Chapter 26
Relativity
Quick Quizzes
1.
(a). Less time will have passed for you in your frame of reference than for your employer
back on Earth. Thus, to maximize your paycheck, you should choose to have your pay
calculated according to the elapsed time on a clock on Earth.
2.
No. Your sleeping cabin is at rest in your reference frame, thus, it will have its proper
length. There will be no change in measured lengths of objects within your spacecraft.
Another observer, on a spacecraft traveling at a high speed relative to yours, will claim
that you are able to fit into a shorter sleeping cabin (if your body is oriented in a direction
parallel to your direction of travel relative to the other observer).
3.
(a), (e). The outgoing rocket will appear to have a shorter length and a slower clock. The
answers are the same for the incoming rocket. Length contraction and time dilation
depend only on the magnitude of the relative velocity, not on the direction.
4.
(a) False
(b) False
(c) True
(d) False
A reflected photon does exert a force on the surface. Although a photon has zero mass, a
photon does carry momentum. When it reflects from a surface, there is a change in the
momentum, just like the change in momentum of a ball bouncing off a wall. According to
the momentum interpretation of Newton’s second law, a change in momentum results in
a force on the surface. This concept is used in theoretical studies of space sailing. These
studies propose building non-powered spacecraft with huge reflective sails oriented
perpendicularly to the rays from the Sun. The large number of photons from the Sun
reflecting from the surface of the sail will exert a force which, although small, will provide
a continuous acceleration. This would allow the spacecraft to travel to other planets
without fuel.
5.
(a). The downstairs clock runs more slowly because it is closer to Earth and hence
experiences a stronger gravitational field than the upstairs clock does.
363
364
CHAPTER 26
Answers to Even Numbered Conceptual Questions
2.
To be strictly correct, the equation should be written as E = γ mc 2 where γ = 1
1 − (v c) .
2
When the object is at rest v = 0 and γ = 1 , so the equation reverts to the popular form
E = mc 2 . When v > 0 , E > ER = mc 2 and the difference KE = E − ER = ( γ − 1) mc 2 accurately
accounts for the kinetic energy of the moving mass.
4.
The two observers will agree on the speed of light and on the speed at which they move
relative to one another.
6.
Special relativity describes inertial reference frames -- that is , reference frames that are not
accelerating. General relativity describes all reference frames.
8.
You would see the same thing that you see when looking at a mirror when at rest. The
theory of relativity tells us that all experiments will give the same results in all inertial
frames of reference.
10.
The clock in orbit will run more slowly. The extra centripetal acceleration of the orbiting
clock makes its history fundamentally different from that of the clock on Earth.
12.
The 8 lightyears represents the proper length of a rod from Earth to Sirius, measured by
an observer seeing both the rod and Sirius nearly at rest. The astronaut sees Sirius coming
toward her at 0.8c but also sees the distance contracted to
d = ( 8 ly ) 1 − ( v c ) = ( 8 ly ) 1 − ( 0.8 ) = 5 ly
2
2
So, the travel time measured on her clock is t =
d 5 ly ( 5 yr ) c
=
=
= 6 yr .
0.8 c
v 0.8 c
Relativity
Answers to Even Numbered Problems
2.
(a)
1.9 × 10 −15 s
(b)
2.0 fringe shifts
4.
(a)
1.38 yr
(b)
1.31 lightyears
6.
(a)
70 beats min
(b)
31 beats min
8.
(a)
65.0 min
(b)
10.6 min
10.
(a) a rectangular box
(b) sides perpendicular to the motion are 2.0 m long, sides parallel to the motion are
1.2 m long
12.
(a)
0.297 L0
(b)
0.088
14.
(a)
17.4 m
(b)
3.30° with respect to the direction of motion
16.
(a)
21.0 yr
18.
(a)
2.7 × 10 −24 kg ⋅ m s
(c)
5.6 × 10 −22 kg ⋅ m s
20.
(a)
0.141c
22.
0.94 c to the left
24.
0.696 c
26.
0.161 Hz
28.
(a)
30.
0.980 c
32.
(a)
34.
m faster = 2.51 × 10 −28 kg, mslower = 8.84 × 10 −28 kg
36.
(a)
38.
0.64 MeV, 3.4 × 10 −22 kg ⋅ m s for each photon
40.
(a)
2.50 MeV c
(b)
4.60 GeV c
42.
(a)
25.0 yr
(b)
15.0 yr
939 MeV
0.582 MeV
3.52 MeV
(b)
14.7 ly
(c)
10.5 ly
(b)
1.6 × 10 −22 kg ⋅ m s
(b)
0.436 c
(b)
3.01 GeV
(b)
2.45 MeV
(b)
(d)
35.7 yr
(c)
2.07 GeV
(c)
12.0 ly
8.50 × 10 20 Hz
365
366
CHAPTER 26
44.
1.2 × 1013 m
46.
0.614 c away from Earth
48.
(a)
~10 2 s
(b)
~108 km
50.
(a)
4.38 × 1011 J
(b)
4.38 × 1011 J
52.
(b)
0.943 c
54.
(a)
t=
(b)
t=
56.
1.47 km
58.
(a)
(b)
$2.12 × 109
2d
c+v
$800
2d c − v
c c+v
367
Relativity
Problem Solutions
26.1
G
G
G
(a) As the plane flies from O to B along path I, v ground = v air + v wind gives
tOB =
L
=
v ground
ur
vair
m
m
m
+ 20.0
= 120
, and
s
s
s
v ground = 100
ur
vwind
ur
vground
200 × 10 3 m
= 1.67 × 10 3 s
120 m s
G
G
G
G
G
For the plane following path II from O to A, v ground = v air + v wind with v ground and v wind
perpendicular to each other. The Pythagorean theorem then yields
2
v ground = v − v
2
air
and
2
wind
L
tOA =
v ground
2
m 
m
m

