chapter 1

Chap 1 - 1Chap 1 - 1 Chap 1 - 2Chap 1 - 2

The Birth of Modern PhysicsThe Birth of Modern Physics9 Classical Physics of the 1890s

Mechanics Electromagnetism Thermodynamics

9 Unresolved Questions of 1895 Existence of Atoms

The atomic theory controversy raises fundamental questions. The structure of matter remained unknown with certainty.

Fundamental problems: The question of the existence of an electromagnetic medium The problem of observed differences in the electric and magnetic field

between stationary and moving reference systems The failure of classical physics to explain blackbody radiation.

The more important fundamental laws and facts of physical science have all been

discovered.

Chap 1 - 3Chap 1 - 3

9 Additional Discoveries Contribute to the Complications Discovery of X-rays Discovery of radioactivity Discovery of the electron Discovery of the Zeeman effect

The Birth of Modern PhysicsThe Birth of Modern Physics

These new discoveries and the many resulting complications required a revision of the fundamental physical assumptions.

modern theory of relativity and quantum mechanics


CHAPTER 1:SPECIAL THEORY OF RELATIVITYSPECIAL THEORY OF RELATIVITY

Albert Einstein (1879-1955)

1.1 Introduction1.2 Frame of reference1.3 Newtonian Relativity1.4 Transformation laws1.5 The Galilean transformations1.6 The speed of light and the Ether1.7 The Michelson-Morley experiment 1.8 Einsteins principle of relativity1.9 The Lorentz transformations1.10 The length contraction1.11 The time dilation 1.12 The relativity of simultaneity 1.13 The combination of velocities1.14 Relativistic dynamics1.15 Relativistic energy1.16 The equivalence of energy and mass


1.1 Introduction (I)Classical or Newtonian mechanics: 9 Deals with the motions of bodies travelling at velocities that are

very much less than the velocity of light.9 Three fundamental concepts of Physics, i.e., space, time and

mass are all absolute and invariant.1. The length of an object is independent of the conditions under which

it is measured.2. The interval of time between two events has the same value for all

observers.3. If two events are simultaneous for an observer, they are

simultaneous for all observers.4. The mass of a body does not depend on the velocity of its motion.5. The mass of an isolated system of bodies does not change with any

processes occurring within the system (law of conservation of mass).


1.1 Introduction (II)Relativistic mechanics: 9 Deals with the motions of bodies travelling at velocity of light

(c) or v c.9 The space, time, and mass are becomes state-of-motion

dependent (via ).9 The speed of light in free space (vacuum) is constant.

(a) Length Contraction(b) Time Dilation

(c) Simultaneity


Question???

The Classical or Newtonian mechanics are out of date.

Why still need to learn/take the ZCT 101 Mechanics???


1.1 Introduction (III) Coordinate system9 Space is 3-D in the sense that it is possible to describe

the position of a point by 3 coordinates numbers.9 In special theory of relativity (STR), the space and

time coordinates cannot be treated separately.E.g., LT, time dilation, length contraction formula.

9 A convenient way to express the results of STR is to regard events as occurring in a 4-D space-time in which the usual 3 coordinates x, y, z refer to space and a 4th coordinate ict refers to time, where i2 = -1.

9 In general, we cannot visualize space-time (i.e., just an imaginary).


1.1 Introduction (IV)9 We choose ict as the time coordinate instead of just t

because the quantity (length squared): s2 = x2 + y2 + z2 - (ct)2

is invariant under Lorentz transformation. (Do it as an (Do it as an exercise).exercise).

9 In this case, the quantity of s has the dimension of length and is called the space-time. It is analogous to distance in classical mechanics.

9 Minkowki diagrams can be used to classify the entire universe of space-time and clarify whether or not one event could be the cause of another.Pls. refer to Arthur Beiser, Concepts of Modern physics,

Appendix 2.

Chap 1 - 10Chap 1 - 10

Space-time diagram and Causality

Classification of one-dimensional space-time into past, future, and elsewhere regions. A particle with world line passing through O cannot reach regions marked elsewhere.

Three pairs of events in space-time: V,W; A,B; C,D.

V could cause W.A could cause B.C could not cause D.

Chap 1 - 11Chap 1 - 11

Frame of referenceChap 1 - 12Chap 1 - 12

Frame of reference

Chap 1 - 13Chap 1 - 13

1.2 Relativity & Frame of reference9 A system of coordinate axes which

defines the position of a particle in 2D or 3D space is called a frame of reference. E.g.: Cartesian system of coordinates.

xz

y

9 STR and the Galilean-Newtonian Relativity deals with how we observe events, particularly how objects and events are observed from different frames of reference. EVERYTHING IS RELATIVE NOTHING IS

ABSOLUTE.

Chap 1 - 14Chap 1 - 14

EVERYTHING IS RELATIVE NOTHING IS ABSOLUTE.

Chap 1 - 15Chap 1 - 15

1.3 Newtonian Relativity (Classic theory)

9 If Newtons laws are valid in one reference frame, then they are also valid in another reference frame moving at a uniform velocity relative to the first system. This reference frame is referred as inertial frame. This means that the all the Newtons laws are identical in all

inertial frames of reference.

9 This is referred to as the Newtonian principle of relativity or Galilean invariance.

Chap 1 - 16Chap 1 - 16

Inertial frame9 A reference frame in uniform motion of translation relative to one

another is Galilean frame or inertial frame. All the Newtons laws are valid in that frame. This is referred to as the Newtonian principle of relativity or Galilean

invariance9 An inertial frame is one in which a free body, subject to no forces,

travels in a straight line at constant velocity, i.e., Newtons 1stlaw holds.

