slide 12
- Slide 13
- Slide 14
- Slide 15
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- Slide 22
- Slide 23
- Slide 24
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- Slide 26
- Slide 27
- Slide 28
- Slide 29
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- Slide 31
- Slide 32
- Slide 33
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- Slide 35
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- Slide 37
- Slide 38
- Slide 39
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- Slide 41
- Slide 42
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- Slide 79
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- Slide 81
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- Slide 83
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- Slide 85
- Slide 86
- Slide 87
- Slide 88
- Slide 89
- Slide 90
- Slide 91
- Slide 92
- Slide 93
- Slide 94
- Slide 95
- Slide 96
- Slide 97
- Slide 98
- Slide 99
- Slide 100
- Slide 101
- Slide 102
- Slide 103
- Slide 104
- Slide 105
- Slide 106
- Slide 107
- Slide 108
- Slide 109
- Slide 110
- Slide 111
- Slide 112
- Slide 113
- Slide 114
- Slide 115
- Slide 116
- Slide 117
- Slide 118
- Slide 119
- Slide 120
- Slide 121
- Slide 122
- Slide 123
-
Schwarzschild 3 33
3 (3)(2 30 )
4 (4)(314)(297 3182 19
)og
M kgD
R m
kg
m
(403) 1
Kanarev
3 33
3 (3)(2 30 )
4 (4)(314)(439 )565 27red
g
M kgD
R m
kg
m
(404) 2
Kanarev
3 33 3
3 (3)(2 30 )276 35 (
4 (4)(314)(553 32 Kanrev error
0012)
)lightg
M kg kgD
R m m
kg
m
(405) 3
4
Kanarevrsquos 3 33
3 (3)(2 30)
4 (4)(314)(154 8)013 54gamma
g
MD
R
kg
m
(406) 5
6
Fernandesrsquo (18 36 9) 3 3
3 (3)(2 30)265 81
4 (4)(314)(18 17)rg
kg
m
MD
R
(407)
7
Nuclear density is around3
(12 24) 17kg
m 8
If the Sun were compressed to ( )( 439 )g redR m (equation 400) its gravitational attraction would
detain only radiation of the far infra-red spectrum Shorter wavelengths will not be detained If photons of gamma radiation were detained the gravitational radius of the Sun would have to
be ( )( 154 8 )g gammaR m (equation 402) This would not be possible because the density of
the Sun (equation 406) would be 37 orders more dense than the nucleus of atoms
Only one type of Black-hole can be all black That is one with a gravitational attraction strong enough to detain the highest energy gamma photons All other Black-holes should have colors which vary with the change of wavelengths of the photons which cannot be detained The order should be infra-red light ultra-violet x-ray and finally gamma photons
Mass is Frequency of an Elementary Particle All known experiments have produced answers that agree with the following equations ( ) ( )( )(distance)mad Action mass acceleration
( )mad
( ) ( ) (constant)(frequency)mad hf
(1) 2( ) ( )(constant)mad mc mass
(2)
c f rf (3)
Equating (1) and (2) 2( )m h c f Mass = (constant)(frequency of one particle)
Plankrsquos constant 6626069 34 HBCP 662603(11) 57(29)69 34 codatah Js h Js
2
222
2
6626069 34( )( ) ( )6626069 347372496 51
(299792458 8 ) 898755179 16
Joule
Kgmm sh Js s kgs
m mcs s
Equating (1)(2) and (3)
2 2
constant
expexp exp
1 6626069 34 22102 42
299792458 8
h h c h h JsM f mr kgm
mr c r cc cs
All mass is built from one elementary particle a ( 737 51 )kg particle
All mass is in relative motion with other mass Time is a description invented to measure the relative movement between different particles of
mass Energy is a term invented to describe what happens when masses interact Energy is a
description Mass does not convert to energy and energy does not convert to mass The word ldquoEnergyrdquo will be replaced with an Action involving mass acceleration and distance ( )mad
When the motion of masses interacts they change to different motions of masses These different mass forms can be measured and their changes in momentum can be measured When the change produces photon particles they have been incorrectly labeled as ldquoEnergyrdquo
Mass Frequency Frequency is a rotation per second
If ( )f is set to one rotation per second (737249638 51 )(1kgs rot s the equation predicts that the
7372496 51kg particle is the elementary mass out of which all mass is made
27372496 51 07372496 50
hKgs kg
c
2 07372496 50 second
m m rotationsf
h kgsc
22 392984064 11
0737249638 50 s
kg rotf
kgs
Kanarev Smallest measurable photon
24241 35328803 15
07372496 50 s
kg rotf
kgs
13598443ev photon 2424143lt35kg
91093897 31123559 20
07372496 50 s
kg rotf
kgs
05109906 6eV electron
9109235lt31kg
16726230 27226873 23
07372496 50 s
kg rotf
kgs
93827241gt6eV proton 1672623lt27kg
16749286 272271858 23
07372496 50 s
kg rotf
kgs
9395656gt6eV neutron 1674928lt27kg
334357629 27226750117 23
(07372496 50 ) )2( s
kg rotf
kgs
(Deuterium nucleus) Proton+Neutron
22102 42mr kgm
334357629 274535202583 23
(07372496 50 ( ) s1)
kg rotf
kgs
Proton+Neutron
22102 42mr kgm
223368 253029747 25
07372496 50 s
kg rotf
kgs
1243gt9eV Higgs Boson 2215849lt25 kg
256482lt25347890 25
07372496 50 s
kg rotf
kgs
144gt11eV Fermi Lab particle 2567033lt25kg
5703889152 1677367 34
07372496 50 s
kg rotf
kgs
320gt20eV highest eV particle measured 5704518lt16kg
1859222909 925218 41
07372496 50 s
kg rotf
kgs
10429476 27eV Fernandes 1859223lt9kg
2176450474 8295212 42
07372496 50 s
kg rotf
kgs
Planck limit
78434 28106387 23
07372496 50 s
kg rotf
kgs
classical radius of an electron
This frequency and radius are not a particle radius
This mass does not exist
308 Is there experimental evidences that an electron has the so-called classic radius of an electron A No experiments do not exist From Questions and Answers of Kanarev Page 95 Electrons-Protons-Neutrons JK ET and Rewrite The handbook of Chemistry and Physics shows the classical radius of the electron as (2817 15 )m Kanarevs theory shows this to be the radius of approach of the magnetic force lines around the torus ring of the electron ( )magr This is confirmed by calculating the dimensionless fine structure constant
( ) (which is equal to the circumference of the inner circle (2 )magr divided by the experimentally
measured radius of the electron exp( )R
exp
2 (2)(314)(2817 15 ) 10007297352568
2426 12 137036magr m
R m
(191)
Kg Rotationsec meters experimental classical rmag mass frequency radius r=cf mr mr(11615lt3) 427E-41 5791781E+09 517617E-02 221022E-42 256717E-45 smallest photon
242414E-35 328809E+15 911753E-08 221022E-42 256717E-45 136eV photon 91094E-31 123559E+20 242631E-12 221022E-42 256717E-45 electron 0511gt6eV 891331E-30 120899E+21 247968E-13 221022E-42 256717E-45 up quark
178266E-29 241799E+21 123984E-13 221022E-42 256717E-45 down quark 188356E-28 255485E+22 117343E-14 221022E-42 256717E-45 muon
240606E-28 326356E+22 918606E-15 221022E-42 256717E-45 meson 248806E-28 337479E+22 88833E-15 221022E-42 256717E-45 meson 356532E-28 483598E+22 619921E-15 221022E-42 256717E-45 strange quark
880118E-28 119379E+23 251128E-15 221022E-42 256717E-45 meson 887249E-28 120346E+23 249109E-15 221022E-42 256717E-45 meson 887177E-28 120336E+23 249129E-15 221022E-42 256717E-45 meson 978325E-28 132699E+23 225919E-15 221022E-42 256717E-45 meson 16726E-27 22687E+23 132143E-15 221022E-42 256717E-45 Proton 938gt6eV
167491E-27 227183E+23 131961E-15 221022E-42 256717E-45 Neutron 939gt6eV 198872E-27 269748E+23 111138E-15 221022E-42 256717E-45 hyperon
212033E-27 2876E+23 104239E-15 221022E-42 256717E-45 hyperon 212579E-27 28834E+23 103972E-15 221022E-42 256717E-45 21347E-27 289549E+23 103538E-15 221022E-42 256717E-45
231746E-27 314339E+23 953725E-16 221022E-42 256717E-45 charm quark 234367E-27 317893E+23 943061E-16 221022E-42 256717E-45
235543E-27 319489E+23 93835E-16 221022E-42 256717E-45 298061E-27 404288E+23 741532E-16 221022E-42 256717E-45 hyperon
802198E-27 10881E+24 27552E-16 221022E-42 256717E-45 bottom quark 143201E-25 194237E+25 154344E-17 221022E-42 256717E-45 W
162556E-25 220489E+25 135967E-17 221022E-42 256717E-45 Z 223368E-25 302974E+25 989499E-18 221022E-42 256717E-45 Higgs125gt9eV
256482E-25 347890E+25 861745E-18 221022E-42 256717E-45 Fermi labs 144gt9eV 311966E-25 423148E+25 708481E-18 221022E-42 256717E-45 top quark
570389E-16 77367E+34 387494E-27 221022E-42 256717E-45 Gamma 320gt20eV 185922E-09 25218E+41 118879E-33 221022E-42 256717E-45 1043gt27eV 186lt9kg
217645E-08 29521E+42 101552E-34 221022E-42 256717E-45 Planck limits
mass frequency radius r=cf mr mr(11615lt3)
mass frequency radius r=cf mr2f 222E-39 298E+11 100E-03 6626069E-34 smallest photon
242414E-35 328809E+15 911753E-08 6626069E-34 Photon 136eV 91094E-31 123559E+20 242631E-12 6626069E-34 electron 0511gt6eV 891331E-30 120899E+21 247968E-13 6626069E-34 up quark
178266E-29 241799E+21 123984E-13 6626069E-34 down quark 188356E-28 255485E+22 117343E-14 6626069E-34 muon