=  100
 −  20.0
 = 98.0
s  
s 
s

=
200 × 10 3 m
= 2.04 × 10 3 s
98.0 m s
ur
vwind
ur
vair
ur
vground
G
G
G
(b) For the return flight along path I, v ground = v air + v wind gives
v ground = 100
m
m
m
− 20.0
= 80.0
s
s
s
tBO =
so
L
v ground
200 × 10 3 m
=
= 2.50 × 10 3 s
80.0 m s
ur
vair
ur
ur
vwind vground
G
G
As the plane flies from A to O along path II, v ground and v wind are again
perpendicular to each other. The Pythagorean theorem gives
2
2
v ground = vair
− vwind
= 98.0
and
tAO =
L
v ground
m
s
= 2.04 × 10 3 s
ur
vair
ur
vground
ur
vwind
368
CHAPTER 26
(c) The total times of flight are
tI = tOB + tBO = 1.67 × 10 3 s + 2.50 × 10 3 s = 4.17 × 10 3 s
tII = tOA + tAO = 2 ( 2.04 × 10 3 s ) = 4.08 × 10 3 s
and
The difference in total flight times is ∆t = tI − tII = 90 s
4
2Lv 2 2 ( 28 m ) ( 3.0 × 10 m s )
= 3 =
= 1.9 × 10 −15 s
3
8
c
( 3.0 × 10 m s )
2
26.2
(a)
∆tnet
(b) Since a fringe shift occurs for every half-wavelength change made in the optical
path length, the number of fringe shifts expected is
N=
=
∆dnet 2 c ( ∆tnet )
=
λ 2
λ
2 ( 3.0 × 108 m s )( 1.9 × 10 −15 s )
550 × 10 −9 m
∆tp
3.0 s
= 2.0 fringe shifts
26.3
∆t = γ ∆tp =
26.4
(a) Observers on Earth measure the time for the astronauts to reach Alpha Centauri as
∆tE = 4.42 yr . But these observers are moving relative to the astronaut’s internal
biological clock and hence, experience a dilated version of the proper time interval
∆tp measured on that clock. From ∆tE = γ ∆tp , we find
1 − (v c)
2
=
1- ( 0.80 )
2
= 5.0 s
∆tp = ∆tE γ = ∆tE 1 − ( v c ) = ( 4.42 yr ) 1 − ( 0.950 ) = 1.38 yr
2
2
(b) The astronauts are moving relative to the span of space separating Earth and Alpha
Centauri. Hence, they measure a length contracted version of the proper distance,
Lp = 4.20 ly . The distance measured by the astronauts is
L = Lp γ = Lp 1 − ( v c ) = ( 4.20 ly ) 1 − ( 0.950 ) = 1.31 ly
2
2
Relativity
26.5
369
(a) Since your ship is identical to his, the proper length of your ship (that is, the length
you measure for your ship at rest relative to you) must be the same as the proper
length that he measured for his own ship. Thus, both ships must be 20 m meters
long when measured by observers at rest relative to them.
(b) His ship moves relative to you at the same speed your ship moves relative to him.
Thus, you observe the same length contraction of his ship as he observed for your
ship and measure his ship to be 19 m in length.
(c) From L = Lp γ = Lp 1 − ( v c ) with Lp = 20 m and L = 19 m , we find
2
(
v = c 1 − L Lp
26.6
)
2
= c 1 − ( 19 m 20 m ) = 0.31 c
2
(a) The time for 70 beats, as measured by the astronaut and any observer at rest with
respect to the astronaut, is ∆tp = 1.0 min . The observer in the ship then measures a
rate of 70 beats min .
(b) The observer on Earth moves at v = 0.90 c relative to the astronaut and measures the
time for 70 beats as
∆t = γ ∆tp =
∆tp
1 − (v c)
2
=
1.0 min
1- ( 0.90 )
2
= 2.3 min
This observer then measures a beat rate of
26.7
(a)
∆t = γ ∆tp =
∆tp
1 − (v c)
2
=
2.6 × 10 −8 s
1- ( 0.98 )
2
70 beats
= 31 beats min
2.3 min
= 1.3 × 10 −7 s
(b) d = v ( ∆t ) = 0.98 ( 3.0 × 10 8 m s )  ( 1.3 × 10 −7 s ) = 38 m
(c)
( )
d′ = v ∆tp =  0.98 ( 3.0 × 10 8 m s )  ( 2.6 × 10 −8 s ) = 7.6 m
370
CHAPTER 26
26.8
(a) As measured by observers in the ship (that is, at rest relative to the astronaut), the
time required for 75.0 pulses is ∆tp = 1.00 min .
The time interval required for 75.0 pulses as measured by the Earth observer is
1.00 min
∆t = γ ∆tp =
, so the Earth observer measures a pulse rate of
2
1 − ( 0.500 )
75.0 75.0 1 − ( 0.500 )
rate =
=
= 65.0 min
∆t
1.00 min
2
(b) If v = 0.990 c , then ∆t = γ ∆tp =
1.00 min
1 − ( 0.990 )
2
and the pulse rate observed on Earth is
75.0 75.0 1 − ( 0.990 )
rate =
=
= 10.6 min
∆t
1.00 min
2
That is, the life span of the astronaut (reckoned by the total number of his
heartbeats) is much longer as measured by an Earth clock than by a clock aboard
the space vehicle.
26.9
As seen by an Earth based observer, the time for the muon to travel 4.6 km is
∆t =
d
4.6 × 10 3 m
=
v 0.99 ( 3.0 × 108 m s )
(a) In the rest frame of the muon, this time (the proper lifetime) is
∆tp =


