39
Relativity
CHAPTER OUTLINE
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
and the Relativistic Form of
Newton’s Laws
39.8 Relativistic Energy
39.9 Mass and Energy
39.10 The General Theory of
Relativity
ANSWERS TO QUESTIONS
39.1
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 would
be measured to be scrunched in.
Q39.3
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.
Q39.4
The clock in orbit runs slower. No, they are not synchronized.
Although they both tick at the same rate after return, a time
difference 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 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
(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, 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.
Q39.8
Nothing physically unusual. An observer riding on the clock does not think that you are really
strange, 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 he
turns around and begins his trip home.
Q39.10
According to p = γmu , doubling the speed u will make the momentum of an object increase by the
factor 2
LM c − u OP
N c − 4u Q
2
2
2
2
12
.
Q39.11
As the object approaches the speed of light, its kinetic energy grows without limit. It would take an
infinite 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
E
without limit, its speed approaches the speed of light and its momentum approaches .
c
Q39.13
Recall that when a spring of force constant k is compressed or stretched from its relaxed position a
1
distance x, it stores elastic potential energy U = kx 2 . According to the special theory of relativity,
2
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
U
by an amount 2 . The fractional change is typically unobservably small for a mechanical spring.
c
Q39.14
You see no change in your reflection at any speed you can attain. You cannot attain the speed of
light, 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 to
give it momentum.
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.
Q39.18
The point of intersection moves to the right. To state the problem precisely, let us assume that each
of the two cards moves toward the other parallel to the long dimension of the picture, with velocity
2v
= 2 v cot φ , where φ is the
of magnitude v. The point of intersection moves to the right at speed
tan φ
small angle between the cards. As φ approaches zero, cot φ approaches infinity. Thus the point of
intersection can move with a speed faster than c if v is sufficiently large and φ sufficiently small. For
example, take v = 500 m s and φ = 0.000 19° . If you are worried about holding the cards steady
enough to be sure of the angle, cut the edge of one card along a curve so that the angle will
necessarily 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 circular
screen of radius R. The linear speed of the spot on the screen is given by v = ω R , where ω is the
angular speed of rotation of the flashlight. With sufficiently large ω and R, the speed of the spot
moving 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 point
of intersection: in the first case, the intersection of two cards and in the second case, the intersection
of a light beam with a screen. A point of intersection is not made of matter so it has no mass, and
hence 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 so
soon as the intersection point reaches him. The second bug would have to wait for sound or light to
travel 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 the
intersecting-cards method. Can you get a visible dot to run across a computer screen faster than
light? Want’a see it again?
Q39.19
In this case, both the relativistic and Galilean treatments would yield the same result: it is that the
experimentally 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 and
mass predicts that each molecule of the hotter object will, on average, have a larger mass than those
in the cooler object. This implies that given the same net applied force, the cooler object would have
a larger acceleration than the hotter object would experience. In a controlled experiment, the
difference will likely be too small to notice.
Q39.21
Special relativity describes inertial reference frames: that is, reference frames that are not
accelerating. 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 stronger
gravitational field than the upstairs clock.
Q39.23
The ants notice that they have a stronger sense of being pushed outward when they venture closer
to the rim of the merry-go-round. If they wish, they can call this the effect of a stronger gravitational
field produced by some mass concentration toward the edge of the disk. An ant named Albert
figures out that the strong gravitational field makes measuring rods contract when they are near the
rim of the disk. He shows that this effect precisely accounts for the discrepancy.
SOLUTIONS TO PROBLEMS
Section 39.1
P39.1
The Principle of Galilean Relativity
In the rest frame,
pi = m1 v1i + m 2 v 2i = 2 000 kg 20.0 m s + 1 500 kg 0 m s = 4.00 × 10 4 kg ⋅ m s
b
g
b
b
gb
g
g b
gb
g
p f = m1 + m 2 v f = 2 000 kg + 1 500 kg v f
Since pi = p f ,
vf =
4.00 × 10 4 kg ⋅ m s
= 11.429 m s .
2 000 kg + 1 500 kg
In the moving frame, these velocities are all reduced by +10.0 m/s.
b
g
v1′ i = v1i − v ′ = 20.0 m s − +10.0 m s = 10.0 m s
b
g
v ′2i = v 2i − v ′ = 0 m s − +10.0 m s = −10.0 m s
b
g
v ′f = 11.429 m s − +10.0 m s = 1.429 m s
Our initial momentum is then
b
gb
g b
b
gb
gb
g
pi′ = m1 v1′ i + m 2 v ′2i = 2 000 kg 10.0 m s + 1 500 kg −10.0 m s = 5 000 kg ⋅ m s
and our final momentum is
b
g
g
p ′f = 2 000 kg + 1 500 kg v ′f = 3 500 kg 1.429 m s = 5 000 kg ⋅ m s .
436
P39.2
P39.3
Relativity
(a)
v = vT + v B = 60.0 m s
(b)
v = vT − v B = 20.0 m s
(c)
v = vT2 + v B2 = 20 2 + 40 2 = 44.7 m s
The first observer watches some object accelerate under applied forces. Call the instantaneous
velocity of the object v 1 . The second observer has constant velocity v 21 relative to the first, and
measures the object to have velocity v 2 = v 1 − v 21 .
dv 2 dv 1
=
.
dt
dt
This is the same as that measured by the first observer. In this nonrelativistic case, they measure the
same forces as well. Thus, the second observer also confirms that ∑ F = ma .
a2 =
The second observer measures an acceleration of
P39.4
The laboratory observer notes Newton’s second law to hold:
F1 = ma 1
(where the subscript 1 implies the measurement was made in the laboratory frame of reference). The
observer in the accelerating frame measures the acceleration of the mass as a 2 = a 1 − a ′
(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 the
laboratory frame of reference). If Newton’s second law held for the accelerating frame, that observer
would then find valid the relation
F2 = ma 2
or
F1 = ma 2
(since F1 = F2 and the mass is unchanged in each). But, instead, the accelerating frame observer will
find that F2 = ma 2 − ma ′ 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 = Lp
Taking L =
P39.6
∆t =
v=c
p
Lp
2
where L p = 1.00 m gives
∆t p
b g
1− v c
2 12
so
v=c
∆t = 2 ∆t p
p
p
L F ∆t I OP
v = c M1 − G
MN H ∆t JK PQ
L F ∆t I OP
v = c M1 − G
MN H 2 ∆t JK PQ
2
= c 1−
1
= 0.866 c .
4
2 12
p
.
2 12
For
FLI
1−G J
HL K
F L 2I
1−G
H L JK
2
v2
1− 2
c
p
p
LM
N
= c 1−
1
4
OP
Q
12
= 0.866 c .
Chapter 39
P39.7
(a)
1
γ =
b g
1− v c
2
=
a
1
1 − 0.500
f
2
=
437
2
3
The time interval between pulses as measured by the Earth observer is
2 60.0 s
∆t = γ∆t p =
= 0.924 s .
3 75.0
60.0 s min
= 64.9 min .
Thus, the Earth observer records a pulse rate of
0.924 s
FG
H
(b)
IJ
K
At a relative speed v = 0.990 c , the relativistic factor γ 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.
*P39.8
(a)
The 0.8 c and the 20 ly are measured in the Earth frame,
x 20 ly 20 ly 1c
∆t = =
=
= 25.0 yr .
so in this frame,
v 0.8 c
0.8 c 1 ly yr
(b)
We see a clock on the meteoroid moving, so we do not measure proper time; that clock
measures proper time.