240606E-28 326356E+22 918606E-15 6626069E-34 meson 248806E-28 337479E+22 88833E-15 6626069E-34 meson 356532E-28 483598E+22 619921E-15 6626069E-34 strange quark
880118E-28 119379E+23 251128E-15 6626069E-34 meson 887249E-28 120346E+23 249109E-15 6626069E-34 meson 887177E-28 120336E+23 249129E-15 6626069E-34 meson 978325E-28 132699E+23 225919E-15 6626069E-34 meson 16726E-27 22687E+23 132143E-15 6626069E-34 Proton 938gt6eV
167491E-27 227183E+23 131961E-15 6626069E-34 Neutron 939gt6eV 198872E-27 269748E+23 111138E-15 6626069E-34 hyperon
212033E-27 2876E+23 104239E-15 6626069E-34 hyperon 212579E-27 28834E+23 103972E-15 6626069E-34 21347E-27 289549E+23 103538E-15 6626069E-34
231746E-27 314339E+23 953725E-16 6626069E-34 charm quark 234367E-27 317893E+23 943061E-16 6626069E-34
235543E-27 319489E+23 93835E-16 6626069E-34 298061E-27 404288E+23 741532E-16 6626069E-34 hyperon
802198E-27 10881E+24 27552E-16 6626069E-34 bottom quark 143201E-25 194237E+25 154344E-17 6626069E-34 W
162556E-25 220489E+25 135967E-17 6626069E-34 Z 223368E-25 302974E+25 989499E-18 6626069E-34 Higgs 125eV
256482E-25 347890E+25 861745E-18 6626069E-34 Fermi labs 144gt9eV 311966E-25 423148E+25 708481E-18 6626069E-34 top quark
570389E-16 77367E+34 387493E-27 6626069E-34 Gamma 320gt20eV 185922E-09 25218E+41 118879E-33 6626069E-34 186ltkg 1043gt27eV 217645E-08 29521E+42 101552E-34 6626071E-34 Planck limits
mass frequency radius r=cf mr2f
frequency radius r=cf rf=C 299792458 8c m s 5791781E+09 517617E-02 2997925E+08 smallest photon
328809E+15 911753E-08 2997925E+08 136eV photon 123559E+20 242631E-12 2997925E+08 electron 0511gt6eV 120899E+21 247968E-13 2997925E+08 up quark
241799E+21 123984E-13 2997925E+08 down quark 255485E+22 117343E-14 2997925E+08 muon
326356E+22 918606E-15 2997925E+08 meson 337479E+22 88833E-15 2997925E+08 meson 483598E+22 619921E-15 2997925E+08 strange quark
119379E+23 251128E-15 2997925E+08 meson 120346E+23 249109E-15 2997925E+08 meson 120336E+23 249129E-15 2997925E+08 meson 132699E+23 225919E-15 2997925E+08 meson 22687E+23 132143E-15 2997925E+08 Proton 938gt6eV
227183E+23 131961E-15 2997925E+08 Neutron 939gt6eV 269748E+23 111138E-15 2997925E+08 hyperon
2876E+23 104239E-15 2997925E+08 hyperon 28834E+23 103972E-15 2997925E+08 289549E+23 103538E-15 2997925E+08
314339E+23 953725E-16 2997925E+08 charm quark 317893E+23 943061E-16 2997925E+08
319489E+23 93835E-16 2997925E+08 404288E+23 741532E-16 2997925E+08 hyperon
10881E+24 27552E-16 2997925E+08 bottom quark 194237E+25 154344E-17 2997925E+08 W
220489E+25 135967E-17 2997925E+08 Z 302974E+25 989499E-18 2997925E+08 Higgs 125eV
347890E+25 861745E-18 2997925E+08 Fermi labs 144gt9eV 423148E+25 708481E-18 2997925E+08 top quark
77367E+34 387493E-27 2997925E+08 Gamma 320gt20eV 25218E+41 118879E-33 2997925E+08 1043gt27eV 186lt9kg
29521E+42 101552E-34 2997925E+08 Planck limits
frequency radius r=cf rf=C
Savior Borg
Two helix makes a mass particle
Fernandes
Kanarevrsquos Photon Model
My website Symmetrymathcomis my interpretation of the Authorrsquos I have referenced
I do not claim that any of these authors will agree withmy interpretations
I leave it to anyone who visits my site to decide the logicof my interpretation
- SYMMETRY MATH (SM) By Jack Kuykendall
- Why is Symmetry Math Needed
- From ldquoWhat Is Mathematicsrdquo by Courant and Robbins ndash page 55 Th
- Slide 4
- Slide 5
- Slide 6
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-
If the Sun were compressed to ( )( 439 )g redR m (equation 400) its gravitational attraction would
detain only radiation of the far infra-red spectrum Shorter wavelengths will not be detained If photons of gamma radiation were detained the gravitational radius of the Sun would have to
be ( )( 154 8 )g gammaR m (equation 402) This would not be possible because the density of
the Sun (equation 406) would be 37 orders more dense than the nucleus of atoms
Only one type of Black-hole can be all black That is one with a gravitational attraction strong enough to detain the highest energy gamma photons All other Black-holes should have colors which vary with the change of wavelengths of the photons which cannot be detained The order should be infra-red light ultra-violet x-ray and finally gamma photons
Mass is Frequency of an Elementary Particle All known experiments have produced answers that agree with the following equations ( ) ( )( )(distance)mad Action mass acceleration
( )mad
( ) ( ) (constant)(frequency)mad hf
(1) 2( ) ( )(constant)mad mc mass
(2)
c f rf (3)
Equating (1) and (2) 2( )m h c f Mass = (constant)(frequency of one particle)
Plankrsquos constant 6626069 34 HBCP 662603(11) 57(29)69 34 codatah Js h Js
2
222
2
6626069 34( )( ) ( )6626069 347372496 51
(299792458 8 ) 898755179 16
Joule
Kgmm sh Js s kgs
m mcs s
Equating (1)(2) and (3)
2 2
constant
expexp exp
1 6626069 34 22102 42
299792458 8
h h c h h JsM f mr kgm
mr c r cc cs
All mass is built from one elementary particle a ( 737 51 )kg particle
All mass is in relative motion with other mass Time is a description invented to measure the relative movement between different particles of
mass Energy is a term invented to describe what happens when masses interact Energy is a
description Mass does not convert to energy and energy does not convert to mass The word ldquoEnergyrdquo will be replaced with an Action involving mass acceleration and distance ( )mad
When the motion of masses interacts they change to different motions of masses These different mass forms can be measured and their changes in momentum can be measured When the change produces photon particles they have been incorrectly labeled as ldquoEnergyrdquo
Mass Frequency Frequency is a rotation per second
If ( )f is set to one rotation per second (737249638 51 )(1kgs rot s the equation predicts that the
7372496 51kg particle is the elementary mass out of which all mass is made
27372496 51 07372496 50
hKgs kg
c
2 07372496 50 second
m m rotationsf
h kgsc
22 392984064 11
0737249638 50 s
kg rotf
kgs
Kanarev Smallest measurable photon
24241 35328803 15
07372496 50 s
kg rotf
kgs
13598443ev photon 2424143lt35kg
91093897 31123559 20
07372496 50 s
kg rotf
kgs
05109906 6eV electron
9109235lt31kg
16726230 27226873 23
07372496 50 s
kg rotf
kgs
93827241gt6eV proton 1672623lt27kg
16749286 272271858 23
07372496 50 s
kg rotf
kgs
9395656gt6eV neutron 1674928lt27kg
334357629 27226750117 23
(07372496 50 ) )2( s
kg rotf
kgs
(Deuterium nucleus) Proton+Neutron
22102 42mr kgm
334357629 274535202583 23
(07372496 50 ( ) s1)
kg rotf
kgs
Proton+Neutron
22102 42mr kgm
223368 253029747 25
07372496 50 s
kg rotf
kgs
1243gt9eV Higgs Boson 2215849lt25 kg
256482lt25347890 25
07372496 50 s
kg rotf
kgs
144gt11eV Fermi Lab particle 2567033lt25kg
5703889152 1677367 34
07372496 50 s
kg rotf
kgs
320gt20eV highest eV particle measured 5704518lt16kg
1859222909 925218 41
07372496 50 s
kg rotf
kgs
10429476 27eV Fernandes 1859223lt9kg
2176450474 8295212 42
07372496 50 s
kg rotf
kgs
Planck limit
78434 28106387 23
07372496 50 s
kg rotf
kgs
classical radius of an electron
This frequency and radius are not a particle radius
This mass does not exist
308 Is there experimental evidences that an electron has the so-called classic radius of an electron A No experiments do not exist From Questions and Answers of Kanarev Page 95 Electrons-Protons-Neutrons JK ET and Rewrite The handbook of Chemistry and Physics shows the classical radius of the electron as (2817 15 )m Kanarevs theory shows this to be the radius of approach of the magnetic force lines around the torus ring of the electron ( )magr This is confirmed by calculating the dimensionless fine structure constant
( ) (which is equal to the circumference of the inner circle (2 )magr divided by the experimentally
measured radius of the electron exp( )R
exp
2 (2)(314)(2817 15 ) 10007297352568
2426 12 137036magr m
R m
(191)
Kg Rotationsec meters experimental classical rmag mass frequency radius r=cf mr mr(11615lt3) 427E-41 5791781E+09 517617E-02 221022E-42 256717E-45 smallest photon
242414E-35 328809E+15 911753E-08 221022E-42 256717E-45 136eV photon 91094E-31 123559E+20 242631E-12 221022E-42 256717E-45 electron 0511gt6eV 891331E-30 120899E+21 247968E-13 221022E-42 256717E-45 up quark
178266E-29 241799E+21 123984E-13 221022E-42 256717E-45 down quark 188356E-28 255485E+22 117343E-14 221022E-42 256717E-45 muon
240606E-28 326356E+22 918606E-15 221022E-42 256717E-45 meson 248806E-28 337479E+22 88833E-15 221022E-42 256717E-45 meson 356532E-28 483598E+22 619921E-15 221022E-42 256717E-45 strange quark
880118E-28 119379E+23 251128E-15 221022E-42 256717E-45 meson 887249E-28 120346E+23 249109E-15 221022E-42 256717E-45 meson 887177E-28 120336E+23 249129E-15 221022E-42 256717E-45 meson 978325E-28 132699E+23 225919E-15 221022E-42 256717E-45 meson 16726E-27 22687E+23 132143E-15 221022E-42 256717E-45 Proton 938gt6eV
