239

Chapter 39 Relativity Conceptual Problems 1 • [SSM] The approximate total energy of a particle of mass m moving at speed u << c is (a) mc2 + 1

2 mu 2 , (b) 12 mu 2 , (c) cmu, (d) mc2, (e) 1

2 cmu . Determine the Concept The total relativistic energy E of a particle is defined to be the sum of its kinetic and rest energies. The sum of the kinetic and rest energies of a particle is given by:

22212 mcmumcKE +=+=

and )(a is correct.

2 • A set of twins work in an office building. One twin works on the top floor and the other twin works in the basement. Considering general relativity, which twin will age more quickly? (a) They will age at the same rate. (b) The twin who works on the top floor will age more quickly. (c) The twin who works in the basement will age more quickly. (d) It depends on the speed of the office building. (e) None of the above. Determine the Concept The gravitational field of Earth is slightly greater in the basement of the office building than it is at the top floor. Because clocks run more slowly in regions of low gravitational potential, clocks in the basement will run more slowly than clocks on the top floor. Hence, the twin who works on the top floor will age more quickly. )(b is correct.

3 • True or false: (a) The speed of light is the same in all reference frames. (b) The time interval between two events is never shorter than the proper time

interval between the two events. (c) Absolute motion can be determined by means of length contraction. (d) The light-year is a unit of distance. (e) Simultaneous events must occur at the same place. (f) If two events are not simultaneous in one frame, they cannot be

simultaneous in any other frame. (g) The mass of a system that consists of two particles that are tightly bound

together by attractive forces is less than the sum of the masses of the individual particles when separated.

(a) True

Chapter 39

240

(b) True (c) False. The shortening of the length of an object in the direction in which it is moving is independent of the velocity of the frame of reference from which it is observed. (d) True (e) False. Consider two explosions equidistant, but in opposite directions, from an observer in the observer’s frame of reference. (f) False. Whether events appear to be simultaneous depends on the motion of the observer. (g) True 4 • An observer sees a system moving past her that consists of a mass oscillating on the end of a spring and measures the period T of the oscillations. A second observer, who is moving with the mass–spring system, also measures its period. The second observer will find a period that is (a) equal to T, (b) less than T, (c) greater than T, (d) either (a) or (b) depending on whether the system was approaching or receding from the first observer, (e) Not enough information is given to answer the question. Determine the Concept Because the clock is moving with respect to the first observer, a time interval will be longer for this observer than for the observer moving with the spring-and-mass oscillator. Hence, the observer moving with the system will measure a period that is less than T. )(b is correct.

5 • The Lorentz transformation for y and z is the same as the classical result: y = y′ and z = z′. Yet the relativistic velocity transformation does not give the classical result yy u'u = and zz u'u = . Explain why this result occurs. Determine the Concept Although Δy = Δy′, Δt ≠Δt′. Consequently, uy = Δy/Δt′ ≠ Δy′/Δt′ = uy′. Estimation and Approximation 6 •• The Sun radiates energy at the rate of approximately 4 × 1026 W. Assume that this energy is produced by a reaction whose net result is the fusion of four protons to form a single 4He nucleus and the release of 25 MeV of energy that is radiated into space. Calculate the Sun’s loss of mass per day.

Relativity

241

Picture the Problem We can calculate the Sun’s loss of mass per day from the number of reactions per second and the loss of mass per reaction. Express the rate at which the Sun loses mass:

mNt

MΔ=

ΔΔ

⇒ tmNM ΔΔ=Δ

where N is the number of reactions per second and Δm is the loss of mass per reaction.

The number of reactions per second, N is given by:

reaction/EPN =

Substitute for N to obtain: tm

EPM ΔΔ=Δ

reaction/

The loss of mass per reaction Δm is: 2

reactionc

Em =Δ

Substituting for Δm and simplifying yields:

tcP

tc

EE

PM

Δ=

Δ⎟⎠⎞

⎜⎝⎛⎟⎠⎞

⎜⎝⎛=Δ

2

2

reaction/reaction/

Substitute numerical values and evaluate ΔM: ( )

kg 104

dks4.86d1

m/s 10998.2 W104

14

28

26

×≈

⎟⎠⎞

⎜⎝⎛ ×

×

×=ΔM

7 •• [SSM] The most distant galaxies that can be seen by the Hubble telescope are moving away from us and have a redshift parameter of about z = 5. (The redshift parameter z is defined as (f – f’)/f’, where f is the frequency measured in the rest frame of the emitter, and f’ is the frequency measured in the rest frame of the receiver.) (a) What is the speed of these galaxies relative to us (expressed as a fraction of the speed of light)? (b) Hubble’s law states that the recession speed is given by the expression v = Hx, where v is the speed of recession, x is the distance, and H, the Hubble constant, is equal to 75 km/s/Mpc , where 1 pc = 3.26 c⋅y. (The abbreviation for parsec is pc.) Estimate the distance of such a galaxy from us using the information given.

Chapter 39

242

Picture the Problem (a) We can use the definition of the redshift parameter and the relativistic Doppler shift equation to show that, for light that is Doppler-

shifted with respect to an observer, ⎟⎟⎠

⎞⎜⎜⎝

⎛+−

=11

2

2

uucv , where u = z + 1, and to find the

ratio of v to c. In Part (b) we can solve Hubble’s law for x and substitute our result from Part (a) to estimate the distance to the galaxy. (a) The red-shift parameter is defined to be:

f''ffz −

= 0

The relativistic Doppler shift for recession is given by: cv

cvf'f+−

=11

0

Substitute for f ′ and simplify to obtain:

111

11

11

0

00

−−+

=

+−

+−

−=

cvcv

cvcvf

cvcvff

z

Letting u = z + 1 and simplifying yields: cv

cvzu−+

=+=111 ⇒

11

2

2

+−

=uu

cv

Substitute for u to express v/c as a function of z:

( )( ) 11

112

2

++−+

=zz

cv

Substituting the numerical value of z and evaluating v/c gives:

( )( )

946.0115115

2

2

=++−+

=cv

(b) Solving Hubble’s law for x yields:

Hvx =

Substitute numerical values and evaluate x:

yG3.12

Mpcy103.26

Mpckm/s75

946.0946.0 6

⋅=

⋅××==

c

ccH

cx

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243

Time Dilation and Length Contraction 8 • The proper mean lifetime of a muon is 2.2 μs. Muons in a beam are traveling through a laboratory at 0.95c. (a) What is their mean lifetime as measured in the laboratory? (b) How far do they travel, on average, before they decay? Picture the Problem We can find the mean lifetime of a muon as measured in the

laboratory using tt' γ= where ( )211 cv−=γ and t is the proper mean lifetime of the muon. The distance L that the muon travels is the product of its speed and its mean lifetime in the laboratory. (a) The mean lifetime of the muon, as measured in the laboratory, is given by:

2

1 ⎟⎠⎞

⎜⎝⎛−

=

cv

tt'

Substitute numerical values and evaluate t′:

s0.7

s046.795.01

s2.22

μ

μμ

=

=

⎟⎠⎞

⎜⎝⎛−

=

cc

t'

(b) The distance L that the muon travels is related to its mean lifetime in the laboratory:

ct'vt'L 95.0==

Substitute numerical values and evaluate L:

( )( )km0.2

s046.7m/s10998.295.0 8

=

×= μL

9 •• In the Stanford linear collider, small bundles of electrons and positrons are fired at each other. In the laboratory’s frame of reference, each bundle is approximately 1.0 cm long and 10 μm in diameter. In the collision region, each particle has an energy of 50 GeV, and the electrons and the positrons are moving in opposite directions. (a) How long and how wide is each bundle in its own reference frame? (b) What must be the minimum proper length of the accelerator for a bundle to have both its ends simultaneously in the accelerator in its own reference frame? (The actual proper length of the accelerator is less than 1000 m.) (c) What is the length of a positron bundle in a reference frame that moves with the electron bundle?

Chapter 39

244

Picture the Problem The proper length Lp of the beam is its length as measured in a reference frame in which it is not moving. The proper length is related to its length in the frame in which it is measured by LL γ=p .

(a) Relate the proper length Lp of the beam to its length L in the laboratory frame of reference:

LL γ=p

The energy of the beam also depends on γ :

2mcE γ=

Solve for and evaluate γ : 42 10785.9

MeV511.0GeV50

×===mcEγ

Substitute numerical values and evaluate Lp:

( )( )km0.98m978.5

cm0.110785.9 4p

==

×=L

and the width w of the beam is unchanged.

(b) Express the length of the accelerator in the electron beam’s frame of reference:

γpacc,

acc

LL =

Set Lacc = Lp: γ

pacc,p

LL = ⇒ ppacc, LL γ=

Substitute numerical values and evaluate pacc,L :

( )( )m106.9

m5.97810785.97

4pacc,

×=

×=L

(c) The length of the positron bundle in the electron’s frame of reference is:

γLL =pos

Substitute numerical values and evaluate posL :

m10.010785.9

cm0.14pos μ=

×=L

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245

10 • Use the binomial expansion equation

( ) ( ) 1 ,12

111 2 <<+≈⋅⋅⋅+−

++=+ xnxxnnnxx n

to derive the following results for the case when v is much less than c, and use the results when applicable in the following problems:

(a) 2

2

211

cv

+≈γ

(b) 2

2

2111

cv

−≈γ

(c) 2

2

21111

cv

≈−≈−γ

γ

Picture the Problem We can use the definition of γ and the binomial expansion of (1 + x)n to show that each of these relationships holds provided v << c. (a) Express the gamma factor: 21

2

2

2

21

1

1−

⎟⎟⎠

⎞⎜⎜⎝

⎛−=

−

=cv

cv

γ

Expand the radical factor binomially to obtain:

sorder termhigher 211

1

2

2

21

2

2

+⎟⎟⎠

⎞⎜⎜⎝

⎛−⎟

⎠⎞

⎜⎝⎛−+=

⎟⎟⎠

⎞⎜⎜⎝

⎛−=

−

cv

cvγ

For v << c:

2

2

211

cv

+≈γ

(b) Express the reciprocal of γ :

2

2

11cv

−=γ

Expand the radical binomially to obtain:

sorder termhigher 211

11

2

2

21

2

2

+⎟⎟⎠

⎞⎜⎜⎝

⎛−⎟

⎠⎞

⎜⎝⎛+=

⎟⎟⎠

⎞⎜⎜⎝

⎛−=

cv

cv

γ

For v << c:

2

2

2111

cv

−≈γ

Chapter 39

246

(c) Express the gamma factor: 21

2

2

2

21

1

1−

⎟⎟⎠

⎞⎜⎜⎝

⎛−=

−

=cv

cv

γ

Subtract one from both sides of the equation to obtain:

11121

2

2

−⎟⎟⎠

⎞⎜⎜⎝

⎛−=−

−

cvγ

Expand the radical binomially to obtain:

sorder termhigher

12111 2

2

+

−⎟⎟⎠

⎞⎜⎜⎝

⎛−⎟

⎠⎞

⎜⎝⎛−+=−

cvγ

For v << c:

2

2

211

cv

≈−γ

11 •• Star A and Star B are at rest relative to Earth. Star A is 27 c⋅y from Earth, and as viewed from Earth, Star B is located beyond (behind) Star A. (a) A spaceship is making a trip from Earth to Star A at a speed such that the trip from Earth to Star A takes 12 y according to clocks on the spaceship. At what speed, relative to Earth, must the spaceship travel? (Assume that the times for the accelerations are very short compared to the overall trip time.) (b) Upon reaching Star A, the spaceship speeds up and departs for Star B at a speed such that the gamma factor, γ, is twice that of Part (a). The trip from Star A to Star B takes 5.0 y (spaceship’s time). How far, in c⋅y, is Star B from Star A in the rest frame of Earth and the two stars? (c) Upon reaching Star B, the spaceship departs for Earth at the same speed as in Part (b). It takes it 10 y (spaceship’s time) to return to Earth. If you were born on Earth the day the ship left Earth and you remain on Earth, how old are you on the day the ship returns to Earth? Picture the Problem We can use the time-dilation relationship to find the speed of the spacecraft. The distance to the second star is the product of the new gamma factor, the speed of the spacecraft, and the elapsed time. Finally, the time that has elapsed on Earth (your age) is the sum of the elapsed times for the three legs of the journey. (a) From the point of view of an observer on Earth, the elapsed time for the trip will be:

vLt =Δ

From the point of view of an observer on the spaceship, the elapsed time for the trip will be:

vLtt'γγ

=Δ

=Δ

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247

Substitute for γ to obtain: 2

2

1cv

vLt' −=Δ ⇒

( )222 t'cL

LcvΔ+

=

Substitute numerical values and evaluate v:

( )( ) ( )

c

ccc

ccv

91.0

914.0y12y27

y27222

=

=+⋅

⋅=

Note that from the point of view of an Earth observer, this part of the trip has taken 27 c⋅ y/0.914c = 29.5 y. (b) The distance the ship travels, from the point of view of an Earth observer, in 5 y is:

tvLL' Δ=Δ=Δ γγ 22 where γ is the gamma factor for the first part of the trip.

