THE SPECIAL THEORY OF RELATIVITY
26
Responses to Questions
1.
No. Since the windowless car in an exceptionally smooth train moving at a constant velocity is an
inertial reference frame and the basic laws of physics are the same in all inertial reference frames, there
is no way for you to tell if you are moving or not. The first postulate of the special theory of relativity
can be phrased as “No experiment can tell you if an inertial reference frame is at rest or moving
uniformly at constant velocity.”
2.
The fact that you instinctively think you are moving is consistent with the relativity principle applied to
mechanics. Even though you are at rest relative to the ground, when the car next to you creeps forward,
you are moving backward relative to that car.
3.
The ball will land back in the worker’s hand. Both the ball and the car are already moving forward
(relative to the ground), so when the ball is thrown straight up into the air with respect to the car, it will
continue to move forward at the same rate as the car and fall back down to land in his hand.
4.
Whether you say that the Earth goes around the Sun or the Sun goes around the Earth depends on your
reference frame. It is valid to say either one, depending on which frame you choose. The laws of
physics, though, won’t be the same in each of these reference frames, since the Earth is accelerating as
it goes around the Sun. The Sun is nearly an inertial reference frame, but the Earth is not.
5.
The starlight would pass you at a speed of c. The speed of light has the same speed in any reference
frame in empty space, according to the second postulate of special relativity.
6.
The clocks are not at fault and they are functioning properly. Time itself is actually measured to pass
more slowly in moving reference frames when compared to a rest frame. Any measurement of time
(heartbeats or decay rates, for instance) would be measured as slower than normal when viewed by an
observer outside the moving reference frame.
7.
Time actually passes more slowly in the moving reference frame, including aging and other life
processes. It is not just that it seems this way—time has been measured to pass more slowly in the
moving reference frame, as predicted by special relativity.
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26-1
26-2
Chapter 26
8.
This is an example of the “twin paradox.” This situation would be possible if the woman was traveling
at high enough speeds during her trip. Time would have passed more slowly for her so she would have
aged less than her son, who stayed on Earth. (Note that the situations of the woman and son are not
symmetric; she must undergo acceleration during her journey.)
9.
You would not notice a change in your own heartbeat, mass, height, or waistline. No matter how fast
you are moving relative to Earth, you are at rest in your own reference frame. Thus, you would not
notice any changes in your own characteristics. To observers on Earth, you are moving away at 0.6c,
which gives γ = 1.25. If we assume that you are standing up, so that your body is perpendicular to the
direction of motion, then to the observers on Earth, it would appear that your heartbeat has slowed by a
factor of 1/1.25 = 0.80, that your mass has increased by a factor of 1.25, and that your waistline has
decreased by a factor of 0.80 (all due to the relativity equations for time dilation, mass increase, and
length contraction), but that your height would be unchanged (since there is no relative motion
between you and Earth in that direction). Also see the section of the text on “rest mass and relativistic
mass” for comments about mass change and relativity.
10.
Yes. However, at a speed of only 90 km/h, υ /c is very small, and therefore γ is very close to 1, so the
effects would not be noticeable.
11.
Length contraction and time dilation would not occur. If the speed of light were infinite, then υ /c would
be zero for all finite values of υ , and therefore γ would always be 1, resulting in Δt = Δt0 and A = A 0 .
12.
Both the length contraction and time dilation formulas include the term 1 − υ 2 /c 2 . If c were not the
limiting speed in the universe, then it would be possible to have a situation with υ > c. However, this
would result in a negative number under the square root, which gives an imaginary number as a result,
indicating that c must be the limiting speed. Also, assuming that the relativistic formulas were still
correct, as υ gets very close to c, an outside observer should be able to show that A = A o 1 − υ 2 /c 2 is
getting smaller and smaller and that the limit as υ → c is A → 0. This would show that c is a limiting
speed, since nothing can get smaller than having a length of 0. A similar analysis for time dilation
Δto
shows that Δt =
is getting longer and longer and that the limit as υ → c is Δt → ∞. This
1 − υ 2 /c 2
would show that c is a limiting speed, since the slowest that time can pass is that it comes to a stop.
13.
If the speed of light were 25 m/s, then we would see relativistic effects all the time, something like the
Chapter-Opening Figure or Figure 26–12 with Question 14. Everything moving relative to us would
look length contracted, and time dilation would have to be taken into account for many events. There
would be no “absolute time” on which we would all agree, so it would be more difficult, for instance,
to plan to meet friends for lunch at a certain time. Many “twin paradox” kind of events would occur,
and the momentum of moving objects would become very large, making it very difficult to change
their motion. One of the most unusual changes for today’s modern inhabitants of Earth would be that
nothing would be able to move faster than 25 m/s, which is only about 56 mi/h.
14.
Mr Tompkins appears shrunk in the horizontal direction, since that is the direction of his motion, and
normal size in the vertical direction, perpendicular to his direction of motion. This length contraction is
a result of the fact that, to the people on the sidewalk, Mr Tompkins is in a moving frame of reference.
If the speed of light were only 20 mi/h, then the amount of contraction, which depends on γ, would be
enough to be noticeable. Therefore, Mr Tompkins and his bicycle appear very skinny. (Compare to the
Chapter-Opening Figure, which is shown from Mr Tompkins’ viewpoint. In this case, Mr Tompkins
sees himself as “normal,” but all the objects moving with respect to him are contracted.)
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The Special Theory of Relativity
26-3
mυ
15.
No. The relativistic momentum of the electron is given by p = γ mυ =
16.
No. Accelerating a particle with nonzero mass up to the speed of light would require an infinite amount
mc 2
of energy, since KE =
− mc 2 , so that is not possible.
2 2
1 − υ /c
17.
No, E = mc 2 does not conflict with the conservation of energy; it actually completes it. Since this
equation shows us that mass and energy are interconvertible, it says it is now necessary to include mass
as a form of energy in the analysis of physical processes.
18.
Yes. One way to describe the energy stored in the compressed spring is to say it is a mass increase
(although it would be so small that it could not be measured). This mass will convert back to energy
when the spring is uncompressed.
19.
Matter and energy are interconvertible (matter can be converted into energy and energy can be
converted into matter), thus we should say “energy can neither be created nor destroyed.”
20.
No, our intuitive notion that velocities simply add is not completely wrong. Our intuition is based on
our everyday experiences, and at these everyday speeds our intuition regarding how velocities add is
correct. Our intuition does break down, though, at very high speeds, where we have to take into
account relativistic effects. Relativity does not contradict classical mechanics, but it is a more general
theory, whereas classical mechanics is a limiting case.
. At low speeds
1 − υ 2 /c 2
(compared to c), this reduces to the classical momentum, p = mυ. As υ approaches c, γ approaches
infinity, so there is no upper limit to the electron’s momentum.
Responses to MisConceptual Questions
1.
(a)
Proper length is the length measured by a person at rest with the object measured. The ship’s
captain is at rest with the ship, so that measurement is the proper length.
2.
(c)
A common misconception is that the distance between the objects is important when measuring
relativistic effects. The important parameter is the relative velocity. The ship’s captain is at rest
with the flashlight and therefore measures the proper time. The space-dock personnel measure
the dilated time, which is always longer than the proper time. Since the dilated time is 1.00 s, the
proper time must be shorter, or 0.87 s.