167491E-27 227183E+23 131961E-15 221022E-42 256717E-45 Neutron 939gt6eV 198872E-27 269748E+23 111138E-15 221022E-42 256717E-45 hyperon
212033E-27 2876E+23 104239E-15 221022E-42 256717E-45 hyperon 212579E-27 28834E+23 103972E-15 221022E-42 256717E-45 21347E-27 289549E+23 103538E-15 221022E-42 256717E-45
231746E-27 314339E+23 953725E-16 221022E-42 256717E-45 charm quark 234367E-27 317893E+23 943061E-16 221022E-42 256717E-45
235543E-27 319489E+23 93835E-16 221022E-42 256717E-45 298061E-27 404288E+23 741532E-16 221022E-42 256717E-45 hyperon
802198E-27 10881E+24 27552E-16 221022E-42 256717E-45 bottom quark 143201E-25 194237E+25 154344E-17 221022E-42 256717E-45 W
162556E-25 220489E+25 135967E-17 221022E-42 256717E-45 Z 223368E-25 302974E+25 989499E-18 221022E-42 256717E-45 Higgs125gt9eV
256482E-25 347890E+25 861745E-18 221022E-42 256717E-45 Fermi labs 144gt9eV 311966E-25 423148E+25 708481E-18 221022E-42 256717E-45 top quark
570389E-16 77367E+34 387494E-27 221022E-42 256717E-45 Gamma 320gt20eV 185922E-09 25218E+41 118879E-33 221022E-42 256717E-45 1043gt27eV 186lt9kg
217645E-08 29521E+42 101552E-34 221022E-42 256717E-45 Planck limits
mass frequency radius r=cf mr mr(11615lt3)
mass frequency radius r=cf mr2f 222E-39 298E+11 100E-03 6626069E-34 smallest photon
242414E-35 328809E+15 911753E-08 6626069E-34 Photon 136eV 91094E-31 123559E+20 242631E-12 6626069E-34 electron 0511gt6eV 891331E-30 120899E+21 247968E-13 6626069E-34 up quark
178266E-29 241799E+21 123984E-13 6626069E-34 down quark 188356E-28 255485E+22 117343E-14 6626069E-34 muon
240606E-28 326356E+22 918606E-15 6626069E-34 meson 248806E-28 337479E+22 88833E-15 6626069E-34 meson 356532E-28 483598E+22 619921E-15 6626069E-34 strange quark
880118E-28 119379E+23 251128E-15 6626069E-34 meson 887249E-28 120346E+23 249109E-15 6626069E-34 meson 887177E-28 120336E+23 249129E-15 6626069E-34 meson 978325E-28 132699E+23 225919E-15 6626069E-34 meson 16726E-27 22687E+23 132143E-15 6626069E-34 Proton 938gt6eV
167491E-27 227183E+23 131961E-15 6626069E-34 Neutron 939gt6eV 198872E-27 269748E+23 111138E-15 6626069E-34 hyperon
212033E-27 2876E+23 104239E-15 6626069E-34 hyperon 212579E-27 28834E+23 103972E-15 6626069E-34 21347E-27 289549E+23 103538E-15 6626069E-34
231746E-27 314339E+23 953725E-16 6626069E-34 charm quark 234367E-27 317893E+23 943061E-16 6626069E-34
235543E-27 319489E+23 93835E-16 6626069E-34 298061E-27 404288E+23 741532E-16 6626069E-34 hyperon
802198E-27 10881E+24 27552E-16 6626069E-34 bottom quark 143201E-25 194237E+25 154344E-17 6626069E-34 W
162556E-25 220489E+25 135967E-17 6626069E-34 Z 223368E-25 302974E+25 989499E-18 6626069E-34 Higgs 125eV
256482E-25 347890E+25 861745E-18 6626069E-34 Fermi labs 144gt9eV 311966E-25 423148E+25 708481E-18 6626069E-34 top quark
570389E-16 77367E+34 387493E-27 6626069E-34 Gamma 320gt20eV 185922E-09 25218E+41 118879E-33 6626069E-34 186ltkg 1043gt27eV 217645E-08 29521E+42 101552E-34 6626071E-34 Planck limits
mass frequency radius r=cf mr2f
frequency radius r=cf rf=C 299792458 8c m s 5791781E+09 517617E-02 2997925E+08 smallest photon
328809E+15 911753E-08 2997925E+08 136eV photon 123559E+20 242631E-12 2997925E+08 electron 0511gt6eV 120899E+21 247968E-13 2997925E+08 up quark
241799E+21 123984E-13 2997925E+08 down quark 255485E+22 117343E-14 2997925E+08 muon
326356E+22 918606E-15 2997925E+08 meson 337479E+22 88833E-15 2997925E+08 meson 483598E+22 619921E-15 2997925E+08 strange quark
119379E+23 251128E-15 2997925E+08 meson 120346E+23 249109E-15 2997925E+08 meson 120336E+23 249129E-15 2997925E+08 meson 132699E+23 225919E-15 2997925E+08 meson 22687E+23 132143E-15 2997925E+08 Proton 938gt6eV
227183E+23 131961E-15 2997925E+08 Neutron 939gt6eV 269748E+23 111138E-15 2997925E+08 hyperon
2876E+23 104239E-15 2997925E+08 hyperon 28834E+23 103972E-15 2997925E+08 289549E+23 103538E-15 2997925E+08
314339E+23 953725E-16 2997925E+08 charm quark 317893E+23 943061E-16 2997925E+08
319489E+23 93835E-16 2997925E+08 404288E+23 741532E-16 2997925E+08 hyperon
10881E+24 27552E-16 2997925E+08 bottom quark 194237E+25 154344E-17 2997925E+08 W
220489E+25 135967E-17 2997925E+08 Z 302974E+25 989499E-18 2997925E+08 Higgs 125eV
347890E+25 861745E-18 2997925E+08 Fermi labs 144gt9eV 423148E+25 708481E-18 2997925E+08 top quark
77367E+34 387493E-27 2997925E+08 Gamma 320gt20eV 25218E+41 118879E-33 2997925E+08 1043gt27eV 186lt9kg
29521E+42 101552E-34 2997925E+08 Planck limits
frequency radius r=cf rf=C
Savior Borg
Two helix makes a mass particle
Fernandes
Kanarevrsquos Photon Model
My website Symmetrymathcomis my interpretation of the Authorrsquos I have referenced
I do not claim that any of these authors will agree withmy interpretations
I leave it to anyone who visits my site to decide the logicof my interpretation
- SYMMETRY MATH (SM) By Jack Kuykendall
- Why is Symmetry Math Needed
- From ldquoWhat Is Mathematicsrdquo by Courant and Robbins ndash page 55 Th
- Slide 4
- Slide 5
- Slide 6
- Slide 7
- Slide 8
- Slide 9
- Slide 10
- Slide 11
- Slide 12
- Slide 13
- Slide 14
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- Slide 35
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- Slide 37
- Slide 38
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- Slide 41
- Slide 42
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- Slide 44
- Slide 45
- Slide 46
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-
Mass is Frequency of an Elementary Particle All known experiments have produced answers that agree with the following equations ( ) ( )( )(distance)mad Action mass acceleration
( )mad
( ) ( ) (constant)(frequency)mad hf
(1) 2( ) ( )(constant)mad mc mass
(2)
c f rf (3)
Equating (1) and (2) 2( )m h c f Mass = (constant)(frequency of one particle)
Plankrsquos constant 6626069 34 HBCP 662603(11) 57(29)69 34 codatah Js h Js
2
222
2
6626069 34( )( ) ( )6626069 347372496 51
(299792458 8 ) 898755179 16
Joule
Kgmm sh Js s kgs
m mcs s
Equating (1)(2) and (3)
2 2
constant
expexp exp
1 6626069 34 22102 42
299792458 8
h h c h h JsM f mr kgm
mr c r cc cs
All mass is built from one elementary particle a ( 737 51 )kg particle
All mass is in relative motion with other mass Time is a description invented to measure the relative movement between different particles of
mass Energy is a term invented to describe what happens when masses interact Energy is a
description Mass does not convert to energy and energy does not convert to mass The word ldquoEnergyrdquo will be replaced with an Action involving mass acceleration and distance ( )mad
When the motion of masses interacts they change to different motions of masses These different mass forms can be measured and their changes in momentum can be measured When the change produces photon particles they have been incorrectly labeled as ldquoEnergyrdquo
Mass Frequency Frequency is a rotation per second
If ( )f is set to one rotation per second (737249638 51 )(1kgs rot s the equation predicts that the
7372496 51kg particle is the elementary mass out of which all mass is made
27372496 51 07372496 50
hKgs kg
c
2 07372496 50 second
m m rotationsf
h kgsc
22 392984064 11
0737249638 50 s
kg rotf
kgs
Kanarev Smallest measurable photon
24241 35328803 15
07372496 50 s
kg rotf
kgs
13598443ev photon 2424143lt35kg
91093897 31123559 20
07372496 50 s
kg rotf
kgs
05109906 6eV electron
9109235lt31kg
16726230 27226873 23
07372496 50 s
kg rotf
kgs
93827241gt6eV proton 1672623lt27kg
16749286 272271858 23
07372496 50 s
kg rotf
kgs
9395656gt6eV neutron 1674928lt27kg
334357629 27226750117 23
(07372496 50 ) )2( s
kg rotf
kgs
(Deuterium nucleus) Proton+Neutron
22102 42mr kgm
334357629 274535202583 23
(07372496 50 ( ) s1)
kg rotf
kgs
Proton+Neutron
22102 42mr kgm
223368 253029747 25
07372496 50 s
kg rotf
kgs
1243gt9eV Higgs Boson 2215849lt25 kg
256482lt25347890 25
07372496 50 s
kg rotf
kgs
144gt11eV Fermi Lab particle 2567033lt25kg
5703889152 1677367 34
07372496 50 s
kg rotf
kgs
320gt20eV highest eV particle measured 5704518lt16kg
1859222909 925218 41
07372496 50 s
kg rotf
kgs
10429476 27eV Fernandes 1859223lt9kg
2176450474 8295212 42
07372496 50 s
kg rotf
kgs
Planck limit
78434 28106387 23
07372496 50 s
kg rotf
kgs
classical radius of an electron
This frequency and radius are not a particle radius
This mass does not exist
308 Is there experimental evidences that an electron has the so-called classic radius of an electron A No experiments do not exist From Questions and Answers of Kanarev Page 95 Electrons-Protons-Neutrons JK ET and Rewrite The handbook of Chemistry and Physics shows the classical radius of the electron as (2817 15 )m Kanarevs theory shows this to be the radius of approach of the magnetic force lines around the torus ring of the electron ( )magr This is confirmed by calculating the dimensionless fine structure constant