25.0 yr
∆t
∆t p =
=
= 25.0 yr 1 − 0.8 2 = 25.0 yr 0.6 = 15.0 yr
∆t = γ∆t p :
γ 1 1 - v2 c2
a 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
Lp
20 ly
20 ly
=
=
= 12.0 ly .
L=
γ
1 1 − 0.8 2 1.667
Method two: The tourist sees the Earth approaching at 0.8 c
0.8 ly yr 15 yr = 12.0 ly .
b
gb
g
Not only do distances and times differ between Earth and meteoroid reference frames, but
within the Earth frame apparent distances differ from actual distances. As we have
interpreted it, the 20-lightyear actual distance from the Earth to the meteoroid at the time of
discovery must be a calculated result, different from the distance measured directly. Because
of the finite maximum speed of information transfer, the astronomer sees the meteoroid as it
was years previously, when it was much farther away. Call its apparent distance d. The time
d
required for light to reach us from the newly-visible meteoroid is the lookback time t = .
c
The astronomer calculates that the meteoroid has approached to be 20 ly away as it moved
with constant velocity throughout the lookback time. We can work backwards to reconstruct
her calculation:
0.8 cd
d = 20 ly + 0.8 ct = 20 ly +
c
0.2d = 20 ly
d = 100 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 material
emanating from quasars, because they happen to be pointed nearly toward the Earth. If we
can watch events unfold on the meteoroid, we see them slowed by relativistic time dilation,
but also greatly speeded up by the Doppler effect.
438
P39.9
P39.10
Relativity
∆t p
∆t = γ∆t p =
∆t p =
so
1 − v2 c2
F
GG
H
I F v I
J ∆t ≅ GH1 − 2 c JK ∆t
c JK
F v I ∆t .
=G
H 2c JK
1−
v2
2
2
2
2
and
∆t − ∆t p
If
v = 1 000 km h =
then
v
= 9.26 × 10 −7
c
and
e∆t − ∆t j = e4.28 × 10 jb3 600 sg = 1.54 × 10
For
2
1.00 × 10 6 m
= 277.8 m s
3 600 s
−13
p
−9
v
= 0.990 , γ = 7.09
c
(a)
The muon’s lifetime as measured in the Earth’s rest frame is
∆t =
4.60 km
0.990 c
and the lifetime measured in the muon’s rest frame is
∆t p =
(b)
P39.11
∆t
γ
=
L = Lp 1 −
LM
MN
1
4.60 × 10 3 m
7.09 0.990 3.00 × 10 8 m s
FG v IJ
H cK
2
e
=
Lp
γ
=
OP
=
j PQ
2.18 µs .
4.60 × 10 3 m
= 649 m
7.09
The spaceship is measured by the Earth observer to be length-contracted to
L = Lp 1 −
v2
c2
or
F
GH
L2 = L2p 1 −
v2
c
2
I.
JK
Also, the contracted length is related to the time required to pass overhead by:
L = vt
or
v2
L2 = v 2 t 2 =
v2
c
2
af
Equating these two expressions gives
L2p − L2p
or
L2p + ct
Using the given values:
L p = 300 m
this becomes
e1.41 × 10
giving
v = 0.800 c .
c2
af
= ct
v2
2
c
5
actf
2
m2
2
2
.
v2
c2
= L2p .
and
j vc
2
2
t = 7.50 × 10 −7 s
= 9.00 × 10 4 m 2
s = 1.54 ns .
Chapter 39
P39.12
(a)
439
The spaceship is measured by Earth observers to be of length L, where
L = Lp 1 −
v2
c2
v∆ t = L p 1 −
F
v G ∆t
H
2
Solving for v,
2
L2p
+
c
2
v2
c2
and
L = v∆t
and
v 2 ∆t 2 = L2p 1 −
F
GH
I =L
JK
2
p
The tanks move nonrelativistically, so we have v =
(c)
For the data in problem 11,
v=
a
f
c 300 m
e3 × 10
8
ms
j e0.75 × 10 sj + a300 mf
2
−6
cL p
v=
(b)
2
2
c 2 ∆t 2 + L2p
a
f
c 300 m
=
a f
m sj a75 sf + a300 mf
c 300 m
e3 × 10
8
2
2
2
225 2 + 300 2 m
a
c 300 m
=
.
300 m
= 4.00 m s .
75 s
in agreement with problem 11. For the data in part (b),
v=
I
JK
v2
.
c2
e2.25 × 10 j
10 2
f
= 0.800 c
= 1.33 × 10 −8 c = 4.00 m s
+ 300 2 m
in agreement with part (b).
P39.13
GMm
We find Cooper’s speed:
r2
=
mv 2
.
r
L GM OP = LM e6.67 × 10 je5.98 × 10 j OP
v=M
N aR + hf Q MN e6.37 × 10 + 0.160 × 10 j PQ
2π a R + hf 2π e6.53 × 10 j
T=
=
= 5. 25 × 10 s .
12
Solving,
−11
24
6
6
12
= 7.82 km s .
6
Then the time period of one orbit,
(a)
v
7.82 × 10
LF v I
The time difference for 22 orbits is ∆t − ∆t = bγ − 1g∆t = MG 1 − J
MNH c K
F 1 v − 1I a22T f = 1 F 7.82 × 10 m s I 22e5.25 × 10 sj =
∆t − ∆t ≈ G 1 +
2 GH 3 × 10 m s JK
H 2 c JK
2
p
p
P39.14
γ =
2
8
For one orbit, ∆t − ∆t p =
1
e
1 − v2 c2
j
= 1.01
p
3
2
(b)
3
3
2
−1 2
OP
PQa f
− 1 22T
2
3
39.2 µs .
39.2 µs
= 1.78 µs . The press report is accurate to one digit .
22
so
v = 0.140 c
440
P39.15
*P39.16
Relativity
(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 = Lp 1 −
from which
L 19.0 m
v2
=
= 0.950 = 1 − 2 and v = 0.312 c .
L p 20.0 m
c
v2
c2
In the Earth frame, Speedo’s trip lasts for a time
∆t =
20.0 ly
∆x
=
= 21.05 yr .
v
0.950 ly yr
Speedo’s age advances only by the proper time interval
∆t p =
∆t
γ
= 21.05 yr 1 − 0.95 2 = 6.574 yr during his trip.
Similarly for Goslo,
∆t p =
20.0 ly
∆x
v2
1− 2 =
1 − 0.75 2 = 17.64 yr .
v
0.750 ly yr
c
While Speedo has landed on Planet X and is waiting for his brother, he ages by
20.0 ly
20.0 ly
−
= 5.614 yr .
0.750 ly yr 0.950 ly yr
b
g
Then Goslo ends up older by 17.64 yr − 6.574 yr + 5.614 yr = 5.45 yr .
P39.17
(a)
∆t = γ∆t p =
∆t p
=
15.0 yr
= 21.0 yr
(b)
b g 1 − a0.700f
d = va ∆t f = 0.700 c b 21.0 yr g = a0.700fb1.00 ly yr g b 21.0 yr g =
(c)
The astronauts see Earth flying out the back window at 0.700 c :
e j
1− v c
2
b
2
g a
fb
gb
14.7 ly
g
d = v ∆t p = 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 years
after the battery stops powering the transmitter, 14.7 ly away:
21.0 yr + 14.7 yr = 35.7 yr
Chapter 39
P39.18
The orbital speed of the Earth is as described by
v=
GmS
=
r
e6.67 × 10
−11
∑ F = ma :
je
N ⋅ m 2 kg 2 1.99 × 10 30 kg
1.496 × 10
11
m
GmS m E
r2
j = 2.98 × 10
4
=
441
mE v 2
r
m s.