3.
(c)
The proper time interval is measured at rest with respect to the flashlight, or the 0.87-s
measurement.
4.
(b)
The ship’s captain will have aged less. While the ship is moving at constant speed, both the
captain and the space-dock personnel record that the other’s clock is running slowly. However,
the ship must accelerate (and therefore change frames of reference) when it turns around. The
space-dock personnel do not accelerate; therefore, they are correct in their measurement that they
have aged more than the ship’s captain.
5.
(f)
A common misconception is that the person on the spacecraft would observe the change in her
measurement of time, or that the person on the spacecraft would see the Earth clocks running
fast. Actually, the person on the spacecraft would observe her time running normally. Since the
spacecraft and Earth are moving relative to each other, observers on the spacecraft and observers
on the Earth would both see the others’ clocks running slowly.
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26-4
Chapter 26
6.
(c)
It might be assumed that the speed of the ship should be added to the speed of light to obtain a
relative speed of 1.5c. However, the second postulate of relativity is that the speed of light is
1.0c, as measured by any observer.
7.
(d)
A common misconception is that the relativistic formulas are only valid for speeds close to the
speed of light. The relativistic formulas are always valid. However, for speeds much smaller than
the speed of light, time dilations, length contractions, and changes in mass are insignificant.
8.
(d, e) Due to relativistic effects, the observers will not necessarily agree on the time an event occurs,
the distance between events, the time interval between events, or the simultaneity of two events.
However, the postulates of relativity state that both observers agree on the validity of the laws of
physics and on the speed of light.
9.
(d)
It is common to think that one frame of reference is preferable to another. Due to relativistic
effects, observers in different frames of reference may make different time measurements.
However, each measurement is correct in that frame of reference.
10.
(e)
Even though the rocket ship is going faster and faster relative to some stationary frame, you
remain at rest relative to the rocket ship. Therefore, in your frame of reference, your mass,
length, and time remain the same.
11.
(d)
A common misconception is that there is a stationary frame of reference and that motion relative
to this frame can be measured. Special relativity demonstrates that a stationary frame does not
exist, but all motion is relative. If the spaceship has no means to observe the outside (and make a
reference to another object), then there is no way to measure your velocity.
12.
(b)
To an observer on Earth, the pendulum would appear to be moving slowly. Therefore, the period
would be longer than 2.0 seconds.
13.
(a)
A common misconception is to add the velocities together to obtain a speed of 1.5c. However,
due to relativistic effects, the relative speed between two objects must be less than the speed of
light. Since the objects are moving away from each other the speed must be greater than the
speed of each object, so it must be greater than 0.75c. The only option that is greater than 0.75c
and less than 1.0c is 0.96c.
Solutions to Problems
1.
You measure the contracted length. Find the rest length from Eq. 26–3a.
A0 =
2.
A
1 − υ 2 /c 2
=
44.2 m
1 − (0.850) 2
= 83.9 m
We find the lifetime at rest from Eq. 26–1a.
2
⎛ 2.70 × 108 m/s ⎞
Δt0 = Δt 1 − υ 2 /c 2 = (4.76 × 10−6 s) 1 − ⎜
= 2.07 × 10−6 s
⎜ 3.00 × 108 m/s ⎟⎟
⎝
⎠
3.
The measured distance is the contracted length. Use Eq. 26–3a.
2
⎛ 2.90 × 108 m/s ⎞
A = A 0 1 − υ 2 /c 2 = (135 ly) 1 − ⎜
= 34.6 ly ≈ 35 ly
⎜ 3.00 × 108 m/s ⎟⎟
⎝
⎠
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The Special Theory of Relativity
4.
The speed is determined from the time dilation relationship, Eq. 26–1a.
Δt0 = Δt 1 − υ 2 /c 2
→
2
⎛ 2.60 × 10−8
⎛ Δt ⎞
υ = c 1 − ⎜ 0 ⎟ = c 1 − ⎜⎜
−8
⎝ Δt ⎠
⎝ 4.40 × 10
5.
2
s⎞
8
⎟ = 0.807c = 2.42 × 10 m/s
s ⎟⎠
The speed is determined from the length contraction relationship, Eq. 26–3a.
2
A = A 0 1 − υ 2 /c 2
6.
26-5
2
⎛ A ⎞
⎛ 35 ly ⎞
8
→ υ = c 1− ⎜ ⎟ = c 1− ⎜
⎟ = 0.70c = 2.1× 10 m/s
⎝ 49 ly ⎠
⎝ A0 ⎠
The speed is determined from the length contraction relationship, Eq. 26–3a.
2
A = A 0 1 − υ 2 /c 2
7.
(a)
⎛ A ⎞
→ υ = c 1 − ⎜ ⎟ = c 1 − (0.900) 2 = 0.436c = 1.31× 108 m/s
⎝ A0 ⎠
We use Eq. 26–3a for length contraction with the contracted length 99.0% of the rest length.
2
⎛ A ⎞
→ υ = c 1 − ⎜ ⎟ = c 1 − (0.990) 2 = 0.141c
⎝ A0 ⎠
A = A 0 1 − υ 2 /c 2
(b)
We use Eq. 26–1a for time dilation with the time as measured from a relative moving frame
1.00% greater than the rest time.
Δt0 = Δt 1 − υ 2 /c 2
2
2
⎛ Δt ⎞
⎛ 1 ⎞
→ υ = c 1− ⎜ 0 ⎟ = c 1− ⎜
⎟ = 0.140c
⎝ 1.0100 ⎠
⎝ Δt ⎠
We see that a speed of 0.14c results in about a 1% relativistic effect.
8.
The speed is determined from the length contraction relationship, Eq. 26–3a. Then the time is found
from the speed and the contracted distance.
A = A 0 1 − υ 2 /c 2
→
2
⎛ A ⎞
A
υ = c 1− ⎜ ⎟ ; t = =
υ
⎝ A0 ⎠
A
⎛ A ⎞
c 1− ⎜ ⎟
⎝ A0 ⎠
2
25 ly
=
⎛ 25 ly ⎞
c 1− ⎜
⎟
⎝ 62 ly ⎠
2
=
(25 yr)c
= 27 yr
c(0.915)
That is the time according to you, the traveler. The time according to an observer on Earth would be as
follows.
t=
9.
(a)
A0
υ
=
(62 yr)c
= 68 yr
c(0.915)
The measured length is the contracted length. We find the rest length from Eq. 26–3a.
A0 =
A
1 − υ 2 /c 2
=
4.80 m
1 − (0.720) 2
= 6.92 m
Distances perpendicular to the motion do not change, so the rest height is 1.35 m .
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26-6
Chapter 26
(b)
The time in the spacecraft is the proper time, found from Eq. 26–1a.
Δt0 = Δt 1 − υ 2 /c 2 = (20.0 s) 1 − (0.720) 2 = 13.9 s
10.
(c)
To your friend, you moved at the same relative speed: 0.720c .
(d)
She would measure the same time dilation: 13.9 s .
