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Relativity
CHAPTER OUTLINE 39.1 The Principle of Galilean Relativity 39.2 The Michelson-Morley Experiment 39.3 Einstein’s Principle of Relativity 39.4 Consequences of the Special Theory of Relativity 39.5 The Lorentz Transformation Equations 39.6 The Lorentz Velocity Transformation Equations 39.7 Relativistic Linear Momentum 39.8 Relativistic Energy 39.9 Mass and Energy 39.10 The General Theory of Relativity
ANSWERS TO QUESTIONS
Q39.1 No. The principle of relativity implies that nothing can travel faster than the speed of light in a vacuum, which is 300 Mm/s. The electron would emit light in a conical shock wave of Cerenkov radiation.
*Q39.2 Answer (c). The dimension parallel to the direction of motion is reduced by the factor and the other dimensions are unchanged.
*Q39.3 Answer (c). An oblate spheroid. The dimension in the direction of motion would be measured to be scrunched in.
*Q39.4 Answer (e). The relativistic time dilation effect is symmetric between the observers.
Q39.5 Suppose a railroad train is moving past you. One way to measure its length is this: You mark the tracks at the cowcatcher forming the front of the moving engine at 9:00:00 AM, while your assistant marks the tracks at the back of the caboose at the same time. Then you find the distance between the marks on the tracks with a tape measure. You and your assistant must make the marks simultaneously in your frame of reference, for otherwise the motion of the train would make its length different from the distance between marks.
*Q39.6 (i) Answer (c). The Earth observer measures the clock in orbit to run slower.
(ii) Answer (b). They are not synchronized. They both tick at the same rate after return, but a time difference has developed between the two clocks.
Q39.7 (a) Yours does.
(b) His does. (c) If the velocity of relative motion is constant, both observers have equally valid views.
405
Note : In chapters 39, 40, and 41 we use u to represent the speed of a particle with mass, reserving v for the speeds associated with reference frames, wave functions, and photons.
406 Chapter 39
Q39.8 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, but rather to make the blocks get shorter. Big Doppler shifts in wave frequencies make red lights look green as you approach them and make car horns and car radios useless. High-speed transportation is very expensive, requiring huge fuel purchases. And it is dangerous, as a speeding car can knock down a building. Having had breakfast at home, you return hungry for lunch, but you find you have missed dinner. There is a five-day delay in transmission when you watch the Olympics in Australia on live television. It takes ninety-five years for sunlight to reach Earth. We cannot see the Milky Way; the fireball of the Big Bang surrounds us at the distance of Rigel or Deneb.
Q39.9 By a curved line. This can be seen in the middle of Speedo’s world-line in Figure 39.11, where he turns around and begins his trip home.
Q39.10 A microwave pulse is reflected from a moving object. The waves that are reflected back are Doppler shifted in frequency according to the speed of the target. The receiver in the radar gun detects the reflected wave and compares its frequency to that of the emitted pulse. Using the frequency shift, the speed can be calculated to high precision. Be forewarned: this technique works if you are either traveling toward or away from your local law enforcement agent!
Q39.11 This system would be seen as a star moving in an elliptical path. Just like the light from a star in a binary star system, the spectrum of light from the star would undergo a series of Doppler shifts depending on the star’s speed and direction of motion relative to the observer. The repetition rate of the Doppler shift pattern is the period of the orbit. Information about the orbit size can be calculated from the size of the Doppler shifts.
Q39.12 According to
p =γ m u , doubling the speed u will make the momentum of an object increase by
the factor 2 4
2 2 2 2
1 2 c u c u
*Q39.13 From E^2 p^2 c^2 m^2 c^4 ( mc^2 K )^2 we consider pc = 2 Kmc^2 + K^2. For a photon, pc E. For particles with mass, the greater the mass the greater the momentum if K is always 1 MeV. The ranking is d > b > c > a.
Q39.14 As the object approaches the speed of light, its kinetic energy grows without limit. It would take an infi nite investment of work to accelerate the object to the speed of light.
*Q39.15 (i) Answer (a).
(ii) (c) and (iii) (d). There is no upper limit on the momentum or energy 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 E c
*Q39.16 Answer (b). Quasar light moves at three hundred million meters per second, just like the light from a firefly at rest.
