Chapter 20 The special theory of relativity

Albert Einstein ( 1879 ~ 1955 )

20-1 Troubles with classical physics

The kinematics developed by Galileo and the mechanics developed by Newton, which form the basis of what we call “classical physics”, had many triumphs. However, a number of experimental phenomena can not be understoodwith these otherwise successful classical theories.

1. Troubles with our ideas about timeThe pions ( or ) created at rest are observed

to decay ( to other particles ) with an average lifetime of only .

In one particular experiment, pions were created in motion at a speed of . In this case they were observed to travel in the laboratory an average distance of before decaying, from which we conclude that they decay in a time given by , much larger than the lifetime measured for pions at rest.

This effect, called “time dilation”, which cannot be explained by Newtonian physics. In Newtonian physics time is a universal coordinate having identical values for all observers.

ns0.26

cv 913.0

nsvD 7.63

mD 4.17

2. Trouble with our ideas about lengthSuppose an observer in the above laboratory placed one marker at the location of the pion’s formation and another at the location of its decay.

The distance between the markers is measured to be 17.4m. Now consider the observer who is traveling along with the pion at a speed of u=0.913c. This observer, to whom the pion appear to be at rest, measures its lifetime to be 26.0ns, and the distance between the markers is

Thus two observers measure different value for the same length interval.

mc 1.7)100.26)(913.0( 9

3. Troubles with our ideas about light

20-2 The postulates of special relativity

1. Einstein offered two postulates that form the basis of his special theory of relativity.(I) The principle of relativity: “The laws of physics are the same in all inertial reference frames.”(II) The principle of the constancy of the speed of light : “ The speed of light in free space has the same value c in all inertial reference frames.”

2. The first postulatefirst postulate declares that the laws of physics are absolute, universal, and same for all inertial observers.

The Second postulateSecond postulate is much more difficult to accept, because it violates our “ common sense”, which is firmly grounded in the Galilean kinematics that we have learned from everyday experiences.

It implies that “it is impossible to accelerate a particle to a speed greater than c”.

20-3 Consequences of Einstein’s postulates

1.The relativity of time

We consider two observers: S is at rest on the ground, and S’ is in a train moving on a long straight track at constant speed u relative to S.

The observers carry identical timing devices, illustrated in Fig 20-4, consisting of a flashing light bulb F attached to a detector D and separated by a distance from a mirror M.

The bulb emits a flash of light that travels to the mirror, when the reflected light returns to D, the clock ticks and another flash is triggered.

0L

The time interval between ticks is: (20-1)The interval is observed by either S or S’ when the clock is at rest respect to that observer.

M

F D

Fig 20-4

0L

0t

cLt 0

02

0t

We now consider the situation when one observer looks at a clock carried by the other. Fig 20-5 shows that S observes on the clock carried by S’ on the moving train.

F D

S

A B C

L L

'S'S'S

tu

Fig 20-5

F F DD

According to S, the flash is emitted at A, reflected at B, and detected at C.This interval is

(20-20)Substituting for from Eq(20-1) and solving Eq(20-2) for gives

(20-3)

c

tuL

cLt

220 )2(22

2

0

)(1 cut

t

t

t0L

The time interval measured by the observer (S’) relative to whom the clock is at rest is called the “proper time(正确时间 ) ”, and .

That is, the observer relative to whom the clock is in motion measures a greater interval between ticks. This effect is called “time dilation”. All observer in motion relative to the clock measure “longer intervals”.

0t

tt 0

Eq(20-3) is valid for any direction of the relative motion of S and S’.

2. The relativity of length Fig 20-6 shows the sequence of events as observed by S for the moving clock which is on the train sideway, so that the light now travels along the direction of motion of the train.

According to S the length of the clock is L, which is different from the length measured by S’, relative to whom the clock is at rest.

0L

(A) (B)L

S(C)

S’S’

S’

u

u

1tu

2tu2tc

22 tuLtc

11 tctuL

Fig 20-6

u

M

F D

0L

FD

FD

FD

In the process from A, B to C, the total time taken is (20-6)From Eq(20-3), setting (20-7)

Setting Eqs(20-6) and (20-7) equal to one another and solving, we obtain (20-8)

221

)(1

12

cuc

LucL

ucLttt

2

0

2

0

)(1

12

)(1 cuc

L

cut

t

cL

t 00

2

20 )(1 c

uLL

Eq(20-8) summarizes the effect known as “length contraction”.