(a)
To an observer on Earth, 21.6 ly is the rest length, so the time will be the distance divided by the
speed.
tEarth =
(b)
A0
υ
=
(21.6 ly)
= 22.74 yr ∼ 22.7 yr
0.950c
We find the proper time (the time that passes on the spacecraft) from Eq. 26–1a.
Δt0 = Δt 1 − υ 2 /c 2 = (22.74 yr) 1 − (0.950) 2 = 7.101 yr ≈ 7.10 yr
(c)
To the spacecraft observer, the distance to the star is contracted. Use Eq. 26–3a.
A = A 0 1 − υ 2 /c 2 = (21.6 ly) 1 − (0.950) 2 = 6.7446 ly ≈ 6.74 ly
(d)
To the spacecraft observer, the speed of the spacecraft is as follows:
υ=
11.
(a)
A (6.745 ly)
=
= 0.95c , as expected.
7.101 yr
Δt
In the Earth frame, the clock on the Enterprise will run slower. Use Eq. 26–1a.
Δt0 = Δt 1 − υ 2 /c 2 = (5.0 yr) 1 − (0.70) 2 = 3.6 yr
(b)
Now we assume that the 5.0 years is the time as measured on the Enterprise. Again use Eq. 26–1a.
Δt0 = Δt 1 − υ 2 /c 2
12.
→ Δt =
Δt0
2
1 − υ /c
2
=
(5.0 yr)
1 − (0.70) 2
= 7.0 yr
The dimension along the direction of motion is contracted, and the other two dimensions are
unchanged. Use Eq. 26–3a to find the contracted length.
A = A 0 1 − υ 2 /c 2 ; V = A(A 0 ) 2 = (A 0 )3 1 − υ 2 /c 2 = (2.6 m)3 1 − (0.80) 2 = 10.55 m3 ≈ 11 m3
13.
The change in length is determined from the length contraction relationship, Eq. 26–3a. The speed is
very small compared to the speed of light, so we use the binomial expansion.
A = A 0 1 − υ 2 /c 2
→
⎛ υ2
A
= 1 − υ 2 /c 2 = ⎜ 1 − 2
⎜ c
A0
⎝
1/2
⎞
⎟⎟
⎠
≈ 1 − 12
2
3
⎛
⎞
−10
1 11.2 × 10 m/s
=
−
1
⎜
⎟⎟ = 1 − 6.97 × 10
2⎜
2
8
c
⎝ 3.00 × 10 m/s ⎠
υ2
So the percent decrease is (6.97 × 10−8 )% .
14.
We find the speed of the particle in the lab frame and use that to find the rest frame lifetime and
distance.
υ=
Δxlab
1.00 m
=
= 2.941× 108 m/s = 0.9803c
Δtlab 3.40 × 10−9 s
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The Special Theory of Relativity
(a)
26-7
Find the rest frame lifetime from Eq. 26–1a.
Δt0 = Δtlab 1 − υ 2 /c 2 = (3.40 × 10−9 s) 1 − (0.9803)2 = 6.715 × 10−10 s ≈ 6.7 × 10−10 s
(b)
In its rest frame, the particle will travel the distance given by its speed and the rest lifetime.
Δx0 = υΔt0 = (2.941× 108 m/s)(6.715 × 10−10 s) = 0.20 m
This could also be found from the length contraction relationship: Δx0 =
15.
Δxlab
1 − υ 2 /c 2
.
In the Earth frame, the average lifetime of the pion will be dilated according to Eq. 26–1a. The speed
of the pion will be the distance moved in the Earth frame times the dilated time.
υ=
d
d
=
1 − υ 2 /c 2
Δt Δt0
υ =c
1
⎛ cΔt0 ⎞
1+ ⎜
⎟
⎝ d ⎠
2
→
1
=c
⎛ (3.00 × 108 m/s)(2.6 × 10−8 s) ⎞
1+ ⎜
⎟⎟
⎜
(32 m)
⎝
⎠
2
= 0.9716c
Note that the significant figure addition rule gives 4 significant figures for the value under the radical sign.
16.
The momentum of the proton is given by Eq. 26–4.
p = γ mυ =
17.
(a)
mυ
2
1 − υ /c
2
=
(1.67 × 10−27 kg)(0.68)(3.00 × 108 m/s)
= 4.6 × 10−19 kg ⋅ m/s
We compare the classical momentum to the relativistic momentum (Eq. 26–4).
pclassical
mυ
=
prelativistic ⎛
mυ
⎜
⎜
2 2
⎝ 1 − υ /c
(b)
1 − 0.68
2
⎞
⎟
⎟
⎠
= 1 − υ 2 /c 2 = 1 − (0.15) 2 = 0.989
The classical momentum is about 21.1% in error. The negative sign means that the classical
momentum is smaller than the relativistic momentum.
We again compare the two momenta.
mυ
mυ
pclassical
=
prelativistic
= 1 − υ 2 /c 2 = 1 − (0.75) 2 = 0.66
1 − υ 2 /c 2
The classical momentum is 234, in error.
18.
The momentum at the higher speed is to be twice the initial momentum. We designate the initial state
with the subscript 0 and the final state with the subscript f.
⎛
mυf
⎜
2 2
⎜
pf ⎝ 1 − υf /c
=
p0 ⎛
mυ0
⎜
⎜ 1 − υ 2 /c 2
0
⎝
⎞
⎛ υf2
⎞
⎟
⎜
⎟
2 2⎟
⎜
⎟
⎠ = 2 → ⎝ 1 − υ f /c ⎠ = 4 →
⎞
⎛ υ02
⎞
⎟
⎜⎜
⎟
2 2⎟
⎟
⎝ 1 − υ 0 /c ⎠
⎠
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26-8
Chapter 26
⎛ υf2
2 2
⎜⎜
⎝ 1 − υ f /c
19.
⎛ υ02
⎞
⎟⎟ = 4 ⎜⎜
2 2
⎠
⎝ 1 − υ 0 /c
⎞
⎡ (0.22c) 2 ⎤
⎛ 0.203 ⎞ 2
= 0.203c 2 → υf2 = ⎜
⎟ = 4⎢
⎟ c → υf = 0.41c
2⎥
⎟
⎝ 1.203 ⎠
⎢⎣1 − (0.22) ⎥⎦
⎠
The two momenta, as measured in the frame in which the particle was initially at rest, will be equal to
each other in magnitude. The lighter particle is designated with the subscript 1 and the heavier particle
with the subscript 2.
m1υ1
p1 = p2 →
υ12
(1 − υ12 /c2 )
1 − υ12 /c 2
⎛m ⎞
=⎜ 2 ⎟
⎝ m1 ⎠
m2υ2
=
2
1 − υ22 /c 2
υ22
→
2
(1 − υ22 /c2 )
⎛ 6.68 × 10−27 kg ⎞ ⎡ (0.60c) 2 ⎤
=⎜
= 9.0c 2
⎜ 1.67 × 10−27 kg ⎟⎟ ⎢⎢1 − (0.60) 2 ⎥⎥
⎝
⎠ ⎣
⎦
→
υ1 = 0.90c = 0.95c
20.