The gamma factor in Part (a) is:

46.2

914.01

1

1

122

=

⎟⎠⎞

⎜⎝⎛−

=

⎟⎠⎞

⎜⎝⎛−

=

cc

cv

γ

Substitute numerical values and evaluate ΔL′:

( )( )( )y22y5.22

y0.5914.046.22

⋅=⋅=

=Δ

cc

cL'

(c) The elapsed time Δt on Earth (your age) is the sum of the times for the spacecraft to travel to the star 27 c⋅ y away, to the second star, and to return home from the second star:

home returningy5.22y5.29 tt Δ++=Δ

The elapsed time on Earth while the spacecraft is returning to Earth is:

( )( )y2.49

y1046.22

2 timeship'shome

returning

==

Δ=Δ tt γ

Substitute for

home returningtΔ and evaluate Δt:

y101

y2.49y5.22y5.29

=

++=Δt

12 • A spaceship travels to a star 35 c⋅y away at a speed of 2.7 × 108 m/s. How long does the spaceship take to get to the star (a) as measured on Earth and (b) as measured by a passenger on the spaceship?

Chapter 39

248

Picture the Problem We can use Δt = L/v, where L is the distance to the star and v is the speed of the spaceship to find the time Δt for the trip as measured on Earth. The travel time as measured by a passenger on the spaceship can be found using Δt′ = Δt/γ. (a) The travel time as measured on Earth is the ratio of the distance traveled L to speed of the spaceship:

vLt =Δ

Substitute numerical values and evaluate Δt:

y39y9.389.0y35

m/s107.2y35

m/s107.2y35

88

===

×=

×⋅

=Δ

c

ct

(a) The travel time as measured by a passenger on the spaceship is given by:

2

1 ⎟⎠⎞

⎜⎝⎛−Δ=

Δ=Δ

cvttt'

γ

Substitute numerical values and evaluate Δt′:

( ) ( ) y1790.01y9.38 2 =−=Δt'

13 •• [SSM] Unobtainium (Un) is an unstable particle that decays into normalium (Nr) and standardium (St) particles. (a) An accelerator produces a beam of Un that travels to a detector located 100 m away from the accelerator. The particles travel with a velocity of v = 0.866c. How long do the particles take (in the laboratory frame) to get to the detector? (b) By the time the particles get to the detector, half of the particles have decayed. What is the half-life of Un? (Note: half-life as it would be measured in a frame moving with the particles) (c) A new detector is going to be used, which is located 1000 m away from the accelerator. How fast should the particles be moving if half of the particles are to make it to the new detector? Picture the Problem The time required for the particles to reach the detector, as measured in the laboratory frame of reference is the ratio of the distance they must travel to their speed. The half life of the particles is the trip time as measured in a frame traveling with the particles. We can find the speed at which the particles must move if they are to reach the more distant detector by equating their half life to the ratio of the distance to the detector in the particle’s frame of reference to their speed.

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249

(a) The time required to reach the detector is the ratio of the distance to the detector and the speed with which the particles are traveling:

cx

vxt

866.0Δ

=Δ

=Δ

Substitute numerical values and evaluate Δt:

( )s385.0

m/s10998.2866.0m100Δ 8

μ=

×=t

(b) The half life is the trip time as measured in a frame traveling with the particles:

2

1 ⎟⎠⎞

⎜⎝⎛−Δ=

Δ=Δ

cvttt'

γ

Substitute numerical values and evaluate Δt′: ( )

s193.0

866.01s385.02

μ

μ

=

⎟⎠⎞

⎜⎝⎛−=Δ

cct'

(c) In order for half the particles to reach the detector: v

cvx'

vx't'

2

1 ⎟⎠⎞

⎜⎝⎛−Δ

=Δ

=Δγ

where Δx′ is the distance to the new detector.

Rewrite this expression to obtain: t'

x'

cv

vΔΔ

=

⎟⎠⎞

⎜⎝⎛−

2

1

Squaring both sides of the equation yields:

2

2

2

1⎟⎠⎞

⎜⎝⎛ΔΔ

=

⎟⎠⎞

⎜⎝⎛−

t'x'

cv

v

Substitute numerical values for Δx′ and Δt′ and simplify to obtain:

( )22

2

2

3.17s193.0

m1000

1c

cv

v=⎟⎟

⎠

⎞⎜⎜⎝

⎛=

⎟⎠⎞

⎜⎝⎛−

μ

Chapter 39

250

Divide both sides of the equation by c2 to obtain:

( )22

2

2

3.171

=

⎟⎠⎞

⎜⎝⎛−

cv

cv

Solving this equation for v2/c2 gives:

( )( )

9967.03.171

3.172

2

2

2

=+

=cv

Finally, solving for v yields: cv 998.0=

14 •• A clock on Spaceship A measures the time interval between two events, both of which occur at the location of the clock. You are on Spaceship B. According to your careful measurements, the time interval between the two events is 1.00 percent longer than that measured by the two clocks on Spaceship A. How fast is Spaceship A moving relative to Spaceship B. (Hint: Use one or more of the results of Problem 10.) Picture the Problem We can express the fractional difference in your time-interval measurements as a function of γ and solve the resulting equation for the relative speed of the two spaceships. Express the fractional difference in the time-interval measurements of the two observers:

0100.01 =ΔΔ

−=ΔΔ−Δ

tt'

tt't

Since Δt′/Δt = 1/γ: 0100.011 =−=ΔΔ−Δ

γtt't

From Problem 10(b) we have: 2

2

2111

cv

−≈γ

Substitute to obtain:

0100.0211111 2

2

=⎟⎟⎠

⎞⎜⎜⎝

⎛−−≈−

cv

γ

or

0100.021

2

2

=cv

Solve for v to obtain:

m/s1023.4

141.00200.07×=

== ccv

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251

15 •• If a plane flies at a speed of 2000 km/h, how long must the plane fly before its clock loses 1.00 s because of time dilation? (Hint: Use one or more of the results of Problem 10.) Picture the Problem We can use the time dilation equation to relate the time lost by the clock to the speed of the plane and the time it must fly. Express the time δ t lost by the clock: ⎟⎟

⎠

⎞⎜⎜⎝

⎛−Δ=

Δ−Δ=Δ−Δ=

γγδ 11p tttttt

Because V << c, we can use Part (b) of Problem 10:

2

2

2111

cV

−≈γ

Substitute to obtain: t

cV

cVtt Δ=⎥

⎦

⎤⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛−−Δ= 2

2

2

2

21

2111δ

Solving for Δt gives:

2

22V

ctt δ=Δ

Substitute numerical values and evaluate Δt:

( )( )( )

y1085.1

Ms31.56y1s10824.5

sh/36001km/h2000m/s10998.2s00.12

4

11

2

28

×=

××=

××

=tΔ

The Lorentz Transformation, Clock Synchronization, and Simultaneity 16 •• Show that when v << c the relativistic transformation equations for x, t, and ux reduce to the classical transformation equations. Picture the Problem We can use the inverse Lorentz transformations and the result of Problem 10(c) to show that when u << c the transformation equations for x, t, and u reduce to the Galilean equations. The inverse transformation for x is:

( )vtxx' −= γ

From Problem 10(c): 2

2

211

cv

+=γ

Chapter 39

252

Substitute for γ and expand to obtain: ( )

tcvx

cvvtx

vtxcvx'

2

3

2

2

2

2

21

21

211

−+−=

−⎟⎟⎠

⎞⎜⎜⎝

⎛+=

When v << c:

vtxx' −≈

The inverse transformation for t is: ⎟

⎠⎞

⎜⎝⎛ −= 2c

vxtt' γ

Substitute for γ and expand to obtain:

xcvt

cv

cvxt

cvxt

cvt'

4

3

2

2

2

22

2

21

21

211

−+−=

⎟⎠⎞

⎜⎝⎛ −⎟⎟⎠

⎞⎜⎜⎝

⎛+=

When v << c:

tt' ≈

The inverse velocity transformation for motion in the x direction is: 21

cvu

vu'ux

xx

−

−=

When v << c: vu'u xx −≈

17 •• [SSM] A spaceship of proper length Lp = 400 m moves past a transmitting station at a speed of 0.760c. (The transmitting station broadcasts signals that travel at the speed of light.) A clock is attached to the nose of the spaceship and a second clock is attached to the transmitting station. The instant that the nose of the spaceship passes the transmitter, the clock attached to the transmitter and the clock attached to the nose of the spaceship are set equal to zero. The instant that the tail of the spaceship passes the transmitter a signal is sent by the transmitter that is subsequently detected by a receiver in the nose of the spaceship. (a) When, according to the clock attached to the nose of spaceship, is the signal sent? (b) When, according to the clocks attached to the nose of spaceship, is the signal received? (c) When, according to the clock attached to the transmitter, is the signal received by the spaceship? (d) According to an observer that works at the transmitting station, how far from the transmitter is the nose of the spaceship when the signal is received? Picture the Problem Let S be the reference frame of the spaceship and S′ be that of Earth (transmitter station). Let event A be the emission of the light pulse and event B the reception of the light pulse at the nose of the spaceship. In (a) and (c)

Relativity

253

we can use the classical distance, rate, and time relationship and in (b) and (d) we can apply the inverse Lorentz transformations. (a) In both S and S′ the pulse travels at the speed c. Thus:

s76.10.760c

m400pA μ===

vL

t

(b) The inverse time transformation is:

⎟⎠⎞

⎜⎝⎛ −= 2B c

vxt't γ

where

( )54.1

760.01

1

1

1

2

2

2

2=

−

=

−

=

cc

cv

γ

Substitute numerical values and evaluate 'tB :

( ) ( )( )

( ) ( )( )

s32.6

m/s10998.2m400760.0s09.354.1

m400760.0s09.354.1

8

2B

μ

μ

μ

=

⎟⎟⎠

⎞⎜⎜⎝

⎛×

−−=

⎟⎠⎞

⎜⎝⎛ −

−=cc't

(c) The elapsed time, according to the clock on the ship is:

Aship oflength

travel topulseB ttt +=

Find the time of travel of the pulse to the nose of the ship: s33.1

m/s102.998m400

8ship oflength

travel topulse

μ=×

=t

Substitute numerical values and evaluate tB:

s09.3s76.1s33.1B μμμ =+=t

(d) The inverse transformation for x is:

( )vtxx' −= γ

Substitute numerical values and evaluate x′:

( ) ( )( )( )[ ] km70.1s1009.3m/s10998.2760.0m40054.1 68 =××−−= −x'

Chapter 39

254

18 •• In frame S, event B occurs 2.0 μs after event A, and event A occurs at the origin whereas event B occurs on the x axis at x = 1.5 km. How fast and in what direction must an observer be traveling along the x axis so that events A and B occur simultaneously? Is it possible for event B to precede event A for some observer? Picture the Problem We can use Equation 39-12, the inverse time transformation equation, to find the required speed of the observer. Use Equation 39-12 to obtain: ( ) ( )

⎟⎠⎞

⎜⎝⎛ Δ

−Δ=

⎥⎦⎤

⎢⎣⎡ −−−=

−=Δ

2

2

'

cxvt

xxcvtt

't'tt

ABAB

AB

γ

γ

where AB ttt −=Δ and AB xxx −=Δ .