4.6 × 10 3 m
2
 1 − ( 0.99 ) = 2.2 × 10 −6 s = 2.2 µ s
=
8
γ  0.99 ( 3.0 × 10 m s ) 
∆t
(b) The muon is at rest in “its frame” and thus travels zero distance as measured in this
frame. However, during this time interval, the muon sees Earth move toward it by a
distance of
( )
d = v ∆tp = 0.99 ( 3.0 × 10 8 m s )  ( 2.2 × 10 −6 s )
= 6.5 × 10 2 m = 0.65 km
371
Relativity
26.10
Rest Frame
of Box
v = 0.80c
Length contraction occurs only in the dimension parallel to the
motion.
(a) The sides labeled L2 and L3 in the figure at the right are
unaffected, but the side labeled L1 will appear contracted
L3
giving the box a rectangular shape.
(b) The dimensions of the box, as measured by the observer
moving at v = 0.80 c relative to it, are
L2
L1
L2 = L2 p = 2.0 m , L3 = L3 p = 2.0 m , and
L1 = L1 p 1 − ( v c ) = ( 2.0 m ) 1 − ( 0.80 ) = 1.2 m
2
26.11
2
(
)
The proper length of the faster ship is three times that of the slower ship Lpf = 3Lps , yet
they both appear to have the same contracted length, L. Thus,
(
L = Lps 1 − ( vs c ) = 3Lps
2
This gives v f =
26.12
c 8 + ( vs c )
3
)
(
1− vf c
8 + ( 0.350 )
2
=
3
)
2
(
, or 1 − ( vs c ) = 9 − 9 v f c
2
)
2
2
c = 0.950 c
(a) Observer A measures the proper length, LpA = L0 , of the rod that is at rest relative to
her. However, observer B is moving at v = 0.955 c relative to this rod and will
measure the contracted length LA for it, where
LA = LpA γ = L0 1 − ( v c ) = L0 1 − ( 0.955 ) = 0.297 L0
2
2
372
CHAPTER 26
(b) Observer A is moving at v = 0.955 c relative to the rod at rest in B’s reference frame,
and she measures the contracted length LB = L0 for the length of this rod. This rod’s
proper length LpB (the length measured by observer B who is at rest relative to this
rod) is found from LB = LpB 1 − ( v c ) as
2
LpB =
LB
1 − (v c)
2
=
L0
1 − ( 0.955 )
2
= 3.37 L0
Thus, the ratio of the length of A’s rod to the length of B’s rod, as measured by
observer B, is
ratio =
26.13
LA 0.297 L0
=
= 0.088
LpB
3.37 L0
The trackside observer sees the supertrain length-contracted as
L = Lp 1 − ( v c ) = ( 100 m ) 1 − ( 0.95 ) = 31 m
2
2
The supertrain appears to fit in the tunnel
with 50 m − 31 m = 19 m to spare
26.14
Note: Excess digits are retained in some steps given
below to more clearly illustrate the method of
solution.
We are given that L = 2.00 m , and θ = 30.0° (both
measured in the observer’s rest frame). The
components of the rod’s length as measured in the
observer’s rest frame are
Lx = L cosθ = ( 2.00 m ) cos 30.0° = 1.732 m
and
Observer’s rest frame
rod’s rest frame
v = 0.995 c
Ly
Lpy
Lp
qp
Lpx
Lx
Ly = L sin θ = ( 2.00 m ) sin 30.0° = 1.00 m
The component of length parallel to the motion has been contracted, but the component
perpendicular to the motion is unaltered. Thus, Lpy = Ly = 1.00 m and
Lpx =
Lx
1 − (v c)
2
=
1.732 m
1 − ( 0.995 )
2
= 17.34 m
Relativity
(a) The proper length of the rod is then
Lp = L2px + L2py =
2
2
( 17.34 m ) + (1.00 m ) = 17.4 m
(b) The orientation angle in the rod’s rest frame is
 Lpy
 Lpx

θ p = tan −1 
26.15

 1.00 m 
 = tan −1 
 = 3.30°

17.34
m



The centripetal acceleration is provided by the gravitational force, so
mv 2 GMm
 GM 
=
, giving Cooper’s speed as v = 
2
r
r
 r 
1/2
 GM 
=

 (RE + h) 
1
or
 ( 6.67 × 10 − 11 N ⋅ m 2 kg 2 ) ( 5.98 × 10 24 kg )  2
 = 7.81 × 10 3 m s
v=
6
6
6.38
10
0.160
10
m
×
+
×


(
)
Then the time period of one orbit is,
6
2π ( RE + h) 2π ( 6.54 × 10 m )
T=
=
= 5.26 × 10 3 s
v
7.81 × 10 3 m s
(a) The time difference for 22 orbits is
− 1/2
∆t − ∆tp = (γ − 1) ∆tp = ( 1 − v 2 c 2 )
− 1 ( 22T )