We find the proton’s momenta using Eq. 26–4.
mpυ1
p0.45 =
1−
p0.98 =
mp (0.45c)
1 − (0.45) 2
= 0.5039mp c;
p0.85 =
υ22
=
mp (0.98c)
1 − (0.98) 2
mpυ2
1−
c2
mpυ2
1−
21.
υ12
=
υ22
=
mp (0.85c)
1 − (0.85) 2
= 1.6136mp c
c2
= 4.9247 mp c
c2
(a)
⎛ 1.6136mp c − 0.5039mp c ⎞
⎛ p2 − p1 ⎞
⎟100 = 220.2 ≈ 220%
⎜
⎟100 = ⎜⎜
⎟
0.5039mp c
⎝ p1 ⎠
⎝
⎠
(b)
⎛ 4.9247mp c − 1.6136mp c ⎞
⎛ p2 − p1 ⎞
⎟100 = 205.2 ≈ 210%
⎜
⎟100 = ⎜⎜
⎟
1.6136mp c
⎝ p1 ⎠
⎝
⎠
The rest energy of the electron is given by Eq. 26–7.
E = mc 2 = (9.11× 10−31 kg)(3.00 × 108 m/s) 2 = 8.199 × 10−14 J ≈ 8.20 × 10−14 J
=
(8.199 × 10−14 J)
(1.60 × 10−13 J/MeV)
= 0.5124 MeV ≈ 0.512 MeV
This does not agree with the value inside the front cover of the book because of significant figures. If
more significant digits were used for the given values, then a value of 0.511 MeV would be obtained.
22.
We find the loss in mass from Eq. 26–7.
m=
23.
E
c
2
=
(200 MeV)(1.60 × 10−13 J/MeV)
8
(3.00 × 10 m/s)
2
= 3.56 × 10−28 kg ≈ 4 × 10−28 kg
We find the mass conversion from Eq. 26–7.
m=
E
c
2
=
(1× 1020 J)
(3.00 × 108 m/s) 2
= 1111 kg ≈ 1000 kg
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No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
The Special Theory of Relativity
24.
26-9
We calculate the mass from Eq. 26–7.
m=
E
c
2
=
1
c
2
(mc 2 ) =
1 (1.67 × 10−27 kg)(3.00 × 108 m/s) 2
c
2
(1.60 × 10
−13
J/MeV)
= 939 MeV/c 2
To get the same answer as inside the front cover of the book, we would need more significant figures
for each constant.
25.
This increase in mass is found from Eq. 26–8.
Δm =
ΔE
c2
=
(4.82 × 104 J)
(3.00 × 108 m/s) 2
= 5.36 × 10−13 kg
Note that this is so small, most chemical reactions are considered to have mass conserved.
26.
We find the speed in terms of c. The kinetic energy is given by Eq. 26–5b and the momentum by
Eq. 26–4.
υ=
(2.90 × 108 m/s)
(3.00 × 108 m/s)
= 0.9667c
⎛
⎞
1
= (γ − 1)mc 2 = ⎜
− 1⎟ (938.3 MeV) = 2728.2 MeV ≈ 2.7 GeV
⎜
⎟
2
⎝ 1 − 0.9667
⎠
1
p = γ mυ =
(938.3 MeV/c 2 )(0.9667 c) = 3544 MeV/c ≈ 3.5 GeV/c
2
1 − 0.9667
KE
27.
The total energy of the proton is the kinetic energy plus the mass energy. Use Eq. 26–9 to find the
momentum.
E = KE + mc 2 ;
(pc) 2 = E 2 − (mc 2 ) 2 = ( KE + mc 2 ) 2 − (mc 2 ) 2 = KE 2 + 2KE(mc 2 )
pc =
KE
2
+ 2KE(mc 2 ) = KE 1 + 2
mc 2
KE
= (950 MeV) 1 + 2
938.3 MeV
= 1638 MeV
950 MeV
p = 1638 MeV/c ≈ 1.6 GeV/c
28.
We use Eq. 26–5b to find the speed from the kinetic energy.
KE
⎛
⎞
1
= (γ − 1)mc 2 = ⎜
− 1⎟ mc 2
⎜
⎟
2 2
⎝ 1 − υ /c
⎠
υ = c 1−
29.
(a)
1
⎛ KE
⎞
⎜ 2 + 1⎟
⎝ mc
⎠
2
= c 1−
→
1
⎛ 1.12 MeV
⎞
+ 1⎟
⎜
⎝ 0.511 MeV ⎠
2
= 0.9497c
The work is the change in kinetic energy. Use Eq. 26–5b. The initial kinetic energy is 0.
⎛
⎞
1
− 1⎟ (938.3 MeV) = 4.499 ×103 MeV
W = ΔKE = KE final = (γ − 1)mc 2 = ⎜
⎜
⎟
2
⎝ 1 − 0.985
⎠
≈ 4.5 GeV
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26-10
Chapter 26
(b)
The momentum of the proton is given by Eq. 26–4.
1
p = γ mυ =
30.
(938.3 MeV/c 2 )(0.985c) = 5.36 × 103 MeV/c ≈ 5.4 GeV/c
The relationship between kinetic energy and mass energy is given in Eq. 26–5b.
KE
31.
1 − 0.985
2
= (γ − 1)mc 2 = 0.33mc 2
1
→
2
1 − υ /c
2
− 1 = 0.33 → υ = c 1 −
1
(1.33) 2
= 0.659c
The kinetic energy is equal to the rest energy. We use Eq. 26–5b.
KE
= (γ − 1)mc 2 = mc 2
→
1
2
1 − υ /c
2
−1 = 1 → υ = c 1−
1
(2) 2
= 0.866c
The momentum is found from Eq. 26–6a and Eq. 26–9.
E = KE + mc 2 = 2mc 2
E 2 = p 2 c 2 + m2 c 4
→ E 2 = 4m 2 c 4
→
p 2 c 2 = E 2 − m 2 c 4 = 3m 2 c 4
→
p = 3mc = 3(9.11× 10−31 kg)(3.00 × 108 m/s) = 4.73 × 10−22 kg ⋅ m/s
⎛ 3.0 × 108 m/s ⎞ ⎛
1 eV
4.73 × 10−22 kg ⋅ m/s ⎜
⎟⎟ ⎜⎜
−19
⎜
c
⎝
⎠ ⎝ 1.60 × 10
32.
⎞ 1 MeV
= 0.887 MeV/c
⎟
J ⎟⎠ 106 eV
Use Eq. 26–5b to calculate the kinetic energy of the proton. Note that the classical answer would be
that a doubling of speed would lead to a fourfold increase in kinetic energy. Subscript 1 represents the
(
)
(
)
lower speed υ = 13 c and subscript 2 represents the higher speed υ = 23 c .
KE
33.
= (γ − 1)mc 2
→
KE 2
KE1
⎛1 −
⎜
(γ 2 − 1)mc
⎝
=
=
2
(γ 1 − 1)mc
⎛1 −
⎜
⎝
2
( 32 )
2 ⎞ −1/2
⎟
⎠
−1/2
1 2⎞
3 ⎟
()
⎠
−1
=
−1
(5/9) −1/2 − 1
(8/9)
−1/2
−1
=
0.3416
= 5.6
0.0607
We find the energy equivalent of the mass from Eq. 26–7.