Q39.17 Any physical theory must agree with experimental measurements within some domain. Newtonian mechanics 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 must and do agree with each other for ordinary nonrelativistic objects. Both statements given in the question are formally correct, but the first is clumsily phrased. It seems to suggest that relativistic mechanics applies only to fast-moving objects.
408 Chapter 39
P39.3 In the rest frame,
p (^) i = m 1 (^) v 1 (^) i + m 2 v 2 (^) i = ( 2 000 kg)( 20 0. m s) + ( 1 500kg) 00 4 00 10 2 000
4
1 2
m s kg m s kg
( ) = × ⋅ = (^) ( + ) = +
p (^) f m m v (^) f ( 1 1 500 kg) v (^) f
Since pi = pf , v (^) f =
× ⋅
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
1 1 2
i i 20 0^ 10 0^ 10 0 i
. m s. m s. m s vv v v
2 0 10 0^ 10 0
i f
− ′ = − +( ) = − ′ =
m s. m s. m s
. mm s − +( 10 0. m s ) =1 429. m s Our initial momentum is then p ′ = (^) i m 1 (^) v 1 ′ + (^) i m 2 v ′ = ( 2 (^) i 2 000 kg)( 10 0. m s) + (1 500 kgg ) −( 10 0. m s ) = 5 000kg m s⋅ and our fi nal momentum has the same value: p ′ = (^) f ( 2 000 kg +1 500 kg ) v ′ = ( f 3 500 kg )(1 429. m s)) = 5 000 kg m s⋅
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
P39.4 L L
p c
2 2
v v = −
c
L
L (^) p
2
Taking L
L (^) p
2 where L (^) p = 1 00. m gives v = −
⎠⎟^
c = − =
L
L
p c c p
2 .
P39.5 ∆
t
t c
= p ⎡⎣ 1 − ( (^) )^2 ⎤⎦ 1 2 v
so v = − ⎛ ⎝⎜^
c
t t 1 p
∆^2 1 2
For ∆ t = 2 ∆ t (^) p v = −
⎣⎢^
c
t t
p c p
2 1 2^ 1 2 ∆ ∆ == 0 866. c
P39.6 (a) γ = − ( (^) )
v c^2.^23 The time interval between pulses as measured by the Earth observer is
∆ t = γ∆ t (^) p = ⎛⎝ ⎞⎠ =
s
s
Thus, the Earth observer records a pulse rate of
s min s = min.
(b) At a relative speed v = 0 990. c , the relativistic factor g increases to 7.09 and the pulse rate recorded 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 as measured by an Earth clock than by a clock aboard the space vehicle.
Relativity 409
P39.7 ∆ ∆
t t
t c p = = p −
γ 1 v^2
so ∆ t ∆ ∆ c
t c p =^ − t
⎟ ≅^ −
⎝⎜^
2 2
2 2
v v
and ∆ t ∆ t ∆ c
− (^) p = t
⎝⎜^
v^2 22 If v = 1 000 = 1 00^ ×^10 =277 8
6 km h m 3 600 s
. (^). m s
then v c
= 9 26. × 10 − 7
and ( ∆ t −∆ t p ) = ( 4 28. × 10 − 13 )( 3 600 s ) = 1 54. × 10 −^9 s = 1 ..54 ns
P39.8 For v c
= 0 990. , γ = 7 09.
(a) The muon’s lifetime as measured in the Earth’s rest frame is
∆ t c
= 4 60.^ km
and the lifetime measured in the muon’s rest frame is
∆ t (^) p = ∆ t = × ( × )
γ ⎣
. m 0.990 3.00 10 8 m s
= 2 18. μs
(b) L L c
L
p = − ⎛ p ⎝
⎠ =^ =^
1^ v^2 4 60^ ×^10 3 = 649 γ
. m 7.