The maximum frequency received by the extraterrestrials is
f obs = fsource
1+v c
= 57.0 × 10 6 Hz
1−v c
e
j 1 − ee2.98 × 10
j e3.00 × 10
m sj e3.00 × 10
1 + 2.98 × 10 4 m s
4
8
8
j = 57.005 66 × 10
m sj
6
Hz .
j = 56.994 34 × 10
m sj
6
Hz .
ms
The minimum frequency received is
f obs = fsource
1−v c
= 57.0 × 10 6 Hz
1+v c
e
j 1 + ee2.98 × 10
j e3.00 × 10
m sj e3.00 × 10
1 − 2.98 × 10 4 m s
4
8
8
ms
The difference, which lets them figure out the speed of our planet, is
b57.005 66 − 56.994 34g × 10
P39.19
(a)
6
Hz = 1.13 × 10 4 Hz .
Let f c be the frequency as seen by the car. Thus,
f c = f source
and, if f is the frequency of the reflected wave,
f = fc
(c)
which gives
a c + vf .
a c − vf
f a c − vf = f
a c + vf
b f − f gc = b f + f g v ≈ 2 f
The beat frequency is then
f beat = f − fsource =
Using the above result,
f beat
λ=
(d)
c+v
.
c−v
f = f source
Combining gives
(b)
c+v
c−v
v=
source
source
a2fb30.0 m sge10.0 × 10
=
8
3.00 × 10 m s
c
fsource
f beat λ
2
=
3.00 × 10 8 m s
10.0 × 10 9 Hz
so
∆v =
9
Hz
j = a2fb30.0 m sg = 2 000 Hz =
b0.030 0 mg
source
2 f source v
2v
.
=
λ
c
2.00 kHz
= 3.00 cm
a
fb
g
source v .
5 Hz 0.030 0 m
∆f beat λ
=
= 0.075 0 m s ≈ 0.2 mi h
2
2
442
P39.20
Relativity
(a)
When the source moves away from an observer, the observed frequency is
f obs = fsource
FG c − v IJ
Hc+v K
12
s
where v s = v source .
s
When v s << c , the binomial expansion gives
FG c − v IJ
Hc+v K
s
12
12
s
f obs ≈ fsource
So,
LM FG v IJ OP LM1 + FG v IJ OP
N H c KQ N H c KQ
FG 1 − v IJ .
H cK
s
= 1−
FG
H
−1 2
s
≈ 1−
vs
2c
IJ FG 1 − v IJ ≈ FG1 − v IJ .
K H 2c K H c K
s
s
s
The observed wavelength is found from c = λ obs fobs = λfsource :
λ obs =
λf source
λfsource
λ
≈
=
f obs
1 − vs c
f source 1 − v s c
b
∆λ = λ obs − λ = λ
Since 1 −
(b)
*P39.21
∆λ
vs
≈1,
c
v source = c
λ
s
c
I F v cI
JK GH 1 − v c JK
−1 = λ
s
s
v source
.
c
FG ∆λ IJ = cFG 20.0 nm IJ =
H λ K H 397 nm K
0.050 4c
For the light as observed
1+v c
1+v c
c
c
=
fobs =
fsource =
λ obs
1−v c
1 − v c λ source
1 + v c λ source 650 nm
=
=
λ obs
520 nm
1− v c
v
v
= 1.562 − 1.562
c
c
v = 0.220 c = 6.59 × 10 7 m s
1+
Section 39.5
*P39.22
≈
F 1
GH 1 − v
g
1+v c
= 1.25 2 = 1.562
1−v c
v 0.562
=
= 0. 220
c 2.562
The Lorentz Transformation Equations
Let Suzanne be fixed in reference from S and see the two light-emission events with coordinates
x 1 = 0 , t1 = 0 , x 2 = 0 , t 2 = 3 µs . Let Mark be fixed in reference frame S ′ and give the events
coordinate x 1′ = 0 , t1′ = 0 , t 2′ = 9 µs .
(a)
Then we have
v
t 2′ = γ t 2 − 2 x 2
c
FG
H
1
9 µs =
v2
c
(b)
2
=
IJ
K
1− v
2
c
2
b 3 µs − 0 g
8
9
b
1−
v2
c
2
=
1
3
v = 0.943 c
g e
x 2′ = γ x 2 − vt 2 = 3 0 − 0.943 c × 3 × 10 −6 s
F
jGH 3 × 10c
8
ms
I=
JK
2.55 × 10 3 m
443
Chapter 39
P39.23
1
γ =
1− v
2
c
2
1
=
= 10.0
1 − 0.995 2
We are also given: L1 = 2.00 m , and θ = 30.0° (both measured in a
reference frame moving relative to the rod).
a fa f
= a 2.00 mfa0.500 f = 1.00 m
Thus,
L1 x = L1 cos θ 1 = 2.00 m 0.867 = 1.73 m
and
L1 y = L1 sin θ 1
L 2 x is a proper length, related to L1 x by L1 x =
L2x
.
γ
= 10.0L1 x = 17.3 m
Therefore,
L2x
and
L 2 y = L1 y = 1.00 m .
FIG. P39.23
(Lengths perpendicular to the motion are unchanged).
*P39.24
bL g + eL j
(a)
L2 =
(b)
θ 2 = tan −1
2
2x
2
2y
L2y
L2x
gives
L 2 = 17.4 m
gives
θ 2 = 3.30°
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
FG
H
t′ = γ t −
v
c
2
IJ
K
x =
1
1 − 0.8
2
FG 0 − 0.8 c a100 mfIJ = 1.667FG − 80 m IJ = −4.44 × 10
H c
K
H 3 × 10 m s K
2
8
−7
s.
The time elapsing before A occurs is 444 ns .
P39.25
(a)
From the Lorentz transformation, the separations between the blue-light and red-light
events are described by
a
∆x ′ = γ ∆x − v∆t
v=
(b)
2.00 m
8.00 × 10
−9
s
f
0 = γ 2.00 m − v 8.00 × 10 −9 s
= 2.50 × 10 8 m s
γ =
e
j
1
e
8
1 − 2.50 × 10 m s
a
f
j e3.00 × 10
2
8
ms
j
2
= 1.81 .
Again from the Lorentz transformation, x ′ = γ x − vt :
e
je
x ′ = 1.81 3.00 m − 2.50 × 10 8 m s 1.00 × 10 −9 s
j
x ′ = 4.97 m .
(c)
F v xIJ :
t′ = γ G t −
H c K
2
LM
t ′ = 1.81 1.00 × 10
MM
N
−9
t ′ = −1.33 × 10 −8 s
e2.50 × 10
s−
e3.00 × 10
8
8
j a3.00 mfOP
PP
m sj
Q
ms
2
444
Relativity
Section 39.6
P39.26
The Lorentz Velocity Transformation Equations
u x = Enterprise velocity
v = Klingon velocity
From Equation 39.16
u x′ =
ux − v
1 − ux v c
2
=
0.900 c − 0.800 c
= 0.357 c .
1 − 0.900 0.800
a
fa
f
FIG. P39.26
P39.27
u ′x =
ux − v
1 − ux v c 2
=
−0.750 c − 0.750 c
= −0.960 c
1 − −0.750 0.750
a
fa
f
FIG. P39.27
*P39.28
Let frame S be the Earth frame of reference. Then
The components of the velocity of the first spacecraft are
v = −0.7 c .
u x = 0.6 c cos 50° = 0.386 c
and
uy
a f
= a0.6 c f sin 50° = 0.459 c .