9 Any reference frame moving with constant velocity with respect to an inertial frame is also an inertial frame.

9 Example of inertial frame: Earth is an inertial frame. A train moving at constant velocity with respect to the ground (earth) is

also an inertial frame.

By invariance of an equation it is meant that the equation will have the same form when determined by two observers.

Chap 1 - 17Chap 1 - 17

Inertial framesChap 1 - 18Chap 1 - 18

Non-inertial frames

Chap 1 - 19Chap 1 - 19EVERYTHING IS RELATIVE NOTHING IS ABSOLUTE

Chap 1 - 20Chap 1 - 20

Which one is moving???

Chap 1 - 21Chap 1 - 21EVERYTHING IS RELATIVE NOTHING IS ABSOLUTE

2nd Initial Frame

Rocket frame(Moving frame)

Constant vx1

1st Initial Frame

Lab frame(stationary frame)

3rd Initial Frame

Rocket frame(Moving frame)

Constant vx2

Chap 1 - 22Chap 1 - 22

1.4 Transformation laws9 Measurements done by any observers from all frame of

reference are equally valid, and are all equivalent. 9 Transformation laws such as Galilean transformation

(GT) and Lorentz transformation (LT) can be used to relate the measurements done in one frame to another.

9 In other words, once you know the values of a measurement in one frame, you can calculate the equivalent values as would be measured in other frames.

Chap 1 - 23Chap 1 - 23

View from different frames

9 In the Lab frame, the observer on the ground sees a parabolic trajectory

9 In the Rocket frame, the pilot sees the projectile to follow a vertically straight line downwards

Chap 1 - 24Chap 1 - 24

Transformation law for the coordinates

9 In Lab frame9 y = (gt2)/2 + H9 x = vxt

9 In rocket frame9 y = (gt2)/29 x = 0

Hy

y

9 Transformation law relating the coordinates of projectile in both frames is

9 y = y - H , x = x - vxt

y = 0

y = 0

Chap 1 - 25Chap 1 - 25

1.5 The Galilean transformations (I)9 S and S are two inertial frames.9 S at rest and S move with uniform velocity v along the

positive X direction.9 Assume that v


Galilean addition law for velocities - Galilean transformations (in terms of velocities)9 Galilean Transformation equations from S to S are given

by:

9 Herevx = velocity measured in system Svx = velocity measured in system SV = velocity of system S relative to system S

vx = vx-V vy = vy vz = vz t = t

Chap 1 - 31Chap 1 - 31

Example 1.1:Two cars are traveling at constant speed along a road in the same direction. Car A moves at 60 km/h and car B moves at 40 km/h, each measured relative to an observer on the ground (Figure 1.0a). What is the speed of car A relative to car B?

FIGURE 1.0a As observed by O at rest on the ground.

Chap 1 - 32Chap 1 - 32

Example 1.1:Two cars are traveling at constant speed along a road in the same direction. Car A moves at 60 km/h and car B moves at 40 km/h, each measured relative to an observer on the ground (Figure 1.0a). What is the speed of car A relative to car B?

Chap 1 - 33Chap 1 - 33

Solution:9 Let O be the observer on the ground, who observes car A to move

at vx = 60 km/h. 9 Assume O' to be moving with car B at u = 40 km/h. Then

vx = vx - u = 60 km/h - 40 km/h = 20 km/h9 Figure 1.0b shows the situation as observed by O'.

FIGURE 1.0 (b) As observed by O' in car B.

Chap 1 - 34Chap 1 - 34

Quick test:In Example 1.1, What is the speed of car B relative to car A?

Chap 1 - 35Chap 1 - 35

Quick test:In Example 1.1, What is the speed of car B relative to car A?

1.The Lab Frame (or stationary frame) is: (O, Car A, or Car B).

2.The Rocket Frame (or moving frame) is: (O, Car A, or Car B).

3.The speed of car B relative to the car A is:.. (vx, vx, or V).

4.The speed of the car A namely, the 60 km/h is:.. (vx, vx, or V)

5.The speed of the car B namely, the 40 km/h is :.. (vx, vx, or V)

Chap 1 - 36Chap 1 - 36

Example 1.1a:Two cars are traveling at constant speed along a road in the same direction. Car A moves at 60 km/h to the right and car B moves at 40 km/h to the left, each measured relative to an observer, O on the ground. What is the speed of car A relative to car B?

Chap 1 - 37Chap 1 - 37

Example 1.1a:Two cars are traveling at constant speed along a road in the same direction. Car A moves at 60 km/h to the right and car B moves at 40 km/h to the left, each measured relative to an observer, O on the ground. What is the speed of car A relative to car B?

1.The Lab Frame (or stationary frame) is: (O, Car A, or Car B).

2.The Rocket Frame (or moving frame) is: (O, Car A, or Car B).

3.The speed of car B relative to the car A is:.. (vx, vx, or V).

4.The speed of the car A namely, the 60 km/h is:.. (vx, vx, or V)

5.The speed of the car B namely, the 40 km/h is :.. (vx, vx, or V)

Chap 1 - 38Chap 1 - 38

Example Extra example:A spaceship moving away from the Earth at a speed of 0.80c fires a missile parallel to its direction of motion (Figure 2.0). The missile moves at a speed of 0.60c relative to the ship. What is the speed of the missile as measured by an observer on the Earth?

FIGURE 2.0. A spaceship moves away from Earth at a speed of 0.80c. An observer O' on the spaceship fires a missile and measures its speed to be 0.60crelative to the ship.

