39

CHAPTER OUTLINE39.1 The Principle of Galilean Relativity39.2 The Michelson-Morley Experiment39.3 Einstein’s Principle of Relativity39.4 Consequences of the Special Theory of Relativity39.5 The Lorentz Transformation Equations39.6 The Lorentz Velocity Transformation Equations

and the Relativistic Form of

39.7 Relativistic Linear Momentum

Newton’s Laws 39.8 Relativistic Energy39.9 Mass and Energy

Relativity39.10 The General Theory of

Relativity

ANSWERS TO QUESTIONS

Q39.1 The speed of light c and the speed v of their relative motion.

Q39.2 An ellipsoid. The dimension in the direction of motion wouldbe measured to be scrunched in.

Q39.3 No. The principle of relativity implies that nothing can travelfaster than the speed of light in a vacuum, which is 300 Mm/s.The electron would emit light in a conical shock wave ofCerenkov radiation.

Q39.4 The clock in orbit runs slower. No, they are not synchronized.Although they both tick at the same rate after return, a timedifference has developed between the two clocks.

Q39.5 Suppose a railroad train is moving past you. One way to measure its length is this: You mark thetracks at the cowcatcher forming the front of the moving engine at 9:00:00 AM, while your assistantmarks the tracks at the back of the caboose at the same time. Then you find the distance between themarks on the tracks with a tape measure. You and your assistant must make the markssimultaneously in your frame of reference, for otherwise the motion of the train would make itslength different from the distance between marks.

Q39.6 (a) Yours does.

(b) His does.

(c) If the velocity of relative motion is constant, both observers have equally valid views.

Q39.7 Get a Mr. Tompkins book by George Gamow for a wonderful fictional exploration of this question.Driving home in a hurry, you push on the gas pedal not to increase your speed by very much, butrather to make the blocks get shorter. Big Doppler shifts in wave frequencies make red lights lookgreen as you approach them and make car horns and car radios useless. High-speed transportationis very expensive, requiring huge fuel purchases. And it is dangerous, as a speeding car can knockdown a building. Having had breakfast at home, you return hungry for lunch, but you find youhave missed dinner. There is a five-day delay in transmission when you watch the Olympics inAustralia on live television. It takes ninety-five years for sunlight to reach Earth. We cannot see theMilky Way; the fireball of the Big Bang surrounds us at the distance of Rigel or Deneb.

Q39.8 Nothing physically unusual. An observer riding on the clock does not think that you are reallystrange, either.

433

434 Relativity

Q39.9 By a curved line. This can be seen in the middle of Speedo’s world-line in Figure 39.12, where heturns around and begins his trip home.

Q39.10 According to p u= γm , doubling the speed u will make the momentum of an object increase by the

factor 24

2 2

2 2

1 2c u

c u−−

LNM

OQP

.

Q39.11 As the object approaches the speed of light, its kinetic energy grows without limit. It would take aninfinite investment of work to accelerate the object to the speed of light.

Q39.12 There is no upper limit on the momentum of an electron. As more energy E is fed into the object

without limit, its speed approaches the speed of light and its momentum approaches Ec

.

Q39.13 Recall that when a spring of force constant k is compressed or stretched from its relaxed position a

distance x, it stores elastic potential energy U kx=12

2 . According to the special theory of relativity,

any change in the total energy of the system is equivalent to a change in the mass of the system.Therefore, the mass of a compressed or stretched spring is greater than the mass of a relaxed spring

by an amount Uc2

. The fractional change is typically unobservably small for a mechanical spring.

Q39.14 You see no change in your reflection at any speed you can attain. You cannot attain the speed oflight, for that would take an infinite amount of energy.

Q39.15 Quasar light moves at three hundred million meters per second, just like the light from a firefly at rest.

Q39.16 A photon transports energy. The relativistic equivalence of mass and energy means that is enough togive it momentum.

Q39.17 Any physical theory must agree with experimental measurements within some domain. Newtonianmechanics agrees with experiment for objects moving slowly compared to the speed of light.Relativistic mechanics agrees with experiment for objects at all speeds. Thus the two theories mustand do agree with each other for ordinary nonrelativistic objects. Both statements given in thequestion are formally correct, but the first is clumsily phrased. It seems to suggest that relativisticmechanics applies only to fast-moving objects.

Q39.18 The point of intersection moves to the right. To state the problem precisely, let us assume that eachof the two cards moves toward the other parallel to the long dimension of the picture, with velocity

of magnitude v. The point of intersection moves to the right at speed 2

2v

vtan

cotφ

φ= , where φ is the

small angle between the cards. As φ approaches zero, cotφ approaches infinity. Thus the point ofintersection can move with a speed faster than c if v is sufficiently large and φ sufficiently small. Forexample, take v = 500 m s and φ = °0 000 19. . If you are worried about holding the cards steadyenough to be sure of the angle, cut the edge of one card along a curve so that the angle willnecessarily be sufficiently small at some place along the edge.

Let us assume the spinning flashlight is at the center of a grain elevator, forming a circularscreen of radius R. The linear speed of the spot on the screen is given by v R=ω , where ω is theangular speed of rotation of the flashlight. With sufficiently large ω and R, the speed of the spotmoving on the screen can exceed c.

continued on next page

Chapter 39 435

Neither of these examples violates the principle of relativity. Both cases are describing a pointof intersection: in the first case, the intersection of two cards and in the second case, the intersectionof a light beam with a screen. A point of intersection is not made of matter so it has no mass, andhence no energy. A bug momentarily at the intersection point could yelp, take a bite out of one card,or reflect the light. None of these actions would result in communication reaching another bug sosoon as the intersection point reaches him. The second bug would have to wait for sound or light totravel across the distance between the first bug and himself, to get the message.

As a child, the author used an Erector set to build a superluminal speed generator using theintersecting-cards method. Can you get a visible dot to run across a computer screen faster thanlight? Want’a see it again?

Q39.19 In this case, both the relativistic and Galilean treatments would yield the same result: it is that theexperimentally observed speed of one car with respect to the other is the sum of the speeds of the cars.

Q39.20 The hotter object has more energy per molecule than the cooler one. The equivalence of energy andmass predicts that each molecule of the hotter object will, on average, have a larger mass than thosein the cooler object. This implies that given the same net applied force, the cooler object would havea larger acceleration than the hotter object would experience. In a controlled experiment, thedifference will likely be too small to notice.

Q39.21 Special relativity describes inertial reference frames: that is, reference frames that are notaccelerating. General relativity describes all reference frames.

Q39.22 The downstairs clock runs more slowly because it is closer to the Earth and hence in a strongergravitational field than the upstairs clock.

Q39.23 The ants notice that they have a stronger sense of being pushed outward when they venture closerto the rim of the merry-go-round. If they wish, they can call this the effect of a stronger gravitationalfield produced by some mass concentration toward the edge of the disk. An ant named Albertfigures out that the strong gravitational field makes measuring rods contract when they are near therim of the disk. He shows that this effect precisely accounts for the discrepancy.

SOLUTIONS TO PROBLEMS

Section 39.1 The Principle of Galilean Relativity

P39.1 In the rest frame,p m v m v

p m m v vi i i

f f f

= + = + = × ⋅

= + = +1 1 2 2

4

1 2

2 000 20 0 1 500 0 4 00 10

2 000 1 500

kg m s kg m s kg m s

kg kg

b gb g b gb gb g b g

. .