As measured from the S ′ frame of the second spacecraft,
The magnitude of u ′ is
a f = 1.086 c = 0.855 c
1.27
1−u v c
1 − a0.386 c fa −0.7 c f c
u
0.459 c 1 − a0.7 f
0.459 ca0.714f
=
=
= 0.258 c
u′ =
1.27
γ e1 − u v c j 1 − a0.386 fa −0.7f
a0.855 cf + a0.285cf = 0.893c
and its direction is at
tan −1
u ′x =
ux − v
x
2
0.386 c − −0.7 c
=
2
2
y
y
x
Section 39.7
P39.29
(a)
2
2
2
0. 258 c
= 16.8° above the x ′ -axis .
0.855 c
Relativistic Linear Momentum and the Relativistic Form of Newton’s Laws
p = γ mu ; for an electron moving at 0.010 0 c,
1
γ =
b g
1− u c
Thus,
e
2
1
=
p = 1.00 9.11 × 10 −31
p = 2.73 × 10 −24 kg ⋅ m s .
(b)
(c)
= 1.000 05 ≈ 1.00 .
b g
kg jb0.010 0 ge3.00 × 10
1 − 0.010 0
Following the same steps as used in part (a),
we find at 0.500 c, γ = 1.15 and
p = 1.58 × 10 −22 kg ⋅ m s .
At 0.900 c, γ = 2.29 and
p = 5.64 × 10 −22 kg ⋅ m s .
2
8
ms
j
Chapter 39
P39.30
p=
we find the difference ∆p from the classical momentum, mu:
∆p = γ mu − mu = γ − 1 mu .
b g
The difference is 1.00% when γ − 1 mu = 0.010 0γ mu :
thus 1 −
(b)
FG u IJ = a0.990f
H cK
2
2
γ =
b g
FG u IJ = a0.900f
H cK
2
2
γ =
γ −1=
1
−1 ≈1+
b g
p − mu 1 F 90.0 m s I
= G
2 H 3.00 × 10 m s JK
mu
1− u c
2
FG IJ
HK
mu
b g
1− u c
1−
becomes
2
P39.33
c2
1 = u2
which gives:
or c 2 = u 2
u2
Fm c
GH p
2 2
2
I
JK
+1
b g
1
1
=
0.990
1− u c
b g
1
1
=
0.900
1− u c
b g
−1=
2
2
FG IJ
HK
1 u
2 c
2
2
= 4.50 × 10 −14
m 2u 2
p2
Fm
GH p
2
2
+
1
c2
I
JK
c
u=
and
=
2
1 u
2 c
8
p=
2
u = 0.436 c .
and
p − mu γ mu − mu
=
=γ −1:
mu
mu
b g
1− u c
u = 0.141c .
, and
The difference is 10.0% when γ − 1 mu = 0.100γ mu :
thus 1 −
P39.32
= γ mu
Using the relativistic form,
(a)
P39.31
mu
445
em c
2 2
j
p2 + 1
.
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 = p1
or
γ 2 m 2u 2 = γ 1 m1u1 =
a
1 − 0.893
e1.67 × 10 kgju = e4.960 × 10
1 − bu c g
F 1.67 × 10 u I = 1 − u
GH 4.960 × 10 c JK
c
−27
or
2.50 × 10 −28 kg
2
2
2
Proceeding to solve, we find
−27
−28
12.3
u22
c2
2
2
2
2
2
= 1 and u 2 = 0.285 c .
f
2
−28
a
× 0.893 c
j
kg c .
f
446
Relativity
Section 39.8
P39.34
Relativistic Energy
b
g
∆E = γ 1 − γ 2 mc 2
mc 2 = 0.511 MeV
For an electron,
F 1 − 1 I mc = 0.582 MeV
GH a1 − 0.810f a1 − 0.250f JK
F
I
1
1
(b)
∆E = G
−
GH 1 − a0.990f 1 − 0.810 JJK mc = 2.45 MeV
I F
F
I
1
1
J
G
JJ mc
G
− 1J mc −
∑W = K − K = G
G
GH 1 − dv ci JK H 1 − bv cg K
I
F
1
1
JJ mc
G
−
or ∑ W = G
GH 1 − dv ci 1 − bv cg JK
(a)
2
∆E =
2
2
P39.35
f
2
i
2
2
2
i
f
2
2
2
i
f
(a)
∑W =
F
GG
H
∑W =
(b)
P39.36
a
1
1 − 0.750
f
2
−
I
J e1.673 × 10
1 − a0.500 f JK
−27
kg 2.998 × 10 8 m s
−
I
J e1.673 × 10
1 − a0.500 f JK
−27
kg 2.998 × 10 8 m s
1
2
je
j
2
je
j
2
5.37 × 10 −11 J
F
H
∑ W = GG
1 − 0.995
∑W =
1.33 × 10 −9 J
a
1
f
2
1
2
The relativistic kinetic energy of an object of mass m and speed u is K r =
For u = 0.100 c ,
The classical equation K c =
different by
1
mu 2 gives
2
Kr =
F
GH
Kc =
1
m 0.100 c
2
1
1 − 0.010 0
a
f
2
F
GG
H
1 − u2 c 2
I
JK
= 0.005 000mc 2
I
JJ
K
− 1 mc 2 .
− 1 mc 2 = 0.005 038mc 2 .
0.005 038 − 0.005 000
= 0.751% .
0.005 038
For still smaller speeds the agreement will be still better.
1
Chapter 39
P39.37
E = γ mc 2 = 2mc 2 or
FG IJ
H K
1
u
Thus, = 1 −
γ
c
γ =2.
2
3
or
2
=
u=
c 3
.
2
F c 3 I = F mc I 3
GH 2 JK GH c JK
F 938.3 MeV IJ 3 = 1.63 × 10
p=G
H c K
2
p = γ mu = 2m
The momentum is then
MeV
.
c
3
P39.38
447
K=
(a)
Using the classical equation,
(b)
Using the relativistic equation, K =
LM
K=M
MMN
b
ge
1
1
mv 2 = 78.0 kg 1.06 × 10 5 m s
2
2
F
GG
H
1
b g
1− v c
2
j
2
= 4.38 × 10 11 J .
I
JJ
K
− 1 mc 2 .
OP
− 1Pb78.0 kg ge 2.998 × 10 m sj = 4.38 × 10 J
PPQ
1 − e1.06 × 10 2.998 × 10 j
LM1 − FG v IJ OP ≈ 1 + 1 FG v IJ .
v
When << 1 , the binomial series expansion gives
c
2 H cK
MN H c K PQ
LM1 − FG v IJ OP − 1 ≈ 1 FG v IJ .
Thus,
2H cK
MN H c K PQ
1 F vI
1
and the relativistic expression for kinetic energy becomes K ≈ G J mc = mv . That is, in
H
K
2 c
2
1
2
8
11
8 2
5
2 −1 2
2
2 −1 2
2
2
2
2
the limit of speeds much smaller than the speed of light, the relativistic and classical
expressions yield the same results.