Chap 1 - 39Chap 1 - 39

Example Extra example:

1) The Lab Frame (or stationary frame) is:

(observer on Earth, spaceship, or missile).

2) The Rocket Frame (or moving frame) is:

(observer on Earth, spaceship, or missile)..

3) The speed of the missile as measured by an observer on the Earth is:

.. (vx, vx, or V).

4) The speed of the spaceship namely, the 0.80c is:.. (vx, vx, or V)

5) The speed of the missile namely, the 0.60c is :.. (vx, vx, or V)

Chap 1 - 40Chap 1 - 40

Solution:9 Need to find out:

9 The speed of the missile as measured by an observer on the Earth, i.e., vx.9 Here,

9 Earth = lab frame9 Spaceship = moving frame.

9 Given that:9 The speed of the spaceship relative to the Earth = 0.80c = V. 9 The speed of the missile relative to the ship = 0.60c = vx .

9 By using: vx = vx + V , then, vx = 0.60c + 0.80c = 1.40c

Chap 1 - 41Chap 1 - 41

The mystery of the speed of lightThe mystery of the speed of lightChap 1 - 42Chap 1 - 42

The mystery of the speed of lightThe mystery of the speed of light

Chap 1 - 43Chap 1 - 43

The Transition to Modern Relativity

9 Although Newtons laws of motion had the same form under the Galilean transformation, Maxwells equations did not.

9 In Maxwells theory the speed of light, in terms of the permeability and permittivity of free space, was given by:

i.e., the velocity of light is a constant ( c 3.00 108m s-1.)

001 =c

Chap 1 - 44Chap 1 - 44

The mystery of the speed of lightThe mystery of the speed of light

Chap 1 - 45Chap 1 - 45

The mystery of the speed of lightThe mystery of the speed of lightChap 1 - 46Chap 1 - 46

1.6 The speed of light and the Ether9 The laws of mechanics are the same in all inertial reference frames.9 Naturally, it is assume that the laws of electricity and

magnetism also are the same in all inertial reference frames. This assumption leads to a paradox concerning the velocity of light.

9 From the Maxwells theory, the speed of light is a constant.9 This deduction stands in contradiction to the Galilean

transformation: The velocity of light ought not to be the same in all reference frames.

9 Ether was proposed as an absolute reference system in which the speed of light was a constant and from which other measurements could be made.

9 As an direct analogy to mechanical wave propagating in elastic medium such as sound wave in air light is thought to be propagating in a medium called ether.

Chap 1 - 47Chap 1 - 47

1.7 The Michelson-Morley experiment (I)9 The Michelson-Morley (MM)

experiment was an attempt to show the existence of ether.

Edward Morley(1838-1923)

Albert Michelson(1852-1931)

Albert Michelson was the first U.S. citizen to receive the Nobel Prize for Physics (1907), and built an extremely precise device called an interferometer to measure the minute phase difference between two light waves traveling in mutually orthogonal directions.

Chap 1 - 48Chap 1 - 48

The Michelson Interferometer

Chap 1 - 49Chap 1 - 49

1.7 The MM experiment (II)

9 Michelson and Morley realized that the earth could not always be stationary with respect to the ether. And light would have a different path length and phase shift depending on whether it propagated parallel and anti-parallel or perpendicular to the ether.

Perpendicular propagation

Mirror

Supposed velocity of earth through the ether

Parallel and anti-parallel propagation

Beam-splitter

Mirror

Chap 1 - 50Chap 1 - 50

1.7 The MM experiment (III)

9 If light requires a medium, then its velocity depends on the velocity of the medium. Velocity vectors add.

Parallel velocities Anti-parallel velocities

vetherr v ether

r

v v+v ligtotal ht ether=totalv

rvlightr

v v-v ligtotal ht ether=

vtotalr

vlightr

Parallel velocities Anti-parallel velocities

vetherr v ether

r

v v+v ligtotal ht ether= v v+v ligtotal ht ether=totalv

rvrvlightr

v v-v ligtotal ht ether= v v-v ligtotal ht ether=

vtotalr

vlightr

9 In the other arm of the interferometer, the total velocity must be perpendicular, so light must propagate at an angle.

Perpendicular velocity to mirror

Perpendicular velocity after mirror

v etherr

vlightr vtotal

rv etherr

vlightr

vtotalr

22etherlighttotal vvv =

Perpendicular velocity to mirror

Perpendicular velocity after mirror

v etherrv etherv etherr

vlightrvlightv lightr vtotal

rvtotalr

v etherrv etherv etherr

vlightrvlightvlightr

vtotalrvtotalr

22etherlighttotal vvv =

Chap 1 - 51Chap 1 - 51

1.7 The MM experiment (IV)

9 Let c be the speed of light, and v be the velocity of the ether.

9 The delays for the two arms depend differently on the velocity of the ether!

Perpendicular propagation

Parallel and anti-parallel propagation v ether

r

2 2 2 2

2 2

2 2

v v( v) ( v)

v v2

v2 1

[1 v / ]

L Ltc cL c L cc c

Lcc

Lc c

= + ++ = +

= =

||

2 2

2 2

2v

2 11 v /

LtcLc c

= =

Chap 1 - 52Chap 1 - 52

1.7 The MM experiment (V)

9 Because we dont know the direction of the ether velocity, Michelson and Morley did the measurement twice, the second time after rotating the apparatus by 90.