E = mc 2 = (1.0 × 10−3 kg)(3.00 × 108 m/s) 2 = 9.0 × 1013 J
We assume that this energy is used to increase the gravitational potential energy.
E = mgh → m =
34.
9.0 × 1013 J
E
=
= 9.2 × 109 kg
hg (1.0 × 103 m)(9.80 m/s 2 )
The work is the change in kinetic energy. Use Eq. 26–5b. The initial kinetic energy is 0.
W1 = (γ 0.90 − 1)mc 2 ;
W2 = KE 0.99c − KE 0.90c = (γ 0.99 − 1)mc 2 − (γ 0.90 − 1)mc 2
1
2
2
γ
−γ
W2 (γ 0.99 − 1)mc − (γ 0.90 − 1)mc
=
= 0.99 0.90 = 1 − 0.99
2
1
W1
γ
(γ 0.90 − 1)mc
0.90 − 1
2
−
1
1 − 0.902 = 3.7
−1
1 − 0.902
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The Special Theory of Relativity
35.
26-11
Each photon has momentum 0.65 MeV/c. Thus each photon has an energy of 0.65 MeV. Assuming
that the photons have opposite initial directions, then the total momentum is 0, so the product mass will
not be moving. Thus all of the photon energy can be converted into the mass energy of the particle.
Thus the heaviest particle would have a mass of 1.30 MeV/c 2 , which is 2.32 × 10−30 kg.
36.
Since the proton was accelerated by a potential difference of 165 MV, its potential energy decreased
by 165 MeV, so its kinetic energy increased from 0 to 165 MeV. Use Eq. 26–5b to find the speed from
the kinetic energy.
KE
⎛
⎞
1
= (γ − 1)mc 2 = ⎜
− 1⎟ mc 2
⎜
⎟
2 2
⎝ 1 − υ /c
⎠
υ = c 1−
37.
⎛ KE
⎞
⎜ 2 + 1⎟
⎝ mc
⎠
2
= c 1−
1
⎛ 165 MeV
⎞
+ 1⎟
⎜
⎝ 938.3 MeV ⎠
= 0.5261c
2
Since the electron was accelerated by a potential difference of 31 kV, its potential energy decreased by
31 keV, so its kinetic energy increased from 0 to 31 keV. Use Eq. 26–5b to find the speed from the
kinetic energy.
KE
⎛
⎞
1
= (γ − 1)mc 2 = ⎜
− 1⎟ mc 2
⎜
⎟
2 2
⎝ 1 − υ /c
⎠
υ = c 1−
38.
1
→
1
⎛ KE
⎞
⎜ 2 + 1⎟
⎝ mc
⎠
2
= c 1−
→
1
⎛ 0.031 MeV ⎞
+ 1⎟
⎜
⎝ 0.511 MeV ⎠
= 0.333c
2
We use Eqs. 26–6a and 26–9 in order to find the mass.
E 2 = p 2 c 2 + m 2 c 4 = (KE + mc 2 ) 2 = KE 2 + 2KE mc 2 + m 2 c 4
m=
2 2
p c − KE
2 KE c
2
2
=
2 2
(121 MeV/c) c − (45 MeV)
2
→
= 140 MeV/c 2 ≈ 2.5 × 10−28 kg
2(45 MeV)c 2
The particle is most likely a π 0 meson.
39.
(a)
(b)
Since the kinetic energy is half the total energy and the total energy is the kinetic energy plus the
rest energy, the kinetic energy must be equal to the rest energy. We also use Eq. 26–5b.
KE
= 12 E = 12 (KE + mc 2 ) → KE = mc 2
KE
= (γ − 1)mc 2 = mc 2
1
→ γ =2=
2
1 − υ /c
→ υ=
2
3
4
c = 0.866c
In this case, the kinetic energy is half the rest energy.
KE
= (γ − 1)mc 2 = 12 mc 2
→ γ =
3
2
=
1
2
1 − υ /c
2
→ υ=
5
9
c = 0.745c
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26-12
40.
Chapter 26
We use Eq. 26–5b for the kinetic energy and Eq. 26–4 for the momentum.
υ
c
=
KE
8.65 × 107 m/s
= 0.2883
3.00 × 108 m/s
⎛
⎞
⎛
⎞
1
1
= (γ − 1)mc 2 = ⎜
− 1⎟ mc 2 = ⎜
− 1⎟ (938.3 MeV) = 41.6 MeV
⎜
⎟
2 2
⎜ 1 − (0.2883) 2
⎟
⎝ 1 − υ /c
⎠
⎝
⎠
p = γ mυ =
mυ
1 − υ 2 /c 2
=
1 mc 2 (υ /c) 1 (938.3 MeV)(0.2883)
=
= 283 MeV/c
c 1 − υ 2 /c 2 c
1 − (0.2883) 2
Evaluate with the classical expressions.
2
⎛υ ⎞
= 12 mυ 2 = 12 mc 2 ⎜ ⎟ = 12 (938.3 MeV)(0.2883) 2 = 39.0 MeV
⎝c⎠
1 2 ⎛υ ⎞
pc = mυ = mc ⎜ ⎟ = (938.3 MeV/c)(0.2883) = 271 MeV/c
c
⎝c⎠
KE c
Calculate the percent error.
− KE
39.0 − 41.6
× 100 = −6.3%
41.6
p −p
271 − 283
Errorp = c
× 100 =
× 100 = −4.2%
283
p
ErrorKE =
41.
(a)
KE c
KE
× 100 =
The kinetic energy is found from Eq. 26–5b.
KE
⎛
⎞
⎛
⎞
1
1
= (γ − 1)mc 2 = ⎜
− 1⎟ mc 2 = ⎜
− 1⎟ (1.7 × 104 kg)(3.00 × 108 m/s) 2
⎜
⎟
⎜
⎟
2 2
2
⎝ 1 − υ /c
⎠
⎝ 1 − 0.15
⎠
= 1.751× 1019 J ≈ 1.8 × 1019 J
(b)
Use the classical expression and compare the two results.
KE
= 12 mυ 2 = 12 (1.7 × 104 kg)[(0.15)(3.00 × 108 m/s)]2 = 1.721× 1019 J
% error =
(1.721× 1019 J) − (1.751× 1019 J)
(1.751× 1019 J)
× 100 = −1.7%
The classical value is 1.7% too low.
42.
All of the energy, both rest energy and kinetic energy, becomes electromagnetic energy. We use
Eq. 26–6b. The masses are the same.
⎛
1
1
+
Etotal = E1 + E2 = γ 1mc 2 + γ 2 mc 2 = (γ 1 + γ 2 )mc 2 = ⎜
⎜
2
1 − 0.652
⎝ 1 − 0.53
⎞
⎟ (105.7 MeV)
⎟
⎠
= 263.7 MeV ≈ 260 MeV
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The Special Theory of Relativity
43.
26-13
We let M represent the mass of the new particle. The initial energy is due to both incoming particles,
and the final energy is the rest energy of the new particle. Use Eq. 26–6b for the initial energies.