m
P39.9 The spaceship is measured by the Earth observer to be length-contracted to
L L p c
2 2
v (^) or L L p c 2 2 2 = 1 − 2
⎝⎜^
v
Also, the contracted length is related to the time required to pass overhead by:
L = vt or L t c
(^2 2 2) ct 2 2
= v = v ( )^2
Equating these two expressions gives L L c
ct p p c 2 2 2 2
2 2
v v
or L ct c
(^2) p (^)^2 Lp 2 2
⎡ + ( )^2
v
Using the given values: L (^) p = 300 m and t = 7 50. × 10 − 7 s
this becomes 1 41 10 5 9 00 10
2 2 (.^ × m^2 ) v =^. ×^4 m^2 c giving v = 0 800. c
Relativity 411
P39.13 (a) Since your ship is identical to his, and you are at rest with respect to your own ship, its
length 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 L p c
2 2
v
from which
L
L (^) p c
2 2
m 20.0 m
v and v = 0 312. c
P39.14 In the Earth frame, Speedo’s trip lasts for a time
∆
t x = = = v
ly 0.950 ly yr yr
Speedo’s age advances only by the proper time interval
∆
t t p =^ γ =^ 21 05.^ yr^1 −^ 0.95^ =6 574. yr (^2) during his trip
Similarly for Goslo,
∆
t x p (^) c
v
v 1
2 2
. ly. 2. 0.750 ly yr 4 4 yr
While Speedo has landed on Planet X and is waiting for his brother, he ages by 20 0 20 0 5 614
ly 0.750 ly yr
ly 0.950 ly yr − = yrr
Then Goslo ends up older by 17 64. yr − ( 6 574. yr +5 614. yr ) = 5 45. yr.
P39.15 The orbital speed of the Earth is as described by Σ F = ma : Gm m r
m r
S E E 2
2
v
v = = Gm ( ×^ − ⋅ ) ( × ) r
S 6 67^10 1 99^10
. 11.^30
N m 2 kg 2 kg 4496 10
11 2 98^104
×
= ×
m
. m s
The maximum frequency received by the extraterrestrials is
f f c obs source (^) c = + Hz −
= (^) ( × )
1 +^ ×
v 6 1 2 98^10 v
4 8
m s m s m s
( ) ( × ) − (^) ( × ) ×
.. m s Hz
( ) = 57 005 66. × 10 6
The minimum frequency received is
f f c obs source (^) c = Hz
= (^) ( × )
1 −^ ×
v 6 1 2 98^10 v
4 8
m s m s m s
( ) ( × )
.. m s Hz
( ) = 56 994 34. × 10 6
The difference, which lets them figure out the speed of our planet, is
(57 005 66. −56 994 34. ) × 10 6 Hz = 1 13. × 104 Hz
412 Chapter 39
P39.16 (a) Let f (^) c be the frequency as seen by the car. Thus, f f c c (^) c
source −
v v
and, if f is the frequency of the reflected wave, (^) f f c c c
v v
Combining gives (^) f f c c
= (^ + )
source( − )
v v
(b) Using the above result, f ( c − v ) = f source( c + v )
which gives (^) ( f − f (^) source ) c = (^) ( f + f (^) source ) v ≈ 2 f source v
The beat frequency is then (^) f f f f beat source c = − = 2 source v^ =^2 v λ
(c) f beat m s 109 Hz m s
( )( ) ( × ) ×
( )( ) ( )
m s m
Hz = 2.00 kHz
λ = = × ×
c = f source
m s Hz
3 00 (^10) cm 10 0 10
8 9
(d) v = f beat^ λ 2
so ∆
v = = ( )( ) = ≈ f beat Hz m m s λ 2
.. mi h
P39.17 (a) When the source moves away from an observer, the observed frequency is
f f c c
s s obs =^ source
v v
1 2 where (^) v (^) s = v source
When vs << c , the binomial expansion gives
c c c c
s s
− s
⎠⎟^
⎣⎢^
⎦⎥^
v + ⎛⎝ ⎞⎠ v
v v 1 2 (^) 1 2 1 ⎡⎡ 1 s ⎣⎢^
⎦⎥^
−1 2 1 2
v (^) s v (^) 1 vs c c c
s
So, (^) f f c
s obs ≈^ source −
v
The observed wavelength is found from c = λ (^) obs f (^) obs =λ f source :
λ λ λ λ obs source obs
source source
( − )
f f
f f 1 vs c 1 −−
= − = −
v
v
v v
s s
c
c
c c ∆λ λ (^) obs λ λ λ
s^1 s
Since 1 −^ vs ≈ 1 c
, ∆λ λ
≈ v source c
(b) v source nm 397 nm
⎝⎜^
c^ ∆λ c ⎛⎝ ⎞⎠ = λ
20 0. (^) 0 050. 44 c
414 Chapter 39
P39.21 Einstein’s reasoning about lightning striking the ends of a train shows that the moving observer sees the event toward which she is moving, event B, as occurring first. The S -frame coordinates of the events we may take as ( x = 0 , y = 0, z = 0 , t = 0) and ( x = 100 m, y = 0, z = 0, t = 0 ). Then the coordinates 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
′ = ⎛⎝ − ⎞⎠ = −
t t ⎛⎝ − ( )⎞⎠ =
c x c c γ v 2 2 2
m .. 667.