Since p pi f= , v f =× ⋅

+=

4 00 102 000 1 500

11 4294.

. kg m s

kg kg m s .

In the moving frame, these velocities are all reduced by +10.0 m/s.

′ = − ′ = − + =

′ = − ′ = − + = −

′ = − + =

v v v

v v v

v

i i

i i

f

1 1

2 2

20 0 10 0 10 0

0 10 0 10 0

11 429 10 0 1 429

. . .

. .

. . .

m s m s m s

m s m s m s

m s m s m s

b gb g

b gOur initial momentum is then

′ = ′ + ′ = + − = ⋅p m v m vi i i1 1 2 2 2 000 10 0 1 500 10 0 5 000 kg m s kg m s kg m sb gb g b gb g. .and our final momentum is

′ = + ′ = = ⋅p vf f2 000 1 500 3 500 1 429 5 000 kg kg kg m s kg m sb g b gb g. .

436 Relativity

P39.2 (a) v v vT B= + = 60 0. m s

(b) v v vT B= − = 20 0. m s

(c) v v vT B= + = + =2 2 2 220 40 44 7. m s

P39.3 The first observer watches some object accelerate under applied forces. Call the instantaneousvelocity of the object v1 . The second observer has constant velocity v21 relative to the first, andmeasures the object to have velocity v v v2 1 21= − .

The second observer measures an acceleration of av v

22 1= =

ddt

ddt

.

This is the same as that measured by the first observer. In this nonrelativistic case, they measure thesame forces as well. Thus, the second observer also confirms that F a∑ = m .

P39.4 The laboratory observer notes Newton’s second law to hold: F a1 1= m(where the subscript 1 implies the measurement was made in the laboratory frame of reference). Theobserver in the accelerating frame measures the acceleration of the mass as a a a2 1= − ′(where the subscript 2 implies the measurement was made in the accelerating frame of reference,and the primed acceleration term is the acceleration of the accelerated frame with respect to thelaboratory frame of reference). If Newton’s second law held for the accelerating frame, that observerwould then find valid the relation

F a2 2= m or F a1 2= m

(since F F1 2= and the mass is unchanged in each). But, instead, the accelerating frame observer willfind that F a a2 2= − ′m m which is not Newton’s second law.

Section 39.2 The Michelson-Morley Experiment

Section 39.3 Einstein’s Principle of Relativity

Section 39.4 Consequences of the Special Theory of Relativity

P39.5 L Lvcp

= −12

2 v cL

Lp= −

FHGIKJ1

2

Taking LLp

=2

where Lp = 1 00. m gives v cL

Lc cp

p= −

FHGIKJ = − =1

21

14

0 866

2

. .

P39.6 ∆∆

tt

v c

p=−1 2

1 2b gso v c

t

tp= −

FHGIKJ

LNMM

OQPP1

2 1 2∆

∆.

For ∆ ∆t tp= 2 v ct

tc cp

p= −FHGIKJ

L

NMM

O

QPP = −LNMOQP =1 2 1

14

0 866

2 1 2 1 2∆

∆. .

Chapter 39 437

P39.7 (a) γ =−

=−

=1

1

1

1 0 500

232 2v cb g a f.

The time interval between pulses as measured by the Earth observer is

∆ ∆t tp= =FHGIKJ =γ

23

60 00 924

..

s75.0

s .

Thus, the Earth observer records a pulse rate of 60 0

0 92464 9

..

. s min

smin= .

(b) At a relative speed v c= 0 990. , the relativistic factor γ increases to 7.09 and the pulse raterecorded by the Earth observer decreases to 10 6. min . That is, the life span of the

astronaut (reckoned by the duration of the total number of his heartbeats) is much longer asmeasured by an Earth clock than by a clock aboard the space vehicle.

*P39.8 (a) The 0 8. c and the 20 ly are measured in the Earth frame,

so in this frame, ∆txv c c

c= = = =

200 8

200 8

11

25 0 ly ly

ly yr yr

. .. .

(b) We see a clock on the meteoroid moving, so we do not measure proper time; that clockmeasures proper time.

∆ ∆t tp= γ : ∆∆

tt

v cp = = = − = =γ

25 025 0 1 0 8 25 0 15 0

2 2

2. . . . . yr

1 1 - yr yr 0.6 yra f

(c) Method one: We measure the 20 ly on a stick stationary in our frame, so it is proper length.The tourist measures it to be contracted to

LLp

= =−

= =γ

20 2012 0

ly

1 1 0.8

ly1.667

ly2

. .

Method two: The tourist sees the Earth approaching at 0 8. c0 8 15 12 0. . ly yr yr lyb gb g = .

Not only do distances and times differ between Earth and meteoroid reference frames, butwithin the Earth frame apparent distances differ from actual distances. As we haveinterpreted it, the 20-lightyear actual distance from the Earth to the meteoroid at the time ofdiscovery must be a calculated result, different from the distance measured directly. Becauseof the finite maximum speed of information transfer, the astronomer sees the meteoroid as itwas years previously, when it was much farther away. Call its apparent distance d. The time

required for light to reach us from the newly-visible meteoroid is the lookback time tdc

= .

The astronomer calculates that the meteoroid has approached to be 20 ly away as it movedwith constant velocity throughout the lookback time. We can work backwards to reconstructher calculation:

d ctcd

cd

d

= + = +

==

20 0 8 200 8

0 2 20100

ly ly

ly ly

..

.

Thus in terms of direct observation, the meteoroid we see covers 100 ly in only 25 years.Such an apparent superluminal velocity is actually observed for some jets of materialemanating from quasars, because they happen to be pointed nearly toward the Earth. If wecan watch events unfold on the meteoroid, we see them slowed by relativistic time dilation,but also greatly speeded up by the Doppler effect.

438 Relativity

P39.9 ∆ ∆∆

t tt

v cp

p= =−

γ1 2 2

so ∆ ∆ ∆tvc

tvc

tp = −FHGG

IKJJ

≅ −FHG

IKJ1 1 2

2

2

2

2

and ∆ ∆ ∆t tvc

tp− =FHGIKJ

2

22.

If v = =×

=1 0001 00 10

277 86

km h m

3 600 s m s

..

thenvc= × −9 26 10 7.

and ∆ ∆t tp− = × = × =− −e j e jb g4 28 10 3 600 1 54 10 1 5413 9. . . s s ns .

P39.10 For vc= 0 990. , γ = 7 09.

(a) The muon’s lifetime as measured in the Earth’s rest frame is

∆tc

=4 60. km0.990

and the lifetime measured in the muon’s rest frame is

∆∆

tt

p = =×

×

L

NMM

O

QPP =γ µ

17 09

4 60 102 18

3

..

. m

0.990 3.00 10 m s s

8e j.

(b) L Lvc

Lp

p= − FHG

IKJ = =

×=1

4 60 10649

2 3

γ. m

7.09 m

P39.11 The spaceship is measured by the Earth observer to be length-contracted to

L Lvcp

= −12

2 or L Lvcp

2 22

21= −FHGIKJ .

Also, the contracted length is related to the time required to pass overhead by:

L vt= or L v tvc

ct2 2 22

22= = a f .

Equating these two expressions gives L Lvc

ctvcp p

2 22

22

2

2− = a f

or L ctvc

Lp p2 2

2

22+ =a f .