E = 2(γ mc 2 ) = Mc 2
→ M = 2γ m =
2m
1 − υ 2 /c 2
We assumed that energy is conserved, so there was no loss of energy in the collision.
The final kinetic energy is 0, so all of the kinetic energy was lost.
KE lost
44.
⎛
⎞
1
= KE initial = 2(γ − 1)mc 2 = ⎜
− 1⎟ 2mc 2
⎜
⎟
2 2
⎝ 1 − υ /c
⎠
By conservation of energy, the rest energy of the americium nucleus is equal to the rest energies of the
other particles plus the kinetic energy of the alpha particle.
mAm c 2 = (mNp + mα )c 2 + KEα
mNp = mAm − mα −
45.
KEα
c
→
= 241.05682 u − 4.00260 u −
2
⎞
5.5 MeV ⎛
1u
⎜⎜
⎟ = 237.04832 u
2
2⎟
c
⎝ 931.49 MeV/c ⎠
We use Eqs. 26–6a and 26–9.
E = KE + mc 2 ; (pc) 2 = E 2 − (mc 2 ) 2 = (KE + mc 2 ) 2 − (mc 2 )2 = KE 2 + 2KE(mc 2 ) →
p=
46.
KE
2
+ 2KE(mc 2 )
c
The kinetic energy of 998 GeV is used to find the speed of the protons. Since the energy is 1000 times
the mass, we expect the speed to be very close to c. Use Eq. 26–5b.
KE
⎛
⎞
1
= (γ − 1)mc 2 = ⎜
− 1⎟ mc 2
⎜
⎟
2 2
⎝ 1 − υ /c
⎠
υ = c 1−
1
⎛ KE
⎞
⎜ 2 + 1⎟
⎝ mc
⎠
2
= c 1−
→
1
⎛ 998 GeV
⎞
+ 1⎟
⎜
0.938
GeV
⎝
⎠
2
= c (to 7 significant figures)
⎛ 998 GeV
⎞
⎛ K
⎞
− 1⎟ (1.673 × 10−27 kg)(3.00 × 108 m/s)
− 1⎟ mc ⎜
⎜
2
0.938
GeV
γ mυ
γ mυ ⎝ mc
⎠
⎠
=
≈
=⎝
= 3.3 T
B=
rqυ
rq
rq
(1.0 × 103 m)(1.60 × 10−19 C)
2
47.
Take the positive direction to be the direction of motion of the person’s rocket. Consider the Earth as
reference frame S and the person’s rocket as reference frame S′. Then the speed of the rocket relative
to Earth is υ = 0.40c, and the speed of the meteor relative to the rocket is u ′ = 0.40c. Calculate the
speed of the meteor relative to the rocket using Eq. 26–10.
u=
(υ + u ′)
(0.40c + 0.40c)
=
= 0.69c
⎛ υ u ′ ⎞ [1 + (0.40)(0.40)]
1
+
⎜
⎟
c2 ⎠
⎝
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26-14
48.
Chapter 26
(a)
Take the positive direction to be the direction of motion of spaceship 1. Consider spaceship 2 as
reference frame S and the Earth as reference frame S′. The velocity of the Earth relative to
spaceship 2 is υ = 0.60c. The velocity of spaceship 1 relative to the Earth is u′ = 0.60c. Solve
for the velocity of spaceship 1 relative to spaceship 2, u, using Eq. 26–10.
u=
(b)
(υ + u′)
(0.60c + 0.60c)
=
= 0.882c
u
′
υ
[1
+ (0.60)(0.60)]
⎛
⎞
⎜1 + 2 ⎟
c ⎠
⎝
Now consider spaceship 1 as reference frame S. The velocity of the Earth relative to spaceship 1
is υ = − 0.60c. The velocity of spaceship 2 relative to the Earth is u′ = −0.60c. Solve for the
velocity of spaceship 2 relative to spaceship 1, u, using Eq. 26–10.
u=
(υ + u′)
(−0.60c − 0.60c)
=
= −0.882c
⎛ υ u′ ⎞ [1 + (−0.60)(−0.60)]
1
+
⎜
⎟
c2 ⎠
⎝
As expected, the two relative velocities are the opposite of each other.
49.
(a)
We take the positive direction in the direction of the first spaceship. We choose reference frame
S as the Earth and reference frame S′ as the first spaceship. So υ = 0.65c. The speed of the
second spaceship relative to the first spaceship is u ′ = 0.82c. We use Eq. 26–10 to solve for the
speed of the second spaceship relative to the Earth, u.
u=
(b)
(υ + u′)
(0.82c + 0.65c)
=
= 0.959c
⎛ υ u′ ⎞ [1 + (0.65)(0.82)]
⎜1 + 2 ⎟
c ⎠
⎝
The only difference is now that u′ = −0.82c.
u=
(υ + u′)
(−0.82c + 0.65c)
=
= −0.36c
⎛ υ u′ ⎞ [1 + (0.65)(−0.82)]
1
+
⎜
⎟
c2 ⎠
⎝
The Problem asks for the speed, which is 0.36c .
50.
We take the positive direction as the direction of the Enterprise. Consider the alien vessel as reference
frame S and the Earth as reference frame S′. The velocity of the Earth relative to the alien vessel is
υ = −0.60c. The velocity of the Enterprise relative to the Earth is u′ = 0.90c. Solve for the velocity of
the Enterprise relative to the alien vessel, u, using Eq. 26–10.
u=
(υ + u′)
(0.90c − 0.60c)
=
= 0.65c
⎛ υ u′ ⎞ [1 + (−0.60)(0.90)]
⎜1 + 2 ⎟
c ⎠
⎝
We could also have made the Enterprise reference frame S, with υ = −0.90c, and the velocity of the
alien vessel relative to the Earth as u′ = 0.60c. The same answer would result.
Choosing the two spacecraft as the two reference frames would also work. Let the alien vessel be
reference frame S and the Enterprise be reference frame S′. Then we have the velocity of the Earth
relative to the alien vessel as u = −0.60c and the velocity of the Earth relative to the Enterprise as
u′ = −0.90c. We solve for υ , the velocity of the Enterprise relative to the alien vessel.
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The Special Theory of Relativity
u=
51.
→ υ=
u − u′
(−0.60c) − (−0.90c)
=
= 0.65c
⎛ u′u ⎞ ⎛ (−0.90c)(−0.60c) ⎞
⎜1 − 2 ⎟ ⎜1 −
⎟
c ⎠ ⎝
c2
⎝
⎠
We take the positive direction as the direction of the motion of the second pod. Consider the first pod
as reference frame S and the spacecraft as reference frame S′. The velocity of the spacecraft relative to
the first pod is υ = 0.70c. The velocity of the second pod relative to the spacecraft is u′ = 0.80c. Solve
for the velocity of the second pod relative to the first pod, u, using Eq. 26–10.
u=
52.
