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PART TWO: RELATIVISTICMECHANICS
PHYS 141: PHYS 141: Principles of MechanicsPrinciples of Mechanics
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I. Basic Principles A. Spacetime1. Graphical Depiction: Before we look at how
space and time are connected through special relativity, let’s establish how we can describe basic motion: spacetime diagramsspacetime diagrams.
x
t1-D motion in time asmeasuredfrom a ParticularReferenceFrame
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I. Basic Principles A. Spacetime1. Graphical Depiction: Before we look at how
space and time are connected through special relativity, let’s establish how we can describe basic motion: spacetime diagramsspacetime diagrams.
x
t
“PAST”
“FUTURE”
“PRESENT”
1-D motion in time asmeasuredfrom a ParticularReferenceFrame
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I. Basic Principles A. Spacetime1. Graphical Depiction: Before we look at how
space and time are connected through special relativity, let’s establish how we can describe basic motion: spacetime diagramsspacetime diagrams.
x
t
“SOMEWHEREAHEAD”“HERE”
“SOMEWHEREBEHIND”
1-D motion in time asmeasuredfrom a ParticularReferenceFrame
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I. Basic Principles A. Spacetimea. Event: a particular location and time.
x
t
“event”(x,t)
“OVER THERE,THEN”1-D motion
in time asmeasuredfrom a ParticularReferenceFrame
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I. Basic Principles A. Spacetimea. Event: a particular location and time.
x
t
“event 1”(x1,t1)
“event 2”(x2,t2)
1-D motion in time asmeasuredfrom a ParticularReferenceFrame
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I. Basic Principles A. Spacetimeb. World Lines are Spacetime trajectories: How
events are ordered.
x
t
“event 1”(x1,t1)
“event 2”(x2,t2)
“Stay at x = x1
for t2-t1 seconds”
1-D motion in time asmeasuredfrom a ParticularReferenceFrame
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I. Basic Principles A. Spacetimeb. World Lines: How events are ordered.
x
t
“event 1”(x1,t1)
“event 2”(x2,t2)
“Instantaneously moveFrom
x1 to x2.” Note: this is impossible.
1-D motion in time asmeasuredfrom a ParticularReferenceFrame
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I. Basic Principles A. Spacetimec.c. Normalized Spacetime diagramsNormalized Spacetime diagrams. Let w = ct.
Then the slope of a trajectory in spacetime is dw/dx = cdt/dx = c/v, with c = speed of light.
x
w = ct
“event 1”(x1,w1)
“event 2”
(x2,w2)
Speed from E1 to E2 is
dw/dx = c/v =1,so
v/c = 1.
1-D motion in time asmeasuredfrom a ParticularReferenceFrame A
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I. Basic Principles A. Spacetimec.c. Normalized Spacetime diagramsNormalized Spacetime diagrams. Let w = ct.
Then the slope of a trajectory in spacetime is dw/dx = cdt/dx = c/v, with c = speed of light.
x
w = ct1-D motion in time asmeasuredfrom a ParticularReferenceFrame A
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I. Basic Principles A. SpacetimeNormalized Spacetime diagrams. Let w = ct. Then the slope of a trajectory in spacetime is dw/dx = cdt/dx = c/v, with c = speed of light.
x
w=ct
v/c = 1
dw/dx = c/v >1, sov/c <1
dw/dx = c/v <1, sov/c >1
1-D motion in time asmeasuredfrom a ParticularReferenceFrame A
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I. Basic Principles A. Spacetimed. Lightcones:Lightcones: Regions of spacetime in which
events are connected by paths with speed v/c ≤ 1.
x
w
“Futurelight cone”
for A
“Past light cone”
for A
v/c = 1
v/c = 1
1-D motion in time asmeasuredfrom a ParticularReferenceFrame A
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I. Basic Principles A. Spacetimed. Lightcones:Lightcones: Regions of spacetime in which
events are connected by paths with speed v/c ≤ 1.
x
wv/c =
1. “Futurelight cone”
for A
“Past light cone”
for A
Inaccessible to A
according torelativity
Inaccessible to A
according torelativity
v/c = 1.
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Constant speed(x’,y’,z’,t’)
I. Basic Principles B. Galilean Transformations
For inertial systems, the Galilean TransformationGalilean Transformation allows us to translate between frames: all we have to do is (basically) subtract out the motion of primed frame (the train in this case).