9 The delay reverses, and any fringe shift seen in this second experiment will be opposite that of the first.

9 Upon rotating the apparatus by 90, the optical path lengths are interchanged producing the opposite change in time. Thus the time difference between path differences is given by:

v etherrv etherrv etherrv etherrv etherr

9 By assuming v

Chap 1 - 53Chap 1 - 53

MM Experimental Prediction:

9 The phase shift is w times this relative delay:

9 The Earths orbital speed is: v = 3 104 m/s9 and the interferometer size is: L = 1.2 m9 So the time difference becomes: 8 1017 s9 which, for visible light, is a phase shift of: 0.2 rad = 0.03 periods9 Although the time difference was a very small number, it was

well within the experimental range of measurement for visible light in the Michelson interferometer, especially with a folded path.

or:2

3v2 Lc

22v4 L c

( ) 23v2 2t t L c || vChap 1 - 54Chap 1 - 54

Typical interferometer fringe pattern expected when the system is rotated by 90

Interference fringe schematic showing (a) fringes before rotation and(b) expected fringe shift after a rotation of the interferometer by 90.

Chap 1 - 55Chap 1 - 55

MM experiment results9 MM expected a fringe shift of about 0.4 in their apparatus when it

was rotated through 90 and their believed that they could detect a shift as small as 0.01 of a fringe.

9 But no displacement of the fringes was observed. 9 They repeated the experiment at different points on the earths

surface and at different seasons of the year without detecting any measurable shift in fringes.

9 This negative result suggests that it is impossible to measure the speed of the earth relative to the ether.

9 Thus, ether seems not to exist! 9 All attempts to make ether as a

fixed frame of reference failed.

Their apparatus

Chap 1 - 56Chap 1 - 56

Explanations for MMs null result (I)9 The negative result of the MM experiment can be explained by

the following three explanations:(a) Ether drag hypothesis.(b) Lorentz-FitzGerald Contraction.(c) Principle of constancy of the speed of light.

1. Ether drag hypothesis: This hypothesis suggested that the Earth dragged the ether along as it rotates

on its axis and revolves about the sun. This was contradicted by stellar aberration wherein telescopes had to be

tilted to observe starlight due to the Earths motion. If ether were dragged along, this tilting would not occur.

Chap 1 - 57Chap 1 - 57

The shift in the observed position of the stars caused by the ether wind.

Chap 1 - 58Chap 1 - 58

Explanations for MMs null result (II)2. Lorentz-FitzGerald Contraction: Proposed independently by Lorentz

and FitzGerald. It was suggested that there was

contraction of bodies along the direction of their motion through the ether.

George F. FitzGerald

(1851-1901)

Hendrik A. Lorentz (1853-1928)

2 21 v / cvelocity of frame velocity of light

v

If a body is moving with a speed v parallel to its length, then the new length will be contracted by a factor of:

Consequently, there will be no time difference expected. But there was no insight as to why such a contraction should

occur.

Chap 1 - 59Chap 1 - 59

3. Principle of constancy of the speed of light:


Explanations for MMs null result (III)

Proposed independently by Albert Einstein.

He concluded that the velocity of light in space is a universal constant.

Hence, the speed of light is c rather than |(c + v)| in any frame.

Consequently, ether hypothesis is not true and can be disposed.

Chap 1 - 60Chap 1 - 60

Two postulates of special theory of relativity (STP):


1.8 Einsteins principle of relativity

(1) The principle of relativity: All the laws of physics are the same in all

inertial frames of reference. There is no way to detect absolute motion, and

no preferred inertial frame of reference exists.

(2) The constancy of the speed of light: The speed of light in free space is constant.

It is independent of the relative motion of the source and the observer.

9 The STP eliminates the Galilean-Newtonian concept of absolute space and time, and considers both as being relative to the observer.

Chap 1 - 61Chap 1 - 61

1.9 The Lorentz transformations (I)9 The constancy of the speed of light is not compatible

with Galilean transformations.9 Hence, a new transformation equations is introduced.

This new transformation equations is known as Lorentz transformations (LT).

It was discovered by Hendrik A. Lorentz (1853-1928, Dutch physicist).

It is consistent with the new concept of the invariance of light velocity in free space.

9 Although the LT was discovered by Lorentz, but its real significance in physics theory transcending electromagnetism was first recognized by Einstein.

Chap 1 - 62Chap 1 - 62

1.8 The Lorentz transformations (II)9 Consider two observers O and O in two systems S and S.9 S at rest and S is moving with a constant velocity v relative to

system S along the positive X axis.9 Let the origins of the two frames coincide at t = 0.9 Suppose a light pulse is emitted when t = 0. 9 The light signal travels as a spherical wave at a constant speed c in

both frames.

Chap 1 - 63Chap 1 - 63

1.8 The Lorentz transformations (III)

9 Since there is no motion in the Y and Z directions, y = y = 0, and z= z = 0, Eqs. (1) and (2) become:

22222 tczyx =++22222 tczyx =++

9 After a time t (t), the origin of the wave centred at O (O) has a radius r = ct (r = ct) , where r2 = x2 + y2 + z2 (r 2 = x 2 + y 2 + z 2).

Spherical wave-front in S: (1)

(2)Spherical wave-front in S:

9 Note that this cannot occur in Galilean transformations, because by substituting x = x + vt, there are a couple of extra terms (2xvt + v2t2) in the primed frame.

0222 = tcx 0222 = tcx9 Subsequently: 222222 tcxtcx = (3)

22222222222 )2( tcxtctvxvtxtcx ++=

Chap 1 - 64Chap 1 - 64

1.8 The Lorentz transformations (IV)9 Clearly, a different transformation is required if the postulates of

STP are to be satisfied. i.e., the spherical wave-fronts in both frames are conserved.