1 v2
∆t − ∆tp ≈  1 +
− 1  ( 22 T )
2
2c


2
1  7.81 × 10 3 m/s 
3
= 
 22 ( 5.26 × 10 s ) = 39.2 µ s
2  3.00 × 10 8 m/s 
(b) For one orbit, ∆t − ∆tp =
39.2 µ s
= 1.78 µ s
22
The press report is accurate to one digit
1/2
373
374
CHAPTER 26
26.16
(a)
∆t = γ ∆tp =
∆tp
1 − (v c)
2
=
15.0 yr
1 − ( 0.700 )
2
= 21.0 yr
(b) d = v ( ∆t ) = [ 0.700 c ] ( 21.0 yr ) = ( 0.700 ) ( 1.00 ly yr )  ( 21.0 yr ) = 14.7 ly
(c) The astronauts see Earth flying out the back window at 0.700 c :
( )
d = v ∆tp = [ 0.700 c ] ( 15.0 yr ) = ( 0.700 ) ( 1.00 ly yr )  ( 15.0 yr ) = 10.5 ly
(d) Mission control gets signals for 21.0 yr while the battery is operating and then for
14.7 yr after the battery stops powering the transmitter, 14.7 ly away:
21.0 yr + 14.7 yr = 35.7 yr
26.17
The momentum of the electron is
pe = γ me v =
( 9.11 × 10
−31
kg ) ( 0.90 c )
1 − ( 0.90 )
2
= ( 1.9 × 10 −30 kg ) c
If the proton has the same momentum, then
pp
(1.67 × 10
=γm v=
p
−27
kg ) v
1 − (v c)
2
= ( 1.9 × 10 −30 kg ) c
which reduces to ( 8.9 × 10 2 ) ( v c ) = 1 − ( v c ) and yields
2
v = ( 1.1 × 10 −3 ) c = ( 1.1 × 10 −3 ) ( 3.0 × 10 8 m s ) = 3.3 × 10 5 m s
26.18
The momentum of the electron is pe = γ me v =
( 9.11 × 10
−31
kg ) v
1 − (v c)
2
.
(a) When v = 0.010 c ,
pe =
( 9.11 × 10
−31
kg ) ( 0.010 ) ( 3.0 × 108 m s )
1 − ( 0.010 )
2
(b) If v = 0.50 c , pe = 1.6 × 10 −22 kg ⋅ m s , and
= 2.7 × 10 −24 kg ⋅ m s
Relativity
375
(c) When v = 0.90 c , pe = 5.6 × 10 −22 kg ⋅ m s
26.19
Momentum must be conserved, so the momenta of the two fragments must add to zero.
Thus, their magnitudes must be equal, or
p2 = p1 = γ 1 m1v1 =
( 2.50 × 10
For the heavier fragment,
−28
kg ) ( 0.893 c )
1 − ( 0.893 )
(1.67 × 10
−27
2
kg ) v
1 − (v c)
= ( 4.96 × 10 −28 kg ) c
= ( 4.96 × 10 −28 kg ) c
2
which reduces to 3.37 ( v c ) = 1 − ( v c ) and yields v = 0.285 c
2
26.20
mv
Using the relativistic form, p =
1 − (v c)
2
= γ mv
we find the difference ∆p from the classical momentum, mv :
∆p = γ mv − mv = (γ − 1)mv
(a) The difference is 1.00% when (γ − 1)mv = 0.010 0 γ mv :
γ=
1
1
=
2
0.990
1 − (v c)
⇒ 1 − ( v c ) = ( 0.990 )
2
2
or
v = 0.141 c
or
v = 0.436 c
(b) The difference is 10.0% when (γ − 1)mv = 0.100 γ mv :
γ=
26.21
1
1
=
2
0.900
1 − (v c)
⇒ 1 − ( v c ) = ( 0.900 )
2
2
Taking toward the right as the positive direction, with vpL = velocity of proton relative
to laboratory, vpe = velocity of proton relative to electron, and veL = velocity of electron
relative to laboratory, the relativistic velocity addition equation gives
vpL =
vpe + veL
−0.70 c + 0.90 c
=
= + 0.54 c
vpe ⋅ veL
−0.70 c )( 0.90 c )
(
1+
1+
c2
c2
376
CHAPTER 26
26.22
Taking toward the right as positive, with vLR = velocity of L relative to R, vLE = velocity
of L relative to Earth, vRE = velocity of R relative to Earth, and vER = −vRE = velocity of
Earth relative to R, the relativistic velocity addition equation gives
vLR =
26.23
−0.70 c + ( −0.70 c )
vLE + vER
=
= − 0.94 c
vLE ⋅ vER
−0.70 c )( −0.70 c )
(
1+
1+
c2
c2
vEO = velocity of Enterprise
relative to Observers on
Earth.
vEO = 0.900 c
vKO = 0.800 c
vKO = velocity of Klingon ship
relative to Observers on
Earth.
vOK = −vKO = velocity of Observers on Earth relative to Klingon ship
The relativistic velocity addition equation yields
vEK =
26.24
With vRS = velocity of rocket relative to ship, vRE = velocity of rocket relative to Earth,
vSE = velocity of ship relative to Earth, and vES = −vSE = velocity of Earth relative to ship,
the relativistic velocity addition equation gives
vRS =
26.25
0.900 c + ( −0.800 c )
vEO + vOK
=
= + 0.357 c
v ⋅v
1 + EO 2 OK 1 + ( 0.900 c )( −0.800 c )
c
c2
0.950 c + ( −0.750 c )
vRE + vES
=
= + 0.696 c
v ⋅v
1 + RE 2 ES 1 + ( 0.950 c )( −0.750 c )
c
c2
Taking toward the right as positive, with vBA = velocity of B relative to A, vRA = velocity
of rocket relative to A, vRB = velocity of rocket relative to B, and vBR = −vRB = velocity of
B relative to rocket, the relativistic velocity addition equation gives
vBA =
− ( −0.95 c ) + ( 0.92 c )
vBR + vRA
=
= + 0.998 c
vBR ⋅ vRA


c
c
0.95
0.92
−
−
(
)
(
)
1+

1+ 
c2
c2
Relativity
26.26
377
First, determine the velocity of the pulsar relative to the rocket.
Taking toward Earth as positive, with vPR = velocity of pulsar relative to rocket, vPE =
velocity of pulsar relative to Earth, vRE = velocity of rocket relative to Earth, and
vER = −vRE = velocity of Earth relative to rocket, the relativistic velocity addition equation
gives
vPR =
0.950 c +  − ( − 0.995 c ) 
vPE + vER
=
= 0.999 87 c
v ⋅v
0.950 c )( 0.995 c )
(
1 + PE 2 ER
1+
c
c2
The period of the pulsar in its own reference frame is ∆tp = 0.100 s , and its period in the
rocket’s frame of reference is
( )
∆t = γ ∆tp =
∆t p
1 − ( vPR c )
2
=
0.100 s
1 − ( 0.999 87 )
= 6.20 s
1
1
=
= 0.161 Hz
∆t 6.20 s
and the frequency in the rocket’s frame is f =
26.27
2
The instructors measure a proper time of ∆tp = 50 min on their clock.
(a) First determine the velocity of the students relative to the instructors (and hence
relative to the official clock). Taking toward the right in Figure P26.27 as the
positive direction, with vSI = velocity of students relative to instructors, vSE =
velocity of students relative to Earth, vIE = velocity of instructors relative to Earth,
and vEI = −vIE = velocity of Earth relative to instructors, we find
vSI =
0.60 c + ( − 0.28 c )
vSE + vEI
=
= 0.385 c
vSE ⋅ vEI
−
c
c
0.60
0.28
(
)(
)
1+
1+
c2
c2
The elapsed time on the student’s clock is then
∆t = γ ∆tp =
∆tp
1 − ( vSI c )
2
=
50 min
1 − ( 0.385 )
2
= 54 min
(b) The elapsed time on Earth is
∆t = γ ∆tp =
∆t p
1 − ( vEI c )
2
=
50 min
1 − ( − 0.28 )
2
= 52 min
378
CHAPTER 26
26.28
(a)
2
1 MeV 
ER = mc 2 = ( 1.67 × 10 −27 kg )( 3.00 × 108 m s ) 
 = 939 MeV
-13
 1.60 × 10 J 
(b) E = γ mc 2 = γ ER =
=
(c)
26.29
939 MeV
1 − ( 0.950 )
1 − (v c)
2
= 3.01 × 10 3 MeV = 3.01 GeV
KE = E − ER = 3.01 × 10 3 MeV − 939 MeV = 2.07 × 10 3 MeV = 2.07 GeV
If KE = ER , then KE = E − ER = ( γ − 1)ER = ER giving γ =
Therefore,
26.30
2
ER
v = c 1−1 γ 2 = c 3 4 = c
(
3 2
1
1 − (v c)
=2
2
)
The work done accelerating a free particle from rest to speed v equals the kinetic energy
given that particle. Thus,
KE = E − ER = γ ER − ER = 3 750 MeV
2
1 MeV
For a proton, ER = mc 2 = ( 1.67 × 10 −27 kg ) ( 3.00 × 108 ) 
-13
 1.60 × 10
Thus, γ − 1 =
or
3 750 MeV
= 3.99
939 MeV
v = c 1 − ( 1 4.99 ) = 0.980 c
2
giving
γ=