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I. Basic Principles B. Galilean Transformations
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I. Basic Principles B. Galilean Transformations
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I. Basic Principles B. Galilean Transformations
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I. Basic Principles B. Galilean Transformations
For inertial systems, the Galilean TransformationGalilean Transformation allows us to translate between frames: all we have to do is (basically) subtract out the motion of primed frame (the train in this case).
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1. Coordinate Transformations: transformations between two reference frames. Consider a primed reference in motion relative to an unprimed one (let v be along the positive x-direction)
x
y
a) Galilean transformations:coordinates for a point P in the
movingprimed frame as seen from the
(unmoving) unprimed frame (and vice versa).x = x’ + vt’. (I.B.1-4)
y = y’.z = z’.t = t’.
I. Basic Principles B. Galilean Transformations
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x
y x = x’ + vt’.y = y’.z = z’.t = t’.
x’
y’v
P
x
x’
I. Basic Principles B. Galilean Transformations
1. Coordinate Transformations: transformations between two reference frames. Consider a primed reference in motion relative to an unprimed one (let v be along the positive x-direction) a) Galilean transformations:
coordinates for a point P in the moving
primed frame as seen from the (unmoving)
unprimed frame (and vice versa).
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x
y x = x’ + vt’.y = y’.z = z’.t = t’.
x’
y’v
P
x
vt’ x’
I. Basic Principles B. Galilean Transformations
1. Coordinate Transformations: transformations between two reference frames. Consider a primed reference in motion relative to an unprimed one (let v be along the positive x-direction) a) Galilean transformations:
coordinates for a point P in the moving
primed frame as seen from the (unmoving)
unprimed frame (and vice versa).
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x
y x = x’ + vt’.y = y’.z = z’.t = t’.
x’
y’v
P
x
vt’ x’
I. Basic Principles B. Galilean Transformations
1. Coordinate Transformations: transformations between two reference frames. Consider a primed reference in motion relative to an unprimed one (let v be along the positive x-direction) a) Galilean transformations:
coordinates for a point P in the moving
primed frame as seen from the (unmoving)
unprimed frame (and vice versa).
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x
y x = x’ + vt’.y = y’.z = z’.t = t’.
x’
y’v
P
x
vt’ x’
I. Basic Principles B. Galilean Transformations
1. Coordinate Transformations: transformations between two reference frames. Consider a primed reference in motion relative to an unprimed one (let v be along the positive x-direction) a) Galilean transformations:
coordinates for a point P in the moving
primed frame as seen from the (unmoving)
unprimed frame (and vice versa).
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x
y x = x’ + vt’.y = y’.z = z’.t = t’.
x’
y’v
P
x
vt’ x’
I. Basic Principles B. Galilean Transformations
1. Coordinate Transformations: transformations between two reference frames. Consider a primed reference in motion relative to an unprimed one (let v be along the positive x-direction) a) Galilean transformations:
coordinates for a point P in the moving
primed frame as seen from the (unmoving)
unprimed frame (and vice versa).
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II. Simultaneity A. Set Up1. Consider two observers, A&B, stationary with
respect to each other and reference frame (w,x). How do we calibrate their identical clocks?
x
w
A B
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II. Simultaneity A. Set Up1. Consider two observers, A&B, stationary with
respect to each other and reference frame (w,x). How do we calibrate their identical clocks?
x
w
A B
wB
Eventsare simultaneous(clocks calibrated)By backtracking
To the equidistantSpacetime position
Between light paths.
Events WA and WB
are simultaneous in this frame.
wA
Note:dotted lineindicates
same TIMEfor unprimed
frame
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II. Simultaneity A. Set Up2. What happens to the axes for a Galilean
Transformation?
x
w
(x,t)
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II. Simultaneity A. Set Up2. What happens to the axes for a Galilean
Transformation? Now consider a frame moving with speed v wrt the original frame.
x
t
(x,t)
x=(x’ + vt’)
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II. Simultaneity A. Set Up2. What happens to the axes for a Galilean
Transformation? Now consider a frame moving with speed v wrt the original frame.
x
t
(x,t)
x=(x’ + vt’)
x=(x’ + v(2t’))
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II. Simultaneity A. Set Up2. What happens to the axes for a Galilean
Transformation? Now consider a frame moving with speed v wrt the original frame.
x
t
(x,t)
x=(x’ + v(3t’))
x=(x’ + vt’)
x=(x’ + v(2t’))
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II. Simultaneity A. Set Up2. What happens to the axes for a Galilean
Transformation? Now consider a frame moving with speed v wrt the original frame.