( ) (which is equal to the circumference of the inner circle (2 )magr divided by the experimentally
measured radius of the electron exp( )R
exp
2 (2)(314)(2817 15 ) 10007297352568
2426 12 137036magr m
R m
(191)
Kg Rotationsec meters experimental classical rmag mass frequency radius r=cf mr mr(11615lt3) 427E-41 5791781E+09 517617E-02 221022E-42 256717E-45 smallest photon
242414E-35 328809E+15 911753E-08 221022E-42 256717E-45 136eV photon 91094E-31 123559E+20 242631E-12 221022E-42 256717E-45 electron 0511gt6eV 891331E-30 120899E+21 247968E-13 221022E-42 256717E-45 up quark
178266E-29 241799E+21 123984E-13 221022E-42 256717E-45 down quark 188356E-28 255485E+22 117343E-14 221022E-42 256717E-45 muon
240606E-28 326356E+22 918606E-15 221022E-42 256717E-45 meson 248806E-28 337479E+22 88833E-15 221022E-42 256717E-45 meson 356532E-28 483598E+22 619921E-15 221022E-42 256717E-45 strange quark
880118E-28 119379E+23 251128E-15 221022E-42 256717E-45 meson 887249E-28 120346E+23 249109E-15 221022E-42 256717E-45 meson 887177E-28 120336E+23 249129E-15 221022E-42 256717E-45 meson 978325E-28 132699E+23 225919E-15 221022E-42 256717E-45 meson 16726E-27 22687E+23 132143E-15 221022E-42 256717E-45 Proton 938gt6eV
167491E-27 227183E+23 131961E-15 221022E-42 256717E-45 Neutron 939gt6eV 198872E-27 269748E+23 111138E-15 221022E-42 256717E-45 hyperon
212033E-27 2876E+23 104239E-15 221022E-42 256717E-45 hyperon 212579E-27 28834E+23 103972E-15 221022E-42 256717E-45 21347E-27 289549E+23 103538E-15 221022E-42 256717E-45
231746E-27 314339E+23 953725E-16 221022E-42 256717E-45 charm quark 234367E-27 317893E+23 943061E-16 221022E-42 256717E-45
235543E-27 319489E+23 93835E-16 221022E-42 256717E-45 298061E-27 404288E+23 741532E-16 221022E-42 256717E-45 hyperon
802198E-27 10881E+24 27552E-16 221022E-42 256717E-45 bottom quark 143201E-25 194237E+25 154344E-17 221022E-42 256717E-45 W
162556E-25 220489E+25 135967E-17 221022E-42 256717E-45 Z 223368E-25 302974E+25 989499E-18 221022E-42 256717E-45 Higgs125gt9eV
256482E-25 347890E+25 861745E-18 221022E-42 256717E-45 Fermi labs 144gt9eV 311966E-25 423148E+25 708481E-18 221022E-42 256717E-45 top quark
570389E-16 77367E+34 387494E-27 221022E-42 256717E-45 Gamma 320gt20eV 185922E-09 25218E+41 118879E-33 221022E-42 256717E-45 1043gt27eV 186lt9kg
217645E-08 29521E+42 101552E-34 221022E-42 256717E-45 Planck limits
mass frequency radius r=cf mr mr(11615lt3)
mass frequency radius r=cf mr2f 222E-39 298E+11 100E-03 6626069E-34 smallest photon
242414E-35 328809E+15 911753E-08 6626069E-34 Photon 136eV 91094E-31 123559E+20 242631E-12 6626069E-34 electron 0511gt6eV 891331E-30 120899E+21 247968E-13 6626069E-34 up quark
178266E-29 241799E+21 123984E-13 6626069E-34 down quark 188356E-28 255485E+22 117343E-14 6626069E-34 muon
240606E-28 326356E+22 918606E-15 6626069E-34 meson 248806E-28 337479E+22 88833E-15 6626069E-34 meson 356532E-28 483598E+22 619921E-15 6626069E-34 strange quark
880118E-28 119379E+23 251128E-15 6626069E-34 meson 887249E-28 120346E+23 249109E-15 6626069E-34 meson 887177E-28 120336E+23 249129E-15 6626069E-34 meson 978325E-28 132699E+23 225919E-15 6626069E-34 meson 16726E-27 22687E+23 132143E-15 6626069E-34 Proton 938gt6eV
167491E-27 227183E+23 131961E-15 6626069E-34 Neutron 939gt6eV 198872E-27 269748E+23 111138E-15 6626069E-34 hyperon
212033E-27 2876E+23 104239E-15 6626069E-34 hyperon 212579E-27 28834E+23 103972E-15 6626069E-34 21347E-27 289549E+23 103538E-15 6626069E-34
231746E-27 314339E+23 953725E-16 6626069E-34 charm quark 234367E-27 317893E+23 943061E-16 6626069E-34
235543E-27 319489E+23 93835E-16 6626069E-34 298061E-27 404288E+23 741532E-16 6626069E-34 hyperon
802198E-27 10881E+24 27552E-16 6626069E-34 bottom quark 143201E-25 194237E+25 154344E-17 6626069E-34 W
162556E-25 220489E+25 135967E-17 6626069E-34 Z 223368E-25 302974E+25 989499E-18 6626069E-34 Higgs 125eV
256482E-25 347890E+25 861745E-18 6626069E-34 Fermi labs 144gt9eV 311966E-25 423148E+25 708481E-18 6626069E-34 top quark
570389E-16 77367E+34 387493E-27 6626069E-34 Gamma 320gt20eV 185922E-09 25218E+41 118879E-33 6626069E-34 186ltkg 1043gt27eV 217645E-08 29521E+42 101552E-34 6626071E-34 Planck limits
mass frequency radius r=cf mr2f
frequency radius r=cf rf=C 299792458 8c m s 5791781E+09 517617E-02 2997925E+08 smallest photon
328809E+15 911753E-08 2997925E+08 136eV photon 123559E+20 242631E-12 2997925E+08 electron 0511gt6eV 120899E+21 247968E-13 2997925E+08 up quark
241799E+21 123984E-13 2997925E+08 down quark 255485E+22 117343E-14 2997925E+08 muon
326356E+22 918606E-15 2997925E+08 meson 337479E+22 88833E-15 2997925E+08 meson 483598E+22 619921E-15 2997925E+08 strange quark
119379E+23 251128E-15 2997925E+08 meson 120346E+23 249109E-15 2997925E+08 meson 120336E+23 249129E-15 2997925E+08 meson 132699E+23 225919E-15 2997925E+08 meson 22687E+23 132143E-15 2997925E+08 Proton 938gt6eV
227183E+23 131961E-15 2997925E+08 Neutron 939gt6eV 269748E+23 111138E-15 2997925E+08 hyperon
2876E+23 104239E-15 2997925E+08 hyperon 28834E+23 103972E-15 2997925E+08 289549E+23 103538E-15 2997925E+08
314339E+23 953725E-16 2997925E+08 charm quark 317893E+23 943061E-16 2997925E+08
319489E+23 93835E-16 2997925E+08 404288E+23 741532E-16 2997925E+08 hyperon
10881E+24 27552E-16 2997925E+08 bottom quark 194237E+25 154344E-17 2997925E+08 W
220489E+25 135967E-17 2997925E+08 Z 302974E+25 989499E-18 2997925E+08 Higgs 125eV
347890E+25 861745E-18 2997925E+08 Fermi labs 144gt9eV 423148E+25 708481E-18 2997925E+08 top quark
77367E+34 387493E-27 2997925E+08 Gamma 320gt20eV 25218E+41 118879E-33 2997925E+08 1043gt27eV 186lt9kg
29521E+42 101552E-34 2997925E+08 Planck limits
frequency radius r=cf rf=C
Savior Borg
Two helix makes a mass particle
Fernandes
Kanarevrsquos Photon Model
My website Symmetrymathcomis my interpretation of the Authorrsquos I have referenced
I do not claim that any of these authors will agree withmy interpretations
I leave it to anyone who visits my site to decide the logicof my interpretation
- SYMMETRY MATH (SM) By Jack Kuykendall
- Why is Symmetry Math Needed
- From ldquoWhat Is Mathematicsrdquo by Courant and Robbins ndash page 55 Th
- Slide 4
- Slide 5
- Slide 6
- Slide 7
- Slide 8
- Slide 9
- Slide 10
- Slide 11
- Slide 12
- Slide 13
- Slide 14
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- Slide 95
- Slide 96
- Slide 97
- Slide 98
- Slide 99
- Slide 100
- Slide 101
- Slide 102
- Slide 103
- Slide 104
- Slide 105
- Slide 106
- Slide 107
- Slide 108
- Slide 109
- Slide 110
- Slide 111
- Slide 112
- Slide 113
- Slide 114
- Slide 115
- Slide 116
- Slide 117
- Slide 118
- Slide 119
- Slide 120
- Slide 121
- Slide 122
- Slide 123
-
Equating (1)(2) and (3)
2 2
constant
expexp exp
1 6626069 34 22102 42
299792458 8
h h c h h JsM f mr kgm
mr c r cc cs
All mass is built from one elementary particle a ( 737 51 )kg particle
All mass is in relative motion with other mass Time is a description invented to measure the relative movement between different particles of
mass Energy is a term invented to describe what happens when masses interact Energy is a
description Mass does not convert to energy and energy does not convert to mass The word ldquoEnergyrdquo will be replaced with an Action involving mass acceleration and distance ( )mad
When the motion of masses interacts they change to different motions of masses These different mass forms can be measured and their changes in momentum can be measured When the change produces photon particles they have been incorrectly labeled as ldquoEnergyrdquo
Mass Frequency Frequency is a rotation per second
If ( )f is set to one rotation per second (737249638 51 )(1kgs rot s the equation predicts that the
7372496 51kg particle is the elementary mass out of which all mass is made
27372496 51 07372496 50
hKgs kg
c
2 07372496 50 second
m m rotationsf
h kgsc
22 392984064 11
0737249638 50 s
kg rotf
kgs
Kanarev Smallest measurable photon
24241 35328803 15
07372496 50 s
kg rotf
kgs
13598443ev photon 2424143lt35kg
91093897 31123559 20
07372496 50 s
kg rotf
kgs
05109906 6eV electron
9109235lt31kg
16726230 27226873 23
07372496 50 s
kg rotf
kgs
93827241gt6eV proton 1672623lt27kg
16749286 272271858 23
07372496 50 s
kg rotf
kgs
9395656gt6eV neutron 1674928lt27kg
334357629 27226750117 23
(07372496 50 ) )2( s
kg rotf
kgs
(Deuterium nucleus) Proton+Neutron
22102 42mr kgm
334357629 274535202583 23
(07372496 50 ( ) s1)
kg rotf
kgs
Proton+Neutron
22102 42mr kgm
223368 253029747 25
07372496 50 s
kg rotf
kgs
1243gt9eV Higgs Boson 2215849lt25 kg
256482lt25347890 25
07372496 50 s
kg rotf
kgs
144gt11eV Fermi Lab particle 2567033lt25kg