Setting 'tΔ equal to zero gives:

02 =Δ

−Δc

xvt ⇒ cxtcv

ΔΔ

=

Substitute numerical values and evaluate v:

( )( )

0.40c

km5.1s0.2m/s10998.2 8

=

×= cv μ

Because :s0.2AB μ=−=Δ ttt

direction. in the moving isobserver The

x+

If B occurs before A, then 'tΔ < 0 : 0' 2 <⎟

⎠⎞

⎜⎝⎛ Δ

−Δ=Δc

xvtt γ

and

02 <Δ

−Δc

xvt

Solving for v yields: c

xtcv 40.0

2

=ΔΔ

>

Yes. Event B will precede event A for some observer ( Bt' will be less than At' ) provided v > 0.40c. 19 •• Observers in reference frame S see an explosion located on the x axis at x1 = 480 m. A second explosion occurs, 5.0 μs later, at x2 = 1200 m. In

Relativity

255

reference frame S′, which is moving along the x axis in the +x direction at speed v, the two explosions occur at the same point in space. What is the separation in time between the two explosions as measured in S′? Picture the Problem We can use Equation 39-12, the inverse time transformation equation, to express the separation in time t'Δ between the two explosions as measured in S′ as a function of the speed v of the observer and Equation 39-11, the inverse position transformation equation, to find the speed of the observer. Use Equation 39-12 to express the separation in time between the two explosions as measured in S′:

2

2

2

1 ⎟⎠⎞

⎜⎝⎛−

Δ−Δ=

⎥⎦⎤

⎢⎣⎡ Δ−Δ=Δ

cv

xcvt

xcvtt' γ

(1)

From Equation 39-11:

( )tvxx' Δ−Δ=Δ γ

Because the explosions occur at the same point in space, Δx′ = 0:

( )tvx Δ−Δ= γ0 ⇒txv

ΔΔ

=

Substitute numerical values and evaluate v:

m/s101.44s0.5

m480m1200 8×=−

=μ

v

Substitute numerical values in equation (1) and evaluate Δt′:

( )( )

s4.4

m/s10998.2m/s1044.11

m480m1200m/s10998.2m/s1044.1s0.5

Δ2

8

8

28

8

μμ

=

⎟⎟⎠

⎞⎜⎜⎝

⎛××

−

−×

×−

=t'

20 ••• In reference frame S, events 1 and 2 are separated by a distance D = x2 – x1 and a time T = t2 – t1. (a) Use the Lorentz transformation to show that in frame S′, which is moving along the x axis with speed v relative to S, the time separation is

′t2 − ′t1 = γ T − vD / c2( ). (b) Show that the events can be

simultaneous in frame S′ only if D is greater than cT. (c) If one of the events is the cause of the other, the separation D must be less than cT, because D/c is the smallest time that a signal can take to travel from x1 to x2 in frame S. Show that if D is less than cT, ′t2 is greater than ′t1 in all reference frames. This shows that if the cause precedes the effect in one frame, it must precede it in all reference

Chapter 39

256

frames. (d) Suppose that a signal could be sent with speed c′ > c so that in frame S the cause precedes the effect by the time T = D/c′. Show that there is then a reference frame moving with speed v less than c in which the effect precedes the cause. Picture the Problem We can use Equation 39-12, the inverse time transformation equation, to establish the results called for in this problem. (a) Use Equation 39-12 to obtain:

( ) ( )

⎟⎠⎞

⎜⎝⎛ −=

⎥⎦⎤

⎢⎣⎡ −−−=−

2

1221212

cvDT

xxcvtt't't

γ

γ

where T = t2 − t1 and D = x2 − x1.

(b) Events 1 and 2 are simultaneous in S′ if:

't't 12 = or

02 =−cvDT ⇒

vTcD

2

=

Because v ≤ c:

cTD ≥

(c) If D < cT, then 12 't't > and the events are not simultaneous in S′. (d) If D = c′T > cT, then:

't'tcc'

cvT

cvDT 122 1 −=⎥⎦

⎤⎢⎣⎡ −=−

In this case, 't't 12 > could be negative; i.e., 't2 could be less than 't1 , or the effect could precede the cause. 21 ••• A rocket that has a proper length of 700 m is moving to the right at a speed of 0.900c. It has two clocks—one in the nose and one in the tail—that have been synchronized in the frame of the rocket. A clock on the ground and the clock in the nose of the rocket both read zero as they pass by each other. (a) At the instant the clock on the ground reads zero, what does the clock in the tail of the rocket read according to observers on the ground? When the clock in the tail of the rocket passes the clock on the ground, and (b) what does the clock in the tail read according to observers on the ground, (c) what does the clock in the nose read according to observers on the ground, and (d) what does the clock in the nose read according to observers on the rocket? (e) At the instant the clock in the nose of the rocket reads 1.00 h, a light signal is sent from the nose of the rocket to an observer standing by the clock on the ground. What does the clock on the ground read when the observer on the ground receives this signal? (f) When the observer

Relativity

257

on the ground receives the signal, he immediately sends a return signal to the nose of the rocket. What is the reading of the clock in the nose of the rocket when this signal is received at the nose of the rocket? Picture the Problem Let S be the ground reference frame, S′ the reference frame of the rocket, and v = 0.900c be the speed of the rocket relative to S. Denote the tail and nose of the rocket by T and N, respectively. The initial conditions in S′ are

0'=Nt , 0'=Nx , and m700'' −=−= LxT . (a) The reading of the tail clock is given by:

22 cvx

cvxt't TT

TTγγ −=⎟

⎠⎞

⎜⎝⎛ −=

because tT = 0

We can find xT using the length contraction equation: γ

L'xT −=

Substitute for Tx to obtain:

2cvL''tT =

Substitute numerical values and evaluate :'Tt

( )( )

s10.2

s101.2m/s10998.2m700900.0

8

μ

μ

=

=×

='tT

(b) The time for the rocket to move a distance L′ is given by:

cL'

vL''tT 9.0

==

Substitute numerical values and evaluate :'Tt ( )( )

s59.2s594.2

m/s10998.2900.0m700

8

μμ ==

×='tT

(c) As seen by an observer on the ground:

s49.0

s2.10s59.2

μ

μμ

=

−=Δ= t'tN

(d) Because the clocks are synchronized in S′:

s59.2 μ== 't't TN

Chapter 39

258

(e) The time the signal is received on the ground is the sum of the time when the signal is sent and the time for it to travel to the ground:

travelrec ttt Δ+Δ=

Find Δt, the time the signal is sent: ( )

h294.20.9001

h00.1

2

2p =

−

=Δ=Δ

cc

tt γ

Find traveltΔ , the time for the signal to travel to the ground:

( )( )

h065.2

900.0h294.2travel

=

==c

ccxt Δ

Δ

Substitute for Δt and traveltΔ and evaluate trec:

h36.4

h359.4h065.2h294.2rec

=

=+=t

(f) Find Δx when the signal is sent:

( )( ) h923.3900.0h359.4 ⋅== ccxΔ

In S, the signal arrives at 0.1c relative to the rocket. The time required for the signal to travel to the rocket is:

h23.390.100

h923.3100.0

=⋅

==c

cc

xt ΔΔ

Find the time when the signal reaches the rocket:

h 43.15h 3.923h 23.39 =+=t

Finally, use the time dilation equation to find 'tN : ( ) ( )

h8.18

900.01h15.43 2

2

=

−==c

ct'tN γ

The Velocity Transformation 22 •• A spaceship, at rest in a certain reference frame S, is given a speed increase of 0.50c (call this increase boost 1). Relative to its new rest frame, the spaceship is given a further 0.50c increase 10 seconds later (as measured in its new rest frame; call this increase boost 2). This process is continued indefinitely, at 10-s intervals, as measured in the rest frame of the spaceship. (Assume that the boosts take a very short time compared to 10 s.) (a) Using a spreadsheet program, calculate and graph the speed of the spaceship in reference frame S as a

Relativity

259

function of the boost number for boost 1 to boost 10. (b) Graph the gamma factor the same manner. (c) How many boosts does it take until the speed of the ship in S is greater than 0.999c? (d) How far does the spaceship move between boost 1 and boost 6, as measured in reference frame S? What is the average speed of the spaceship between boost 1 and boost 6, as measured in S? Picture the Problem We’ll let the speed (in S) of the spaceship after the ith boost be vi and derive an expression for the ratio of v to c after the spaceship’s (i + 1)th boost as a function of the number of boosts N. We can use the definition of γ, in terms of v/c to plot γ as a function of N. (a) and (b) The speed of the spaceship after the (i + 1)th boost is given by relativistic velocity addition equation:

( )2

1 50.01

50.0

cvccvv

i

ii

+

+=+

Factor c from both the numerator and denominator to obtain:

( )cv

cv

vi

i

i

50.01

50.01

+

+=+

γi is given by:

⎟⎠⎞

⎜⎝⎛−

=

cvi

i

1

1γ

A spreadsheet program to calculate v/c and γ as functions of the number of boosts N is shown below. The formulas used to calculate the quantities in the columns are as follows: Cell Content/Formula Algebraic Form A3 0 N B2 0 v0 B3 (B2+0.5)/(1+0.5*B2) vi+1 C1 1/(1−B2^2)^0.5 γ

A B C 1 N v/c γ 2 0 0.000 1.00 3 1 0.500 1.15 4 2 0.800 1.67 5 3 0.929 2.69 6 4 0.976 4.56 7 5 0.992 7.83 8 6 0.997 13.52 9 7 0.999 23.39

Chapter 39

260

10 8 1.000 40.51 11 9 1.000 70.15 12 10 1.000 121.50

A graph of v/c as a function of N follows:

0.0

0.2

0.4

0.6

0.8

1.0

0 2 4 6 8 10

N

v /c

A graph of γ as a function of N follows:

0

20

40

60

80

100

120

0 2 4 6 8 10

N

γ

(c) Examination of the spreadsheet or of the graph of v/c as a function of N indicates that, after 8 boosts, the speed of the spaceship is greater than 0.999c.

Relativity

261

(d) After 5 boosts, the spaceship has traveled a distance Δx, measured in the Earth frame of reference (S), given by:

( )( ) ( )( ) ( )( )( )( ) ( )( )

( )( )( ) ( )( )( ) ( )( )( )( )( )( ) ( )( )( )

s166

83.7s10992.056.4s10976.069.2s10929.067.1s108.015.1s105.0

s10992.0s10976.0s10929.0s108.0s105.0

6554

433221

6554433221

⋅=

++++=

++++=Δ+Δ+Δ+Δ+Δ=Δ

→→

→→→

→→→→→

c

ccccc

ccccc

xxxxxx

γγγγγ

The average speed of the spaceship, between boost 1 and boost 5, as measured in S is given by:

txvΔΔ

=av

where Δt is the travel time as measured in the Earth frame of reference.

Express Δt as the sum of the times the spaceship travels during each 10-s interval following a boost in its speed:

( ) ( ) ( ) ( ) ( )( )( )6554433221

6554433221

6554433221

s10s10s10s10s10s10

→→→→→

→→→→→

→→→→→

++++=++++=

Δ+Δ+Δ+Δ+Δ=Δ

γγγγγγγγγγ

tttttt

Substitute numerical values and evaluate Δt:

( )( ) s17983.756.469.267.115.1s10 =++++=Δt

Substitute for Δx and Δt and evaluate vav:

ccv 927.0s179s166

av =⋅

=

Remarks: This result seems to be reasonable. Relativistic time dilation implies that the spacecraft will be spending larger amounts of time at high speed (as seen in reference frame S). The Relativistic Doppler Shift 23 • Light is emitted by a sodium sample that is moving toward Earth with speed v. The wavelength of the light is 589 nm in the rest frame of the sample. The wavelength measured in the frame of Earth is 547 nm. Find v. Picture the Problem We can substitute, using c = fλ, in the relativistic Doppler shift equation and solve for the speed of the source.