x
t
(x,t)
x=(x’ + v(3t’))
x=(x’ + vt’)
x=(x’ + v(2t’))
x’
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II. Simultaneity A. Set Up2. What happens to the axes for a Galilean
Transformation? Now consider a frame moving with speed v wrt the original frame.
x
t
(x,t)
x’
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II. Simultaneity A. Set Up2. What happens to the axes for a Galilean
Transformation? Now consider a frame moving with speed v wrt the original frame.
x
t
(x,t)
x’
t’=t
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II. Simultaneity A. Set Up2. What happens to the axes for a Galilean
Transformation? Now consider a frame moving with speed v wrt the original frame.
x
t
x’
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II. Simultaneity A. Set Up3. Why? Newtonian addition of speeds.
x
tx’
t’
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II. Simultaneity A. Set Up2. What happens to the axes for a Galilean
Transformation? Now consider a frame moving with speed v wrt the original frame.
x
t
+=
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I. Basic Principles C. The ‘Principle of Relativity’
“The Laws of Mechanics are the same in every inertial frame, and the Galilean Transformation is valid.”
Problem: Electromagnetism (1850)
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I. Basic Principles D. Postulates of Special Relativity
1. The speed of light in vacuum is a constant, independent of the motion of the source, the observer, or both.
2. The Laws of Physics are everywhere the same for inertial frames, and the connection between frames is the Lorentz Transformation.
3. When v/c is small, then the LT reduces to the GT: x’ = (x - vt) & t’ = t.
x’x
v
P
x’ = (x - vt)/{1 - (v/c)2}1/2.y’ = y.z’ = z.
t’ = t - vx/c2)/{1 - (v/c)2}1/2.(I.D.1-4)
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II. Simultaneity A. Set Up2. What happens to the axes for a Lorentz
Transformation?
x
w
(x,w)
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II. Simultaneity A. Set Up
x
w
(x’,w’)
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II. Simultaneity A. Set Up
x
w
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II. Simultaneity B. Result1. Now what is measured by another observer moving
with respect to the original frame with speed (v’)?
x
w
A B
w’
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1. Now what is measured by another observer moving with respect to the original frame with speed (v’)?
x
w
A B
w’
x’
II. Simultaneity B. Result
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II. Simultaneity B. Results1. Now what is measured by another observer moving
with respect to the original frame with speed (v’)?
x
w
A B
w’
x’
wB’
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II. Simultaneity B. Results1. Now what is measured by another observer moving
with respect to the original frame with speed (v’)?
x
w
A B
w’
wA’
SimultaneityIs frame
dependent
wB’
Events wA’ and wB’are simultaneous in
the red frame.
However, now the events are NOT
simultaneous in theunprimed frame.
Note:dotted lineindicates
same TIMEfor primed
frame
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II. Simultaneity B. Results1. Now what is measured by another observer moving
with respect to the original frame with speed (v)?
x
w
t3t2t1
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II. Simultaneity B. Results1. Now what is measured by another observer moving
with respect to the original frame with speed (v)?
x
w w’
x’
t3’
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II. Simultaneity B. Results1. Now what is measured by another observer moving
with respect to the original frame with speed (v)?
x
w w’
x’
t3’
Example:x = 1 lys, t=0s, v/c = 0.5
x’= (1 lys - 0.5c(0s))/{1-0.52}1/2
= 1.15*(5.5) = 1.15 lys
t’ = (0 s - 0.5c(1 ly-s)/c2)/{1-0.52}1/2
=1.15*(-.5) = -0.6 second.
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II. Simultaneity B. Results1. Now what is measured by another observer moving
with respect to the original frame with speed (v)?
x
w w’
x’
t3’
Example:x = 5 lys, t=0s, v/c = 0.5
x’= (5 lys – 0.5c(0s))/{1-0.52}1/2
= 1.15*(5) = 5.8 lys
t’ = (0 s – 0.5c(5 lys)/c2)/{1-0.52}1/2
=1.15*(-2.5) = -2.9 second.
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II. Simultaneity B. Results1. Now what is measured by another observer moving
with respect to the original frame with speed (v’)?
x
w w’
x’
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II. Simultaneity B. Results1. Now what is measured by another observer moving
with respect to the original frame with speed (v’)?
x
ww’
x’
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2 2 2 2( ) .c t x y z
(II.C.1)
Invariant: independent of frame that is
measured
What is the space time interval on a lightcone? 2
2
2
0
0
0
s timelike
s lightlike
s spacelike
II. Simultaneity C. Spacetime Interval
Consider a light pulse that travels in a sphere.