5703889152 1677367 34
07372496 50 s
kg rotf
kgs
320gt20eV highest eV particle measured 5704518lt16kg
1859222909 925218 41
07372496 50 s
kg rotf
kgs
10429476 27eV Fernandes 1859223lt9kg
2176450474 8295212 42
07372496 50 s
kg rotf
kgs
Planck limit
78434 28106387 23
07372496 50 s
kg rotf
kgs
classical radius of an electron
This frequency and radius are not a particle radius
This mass does not exist
308 Is there experimental evidences that an electron has the so-called classic radius of an electron A No experiments do not exist From Questions and Answers of Kanarev Page 95 Electrons-Protons-Neutrons JK ET and Rewrite The handbook of Chemistry and Physics shows the classical radius of the electron as (2817 15 )m Kanarevs theory shows this to be the radius of approach of the magnetic force lines around the torus ring of the electron ( )magr This is confirmed by calculating the dimensionless fine structure constant
( ) (which is equal to the circumference of the inner circle (2 )magr divided by the experimentally
measured radius of the electron exp( )R
exp
2 (2)(314)(2817 15 ) 10007297352568
2426 12 137036magr m
R m
(191)
Kg Rotationsec meters experimental classical rmag mass frequency radius r=cf mr mr(11615lt3) 427E-41 5791781E+09 517617E-02 221022E-42 256717E-45 smallest photon
242414E-35 328809E+15 911753E-08 221022E-42 256717E-45 136eV photon 91094E-31 123559E+20 242631E-12 221022E-42 256717E-45 electron 0511gt6eV 891331E-30 120899E+21 247968E-13 221022E-42 256717E-45 up quark
178266E-29 241799E+21 123984E-13 221022E-42 256717E-45 down quark 188356E-28 255485E+22 117343E-14 221022E-42 256717E-45 muon
240606E-28 326356E+22 918606E-15 221022E-42 256717E-45 meson 248806E-28 337479E+22 88833E-15 221022E-42 256717E-45 meson 356532E-28 483598E+22 619921E-15 221022E-42 256717E-45 strange quark
880118E-28 119379E+23 251128E-15 221022E-42 256717E-45 meson 887249E-28 120346E+23 249109E-15 221022E-42 256717E-45 meson 887177E-28 120336E+23 249129E-15 221022E-42 256717E-45 meson 978325E-28 132699E+23 225919E-15 221022E-42 256717E-45 meson 16726E-27 22687E+23 132143E-15 221022E-42 256717E-45 Proton 938gt6eV
167491E-27 227183E+23 131961E-15 221022E-42 256717E-45 Neutron 939gt6eV 198872E-27 269748E+23 111138E-15 221022E-42 256717E-45 hyperon
212033E-27 2876E+23 104239E-15 221022E-42 256717E-45 hyperon 212579E-27 28834E+23 103972E-15 221022E-42 256717E-45 21347E-27 289549E+23 103538E-15 221022E-42 256717E-45
231746E-27 314339E+23 953725E-16 221022E-42 256717E-45 charm quark 234367E-27 317893E+23 943061E-16 221022E-42 256717E-45
235543E-27 319489E+23 93835E-16 221022E-42 256717E-45 298061E-27 404288E+23 741532E-16 221022E-42 256717E-45 hyperon
802198E-27 10881E+24 27552E-16 221022E-42 256717E-45 bottom quark 143201E-25 194237E+25 154344E-17 221022E-42 256717E-45 W
162556E-25 220489E+25 135967E-17 221022E-42 256717E-45 Z 223368E-25 302974E+25 989499E-18 221022E-42 256717E-45 Higgs125gt9eV
256482E-25 347890E+25 861745E-18 221022E-42 256717E-45 Fermi labs 144gt9eV 311966E-25 423148E+25 708481E-18 221022E-42 256717E-45 top quark
570389E-16 77367E+34 387494E-27 221022E-42 256717E-45 Gamma 320gt20eV 185922E-09 25218E+41 118879E-33 221022E-42 256717E-45 1043gt27eV 186lt9kg
217645E-08 29521E+42 101552E-34 221022E-42 256717E-45 Planck limits
mass frequency radius r=cf mr mr(11615lt3)
mass frequency radius r=cf mr2f 222E-39 298E+11 100E-03 6626069E-34 smallest photon
242414E-35 328809E+15 911753E-08 6626069E-34 Photon 136eV 91094E-31 123559E+20 242631E-12 6626069E-34 electron 0511gt6eV 891331E-30 120899E+21 247968E-13 6626069E-34 up quark
178266E-29 241799E+21 123984E-13 6626069E-34 down quark 188356E-28 255485E+22 117343E-14 6626069E-34 muon
240606E-28 326356E+22 918606E-15 6626069E-34 meson 248806E-28 337479E+22 88833E-15 6626069E-34 meson 356532E-28 483598E+22 619921E-15 6626069E-34 strange quark
880118E-28 119379E+23 251128E-15 6626069E-34 meson 887249E-28 120346E+23 249109E-15 6626069E-34 meson 887177E-28 120336E+23 249129E-15 6626069E-34 meson 978325E-28 132699E+23 225919E-15 6626069E-34 meson 16726E-27 22687E+23 132143E-15 6626069E-34 Proton 938gt6eV
167491E-27 227183E+23 131961E-15 6626069E-34 Neutron 939gt6eV 198872E-27 269748E+23 111138E-15 6626069E-34 hyperon
212033E-27 2876E+23 104239E-15 6626069E-34 hyperon 212579E-27 28834E+23 103972E-15 6626069E-34 21347E-27 289549E+23 103538E-15 6626069E-34
231746E-27 314339E+23 953725E-16 6626069E-34 charm quark 234367E-27 317893E+23 943061E-16 6626069E-34
235543E-27 319489E+23 93835E-16 6626069E-34 298061E-27 404288E+23 741532E-16 6626069E-34 hyperon
802198E-27 10881E+24 27552E-16 6626069E-34 bottom quark 143201E-25 194237E+25 154344E-17 6626069E-34 W
162556E-25 220489E+25 135967E-17 6626069E-34 Z 223368E-25 302974E+25 989499E-18 6626069E-34 Higgs 125eV
256482E-25 347890E+25 861745E-18 6626069E-34 Fermi labs 144gt9eV 311966E-25 423148E+25 708481E-18 6626069E-34 top quark
570389E-16 77367E+34 387493E-27 6626069E-34 Gamma 320gt20eV 185922E-09 25218E+41 118879E-33 6626069E-34 186ltkg 1043gt27eV 217645E-08 29521E+42 101552E-34 6626071E-34 Planck limits
mass frequency radius r=cf mr2f
frequency radius r=cf rf=C 299792458 8c m s 5791781E+09 517617E-02 2997925E+08 smallest photon
328809E+15 911753E-08 2997925E+08 136eV photon 123559E+20 242631E-12 2997925E+08 electron 0511gt6eV 120899E+21 247968E-13 2997925E+08 up quark
241799E+21 123984E-13 2997925E+08 down quark 255485E+22 117343E-14 2997925E+08 muon
326356E+22 918606E-15 2997925E+08 meson 337479E+22 88833E-15 2997925E+08 meson 483598E+22 619921E-15 2997925E+08 strange quark
119379E+23 251128E-15 2997925E+08 meson 120346E+23 249109E-15 2997925E+08 meson 120336E+23 249129E-15 2997925E+08 meson 132699E+23 225919E-15 2997925E+08 meson 22687E+23 132143E-15 2997925E+08 Proton 938gt6eV
227183E+23 131961E-15 2997925E+08 Neutron 939gt6eV 269748E+23 111138E-15 2997925E+08 hyperon
2876E+23 104239E-15 2997925E+08 hyperon 28834E+23 103972E-15 2997925E+08 289549E+23 103538E-15 2997925E+08
314339E+23 953725E-16 2997925E+08 charm quark 317893E+23 943061E-16 2997925E+08
319489E+23 93835E-16 2997925E+08 404288E+23 741532E-16 2997925E+08 hyperon
10881E+24 27552E-16 2997925E+08 bottom quark 194237E+25 154344E-17 2997925E+08 W
220489E+25 135967E-17 2997925E+08 Z 302974E+25 989499E-18 2997925E+08 Higgs 125eV
347890E+25 861745E-18 2997925E+08 Fermi labs 144gt9eV 423148E+25 708481E-18 2997925E+08 top quark
77367E+34 387493E-27 2997925E+08 Gamma 320gt20eV 25218E+41 118879E-33 2997925E+08 1043gt27eV 186lt9kg
29521E+42 101552E-34 2997925E+08 Planck limits
frequency radius r=cf rf=C
Savior Borg
Two helix makes a mass particle
Fernandes
Kanarevrsquos Photon Model
My website Symmetrymathcomis my interpretation of the Authorrsquos I have referenced
I do not claim that any of these authors will agree withmy interpretations
I leave it to anyone who visits my site to decide the logicof my interpretation
- SYMMETRY MATH (SM) By Jack Kuykendall
- Why is Symmetry Math Needed
- From ldquoWhat Is Mathematicsrdquo by Courant and Robbins ndash page 55 Th
- Slide 4
- Slide 5
- Slide 6
- Slide 7
- Slide 8
- Slide 9
- Slide 10
- Slide 11
- Slide 12
- Slide 13
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-
27372496 51 07372496 50
hKgs kg
c
2 07372496 50 second
m m rotationsf
h kgsc
22 392984064 11
0737249638 50 s
kg rotf
kgs
Kanarev Smallest measurable photon
24241 35328803 15
07372496 50 s
kg rotf
kgs
13598443ev photon 2424143lt35kg
91093897 31123559 20
07372496 50 s
kg rotf
kgs
05109906 6eV electron
9109235lt31kg
16726230 27226873 23
07372496 50 s
kg rotf
kgs
93827241gt6eV proton 1672623lt27kg
16749286 272271858 23
07372496 50 s
kg rotf
kgs
9395656gt6eV neutron 1674928lt27kg
334357629 27226750117 23
(07372496 50 ) )2( s
kg rotf
kgs
(Deuterium nucleus) Proton+Neutron
22102 42mr kgm
334357629 274535202583 23
(07372496 50 ( ) s1)
kg rotf
kgs
Proton+Neutron
22102 42mr kgm
223368 253029747 25
07372496 50 s
kg rotf
kgs
1243gt9eV Higgs Boson 2215849lt25 kg
256482lt25347890 25
07372496 50 s
kg rotf
kgs
144gt11eV Fermi Lab particle 2567033lt25kg
5703889152 1677367 34
07372496 50 s
kg rotf
kgs
320gt20eV highest eV particle measured 5704518lt16kg
1859222909 925218 41
07372496 50 s
kg rotf
kgs
10429476 27eV Fernandes 1859223lt9kg
2176450474 8295212 42
07372496 50 s
kg rotf
kgs
Planck limit
78434 28106387 23
07372496 50 s
kg rotf
kgs
classical radius of an electron
This frequency and radius are not a particle radius
This mass does not exist
308 Is there experimental evidences that an electron has the so-called classic radius of an electron A No experiments do not exist From Questions and Answers of Kanarev Page 95 Electrons-Protons-Neutrons JK ET and Rewrite The handbook of Chemistry and Physics shows the classical radius of the electron as (2817 15 )m Kanarevs theory shows this to be the radius of approach of the magnetic force lines around the torus ring of the electron ( )magr This is confirmed by calculating the dimensionless fine structure constant