Chapter 39

262

Using the expression for the relativistic Doppler shift, express f ′ as a function of v:

cvcv

ff'−+

=11

Substitute using c = fλ and simplify to obtain: cv

cvc'

c−+

=11

λλ ⇒

cvcv

' −+

=11

λλ

or

cvcv

' −+

=⎟⎠⎞

⎜⎝⎛

112

λλ

Solving for v gives:

c

'

'v

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

+⎟⎠⎞

⎜⎝⎛

−⎟⎠⎞

⎜⎝⎛

=1

12

2

λλλλ

Substitute numerical values and evaluate v:

( )

m/s1022.2

m/s10998.2

1nm547nm589

1nm547nm589

7

82

2

×=

×

⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢

⎣

⎡

+⎟⎟⎠

⎞⎜⎜⎝

⎛

−⎟⎟⎠

⎞⎜⎜⎝

⎛

=v

24 • A distant galaxy is moving away from us at a speed of 1.85 × 107 m/s. Calculate the fractional redshift (λ′ – λ0)/λ0 that we observe the light from this galaxy to have. Picture the Problem We can use the relativistic Doppler shift, when the source and the receiver are receding, to relate the frequencies of the two wavelengths and c = fλ to express the ratio of the wavelengths as a function of the speed of the galaxy. When the source and receiver are moving away from each other, the relativistic Doppler shift is given by:

0

1

1' f

cvcv

f+

−=

Use the relationship between the wavelength and frequency to obtain: 01

1

λλc

cvcv

'c

+

−= ⇒

cvcv

' +

−=

1

10

λλ

Relativity

263

Solving for λ′/λ0 yields:

cvcv

'

−

+=

1

1

0λλ

Express the fractional redshift: 1

1

11

00

0 −−

+=−=

−

cvcv

''λλ

λλλ

Substitute numerical values and evaluate (λ′−λ0)/ λ0:

0637.0

1

m/s10998.2m/s1085.11

m/s10998.2m/s1085.11

8

7

8

7

0

0

=

−

××

−

××

+=

−λλλ'

25 •• Derive

′f = f0 1− v 2 / c2( ) 1− v / c( )⎡⎣ ⎤⎦ (Equation

Error! Reference source not found.a) for the frequency received by an observer moving with speed v toward a stationary source of electromagnetic waves. Picture the Problem We can begin the derivation by expressing the number of waves received by the observer, in the rest frame of the source, in a time interval Δt. We can then relate this time interval to the time interval in the rest frame of the observer to complete the derivation of Equation 39-16a. Express the number of waves n encountered by the observer, in the rest frame of the source, in a time interval Δts:

( ) ( )

so

sos

1 tcvf

ctfvctvcn

Δ⎟⎠⎞

⎜⎝⎛ +=

Δ+=

Δ+=

λ

This time interval in the rest frame of the observer is given by:

γs

ott Δ

=Δ

Express the frequency received by the observer and simplify to obtain:

o

2

2

oo

2

2oo 1

1

1

1

1

11' f

cvcv

f

cvcv

f

cvcv

fcv

tnf

−

−=

−

+=

−

+=⎟

⎠⎞

⎜⎝⎛ +== γ

Δ

Chapter 39

264

26 • Show that if v is much less than c, the Doppler shift is given approximately by cvff ±≈Δ . Picture the Problem We can use the expression for the relativistic Doppler shift to show that, to a good approximation, Δf/f ≈ ±v/c. Express the fractional Doppler shift in terms of f and f0:

100

0

0

−=−

=Δ

ff

fff

ff

When the source and receiver are approaching each other, the relativistic Doppler shift is given by:

0

1

1f

cvcv

f−

+= ⇒

cvcv

ff

−

+=

1

1

0

Substitute in the expression for Δf/f0 to obtain:

111

11

1

2121

0

−⎟⎠⎞

⎜⎝⎛ −⎟

⎠⎞

⎜⎝⎛ +=

−−

+=

Δ

−

cv

cv

cvcv

ff

Keeping just the lowest order terms in v/c, expand binomially to obtain:

cv

cv

cv

cv

ff

=−+≈

−⎟⎠⎞

⎜⎝⎛ +⎟⎠⎞

⎜⎝⎛ +=

Δ

11

1211

211

0

The sign depends on whether the source and receiver are approaching or receding. Here we have assumed that they are approaching. 27 •• [SSM] A clock is placed in a satellite that orbits Earth with an orbital period of 90 min. By what time interval will this clock differ from an identical clock on Earth after 1.0 y? (Assume that special relativity applies and neglect general relativity.) Picture the Problem Due to its motion, the orbiting clock will run more slowly than the Earth-bound clock. We can use Kepler’s third law to find the radius of the satellite’s orbit in terms of its period, the definition of speed to find the orbital speed of the satellite from the radius of its orbit, and the time dilation equation to find the difference δ in the readings of the two clocks.

Relativity

265

Express the time δ lost by the clock: ⎟⎟⎠

⎞⎜⎜⎝

⎛−Δ=

Δ−Δ=Δ−Δ=

γγδ 11p ttttt

Because v << c, we can use Part (b) of Problem 10:

2

2

2111

cv

−≈γ

Substitute to obtain: t

cv

cvt Δ=⎥

⎦

⎤⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛−−Δ= 2

2

2

2

21

2111δ (1)

Express the square of the speed of the satellite in its orbit:

2

2222 42

Tr

Trv ππ

=⎟⎠⎞

⎜⎝⎛= (2)

where T is its period and r is the radius of its (assumed) circular orbit.

Use Kepler’s third law to relate the period of the satellite to the radius of its orbit about Earth:

32E

23

E

22 44 r

gRr

GMT ππ

== ⇒ 32

22E

4πTgRr =

Substitute numerical values and evaluate r:

( )( ) ( ) m1065.64

s/min60min90km6370m/s81.9 632

222

×=×

=π

r

Substitute numerical values in equation (2) and evaluate v2:

( )( )

227

2

2622

s/m1099.5s/min60min90

m1065.64

×=

××

=πv

Finally, substitute for v2 in equation (1) and evaluate δ:

( )( )( )

ms11m/s10998.2

Ms/y31.56y0.1s/m1099.521

28

227

=×

××=δ

28 •• For light that is Doppler-shifted with respect to an observer, we define the redshift parameter z = f − ′f( ) ′f , where f is the frequency of the light measured in the rest frame of the emitter, and f′ is the frequency measured in the rest frame of the receiver. If the emitter is moving directly away from the receiver, show that the relative velocity between the emitter and the receiver is

v = c u2 −1( ) u 2 +1( ), where u = z + 1.

Chapter 39

266

Picture the Problem We can use the definition of the redshift parameter and the

relativistic Doppler shift equation to show that ⎟⎟⎠

⎞⎜⎜⎝

⎛+−

=11

2

2

uucv , where u = z + 1.

The red-shift parameter is defined to be:

f''ffz −

= 0

The relativistic Doppler shift for recessional motion is given by: cv

cvf'f+−

=11

0

Substitute for f ′ and simplify to obtain:

111

11

11

0

00

−−+

=

+−

+−

−=

cvcv

cvcvf

cvcvff

z

Letting u = z + 1:

cvcvzu

−+

=+=111 ⇒ ⎟⎟

⎠

⎞⎜⎜⎝

⎛+−

=11

2

2

uucv

29 • A light beam moves along the y′ axis with speed c in frame S′, which is moving to the in the +x direction with speed v relative to frame S. (a) Find the x and y components of the velocity of the light beam in frame S. (b) Show that, according to the velocity transformation equations, the magnitude of the velocity of the light beam in S is c. Picture the Problem We can use the velocity transformation equations for the x and y directions to express the x and y components of the velocity of the light beam in frame S. (a) The x and y components of the velocity of the light beam in frame S are: 21

c'vu

v'uux

xx

+

+= and

⎟⎠⎞

⎜⎝⎛ +

=

21c

'vu'u

ux

yy

γ

Because ux′ = 0: vux = and

γcuy =

Relativity

267

(b) The magnitude of the velocity of the light beam in S is given by:

cccvv

cvuuu yx

=⎟⎟⎠

⎞⎜⎜⎝

⎛−+=

+=+=

22

22

2

2222

1

γ

30 •• A spaceship is moving east at speed 0.90c relative to Earth. A second spaceship is moving west at speed 0.90c relative to Earth. What is the speed of one spaceship relative to the other spaceship? Picture the Problem Let S be the Earth reference frame and S′ be that of the ship traveling East (+x direction). Then in the reference frame S′, the velocity of S is directed West, i.e., v = −ux. We can apply the inverse velocity transformation equation to express xu' in terms of ux. Apply the inverse velocity transformation equation to obtain: 21

cvu

vu'ux

xx

−

−=

Substituting for v gives:

2

2

2

2

1

2

1cu

u

cuuu'u

x

x

x

xxx

+=

+

+=

Because ux = 0.90c: ( )

( )c

ccc'ux 99.0

90.01

90.02

2

2 =+

=

31 •• [SSM] A particle moves with speed 0.800c in the + ′′x direction along the ′′x axis of frame ′′S , which moves with the same speed and in the same direction along the x′ axis relative to frame S′. Frame S′ moves with the same speed and in the same direction along the x axis relative to frame S. (a) Find the speed of the particle relative to frame S′. (b) Find the speed of the particle relative to frame S. Picture the Problem We can apply the inverse velocity transformation equation to express the speed of the particle relative to both frames of reference. (a) Express 'ux in terms of :''ux

21c

''vuv''u'ux

xx

+

+=

where v of S ′, relative to S″, is 0.800c.

Chapter 39

268

Substitute numerical values and evaluate 'ux : ( )

c

c

cc

cc'ux

976.0

64.160.1

800.01

800.0800.0

2

2

=

=+

+=

(b) Express ux in terms of 'ux :

21c

'vuv'uux

xx

+

+= where v, the speed of S,

relative to S ′, is 0.800c.

Substitute numerical values and evaluate ux: ( )( )

c

c

ccc

ccux

997.0

781.1776.1

976.0800.01

800.0976.0

2

=

=+

+=

Relativistic Momentum and Relativistic Energy 32 • A proton that has a rest energy equal to 938 MeV has a total energy of 2200 MeV. (a) What is its speed? (b) What is its momentum? Picture the Problem We can use the relation for the total energy, momentum, and rest energy to find the momentum of the proton and Equation 39-25 to relate the speed of the proton to its energy and momentum. Relate the energy of the proton to its momentum:

( )22222 mccpE +=

(b) Solving for p gives: ( )

2

222

cmcEp −

=

Substitute numerical values and evaluate p:

( ) ( )

c

cp

GeV99.1

MeV938MeV2200 22

=

−=

(a) From Equation 39-25 we have: E

pccv= ⇒ c

Epcv =

Substitute numerical values and evaluate v: ccv 905.0

MeV2200GeV99.1

==

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269

33 • If the kinetic energy of a particle equals twice its rest energy, what percentage error is made by using p = mu for the magnitude of its momentum? Picture the Problem We can use ( )22222 mccpE += (Equation R-17) to find the relativistic momentum of the particle in terms of γ and the fact that the kinetic energy of the particle equals twice its rest energy to find the error made in using mv for the momentum of the particle. Express the error e in using p′ = mv for the momentum of the particle:

pp'

pp'pe −=

−= 1 (1)

The relationship between the total energy E, momentum p, and rest energy mc2 of the particle is:

( )22222 mccpE += R-17

Solve for p and simplify to obtain: ⎟⎟⎠

⎞⎜⎜⎝

⎛−=−= 142

22222

2

2

cmEcmcm

cEp

Because E = γmc2: 12 −= γmcp

Substitute for p and p′ in equation (1) to obtain:

11

11

22 −−=

−−=

γγ cu

mcmue (2)

From the definition of γ: 2

11γ

−=cu

Eliminate v/c in equation (2) to obtain: γγγ

11111

11 22−=−

−−=e (3)

The kinetic energy of the particle is related to its rest energy:

( ) 21 mcK −= γ ⇒ 21mcK

+=γ

Because the kinetic energy of the particle is twice its rest energy:

321 2

2

=+=mcmcγ

Substitute for γ in equation (3) and evaluate e: %7.66667.0

311 ==−=e

Chapter 39

270

34 •• In a certain reference frame, a particle has momentum of 6.00 MeV/c and total energy of 8.00 MeV. (a) Determine the mass of the particle. (b) What is the total energy of the particle in a reference frame in which its momentum is 4.00 MeV/c? (c) What is the relative speed of the two reference frames? Picture the Problem (a) The mass of the particle is given by m = E0/c2. (b) We can use Equation R-17 [ ( )22222 mccpE += ] to find the total energy of the particle in a reference frame in which its momentum is 4.00 MeV/c. (c) We can apply the inverse velocity transformation equation to find the relative speeds of the two reference frames. (a) The mass of the particle is given by:

20

cEm =

Substituting numerical values yields:

( )( )( )

kg 1043.1

m/s 10998.2eV/J10602.1MeV 00.8

29

28

19

−

−

×=

×

×=m

(b) Solving equation R-17 for E gives:

( ) 20

222222 EcpmccpE +=+= (1)

We can use Equation R-17 to find the energy of the particle in the reference frame in which it has momentum of 6.00 MeV/c and total energy of 8.00 MeV:

20

222 EcpE += ⇒ 2220 cpEE −=

Substitute numerical values and evaluate E0:

( ) ( )( ) ( )

MeV292.5MeV00.6MeV00.8

MeV/00.6MeV00.822

2220

=

−=

−= ccE

Substitute numerical values in equation (1) and evaluate the total energy E of the particle:

( ) ( )MeV63.6

MeV292.5MeV/00.4 222

=

+= ccE

(c) The inverse speed transformation is: 21

cvu

vuua

ab

−

−=

where the subscripts refer to the speeds in Parts (a) and (b) of the problem.

Relativity

271

Solving for v gives:

21cuuuuv

ba

ba

−

−= (1)

Relate the relativistic energy of the particle in (a) to its speed: 2

20

1cu

EEa−

= ⇒ 2

01 ⎟⎠⎞

⎜⎝⎛−=

EEcua

Substitute numerical values and evaluate ua: ccua 7499.0

MeV00.8MeV292.51

2

=⎟⎟⎠

⎞⎜⎜⎝

⎛−=

Relate the relativistic momentum of the particle in (b) to its speed: 2

2

0

1cu

ump

b

b

−

= ⇒ 2

02 cmp

pcub+

=

Substitute numerical values and evaluate ub:

( )( ) ( )

ccc

ccub

6030.0MeV/292.5MeV/00.4

MeV/00.422

=

+=

Substitute in equation (1) and evaluate v: ( )( ) c

ccc

ccv 268.06030.07499.01

6030.07499.0

2

=−

−=

35 •• Show that ducum

cuumd

23

2

2

0220 1

1

−

⎟⎟⎠

⎞⎜⎜⎝

⎛−=

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛

−. Note: This relationship

was used to derive the relativistically correct expression for kinetic energy (Equation 39-22). Picture the Problem We can use the rule for the derivative of a quotient to establish the result given in the problem statement. Use the expression for the derivative of a quotient to obtain:

2

2

2

22

2

2

2

221

1

11

1cu

cuc

mumcu

cumu

dud

−

−

+−

=⎟⎟

⎠

⎞

⎜⎜

⎝

⎛

−

Chapter 39

272

Multiply the numerator and denominator of the right-hand side of this expression

by 2

2

1cu

− and simplify to obtain:

23

2

2

23

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

22

1

1

1

11

1

111

1

−

⎟⎟⎠

⎞⎜⎜⎝

⎛−=

⎟⎟⎠

⎞⎜⎜⎝

⎛−

+⎟⎟⎠

⎞⎜⎜⎝

⎛−

=

−⎟⎟⎠

⎞⎜⎜⎝

⎛−

−

−+−−

=⎟⎟

⎠

⎞

⎜⎜

⎝

⎛

−

cum

cu

cmum

cu

cu

cu

cucu

cmum

cu

cu

cumu

dud

and

ducum

cumud

23

2

2

221

1

−

⎟⎟⎠

⎞⎜⎜⎝

⎛−=

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛

−

36 •• The K0 particle has a mass of 497.7 MeV/c2. It decays into a π– and π+, each having mass 139.6 MeV/c2. Following the decay of a K0, one of the pions is at rest in the laboratory. Determine the kinetic energy of the other pion after the decay and of the K0 prior to the decay. Picture the Problem We will first consider the decay process in the center of mass reference frame and then transform to the laboratory reference frame in which one of the pions is at rest. Apply energy conservation in the center of mass frame of reference to obtain:

22K 00

2 cmcm γπ= ⇒0

0

2K

π

γm

m=

Substitute numerical values and evaluate γ :

( ) 783.1MeV/6.1392MeV/7.497

2

2

==ccγ

Because one of the pions is at rest in the laboratory frame, γ = 1.78 for the transformation to the laboratory frame. The kinetic energy of the K0 particle is:

( )( )( )

MeV388

MeV2.388MeV7.4971783.1

10K

=

=−=

−= EK γ

Relativity

273

The total initial energy in the laboratory frame is:

MeV9.885MeV388.2MeV7.497

=+=E

Express the energy of the other pion:

202 cmEE ππ −=

Substitute numerical values and evaluate Eπ:

( )MeV607

MeV6.1392MeV9.885

=

−=πE

37 •• [SSM] In reference frame S’, two protons, each moving at 0.500c, approach each other head-on. (a) Calculate the total kinetic energy of the two protons in frame S′. (b) Calculate the total kinetic energy of the protons as seen in reference frame S, which is moving with one of the protons. Picture the Problem The total kinetic energy of the two protons in Part (a) is the sum of their kinetic energies and is given by ( ) 012 EK −= γ . Part (b) differs from Part (a) in that we need to find the speed of the moving proton relative to frame S. (a) The total kinetic energy of the protons in frame S′ is given by:

( ) 012 EK −= γ

Substitute for γ and E0 and evaluate K:

( )( )

MeV290

MeV28.9381500.01

12

2

2

=

⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜

⎝

⎛

−

−

=

cc

K

(b) The kinetic energy of the moving proton in frame S is given by:

( ) 01 EK −= γ (1) where

21

1

cuv

−=γ

Express the speed u of the proton in frame S:

21c

'vuv'uux

x

+

+=

Substitute numerical values and evaluate u: ( )( ) c

ccc

ccu 800.0500.0500.01

500.0500.0

2

=+

+=

Chapter 39

274

Evaluate γ: ( )( )

67.1800.0800.01

1

2

=−

=

ccc

γ

Substitute numerical values in equation (1) and evaluate K:

( )( )MeV629

MeV28.938167.1

=

−=K

38 •• An antiproton p has the same mass m as a proton p. The antiproton is created during the reaction p + p → p + p + p + p . During an experiment, protons at rest in the laboratory are bombarded with protons of kinetic energy KL, which must be great enough so that an amount of kinetic energy equal to 2mc2 can be converted into the rest energy of the two particles. In the frame of the laboratory, the total kinetic energy cannot be converted into rest energy because of conservation of momentum. However, in the zero-momentum reference frame in which the two initial protons are moving toward each other with equal speed u, the total kinetic energy can be converted into rest energy. (a) Find the speed of each proton u so that the total kinetic energy in the zero-momentum frame is 2mc2. (b) Transform to the laboratory’s frame in which one proton is at rest, and find the speed u′ of the other proton. (c) Show that the kinetic energy of the moving proton in the laboratory’s frame is KL = 6mc2. Picture the Problem (a) We can find the speed of each proton by equating their total relativistic kinetic energy to 2mc2. In Part (b) we can use the inverse velocity transformation with V = u and ux = −u to find ux′. In Part (c) we’ll need to evaluate γ ′ for the kinetic energy transformation ( ) 0L 1 E'K −= γ . (a) Set the relativistic kinetic energy of the protons equal to 2mc2 to obtain:

( ) 20 212 mcE =−γ ⇒ 2=γ

Substitute for γ : 2

1

1

2

2=

−cu

⇒ ccu 866.023

==

Relativity

275

(b) Use the inverse velocity transformation with v = u and ux = −u to find ux′:

ccc

c

cc

cvu

vu'ux

xx

990.07

34

c23

23

1

23

23

1

2

2

−=−

=

⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛−

−

−−=

−

−=

(c) The kinetic energy of the moving proton in the laboratory’s frame is given by:

( ) 0L 1 E'K −= γ where

( )7

734

1

1

1

1

2

2

2

2=

⎟⎟⎠

⎞⎜⎜⎝

⎛−

−

=

−

=

c

cc'u

'x

γ

Substitute for γ ′ and E0 and evaluate KL:

( ) 22L 617 mcmcK =−=

39 ••• A particle of mass 1.0 MeV/c2 and kinetic energy 2.0 MeV collides with a stationary particle of mass 2.0 MeV/c2. After the collision, the particles stick together. Find (a) the speed of the first particle before the collision, (b) the total energy of the first particle before the collision, (c) the initial total momentum of the system, (d) the total kinetic energy after the collision, and (e) the mass of the system after the collision. Picture the Problem Parts (a) and (b) The initial speed of the particle can be found from its total energy and its total energy found using .00 EEKE γ=+= (c)

We can solve ( )22222 mccpE += for the initial momentum of the system. In Parts (d) and (e) we can use conservation of energy and conservation of momentum to find the total kinetic energy after the collision and the mass of the system after the collision. (a) Express the total energy of the particle:

00 EEKE γ=+=

Because the kinetic energy of the particle is twice its energy:

0002 EEE γ=+ and 3=γ

Chapter 39

276

Solving ( )21

1

cu−=γ γ for u

yields:

211γ

−= cu

Substituting the numerical value of γ and evaluating u gives: cccu 943.0

98

311 2 ==−=

(b) The total energy of the particle is:

000 3EEEKE ==+= γ

Substitute for E0 and evaluate E: ( ) MeV0.3MeV0.13 ==E

(c) The initial momentum of the incoming particle is related to its energy and mass according to:

( )22222 mccpE +=

Solving for p yields:

( )2221 mcEc

p −=

Substitute for E and mc2 and simplify to obtain:

( ) ( )cEEE

cp 02

02

0831

=−=

Substitute for E0 and evaluate p:

( ) cc

p MeV/8.2MeV0.18==

(d) and (e) From conservation of energy we have:

MeV0.5if == EE

From conservation of momentum we have:

if pp =

The final momentum of the system is related to its energy and mass according to:

2f0

22f

2f EcpE += ⇒ 22

f2ff0 cpEE −=

Substitute numerical values and evaluate Ef0:

( ) ( )MeV12.4

MeV/83.2MeV0.5 222f0

=

−= ccE

Because :2

f0f0 cmE = 22f0

f0 MeV/1.4 ccEm ==

Relativity

277

The total kinetic energy after the collision is given by: MeV0.9

MeV4.122MeV0.5f0ff

=

−=−= EEK

General Relativity 40 •• Light traveling in the direction of increasing gravitational potential undergoes a frequency redshift. Calculate the shift in wavelength if a beam of light of wavelength λ = 632.8 nm is sent up a vertical shaft of height L = 100 m. Picture the Problem Let m represent the mass equivalent of a photon. We can equate the change in the gravitational potential energy of a photon as it rises a distance L in the gravitational field to hΔf and then express the wavelength shift in terms of the frequency shift. The speed of the photons in the light beam are related to their frequency and wavelength:

λfc = ⇒ λcf =

Differentiate this expression with respect to λ to obtain: 2

2

λλ

λcc

ddf

−=−= −

Approximate df/dλ by Δf/Δλ and solve for Δf:

λλ

ΔΔ 2

cf −=

Divide both sides of this equation by f to obtain: λ

λ

λ

λλ ΔΔΔ

−=−

=c

c

ff 2

Solving for Δλ gives: f

fΔ−=Δ λλ (1)

The change in the energy of the photon as it rises a distance L in a gravitational field is given by:

mgLUE =Δ=Δ

Because ΔE = hΔf : mgLfh =Δ (2)