×
⎝⎜^
= − × −
m 3 10 m s 8 s The time elapsing before A occurs is 444 ns.
P39.22 (a) From the Lorentz transformation, the separations between the blue-light and red-light events are described by ∆ x ′ = γ ( ∆ x − v ∆ t ) 0 = γ ⎡⎣2 00. m − v (8 00. × 10 −^9 s)⎤⎦
v = ×
− =^ ×
9 2 50^108
m s m s γ = − (^) ( × ) ( × ) =
. 8 2.^8
m s m s
(b) Again from the Lorentz transformation,
x ′ = γ ( x − vt ):
x ′ = 1 81 3 00. ⎡⎣. m − (^) ( 2 50. × 10 8 m s (^) ) ( 1 00. × 10 −^9 s) ⎤⎦⎦
x ′ = 4 97. m
(c)^ t^ ′ =^ γ^ ⎛⎝ t^ − c x ⎞⎠
v 2 :^ ′ =^ ×^ −^
( × ) ×
t 1 81 1 00 10 − 2 50^10 3 00 10
9 8
.. (^8)
s
m s (( m s)
2 3 00.^ m
t ′ = − 1 33. × 10 −^8 s
Section 39.6 The Lorentz Velocity Transformation Equations
P39.23 ′ = −
= −^ −
u u u c
c c x x x
v 1 v
= −. c
speed = 0.960 c
P39.24 u (^) x = Enterprise velocity
v = Klingon velocity From Equation 39.
′ = − −
u u u c
c c x x x
v 1 v
= 0 357. c
FIG. P39.
FIG. P39.
Relativity 415
Section 39.7 Relativistic Linear Momentum
P39.25 (a) p =γ mu ; for an electron moving at 0.010 0 c ,
γ = − ( )
− ( )
2 2 1 000 05^ 1 00
u c.
Thus, p = 1 00 9 11. (^) (. × 10 − 31 kg)( 0 010 0. (^) ) ( 3 00. × 108 m s)
p = 2 73. × 10 −^24 kg m s⋅
(b) Following the same steps as used in part (a),
we fi nd at 0.500 c , γ = 1 15. and p = 1 58. × 10 −^22 kg m s⋅
(c) At 0.900 c , γ = 2 29. and p = 5 64. × 10 −^22 kg m s⋅
P39.26 Using the relativistic form, (^) p mu u c
= mu − ( )
γ
we fi nd the difference ∆ from the classical momentum, mu : ∆ p = γ mu − mu = ( γ − 1 ) mu
(a) The difference is 1.00% when( γ − 1 ) mu =0 010 0. γ mu : γ = = − ( )
. (^1) u c^2
thus 1 0 990
2 − ⎛^2 ⎝
⎠ = (^ )
u c
. , and u = 0 141. c
(b) The difference is 10.0% when( γ − 1 ) mu =0 100. γ mu : γ = = − ( )
. (^1) u c^2
thus 1 0 900
2 − ⎛^2 ⎝
⎠ = (^ )
u c
. and u = 0 436. c
P39.27 p^ mu mu
mu mu mu
− (^) = γ − (^) = γ −1: γ − = − ( )
2 2
u c
u c
u c
p mu mu
×
⎝⎜^
⎠⎟^
= × −
8 4 50^10
.^2
m s m s
114
Relativity 417
P39.29 Relativistic momentum of the system of fragments must be conserved. For total momentum to be zero after as it was before, we must have, with subscript 2 referring to the heavier fragment, and subscript 1 to the lighter, p (^) 2 = p 1
or γ 2 2 2 γ 1 1 1
28 2
m u = m u = 0 893
×
×
kg
( cc )
or
27 2 2
2
(.^ × ). 28 − ( )
= (^) ( × )
− kg u − kg u c
c
Proceeding to solve, we find
27 28 2
2 2 2 2
. ×
×
⎝⎜^
⎠⎟^
− 4.960 10 −
u c
u c
2
. (^2) u c = and u (^) 2 = 0 285. c
Section 39.8 Relativistic Energy
*P39.30 (a) K = E − E (^) R = 5 ER
E = 6 E (^) R = 6 9 11( × 10 − (^31) ) ( 3 00 × (^10 8) ) = 4 92 × 1 2
. kg. m s. 00 −^13 J = 3 07. MeV
(b) E = γ mc^2 =γ E (^) R
Thus γ = = = −
E
E (^) R u c
which yields u = 0 986. c
P39.31 Σ W K K
c
mc c
f i f i
− ( )
2 2 v v
mc^2
or Σ W c c
mc f i
− ( )
2 v v
(a) Σ W =
⎟ ×^ −
2 2 1 673^1027
(.^ kg^ ) ( 2 998^ ×^10 8 m s)
2 .