( ) (which is equal to the circumference of the inner circle (2 )magr divided by the experimentally
measured radius of the electron exp( )R
exp
2 (2)(314)(2817 15 ) 10007297352568
2426 12 137036magr m
R m
(191)
Kg Rotationsec meters experimental classical rmag mass frequency radius r=cf mr mr(11615lt3) 427E-41 5791781E+09 517617E-02 221022E-42 256717E-45 smallest photon
242414E-35 328809E+15 911753E-08 221022E-42 256717E-45 136eV photon 91094E-31 123559E+20 242631E-12 221022E-42 256717E-45 electron 0511gt6eV 891331E-30 120899E+21 247968E-13 221022E-42 256717E-45 up quark
178266E-29 241799E+21 123984E-13 221022E-42 256717E-45 down quark 188356E-28 255485E+22 117343E-14 221022E-42 256717E-45 muon
240606E-28 326356E+22 918606E-15 221022E-42 256717E-45 meson 248806E-28 337479E+22 88833E-15 221022E-42 256717E-45 meson 356532E-28 483598E+22 619921E-15 221022E-42 256717E-45 strange quark
880118E-28 119379E+23 251128E-15 221022E-42 256717E-45 meson 887249E-28 120346E+23 249109E-15 221022E-42 256717E-45 meson 887177E-28 120336E+23 249129E-15 221022E-42 256717E-45 meson 978325E-28 132699E+23 225919E-15 221022E-42 256717E-45 meson 16726E-27 22687E+23 132143E-15 221022E-42 256717E-45 Proton 938gt6eV
167491E-27 227183E+23 131961E-15 221022E-42 256717E-45 Neutron 939gt6eV 198872E-27 269748E+23 111138E-15 221022E-42 256717E-45 hyperon
212033E-27 2876E+23 104239E-15 221022E-42 256717E-45 hyperon 212579E-27 28834E+23 103972E-15 221022E-42 256717E-45 21347E-27 289549E+23 103538E-15 221022E-42 256717E-45
231746E-27 314339E+23 953725E-16 221022E-42 256717E-45 charm quark 234367E-27 317893E+23 943061E-16 221022E-42 256717E-45
235543E-27 319489E+23 93835E-16 221022E-42 256717E-45 298061E-27 404288E+23 741532E-16 221022E-42 256717E-45 hyperon
802198E-27 10881E+24 27552E-16 221022E-42 256717E-45 bottom quark 143201E-25 194237E+25 154344E-17 221022E-42 256717E-45 W
162556E-25 220489E+25 135967E-17 221022E-42 256717E-45 Z 223368E-25 302974E+25 989499E-18 221022E-42 256717E-45 Higgs125gt9eV
256482E-25 347890E+25 861745E-18 221022E-42 256717E-45 Fermi labs 144gt9eV 311966E-25 423148E+25 708481E-18 221022E-42 256717E-45 top quark
570389E-16 77367E+34 387494E-27 221022E-42 256717E-45 Gamma 320gt20eV 185922E-09 25218E+41 118879E-33 221022E-42 256717E-45 1043gt27eV 186lt9kg
217645E-08 29521E+42 101552E-34 221022E-42 256717E-45 Planck limits
mass frequency radius r=cf mr mr(11615lt3)
mass frequency radius r=cf mr2f 222E-39 298E+11 100E-03 6626069E-34 smallest photon
242414E-35 328809E+15 911753E-08 6626069E-34 Photon 136eV 91094E-31 123559E+20 242631E-12 6626069E-34 electron 0511gt6eV 891331E-30 120899E+21 247968E-13 6626069E-34 up quark
178266E-29 241799E+21 123984E-13 6626069E-34 down quark 188356E-28 255485E+22 117343E-14 6626069E-34 muon
240606E-28 326356E+22 918606E-15 6626069E-34 meson 248806E-28 337479E+22 88833E-15 6626069E-34 meson 356532E-28 483598E+22 619921E-15 6626069E-34 strange quark
880118E-28 119379E+23 251128E-15 6626069E-34 meson 887249E-28 120346E+23 249109E-15 6626069E-34 meson 887177E-28 120336E+23 249129E-15 6626069E-34 meson 978325E-28 132699E+23 225919E-15 6626069E-34 meson 16726E-27 22687E+23 132143E-15 6626069E-34 Proton 938gt6eV
167491E-27 227183E+23 131961E-15 6626069E-34 Neutron 939gt6eV 198872E-27 269748E+23 111138E-15 6626069E-34 hyperon
212033E-27 2876E+23 104239E-15 6626069E-34 hyperon 212579E-27 28834E+23 103972E-15 6626069E-34 21347E-27 289549E+23 103538E-15 6626069E-34
231746E-27 314339E+23 953725E-16 6626069E-34 charm quark 234367E-27 317893E+23 943061E-16 6626069E-34
235543E-27 319489E+23 93835E-16 6626069E-34 298061E-27 404288E+23 741532E-16 6626069E-34 hyperon
802198E-27 10881E+24 27552E-16 6626069E-34 bottom quark 143201E-25 194237E+25 154344E-17 6626069E-34 W
162556E-25 220489E+25 135967E-17 6626069E-34 Z 223368E-25 302974E+25 989499E-18 6626069E-34 Higgs 125eV
256482E-25 347890E+25 861745E-18 6626069E-34 Fermi labs 144gt9eV 311966E-25 423148E+25 708481E-18 6626069E-34 top quark
570389E-16 77367E+34 387493E-27 6626069E-34 Gamma 320gt20eV 185922E-09 25218E+41 118879E-33 6626069E-34 186ltkg 1043gt27eV 217645E-08 29521E+42 101552E-34 6626071E-34 Planck limits
mass frequency radius r=cf mr2f
frequency radius r=cf rf=C 299792458 8c m s 5791781E+09 517617E-02 2997925E+08 smallest photon
328809E+15 911753E-08 2997925E+08 136eV photon 123559E+20 242631E-12 2997925E+08 electron 0511gt6eV 120899E+21 247968E-13 2997925E+08 up quark
241799E+21 123984E-13 2997925E+08 down quark 255485E+22 117343E-14 2997925E+08 muon
326356E+22 918606E-15 2997925E+08 meson 337479E+22 88833E-15 2997925E+08 meson 483598E+22 619921E-15 2997925E+08 strange quark
119379E+23 251128E-15 2997925E+08 meson 120346E+23 249109E-15 2997925E+08 meson 120336E+23 249129E-15 2997925E+08 meson 132699E+23 225919E-15 2997925E+08 meson 22687E+23 132143E-15 2997925E+08 Proton 938gt6eV
227183E+23 131961E-15 2997925E+08 Neutron 939gt6eV 269748E+23 111138E-15 2997925E+08 hyperon
2876E+23 104239E-15 2997925E+08 hyperon 28834E+23 103972E-15 2997925E+08 289549E+23 103538E-15 2997925E+08
314339E+23 953725E-16 2997925E+08 charm quark 317893E+23 943061E-16 2997925E+08
319489E+23 93835E-16 2997925E+08 404288E+23 741532E-16 2997925E+08 hyperon
10881E+24 27552E-16 2997925E+08 bottom quark 194237E+25 154344E-17 2997925E+08 W
220489E+25 135967E-17 2997925E+08 Z 302974E+25 989499E-18 2997925E+08 Higgs 125eV
347890E+25 861745E-18 2997925E+08 Fermi labs 144gt9eV 423148E+25 708481E-18 2997925E+08 top quark
77367E+34 387493E-27 2997925E+08 Gamma 320gt20eV 25218E+41 118879E-33 2997925E+08 1043gt27eV 186lt9kg
29521E+42 101552E-34 2997925E+08 Planck limits
frequency radius r=cf rf=C
Savior Borg
Two helix makes a mass particle
Fernandes
Kanarevrsquos Photon Model
My website Symmetrymathcomis my interpretation of the Authorrsquos I have referenced
I do not claim that any of these authors will agree withmy interpretations
I leave it to anyone who visits my site to decide the logicof my interpretation
- SYMMETRY MATH (SM) By Jack Kuykendall
- Why is Symmetry Math Needed
- From ldquoWhat Is Mathematicsrdquo by Courant and Robbins ndash page 55 Th
- Slide 4
- Slide 5
- Slide 6
- Slide 7
- Slide 8
- Slide 9
- Slide 10
- Slide 11
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-
308 Is there experimental evidences that an electron has the so-called classic radius of an electron A No experiments do not exist From Questions and Answers of Kanarev Page 95 Electrons-Protons-Neutrons JK ET and Rewrite The handbook of Chemistry and Physics shows the classical radius of the electron as (2817 15 )m Kanarevs theory shows this to be the radius of approach of the magnetic force lines around the torus ring of the electron ( )magr This is confirmed by calculating the dimensionless fine structure constant
( ) (which is equal to the circumference of the inner circle (2 )magr divided by the experimentally
measured radius of the electron exp( )R
exp
2 (2)(314)(2817 15 ) 10007297352568
2426 12 137036magr m
R m
(191)
Kg Rotationsec meters experimental classical rmag mass frequency radius r=cf mr mr(11615lt3) 427E-41 5791781E+09 517617E-02 221022E-42 256717E-45 smallest photon
242414E-35 328809E+15 911753E-08 221022E-42 256717E-45 136eV photon 91094E-31 123559E+20 242631E-12 221022E-42 256717E-45 electron 0511gt6eV 891331E-30 120899E+21 247968E-13 221022E-42 256717E-45 up quark
178266E-29 241799E+21 123984E-13 221022E-42 256717E-45 down quark 188356E-28 255485E+22 117343E-14 221022E-42 256717E-45 muon
240606E-28 326356E+22 918606E-15 221022E-42 256717E-45 meson 248806E-28 337479E+22 88833E-15 221022E-42 256717E-45 meson 356532E-28 483598E+22 619921E-15 221022E-42 256717E-45 strange quark
880118E-28 119379E+23 251128E-15 221022E-42 256717E-45 meson 887249E-28 120346E+23 249109E-15 221022E-42 256717E-45 meson 887177E-28 120336E+23 249129E-15 221022E-42 256717E-45 meson 978325E-28 132699E+23 225919E-15 221022E-42 256717E-45 meson 16726E-27 22687E+23 132143E-15 221022E-42 256717E-45 Proton 938gt6eV
167491E-27 227183E+23 131961E-15 221022E-42 256717E-45 Neutron 939gt6eV 198872E-27 269748E+23 111138E-15 221022E-42 256717E-45 hyperon
212033E-27 2876E+23 104239E-15 221022E-42 256717E-45 hyperon 212579E-27 28834E+23 103972E-15 221022E-42 256717E-45 21347E-27 289549E+23 103538E-15 221022E-42 256717E-45
231746E-27 314339E+23 953725E-16 221022E-42 256717E-45 charm quark 234367E-27 317893E+23 943061E-16 221022E-42 256717E-45
235543E-27 319489E+23 93835E-16 221022E-42 256717E-45 298061E-27 404288E+23 741532E-16 221022E-42 256717E-45 hyperon