Using the given values: Lp = 300 m and t = ×−7 50 10 7. s

this becomes 1 41 10 9 00 1052

24. .× = × m m2 2e j v

c

giving v c= 0 800. .

Chapter 39 439

P39.12 (a) The spaceship is measured by Earth observers to be of length L, where

L Lvcp

= −12

2 and L v t= ∆

v t Lvcp

∆ = −12

2 and v t Lvcp

2 2 22

21∆ = −FHGIKJ .

Solving for v, v tL

cLp p

2 22

22∆ +

FHG

IKJ= v

cL

c t L

p

p

=+2 2 2∆

.

(b) The tanks move nonrelativistically, so we have v = =300

4 00 m

75 s m s. .

(c) For the data in problem 11,

vc c

c=× × +

=+

=−

300

3 10 0 75 10 300

300

225 3000 800

8 2 6 2 2 2 2

m

m s s m

m

m

a fe j e j a f

a f.

.

in agreement with problem 11. For the data in part (b),

vc c

c=× +

=× +

= × =−300

3 10 75 300

300

2 25 10 3001 33 10 4 00

8 2 2 2 10 2 2

8 m

m s s m

m

m m s

a fe j a f a f

a fe j.

. .

in agreement with part (b).

P39.13 We find Cooper’s speed:GMm

rmv

r22

= .

Solving, vGMR h

=+

LNM

OQP

=× ×

× + ×

L

NMM

O

QPP =

−

a fe je je j

1 2 11 24

6 6

1 26 67 10 5 98 10

6 37 10 0 160 107 82

. .

. .. km s .

Then the time period of one orbit, TR hv

=+

=×

×= ×

2 2 6 53 10

7 82 105 25 10

6

33π πa f e j.

.. s .

(a) The time difference for 22 orbits is ∆ ∆ ∆t t tvc

Tp p− = − = −FHGIKJ −

LNMM

OQPP

−

γ 1 1 1 222

2

1 2

b g a f

∆ ∆t tvc

Tp− ≈ + −FHG

IKJ =

×

×

FHG

IKJ × =1

12

1 2212

7 82 103 10

22 5 25 10 39 22

2

3

8

23a f e j. . . m s

m s s sµ .

(b) For one orbit, ∆ ∆t tp− = =39 2

1 78.

. s

22 s

µµ . The press report is accurate to one digit .

P39.14 γ =−

=1

11 01

2 2v ce j. so v c= 0 140.

440 Relativity

P39.15 (a) Since your ship is identical to his, and you are at rest with respect to your own ship, itslength is 20 0. m .

(b) His ship is in motion relative to you, so you measure its length contracted to 19 0. m .

(c) We have L Lvcp

= −12

2

from whichL

Lvcp

= = = −19 0

0 950 12

2.

. m

20.0 m and v c= 0 312. .

*P39.16 In the Earth frame, Speedo’s trip lasts for a time

∆∆

tx

v= = =

20 021 05

..

ly0.950 ly yr

yr .

Speedo’s age advances only by the proper time interval

∆∆

tt

p = = − =γ21 05 6 574. . yr 1 0.95 yr2 during his trip.

Similarly for Goslo,

∆∆

tx

vvcp

= − = − =120 0

1 0 75 17 642

22. . .

ly0.750 ly yr

yr .

While Speedo has landed on Planet X and is waiting for his brother, he ages by

20 0 20 05 614

. ..

ly0.750 ly yr

ly0.950 ly yr

yr− = .

Then Goslo ends up older by 17 64 6 574 5 614 5 45. . . . yr yr yr yr− + =b g .

P39.17 (a) ∆ ∆∆

t tt

v cp

p= =

−=

−=γ

1

15 021 0

2 2b g a f.

. yr

1 0.700 yr

(b) d v t c= = = =∆a f b g a fb g b g0 700 21 0 0 700 1 00 21 0 14 7. . . . . . yr ly yr yr ly

(c) The astronauts see Earth flying out the back window at 0 700. c :

d v t cp= = = =∆e j b g a fb g b g0 700 15 0 0 700 1 00 15 0 10 5. . . . . . yr ly yr yr ly

(d) Mission control gets signals for 21.0 yr while the battery is operating, and then for 14.7 yearsafter the battery stops powering the transmitter, 14.7 ly away:

21 0 14 7 35 7. . . yr yr yr+ =

Chapter 39 441

P39.18 The orbital speed of the Earth is as described by F ma∑ = : Gm mr

m vr

S E E2

2

=

vGm

rS= =

× ⋅ ×

×= ×

−6 67 10 1 99 10

1 496 102 98 10

11 30

114

. .

..

N m kg kg

m m s

2 2e je j.

The maximum frequency received by the extraterrestrials is

f fv cv cobs source

Hz m s m s

m s m s Hz=

+−

= ×+ × ×

− × ×= ×

11

57 0 101 2 98 10 3 00 10

1 2 98 10 3 00 1057 005 66 106

4 8

4 86.

. .

. ..e j e j e je j e j

.

The minimum frequency received is

f fv cv cobs source

Hz m s m s

m s m s Hz=

−+

= ×− × ×

+ × ×= ×

11

57 0 101 2 98 10 3 00 10

1 2 98 10 3 00 1056 994 34 106

4 8

4 86.

. .

. ..e j e j e je j e j

.

The difference, which lets them figure out the speed of our planet, is

57 005 66 56 994 34 10 1 13 106 4. . .− × = ×b g Hz Hz .

P39.19 (a) Let fc be the frequency as seen by the car. Thus, f fc vc vc

=+−source

and, if f is the frequency of the reflected wave, f fc vc vc

=+−

.

Combining gives f fc vc v

=+−sourcea fa f .

(b) Using the above result, f c v f c v− = +a f a fsourcewhich gives f f c f f v f v− = + ≈source source sourceb g b g 2 .

The beat frequency is then f f ff v

cv

beat sourcesource= − = =

2 2λ

.

(c) fbeat

9 m s 10 Hz

m s

m s

m Hz= 2.00 kHz=

×

×= =

2 30 0 10 0

3 00 10

2 30 0

0 030 02 0008

a fb ge j a fb gb g

. .

.

.

.

λ = =×

×=

cfsource

m s Hz

cm3 00 1010 0 10

3 008

9

..

.

(d) vf

= beatλ

2so ∆

∆v

f= = = ≈beat

Hz m m s mi h

λ2

5 0 030 0

20 075 0 0 2

a fb g.. .

442 Relativity

P39.20 (a) When the source moves away from an observer, the observed frequency is

f fc vc v

s

sobs source=

−+FHGIKJ

1 2

where v vs = source .

When v cs

Chapter 39 443

P39.23 γ =−

=−

=1

1

1

1 0 99510 0

2 2 2v c ..

We are also given: L1 2 00= . m , and θ = °30 0. (both measured in areference frame moving relative to the rod).

Thus, L Lx1 1 1 2 00 0 867 1 73= = =cos . . .θ m ma fa fand L Ly1 1 1 2 00 0 500 1 00= = =sin . . .θ m ma fa fL x2 is a proper length, related to L x1 by L

Lx

x1

2=γ

.

Therefore, L Lx x2 110 0 17 3= =. . m

and L Ly y2 1 1 00= = . m .