2 2 2 2( ') ' ' ' .c t x y z
2 2 2 2 2( ).s c t x y z
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x
w=ct
ds2 = 0,lightlike
ds2>0,timelike,Causally
connected
ds2<0,Spacelike,Causally
unconnected
A
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1. What is measured by another observer moving with respect to the original frame with speed (v’)?
x
w w’
w*
We know thatw = 3w* = 2
both measuredfrom the black
Reference frame clock.
What is w’ measured from red
Reference Frame clock?
III.Time Dilation A. Definition
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1. Now what is measured by another observer moving with respect to the original frame with speed (v’)?
x
w w’
w*
w’ = w*w=w’, and so
w = w*.
III.Time Dilation A. Definition
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1. Now what is measured by another observer moving with respect to the original frame with speed (v’)?
x
w
w’w*
w’ = w*w=w’, and so
w = w*.
w’ = w/w’ < w
Time divisions in theprimed frame are differentthan in the unprimed frames
III.Time Dilation A. Definition
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1. Clocks moving relative to an observer are measured by that observer to run slow compared to a clock at rest.
2. Another thought experiment: Albert on a train w/speed v.
t’ = 0: light
leaves
Note: only one clock is needed to measure emission and return
of light.
III.Time Dilation B. Derivation
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Light travelsa total distance
equal to 2d.
Light requirest’ = 2d/c seconds to reach the floor.
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2. And now Albert’s view from the platform. Now we’ll need two synchronized clocks to measure both events:
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Clock 1 Clock 2
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Clock 1 Clock 2
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Clock 1 Clock 2
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Clock 1 Clock 2
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Clock 1 Clock 2
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Clock 1 Clock 2
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Clock 1 Clock 2
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Clock 1 Clock 2
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Clock 1 Clock 2
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Clock 1 Clock 2
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3. Mathematical formulation (pseudo-classical):
L = vt.
d = ct’.D = (d2 + L2)1/2.
c = D/t = (d2 + L2)1/2 /t;c2 = (d2 + L2)/t2.
t2 = (d/c)2 + (vt/c)2;t2 (1- (v/c)2) = (d/c)2.
However, d = ct’, and sot = t’/{1- (v/c)2}1/2.
t’ = t{1- (v/c)2}1/2.Measured from platformMeasured on train
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t = t’/{1- (v/c)2}1/2. (III.B.1) t’ = t{1- (v/c)2}1/2. (III.B.2)
Note: a) t is the time in the train as measured by the
platform observer.b) t’ is the time in the train as measured by the
train observer.c) t > t’: moving clocks run slow as
measured in the platform frame.
III.Time Dilation
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III. Time Dilation C. Implications
Let = 1/{1- (v/c)2}1/2
(III.B.3) t = t’. (III.B.4)
t’ = t/. (III.B.5)
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1. Verifications & Implicationsa) Muon lifetime
Particle decay time longer as measured in lab (rest) frame
b) Atomic clock experimentsc) The “Twin Paradox”
III. Time Dilation C. Implications
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III. Time DilationTwin “Paradox”
x
w
w’
w*
Rocket and EarthClocks synchronizedAt w’ and w*
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III. Time DilationTwin “Paradox”
x
w**
w’
w*
On return,Rocket and EarthClocks synchronizedAt w’ and w**
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III. Time DilationTwin “Paradox”
x
w**
w’
w*
On return,Rocket and EarthClocks synchronizedAt w’ and w**
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III. Time DilationTwin “Paradox”
x
w’
x’
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III. Time DilationTwin “Paradox”
x
w’
x’
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III. Time DilationTwin “Paradox”
x
w’
x’
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III. Time DilationTwin “Paradox”
x
w’
x’
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III. Time DilationTwin “Paradox”
x
w’
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III. Time DilationTwin “Paradox”
x
w’
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III. Time DilationTwin “Paradox”
x
w’
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2. Example: An astronaut travels to a star 100 lys away with a speed equal to 0.99c. How much time elapses on the ship as measured by someone in the ship?
The time on the ship is given by B.2
t’ = t{1- (v/c)2}1/2.
Note: t, the time elapsed on the Earth is {100 ly/0.99 ly/y} = 101 y, since v = 0.99c. Thus,
t0 =t’ = t{1- (v/c)2}1/2 ~ (101 yrs)(1 - .992)1/2 = 14 yrs.“proper time”
III. Time Dilation D. Implications
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1. Definition: the length of an object L is measured to be shorter when it is moving relative to the observer than when it is at rest.