 = 939 MeV
J
1
1 − (v c)
2
= 4.99
Relativity
26.31
The nonrelativistic expression for kinetic energy is KE =
1
mv 2 , while the relativistic
2
1 − ( v c ) . Thus, when the
expression is KE = E − ER = ( γ − 1) ER = ( γ − 1) mc 2 where γ = 1
2
relativistic kinetic energy is twice the predicted nonrelativistic value, we have


1
1


− 1  mc 2 = 2  mv 2 
2


2

 1 − (v c)

or
  v 2 
2
1 = 1 +    1 − ( v c )
  c  
v
Squaring both sides of the last result and simplifying gives  
c
4
Ignoring the trivial solution v c = 0 , we must have
2
 v  4  v  2 
  +   − 1 = 0
 c   c 

2
v v
  +  −1= 0
c c
This is a quadratic equation of the form x 2 + x − 1 = 0 with x = ( v c ) . Applying the
2
x=
quadratic formula gives
−1 ± 5
2
Since x = ( v c ) , we ignore the negative solution and find
2
−1 + 5
v
x=  =
= 0.618
2
c
2
26.32
which yields
v = c 0.618 = 0.786 c
(
)
The energy input to the electron will be W = E f − Ei = γ f − γ i ER
or


1
W =
 1− vf c

(
)
2
−
1
1 − ( vi c )
2


 ER where ER = 0.511 MeV


(a) If v f = 0.900 c and vi = 0.500 c , then

1
1
W =
−
2
2
 1 − ( 0.900 )
1 − ( 0.500 )

379

 ( 0.511 Mev ) = 0.582 MeV


380
CHAPTER 26
(b) When v f = 0.990 c and vi = 0.900 c , we have

1
1
W =
−
2
 1 − ( 0.990 )2
1 − ( 0.900 )

26.33

 ( 0.511 Mev ) = 2.45 MeV


When a cubical box moves at speed v, the dimension parallel to the motion is length
contracted and other dimensions are unaffected. If all edges of the box had length Lp
when at rest, the volume of the moving box is
( )( ) ( L
V = Lp L p
1 − (v c)
p
2
)=V
p
1 − (v c)
2
The relativistic density is
ρ=
8.00 g
m
m
=
=
= 18.4 g cm 3
V V 1 − ( v c )2 ( 1.00 cm )3 1 − ( 0.900 )2
p
Note that ρ =
26.34
E
m mc 2
= 2 = 2 R as suggested in the problem statement.
V c V c V
Let m1 be the mass of the fragment moving at v1 = 0.868 c , and m2 be the mass moving
at v2 = 0.987 c .
From conservation of mass-energy,
E=
m1c 2
1 − ( 0.868 )
2
+
m2 c 2
1 − ( 0.987 )
2
= mc 2
giving 2.01 m1 + 6.22 m2 = 3.34 × 10 −27 kg
(1)
The momenta of the two fragments must add to zero, so the magnitudes must be equal,
giving p1 = p2 or γ 1 m1v1 = γ 2 m2 v2 . This yields
2.01 m1 ( 0.868 c ) = 6.22 m2 ( 0.987 c ) , or m1 = 3.52 m2
Substituting equation (2) into (1) gives
( 7.07 + 6.22 ) m2 = 3.34 × 10−27 kg , or m2 = 2.51 × 10 −28 kg
Equation (2) then yields m1 = 3.52 ( 2.51 × 10 −28 kg ) = 8.84 × 10 −28 kg
(2)
Relativity
26.35
381
The total kinetic energy equals the energy of the original photon minus the total rest
energy of the pair, or
KE = Eγ − 2 ER = 3.00 MeV − 2 ( 0.511 MeV ) = 1.98 MeV
26.36
(a) The energy of the original photon must equal the total energy (total kinetic energy
plus total rest energy) of the pair.
Eγ = KEtotal + 2 ER = 2.50 MeV + 2 ( 0.511 MeV ) = 3.52 MeV
Thus,
(b)
26.37
f=
 3.52 MeV   1.60 × 10 −13 J 
20
=
 = 8.50 × 10 Hz

h  6.63 × 10 −34 J ⋅ s  1 MeV 
Eγ
To produce minimum energy photons, both members of the proton-antiproton pair
should be at rest (that is, have zero kinetic energy) just before annihilation. Then, the
total momentum is zero both before and after annihilation. This means that the two
photons must have equal magnitude but oppositely directed momenta, and hence, equal
energies.
From conservation of energy, 2 Eγ = 2 ( ER + 0 ) or Eγ = ER for each photon. The rest
energy of a proton is
ER = mp c 2 = ( 1.67 × 10 −27 kg )( 3.00 × 10 8 m s ) ( 1 MeV 1.60 × 10 −13 J ) = 939 MeV
2
f=
and
λ=
Eγ
h
=
939 MeV  1.60 × 10 −13 J 
23