(a) The length measured by an observer who is at rest with respect to the object being measured is called the “rest length” or “proper length”.

(b) All observers in motion relative to S’ measure a shorter length, but only for dimensions along the direction of motion; length measurement transverse to the direction of motion are unaffected.

0L

(c) Under ordinary circumstances, and the effects of length contraction are too small to be observed.

cu

3 The relativistic addition of velocities Let us now modify our timing device, as shown in Fig20-7. The flashing bulb F is moved to the mirror end and is replaced by a device P that emits particles at a speed , as measured by an observer at rest with respect to the device.

light

P

DF

particle

0v

0L

0v

The time interval measured by an observer (such as S’) who is at rest with respect to the device is: (20-9)

0t

CL

vLt 0

0

00

What’s the velocity of the particles, measured by the observer S on the ground?

20

0

1cuvuvv

(20-12)

Eq(20-12) gives one form of the velocity addition law.

(a) According to Galileo and Newton, a projectile fired forward at speed in a train that is moving at speed u should have a speed relative to an observer on the ground.

This clearly permits speeds in excess of c to be realized.

0vuv 0

(b) The Eq(20-12) prevents the relative speed from ever exceeding c.

(a)If

(b)If

Thus, Eq(20-12) is consistent with Einstein’s second postulate

cv 0 uv

cuvuv

v

0

20

0

1

cv 0 c

ccuucv

21

A spaceship is moving away from the Earth at a speed of 0.80c when it fires a missile parallel to the direction of motion of the ship. The missile moves at a peed of 0.60c relative to the ship. What would be the speed of the missile as measured by an observer on the Earth?Compare with the predictions of Galilean kinematics.

Sample Problem 20-2

20-4 The Lorentz transformation

We wish to calculate the coordinates x’ , y’ , Z’, t’ of an event as observed by S’ from the coordinates of x, y, z, t of the same event according to S.

We simplify this problem somewhat, withoutlosing generality by always choosing the x and x’ axes to be along the direction of .

Namely the velocity of O’ is , respective to O.

u

u

The Lorentz transformation equations are

(20-14)

Where the factor is

)()(1

'2

utx

cuutxx

zz 'yy '

1)

cu(1

1

2

)()(1

' 22

2

cuxt

cucuxt

t

(20-15)

It is convenient in relativity equation to introduce the speed parameter , defined as (20-16)

cu

The inverse Lorentz transformation:

)''( utxx

)''( 2cuxtt

'yy 'zz

–u u

(20-17)

Sample problem 20-3In inertial frame S, a red light and a blue light are separated by a distance , with the red light at the larger value of x. The blue light flashes, and later the red light flashes. Frame S’ is moving in the direction of increasing x with a speed of . What is the distance between the two flashes and the time between them as measured in S’ ?

kmx 45.2

s35.5

cu 885.0

Lorentz transformation

Inverse transformation

Interval transformation

Inverse Interval transformation

)(' utxx )''( utxx )(' tuxx

zz '

yy '

zz '

yy '

'zz

'yy

)''( 2cuxtt )(' 2c

uxtt )''( 2cxutt )(' 2c

xutt

)''( tuxx

'zz

'yy

Table 20-2The velocity in x direction of O’ is u, respective to O.

The Lorentz parameter is

From table 20-2

and

928.1)885.0(1

1

)(1

122

cu

kmmssmm

tuxx

08.22078)]1035.5)(/1000.3)(885.0(2450[928.1

)('68

scxutt 15.3)(' 2

Solution:

20-5* Measuring the space-time coordinates of an event

20-6 The transformation of velocities

22 )(

1)(

)('''

ct

xu

utx

cxuttux

txvx

is the velocity of a particle in S referenceis the velocity of S’ reference relative to S in x directionis the velocity of the particle in S’ reference

vu

'v

xvtx

, then

(20-18)

In similar fashion,

(20-19)

Eqs(20-18) and (20-19) give the Lorentz velocity transformation.

)1('

2cuvv

vx

yy

)1('

2cuvv

vx

xz

21'

cuvuv

vx

xx

1 We now show directly that Lorentz velocity transformation gives the result demanded by Einstein’s Second postulate ( the constancy of the speed of light ). Suppose that the common event being observed by S and S’ is the passage of a light beam along the x direction. Observer S measures andcvx

.Using Eqs (20-18) and (20-19)

Thus the speed of light is indeed the same for all observers or all frames.