Σ W = 5 37. × 10 − 11 J
(b) Σ W =
⎟ ×^ −
2 2 1 673^1027
(.^ kg^ ) ( 2 998.^ ×^10 8 m s)^2
Σ W = 1 33. × 10 − 9 J
418 Chapter 39
P39.32 The relativistic kinetic energy of an object of mass m and speed u is K u c r =^ mc −
For u = 0 100. c , K (^) r = mc mc −
⎠⎟^
1 2 0 005 038^2
The classical equation K (^) c = 1 mu 2
(^2) gives K m c mc
c =^ (^ )^ =
0 100. 2 0 005 000.^2
different by 0 005 038^ 0 005 000 0 005 038
For still smaller speeds the agreement will be still better.
P39.33 E = γ mc^2 = 2 mc 2 or γ = 2
Thus, u c
⎠⎟^
1 1 =^3
2 γ or u = c^3 2
The momentum is then p mu m c^ mc c
⎝⎜^
γ 2 3 2
2
p c c
= ⎛⎝^ 938 3.^ MeV^ ⎞⎠ 3 = 1 63. × 10 3 MeV
*P39.34 (a) Using the classical equation, K = mu = ( ) (^) ( × )
= ×
2 5 2
11
kg m s
J
(b) Using the relativistic equation, K u c
= mc − ( )
K =
− (^) ( × × )
⎥ (
. 5.^8
. kg)) (^) ( 2 998 × (^10 8) ) = 4 38 × 10 (^2 ) . m s. J
(c) When u c
<< 1, the binomial series expansion gives 1 1 1 2
2 1 2^2 − ⎛⎝⎜ ⎞⎠⎟
⎥ ≈^ +^
⎝⎜^
− u c
u c
Thus, 1 1 1 2
2 1 2^2 − ⎛⎝⎜ ⎞⎠⎟
⎥ −^ ≈^
⎝⎜^
− u c
u c
and the relativistic expression for kinetic energy becomes K u c
≈ 1 ⎛⎝⎜ ⎞⎠⎟ mc = mu 2
2 (^2 ).
That is, in the limit of speeds much smaller than the speed of light, the relativistic and classical expressions yield the same results. In this situation the two kinetic energy values are experimentally indistinguishable. The fastest-moving macroscopic objects launched by human beings move sufficiently slowly compared to light that relativistic corrections to their energy are negligible.
420 Chapter 39
*P39.39 From K mc u c
= ( − ) = mc −
γ (^1) ⎟
(^22 212) we have K mc (^) u c
K mc (^2 2 2) mc
2 (^1 )
= +^.
2 2
2 4 2 2
2 2
2 2 2 2
( + )
= − ( ) ( + )
=
u c
m c K mc
u c
mc K mc
u c 11
2 2
2 1 2 −
mc K mc
(a) Electron: u = c − ⎛⎝⎜ ⎞⎠⎟ c
⎠⎟^
.^2 1 2
(b) Proton: u = c − ⎛⎝⎜ ⎞⎠⎟ c
⎠⎟^
2 1 2 .
(c) 0 979. c − 0 065 2. c = 0 914. c = 2 74. × 10 8 m s
In this case the electron is moving relativistically, but the classical expression^1 2
mv^2 is accurate to two digits for the proton.