 = 2.27 × 10 Hz
6.63 × 10 −34 J ⋅ s  1 MeV 
c 3.00 × 10 8 m s
=
= 1.32 × 10 −15 m= 1.32 fm
f 2.27 × 10 23 Hz
382
CHAPTER 26
26.38
The total momentum is zero both before and after the annihilation. Thus, the momenta
of the two photons must have equal magnitudes and be oppositely directed. Since
Eγ = pγ c , the photon energies are also equal and conservation of energy gives
1 − (v c)
2 Eγ = 2 ( KE + ER ) = 2 E = 2 ER
ER
Eγ =
pγ =
=
26.39
1 − (v c)
Eγ
c
2
=
0.511 MeV
1 − ( 0.60 )
2
2
or
1 − (v c)
Eγ = ER
= 0.64 MeV
Mev
c
= 0.64
( 0.64 Mev ) ( 1.60 × 10 −13 J 1 MeV )
3.00 × 10 m s
8
= 3.4 × 10 −22 kg ⋅ m s
KE = E − ER = ( γ − 1) ER
so
γ = 1+
KE
1
=
2
ER
1 − (v c)
v = c 1−
giving
1
(1 + KE ER )
2
(a) When KE = q ( ∆V ) = e ( 500 V ) = 500 eV , and ER = 939 MeV , this yields
v = c 1−
1
1 + 500 eV ( 939 × 10 eV ) 


6
2
= 1.03 × 10 −3 c = 3.10 × 10 5 m s
(b) When KE = q ( ∆V ) = e ( 5.00 × 108 V ) = 500 MeV
v = c 1−
26.40
1
(1 + 500 MeV 939 MeV )
2
= 0.758 c
From E2 = ( pc ) + ER2 with E = 5 ER , we find that p =
2
(a) For an electron,
p=
(b) For a proton,
p=
( 0.511 MeV ) 24
c
( 939 MeV ) 24
c
ER 24
c
= 2.50 MeV c
= 4.60 × 10 3
MeV
= 4.60 GeV c
c
2
Relativity
26.41
E = KE + ER = 1.00 MeV + 0.511 MeV = 1.51 MeV , so E2 = ( pc ) + ER2 gives
2
E2 − ER2
p=
26.42
383
c
2
2
( 1.51 MeV ) − ( 0.511 MeV )
=
c
= 1.42 MeV c
(a) The astronomer on Earth measured both the speed and distance in his frame of
reference. Thus, the time to impact on his clock is
∆t =
Lp
v
=
20.0 ly 20.0 c ( 1 yr ) 
=
= 25.0 yr
0.800 c
0.800 c
(c) The observer on the meteoroid sees Earth rushing at him at speed 0.800 c but sees a
length contracted distance of separation given by
L = Lp 1 − ( v c ) = ( 20.0 ly ) 1 − ( 0.800 ) = 12.0 ly
2
2
(b) The time to impact as computed by the observer on the meteoroid is
∆t =
26.43
L 12.0 ly 12.0  c ( 1 yr ) 
=
=
= 15.0 yr
v 0.800 c
0.800 c
(a) Since Ted and Mary are in the same frame of reference, they measure the same
speed for the ball, namely u = 0.80 c .
(b) ∆tp =
Lp
u
=
1.8 × 1012 m
= 7.5 × 10 3 s
8
0.80 ( 3.00 × 10 m s )
(c) The distance of separation, as measured by Jim, is
L = Lp 1 − ( v c ) = ( 1.8 × 1012 m ) 1 − ( 0.60 ) = 1.4 × 1012 m
2
2
Taking toward the right in Figure P26.43 as positive, with vBJ = velocity of ball
relative to Jim, vBT = velocity of ball relative to Ted, and vTJ = velocity of Ted
relative to Jim, the relativistic velocity addition equation gives
vBJ =
vBT + vTJ
−0.80 c + 0.60 c
=
= − 0.38 c
vBT ⋅ vTJ
− 0.80 c )( 0.60 c )
(
1+
1+
c2
c2
Thus, according to Jim, the ball moves with a speed of 0.38 c
384
CHAPTER 26
26.44
The clock, at rest in the ship’s frame of reference, will measure a proper time of
∆tp = 10 h before sounding. Observers on Earth move at v = 0.75c relative to the clock
and measure an elapsed time of
( )
∆t = γ ∆tp =
∆tp
1 − (v c)
2
=
10 h
1- ( 0.75 )
2
= 15 h
The observers on Earth see the clock moving away at 0.75c and compute the distance
traveled before the alarm sounds as
 3600
d = v ( ∆t ) = 0.75 ( 3.0 × 10 8 m s )  ( 15 h ) 
 1h
26.45
With vnL = velocity of nucleus relative to laboratory, veL = velocity of electron relative to
laboratory, ven = velocity of electron relative to nucleus, and vne = −ven = velocity of
nucleus relative to electron, the relativistic velocity addition equation gives
vnL =
26.46
−0.70 c + 0.85 c
vne + veL
=
= + 0.37 c
v ⋅v
1 + ne 2 eL 1 + ( −0.70 c )( 0.85 c )
c
c2
Taking away from Earth as positive, with vJE = velocity of jet of material relative to
Earth, vJQ = velocity of jet relative to quasar, and vQE = velocity of quasar relative to
Earth, the relativistic velocity addition equation gives
vJE =
26.47
s
13
 = 1.2 × 10 m

vJQ + vQE
−0.550 c + 0.870 c
=
= + 0.614 c
vJQ ⋅ vQE
−0.550 c )( 0.870 c )
(
1+
1+
c2
c2
(a) Observers on Earth measure the distance to Andromeda to be
d = 2.00 × 106 ly = ( 2.00 × 106 yr ) c . The time for the trip, in Earth’s frame of
reference, is
( )
∆t = γ ∆tp =
30.0 yr
1 − (v c)
2
Relativity
385
2.00 × 106 yr ) c
(
d
The required speed is then v =
=
∆t 30.0 yr 1- ( v c )2
(1.50 × 10 ) vc =
−5
which gives
1 − (v c)
2
Squaring both sides of this equation and solving for v c yields
2.25 × 10 −10
1
v
=
≈ 1−
= 1 − 1.12 × 10 −10
−
10
c
2
1 + 2.25 × 10
(b) KE = ( γ − 1) mc 2 , and
γ=
1
1 − (v c)
2
≈
1
1 − ( 1 − 1.12 × 10
)
−10 2
=
1
2.24 × 10
−10
= 6.68 × 10 4
Thus,
KE = ( 6.68 × 10 4 − 1)( 1.00 × 106 kg )( 3.00 × 10 8 m s ) = 6.01 × 10 27 J
2
(c)
cost = KE × rate