P39.39
e
je
(a)
ER = mc 2 = 1.67 × 10 −27 kg 2.998 × 10 8 m s
(b)
E = γ mc 2 =
1.50 × 10 −10 J
b
1 − 0.950 c c
2
= 1.50 × 10 −10 J = 938 MeV
= 4.81 × 10 −10 J = 3.00 × 10 3 MeV
K = E − mc 2 = 4.81 × 10 −10 J − 1.50 × 10 −10 J = 3.31 × 10 −10 J = 2.07 × 10 3 MeV
(c)
P39.40
g
2 12
j
The relativistic density is
ER
2
c V
=
mc 2 m
= =
c 2V V
eL jeL jLMNL
p
p
m
8.00 g
=
O
1 − bu c g P a1.00 cmf
Q
2
p
3
a
1 − 0.900
f
2
= 18.4 g cm 3 .
448
P39.41
Relativity
We must conserve both energy and relativistic momentum of the
system of fragments. With subscript 1 referring to the 0.868 c
particle and subscript 2 to the 0.987 c particle,
1
1
γ1 =
= 2.01 and γ 2 =
= 6.22 .
2
2
1 − 0.868
1 − 0.987
a
a
f
f
Conservation of energy gives E1 + E 2 = Etotal
which is
γ 1 m1 c 2 + γ 2 m 2 c 2 = m total c 2
or
2.01m1 + 6. 22m 2 = 3.34 × 10 −27 kg .
This reduces to:
m1 + 3.09m 2 = 1.66 × 10 −27 kg .
(1)
Since the final momentum of the system must equal zero, p1 = p 2
γ 1 m1u1 = γ 2 m 2 u 2
gives
or
a2.01fa0.868cfm = a6.22fa0.987cfm
which becomes
m1 = 3.52m 2 .
1
(2)
Solving (1) and (2) simultaneously, m1 = 8.84 × 10
*P39.42
Energy conservation:
3 108 kg 1 −
1
1 − 02
1 452 kg 1 −
P39.43
a
1 − 0.85 2
900 kg 0.85 c
1 − 0.85
2
−28
900 kgc 2
1 400 kgc 2 +
v2
= M.
c2
Momentum conservation: 0 +
FIG. P39.41
2
f=
kg and m 2 = 2.51 × 10 −28 kg .
Mc 2
=
1 − v2 c2
Mv
1 − v2 c2
v 2 Mv
=
.
c
c2
v 1 452
=
= 0.467
c 3 108
v = 0.467 c .
(a)
Dividing gives
(b)
Now by substitution 3 108 kg 1 − 0.467 2 = M = 2.75 × 10 3 kg .
E = γ mc 2
e
p = γ mu
p = bγ mug
j
F u IF u I
= eγ mc j − bγ mug c = γ F emc j − amc f u I = emc j G 1 − J G 1 − J = emc j
H
K
H c KH c K
E 2 = γ mc
2
2 2
E −p c
2 2
2
2
2 2
2 2
2
2 2
2
2
2 2
2
2
2
2
Q.E.D.
P39.44
(a)
a f
Thus, γ =
(b)
b g
q ∆V = K = γ − 1 m e c 2
b g
1
b g
1− u c
2
=1+
a f e
a f from which
q ∆V
mec 2
je
u = 0.302 c .
j
K = γ − 1 m e c 2 = q ∆V = 1.60 × 10 −19 C 2.50 × 10 4 J C = 4.00 × 10 −15 J
−1
2 2
Chapter 39
P39.45
(a)
E = γ mc 2 = 20.0 GeV with mc 2 = 0.511 MeV for electrons. Thus,
γ =
*P39.46
449
20.0 × 10 9 eV
0.511 × 10 6 eV
1
= 3.91 × 10 4 .
= 3.91 × 10 4 from which u = 0.999 999 999 7 c
(b)
γ =
(c)
L = Lp 1 −
(a)
P=
(b)
The kinetic energy of one electron with v = 0.999 9 c is
b g
1− u c
2
FG u IJ
H cK
2
=
Lp
γ
=
3.00 × 10 3 m
= 7.67 × 10 −2 m = 7.67 cm
4
3.91 × 10
energy
2J
=
= 2.00 × 10 13 W
∆t
100 × 10 −15 s
F
bγ − 1gmc = GG
H
1
2
1 − 0.999 9
2
I
JJ
K
e
− 1 9.11 × 10 −31 kg 3 × 10 8 m s
j
2
e
= 69.7 8.20 × 10 −14 J
j
= 5.72 × 10 −12 J
Then we require
N=
P39.47
0.01
2 J = N 5.72 × 10 −12 J
100
e
j
2 × 10 −4 J
= 3.50 × 10 7 .
5.72 × 10 −12 J
Conserving total momentum of the decaying particle system,
p before decay = p after decay = 0
b
g
p v = p µ = γm µ u = γ 207 m e u .
Conservation of mass-energy for the system gives Eµ + E v = Eπ : γ m µ c 2 + p v c = mπ c 2
b g pc = 273m .
u
γ b 207m g + γ b 207m g = 273m
c
v
γ 207m e +
Substituting from the momentum equation above,
FG
H
IJ
K
u
273
=
= 1.32 :
c
207
1+u c
= 1.74
1−u c
or
γ 1+
Then,
K µ = γ − 1 m µ c 2 = γ − 1 207 m e c 2 :
b g
b g e
j
e
e
e
u
= 0.270 .
c
Kµ =
F
GG
H
a
1
1 − 0.270
f
2
e
I
JJ a
K
− 1 207 0.511 MeV
K µ = 4.08 MeV .
Also,
Ev = Eπ − Eµ :
b
g
Ev = mπ c 2 − γ m µ c 2 = 273 − 207γ m e c 2
F
GG
H
Ev = 273 −
I
J a0.511 MeVf
1 − a0.270 f JK
Ev = 29.6 MeV
207
2
f
450
*P39.48
Relativity
3c
3c
and u 2 = − . Let observer B be at
4
4
3c
v = u1 = .
4
Let observer A hold the unprimed reference frame, with u1 =
rest in the primed frame with u1′ = 0 =
(a)
Then u 2′ =
u2 − v
1 − u2 v c
speed = u2′ =
(b)
2
=
u1 − v
1 − u1 v c 2
− 3c 4 − 3c 4
−1.5 c
=
1 − − 3 c 4 + 3 c 4 1 + 9 16
b
gb
g
3 c 2 24
=
c = 0.960 c .
25 16 25
In the unprimed frame the objects, each of mass m, together have energy
mc 2
γ mc 2 + γ mc 2 = 2
= 3.02mc 2 .
2
1 − 0.75
mc 2
In the primed frame the energy is
2
mc 2
+
2
= 4.57mc 2 , greater by
1−0
1 − 0.96
4.57mc 2
= 1.51 times greater as measured by observer B .
3.02mc 2
Section 39.9
P39.49
P39.50
Mass and Energy
Let a 0.3-kg flag be run up a flagpole 7 m high.
e
j
We put into it energy
mgh = 0.3 kg 9.8 m s 2 7 m ≈ 20 J .
So we put into it extra mass
∆m =
for a fractional increase of
2 × 10 −16 kg
~ 10 −15 .
0.3 kg
E
c2
=
20 J
e3 × 10
8
ms
j
2
= 2 × 10 −16 kg
E = 2.86 × 10 5 J . Also, the mass-energy relation says that E = mc 2 .
Therefore,
m=
2.86 × 10 5 J
E
=
c2
3.00 × 10 8 m s
e
j
2
= 3.18 × 10 −12 kg .