802198E-27 10881E+24 27552E-16 221022E-42 256717E-45 bottom quark 143201E-25 194237E+25 154344E-17 221022E-42 256717E-45 W
162556E-25 220489E+25 135967E-17 221022E-42 256717E-45 Z 223368E-25 302974E+25 989499E-18 221022E-42 256717E-45 Higgs125gt9eV
256482E-25 347890E+25 861745E-18 221022E-42 256717E-45 Fermi labs 144gt9eV 311966E-25 423148E+25 708481E-18 221022E-42 256717E-45 top quark
570389E-16 77367E+34 387494E-27 221022E-42 256717E-45 Gamma 320gt20eV 185922E-09 25218E+41 118879E-33 221022E-42 256717E-45 1043gt27eV 186lt9kg
217645E-08 29521E+42 101552E-34 221022E-42 256717E-45 Planck limits
mass frequency radius r=cf mr mr(11615lt3)
mass frequency radius r=cf mr2f 222E-39 298E+11 100E-03 6626069E-34 smallest photon
242414E-35 328809E+15 911753E-08 6626069E-34 Photon 136eV 91094E-31 123559E+20 242631E-12 6626069E-34 electron 0511gt6eV 891331E-30 120899E+21 247968E-13 6626069E-34 up quark
178266E-29 241799E+21 123984E-13 6626069E-34 down quark 188356E-28 255485E+22 117343E-14 6626069E-34 muon
240606E-28 326356E+22 918606E-15 6626069E-34 meson 248806E-28 337479E+22 88833E-15 6626069E-34 meson 356532E-28 483598E+22 619921E-15 6626069E-34 strange quark
880118E-28 119379E+23 251128E-15 6626069E-34 meson 887249E-28 120346E+23 249109E-15 6626069E-34 meson 887177E-28 120336E+23 249129E-15 6626069E-34 meson 978325E-28 132699E+23 225919E-15 6626069E-34 meson 16726E-27 22687E+23 132143E-15 6626069E-34 Proton 938gt6eV
167491E-27 227183E+23 131961E-15 6626069E-34 Neutron 939gt6eV 198872E-27 269748E+23 111138E-15 6626069E-34 hyperon
212033E-27 2876E+23 104239E-15 6626069E-34 hyperon 212579E-27 28834E+23 103972E-15 6626069E-34 21347E-27 289549E+23 103538E-15 6626069E-34
231746E-27 314339E+23 953725E-16 6626069E-34 charm quark 234367E-27 317893E+23 943061E-16 6626069E-34
235543E-27 319489E+23 93835E-16 6626069E-34 298061E-27 404288E+23 741532E-16 6626069E-34 hyperon
802198E-27 10881E+24 27552E-16 6626069E-34 bottom quark 143201E-25 194237E+25 154344E-17 6626069E-34 W
162556E-25 220489E+25 135967E-17 6626069E-34 Z 223368E-25 302974E+25 989499E-18 6626069E-34 Higgs 125eV
256482E-25 347890E+25 861745E-18 6626069E-34 Fermi labs 144gt9eV 311966E-25 423148E+25 708481E-18 6626069E-34 top quark
570389E-16 77367E+34 387493E-27 6626069E-34 Gamma 320gt20eV 185922E-09 25218E+41 118879E-33 6626069E-34 186ltkg 1043gt27eV 217645E-08 29521E+42 101552E-34 6626071E-34 Planck limits
mass frequency radius r=cf mr2f
frequency radius r=cf rf=C 299792458 8c m s 5791781E+09 517617E-02 2997925E+08 smallest photon
328809E+15 911753E-08 2997925E+08 136eV photon 123559E+20 242631E-12 2997925E+08 electron 0511gt6eV 120899E+21 247968E-13 2997925E+08 up quark
241799E+21 123984E-13 2997925E+08 down quark 255485E+22 117343E-14 2997925E+08 muon
326356E+22 918606E-15 2997925E+08 meson 337479E+22 88833E-15 2997925E+08 meson 483598E+22 619921E-15 2997925E+08 strange quark
119379E+23 251128E-15 2997925E+08 meson 120346E+23 249109E-15 2997925E+08 meson 120336E+23 249129E-15 2997925E+08 meson 132699E+23 225919E-15 2997925E+08 meson 22687E+23 132143E-15 2997925E+08 Proton 938gt6eV
227183E+23 131961E-15 2997925E+08 Neutron 939gt6eV 269748E+23 111138E-15 2997925E+08 hyperon
2876E+23 104239E-15 2997925E+08 hyperon 28834E+23 103972E-15 2997925E+08 289549E+23 103538E-15 2997925E+08
314339E+23 953725E-16 2997925E+08 charm quark 317893E+23 943061E-16 2997925E+08
319489E+23 93835E-16 2997925E+08 404288E+23 741532E-16 2997925E+08 hyperon
10881E+24 27552E-16 2997925E+08 bottom quark 194237E+25 154344E-17 2997925E+08 W
220489E+25 135967E-17 2997925E+08 Z 302974E+25 989499E-18 2997925E+08 Higgs 125eV
347890E+25 861745E-18 2997925E+08 Fermi labs 144gt9eV 423148E+25 708481E-18 2997925E+08 top quark
77367E+34 387493E-27 2997925E+08 Gamma 320gt20eV 25218E+41 118879E-33 2997925E+08 1043gt27eV 186lt9kg
29521E+42 101552E-34 2997925E+08 Planck limits
frequency radius r=cf rf=C
Savior Borg
Two helix makes a mass particle
Fernandes
Kanarevrsquos Photon Model
My website Symmetrymathcomis my interpretation of the Authorrsquos I have referenced
I do not claim that any of these authors will agree withmy interpretations
I leave it to anyone who visits my site to decide the logicof my interpretation
- SYMMETRY MATH (SM) By Jack Kuykendall
- Why is Symmetry Math Needed
- From ldquoWhat Is Mathematicsrdquo by Courant and Robbins ndash page 55 Th
- Slide 4
- Slide 5
- Slide 6
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-
Kg Rotationsec meters experimental classical rmag mass frequency radius r=cf mr mr(11615lt3) 427E-41 5791781E+09 517617E-02 221022E-42 256717E-45 smallest photon
242414E-35 328809E+15 911753E-08 221022E-42 256717E-45 136eV photon 91094E-31 123559E+20 242631E-12 221022E-42 256717E-45 electron 0511gt6eV 891331E-30 120899E+21 247968E-13 221022E-42 256717E-45 up quark
178266E-29 241799E+21 123984E-13 221022E-42 256717E-45 down quark 188356E-28 255485E+22 117343E-14 221022E-42 256717E-45 muon
240606E-28 326356E+22 918606E-15 221022E-42 256717E-45 meson 248806E-28 337479E+22 88833E-15 221022E-42 256717E-45 meson 356532E-28 483598E+22 619921E-15 221022E-42 256717E-45 strange quark
880118E-28 119379E+23 251128E-15 221022E-42 256717E-45 meson 887249E-28 120346E+23 249109E-15 221022E-42 256717E-45 meson 887177E-28 120336E+23 249129E-15 221022E-42 256717E-45 meson 978325E-28 132699E+23 225919E-15 221022E-42 256717E-45 meson 16726E-27 22687E+23 132143E-15 221022E-42 256717E-45 Proton 938gt6eV
167491E-27 227183E+23 131961E-15 221022E-42 256717E-45 Neutron 939gt6eV 198872E-27 269748E+23 111138E-15 221022E-42 256717E-45 hyperon
212033E-27 2876E+23 104239E-15 221022E-42 256717E-45 hyperon 212579E-27 28834E+23 103972E-15 221022E-42 256717E-45 21347E-27 289549E+23 103538E-15 221022E-42 256717E-45
231746E-27 314339E+23 953725E-16 221022E-42 256717E-45 charm quark 234367E-27 317893E+23 943061E-16 221022E-42 256717E-45
235543E-27 319489E+23 93835E-16 221022E-42 256717E-45 298061E-27 404288E+23 741532E-16 221022E-42 256717E-45 hyperon
802198E-27 10881E+24 27552E-16 221022E-42 256717E-45 bottom quark 143201E-25 194237E+25 154344E-17 221022E-42 256717E-45 W
162556E-25 220489E+25 135967E-17 221022E-42 256717E-45 Z 223368E-25 302974E+25 989499E-18 221022E-42 256717E-45 Higgs125gt9eV
256482E-25 347890E+25 861745E-18 221022E-42 256717E-45 Fermi labs 144gt9eV 311966E-25 423148E+25 708481E-18 221022E-42 256717E-45 top quark
570389E-16 77367E+34 387494E-27 221022E-42 256717E-45 Gamma 320gt20eV 185922E-09 25218E+41 118879E-33 221022E-42 256717E-45 1043gt27eV 186lt9kg
217645E-08 29521E+42 101552E-34 221022E-42 256717E-45 Planck limits
mass frequency radius r=cf mr mr(11615lt3)
mass frequency radius r=cf mr2f 222E-39 298E+11 100E-03 6626069E-34 smallest photon
242414E-35 328809E+15 911753E-08 6626069E-34 Photon 136eV 91094E-31 123559E+20 242631E-12 6626069E-34 electron 0511gt6eV 891331E-30 120899E+21 247968E-13 6626069E-34 up quark
178266E-29 241799E+21 123984E-13 6626069E-34 down quark 188356E-28 255485E+22 117343E-14 6626069E-34 muon
240606E-28 326356E+22 918606E-15 6626069E-34 meson 248806E-28 337479E+22 88833E-15 6626069E-34 meson 356532E-28 483598E+22 619921E-15 6626069E-34 strange quark
880118E-28 119379E+23 251128E-15 6626069E-34 meson 887249E-28 120346E+23 249109E-15 6626069E-34 meson 887177E-28 120336E+23 249129E-15 6626069E-34 meson 978325E-28 132699E+23 225919E-15 6626069E-34 meson 16726E-27 22687E+23 132143E-15 6626069E-34 Proton 938gt6eV
167491E-27 227183E+23 131961E-15 6626069E-34 Neutron 939gt6eV 198872E-27 269748E+23 111138E-15 6626069E-34 hyperon
212033E-27 2876E+23 104239E-15 6626069E-34 hyperon 212579E-27 28834E+23 103972E-15 6626069E-34 21347E-27 289549E+23 103538E-15 6626069E-34
231746E-27 314339E+23 953725E-16 6626069E-34 charm quark 234367E-27 317893E+23 943061E-16 6626069E-34
235543E-27 319489E+23 93835E-16 6626069E-34 298061E-27 404288E+23 741532E-16 6626069E-34 hyperon
802198E-27 10881E+24 27552E-16 6626069E-34 bottom quark 143201E-25 194237E+25 154344E-17 6626069E-34 W
162556E-25 220489E+25 135967E-17 6626069E-34 Z 223368E-25 302974E+25 989499E-18 6626069E-34 Higgs 125eV
256482E-25 347890E+25 861745E-18 6626069E-34 Fermi labs 144gt9eV 311966E-25 423148E+25 708481E-18 6626069E-34 top quark
570389E-16 77367E+34 387493E-27 6626069E-34 Gamma 320gt20eV 185922E-09 25218E+41 118879E-33 6626069E-34 186ltkg 1043gt27eV 217645E-08 29521E+42 101552E-34 6626071E-34 Planck limits
mass frequency radius r=cf mr2f
frequency radius r=cf rf=C 299792458 8c m s 5791781E+09 517617E-02 2997925E+08 smallest photon
328809E+15 911753E-08 2997925E+08 136eV photon 123559E+20 242631E-12 2997925E+08 electron 0511gt6eV 120899E+21 247968E-13 2997925E+08 up quark