Letting m represent the mass equivalent of the photon:

hfmcE == 2 (3)

Chapter 39

278

Divide equation (2) by equation (3) to obtain:

2mcmgL

hffh=

Δ ⇒ 2cgL

ff=

Δ

Substitute for Δf/f in equation (1): 2c

gLλλ −=Δ

Substitute numerical values and evaluate Δλ:

( )( )( )( )

nm1090.6

m/s10998.2nm8.632m100m/s81.9Δ

12

28

2

−×−=

×−=λ

41 •• Let us revisit a problem from Chapter 3: Two cannons are pointed directly toward each other, as shown in Figure 39-17. When fired, the cannonballs will follow the trajectories shown. Point P is the point where the trajectories cross each other. Ignore any effects due to air resistance. Using the principle of equivalence, show that if the cannons are fired simultaneously (in the rest frame of the cannons), the cannonballs will hit each other at point P. Picture the Problem In a freely falling reference frame, both cannonballs travel along straight lines, so they must hit each other, as they were pointed at each other when they were fired. On the other hand, in Earth’s homogeneous gravitational field both balls will have the same vertical acceleration and so will hit each other at P. This shows that a homogeneous gravitational field is completely equivalent to a uniformly accelerated reference frame. 42 ••• A horizontal turntable rotates with angular speed ω. There is a clock at the center of the turntable and an identical clock mounted on a turntable a distance r from the center. In an inertial reference frame in which the clock at the center is at rest, the clock at distance r is moving with speed u = rω. (a) Show that from time dilation according to special relativity, the time between ticks, Δt0 for the clock at rest and ΔtR for the moving clock, are related by

Δtr − Δt0

Δt0

= −r2ω 2

2c2 rω = c

(b) In a reference frame rotating with the table, both clocks are at rest. Show that the clock at distance r experiences a pseudoforce Fr = mrω2 in this rotating frame and that this is equivalent to a difference in gravitational potential between r and the origin of φr − φ0 = −

12 r2ω 2 . Use the difference in gravitational potential given

in Part (b) to show that in this frame the difference in time intervals is the same as in the inertial frame. Picture the Problem Consider the turntable to be a giant hollow cylinder in space that is spinning about its axis. Someone on the inside surface of the cylinder would experience a centripetal acceleration caused by the normal force of the surface pushing them toward the rotation axis. Alternatively, they can consider

Relativity

279

that they are not accelerating but a gravitational field r g = ω2r ˆ r is pushing them

away from the axis ( ̂ r is away from the axis). This is the principle of equivalence. From this perspective, up is toward the axis and the points closer to the axis are at the higher gravitational potential. (Just like the electric field points in the direction of decreasing electric potential, the gravitational field points in the direction of decreasing gravitational potential.) (a) From the time dilation equation we have: γ

0r

tt Δ=Δ and 11

0

0r −=ΔΔ−Δ

γttt

Because rω/c << 1:

2

22

2

2

21

2111

cr

cu ω

γ−=−≈

Substitute for γ1

and simplify to

obtain:

2

22

2

22

0

0r

21

21

cr

cr

ttt ωω

−=−−=ΔΔ−Δ

(b) The pseudoforce is given by: maF −=P where a is the acceleration of the non-inertial reference frame.

In this case a is the centripetal acceleration:

2ωra −= ⇒ 2P ωmrFF r ==

To relate this problem to Equation 39-31, point 2 is a distance r from the axis and point 1 is on the axis. The term in parentheses on the right hand side of Equation 39-31 is φ2 − φ1, which translates to φr − φ0. Because φr is at a lower potential than φ0, this term is negative. Hence:

2221

0

2

0

2

00 ˆ ωωωφφ rdrrdrd

rrr

r −=−=⋅−=⋅−=− ∫∫∫ lr

lrr rg

From Equation 39-31: ( ) ( )

2

22

2221

2020

0r

2

11

cr

rcct

ttr

ω

ωφφ

−=

−=−=ΔΔ−Δ

Chapter 39

280

General Problems 43 • How fast must a muon travel so that its mean lifetime is 46 μs if its mean lifetime at rest is 2.2 μs? Picture the Problem We can use the definition of γ and the time dilation equation to find the speed of the muon. (a) From the definition of γ we have: 2

1

1

⎟⎠⎞

⎜⎝⎛−

=

cu

γ ⇒ 211γ

−=cu

Relate the mean lifetime of the muon to its proper lifetime:

ptt Δ=Δ γ ⇒ ptt

ΔΔ

=γ

Substitute in the expression for u/c to obtain:

2p1 ⎟⎟⎠

⎞⎜⎜⎝

⎛ΔΔ

−=tt

cu

Substitute numerical values and evaluate u/c:

999.0s46s2.21

2

=⎟⎟⎠

⎞⎜⎜⎝

⎛−=

μμ

cu

or cu 999.0=

44 • A distant galaxy is moving away from Earth with a speed that results in each wavelength received on Earth being shifted so that λ′ = 2λ0. Find the speed of the galaxy relative to Earth. Picture the Problem We can use the relativistic Doppler shift, when the source and the receiver are receding, to relate the frequencies of the two wavelengths and c = fλ to express the ratio of the wavelengths as a function of the speed of the galaxy. When the source and receiver are moving away from each other, the relativistic Doppler shift is given by:

0

1

1f

cvcv

f'+

−=

Relativity

281

Use the relationship between the wavelength and frequency to obtain:

01

1

λλc

cvcv

'c

+

−= ⇒

cvcv

' +

−=

1

10

λλ

Because λ′ = 2λ0:

cvcv

+

−==

1

1

21

2 0

0

λλ

⇒ cv 600.0=

45 •• [SSM] Frames S and S′ are moving relative to each other along the x and x′ axes (which superpose). Observers at rest in the two frames set their clocks to t = 0 when the two origins coincide. In frame S, event 1 occurs at x1 = 1.0 c⋅y and t1 = 1.00 y and event 2 occurs at x2 = 2.0 c⋅y and t2 = 0.50 y. These events occur simultaneously in frame S′. (a) Find the magnitude and direction of the velocity of S′ relative to S. (b) At what time do both these events occur as measured in S′? Picture the Problem We can use Equation 39-12, the inverse time transformation equation, to relate the elapsed times and separations of the events in the two systems to the velocity of S′ relative to S. We can use this same relationship in Part (b) to find the time at which these events occur as measured in S′. (a) Use Equation 39-12 to obtain:

( ) ( )

⎥⎦⎤

⎢⎣⎡ Δ−Δ=

⎥⎦⎤

⎢⎣⎡ −−−=−=Δ

xcvt

xxcvtt't'tt'

2

1221212

γ

γ

Because the events occur simultaneously in frame S′, Δt′ = 0 and:

xcvt Δ−Δ= 20 ⇒

xtcv

ΔΔ

=2

Substitute for Δt and Δx and evaluate v:

( ) ccc

cv 50.0y1.00y0.2

y1.00y50.02

−=⋅−⋅

−=

Because :y50.012 −=−= tttΔ direction. in the moves ' xS −

Chapter 39

282

(b) Use the inverse time transformation to obtain:

2

2

22

2

22

22

1cvcvxt

cvxt't

−

−=⎟

⎠⎞

⎜⎝⎛ −= γ

Substitute numerical values and evaluate t2′ and t1′:

( )( )

( )

y7.1

50.01

y0.250.0y50.0

2

2

2

12

=

−−

⋅−−

==

cc

ccc

't't

46 •• An interstellar spaceship travels from Earth to a star system 12 light-years away (as measured in Earth’s frame). The trip takes 15 y as measured by clocks on the spaceship. (a) What is the speed of the spaceship relative to Earth? (b) When the spaceship arrives, it sends an electromagnetic signal to Earth. How long after the spaceship leaves Earth will observers on Earth receive the signal? Picture the Problem We can use the relationship between distance, speed, and time and the length contraction relationship to find the speed of the ship relative to Earth. The elapsed time between the departure of the spaceship and the receipt of the signal at Earth is the sum of the travel time to the distant star system and the time it takes the signal to return to Earth. (a) Express the travel time as measured on the spaceship: u

LuL't'

γ==Δ ⇒

t'LuΔ

=γ

Substitute numerical values and evaluate γu:

ccu 80.0y15y12=

⋅=γ

or

uc

cu

80.0

1

1

2

2=

−

⇒ cu 63.0=

(b) The elapsed time T before Earth receives the signal is the sum of the travel time to the distant star system and the time it takes the signal to return:

cL

uLT +=

Substitute numerical values and evaluate T:

y31y12625.0

y12=

⋅+

⋅=

cc

ccT

Relativity

283

47 •• The neutral pion π 0 has a rest mass of 135.0 MeV/c2. This particle can be created in a proton–proton collision: p + p → p + p + π 0. Determine the threshold kinetic energy for the creation of a π0 in a collision of a moving and stationary proton. (See Problem 38.) Picture the Problem We can use conservation of energy to find γ in the CM frame of reference and then use the definition of γ to find the speed u of the projectile proton. We can then use the velocity transformation equation to find the speed and kinetic energy of this proton in the laboratory frame of reference. Use conservation of energy to find γ in the CM frame of reference:

fi EE =γ ⇒ i

f

EE

=γ

Ei and Ef are:

MeV1876MeV938Mev938i =+=E and

MeV0112MeV135.0MeV938Mev938f

=++=E

Substitute Ei and Ef and evaluate γ : 072.1

MeV1876MeV2011

==γ

Express γ as a function of the speed u of the projectile proton: 2

2

1

1

cu

−

=γ ⇒ 211γ

−= cu

Substitute for γ and evaluate u: ( )

ccu 360.0072.111 2 =−=

Transform to the laboratory frame and find u′:

( )( )( )

cc

ccc

cvuvuu'

637.0

c360.0360.01

360.0360.0

1 22

=

−−

−−=

−

−=

Chapter 39

284

The kinetic energy of the moving proton in the laboratory’s frame is given by:

( ) 0LL 1 EK −= γ where

( )

( )297.1

637.01

1

1

1

2

2

2

2L

=

−

=

−

=

cc

cu'

γ

Substitute for γL and E0 and evaluate KL:

( )( ) MeV281MeV9381297.1L =−=K

Remarks: In Problem 51 we show that the threshold kinetic energy of the

projectile is given by( )( )

.target

infinfininth m

cmmmmK

2

2∑ ∑∑ ∑ −+

=

48 •• A rocket that has a proper length of 1000 m moves away from a space station and in the +x direction at 0.60c relative to an observer on the station. An astronaut stands at the rear of the rocket and fires a dart toward the front of the rocket at 0.80c relative to the rocket. How long does it take the dart to reach the front of the rocket (a) as measured in the frame of the rocket, (b) as measured in the frame of the space station, and (c) as measured in the frame of the dart? Picture the Problem Let S be the frame of the space station and let S′ be the frame of the rocket. Then v = +0.60c. In addition, let event A be the dart leaving the gun, event B be the dart arriving at the front of the ship, and L be the rest length of the ship. Then x′A = 0, x′B = L, and t′A = 0. Finally, let ux and u′x be the speed of the dart in reference frames S and S′ respectively. (a) In the reference frame of the rocket:

cL

u'Lt't'

x 80.0AB ==−

or, because At' = 0,

cLt'80.0B =

Substitute numerical values and evaluate Bt' :

( )( ) st' μ2.4m/s10998.280.0

m10008B =

×=

Relativity

285

(b) In the reference frame of the space station:

( )

2

2BB

AB

1 ⎟⎠⎞

⎜⎝⎛−

−+

=−

cv

cx'x'v

u'L

t't' x

or, because t′A = 0, x′A = 0, and x′B = L,

2

2

2

2

B

1

1

1 ⎟⎠⎞

⎜⎝⎛−

⎟⎟⎠

⎞⎜⎜⎝

⎛+

=

⎟⎠⎞

⎜⎝⎛−

+=

cv

Lcv

u'

cv

cvL

u'L

t' xx

Substituting numerical values gives: ( )

s7.7km 313.2

60.01

m 100060.080.01

2

2

B

μ==

⎟⎠⎞

⎜⎝⎛−

⎟⎠⎞

⎜⎝⎛ +

=

c

cc

cc

ct'