(Lengths perpendicular to the motion are unchanged).

FIG. P39.23

(a) L L Lx y2 22

22

= +b g e j gives L2 17 4= . m

(b) θ 21 2

2= −tan

L

Ly

xgives θ 2 3 30= °.

*P39.24 Einstein’s reasoning about lightning striking the ends of a train shows that the moving observer seesthe event toward which she is moving, event B , as occurring first. The S-frame coordinates of theevents we may take as (x = 0 , y = 0 , z = 0, t = 0) and (x = 100 m, y = 0 , z = 0, t = 0). Then thecoordinates in ′S are given by the Lorentz transformation. Event A is at ( ′ =x 0 , ′ =y 0 , ′ =z 0 , ′ =t 0).The time of event B is

′ = −FHGIKJ = −

−FHGIKJ = − ×

FHG

IKJ = − ×

−t tvc

xc

cγ 2 2 2

71

1 0 80

0 8100 1 667

804 44 10

.

.. . m

m3 10 m s

s8a f .

The time elapsing before A occurs is 444 ns .

P39.25 (a) From the Lorentz transformation, the separations between the blue-light and red-lightevents are described by

∆ ∆ ∆′ = −x x v tγ a f 0 2 00 8 00 10 9= − × −γ . . m sve jv =

×= ×−

2 008 00 10

2 50 1098.

..

m s

m s γ =− × ×

=1

1 2 50 10 3 00 101 81

8 2 8 2. ..

m s m se j e j.

(b) Again from the Lorentz transformation, ′ = −x x vtγ a f :′ = − × × −x 1 81 3 00 2 50 10 1 00 108 9. . . . m m s se je j′ =x 4 97. m .

(c) ′ = −FHGIKJt t

vc

xγ 2 : ′ = × −×

×

L

NMMM

O

QPPP

−t 1 81 1 00 102 50 10

3 00 103 009

8

8 2. .

.

.. s

m s

m s m

e je j

a f

′ = − × −t 1 33 10 8. s

444 Relativity

Section 39.6 The Lorentz Velocity Transformation Equations

P39.26 ux = Enterprise velocity

v = Klingon velocity

From Equation 39.16

′ =−

−=

−−

=uu vu v c

c ccx

x

x10 900 0 800

1 0 900 0 8000 3572

. .. .

.a fa f .FIG. P39.26

P39.27 ′ =−

−=

− −− −

= −uu vu v c

c ccx

x

x10 750 0 750

1 0 750 0 7500 9602

. .. .

.a fa f

FIG. P39.27

*P39.28 Let frame S be the Earth frame of reference. Then v c= −0 7. .The components of the velocity of the first spacecraft are u c cx = °=0 6 50 0 386. cos .a fand u c cy = °=0 6 50 0 459. sin .a f .As measured from the ′S frame of the second spacecraft,

′ =−

−=

− −

− −= =

′ =−

=−

− −= =

uu vu v c

c c

c c c

cc

uu

u v c

c cc

xx

x

yy

x

1

0 386 0 7

1 0 386 0 7

1 0861 27

0 855

1

0 459 1 0 7

1 0 386 0 70 459 0 714

1 270 258

2 2

2

2

. .

. .

..

.

. .

. .. .

..

a fa fa f

e ja f

a fa fa f

γ

The magnitude of ′u is 0 855 0 285 0 8932 2. . .c c ca f a f+ =and its direction is at tan

.

..− = ° ′1

0 2580 855

16 8cc

x above the -axis .

Section 39.7 Relativistic Linear Momentum and the Relativistic Form of Newton’s Laws

P39.29 (a) p mu= γ ; for an electron moving at 0.010 0c,

γ =−

=−

= ≈1

1

1

1 0 010 01 000 05 1 00

2 2u cb g b g.. . .

Thus, p = × ×−1 00 9 11 10 0 010 0 3 00 1031 8. . . . kg m se jb ge jp = × ⋅−2 73 10 24. kg m s .

(b) Following the same steps as used in part (a),

we find at 0.500c, γ = 1 15. and p = × ⋅−1 58 10 22. kg m s .

(c) At 0.900c, γ = 2 29. and p = × ⋅−5 64 10 22. kg m s .

Chapter 39 445

P39.30 Using the relativistic form, pmu

u cmu=

−=

12b g

γ

we find the difference ∆p from the classical momentum, mu: ∆p mu mu mu= − = −γ γ 1b g .

(a) The difference is 1.00% when γ γ− =1 0 010 0b gmu mu. : γ = =−

10 990

1

12. u cb g

thus 1 0 9902

2− FHGIKJ =

uc

.a f , and u c= 0 141. .

(b) The difference is 10.0% when γ γ− =1 0 100b gmu mu. : γ = =−

10 900

1

12. u cb g

thus 1 0 9002

2− FHGIKJ =

uc

.a f and u c= 0 436. .

P39.31p mu

mumu mu

mu−

=−

= −γ

γ 1 : γ − =−

− ≈ + FHGIKJ − =

FHGIKJ1

1

11 1

12

1122

2 2

u c

uc

ucb g

p mumu−

=×

FHG

IKJ = ×

−12

90 03 00 10

4 50 108

214.

..

m s m s

P39.32 pmu

u c=

−12b g

becomes 12

2

2 2

2− =uc

m up

which gives: 112

2

2 2= +FHG

IKJu

mp c

or c um c

p2 2

2 2

2 1= +FHG

IKJ and u

c

m c p=

+2 2 2 1e j.

P39.33 Relativistic momentum of the system of fragments must be conserved. For total momentum to bezero after as it was before, we must have, with subscript 2 referring to the heavier fragment, andsubscript 1 to the lighter, p p2 1=

or γ γ2 2 2 1 1 128

2

2 50 10

1 0 8930 893m u m u c= =

×

−×

−.

..

kg

a fa f

or1 67 10

14 960 10

272

22

28.

.×

−= ×

−−

kg kg

e jb g

e ju

u cc .

Proceeding to solve, we find1 67 10

127

282

222

2. ×

×

FHG

IKJ = −

−

−4.960 10uc

uc

12 3 122

2.uc

= and u c2 0 285= . .

446 Relativity

Section 39.8 Relativistic Energy

P39.34 ∆E mc= −γ γ1 22b g

For an electron, mc2 0 511= . MeV

(a) ∆E mc=−

−−

FHG

IKJ =

11 0 810

11 0 250

0 5822. .

.a f a f MeV

(b) ∆E mc=−

−−

FHGG

IKJJ

=1

1 0 990

11 0 810

2 4522

. ..a f MeV

P39.35 W K Kv c

mcv c

mcf if i

∑ = − =−

−

F

HGGG

I

KJJJ

−−

F

HGG

I

KJJ

1

11

1

122

2

2

d i b g

or Wv c v c

mc

f i

∑ =−

−−

F

HGGG

I

KJJJ

1

1

1

12 22

d i b g

(a) W∑ =−

−−

F

HGG

I

KJJ × ×

−1

1 0 750

1

1 0 5001 673 10 2 998 10

2 2

27 8 2

. .. .

a f a f e je j kg m s

W∑ = × −5 37 10 11. J

(b) W∑ =−

−−

F

HGG

I

KJJ × ×

−1

1 0 995

1

1 0 5001 673 10 2 998 10

2 2

27 8 2

. .. .

a f a f e je j kg m s

W∑ = × −1 33 10 9. J

P39.36 The relativistic kinetic energy of an object of mass m and speed u is Ku c

mcr =−

−FHGG

IKJJ

1

11

2 2

2 .