9 A reasonable guess about the dependence of x on x and t is:)( vtxkx = (4)

9 Here k is a dimensionless factor that does not depend on x or t but is some function of v/c.

9 Eq. (4) reduces to Galilean transformation (i.e., v/c 0, and k = 1) for low speed (v

Chap 1 - 65Chap 1 - 65

1.8 The Lorentz transformations (V)9 The solutions for constants k, a and b are:

( )2

2211

cvb

cvak

=

===

(7)

(6)

9 The constant k is known as the Lorentz factor, :9 The complete LT from S to S are:

( )

( )( )222

22

1

1

cvcvxtt

zzyy

cvvtxx

=

==

=

Chap 1 - 66Chap 1 - 66

The inverse LT from S to SStep 1: Replace -v with +v.Step 2: Replace primed quantities with unprimed

and unprimed with primed.

(8)( )

( )( )222

22

1

1

cvcxvtt

zzyy

cvvtxx

+=

==

+=

Chap 1 - 67Chap 1 - 67

LT vs GT9 In the LT, t depends on both t and x. Like wise, t depends on

both t and x.9 In GT, t = t .9 Space and time now becomes state-of-motion dependent (via )

the length and time interval measured become dependent of the state of motion (in terms of ) in contrast to Newtons classical viewpoint

9 LT reduces to GT when v

Chap 1 - 69Chap 1 - 69

9 When v

Chap 1 - 73Chap 1 - 73

1.10 The combination of velocities (IV)

9 The inverse combination laws of velocities:

( )

( )

+=

+=

+=

+=

++=

x

z

x

zz

x

y

x

yy

x

xx

vcv

cvv

vcv

vv

vcv

cvv

vcv

vv

vcv

vvv

2

22

2

2

22

2

2

1

1

1

1

1

1

1

(22)

(23)

(24)

Chap 1 - 74Chap 1 - 74

Comparing the LT of velocity with that of GT

2

''' 1

xx

x

u vdxu u vdtc

= =

Galilean transformation of velocity:

'x xu u v= LT reduces to GT in the limit:

Lorentz transformation of velocity:

2xu v c

Chap 1 - 77

Example: Involved lightA boxcar moves right at a very high speed, V. A green flash of light moves from left to right, and a blue flash from right to left.

v

Qs:1) According to an observer in the boxcar, which flash

will reach him/her first?2)According to an observer on the ground, which flash

will reach the observer in the boxcar first??

Chap 1 - 78Chap 1 - 78

Example 1.7:A spaceship moving away from the Earth at a speed of 0.80c fires a missile parallel to its direction of motion (Figure 2.0). The missile moves at a speed of 0.60c relative to the ship. What is the speed of the missile as measured by an observer on the Earth?

FIGURE 2.0. A spaceship moves away from Earth at a speed of 0.80c. An observer O' on the spaceship fires a missile and measures its speed to be 0.60crelative to the ship.

Chap 1 - 79Chap 1 - 79

Example Extra example:

1) The Lab Frame (or stationary frame) is:

(observer on Earth, spaceship, or missile).

2) The Rocket Frame (or moving frame) is:

(observer on Earth, spaceship, or missile)..

3) The speed of the missile as measured by an observer on the Earth is:

.. (vx, vx, or V).

4) The speed of the spaceship namely, the 0.80c is:.. (vx, vx, or V)

5) The speed of the missile namely, the 0.60c is :.. (vx, vx, or V)

Chap 1 - 80Chap 1 - 80

Solution:9 Here O' is on the ship and O is on Earth; O' moves with a speed of

u = 0.80c relative to O. The missile moves at speed v' = 0.60crelative to O', and we seek its speed v relative to O.

9 From the Lorentz transformation of velocity,

9 According to classical kinematics (the Galilean addition law for velocities ), an observer on the Earth would see the missile moving at 0.60c + 0.80c =1.40c, thereby exceeding the maximum relative speed of c permitted by relativity.

( )( )[ ] ccccc ccvcv

uvvx 95.048.1

40.180.060.0180.060.0

12

2

==++=

++=

Chap 1 - 81Chap 1 - 81

Example E6: Advanced questionTwo rockets are leaving their space station along perpendicular paths, as measured by an observer on the space station. Rocket 1 moves at 0.6c and rocket 2 moves at 0.8c, both measured relative to the space station. What is the velocity of rocket 2 as observed by rocket 1?

Solution:9 Let observer O is the space station

and observer O is rocket 1.9 Since rocket 1 and rocket 2 are

leaving their space station along perpendicular paths, we take these to be the X and Y direction of the reference frame of O, respectively.

9 Thus O observes rocket 2 to have velocity components vx = 0, and vy = 0.8c, as shown in Fig. (a).

Chap 1 - 82Chap 1 - 82

Solution (cont.):

( ) ( )c

cc

ccc

ucv

cvuu

c

ccc

ucv

vuu

x

yy

x

xx

64.006.01

6.018.0

1

1

6.006.01

6.00

1

2

2

2

22

22

=

=

=

=

=

=

9 Consequently, ux = 0, uy = 0.8c, v = 0.6c, ux = ?, uy = ?

9 From the Lorentz transformation of velocity,

9 Thus, according to O , the situation looks like Fig. (b).9 Finally, the speed of rocket 2 according to O is

( ) ( ) ( ) ( ) cccuuu yx 88.064.06.0 2222 =+=+=

Chap 1 - 83Chap 1 - 83

Consequences of the LT9 According to LT, the space and time are state-of-

motion dependent (via ).9 The relativistics factor/indicator.