(u′ + υ )
⎛ υ u′ ⎞
⎜1 + 2 ⎟
c ⎠
⎝
26-15
(υ + u′)
(0.80c + 0.70c)
=
= 0.962c
⎛ υ u′ ⎞ [1 + (0.70)(0.80)]
⎜1 + 2 ⎟
c ⎠
⎝
We take the positive direction as the direction of motion of rocket A. Consider rocket A as reference
frame S and the Earth as reference frame S′. The velocity of the Earth relative to rocket A is
υ = −0.65c. The velocity of rocket B relative to the Earth is u′ = 0.95c. Solve for the velocity of
rocket B relative to rocket A, u, using Eq. 26–10.
u=
(υ + u′)
(0.95c − 0.65c)
=
= 0.78c
⎛ υ u′ ⎞ [1 + (−0.65)(0.95)]
⎜1 + 2 ⎟
c ⎠
⎝
Note that a Galilean analysis would have resulted in u = 0.30c.
53.
We assume that the given speed of 0.90c is relative to the planet that you are approaching. We take the
positive direction as the direction that you are traveling. Consider your spaceship as reference frame S
and the planet as reference frame S′. The velocity of the planet relative to you is υ = −0.90c. The
velocity of the probe relative to the planet is u′ = 0.95c. Solve for the velocity of the probe relative to
your spaceship, u, using Eq. 26–10.
u=
54.
(υ + u′)
(0.95c − 0.90c)
=
= 0.34c ≈ 0.3c
⎛ υ u′ ⎞ [1 + (−0.90)(0.95)]
1
+
⎜
⎟
c2 ⎠
⎝
The kinetic energy is given by Eq. 26–5b.
KE
= (γ − 1)mc 2 = mc 2
→ γ =2=
1
2
1 − υ /c
2
→ υ=
3
c = 0.866c
4
Notice that the mass of the particle does not affect the result.
55.
(a)
To travelers on the spacecraft, the distance to the star is contracted, according to Eq. 26–3a. This
contracted distance is to be traveled in 4.9 years. Use that time with the contracted distance to
find the speed of the spacecraft.
υ=
Δxspacecraft
Δtspacecraft
υ =c
=
ΔxEarth 1 − υ 2 /c 2
Δtspacecraft
1
⎛ cΔtspacecraft ⎞
1+ ⎜
⎟
⎝ ΔxEarth ⎠
2
=c
→
1
⎛ 4.9 ly ⎞
1+ ⎜
⎟
⎝ 4.3 ly ⎠
2
= 0.6596c ≈ 0.66c
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26-16
Chapter 26
(b)
Find the elapsed time according to observers on Earth, using Eq. 26–1a.
Δtspaceship
4.9 yr
ΔtEarth =
=
= 6.5 yr
2 2
1 − υ /c
1 − 0.65962
Note that this agrees with the time found from distance and speed.
tEarth =
56.
ΔxEarth
υ
=
4.3 ly
= 6.5 yr
0.6596c
The numerical values and
graph were generated in a
spreadsheet. The graph is
shown also.
υ /c 0.00 0.01 0.05 0.10 0.20 0.30 0.40 0.50 0.60 0.70 0.80 0.90 0.95 0.99
γ 1.000 1.000 1.001 1.005 1.021 1.048 1.091 1.155 1.250 1.400 1.667 2.294 3.203 7.089
57.
Time dilation effects have changed the proper time for one heartbeat, which is given as 1 second, to a
dilated time of 2.4 seconds as observed on Earth. Use Eq. 26–1a.
58.
(a)
2
2
⎛ 1.0 s ⎞
⎛ Δt ⎞
→ = c 1− ⎜ 0 ⎟ = c 1− ⎜
⎟ = 0.91c
Δ
t
⎝
⎠
⎝ 2.4 s ⎠
Δt0 = Δt 1 − υ 2 /c 2
We find the speed from Eq. 26–5b.
KE
⎛
⎞
1
= (γ − 1)mc 2 = ⎜
− 1⎟ mc 2 = 14,000mc 2
⎜
⎟
2 2
⎝ 1 − υ /c
⎠
⎛
⎞
1
2
c⎛
1
⎞
υ = c 1− ⎜
⎟ ≈c− ⎜
⎟
2 ⎝ 14,001 ⎠
⎝ 14,001 ⎠
2
→
2
→
2
c⎛ 1 ⎞
(3.00 × 108 m/s) ⎛ 1 ⎞
c −υ = ⎜
=
⎟
⎜
⎟ = 0.77 m/s
2 ⎝ 14,001 ⎠
2
⎝ 14,001 ⎠
(b)
The tube will be contracted in the rest frame of the electron, according to Eq. 26–3a.
⎡ ⎛ 1 ⎞2 ⎤
A 0 = A 1 − υ /c = (3.0 × 10 m) 1 − ⎢1 − ⎜
⎟ ⎥ = 0.21 m
⎢⎣ ⎝ 14,001 ⎠ ⎥⎦
2
2
3
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The Special Theory of Relativity
59.
26-17
The minimum energy required would be the energy to produce the pair with no kinetic energy, so the
total energy is their rest energy. They both have the same mass. Use Eq. 26–7.
E = 2mc 2 = 2(0.511 MeV) = 1.022 MeV(1.637 × 10−13 J)
60.
The wattage times the time is the energy required. We use Eq. 26–7 to calculate the mass.
E = Pt = mc 2
61.
c2
=
(75 W)(3.16 × 107 s) ⎛ 1000 g ⎞
−5
⎜
⎟ = 2.6 × 10 g
(3.00 × 108 m/s) 2 ⎝ 1 kg ⎠
= mn c 2 − (mp c 2 + me c 2 + mν c 2 ) = 939.57 MeV − (938.27 MeV + 0.511 MeV + 0) = 0.79 MeV
The increase in kinetic energy comes from the decrease in potential energy.
KE
⎛
⎞
1
= (γ − 1)mc 2 = ⎜
− 1⎟ mc 2
⎜
⎟
2 2
⎝ 1 − υ /c
⎠
⎛
⎜
1
υ = c ⎜⎜1 −
2
⎜ ⎛ KE + 1⎞
⎟
⎜ ⎜ 2
⎠
⎝ ⎝ mc
63.
Pt
The kinetic energy available comes from the decrease in rest energy.
KE
62.
→ m=
(a)
1/2
⎞
⎟
⎟
⎟
⎟
⎟
⎠
→
1/2
⎛
⎞
⎜
⎟
⎜
⎟
1
= c ⎜1 −
⎟
2
−14
⎜ ⎛
⎟
⎞
6.20 × 10
J
+ 1⎟ ⎟
⎜ ⎜⎜
31
8
2
−
⎟
⎜
⎟
⎠ ⎠
⎝ ⎝ (9.11× 10 kg)(3.00 × 10 m/s)
= 0.822c
We find the rate of mass loss from Eq. 26–7.
E = mc 2
→ ΔE = ( Δ m ) c 2
→
26
4 × 10 J/s
Δm 1 ⎛ ΔE ⎞
= 2⎜
= 4.44 × 109 kg/s ≈ 4 × 109 kg/s
⎟=
Δt c ⎝ Δt ⎠ (3.00 × 108 m/s)2
(b)
Find the time from the mass of the Sun and the rate determined in part (a).
Δt =
(c)
We find the time for the Sun to lose all of its mass at this same rate.