For u c= 0 100. , K mc mcr = −−

FHG

IKJ

=1

1 0 010 01 0 005 0382 2

.. .

The classical equation K muc =12

2 gives K m c mcc = =12

0 100 0 005 0002 2. .a f

different by0 005 038 0 005 000

0 005 0380 751%

. ..

.−

= .

For still smaller speeds the agreement will be still better.

Chapter 39 447

P39.37 E mc mc= =γ 2 22 or γ = 2 .

Thus, uc= −FHGIKJ =1

1 32

2

γ or u

c=

32

.

The momentum is then p mu mc mc

c= =

FHGIKJ =FHGIKJγ 2

32

32

pc c

= FHGIKJ = ×

938 33 1 63 103

..

MeV

MeV.

P39.38 (a) Using the classical equation, K mv= = × = ×12

12

78 0 1 06 10 4 38 102 52 11. . . kg m s Jb ge j .

(b) Using the relativistic equation, Kv c

mc=−

−F

HGG

I

KJJ

1

11

2

2

b g.

K =− × ×

−

L

N

MMMM

O

Q

PPPP× = ×

1

1 1 06 10 2 998 101 78 0 2 998 10 4 38 10

5 8 2

8 2 11

. .. . .

e jb ge j kg m s J

When vc

448 Relativity

P39.41 We must conserve both energy and relativistic momentum of thesystem of fragments. With subscript 1 referring to the 0 868. cparticle and subscript 2 to the 0 987. c particle,

γ 1 21

1 0 8682 01=

−=

..

a f and γ 2 2

1

1 0 9876 22=

−=

..

a f.

Conservation of energy gives E E E1 2+ = totalwhich is γ γ1 1

22 2

2 2m c m c m c+ = totalor 2 01 6 22 3 34 101 2

27. . .m m+ = × − kg .

This reduces to: m m1 2273 09 1 66 10+ = × −. . kg . (1)

Since the final momentum of the system must equal zero, p p1 2=

gives γ γ1 1 1 2 2 2m u m u=

or 2 01 0 868 6 22 0 9871 2. . . .a fa f a fa fc m c m=which becomes m m1 23 52= . . (2)

FIG. P39.41

Solving (1) and (2) simultaneously, m1288 84 10= × −. kg and m2

282 51 10= × −. kg .

*P39.42 Energy conservation: 1

1 01 400

900

1 0 85 122

2

2

2

2 2−+

−=

− kg

kgc

c Mc

v c.

3 108 12

2 kg − =vc

M .

Momentum conservation: 0900 0 85

1 0 85 12 2 2+

−=

−

kg .

.

c Mv

v c

a f

1 452 12

2 kg − =vc

Mvc

.

(a) Dividing gives vc= =

1 4523 108

0 467. v c= 0 467. .

(b) Now by substitution 3 108 1 0 467 2 75 102 3 kg kg− = = ×. .M .

P39.43 E mc= γ 2 p mu= γ

E mc2 22

= γe j p mu2 2= γb g

E p c mc mu c mc mc u mcuc

uc

mc2 2 2 22 2 2 2 2 2 2 2 2 2

2

2

2

2

12 21 1− = − = −FH IK = −

FHGIKJ −FHGIKJ =−

γ γ γe j b g e j a f e j e jQ.E.D.

P39.44 (a) q V K m ce∆a f b g= = −γ 1 2

Thus, γ =−

= +1

11

2 2u c

q V

m ceb ga f∆

from which u c= 0 302. .

(b) K m c q Ve= − = = × × = ×− −γ 1 1 60 10 2 50 10 4 00 102 19 4 15b g a f e je j∆ . . . C J C J

Chapter 39 449

P39.45 (a) E mc= =γ 2 20 0. GeV with mc 2 0 511= . MeV for electrons. Thus,

γ =××

= ×20 0 10

3 91 109

4. . eV

0.511 10 eV6.

(b) γ =−

= ×1

13 91 10

2

4

u cb g. from which u c= 0 999 999 999 7.

(c) L Luc

Lp

p= − FHG

IKJ = =

××

= × =−13 00 10

7 67 10 7 672 3

2

γ.

. . m

3.91 10 m cm4

*P39.46 (a) P = =×

= ×−energy J

100 10 s W

∆t2

2 00 101513.

(b) The kinetic energy of one electron with v c= 0 999 9. is

γ − =−

−FHGG

IKJJ × × = ×

= ×

− −

−

11

1 0 999 91 9 11 10 69 7 8 20 10

5 72 10

2

2

31 2 14

12

b g e j e jmc.

. . .

.

kg 3 10 m s J

J

8

Then we require 0 01100

2 5 72 10 12.

. J J= × −Ne j

N =××

= ×−

−2 10

5 72 103 50 10

4

127 J

J.. .

P39.47 Conserving total momentum of the decaying particle system, p pbefore decay after decay= = 0

p p m u m uv e= = =µ µγ γ 207b g .Conservation of mass-energy for the system gives E E Evµ π+ = : γ µ πm c p c m cv

2 2+ =

γ 207 273mpc

mev

eb g+ = .

Substituting from the momentum equation above, γ γ207 207 273m muc

me e eb g b g+ =

or γ 1273207

1 32+FHGIKJ = =

uc

. :11

1 74+−

=u cu c

.uc= 0 270. .

Then, K m c m ceµ µγ γ= − = −1 1 2072 2b g b g e j : Kµ =

−−

F

HGG

I

KJJ

1

1 0 2701 207 0 511

2..

a fa f MeV

Kµ = 4 08. MeV .

Also, E E Ev = −π µ : E m c m c m cv e= − = −π µγ γ2 2 2273 207b g

E

E

v

v

= −−

F

HGG

I

KJJ

=

273207

1 0 2700 511

29 6

2..

.

a fa f MeV

MeV

450 Relativity

*P39.48 Let observer A hold the unprimed reference frame, with uc

134

= and uc

234

= − . Let observer B be at

rest in the primed frame with ′ = =−

−u

u vu v c11

120 1

v uc

= =134

.

(a) Then ′ =−

−=

− −− − +

=−+

uu vu v c

c cc c

c2

2

221

3 4 3 41 3 4 3 4

1 51 9 16b gb g

.

speed = ′ = = =uc

c c23 225 16

2425

0 960. .

(b) In the unprimed frame the objects, each of mass m, together have energy

γ γmc mcmc

mc2 22

2

221 0 75

3 02+ =−

=.

. .

In the primed frame the energy is mc mc

mc2

2

2

2

2

1 0 1 0 964 57

−+

−=

.. , greater by

4 573 02

1 512

2..

.mcmc

= times greater as measured by observer B .

Section 39.9 Mass and Energy

P39.49 Let a 0.3-kg flag be run up a flagpole 7 m high.

We put into it energy mgh = ≈0 3 7 20. kg 9.8 m s m J2e j .

So we put into it extra mass ∆mEc

= =×

= × −2 8 21620

3 102 10

J

m s kg

e j

for a fractional increase of2 10

1016

15×−

− kg0.3 kg

~ .

P39.50 E = ×2 86 105. J . Also, the mass-energy relation says that E mc= 2 .