No, a mass loss of this magnitude (out of a total of 9.00 g) could not be detected .
jb
e
ge
j
a f e
je
e
j
9
7
E P ∆t 0.800 1.00 × 10 J s 3.00 yr 3.16 × 10 s yr
=
=
= 0.842 kg
2
c2
c2
3.00 × 10 8 m s
P39.51
∆m =
P39.52
1 030 kg m3 1.40 × 10 9 10 3 m 4 186 J kg⋅° C 10.0 ° C
ρVc ∆T
E mc ∆T
∆m = 2 =
=
=
2
c
c2
c2
3.00 × 10 8 m s
a f
∆m = 6.71 × 10 8 kg
je
e
jb
ga
3
j
f
Chapter 39
P39.53
P=
3.77 × 10 26 J s
dm
=
= 4.19 × 10 9 kg s
2
8
dt
3.00 × 10 m s
e
j
2m e c 2 = 1.02 MeV
Section 39.10
*P39.55
e j
2
dE d mc
dm
=
= c2
= 3.77 × 10 26 W
dt
dt
dt
Thus,
P39.54
(a)
451
Eγ ≥ 1.02 MeV
The General Theory of Relativity
GM E m
∑ F = ma :
For the satellite
r2
=
FG
H
mv 2 m 2π r
=
r
r
T
GM ET 2 = 4π 2 r 3
F 6.67 × 10
r =G
GH
jb
e
−11
N ⋅ m 2 5.98 × 10 24 kg 43 080 s
kg 2 4π 2
g IJ
JK
2
IJ
K
2
13
r = 2.66 × 10 7 m
e
j
7
2π r 2π 2.66 × 10 m
=
= 3.87 × 10 3 m s
43 080 s
T
(b)
v=
(c)
The small fractional decrease in frequency received is equal in magnitude to the fractional
increase in period of the moving oscillator due to time dilation:
I
F
1
J
G
fractional change in f = −bγ − 1g = − G
− 1J
GH 1 − e3.87 × 10 3 × 10 j JK
F 1 L F 3.87 × 10 I OI
= 1 − G 1 − M− G
GH 2 MN H 3 × 10 JK PPQJJK = −8.34 × 10
3
3
8 2
2
−11
8
(d)
The orbit altitude is large compared to the radius of the Earth, so we must use
GM E m
.
Ug = −
r
∆U g = −
e
j
6.67 × 10 −11 Nm 2 5.98 × 10 24 kg m
= 4.76 × 10
2
7
kg 2.66 × 10 m
7
+
J kg m
∆f ∆U g 4.76 × 10 7 m 2 s 2
= +5. 29 × 10 −10
=
=
2
8
f
mc 2
3 × 10 m s
e
(e)
j
−8.34 × 10 −11 + 5.29 × 10 −10 = +4.46 × 10 −10
e
j
6.67 × 10 −11 Nm 5.98 × 10 24 kg m
6
kg 6.37 × 10 m
452
Relativity
Additional Problems
P39.56
(a)
d earth = vt earth = vγ t astro
FG IJ e
H K
v2
v
1− 2 =
1.50 × 10 −5
c
c
v2
1 = 2 1 + 2.25 × 10 −10
c
v
= 1 − 1.12 × 10 −10
c
e
P39.57
F
GG
H
j
I
JJ
K
K=
(c)
6.00 × 10 27 J = 6.00 × 10 27
(a)
10 13 MeV = γ − 1 m p c 2
1− v
c
2
e
30.0 yr
j
−1 2
=1−
1
2.25 × 10 −10
2
e
F 2.00 × 10 yr − 1I b1 000gb1 000 kg g 3 × 10 m s
e
j
GH 30 yr
JK
F 13¢ IJ FG k IJ FG W ⋅ s IJ FG h IJ = $2.17 × 10
JG
H kWh K H 10 K H J K H 3 600 s K
6
8
b g
= 6.00 × 10 27 J
γ = 10 10
so
t′ =
t
γ
=
10 5 yr
10 10
= 10 −5 yr ~ 10 2 s
(b)
d ′ = ct ′ ~ 10 8 km
(a)
When K e = K p ,
mec 2 γ e − 1 = mp c 2 γ p − 1 .
In this case,
m e c 2 = 0.511 MeV , m p c 2 = 938 MeV
and
γ e = 1 − 0.750
Substituting,
γ p =1+
but
γp=
b
2
g
e
a
f
2 −1 2
j
= 1.511 9 .
g = 1 + a0.511 MeVfb1.511 9 − 1g = 1.000 279
b
me c 2 γ e − 1
mp c
2
938 MeV
1
LM1 − eu cj OP
N
Q
2 12
.
p
(b)
Therefore,
u p = c 1 − γ p−2 = 0.023 6 c .
When p e = p p
γ p m pu p = γ e m eu e or γ pu p =
Thus,
γ p up =
γ e m eu e
.
mp
b1.511 9ge0.511 MeV c ja0.750 cf = 6.177 2 × 10
2
and
which yields
up
c
j
20
3
vp ≈ c
P39.58
c2
c2
v
= 1 + 2.25 × 10 −10
c
so
2
−10
2
c2
− 1 mc 2 =
(b)
2
6
2
j
1
e2.00 × 10 yrjc = v 1 − v1
v e 2.25 × 10 j
v
=
1−
so
938 MeV c
= 6.177 2 × 10 −4 1 −
Fu I
GH c JK
p
2
2
u p = 6.18 × 10 −4 c = 185 km s .
−4
c
Chapter 39
P39.59
(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 ′ =
(c)
L = Lp 1 −
Lp
u x′
=
1.80 × 10 12 m
e
8
0.800 3.00 × 10 m s
v2
c2
e
j
453
= 7.50 × 10 3 s
a0.600cf
j
= 1.80 × 10 12 m 1 −
2
= 1.44 × 10 12 m
c2
Since v = 0.600 c and u ′x = −0.800 c , the velocity Jim measures for the ball is
ux =
(d)
(a)
(b)
P39.61
*P39.62
1 + u ′x v c
2
=
a−0.800 cf + a0.600 cf =
1 + a −0.800 fa0.600f
−0.385 c .
Jim measures the ball and Mary to be initially separated by 1.44 × 10 12 m . Mary’s motion at
0.600c and the ball’s motion at 0.385c 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 =
*P39.60
u ′x + v
1.44 × 10 12 m
e
0.985 3.00 × 10 8 m s
j
= 4.88 × 10 3 s .
b
fb
ga
g
The charged battery stores energy
E = P t = 1.20 J s 50 min 60 s min = 3 600 J
so its mass excess is
∆m =
−14
kg
∆m 4.00 × 10
=
= 1.60 × 10 −12
−3
m
25 × 10 kg
a
f
3 600 J
E
=
2
c
3 × 10 8 m s
e
j
2
= 4.00 × 10 −14 kg .
too small to measure.
∆mc 2 4 938.78 MeV − 3 728.4 MeV
=
× 100% = 0.712%
4 938.78 MeV
mc 2
a
f
e
The energy of the first fragment is given by E12 = p12 c 2 + m1 c 2
E1 = 2.02 MeV .
a
For the second, E22 = 2.00 MeV
(a)
2
2
2
2
2
E2 = 2.50 MeV .