(d) Electron: u = c −
2 1 2 . .
. 77 c
Proton: u = c − c
2 1 2 .
Excess speed (^) = 0 999 999 97. c − 0 948. c = 0 052 3. c = 1 57. × 10 7 m s
As the kinetic energies of both particles become large, the difference in their speeds approaches zero. By contrast, classically the speed difference would become large without any fi nite limit.
P39.40 (a) E = γ mc^2 =20 0. GeV with mc^2 = 0 511. MeV for electrons.
Thus, γ =
×
×
= ×
eV 0.511 10 6 eV
(b) γ = −
1 = ×
2 3 91^104
u c
from which u = 0 999 999 999 7. c
(c) L L u c
L
p = − ⎛ p ⎝⎜^
⎠⎟^ =^ =^
×
×
1 3 00^10 =7 67
(^2 ) γ
. m. 3.91 10 4
×× 10 − 2 m = 7 67. cm
Relativity 421
P39.41 Conserving total momentum of the decaying particle system, p (^) before decay = p after decay= 0
p (^) v = p (^) μ = γ m u μ = γ ( 207 m (^) e ) u
Conservation of mass-energy for the system gives E (^) μ+ E (^) v = E π: γ m c μ^2 + p cv = m c π^2
γ 207 m p 273 c ( (^) e ) +^ v = me
Substituting from the momentum equation above, γ 207 m γ 207 m u 273 c ( (^) e ) +^ ( (^) e ) = me
or γ 1 273 207
⎛⎝ +^ u ⎞⎠ = =1 32 c
u c = u c
. u c
Then, K (^) μ= (γ − 1 ) m c μ^2 = (γ −1 207) (^) ( m ce^2 ): K μ = − ( (^) )
( )
. MeV
K μ = 4 08. MeV
Also, E (^) v = E (^) π − E μ: E (^) v = m c π^2 − γ m c μ^2 = (^) ( 273 − 207 γ) m ce^2
E
E
v
v
⎟ (^ )
MeV
6 6 MeV
*P39.42 K = ( − ) mc = (^) (( − u c (^) ) − ) mc − γ 1 2 1 2 2 1 1 2/ (^2)
. We use the series expansion from Appendix B.5:
K = mc^2 ⎡⎣ 1 + −( 12 ) −( u^2 / c^2 ) + −( 12 ) −( 32 ) (^21)! (− u^2 / c^2 )^2 ++ − 1 ⎤⎦
K mu m u c
4 2
(^2) The actual kinetic energy, given by this relativistic equation, is greater
than the classical (1/2) mu^2. The difference, for m = 1 000 kg and u = 25 m/s, is 38 1 8 9 3 10
000 kg) (25 m/s). m/s)
J
4 × 2 =^ ×^
− ~~ 10 − 9 J
Section 39.9 Mass and Energy
*P39.43 E = 2 86. × 10 5 J leaves the system so the final mass is smaller. The mass-energy relation says
that E = mc^2. Therefore, m E c
= = ×
( × )
2 =^ ×^ −
5 2
2 86. 10 J 3 18. 1012
3.00 10 m s 8 kgg A mass loss of this magnitude, as a fraction of a total of 9.00 g, could not be detected.
P39.44 ∆^
m E ∆ c
t c
( × )( ) 2 2
P 0 800 1 00.^.^109 J s^ 3 00.^ yr^ 3 1.^66
7 8 2
( × ) ( × )
s yr m s
kg .
Relativity 423
Additional Problems
P39.48 (a) d (^) earth = vt (^) earth = v γ t astro so 2 00 10 1 1
(.^ ×^6 ) =^2 2 30 0. −
yr c yr c
v v
1 1 50 10
2 2
− = ⎛^5
⎠ (^ × )
v v − c c
2
2 10 − = (^2) v v ( × − ) c c
2 2 = v ( + × −^10 ) c
. (^) so v c = (^) ( + × −^ ) = − (^) ( × ) − (^) − 1 2 25 10 1
1 2 (^10)
..