Therefore, mEc

= =×

×= × −2

5

2122 86 10 3 18 10

..

J

3.00 10 m s kg

8e j.

No, a mass loss of this magnitude (out of a total of 9.00 g) could not be detected .

P39.51 ∆∆

mEc

tc

= = =× ×

×=2 2

9 7

8 2

0 800 1 00 10 3 00 3 16 10

3 00 100 842

P . . . .

..

J s yr s yr

m s kg

e jb ge je j

P39.52 ∆∆ ∆

mEc

mc T

c

Vc T

c= = = =

× ⋅° °

×2 2 2

9 3 3

8 2

1 030 1 40 10 10 4 186 10 0

3 00 10

a f a f e je je j b ga fe j

ρ kg m m J kg C C

m s

3 . .

.

∆m = ×6 71 108. kg

Chapter 39 451

P39.53 P = = = = ×dEdt

d mc

dtc

dmdt

22 263 77 10

e j. W

Thus, dmdt

=×

×= ×

3 77 10

3 00 104 19 10

26

8 29.

..

J s

m s kg s

e j

P39.54 2 1 022m ce = . MeV Eγ ≥ 1 02. MeV

Section 39.10 The General Theory of Relativity

*P39.55 (a) For the satellite F ma∑ = : GM mr

mvr

mr

rT

E2

2 22= = FHG

IKJ

π

GM T r

r

r

E2 2 3

11 24 2

2

1 3

7

4

6 67 10 5 98 10 43 080

4

2 66 10

=

=× ⋅ ×F

HGG

IKJJ

= ×

−

π

π

. .

.

N m kg s

kg

m

2

2

e jb g

(b) vr

T= =

×= ×

2 2 2 66 10

43 0803 87 10

73π π . .

m

s m s

e j

(c) The small fractional decrease in frequency received is equal in magnitude to the fractionalincrease in period of the moving oscillator due to time dilation:

fractional change in f = − − = −− × ×

−

F

HGGG

I

KJJJ

= − − −×

×

FHG

IKJ

LNMM

OQPP

FHGG

IKJJ = − ×

−

γ 11

1 3 87 10 3 101

1 112

3 87 103 10

8 34 10

3 8 2

3

8

211

b ge j.

..

(d) The orbit altitude is large compared to the radius of the Earth, so we must use

UGM m

rgE= − .

∆

∆ ∆

Um m

m

ff

U

mc

g

g

= −× ×

×+

× ×

×

= ×

= =×

×= + ×

− −

−

6 67 10 5 98 10

2 66 10

6 67 10 5 98 10

6 37 10

4 76 10

4 76 10

3 105 29 10

11 24

7

11 24

6

7

2

7

8 210

. .

.

. .

.

.

..

Nm kg

kg m

Nm kg

kg m

J kg

m s

m s

2

2

2 2

e j e j

e j

(e) − × + × = + ×− − −8 34 10 5 29 10 4 46 1011 10 10. . .

452 Relativity

Additional Problems

P39.56 (a) d vt v tearth earth astro= = γ so 2 00 101

130 06

2 2. .× =

− yr yre jc v

v c

1 1 50 102

25− = FHG

IKJ ×

−vc

vc

.e j 12 25 102

2

2 10

2− =× −v

c

v

c

.e j

1 1 2 25 102

210= + × −

vc

.e j so vc = + × = − ×− − −1 2 25 10 1

12

2 25 10101 2 10. .e j e j

vc= − × −1 1 12 10 10.

(b) Kv c

mc=−

−FHGG

IKJJ =

×−

FHG

IKJ × = ×

1

11

2 00 101 1 000 1 000 3 10 6 00 10

2 2

26

8 2 27. . yr

30 yr kg m s Jb gb ge j

(c) 6 00 10 6 00 1013¢

3 60017 1027 27 20. .

h$2.× = × FHG

IKJFHGIKJ

⋅FHGIKJFHG

IKJ = × J J kWh

k10

W sJ s3

P39.57 (a) 10 113 2 MeV = −γb gm cp so γ = 1010

v cp ≈ ′ = = =−t

tγ

1010 10

55 2 yr

10 yr s10 ~

(b) ′ = ′d ct ~ 108 km

P39.58 (a) When K Ke p= , m c m ce e p p2 21 1γ γ− = −b g e j .

In this case, m ce2 0 511= . MeV , m cp

2 938= MeV

and γ e = − =−

1 0 750 1 511 921 2

. .a f .

Substituting, γγ

pe e

p

m c

m c= +

−= +

−=1

11

0 511 1 511 9 1

9381 000 279

2

2

b g a fb g. ..

MeV

MeV

but γ ppu c

=−LNM

OQP

1

12 1 2e j

.

Therefore, u c cp p= − =−1 0 023 62γ . .

(b) When p pe p= γ γp p p e e em u m u= or γ

γp p

e e e

pu

m um

= .

Thus, γ p puc c

cc= = × −

1 511 9 0 511 0 750

9386 177 2 10

2

24

. . ..

b ge ja f MeV MeV

andu

c

u

cp p= × −

FHGIKJ

−6 177 2 10 142

.

which yields u cp = × =−6 18 10 1854. km s .

Chapter 39 453

P39.59 (a) Since Mary is in the same reference frame, ′S , as Ted, she measures the ball to have thesame speed Ted observes, namely ′ =u cx 0 800. .

(b) ∆ ′ =′=

×

×= ×t

L

up

x

1 80 107 50 10

123. .

m

0.800 3.00 10 m s s

8e j

(c) L Lvc

c

cp= − = × − = ×1 1 80 10 1

0 6001 44 10

2

212

2

212.

.. m me j a f

Since v c= 0 600. and ′ = −u cx 0 800. , the velocity Jim measures for the ball is

uu vu v c

c ccx

x

x

=′ +

+ ′=

− ++ −

= −1

0 800 0 6001 0 800 0 600

0 3852. .

. ..

a f a fa fa f .

(d) Jim measures the ball and Mary to be initially separated by 1 44 1012. × m . Mary’s motion at0.600c and the ball’s motion at 0.385c nibble into this distance from both ends. The gap closesat the rate 0 600 0 385 0 985. . .c c c+ = , so the ball and catcher meet after a time

∆t =×

×= ×

1 44 104 88 10

123. .

m

0.985 3.00 10 m s s

8e j.

*P39.60 (a) The charged battery stores energy E t= = =P 1 20 50 60 3 600. J s min s min Jb ga fb g

so its mass excess is ∆mEc

= =×

= × −2 2143 600 4 00 10

J

3 10 m s kg

8e j. .

(b)∆mm

=×

×= ×

−

−−4 00 10

101 60 10

14

312. .

kg25 kg

too small to measure.

P39.61∆mcmc

2

2

4 938 78 3 728 4100% 0 712%=

−× =

. ..

MeV MeV4 938.78 MeV

a fa f

*P39.62 The energy of the first fragment is given by E p c m c12

12 2

12 2 2 21 75 1 00= + = +e j a f a f. . MeV MeV

E1 2 02= . MeV .

For the second, E22 2 22 00 1 50= +. . MeV MeVa f a f E2 2 50= . MeV .