Energy is conserved, so the unstable object had E = 4.52 MeV . Each component of
momentum is conserved, so the original object moved with
2
2
1.75 MeV
2.00 MeV
+
. Then for it
p 2 = p x2 + p y2 =
c
c
2
3.65 MeV
2
2
2
4.52 MeV = 1.75 MeV + 2.00 MeV + mc 2
m=
.
c2
a
(b)
f + a1.50 MeVf
j = a1.75 MeVf + a1.00 MeVf
FG
H
f a
IJ
K
f
FG
H
a
Now E = γ mc 2 gives 4.52 MeV =
IJ
K
f e j
1
1 − v2 c2
3.65 MeV
1−
v2
= 0.654 , v = 0.589 c .
c2
454
P39.63
Relativity
(a)
Take the spaceship as the primed frame, moving toward the right at v = +0.600 c .
Then u x′ = +0.800 c , and
Lp
ux =
u ′x + v
b g
1 + u′x v c
b
=
2
0.800 c + 0.600 c
= 0.946 c .
1 + 0.800 0.600
a
fa
g 1 − a0.600f
2
f
(b)
L=
(c)
The aliens observe the 0.160-ly distance closing because the probe nibbles into it from one
end at 0.800c and the Earth reduces it at the other end at 0.600c.
L = 0.200 ly
:
γ
time =
Thus,
(d)
K=
F
GG
H
1
2
1−u c
2
I
JJ
K
− 1 mc 2 :
K=
F
GG
H
= 0.160 ly
0.160 ly
= 0.114 yr .
0.800 c + 0.600 c
a
1
1 − 0.946
f
2
I
JJ e
K
je
− 1 4.00 × 10 5 kg 3.00 × 10 8 m s
j
2
K = 7.50 × 10 22 J
P39.64
In this case, the proper time is T0 (the time measured by the students on a clock at rest relative to
∆t = γT0
them). The dilated time measured by the professor is:
where ∆t = T + t Here T is the time she waits before sending a signal and t is the time required for
the signal to reach the students.
T + t = γ T0 .
Thus, we have:
(1)
To determine the travel time t, realize that the distance the students will have moved beyond the
professor before the signal reaches them is:
d = v T+t .
The time required for the signal to travel this distance is:
Solving for t gives:
Substituting this into equation (1) yields:
a f
d F vI
t = = G J aT + t f .
c H cK
bv cgT .
t=
1 − b v cg
bv cgT = γ T
T+
1 − bv cg
0
T
= γ T0 .
1− v c
or
Then T = T0
b g =T
1 − ev c j
1− v c
2
2
0
b g
1 + b v c g 1 − b v cg
1− v c
= T0
b g
1 + b v cg
1− v c
.
Chapter 39
P39.65
455
Look at the situation from the instructors’ viewpoint since they are at rest relative to the clock, and
hence measure the proper time. The Earth moves with velocity v = −0.280 c relative to the
instructors while the students move with a velocity u ′ = −0.600 c relative to Earth. Using the velocity
addition equation, the velocity of the students relative to the instructors (and hence the clock) is:
u=
a
a
f a
fa
f
f
−0. 280 c − 0.600 c
v + u′
=
= −0.753 c (students relative to clock).
1 + vu ′ c 2 1 + −0. 280 c −0.600 c c 2
(a)
With a proper time interval of ∆t p = 50.0 min , the time interval measured by the students is:
∆t = γ ∆t p
with
γ =
1
a
1 − 0.753 c
f
2
c2
= 1.52 .
a
f
Thus, the students measure the exam to last T = 1.52 50.0 min = 76.0 minutes .
(b)
The duration of the exam as measured by observers on Earth is:
∆t = γ ∆t p with γ =
P39.66
1
a
1 − 0. 280 c
f
2
c2
a
f
so T = 1.04 50.0 min = 52.1 minutes .
The energy which arrives in one year is
e
je
j
E = P ∆t = 1.79 × 10 17 J s 3.16 × 10 7 s = 5.66 × 10 24 J .
Thus,
P39.67
m=
5.66 × 10 24 J
E
=
c2
3.00 × 10 8 m s
e
j
2
= 6.28 × 10 7 kg .
The observer measures the proper length of the tunnel, 50.0 m, but measures the train contracted to
length
L = Lp 1 −
shorter than the tunnel by
P39.68
v2
= 100 m 1 − 0.950
c2
a
f
2
= 31.2 m
50.0 − 31.2 = 18.8 m so it is completely within the tunnel. .
If the energy required to remove a mass m from the surface is equal to its rest energy mc 2 ,
then
and
GM s m
= mc 2
Rg
Rg =
GM s
c
2
e6.67 × 10
=
−11
je
N ⋅ m 2 kg 2 1.99 × 10 30 kg
e3.00 × 10
R g = 1.47 × 10 3 m = 1.47 km .
8
ms
j
2
j
456
P39.69
Relativity
(a)
At any speed, the momentum of the particle is given by
mu
p = γ mu =
2
−1 2
2
dp
:
dt
Since F = qE =
.
b g
OP
d L F
u I
qE = MmuG 1 − J
dt M H
PQ
c K
N
F u I du + 1 muF 1 − u I FG 2u IJ du .
qE = mG 1 − J
H c K dt 2 GH c JK H c K dt
L
O
qE du M 1 − u c + u c P
=
m dt M 1 − u c
j PPQ
MN e
du qE F
u I
=
a=
1− J
.
G
dt m H
c K
1− u c
2
−1 2
2
2
2
2
2
So
2
2
32
2
qE
. As u
m
approaches c, the acceleration approaches zero, so that the object can never reach the speed
of light.
(b)
For u small compared to c, the relativistic expression reduces to the classical a =
(c)
ze
0
du
1 − u2 c
j
2 3 2
z
t
x = udt = qEc
0
(a)
=
z
z
t
qE
dt
m
t =0
t
u=
tdt
2 2
2
x=
2 2
m c +q E t
0
qEct
2 2
m c + q2E2t 2
c
qE
FH
m 2 c 2 + q 2 E 2 t 2 − mc
IK
An observer at rest relative to the mirror sees the light travel a distance D = 2d − x , where
x = vtS is the distance the ship moves toward the mirror in time tS . Since this observer
agrees that the speed of light is c, the time for it to travel distance D is
tS =
(b)
2
2 3 2
2
u
P39.70
2
2
and
−3 2
D 2d − vtS
=
c
c
tS =
2d
.
c+v
The observer in the rocket measures a length-contracted initial distance to the mirror of
L = d 1−
v2
c2
and the mirror moving toward the ship at speed v. Thus, he measures the distance the light
vt
is the distance the mirror moves toward the ship before
travels as D = 2 L − y where y =
2
the light reflects from it. This observer also measures the speed of light to be c, so the time
for it to travel distance D is:
b
t=
LM
MN
g
OP
PQ a f
2d
D 2
v 2 vt
=
d 1− 2 −
so c + v t =
c
2
c c
c
ac + vfac − vf or t =
2d
c
c−v
.
c+v
Chapter 39
*P39.71
Take the two colliding protons as the system
E1 = K + mc 2
E2 = mc 2
E12 = p12 c 2 + m 2 c 4
p2 = 0 .
2
In the final state, E f = K f + Mc :
E 2f = p 2f c 2 + M 2 c 4 .
By energy conservation, E1 + E2 = E f , so
E12 + 2E1 E2 + E22 = E 2f
e
j
p12 c 2 + m 2 c 4 + 2 K + mc 2 mc 2 + m 2 c 4
= p 2f c 2 + M 2 c 4
By conservation of momentum,
p1 = p f .
Then
M 2 c 4 = 2Kmc 2 + 4m 2 c 4 =
Mc 2 = 2mc 2 1 +
K
2mc 2
.
4Km 2 c 4
+ 4m 2 c 4
2mc 2
FIG. P39.71
By contrast, for colliding beams we have
E1 = K + mc 2
In the original state,
E2 = K + mc 2 .