241799E+21 123984E-13 2997925E+08 down quark 255485E+22 117343E-14 2997925E+08 muon
326356E+22 918606E-15 2997925E+08 meson 337479E+22 88833E-15 2997925E+08 meson 483598E+22 619921E-15 2997925E+08 strange quark
119379E+23 251128E-15 2997925E+08 meson 120346E+23 249109E-15 2997925E+08 meson 120336E+23 249129E-15 2997925E+08 meson 132699E+23 225919E-15 2997925E+08 meson 22687E+23 132143E-15 2997925E+08 Proton 938gt6eV
227183E+23 131961E-15 2997925E+08 Neutron 939gt6eV 269748E+23 111138E-15 2997925E+08 hyperon
2876E+23 104239E-15 2997925E+08 hyperon 28834E+23 103972E-15 2997925E+08 289549E+23 103538E-15 2997925E+08
314339E+23 953725E-16 2997925E+08 charm quark 317893E+23 943061E-16 2997925E+08
319489E+23 93835E-16 2997925E+08 404288E+23 741532E-16 2997925E+08 hyperon
10881E+24 27552E-16 2997925E+08 bottom quark 194237E+25 154344E-17 2997925E+08 W
220489E+25 135967E-17 2997925E+08 Z 302974E+25 989499E-18 2997925E+08 Higgs 125eV
347890E+25 861745E-18 2997925E+08 Fermi labs 144gt9eV 423148E+25 708481E-18 2997925E+08 top quark
77367E+34 387493E-27 2997925E+08 Gamma 320gt20eV 25218E+41 118879E-33 2997925E+08 1043gt27eV 186lt9kg
29521E+42 101552E-34 2997925E+08 Planck limits
frequency radius r=cf rf=C
Savior Borg
Two helix makes a mass particle
Fernandes
Kanarevrsquos Photon Model
My website Symmetrymathcomis my interpretation of the Authorrsquos I have referenced
I do not claim that any of these authors will agree withmy interpretations
I leave it to anyone who visits my site to decide the logicof my interpretation
- SYMMETRY MATH (SM) By Jack Kuykendall
- Why is Symmetry Math Needed
- From ldquoWhat Is Mathematicsrdquo by Courant and Robbins ndash page 55 Th
- Slide 4
- Slide 5
- Slide 6
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-
mass frequency radius r=cf mr2f 222E-39 298E+11 100E-03 6626069E-34 smallest photon
242414E-35 328809E+15 911753E-08 6626069E-34 Photon 136eV 91094E-31 123559E+20 242631E-12 6626069E-34 electron 0511gt6eV 891331E-30 120899E+21 247968E-13 6626069E-34 up quark
178266E-29 241799E+21 123984E-13 6626069E-34 down quark 188356E-28 255485E+22 117343E-14 6626069E-34 muon
240606E-28 326356E+22 918606E-15 6626069E-34 meson 248806E-28 337479E+22 88833E-15 6626069E-34 meson 356532E-28 483598E+22 619921E-15 6626069E-34 strange quark
880118E-28 119379E+23 251128E-15 6626069E-34 meson 887249E-28 120346E+23 249109E-15 6626069E-34 meson 887177E-28 120336E+23 249129E-15 6626069E-34 meson 978325E-28 132699E+23 225919E-15 6626069E-34 meson 16726E-27 22687E+23 132143E-15 6626069E-34 Proton 938gt6eV
167491E-27 227183E+23 131961E-15 6626069E-34 Neutron 939gt6eV 198872E-27 269748E+23 111138E-15 6626069E-34 hyperon
212033E-27 2876E+23 104239E-15 6626069E-34 hyperon 212579E-27 28834E+23 103972E-15 6626069E-34 21347E-27 289549E+23 103538E-15 6626069E-34
231746E-27 314339E+23 953725E-16 6626069E-34 charm quark 234367E-27 317893E+23 943061E-16 6626069E-34
235543E-27 319489E+23 93835E-16 6626069E-34 298061E-27 404288E+23 741532E-16 6626069E-34 hyperon
802198E-27 10881E+24 27552E-16 6626069E-34 bottom quark 143201E-25 194237E+25 154344E-17 6626069E-34 W
162556E-25 220489E+25 135967E-17 6626069E-34 Z 223368E-25 302974E+25 989499E-18 6626069E-34 Higgs 125eV
256482E-25 347890E+25 861745E-18 6626069E-34 Fermi labs 144gt9eV 311966E-25 423148E+25 708481E-18 6626069E-34 top quark
570389E-16 77367E+34 387493E-27 6626069E-34 Gamma 320gt20eV 185922E-09 25218E+41 118879E-33 6626069E-34 186ltkg 1043gt27eV 217645E-08 29521E+42 101552E-34 6626071E-34 Planck limits
mass frequency radius r=cf mr2f
frequency radius r=cf rf=C 299792458 8c m s 5791781E+09 517617E-02 2997925E+08 smallest photon
328809E+15 911753E-08 2997925E+08 136eV photon 123559E+20 242631E-12 2997925E+08 electron 0511gt6eV 120899E+21 247968E-13 2997925E+08 up quark
241799E+21 123984E-13 2997925E+08 down quark 255485E+22 117343E-14 2997925E+08 muon
326356E+22 918606E-15 2997925E+08 meson 337479E+22 88833E-15 2997925E+08 meson 483598E+22 619921E-15 2997925E+08 strange quark
119379E+23 251128E-15 2997925E+08 meson 120346E+23 249109E-15 2997925E+08 meson 120336E+23 249129E-15 2997925E+08 meson 132699E+23 225919E-15 2997925E+08 meson 22687E+23 132143E-15 2997925E+08 Proton 938gt6eV
227183E+23 131961E-15 2997925E+08 Neutron 939gt6eV 269748E+23 111138E-15 2997925E+08 hyperon
2876E+23 104239E-15 2997925E+08 hyperon 28834E+23 103972E-15 2997925E+08 289549E+23 103538E-15 2997925E+08
314339E+23 953725E-16 2997925E+08 charm quark 317893E+23 943061E-16 2997925E+08
319489E+23 93835E-16 2997925E+08 404288E+23 741532E-16 2997925E+08 hyperon
10881E+24 27552E-16 2997925E+08 bottom quark 194237E+25 154344E-17 2997925E+08 W
220489E+25 135967E-17 2997925E+08 Z 302974E+25 989499E-18 2997925E+08 Higgs 125eV
347890E+25 861745E-18 2997925E+08 Fermi labs 144gt9eV 423148E+25 708481E-18 2997925E+08 top quark
77367E+34 387493E-27 2997925E+08 Gamma 320gt20eV 25218E+41 118879E-33 2997925E+08 1043gt27eV 186lt9kg
29521E+42 101552E-34 2997925E+08 Planck limits
frequency radius r=cf rf=C
Savior Borg
Two helix makes a mass particle
Fernandes
Kanarevrsquos Photon Model
My website Symmetrymathcomis my interpretation of the Authorrsquos I have referenced
I do not claim that any of these authors will agree withmy interpretations
I leave it to anyone who visits my site to decide the logicof my interpretation
- SYMMETRY MATH (SM) By Jack Kuykendall
- Why is Symmetry Math Needed
- From ldquoWhat Is Mathematicsrdquo by Courant and Robbins ndash page 55 Th
- Slide 4
- Slide 5
- Slide 6
- Slide 7
- Slide 8
- Slide 9
- Slide 10
- Slide 11
- Slide 12
- Slide 13
- Slide 14
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frequency radius r=cf rf=C 299792458 8c m s 5791781E+09 517617E-02 2997925E+08 smallest photon
328809E+15 911753E-08 2997925E+08 136eV photon 123559E+20 242631E-12 2997925E+08 electron 0511gt6eV 120899E+21 247968E-13 2997925E+08 up quark
241799E+21 123984E-13 2997925E+08 down quark 255485E+22 117343E-14 2997925E+08 muon
326356E+22 918606E-15 2997925E+08 meson 337479E+22 88833E-15 2997925E+08 meson 483598E+22 619921E-15 2997925E+08 strange quark
119379E+23 251128E-15 2997925E+08 meson 120346E+23 249109E-15 2997925E+08 meson 120336E+23 249129E-15 2997925E+08 meson 132699E+23 225919E-15 2997925E+08 meson 22687E+23 132143E-15 2997925E+08 Proton 938gt6eV
227183E+23 131961E-15 2997925E+08 Neutron 939gt6eV 269748E+23 111138E-15 2997925E+08 hyperon
2876E+23 104239E-15 2997925E+08 hyperon 28834E+23 103972E-15 2997925E+08 289549E+23 103538E-15 2997925E+08
314339E+23 953725E-16 2997925E+08 charm quark 317893E+23 943061E-16 2997925E+08
319489E+23 93835E-16 2997925E+08 404288E+23 741532E-16 2997925E+08 hyperon
10881E+24 27552E-16 2997925E+08 bottom quark 194237E+25 154344E-17 2997925E+08 W
220489E+25 135967E-17 2997925E+08 Z 302974E+25 989499E-18 2997925E+08 Higgs 125eV
347890E+25 861745E-18 2997925E+08 Fermi labs 144gt9eV 423148E+25 708481E-18 2997925E+08 top quark
77367E+34 387493E-27 2997925E+08 Gamma 320gt20eV 25218E+41 118879E-33 2997925E+08 1043gt27eV 186lt9kg
29521E+42 101552E-34 2997925E+08 Planck limits
frequency radius r=cf rf=C
Savior Borg
Two helix makes a mass particle
Fernandes
Kanarevrsquos Photon Model
My website Symmetrymathcomis my interpretation of the Authorrsquos I have referenced
I do not claim that any of these authors will agree withmy interpretations
I leave it to anyone who visits my site to decide the logicof my interpretation
- SYMMETRY MATH (SM) By Jack Kuykendall
- Why is Symmetry Math Needed
- From ldquoWhat Is Mathematicsrdquo by Courant and Robbins ndash page 55 Th
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Savior Borg
Two helix makes a mass particle
Fernandes
Kanarevrsquos Photon Model
My website Symmetrymathcomis my interpretation of the Authorrsquos I have referenced
I do not claim that any of these authors will agree withmy interpretations
I leave it to anyone who visits my site to decide the logicof my interpretation
- SYMMETRY MATH (SM) By Jack Kuykendall
- Why is Symmetry Math Needed
- From ldquoWhat Is Mathematicsrdquo by Courant and Robbins ndash page 55 Th
- Slide 4
- Slide 5
- Slide 6
- Slide 7
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-
My website Symmetrymathcomis my interpretation of the Authorrsquos I have referenced
I do not claim that any of these authors will agree withmy interpretations
I leave it to anyone who visits my site to decide the logicof my interpretation
- SYMMETRY MATH (SM) By Jack Kuykendall
- Why is Symmetry Math Needed
- From ldquoWhat Is Mathematicsrdquo by Courant and Robbins ndash page 55 Th
- Slide 4
- Slide 5
- Slide 6
- Slide 7
- Slide 8
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- Slide 10
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