0 zy vv

ccuuc

cuvuv

vx

xx

11'

2

0'' zy vv

2. When ( or equivalently, when ), Eqs(20-18) and (20-19) reduce to

and (20-20)

which are the Galilean results.

cu

uvv xx 'yy vv 'zz vv '

c

Sample Problem 20-4

A particle is accelerated from rest in the lab until its velocity is 0.60c. As viewed from a frame that is moving with the particle at a speed of 0.60c relative to the laboratory, the particle is then given an additional increment of velocity amounting to 0.60c. Find the final velocity of the particle as measured in the lab frame.

20-7 Consequences of the Lorentz transformation

1.The relativity of timeFig20-15 shows a different view of the time dilation effect.

(a)

(b)

S

…… ……Fig 20-15

'c 'c's 's

'1t

2x

'' 12 tt

1x

2t1t

u

'0x '0x

u

(20-21) , the is a proper time ( )

Note that: the time dilation effect is completely symmetric. If a clock C at rest in S is observed by S’, then S’ concludes that clock C is running slow.

)''( 2cxutt

0'x 't

02

00

)(1' t

cut

ttt

0t

(a) The relativity of simultaneity

Suppose S’ has two clocks at rest, located at and . A flash of light emitted from a point midway between the clocks reaches the two clocks

'2x'1x

)''( 2cxutt

If and , then .0't 0'x 0t

According to S’ ( see Fig 20-16a), . According to S, the light signal reaches clock 1 before it reaches clock 2, and thus the arrival of the light signals at the locations of the two clocks is not simultaneous to S.

0't

S S

1 12 2

(a) (b)

's

u's

u

* *c c

Fig 20-16

(b) The twin paradox

2. The relativity of length

Sample problem 20-5 An observer S is standing on a platform of length on a space station. A rocket passes at a relative speed of 0.8c moving parallel to the edge of the platform. The observer S notes that the front and back of the rocket simultaneously line up with the ends of the platform at a particular instant (Fig 20-29a)(a) According to S, what is the time necessary for the rocket to pass a particular point on the platform?

mD 650

(b) What is the rest length of the rocket?(c) According to S’ on the rocket what is the length

D of the platform?(d) According to S’, how long does it take for

observer S to pass the entire length of the rocket?

(e) According to S, the ends of the rocket simultaneously line up with the ends of the platform. Are these events simultaneous to S’?

0L

Solution:(a)

(b)

(c) is rest length.

65m0.8c

108m

108m

0.8c

0.8c

S

S’

S

S

S’

S’

39m

39m

(a)

(b)

(c)

mmLL 108)8.0(1

6520

mD 650

mcuD

DD 39)(1 2

00

ssm

mcLt 27.0

/104.265

8.0 80

Fig (20-29)

(d)

(e) According to S’, the rocket has a rest length of

and the platform has a contracted length of . From Fig 20-19b and 20-19c, the time interval

or

scmt 45.0

8.0108'

mL 1080 mD 39

sc

mccxut 29.0

8.01

)65)(8.0('222

sc

t 29.08.0

39108'

20-8 Relativistic momentumHere we discuss the relativistic view of linear momentum. Consider the collision shown in Fig 20-20a, viewed from the S frame. Two particles, each of mass m, move with equal and opposite velocity v and –v along the x axis. They collide at the origin, and the distance between their lines of approach has been adjusted so that after the collision the particles move along the y axis with equal and opposite final velocities (Fig 20-20b).

Frame S

Frame S’

y

S S1

21

1

1

2

2

2

Before collision After the collision

(a)

(b)

v

v

v

v

vu

vu

2)(1 cvv

2)(2 cvv

2)(1 cvv

2)(2 cvv

22

1

2

cvv

S’

S’v

v

Fig 20-20(c)

(d)

The collision to be perfectly elastic, in the S frameInitial :Final :

The momentum is conserved in the S frame. According to S’ which moves relative to the S frame with speed (Fig 20-20c) , particle 2is at rest before the collision. Using Eqs (20-18) and (20-19) we can find the transformed x’ and y’ component of the initial and final velocities.