Alternate solution to Part (b) In the reference frame of the dart the speed of the rocket is 0.80c and the length of the ship is:

( )2

2

1c

u'LL' x−=

or, because u′x = 0.80c, ( ) L

ccLL' 60.080.01 2

2

=−=

The time between events A and B (the proper time) is:

xu'L't't' =− AB

Substituting for At' , L′ and xu' gives:

cL

cLt' 75.0

80.060.0

B ==

The speed of the dart in the frame of reference of the space station is given by Equation 39-18a: 21

cvu'

vu'u

x

xx

+

+=

Substituting numerical values and simplifying yields: ( )( ) c

ccc

ccux 9459.060.080.01

80.060.0

2

=+

+=

Chapter 39

286

The time between the events in S is related to the proper time between the events by the time dilation formula:

2

ABAB

1 ⎟⎠⎞

⎜⎝⎛−

−=−

cu

t't'ttx

Substituting for L, At , Bt' , and xu

gives: ( )

s 7.7

km 313.2

9459.01

m 100075.0

2B

μ=

=

⎟⎠⎞

⎜⎝⎛−

=c

cc

ct

(c) In the bullet’s frame of reference, the elapsed time is given by: xu'

L't't' =− AB

or, because At' = 0,

xu'

L't' =B (1)

The length L′ of the rocket in the dart’s frame of reference is given by:

( )2

2

1c

u'LLL' x−==

γ

Substitute for L′ in equation (1) to obtain:

( )

x

x

u'c

u'L

t'2

2

B

1−=

Substituting numerical values gives: ( ) ( )

s5.2c

m 75080.0

80.01m1000 2

2

B

μ=

=−

=c

cc

t'

49 •• [SSM] Using a simple thought experiment, Einstein showed that there is mass associated with electromagnetic radiation. Consider a box of length L and mass M resting on a frictionless surface. Attached to the left wall of the box is a light source that emits a directed pulse of radiation of energy E, which is completely absorbed at the right wall of the box. According to classical electromagnetic theory, this radiation carries momentum of magnitude p = E/c (Equation 32-13). The box recoils when the pulse is emitted by the light source. (a) Find the recoil velocity of the box so that momentum is conserved when the light is emitted. (Because p is small and M is large, you may use classical

Relativity

287

mechanics.) (b) When the light is absorbed at the right wall of the box the box stops, so the total momentum of the system remains zero. If we neglect the very small velocity of the box, the time it takes for the radiation to travel across the box is Δt = L/c. Find the distance moved by the box in this time. (c) Show that if the center of mass of the system is to remain at the same place, the radiation must carry mass m = E/c2. Picture the Problem We can use conservation of energy to express the recoil velocity of the box and the relationship between distance, speed, and time to find the distance traveled by the box in time Δt = L/c. Equating the initial and final locations of the center of mass will allow us to show that the radiation must carry mass m = E/c2. (a) Apply conservation of momentum to obtain: 0i ==+ pMv

cE

⇒McEv −=

(b) The distance traveled by the box in time Δt = L/c is:

cvLtvd =Δ=

Substitute for v from (a) to obtain: 2Mc

LEMcE

cLd −=⎟

⎠⎞

⎜⎝⎛−=

(c) Let x = 0 be at the center of the box and let the mass of the photon be m. Then initially the center of mass is at:

mMmLx+

−= 2

1

CM

When the photon is absorbed at the other end of the box, the center of mass is at: mM

McELLm

McMEL

x+

⎥⎦

⎤⎢⎣

⎡⎟⎠⎞

⎜⎝⎛ −+

−

=22

12

CM

Because no external forces act on the system, these expressions for xCM must be equal:

mMMcELLm

McMEL

mMmL

+

⎥⎦

⎤⎢⎣

⎡⎟⎠⎞

⎜⎝⎛ −+

−

=+

− 221

221

Solving for m yields:

⎟⎠⎞

⎜⎝⎛ −

=

22 1

McEc

Em

Chapter 39

288

Because Mc2 is of the order of 1016 J and E = hf is of the order of 1 J for reasonable values of f, E/Mc2 << 1 and:

2cEm =

50 ••• Using the relativistic conservation of momentum and energy and the relation between energy and momentum for a photon E = pc, prove that a free electron (an electron not bound to an atomic nucleus) cannot absorb or emit a photon. Picture the Problem Without loss of generality, we’ll consider the absorption case. We’ll assume that the electron is initially at rest and that it travels with a speed v after it absorbs the photon. Applying the conservation of energy and the conservation of momentum will lead us to an absurd conclusion that, in turn, will force us to abandon our initial assumption that an electron can absorb a photon. Such an argument is known as a reductio ad absurdum argument. When the electron absorbs a photon, the conservation of relativistic momentum requires that its momentum become:

mvp γ=

From the conservation of energy:

22 mcpcmc γ=+ ⇒ ( )mcp 1−= γ

Equate these expression for p to obtain:

( )mcmv 1−= γγ ⇒ cv ⎟⎟⎠

⎞⎜⎜⎝

⎛ −=

γγ 1 (1)

Squaring both sides of the equation gives:

22

22

22 121 ccv

γγγ

γγ +−

=⎟⎟⎠

⎞⎜⎜⎝

⎛ −= (2)

From the definition of γ :

22

2

2

22

1

1vc

c

cv −

=−

=γ

Solve for v2 to obtain:

22

22 1cv

γγ −

=

Substituting for v2 in equation (2) and simplifying gives:

22

22

2

2 121 ccγγγ

γγ +−

=−

or 121 +−=− γ ⇒ 1=γ

Relativity

289

Substitute for γ in equation (1) and evaluate v: 0

111

=⎟⎠⎞

⎜⎝⎛ −

= cv

Our assumption that an electron can absorb a photon has led to the contradictory conclusion that its speed after the absorption is zero. Hence we must conclude that the electron cannot absorb a photon. 51 ••• [SSM] When a moving particle that has a kinetic energy greater than the threshold kinetic energy Kth strikes a stationary target particle, one or more particles may be created in the inelastic collision. Show that the threshold kinetic energy of the moving particle is given by

Kth =

Σmin + Σmfin( ) Σmfin + Σmin( )c2

2mtarget

Here Σmin is the sum of the masses of the particles prior to the collision, Σmfin is the sum of the masses of the particles following the collision, and mtarget is the mass of the target particle. Use this expression to determine the threshold kinetic energy of protons incident on a stationary proton target for the production of a proton–antiproton pair; compare your result with the result of Problem 38. Picture the Problem Let mi denote the mass of the incident (projectile) particle. Then ∑min = mi + mtarget and we can use this expression to determine the threshold kinetic energy of protons incident on a stationary proton target for the production of a proton–antiproton pair. Consider the situation in the center of mass reference frame. At threshold we have:

∑=− 2fin

222 cmcpE

Note that this is a relativistically invariant expression.

In the laboratory frame, the target is at rest so:

t,0ttarget EEE ==

We can, therefore, write:

( ) ( )22fin

22i

2t,0i ∑=−+ cmcpEE

For the incident particle:

2i,0

22i

2i EcpE =−

and thi,0i KEE +=

where thK is the threshold kinetic energy of the incident particle in the laboratory frame.

Chapter 39

290

Express thK in terms of the rest energies:

( ) ( )22fint,0th

2i.0t,0 2 ∑=++ cmEKEE

where ∑=+ 2

fini.0t,0 cmEE

and 2

targett,0 cmE =

Substitute to obtain:

( ) ( )22fin

2targetth

22fin 2 ∑∑ =+ cmcmKcm

Solving for thK gives:

( )( )target

2infinfinin

th 2mcmmmm

K ∑ ∑∑ ∑ −+=

For the creation of a proton -

antiproton pair in a proton - proton collision:

∑ = pin 2mm , ∑ = pfin 4mm and

ptarget mm =

Substituting for the sums and simplifying yields:

( )( )

( )( ) 2p

p

2pp

p

2pppp

th

62

26

22442

cmm

cmm

mcmmmm

K

==

−+=

in agreement with Problem 38. 52 ••• A particle of mass M decays into two identical particles, each of mass m, where m = 0.30M. Prior to the decay, the particle of mass M has a total energy of 4.0Mc2 in the laboratory reference frame. The velocities of the decay products are along the direction of motion of M. Find the velocities of the decay products in the laboratory reference frame. Picture the Problem We’ll solve the problem for the general case of a particle of rest mass M decaying into two identical particles each of rest mass m. In the center of mass reference frame:

222 22 mcmcMc γ==

Solve for u/c to obtain:

221 ⎟⎠⎞

⎜⎝⎛−=

Mm

cu

where u is the speed of each of the decay particles in the CM frame.

Relativity

291

Next we determine the speed v of the laboratory frame relative to the CM frame. The energy of the particle of rest mass M is:

2CMMcγ where

2

2CM

1

1

cv

−

=γ

and 2CM

CM11γ

β −==cv

Use Equation 39-18a to express ulab, the speeds of the decay products in the laboratory reference frame:

c

cucu

uCM

CM

lab

1 β

β

±

±=

where ± refers to the fact that one of the decay particles will travel in the direction of M, and the other in the direction opposite to that of M.

In this problem we have:

0.4CM =γ , 968.0CM =β , 600.02

0

0 =Mm ,

and 800.0=cu

Substituting numerical values gives:

( )( ) ccu 996.0800.0968.01

800.0968.0lab =

++

=

and

( )( ) ccu 745.0800.0968.01

800.0968.0lab =

−−

=

53 ••• A rod of proper length Lp makes an angle θ with the x axis in frame S. Show that the angle θ ′ made with the x′ axis in frame S′, which is moving in the +x direction with speed v, is given by tan θ′ = γ tan θ and that the length of the

stick in S′ is ′L = Lp γ −2cos2θ + sin 2θ( )1/ 2.

Picture the Problem We can write the components of the stick in its reference frame and then apply the Lorentz length contraction equation to obtain the given result. In its reference frame, the stick has x and y components:

θcospp LL x = and θsinpp LL y =

Only Lpx is Lorentz contracted: γ

xx

L'L p=

Chapter 39

292

Hence, the length in the reference frame S′ is:

( ) ( )[ ]21

22

2

p

2122

sincos⎟⎟⎠

⎞⎜⎜⎝

⎛+=

+=

θγ

θL

'L'LL' yx

The angle that L′ makes with the x′ axis is given by: θγ

γθθθ tancos

sintan ==='L'L

'x

y

54 ••• Show that if a particle moves at an angle θ with the x axis with speed u in frame S, it moves at an angle θ ′ with the x′ axis in S′ given by

( )uv−= − θθγθ cossin'tan 1 Picture the Problem We can express the tangent of the angle u′ makes with the x′ axis and then use the velocity transformation equations to obtain the given result. Express the tangent of the angle u′ makes with the x′ axis:

'u'u

'x

y=θtan

Substitute for uy′ and ux′:

( )vuu

cvu

vucvu

u

'x

y

x

x

x

y

−=

−

−

⎟⎠⎞

⎜⎝⎛ −

=γ

γθ

2

2

1

1tan

Substitute for uy and ux and simplify to obtain: ( ) uvvu

u'−

=−

=−

θθγ

θγθθ

cossin

cossintan

1

55 ••• [SSM] For the special case of a particle moving with speed u along the y axis in frame S, show that its momentum and energy in frame S′ are related to its momentum and energy in S by the transformation equations

⎟⎠⎞

⎜⎝⎛ −= 2

'

cvEpp xx γ yy pp =' zz pp =' ⎟

⎠⎞

⎜⎝⎛ −=

cvp

cE

cE xγ' .

Compare these equations with the Lorentz transformation equations for x′, y′, z′, and t′. Notice that that the quantities px, py, pz and E/c transform in the same way as do x, y, z, and ct.