v c
= 1 − 1 12. × 10 −^10
(b) K c
= mc −
⎟ =^
×
⎝⎜^
6 v
. yr 30 yr ⎠⎠⎟ ( )( ) ( × )
= ×
8 2
27
kg m s
. J
(c) 6 00 10 6 00 10
. × =. × ⎛⎝ ⎞⎠⎛⎝ ⎞⎠
J J
kWh
k 10
W
3
ss J s
⎝⎜^
= ×
h $. 3 600
P39.49 (a) 10 13 MeV = ( γ − 1 ) m cp^2 so γ = 1010
vp ≈ c t ′ = = = − t γ
(^5) yr 5 2 10 10 yr^ ~ s (b) d ′ = ct ′ ~10 8 km
P39.50 (a) When K e = Kp , m ce^2 ( γ e − 1 ) = m cp^2 (γ p − 1 )
In this case, m ce^2 = 0 511. MeV, m cp^2 = 938 MeV
and γ (^) e = ⎡⎣ − ( )⎤⎦ =
− 1 0 750 2 1 511 9
1 2
..
Substituting, γ γ p e e p
m c m c
( − ) = + ( ) ( − ) 1
2
. MeV. 9938 1 000 279
MeV =. but γ (^) p u (^) p c
2 1 2
Therefore, u (^) p = c 1 − γ (^) p −^2 = 0 023 6. c
(b) When p (^) e = pp γ (^) p m up p = γ e m ue e or γ γ p p e^ e^ e p
u m u m
Thus, γ (^) p u (^) p
c c c
(1 511 9 )( 0 511 (^) )( 0 750) 938
2 2
.. MeV. MeV == 6 177 2. × 10 − 4 c
and u c
u c
p (^) = × − ⎛ p ⎝⎜^
6 177 2 10 −^41
2 .
which yields u (^) p = 6 18. × 10 −^4 c = 185 km s
424 Chapter 39
*P39.51 (a) We let H represent K mc / 2. Then H u c
u c H H
2 2 / so /^2 u c H H
H H
H H
u c H
2 2 2
2 2
2 1 1 2 1
and = ++
1 2 H H H
/
(b) u goes to 0 as K goes to 0.
(c) u approaches c as K increases without limit.
(d) a du dt
c
H H
H H
H H
⎝⎜^
1 − ⎡⎣ +^ + ⎤⎦
2
2 2 2
H H H H
H H
[ + ] − ⎡⎣ + ⎤⎦ [ + ]
dd K mc dt
= = +^ +
⎝⎜^
⎠⎟^
[ + ]
a c H^ H H H
H
H mc
2 2
1 2 4
/
P
22 =^ 1 2 + 2 1 2 + 12
P
mcH /^ ( H ) /( H )
(e) When H is small we have approximately a mcH mK
P P
1 2/ 2 1 2 / 2 1 2/ , in agreement with the nonrelativistic case.
(f) When H is large the acceleration approaches P / mcH^3 = P m^2 c^5 / K^3.
(g) As energy is steadily imparted to particle, the particle’s acceleration decreases. It decreases steeply, proportionally to 1/ K^3 at high energy. In this way the particle’s speed cannot reach or surpass a certain upper limit, which is the speed of light in vacuum.
P39.52 (a) Since Mary is in the same reference frame, S ′, as Ted, she measures the ball to have the
same speed Ted observes, namely u (^) x ′ = 0 800. c.
(b) ∆ ′ =^ ′ =^
×
( × )
t =
L
u
p x
1 80. 10 12 m 7 5. 0.800 3.00 10 8 m s
00 × 10 3 s
(c) L L c
c p c = − = (^) ( × ) − (^
1 1 80 10 1 0 600^ = 1 44×
2 2
12 2 2
v (^). m. (^). 11012 m
Since v = 0 600. c and u (^) x ′ = − 0 800. c , the velocity Jim measures for the ball is
u u u c
c c x x x
( − ) + ( )
v 1 v
((. )( )
. c
(d) Jim measures the ball and Mary to be initially separated by 1 44. × 1012 m. Mary’s motion at 0.600 c and the ball’s motion at 0.385 c nibble into this distance from both ends. The gap closes at the rate 0 600. c + 0 385. c = 0 985. c , so the ball and catcher meet after a time
∆ t = × ( × )
1 44 10 = 4 88 × 10
m. 0.985 3.00 10 m s 8 s
P39.53 ∆ mc mc
2 2
= 4 938 78(^.^ MeV) −^ 3 728 4. MeV
4 938.78 MeV(( )^
× 100 % = 0 712. %