 1 kWh
= ( 6.01 × 10 27 J ) 
6
 3.60 × 10

26.48

20
  ( $0.13 kWh ) = $2.17 × 10
J 
The solution to this problem is easier if we start with part (b). First, compute the speed
of the protons.
KE
1013 MeV
KE = ( γ − 1) ER , so γ = 1 +
= 1+
= 1.06 × 1010 , or γ ~ 1010
ER
939 MeV
Thus, v = c 1 − 1 γ 2 ~ c 1 − 10 −20 ≈ c
386
CHAPTER 26
(b) The diameter of the galaxy, as seen in the proton’s frame of reference, is
L = Lp 1 − ( v c ) =
2
Lp
~
γ
10 5 ly
= 10 −5 ly
1010
Since 1 ly = ( 1 yr ) c = ( 3.156 × 107 s )( 3.00 × 108 m s ) ≈ 1016 m
 1016 m   1 km 
8
L ≈ 10 −5 ly 
  3  ~ 10 km
1
ly
10
m




(a) The proton sees the galaxy rushing by at v ≈ c . The time, in the proton’s frame of
reference for the galaxy to pass is
 3.156 × 107 s 
L 10 −5 ly ( 10 yr ) c
=
= 10 −5 yr 
~
 = 316 s
v
c
c
1 yr


−5
∆t =
or ∆t ~ 10 2 s
26.49
The length of the space ship, as measured by observers on Earth, is L = Lp 1 − ( v c ) . In
2
Earth’s frame of reference, the time required for the ship to pass overhead is
L Lp 1 − ( v c )
1 1
∆t = =
= Lp
−
v
v
v2 c2
2
Thus,
2
2
 0.75 × 10 −6 s 
1
1  ∆t 
1
s2
−17
=
+
=
+
=
×
1.74
10




v 2 c 2  Lp  ( 3.00 × 108 m s )2  300 m 
m2
or
v=
1
1.74 × 10 −17
s2
m2

m 
c

=  2.4 × 10 8
 = 0.80 c

8
s   3.00 × 10 m s 

Relativity
26.50
(a) Classically, KE =
 v
(b) γ = 1 −  
  c 
2



−
1
2
387
2
1
1
mv 2 = ( 78.0 kg ) ( 106 × 10 3 m s ) = 4.38 × 1011 J
2
2
  106 × 10 3 m s  2 
= 1 − 
 
8
  3.00 × 10 m s  
Using the approximation ( 1 − x )
−
1
2
≈ 1+
−
1
2
= 1 − 1.25 × 10 −7 
−
1
2
x
for x << 1 gives γ ≈ 1 + 6.24 × 10 −8 , so
2
KE = ( γ − 1)mc 2 ≈ ( 6.24 × 10 −8 ) ( 78.0 kg ) ( 3.00 × 10 8 m s ) = 4.38 × 1011 J
2
In the limit v << c , the classical and relativistic equations yield the same results.
26.51
(a) Taking toward Earth as positive, with vLE = velocity of lander relative to Earth,
vLS = velocity of lander relative to ship, and vSE = velocity of ship relative to Earth,
the relativistic velocity addition equation gives
vLE =
0.800 c + 0.600 c
vLS + vSE
=
= + 0.946 c
v ⋅v
1 + LS 2 SE 1 + ( 0.800 c )( 0.600 c )
c
c2
(b) The length contracted distance of separation as measured in the ship’s frame of
reference is
L = Lp 1 − ( v c ) = ( 0.200 ly ) 1 − ( 0.600 ) = 0.160 ly
2
2
(c) The aliens observe the 0.160-ly distance closing because the probe nibbles into it
from one end at 0.800 c , and Earth reduces it from the other end at 0.600 c . Thus,
time =
0.160 ly
( 0.160 yr ) c = 0.114 yr
=
0.800 c + 0.600 c
1.40 c
(d) KE = ( γ − 1)mc 2
2