( )

( )( )22

2

22

1

1

cvcxvtt

zzyy

cvvtxx

+=

==

+=( )

( )( )222

22

1

1

cvcvxtt

zzyy

cvvtxx

=

==

=

LT from S to SLT from S to SLength

contraction

Simultaneity problems

Time dilation

Chap 1 - 84Chap 1 - 84

Consequences of the LT (continue)

(a) Length Contraction (b) Time Dilation

(c) Simultaneity

Faster means shorter Moving clocks run slow

The simultaneity is relative

Chap 1 - 85Chap 1 - 85

Consequences of the LT

(a) Length Contraction: Faster means shorter.

(b)Time Dilation: Moving clocks run slow. A moving clock ticks more slowly than a clock at rest.

(c) Simultaneity The simultaneity is relative. The simultaneity of two events depends on the reference

frame.

9 The most interesting consequences of the LT:

Chap 1 - 86Chap 1 - 86

1.10 Length contraction (I)9 Consider two inertial frames S and S with their X-axes coinciding

at time t = 0.9 S is moving with a uniform velocity v along the positive X

direction with respect to S.9 Assume that a rod (AB) is at rest relative to S .

Chap 1 - 87Chap 1 - 87

1.10 Length contraction (II)9 The lengths of the rod measured at any instant of time in frames S

and S are:Frame S : L0 = x2 x1 (proper length) (9)Frame S : L = x2 x1 (improper length) (10)

9 Based on the LT, the length of the rod measured in frame S can be deduced as followed:

L0 = x2 - x1 = [(x2 - x1)] or (11)

9 Consequently, the length of the rod measured in frame S is:L = (1/) L0 (12)

9 According to the observer in S, the length of the rod (in S ) has contracted by the factor of .

( )220 1 cvLLL

==

is > 1

Chap 1 - 88Chap 1 - 88

Chap 1 - 89Chap 1 - 89

1.10 Length contraction (III)Notes:

9 The proper length of an object is the length determined by an observer at rest with respect to the object.

9 The shortening or contraction in the length of an object along its direction of motion is known as the Lorentz Fitzgerald contraction.

9 There is no contraction in a direction perpendicular to the direction of motion.

9 The contraction becomes appreciable only when v c. 9 The contraction is reciprocal, i.e., if two identical rods are at rest

one in S and the other in S, each of the observers finds that the other is shorter than the rod of his own system.

Chap 1 - 90Chap 1 - 90

1.10 Length contraction (IV)9 A body which appears to be spherical to an observer at rest relative

to it, will appear to be an oblate spheroid to a moving observer. Similarly, a square and a circle in one appear to the observer in the other to be a rectangle and an ellipse respectively.

9 In conclusion: Length contraction only happens along the direction of motion.

Chap 1 - 91Chap 1 - 91

1.9 Length contraction (V)9 The contraction is reciprocal, i.e., if two identical rods are at rest

one in S and the other in S, each of the observers finds that the other is shorter than the rod of his own system.

(b) From the rocket's frame of reference, the Earth appears contracted.

(a) The Earth views the passing contracted rocket.

9 Final remark: 9Length contraction suggests that objects in motion are

measured to have a shorter length than they do at rest.

Chap 1 - 92Chap 1 - 92

Example E1: A rod 1 m long is moving along its length with a velocity 0.6c. Calculate its length as it appears to an observer (a) on the earth and (b) moving with the rod itself.

Solution:9 Here, 1 m is the proper length (L0) of the rod in its moving frame

of reference.(a) Let L be the length of the rod as it appears to an observer in the stationary

reference frame of the earth.Here, L0 = 1 m, v = 0.60c; L = ?

Hence, the observer on the earth will estimate the length of the rod to be 0.8 m.

(b) For an observer moving with the rod itself, the length of the rod is 1 m.

( ) ( ) m 8.06.01111 22

2

2

00 =

=

==

cc

cvLLL

Chap 1 - 93Chap 1 - 93

Example 1.4 : Advanced Qs.A spaceship in the form of a triangle flies by an observer at 0.95c. When the ship is measured by an observer at rest with respect to the ship (see fig (a)), the distance x and y are found to be 50.0 m and 25.0 m, respectively. What is the shape of the ship as seen by an observer who sees the ship in motion along the direction shown in fig (b)?

Chap 1 - 94Chap 1 - 94

Solution:9 The observer sees the horizontal length of the ship to

be contracted to a length of 9 L = Lp/ = 50 m(1 0.9502) = 15.6 m9 The 25 m vertical height is unchanged because it is

perpendicular to the direction of relative motion between the observer and the spaceship.

Chap 1 - 95Chap 1 - 95

1.11 The time dilation (I)9 A gun is placed at the position (x, y, z) in S.9 Suppose it fires two shots at times t1 and t2 measured with respect to S .9 In S the clock is at rest relative to the observer and the moving clock is always

at x = 0.9 The time interval measured by a clock at rest relative to the observer is called the

proper time interval (t ).9 Hence, t = t2 - t1 is the time interval between the two shots for the observer

in S.9 Since the gun is fixed in S, it has a velocity v with respect to S in the direction of

the positive X-axis.9 Let t = t2 - t1 represent the time interval between the two shots as measured by

an observer in S. (t = improper time)9 From the inverse LT:

9 Hence,

( )( )22

22

21 cv

cvxtt

+=

( )22 1212 1 cvtttt

= ( ) tcvtt = = 1 22 or (13)

( )2211

cv=

( )( ) 1 22

21

1cv

cvxtt

+= and

Chap 1 - 96Chap 1 - 96

1.11 The time dilation (II)9 Since the is > 1, then t > t moving clocks run slow.9 Thus, the time interval, between two events occurring at a given point in the

frame S appears to be longer to the observer in the frame S; 9 Recheck the definition:

t = time interval measured by the observer at rest relative with respect to clock in S (stationary clock)

t = time interval measured by the observer moving with respect to clock in S (moving clock)

9 This effect is called time dilation.9 The time dilation is reciprocal:

observers in S see time travel faster than for those in S. And vice versa!Proper Time (or proper time interval): The time measured by a clock at rest relative to the observer. Or time measured with a clock that stays at one place in its reference frame.Example:If you observe your wristwatch while you are walking around, you are

measuring???proper time intervals

Chap 1 - 97Chap 1 - 97

However, even to get Bob's time to flow at half the rate of Alice's, he must move at around 86% the speed of light. So it is not an issue for life on earth. However,

time dilation has actually been observed, as we'll next encounter.