(a) Energy is conserved, so the unstable object had E = 4 52. MeV. Each component ofmomentum is conserved, so the original object moved with

p p pc cx y

2 2 22 22 00

= + = FHGIKJ +FHG

IKJ

1.75 MeV MeV.. Then for it

4 52 1 75 2 002 2 2 22

. . . MeV MeV MeVa f a f a f e j= + + mc mc

=3.65 MeV

2 .

(b) Now E mc= γ 2 gives 4 521

13 65

2 2. . MeV MeV=

− v c1 0 654

2

2− =vc

. , v c= 0 589. .

454 Relativity

P39.63 (a) Take the spaceship as the primed frame, moving toward the right at v c= +0 600. .

Then ′ = +u cx 0 800. , and uu vu v c

c ccx

x

x

=′ +

+ ′=

++

=1

0 800 0 6001 0 800 0 600

0 9462b g a fa f. .

. .. .

(b) LLp

=γ

: L = − =0 200 1 0 600 0 1602. . . ly lyb g a f

(c) The aliens observe the 0.160-ly distance closing because the probe nibbles into it from oneend at 0.800c and the Earth reduces it at the other end at 0.600c.

Thus, time =+

=0 160

0 6000 114

..

. ly

0.800 yr

c c.

(d) Ku c

mc=−

−FHGG

IKJJ

1

11

2 2

2 : K =−

−F

HGG

I

KJJ × ×

1

1 0 9461 4 00 10 3 00 10

2

5 8 2

.. .

a f e je j kg m s

K = ×7 50 1022. J

P39.64 In this case, the proper time is T0 (the time measured by the students on a clock at rest relative tothem). The dilated time measured by the professor is: ∆t T= γ 0

where ∆t T t= + Here T is the time she waits before sending a signal and t is the time required forthe signal to reach the students.

Thus, we have: T t T+ = γ 0 . (1)

To determine the travel time t, realize that the distance the students will have moved beyond theprofessor before the signal reaches them is: d v T t= +a f .

The time required for the signal to travel this distance is: tdc

vc

T t= = FHGIKJ +a f .

Solving for t gives: tv c T

v c=

−b gb g1 .

Substituting this into equation (1) yields: Tv c T

v cT+

−=

b gb g1 0γ

orTv c

T1 0−

= γ .

Then T Tv c

v cT

v c

v c v cT

v c

v c=

−

−=

−

+ −=

−

+0 2 2 0 01

1

1

1 1

1

1b ge j

b gb g b g

b gb g .

Chapter 39 455

P39.65 Look at the situation from the instructors’ viewpoint since they are at rest relative to the clock, andhence measure the proper time. The Earth moves with velocity v c= −0 280. relative to theinstructors while the students move with a velocity ′ = −u c0 600. relative to Earth. Using the velocityaddition equation, the velocity of the students relative to the instructors (and hence the clock) is:

uv uvu c

c c

c c cc=

+ ′+ ′

=− −

+ − −= −

1

0 280 0 600

1 0 280 0 6000 7532 2

. .

. ..

a f a fa fa f (students relative to clock).

(a) With a proper time interval of ∆tp = 50 0. min , the time interval measured by the students is:

∆ ∆t tp= γ with γ =−

=1

1 0 7531 52

2 2..

c ca f.

Thus, the students measure the exam to last T = =1 52 50 0 76 0. . . min minutesa f .

(b) The duration of the exam as measured by observers on Earth is:

∆ ∆t tp= γ with γ =−

1

1 0 280 2 2. c ca f so T = =1 04 50 0 52 1. . . min minutesa f .

P39.66 The energy which arrives in one year is

E t= = × × = ×P ∆ 1 79 10 3 16 10 5 66 1017 7 24. . . J s s Je je j .

Thus, mEc

= =×

×= ×2

24

275 66 10 6 28 10

..

J

3.00 10 m s kg

8e j.

P39.67 The observer measures the proper length of the tunnel, 50.0 m, but measures the train contracted tolength

L Lvcp

= − = − =1 100 1 0 950 31 22

22 m m. .a f

shorter than the tunnel by 50 0 31 2 18 8. . .− = m so it is completely within the tunnel. .

P39.68 If the energy required to remove a mass m from the surface is equal to its rest energy mc 2 ,

thenGM m

Rmcs

g= 2

and RGM

cgs= =

× ⋅ ×

×

−

2

11 30

8 2

6 67 10 1 99 10

3 00 10

. .

.

N m kg kg

m s

2 2e je je j

Rg = × =1 47 10 1 473. . m km .

456 Relativity

P39.69 (a) At any speed, the momentum of the particle is given by

p mumu

u c= =

−γ

1 2b g.

Since F qEdpdt

= = : qEddt

muuc

= −FHGIKJ

LNMM

OQPP

−

12

2

1 2

qE muc

dudt

muuc

uc

dudt

= −FHGIKJ + −

FHGIKJFHGIKJ

− −

112

122

2

1 2 2

2

3 2

2 .

SoqEm

dudt

u c u c

u c=

− +

−

L

NMMM

O

QPPP

1

1

2 2 2 2

2 2 3 2e j

and adudt

qEm

uc

= = −FHGIKJ1

2

2

3 2

.

(b) For u small compared to c, the relativistic expression reduces to the classical aqEm

= . As u

approaches c, the acceleration approaches zero, so that the object can never reach the speedof light.

(c)du

u c

qEm

dtu

t

t

1 2 23 2

0 0−=z z

=e ju

qEct

m c q E t=

+2 2 2 2 2

x udt qEctdt

m c q E t

t t

= =+

z z0

2 2 2 2 20

xc

qEm c q E t mc= + −FH IK2 2 2 2 2

P39.70 (a) An observer at rest relative to the mirror sees the light travel a distance D d x= −2 , wherex vtS= is the distance the ship moves toward the mirror in time tS . Since this observeragrees that the speed of light is c, the time for it to travel distance D is

tDc

d vtcS

S= =−2

td

c vS=

+2

.

(b) The observer in the rocket measures a length-contracted initial distance to the mirror of

L dvc

= −12

2

and the mirror moving toward the ship at speed v. Thus, he measures the distance the light

travels as D L y= −2b g where y vt=2

is the distance the mirror moves toward the ship before

the light reflects from it. This observer also measures the speed of light to be c, so the timefor it to travel distance D is:

tDc c

dvc

vt= = − −

LNMM

OQPP

21

2

2

2 so c v tdc

c v c v+ = + −a f a fa f2 or t dc

c vc v

=−+

2.

Chapter 39 457

*P39.71 Take the two colliding protons as the system

E K mc12= + E mc2

2=

E p c m c12

12 2 2 4= + p2 0= .

In the final state, E K Mcf f= +2 : E p c M cf f

2 2 2 2 4= + .

By energy conservation, E E E f1 2+ = , so

E E E E E f12

1 2 22 22+ + =

p c m c K mc mc m c

p c M cf

12 2 2 4 2 2 2 4

2 2 2 4

2+ + + +

= +

e j

By conservation of momentum, p p f1 = .

Then M c Kmc m cKm c

mcm c2 4 2 2 4

2 4

22 42 4

42

4= + = +

Mc mcKmc

2 222 1 2

= + .

By contrast, for colliding beams we have

In the original state, E K mc12= +

E K mc22= + .

In the final state, E Mcf =2

E E E f1 2+ = : K mc K mc Mc+ + + =2 2 2

Mc mcKmc

2 222 1 2

= +FHGIKJ .