1
m

22
=
− 1  ( 4.00 × 10 5 kg )  3.00 × 108
 = 7.50 × 10 J
 1 − ( 0.946 )2

s




388
CHAPTER 26
26.52
The kinetic energy gained by the electron will equal the loss of potential energy, so
KE = q ( ∆V ) = e ( 1.02 MV ) = 1.02 MeV
(a) If Newtonian mechanics remained valid, then KE =
1
mv 2 , and the speed attained
2
would be
v=
2 ( 1.02 MeV ) ( 1.60 × 10 −13 J MeV )
2 KE
m
=
= 5.99 × 108
≈ 2c
m
9.11 × 10 -31 kg
s
(b) KE = ( γ − 1) ER , so γ = 1 +
KE
1.02 MeV
= 1+
= 3.00
ER
0.511 MeV
The actual speed attained is
v = c 1 − 1 γ 2 = c 1 − 1 ( 3.00 ) = 0.943 c
2
26.53
(a) When at rest, muons have a mean lifetime of ∆tp = 2.2 µ s . In a frame of reference
where they move at v = 0.95 c , the dilated mean lifetime of the muons will be
τ = γ ( ∆t p ) =
∆tp
1 − (v c)
2
=
2.2 µ s
1 − ( 0.95 )
2
= 7.0 µ s
(b) In a frame of reference where the muons travel at v = 0.95 c , the time required to
travel 3.0 km is
t=
d
3.0 × 10 3 m
=
= 1.05 × 10 −5 s = 10.5 µ s
8
v 0.95 ( 3.00 × 10 m s )
If N 0 = 5.0 × 10 4 muons started the 3.0 km trip, the number remaining at the end is
N = N 0 e − t τ = ( 5.0 × 10 4 ) e − 10.5 µ s 7.0 µ s = 1.1 × 10 4
Relativity
26.54
389
(a) An observer at rest relative to the mirror sees the light travel a distance D = 2d − x ,
where x = vt is the distance the ship moves toward the mirror in time t . Since this
observer agrees that the speed of light is c , the time for it to travel distance D is
D 2 d − vt
.
t= =
c
c
Solving for t yields
t=
2d
c+v
(b) The observer in the rocket sees a length-contracted initial distance to the mirror of
v2
and the mirror moving toward the ship at speed v. Thus, he measures
c2
the distance the light travels as D = 2 ( L − y ) , where y = vt 2 is the distance the
L = d 1−
mirror moves toward the ship before the light reflects off it.
This observer also measures the speed of light to be c, so the time for it to travel
D 2
v 2 vt 
distance D is t = =  d 1 − 2 − 
c c 
c
2 
26.55
2d c − v
c c+v
t=
Solving for t and simplifying yields
The students are to measure a proper time of ∆tp = T0 on the clock at rest relative to
them. The professor will measure a dilated time given by
( )
∆t = γ ∆tp = T + t ,
or
T0
1 − (v c)
2
=T +t
(1)
where T is the time she should wait before sending the light signal, and t is the transit
time for the signal, both measured in Earth’s frame of reference.
The distance, measured in Earth’s frame, the signal must travel to reach the receding
students is d = v ( T + t ) , and the transit time is
t=
d v
= (T + t )
c c
or
t=
(v c)T
1 − (v c)
(2)
390
CHAPTER 26
Substituting equation (2) into (1) yields
T0
1 − (v c)
2

vc 
T
= T 1 +
=
1− v c  1− v c

1

1− v c
Thus, T = T0 
2

 1 − ( v c )
26.56
or
T = T0
1− v c
1+ v c
The work required equals the increase in the gravitational potential energy, or
GMSun m
W=
. If this is to equal the rest energy of the mass removed, then
Rg
mc 2 =
Rg =
26.57
2


2
(1 − v c )
=T 

0

 ( 1 − v c )( 1 + v c ) 

GMSun m
Rg
( 6.67 × 10
Rg =
or
−11
GMSun
c2
N ⋅ m 2 kg 2 )( 1.99 × 10 30 kg )
( 3.00 × 10
8
m s)
2
= 1.47 × 10 3 m = 1.47 km
(a) The components of length, measured in the frame moving with the rod, are
Lpx = L0 cosθ 0
and
Lpy = L0 sin θ 0
The stationary observer will see a length contracted component in the direction
parallel to the motion, with the other component unaffected. Therefore,
Lx = Lpx 1 − ( v c ) = L0 1 − ( v c ) cosθ 0
2
2
and
Ly = Lpy = L0 sin θ 0
The length of the rod, as measured by the stationary observer, is
L = L2x + L2y = L0 cos 2 θ 0 − ( v c ) cos 2 θ 0 + sin 2 θ 0
2
or L = L0 1 − ( v c ) cos 2 θ 0
2
(b) The orientation angle seen by the stationary observer is given by
tan θ =
Ly
Lx
=
L0 sin θ 0
L0 1 − ( v c ) cosθ 0
2
, or tan θ = γ tan θ 0
Relativity
26.58
The speed of light in a vacuum, expressed in km h , is
m  1 km   3 600 s 

9
c =  3.00 × 108
 3 
 = 1.08 × 10 km h
s  10 m  1 h 

At the speed limit, the momentum is
plim =
mvlim
1 − ( vlim c )
2
m ( 90.0 km h )
=
Similarly, at v = 190 km h ,
1 − ( 90.0 km h 1.08 × 109 km h )
2
= m ( 90.0 km h )
p = m ( 190 km h ) , and with the fine proportional to the
excess momentum  Fine = k ( p − plim ) with k = constant  , we find that
k=
Fine
$80.0
$80.0
=
=
p − plim m ( 190 km h − 90.0 km h ) ( 100 m ) km h
(a) When v = 1 090 km h ,
p=
m ( 1 090 km h )
1 − ( 1 090 km h )
2
(1.08 × 10
9
km h )
2
= m ( 1 090 km h )
and the fine is


$80.0
 m 1 090 km h − 90.0 km h )  = $800
Fine = k ( p − plim ) = 
 ( 100 m ) km h   (


(b) When v = 1 000 000 090 km h
p=
m ( 1 000 000 090 km h )
1 − ( 1 000 000 090 km h )
2
(1.08 × 10
9
km h )
2
= m ( 2.65 × 109 km h )
and the fine is


$80.0
 m 2.65 × 109 km h − 90.0 km h )  = $2.12 × 109
Fine = 

 ( 100 m ) km h   (


391
392
CHAPTER 26
26.59
According to Earth-based observers, the times required for the two trips are
For Speedo:
TS =
L0 20.0 yr
=
= 21.05 yr
vS 0.950 c
For Goslo:
TG =
L0 20.0 yr
=
= 26.67 yr
vG 0.750 c
Thus, after Speedo lands, he must wait and age at the same rate as planet-based
observers, for an additional ∆TS = TG − TS = ( 26.67 − 21.05 ) yr = 5.614 yr before Goslo
arrives.
The time required for the trip according to Speedo’s internal biological clock (which
measures the proper time for his aging process during the trip) is
T0 S =
TS
γ
= TS 1 − ( vS c ) = ( 21.05 yr ) 1 − ( 0.950 ) = 6.574 yr
2
2
When Goslo arrives, Speedo has aged a total of
∆tS = T0 S + ∆TS = 6.574 yr + 5.614 = 12.19 yr
The time required for the trip according to Goslo’s internal biological clock (and hence
the amount he ages) is
∆tG = T0G =
TG
γ
= TG 1 − ( vG c ) = ( 26.67 yr ) 1 − ( 0.750 ) = 17.64 yr
2
2
Thus, we see that when he arrives, Goslo is older than Speedo, having aged an
additional
∆tG − ∆tS = 17.64 yr − 12.19 yr = 5.45 yr