Chap 1 - 98Chap 1 - 98

9 According to an accurate clock on a fast-moving train, a person (a) begins dinner at 7:00 and (b) finishes at 7:15. At the beginning of the meal, two observers on Earth set their watches to correspondwith the clock on the train. These observers measure the eating time as 20 minutes.

Chap 1 - 99Chap 1 - 99

Time Dilation

(a) A mirror is fixed to a moving vehicle, and a light pulse is sent out by observer O at rest in the vehicle. (b) Relative to a stationary observer O standing alongside the vehicle, the mirror and O move with a speed v. (c) The right triangle for calculating the relationship between t and tp.

Chap 1 - 100Chap 1 - 100

Observing Muons: Read Yourself

(a) A muon created in the upper atmosphere and moving at 0.99c relative Earth would ordinarily travel only 660 m, on the average, before decaying after 2.2x10-6 s. (b) Because of time dilation, an Earth observer measures a longer muon lifetime, so the muons travel an average of 4.8x103 m before decaying. From the muons point of view, by contrast, their lifetime is only 2.2x10-6 s on the average, but the distance between them and the Earth is contracted, again making it possible for more of them to reach the surface before decaying.

Chap 1 - 101Chap 1 - 101

Example 1.5:9 A clock in a space ship emits signals at intervals of 1

second as observed by an astronaut in the space ship. If the space ship travels with a speed of 3 107 ms-1, what is the interval between successive signals as seen by an observer at the control centre on the ground?

9 Here, t0 = 1 s; v = 3 107 ms-1; and c = 3 108 ms-1; t = ?

( ) ( )( )s 005.1

1031031

11

28

27220 =

== cvtt

Solution:

Chap 1 - 102Chap 1 - 102

Example E4: Suppose that a space ship (S) moves past the earth (S) at speed 0.999 c. Clocks on the earth and on the ship are synchronized at that instant, i.e., t = 0, t = 0. The space ship continues on to a star 100 ly away. Clocks at the star are synchronized with those on earth.(a) According to an observer on earth, how far did the ship travel and how long did it take? (b) According to an observer on the ship, how far did he travel and how long did it take?

Note:1 ly = 1 light year = the distance light travels in vacuum in 1 year.1 ly = 9.46 1015 m

Chap 1 - 103Chap 1 - 103

Solution:9 Consider the arrival of the ship at the star to be an event with

coordinates:(x, t) in frame S and (x, t) in frame S

(a) According to an observer on earth, x = 100 ly, t = x/v = (100 ly)/(0.999 c) = 100.10 years

(b) According to an observer on the ship, x = 0. Because he is still at his origin when he arrives at the star.Or x = (x vt) = 0 where = 22.37

t = t / = (100.1 y)/(22.37) = 4.47 yearsOr

( ) ( ) years 47.4100999.010.10037.22 22 =

=

=c

lyxcycvxt-t

Chap 1 - 104Chap 1 - 104

Example R1: A -meson with an average lifetime of 2 x 10-6 s is created in the upper atmosphere at an elevation of 6000 m. When it is created it has a velocity of 0.998c in a direction toward the earth. What is the average distance that it will travel before decaying, as determined by an observer on the earth?

Solution:9 (Classically, this distance is

d = vt0 = (0.9983 108) (210-6) = 599 mso that -mesons would not, on the average, reach the earth.)

9 As determined by an observer on the earth, the lifetime is increased because of time dilation:

9 The average distance moved by the particle before disintegrating is given by:d = vtearth = (0.9983 108) (31.610-6) = 9470 m

( ) s6.13s 106.3110998.01102

1 6

28

6

2

20

0 ===

==

cv

tttearth

Chap 1 - 105Chap 1 - 105

Example R1 cont: Now, consider an observer at rest with respect to the -meson. How far will he measure the earth to approach him before the -meson disintegrates? Compare this distance with the distance he measures from the point of creation of the -meson to the earth. Solution:9 As determined by an observer at rest with respect to the -meson, the distance

traveled by the earth is:d = vt0 = (0.9983 108) (210-6) = 599 m

9 The initial distance, L, to the earth, however, is shortened because of the Lorentz contraction:

9 Thus, an observer on the -meson determines that, on the average, it will reach the earth, in agreement with the previous result.

( ) m 37910998.01)m 599(1 1 282200 ==== cvLLL

Chap 1 - 106Chap 1 - 106

Twin Paradox: Read yourself

Chap 1 - 107Chap 1 - 107Twin Paradox: read yourself

The Set-upTwins Mary and Frank at age 30 decide on two career paths: Mary decides to become an astronaut and to leave on a trip 8 lightyears (ly) from the Earth at a great speed and to return; Frank decides to reside on the Earth.

The ProblemUpon Marys return, Frank reasons that her clocks measuring her age must run slow. As such, she will return younger. However, Mary claims that it is Frank who is moving and consequently his clocks must run slow.