2. Mathematical formulation:
L = L0{1- (v/c)2}1/2. (IV.D.1)
3. Example: in the previous example, if the ship has a length of 100m as measured in its rest frame, what is its length as seen from the Earth?
L = L0{1- (v/c)2}1/2 = (100m)(1 - .992)1/2 = 14 m.
IV. Length Contraction A. Formulation
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1. Relative velocity is not computed by simply adding (or subtracting velocity vectors.
2. Mathematical Formulation:
u = (v ± u’)/(1 ± vu’/c2).(V.A.1)
v u’
V. Velocity Addition A. Formulation
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3. Example: two particles collide with individual speeds of .9c and .5c, respectively as measured in the lab. What is the relative speed of the second particle as seen by the first particle?
Classically, we would just add these speeds and get 1.4c, which is clearly wrong if SR is valid.
u = (v + u’)/(1 + vu’/c2) = (1.4c)/(1 + .45c/c) = 0.97c.
v u’
V. Velocity Addition A. Formulation
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V. Velocity Addition B. Spacetime1. How does velocity addition appear in ST diagrams?
a) Galilean addition
x
w=ctv/c =
1
= arctan(.5) = 27o
Let a trainmove atv/c = 0.5 wrtthe platform.Let’s considerAn object at reston the trainas it speeds by the platform.
x’
v’/c = 0.5
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V. Velocity Addition B. Spacetime1. How does velocity addition appear in ST diagrams?
a) Galilean addition
x
w=ctv/c =
1
= arctan(.5) = 27o
Let a trainmove atv/c = 0.5 wrtthe platform.Let’s considerAn object at reston the trainas it speeds by the platform.
Now consider someone runningDown the train at u’/c = 0.5
x’
v’/c = 0.5
u’/c = 0.5
Galileo: combined
velocity = c. D’oh!
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V. Velocity Addition B. Spacetime1. How does velocity addition appear in ST diagrams?
a) Lorentz addition
x
w=ctv/c =
1
= arctan(.5) = 27o
Let a trainmove atv/c = 0.5 wrtthe platform.Let’s considerAn object at reston the trainas it speeds by the platform.
Now consider someone runningdown the train at u’/c = 0.5
x’
v’/c = 0.5
u’/c = 0.5
Einstein: combined
velocity = .8c.
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A. Momentum
p = mv/{1 - 2}1/2,= mv = mc. (VI.A.1)
Equation A.1 => v for a massive object can never be ≥ c.
B. Energy1. Relativistic Kinetic Energy:
K = mc2 - mc2 = ( - 1)mc2, or (VI.B.1)K = mc2( 1/{1 - 2}1/2 - 1).
Note: for << 1, then 1/{1 - 2}1/2 ~ 1 + (1/2)2, and
K~ mc2(1 + (1/2)2 - 1) = (1/2)mv2, (VI.B.2)
VI. Momentum & Energy
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2. Total Energy
E = mc2 = mc2 + K, where (B.3)
The quantity mc2 is the rest energy of the object. Using the
momentum equation (A.1), we can also write:
E2 = p2c2 + (mc2)2. (B.4)
The consequences of (B.3, 4) are that mass and energy are equivalent, and that they can be converted into each other.
VI. Momentum & Energy
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3. Example: Suppose our astronaut has a mass of 70kg. Compare her classical and relativistic KE.
K(class.) = 1/2mv2 = 3.2 x 1018 J.K(rel.) = mc2 - mc2 = 3.8 x 1019 J.
4. Example: How much energy would be released if our astronaut were converted completely into energy?
E(rest) = mc2 = 6.3 x 1018 J.E(tot) = E(rest) + K ~ 4 x 1019 J ~ 10,000 Megatons of TNT.
VI. Momentum & Energy
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5. Another example: An electron has a total energy of
10.0 “Mega-electron volts” (MeV) as measured in a particular laboratory frame. What is the value of its K, v?
Note: Eelectron(rest) = melectronc2 = 8.2 x 10-14 J = 0.511 MeV.
Thus, K = E – mc2 = 9.5 MeV.E = mc2; = E/mc2 = 19.6;
2 = {1 – 2}-1 = 384 => 2 = 1 - 1/384 = 0.997.
= 0.999
VI. Momentum & Energy