In the final state,
E f = Mc 2
E1 + E2 = E f :
K + mc 2 + K + mc 2 = Mc 2
FG
H
Mc 2 = 2mc 2 1 +
*P39.72
K
2mc 2
IJ .
K
Conservation of momentum γ mu :
mu
+
1 − u2 c 2
a f
m −u
3 1 − u2 c 2
Mv f
=
1 − v 2f c 2
=
2mu
3 1 − u2 c 2
.
Conservation of energy γ mc 2 :
mc 2
+
1 − u2 c 2
mc 2
3 1 − u2 c 2
Mc 2
=
1 − v 2f c 2
To start solving we can divide: v f =
M
1−u
M=
2
4c
2
=
4m
3 1−u
2
c
2
=
b1 2g
=
4mc 2
3 1 − u2 c 2
.
2u u
= . Then
4 2
M
4 − u2 c 2
2m 4 − u 2 c 2
3 1 − u2 c 2
Note that when v << c , this reduces to M =
4m
, in agreement with the classical result.
3
457
458
P39.73
Relativity
(a)
L20 = L20 x + L20 y and
L2 = L2x + L2y .
The motion is in the x direction: L y = L0 y = L0 sin θ 0
FG v IJ = bL cosθ g 1 − FG v IJ .
H cK
H cK
L F vI O
L F vI
L = L cos θ M1 − G J P + L sin θ = L M1 − G J
MN H c K PQ
MN H c K
L F vI
O .
L = L M1 − G J cos θ P
MN H c K
PQ
2
L x = L0 x 1 −
2
Thus,
2
0
2
0
P39.74
L0 y
2
0
0
0
Ly
0
2
2
or
2
2
0
2
0
2
cos 2 θ 0
OP
PQ
12
2
0
(b)
tan θ =
(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 blast
waves 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 be
simultaneous. We measure the distance we have traveled from the Sun as
Lx
L = Lp 1 −
=
b g
L0 x 1 − v c
2
= γ tan θ 0
FG v IJ = b6.00 lyg 1 − a0.800f
H cK
2
2
= 3.60 ly .
We see the Sun flying away from us at 0.800c while the light from the Sun approaches at
1.00c. Thus, the gap between the Sun and its blast wave has opened at 1.80c, and the time
we calculate to have elapsed since the Sun exploded is
3.60 ly
= 2.00 yr .
1.80 c
We see Tau Ceti as moving toward us at 0.800c, while its light approaches at 1.00c, only
0.200c faster. We measure the gap between that star and its blast wave as 3.60 ly and
growing at 0.200c. We calculate that it must have been opening for
3.60 ly
= 18.0 yr
0.200 c
and conclude that Tau Ceti exploded 16.0 years before the Sun .
Chapter 39
P39.75
Since the total momentum is zero before decay, it is necessary that after the decay
p nucleus = p photon =
Eγ
=
c
14.0 keV
.
c
e j with mc = 8.60 × 10 J = 53.8 GeV .
0 keV I
JK + 1 .
emc + K j = a14.0 keV f + emc j or FGH 1 + mcK IJK = FGH 14.mc
K
F 14.0 keV IJ ≈ 1 + 1 FG 14.0 keV IJ (Binomial Theorem)
= 1+G
1+
H mc K
K
2 H mc
mc
a14.0 keVf = e14.0 × 10 eVj = 1.82 × 10 eV .
K ≈=
2mc
2e53.8 × 10 eV j
Also, for the recoiling nucleus, E 2 = p 2 c 2 + mc 2
2
2
Thus,
2
2
2 2
2
2
P39.76
2
2
2
2
2
3
2
and
2
2
2
So
−9
2
−3
2
9
Take m = 1.00 kg .
Kc =
The classical kinetic energy is
and the actual kinetic energy is
af
u
c
0.000
Kc J
0.000
0.100 0.045 × 10 16
0.200 0.180 × 10 16
0.300 0.405 × 10 16
0.400 0.720 × 10 16
0.500
0.600
0.700
0.800
0.900
0.990
1.13 × 10 16
1.62 × 10 16
2.21 × 10 16
2.88 × 10 16
3.65 × 10 16
4.41 × 10 16
Kr
FG IJ = e4.50 × 10
HK
I
1
− 1J mc = e9.00 × 10
JK
1 − bu c g
1
1
u
mu 2 = mc 2
2
2
c
F
=G
GH
2
16
2
2
16
jFGH uc IJK
F 1
JjG
GH 1 − bu cg
2
J
af
Kr J
0.000
0.0453 × 10 16
0.186 × 10 16
0.435 × 10 16
0.820 × 10 16
1.39 × 10 16
2.25 × 10 16
3.60 × 10 16
FIG. P39.76
6.00 × 10 16
11.6 × 10 16
54.8 × 10 16
FG IJ
HK
1 u
K c = 0.990K r , when
2 c
2
LM
= 0.990 M
NM
OP
− 1P , yielding u =
1 − bu c g
QP
1
2
Similarly,
K c = 0.950K r
when
u = 0.257 c
and
K c = 0.500K r
when
u = 0.786 c .
0.115 c .
2
I
− 1J .
JK
459
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
P39.4
see the solution
P39.46
(a) 20.0 TW; (b) 3.50 × 10 7 electrons
P39.6
0.866 c
P39.48
P39.8
(a) 25.0 yr; (b) 15.0 yr; (c) 12.0 ly
(a) 0.960c; (b) 1.51 times greater as
measured by B.
P39.10
(a) 2.18 µs ; (b) The moon sees the planet
surface moving 649 m up toward it.
P39.50
3.18 × 10 −12 kg , not detectable
P39.52
6.71 × 10 8 kg
P39.54
1.02 MeV
P39.56
(a)
P39.12
(a)
cL p
2
c ∆t
2
+ L2p
; (b) 4.00 m s ;
(c) see the solution
v
= 1 − 1.12 × 10 −10 ; (b) 6.00 × 10 27 J ;
c
(c) $2.17 × 10 20
P39.14
v = 0.140 c
P39.16
5.45 yr, Goslo is older
P39.58
(a) 0.023 6 c ; (b) 6.18 × 10 −4 c
P39.18
11.3 kHz
P39.60
(a) 4.00 × 10 −14 kg ; (b) 1.60 × 10 −12
P39.20
(a) see the solution; (b) 0.050 4 c
P39.62
(a)
P39.22
(a) 0.943 c ; (b) 2.55 km
P39.64
see the solution
P39.24
B occurred 444 ns before A
P39.66
6.28 × 10 7 kg
P39.68
1.47 km
P39.70
(a)
P39.72
M=
P39.74
(a) Tau Ceti exploded 16.0 yr before the
Sun; (b) they exploded simultaneously
P39.76
see the solution, 0.115 c , 0.257 c , 0.786 c
P39.26
0.357 c
P39.28
0.893 c at 16.8° above the x ′ -axis
P39.30
(a) 0.141c ; (b) 0.436 c
P39.32
see the solution
P39.34
(a) 0.582 MeV ; (b) 2.45 MeV
P39.36
see the solution
P39.38
(a) 438 GJ ; (b) 438 GJ
P39.40
18.4 g cm 3
P39.42
(a) 0.467c; (b) 2.75 × 10 3 kg
3.65 MeV
; (b) v = 0.589 c
c2
2d
2d
; (b)
c+v
c
2m
3
c−v
c+v
4 − u2 c 2
1 − u2 c2