0)( vmmvPxi 0yiP0)( vmmvPyf0xfP

vu

Thus in the S’ frame:

,momentum is not conserved. Therefore, if we are to retain the conservation of momentum

22

22

1

2)0()1

2('

cvmvm

cvvmPxi

0'yiP0))(1()(1' 22 c

vvmcvmvPyf

'' xfxi PP

mvmvmvPxf 2'

as a general law consistent with Einstein’s first postulate, we must find a new definition of momentum, that is (20-23)In terms of components,

and (20-24)

There the speed v in the denominator of these expressions is always the speed of the particle as

2)(1 cvvmP

2)(1 cv

mvP xx

2)(1 cv

mvP yy

measured in particular inertial frame. It is not the speed of an inertial frame.

This new definition restores conservation of momentum in the collision. In the S frame, the velocities before and after are equal and opposite, and thus Eq(20-23) again gives zero for the initial and final momenta. In the S’ frame,

(20-25)2

21

2''

cvmvPP xfxi

0'' yfyi PP

20-9 Relativistic energy1. Using the velocity shown in Figs 20-20c and 20-20d, you can show that, with , the total initial and final kinetic energies are

(20-26)

Thus is not equal to . According to S .(a) This situation violates the relativity postulate, we require a new definition of kinetic energy if we

2

21 mvk

22

2

2

)1(

2'

cvmvk i

)2(' 2

22

cvmvk f

'ik 'fk fi kk

are to preserve the law of conservation of energy and the relativity postulate.(b) The classical expression for kinetic energy also violates the Second relativity postulate by allowing speed in excess of the speed of light. These is no limit ( in either classical or relativistic dynamic ) to the energy we can give to a particle.(c) Relativistic kinetic energy is (20-27)2

22

2

1mc

cv

mck

Using Eq(20-27), we can show that kinetic energy is conserved in the S’ frame of the collision of Fig 20-20.2. Energy and mass in special relativityWe can also express Eq(20-27) as (20-30)where the total relativistic energy E is defined as (20-31)

0EEK

22

2

1c

v

mcE

and rest energy (20-32)The rest energy can be regarded as the internal energy of a particle at rest.

(a) According to Eq(20-32), whenever we add energy to a object that remain at rest, we increase its mass by an amount .If we compress a spring and increase its potential energy by an mount , then its mass increases by .

20 mcE

0E

2cEm

U

2cU

0E

E

(b) The total relativistic energy must be conserved in any interaction.

The sun radiates an energy of every second, and the corresponding change in the mass is

J26104

kgsmJ

cEm 9

28

26

2 104)/103(

104

Sample problem 20-8

Two 35g putty balls are thrown toward each other, each with a speed of 1.7m/s. The balls strike each other head-on and stick together. By how much does the mass of the combined ball differ from the sum of the masses of the two original balls?Solution: We treat the two putty balls as an isolated system. No external work is done on this isolated system.

With , where .We have

The corresponding increase in mass is

if KKK 0fK

JsmkgmvKE i 101.0)/7.1)(035.0()21(2 22

0

kgsmJ

cEm 18

2820 101.1

)/103(101.0

Such a tiny increase in mass is beyond our ability to measure.

3. Conservation of total relativistic energyEq(20-30) can be written as

Manipulation of Eqs(20-23) and (20-31) gives a useful relationship among the total energy, momentum, and rest energy

0EKE

222 )()( mcpcE

Sample problem 20-10A certain accelerator produces a beam of neutral Kaons ( ) with kinetic energy 325 Mev. Consider a Kaon that decays in flight two pion( ) . Find the kinetic energy of each pion in the special case in which the pions travel parallel or antiparallel to the direction of the Kaon beam.Solution:From Eq(20-33), the initial total relativistic energy is

Mevcmk 4982

Mevcm 1402

MevMevMevcmKE kk 8234983252

The initial momentum is

The total energy of the final system consisting of the two pions is

(20-35)Thus we have one equation in the two unknowns and , with conservation of momentum.

Mev

cmcpcmcpEEE

823)()()()( 222

2222

121

2P1P

MevcmEcP kkk 655498823)( 22222

thus Using Eqs(20-30) and (20-34)

Mevcp 66811 Mev13

2222 )()( cmcmpcK

MevMevMevMevK 543140140)668( 21

or

Mevcpcpcp k 655211

MevMevMevMevK 6.0140140)13( 22