FIG. P39.71

*P39.72 Conservation of momentum γ mu :

mu

u c

m u

u c

Mv

v c

mu

u c

f

f1 3 1 1

2

3 12 2 2 2 2 2 2 2−+

−

−=

−=

−

a f.

Conservation of energy γ mc 2 :

mc

u c

mc

u c

Mc

v c

mc

u cf

2

2 2

2

2 2

2

2 2

2

2 21 3 1 1

4

3 1−+

−=

−=

−.

To start solving we can divide: vu u

f = =24 2

. Then

M

u c

m

u c

M

u c

Mm u c

u c

1 4

4

3 1 1 2 4

2 4

3 1

2 2 2 2 2 2

2 2

2 2

−=

−=

−

=−

−

b g

Note that when v c

458 Relativity

P39.73 (a) L L Lx y02

02

02= + and L L Lx y

2 2 2= + .

The motion is in the x direction: L L Ly y= =0 0 0sinθ

L Lvc

Lvcx x

= − FHGIKJ = −

FHGIKJ0

2

0 0

2

1 1cosθb g .

Thus, L Lvc

L Lvc

202 2

0

2

02 2

0 02

22

01 1= −FHGIKJ

LNMM

OQPP + = −

FHGIKJ

LNMM

OQPPcos sin cosθ θ θ

or L Lvc

= − FHGIKJ

LNMM

OQPP0

22

0

1 2

1 cos θ .

(b) tan tanθ γ θ= =−

=L

L

L

L v c

y

x

y

x

0

02 01 b g

P39.74 (b) Consider a hermit who lives on an asteroid halfway between the Sun and Tau Ceti,stationary with respect to both. Just as our spaceship is passing him, he also sees the blastwaves from both explosions. Judging both stars to be stationary, this observer concludes that

the two stars blew up simultaneously .

(a) We in the spaceship moving past the hermit do not calculate the explosions to besimultaneous. We measure the distance we have traveled from the Sun as

L Lvcp

= − FHGIKJ = − =1 6 00 1 0 800 3 60

22. . . ly lyb g a f .

We see the Sun flying away from us at 0.800c while the light from the Sun approaches at1.00c. Thus, the gap between the Sun and its blast wave has opened at 1.80c, and the timewe calculate to have elapsed since the Sun exploded is

3 602 00

..

ly1.80

yrc= .

We see Tau Ceti as moving toward us at 0.800c, while its light approaches at 1.00c, only0.200c faster. We measure the gap between that star and its blast wave as 3.60 ly andgrowing at 0.200c. We calculate that it must have been opening for

3 6018 0

..

ly0.200

yrc=

and conclude that Tau Ceti exploded 16.0 years before the Sun .

Chapter 39 459

P39.75 Since the total momentum is zero before decay, it is necessary that after the decay

p pE

c cnucleus photon keV

= = =γ14 0.

.

Also, for the recoiling nucleus, E p c mc2 2 2 22

= + e j with mc 2 98 60 10 53 8= × =−. . J GeV .

Thus, mc K mc22 2 2 214 0+ = +e j a f e j. keV or 1 14 0 12

2

2

2

+FHGIKJ =FHG

IKJ +

Kmc mc

. keV.

So 1 114 0

112

14 02 2

2

2

2

+ = + FHGIKJ ≈ +

FHG

IKJ

Kmc mc mc

. . keV keV (Binomial Theorem)

and Kmc

≈= =×

×= × −

14 0

2

14 0

2 53 8 101 82 10

2

2

2

93. .

..

keV 10 eV

eV eV

3a f e je j

.

P39.76 Take m = 1 00. kg .

The classical kinetic energy is K mu mcuc

ucc

= = FHGIKJ = ×

FHGIKJ

12

12

4 50 102 22

162

. Je j

and the actual kinetic energy is Ku c

mcu c

r =−

−F

HGG

I

KJJ = × −

−F

HGG

I

KJJ

1

11 9 00 10

1

11

2

2 16

2b ge j

b g. J .

uc

K Kc r J J

0.000 0.000 0.0000.100 0.045 0.04530.200 0.180 0.1860.300 0.405 0.4350.400 0.720 0.8200.500 1.13 1.390.600 1.62 2.250.700 2.21 3.600.800 2.88 6.000.900 3.65 11.60.990 54.8

a f a f

× ×× ×× ×× ×× ×× ×× ×× ×× ×

4.41× ×

10 1010 1010 1010 10

10 1010 1010 1010 1010 1010 10

16 16

16 16

16 16

16 16

16 16

16 16

16 16

16 16

16 16

16 16

FIG. P39.76

K Kc r= 0 990. , when 12

0 9901

11

2

2

uc u c

FHGIKJ = −

−L

NMMM

O

QPPP

.b g

, yielding u c= 0 115. .

Similarly, K Kc r= 0 950. when u c= 0 257.

and K Kc r= 0 500. when u c= 0 786. .

460 Relativity

ANSWERS TO EVEN PROBLEMS

P39.2 (a) 60 0. m s ; (b) 20 0. m s ; (c) 44 7. m s P39.44 (a) 0 302. c ; (b) 4 00. fJ

P39.46 (a) 20.0 TW; (b) 3 50 107. × electronsP39.4 see the solution

P39.6 0 866. c P39.48 (a) 0.960c; (b) 1.51 times greater asmeasured by B.

P39.8 (a) 25.0 yr; (b) 15.0 yr; (c) 12.0 lyP39.50 3 18 10 12. × − kg , not detectable

P39.10 (a) 2 18. sµ ; (b) The moon sees the planetsurface moving 649 m up toward it. P39.52 6 71 108. × kg

P39.12 (a) cL

c t L

p

p2 2 2∆ +

; (b) 4 00. m s ; P39.54 1 02. MeV

P39.56 (a) vc= − × −1 1 12 10 10. ; (b) 6 00 1027. × J;(c) see the solution

(c) $2.17 1020×P39.14 v c= 0 140.

P39.58 (a) 0 023 6. c ; (b) 6 18 10 4. × − cP39.16 5.45 yr, Goslo is older

P39.60 (a) 4 00 10 14. × − kg ; (b) 1 60 10 12. × −P39.18 11 3. kHz

P39.20 (a) see the solution; (b) 0 050 4. c P39.62 (a) 3 65

2. MeV

c; (b) v c= 0 589.

P39.22 (a) 0 943. c ; (b) 2.55 kmP39.64 see the solution

P39.24 B occurred 444 ns before AP39.66 6 28 107. × kg

P39.26 0 357. cP39.68 1 47. km

P39.28 0 893. c at 16 8. ° ′ above the -axisx

P39.70 (a) 2d

c v+; (b)

2dc

c vc v−+P39.30 (a) 0 141. c ; (b) 0 436. c

P39.32 see the solutionP39.72 M

m u cu c

=−

−23

41

2 2

2 2P39.34 (a) 0 582. MeV; (b) 2 45. MeV

P39.36 see the solution P39.74 (a) Tau Ceti exploded 16.0 yr before theSun; (b) they exploded simultaneously

P39.38 (a) 438 GJ ; (b) 438 GJP39.76 see the solution, 0 115. c , 0 257. c , 0 786. c

P39.40 18 4. g cm3

P39.42 (a) 0.467c; (b) 2 75 103. × kg