Relativity

293

Picture the Problem We can use the expressions for pr and E in S together with the relations we wish to verify and the inverse velocity transformation equations

to establish the condition ( ) ( ) ( ) 2

222222

γuv'u'u'uu' zyx +=++= and then use this

result to verify the given expressions for px′, py′, pz′ and E′/c. In any inertial frame the momentum and energy are given by: 2

2

1cu

m

−

=upr

r and

2

2

2

1cu

mcE−

=

where ur is the velocity of the particle and u is its speed.

The components of pr in S are:

2

2

1cu

mup xx

−

= ,

2

2

1cu

mup y

y

−

= , and

2

2

1cu

mup zz

−

=

Because ux = uz = 0 and uy = u:

0== zx pp and

2

2

1cu

mupy

−

=

Substituting zeros for px and pz in the relations we are trying to show yields:

220cvE

cvE'px γγ −=⎟

⎠⎞

⎜⎝⎛ −= , yy p'p = ,

0='pz , and

cE

cE

cE' γγ =⎟

⎠⎞

⎜⎝⎛ −= 0

In S′ the momentum components are: 2

2

1cu''mu'p x

x

−

= ,

2

2

1cu'

'mu'p y

y

−

= , and

2

2

1cu''mu'p z

z

−

=

Chapter 39

294

The inverse velocity transformations are: 21

cvuvu'u

x

xx

−

−= ,

21cvu

u'u

y

yy

−

= , and

21cvu

u'uz

zz

−=

Substitute ux = uz = 0 and uy = u to obtain:

v'ux −= , u'uy γ= , and 0='uz

Thus:

( ) ( ) ( )

2

22

2222

γuv

'u'u'uu' zyx

+=

++=

First, we verify that pz′ = pz = 0: ( ) 0

1

0

2

2==

−

= zz p

cu'

m'p

Next, we verify that py′ = py:

y

y

yy

p

cv

cu

cv

cv

cu

cv

p

cv

cu

cv

cv

cu

cu

mu

cu

cv

cu

cu

mu

cu

cvmu

cu'

'mu'p

=

⎟⎟⎠

⎞⎜⎜⎝

⎛−−−

⎟⎟⎠

⎞⎜⎜⎝

⎛−−−

=

⎟⎟⎠

⎞⎜⎜⎝

⎛−−−

⎟⎟⎠

⎞⎜⎜⎝

⎛−⎟⎟

⎠

⎞⎜⎜⎝

⎛−

−

=

−−

−

−

=

−−

=

−

=

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

22

2

2

2

2

2

2

2

22

2

2

2

2

2

11

11

11

11

1

1

1

111γ

γγ

γ

Relativity

295

Next, we verify that ⎟⎠⎞

⎜⎝⎛ −= 2c

vEp'p xx γ :

Ecv

cv

cu

cv

cv

cu

cv

Ecv

cv

cu

cv

cv

cu

Ecv

cu

cv

cu

cu

mccv

cu

cv

mv

cu''mu'p x

x

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

22

2

2

2

2

21

2

2

2

2

22

2

2

2

2

2

11

11

11

11

1

1

111

γ

γγ

γ

γγ

γγ

−=

⎟⎟⎠

⎞⎜⎜⎝

⎛−−−

⎟⎟⎠

⎞⎜⎜⎝

⎛−−−

−=

⎟⎟⎠

⎞⎜⎜⎝

⎛−−−

⎟⎟⎠

⎞⎜⎜⎝

⎛−⎟⎟

⎠

⎞⎜⎜⎝

⎛−

−=

−−

−

−

−=

−−

−=

−

=

−

Finally, we verify that :or , EE'cE

cvp

cE

cE' x γγγ ==⎟

⎠⎞

⎜⎝⎛ −=

E

cv

cu

cv

cv

cu

cv

E

cv

cu

cv

cv

cu

E

cu

cv

cu

E

cu'

cu

cu

mc

cu'

mcE'

γ

γγ

γ

γγ

γγ

=

⎟⎟⎠

⎞⎜⎜⎝

⎛−−−

⎟⎟⎠

⎞⎜⎜⎝

⎛−−−

=

⎟⎟⎠

⎞⎜⎜⎝

⎛−−−

⎟⎟⎠

⎞⎜⎜⎝

⎛−⎟⎟

⎠

⎞⎜⎜⎝

⎛−

=

−−

−=

−

−

−

=

−

=

−−

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

22

2

2

2

2

21

2

2

2

21

2

2

2

2

2

2

11

11

11

11

1

1

1

1

11

The x, y, z, and t transformation equations are:

( )vtxx' −= γ , yy' = , zz' = and

⎟⎠⎞

⎜⎝⎛ −= 2c

vxtt' γ

Chapter 39

296

The x, y, z, and ct transformation equations are:

⎟⎠⎞

⎜⎝⎛ −= ct

cvxx' γ , yy' = , zz' =

and

⎟⎠⎞

⎜⎝⎛ −= x

cvctct' γ

The px, py, pz, and E/c transformation equations are:

⎟⎠⎞

⎜⎝⎛ −=

cE

cvp'p xx γ , yy p'p = , zz p'p =

and

⎟⎠⎞

⎜⎝⎛ −= xp

cv

cE

cE' γ

Note that the transformation equations for x, y, z, and ct and the transformation equations for px, py, pz, and E/c are identical. 56 ••• The equation for the spherical wavefront of a light pulse that begins at the origin at time t = 0 is x2 + y2 + z2 – (ct)2 = 0. Frame S′ moves with velocity v along the x axis. Using the Lorentz transformation, show that such a light pulse also has a spherical wavefront in frame S′ by showing that

′x 2 + ′y 2 + ′z 2 − c ′t( )2 = 0 . Picture the Problem The Lorentz transformation was derived on the basis of the postulate that the speed of light is c in any inertial reference frame. Thus, if the clocks in S and S′ are synchronized at t = t′ = 0, then it follows from the Einstein postulate that r2 = c2t2 and r′2 = c2t′2 or r2 − c2t2 = 0 = r′2 − c2t′2. In other words, the quantity s2 = r2 − c2t2 = 0 is a relativistic invariant, which can also be written as x2 + y2 + z2 − c2t2 = 0. Using the Lorentz transformation equations for x, y, z, and t we have:

x′2 + y′2 + z′2 − (ct′)2 = γ 2(x2 − 2vxt + v 2t2) + y2 + z2 − γ 2(c2t2 − 2vxt + v 2x2/c2) The terms linear in x cancel and the terms in x2 combine to give:

γ 2x2(1 − v 2/c2) = x2

The coefficients of the terms in (ct)2 give:

γ 2(v 2/c2 − 1) = −1

Thus, r2 − c2t2 = r′2 − c2t′2 as required by the Einstein postulate.

Relativity

297

57 ••• In Problem 56, you showed that the quantity x2 + y2 + z2 – (ct)2 has the same value (0) in both S and S′. A quantity that has the same value in all inertial frames is called a Lorentz invariant. From the results of Problem 55, the quantity

px

2 + py2 + pz

2 − E2 / c2 must also be a Lorentz invariant. Show that this quantity

has the value –m2c2 in both the S and S′ reference frames. Picture the Problem We’ll use Equation 39-27 to show that this quantity has the value −mc2 in both the S and S′ reference frames. From Equation 39-27, the relationship between total energy E, momentum p, and rest energy mc2 is:

( )22222 mccpE += or

( )22222 mcEcp −=−

Divide both sides of this equation by c2 to obtain:

( ) 22

2 mccEp −=⎟⎠⎞

⎜⎝⎛− (1)

We can relate p to px, py, and pz: 2222zyx pppp ++=

Substitute for p2 in equation (1) to obtain:

222

222 cmcEppp zyx −=⎟⎠⎞

⎜⎝⎛−++

Because m is the mass of the particle in its rest frame, it is constant. Hence:

22 ⎟

⎠⎞

⎜⎝⎛−

cEp must be a relativistic

invariant.

Also, in Problem 55 we saw that the components of p and the quantity E/c transform like the components of r and the quantity ct. In Problem 56 we demonstrated that r2 − (ct)2 is a relativistic invariant. Consequently, p2 − (E/c)2 must also be relativistically invariant. 58 ••• A long rod that is parallel to the x axis is released from rest. Subsequently, it is in free fall with an acceleration of magnitude g in the –y direction. An observer in a rocket ship moving with speed v parallel to the x axis passes by. Using the Lorentz transformations, show that the observer on the rocket ship will measure the rod to be bent into a parabolic shape. Is the parabola concave upward or concave downward? Picture the Problem We can use the inverse Lorentz transformation for time to show that the observer will conclude that the rod is bent into a parabolic shape.

Chapter 39

298

In frame S where the rod is not moving along the x axis, the height of the rod at time t is:

( ) 221 gtty −=

The inverse Lorentz time transformation is: ⎟

⎠⎞

⎜⎝⎛ += 2c

vxt't γ

Express y′(t) in the moving frame of reference: ( )

2

221 ⎟

⎠⎞

⎜⎝⎛ +−=

cvxt'gty' γ

Evaluate y′(t) at t′ = 0 to obtain: ( ) 2

2

2

2x

cvgty' γ

−=

Because ( ) 22

2

2x

cvgty' γ

−= is the equation of a parabola, we’ve shown that the

moving observer will conclude that the rod is bent into a parabolic shape. Because the coefficient of x2 is negative, the parabola is concave downward. 59 •• Show that if ux′ and v in Equation 39-18a are both less than c, then ux

is less than c. (Hint: Let ( )cux 11' ε−= and ( )cv 21 ε−= , where ε1 and ε2 are small positive numbers that are less than 1.) Picture the Problem We can make the substitutions given in the hint in Equation 39-18a and simplify the resulting expression to show that ux < c. Equation 39-18a gives the x direction relativistic velocity transformation:

2'1

'

cvu

vuux

xx

+

+= or

cvuc

vucu

x

xx

''

+

+=

Make the substitutions given in the hint to obtain:

( ) ( )( ) ( )

( )( )( )

( )( ) 2121

21

12

21

12

21

22

1112

1111

εεε

εε

εε

εεε

εε

ccεcc

ccεcux

++−+−

=

−−++−

=

−−+

−+−=

Relativity

299

Because ε1 and ε2 are small positive numbers that are less than 1, the numerator is less than the denominator and:

cuc

ux

x <⇒<1

60 ••• In reference frame S the acceleration of a particle is

kjia ˆˆˆzyx aaa ++=

r . Derive expressions for the acceleration components ax', ay',

and az' of the particle in reference frame S' that is moving relative to S in the x direction with velocity v. Picture the Problem We can evaluate the differentials of Equations 39-19a, b, and c and 39-12 and express their ratio to obtain expressions for ax', ay', and az'. From Equation 39-19a we have:

( )x

xx

xxxx

x

xx du

cvucv

cvu

ducvvudu

cvu

cvu

vud'du 2

2

2

2

2

2

22

2 1

1

1

1

1 ⎟⎠⎞

⎜⎝⎛ −

⎟⎟⎠

⎞⎜⎜⎝

⎛−

=

⎟⎠⎞

⎜⎝⎛ −

⎟⎠⎞

⎜⎝⎛−+⎟

⎠⎞

⎜⎝⎛ −

=⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

−

−=

From Equation 39-12:

dtcvudt

dtdx

cvdx

cvdt

cvxtddt' x ⎟

⎠⎞

⎜⎝⎛ −==−=⎟

⎠⎞

⎜⎝⎛ −= 2222 1γγγγγ

Divide dux′ by dt′ to obtain:

xx

xx

xx

xx a

dtdu

cvucv

dtcvu

du

cvucv

dt''du'a 333

2

2

2

2

2

2

2

2

1

1

1

1

1

1

δγγγ

=

⎟⎠⎞

⎜⎝⎛ −

⎟⎟⎠

⎞⎜⎜⎝

⎛−

=⎟⎠⎞

⎜⎝⎛ −

⎟⎠⎞

⎜⎝⎛ −

⎟⎟⎠

⎞⎜⎜⎝

⎛−

== where 21cvux−=δ

Proceeding in exactly the same manner, one obtains:

xy

yy ac

vua'a 23322

1δγδγ

+=

and an identical expression for az′ with z replacing y.

Chapter 39

300