The ParadoxWho is younger upon Marys return?

Chap 1 - 108Chap 1 - 108The Resolution

1) Franks clock is in an inertial system during the entire trip; however, Marys clock is not. As long as Mary is traveling at constant speed away from Frank, both of them can argue that the other twin is aging less rapidly.

2) When Mary slows down to turn around, she leaves her original inertial system and eventually returns in a completely different inertial system.

3) Marys claim is no longer valid, because she does not remain in the same inertial system. There is also no doubt as to who is in the inertial system. Frank feels no acceleration during Marys entire trip, but Mary does.

Chap 1 - 109Chap 1 - 109

Example E6: Time dilation at 100 km/h.

A car traveling with constant velocity of 100 km/h covers a certain distance in 10.00 s according to the drivers watch. What does an observer at rest on Earth measure for the time interval?

Chap 1 - 110Chap 1 - 110

Chap 1 - 111Chap 1 - 111

1.11 Lorentz Factor

9 Lorentz factor:

9 Note that as v > 1 is the ultra-relativistic regime where SR is most

pronounce.

2

1 1

1 vc

=

Chap 1 - 112Chap 1 - 112

What happens at high and low speed???

9 At low speed (v

Chap 1 - 113Chap 1 - 113

Space travel with time-dilation9 To an observer on the Earth frame, the person in a

rocket frame traveling near the light speed appears to be in a slow motion mode. This is because, according to the Earth observer, the rate of time flow in the rocket frame appear to be slower as compared to the Earths frame rate of time flow.

9 A journey that takes, say, 10 years to complete, according to a traveler on board (this is his proper time), looks like as if they take 10 yr according to Earth observers.

Chap 1 - 114Chap 1 - 114

Space travel with time-dilation - Example9 A spaceship traveling at speed v = 0.995c is sent to

planet 100 light-year away from Earth9 How long will it takes, according to a Earths

observer?9 t = 100 ly/0.995c = 100.05yr9 But, due to time-dilation effect, according to the

traveler on board, the time taken is only = t/ = t(1-0.9952) = 9.992 yr, not 100.05 yr as the Earthly think

9 So it is still possible to travel a very far distance within ones lifetime ( 50 yr) as long as (or equivalently, v) is large enough


Chap 1 - 117Chap 1 - 117

Example R2:An atomic clock is placed in a jet airplane. The clock measures a time interval of 3600 s when the jet moves with a speed of 400 m/s. How much longer or shorter a time interval does an identical clock held by an observer on the ground measure? (Hint: For v/c

Chap 1 - 121Chap 1 - 121

1.12 The relativity of simultaneity (I)9 Consider two events the explosion of a pair of time bombs that

occur at the same time to an observer O in a reference frame S.9 Let the two events occur at different locations x1 and x2.9 Consider another observer O in S moving with a uniform relative

speed v with respect to S in the positive X-direction.9 To O , the explosion at x1 and t0 occurs at the time:

( )( ) 1 22

210

1cv

cvxtt

=

( )( ) 1 22

220

2cv

cvxtt

=

(14)

and the explosion at x2 and t0 occurs at the time:

(15)

Chap 1 - 122Chap 1 - 122

1.12 The relativity of simultaneity (II)9 The time interval between the two events as observed by the

observer O is:( )( )

( ) 1 222

2112

cv

cvxxttt

==

9 This is not zero.9 This indicates that two events at x1 and x2, which are simultaneous

to the observer in S, do not appear so to the observer in S.9 Therefore, the concept of simultaneity has only a relative and not

an absolute meaning.

(16)

Chap 1 - 123Chap 1 - 123

Example E6: At 6.00pm, the street lights in New York and Boston are switched on simultaneously in the reference frame of the Earth. What is the time difference as reckoned in the reference frame of a spaceship traveling in the direction New York Boston at a speed of 0.9c? The distance between New York and Boston is 290 km.

Solution:9 With the X-axis along the direction New York Boston, the

displacement is 290 km.9 Consequently, x1 x2 = 290 km; v = 0.9c; t = ?9 By using the time interval formula [Eq. (16)],

( )( ) ( )( )( )

( )s 10996.1

9.01

9.010290

1

3

2

2

23

2

2

221 =

=

=

cc

cc

cv

cvxx

t

Chap 1 - 124Chap 1 - 124

Example 1.6: Principle of Relativity AppliedDivide the following items into two lists, On one list, labeled SAME, place items that name properties and laws that are always the same in every frame. On the second list, labeled MAY BE DIFFERENT. place items that name properties that can be different in different frames:

a. the time between two given eventsb. the distance between two given eventsc. the numerical value of Plancks constant hd. the numerical value of the speed of light ce. the numerical value of the charge e on the electronf. the mass of an electron (measured at rest)g. the elapsed time on the wristwatch of a person moving between two given

eventsh. the order of elements in the periodic tablei. Newtons First Law of Motion (A particle initially at rest remains at rest, and

)j. Maxwells equations that describe electromagnetic fields in a vacuum

Chap 1 - 125Chap 1 - 125

SolutionTHE SAME IN ALL FRAMES9 c. numerical value of h9 d. numerical value of c9 e. numerical value of e9 f. mass of electron (at rest)9 g. wristwatch time between two

event (this is the proper time interval between two event)

9 h. order of elements in the periodic table

9 i. Newtons First Law of Motion9 j. Maxwells equations

MAY BE DIFFERENT IN DIFFERENT FRAMES

9 a. time between two given events

9 b. distance between two give events

Chap 1 - 126Chap 1 - 126

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modern theory of relativity

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