Δt =
64.
mEarth
(5.98 × 1024 kg)
=
= 4.27 × 107 yr ≈ 4 × 107 yr
Δm /Δt (4.44 × 109 kg/s)(3.156 × 107 s/yr)
mSun
(1.99 × 1030 kg)
=
= 1.42 × 1013 yr ≈ 1× 1013 yr
Δm /Δt (4.44 × 109 kg/s)(3.156 × 107 s/yr)
The total binding energy is the energy required to provide the increase in rest energy.
E = [(2mp + e + 2mn ) − mHe ]c 2
⎛ 931.5 MeV/c 2 ⎞
= [2(1.00783 u) + 2(1.00867 u) − 4.00260 u]c 2 ⎜
⎟⎟ = 28.32 MeV
⎜
u
⎝
⎠
65.
The momentum is given by Eq. 26–4, and the energy is given by Eq. 26–6b and Eq. 26–9.
p = γ mυ =
γ mc 2υ
c
2
=
Eυ
c
2
→ υ=
pc 2
=
E
pc 2
2 4
2 2
m c +p c
=
pc
2 2
m c + p2
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26-18
66.
Chapter 26
(a)
The magnitudes of the momenta are equal. Use Eq. 26–4.
p = γ mυ =
mυ
1 − υ 2 /c 2
=
1 mc 2 (υ /c) 1 (938.3 MeV)(0.990)
=
= 6585 MeV/c
c 1 − υ 2 /c 2 c
1 − 0.9902
⎛
⎞ ⎛ 1.602 × 10−10 J/GeV ⎞
1c
≈ 6.59 GeV/c = (6.585 GeV/c) ⎜
⎟⎟
⎜ 3.00 × 108 m/s ⎟⎟⎜⎜
1 GeV
⎝
⎠⎝
⎠
= 3.52 × 10−18 kg ⋅ m/s
(b)
Because the protons are moving in opposite directions, the vector sum of the momenta is 0.
(c)
In the reference frame of one proton, the laboratory is moving at 0.990c. The other proton is
moving at + 0.990c relative to the laboratory. We find the speed of one proton relative to the
other, and then find the momentum of the moving proton in the rest frame of the other proton by
using that relative velocity.
(υ + u′)
[0.990c + (0.990c)] ⎛ 2(0.990)c ⎞
=
=⎜
⎟ ≈ 0.99995c
u
υ
′
⎛
⎞ [1 + (0.990)(0.990)] ⎝ 1 + 0.9902 ⎠
⎜1 + 2 ⎟
c ⎠
⎝
⎛ 2(0.990) ⎞
(938.3 MeV) ⎜
⎟
mu
1 mc 2 (u /c) 1
⎝ 1 + 0.9902 ⎠ = 93358 MeV/c
=
=
p = γ mu =
2
1 − u 2 /c 2 c 1 − u 2 /c 2 c
⎛ 2(0.990) ⎞
1− ⎜
⎟
⎝ 1 + 0.9902 ⎠
u=
⎛
⎞ ⎛ 1.602 × 10−10 J/GeV ⎞
1c
≈ 93.4 GeV/c = (93.358 GeV/c) ⎜
⎟⎟
⎟⎟⎜⎜
8
⎜
1 GeV
⎝ 3.00 × 10 m/s ⎠⎝
⎠
= 4.99 × 10−17 kg ⋅ m/s
67.
We find the loss in mass from Eq. 26–8.
Δm =
ΔE
c2
=
484 × 103 J
(3.00 × 108 m/s) 2
= 5.38 × 10−12 kg
Two moles of water has a mass of 36.0 × 10−3 kg. Find the percentage of mass lost.
5.38 ×10−12 kg
36.0 × 10−3 kg
68.
= 1.49 × 10−10 = 1.49 × 10−8 %
Use Eq. 26–5b for kinetic energy and Eq. 26–7 for rest energy.
KE
= (γ − 1)mEnterprise c 2 = mconverted c 2
→
⎛
⎞
⎛
⎞
1
1
− 1⎟ mEnterprise = ⎜
− 1⎟ (6 × 109 kg) = 3 × 107 kg
mconverted = ⎜
⎜
⎟
⎜
⎟
2 2
2
⎝ 1 − υ /c
⎠
⎝ 1 − 0.10
⎠
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The Special Theory of Relativity
69.
(a)
26-19
For a particle of nonzero mass, we derive the
following relationship between kinetic energy and
momentum.
E = KE + mc 2 ;
( pc) 2 = E 2 − (mc 2 ) 2 = ( KE + mc 2 ) 2 − (mc 2 ) 2
KE
→
= ( pc) 2 + (mc 2 )2 − mc 2
This is the relationship that is graphed. But note the
two extreme regions. If the momentum is small, then we have the classical relationship.
KE
⎛ pc ⎞
= −mc 2 + (mc 2 ) 2 + ( pc) 2 = − mc 2 + mc 2 1 + ⎜ 2 ⎟
⎝ mc ⎠
2
2
⎛
⎛ pc ⎞ ⎞ p 2
≈ −mc 2 + mc 2 ⎜ 1 + 12 ⎜ 2 ⎟ ⎟ =
⎜
⎝ mc ⎠ ⎟⎠ 2m
⎝
If the momentum is large, then we have this relationship, which is linear.
KE
70.
= −mc 2 + (mc 2 ) 2 + ( pc) 2 ≈ pc − mc 2
(b)
For a particle of zero mass, the relationship is simply
KE = pc. See the adjacent graph.
(a)
We set the kinetic energy of the spacecraft equal to the rest energy of an unknown mass, m. Use
Eqs. 26–5a and 26–7.
KE
= (γ − 1)mship c 2 = mc 2
→
⎛
⎞
⎛
⎞
1
1
m = (γ − 1)mship = ⎜
− 1⎟ mship = ⎜
− 1⎟ (1.6 × 105 kg) = 6.4 × 104 kg
⎜
⎟
⎜
⎟
2 2
2
⎝ 1 − υ /c
⎠
⎝ 1 − 0.70
⎠
(b)
From the Earth’s point of view, the distance is 35 ly and the speed is 0.70c. Those data are used
to calculate the time from the Earth frame, and then Eq. 26–1a is used to calculate the time in the
spaceship frame.
Δt =
71.
d
υ
=
(35 yr)c
= 50 yr;
0.70c
Δt0 = Δt 1 − υ 2 /c 2 = (50 yr) 1 − 0.702 = 36 yr
We assume that one particle is moving in the negative direction in the laboratory frame and the other
particle is moving in the positive direction. We consider the particle moving in the negative direction
as reference frame S and the laboratory as reference frame S′. The velocity of the laboratory relative to
the negative-moving particle is υ = 0.82c, and the velocity of the positive-moving particle relative to
the laboratory frame is u′ = 0.82c. Solve for the velocity of the positive-moving particle relative to the
negative-moving particle, u. Use Eq. 26–10.
u=
(υ + u′)
(0.82c + 0.82c)
=
= 0.981c
⎛ υ u′ ⎞ [1 + (0.82)(0.82)]
1
+
⎜
⎟
c2 ⎠
⎝
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26-20
72.
Chapter 26
We consider the motion from the reference frame of the spaceship. The passengers will see the trip
distance contracted, as given by Eq. 26–3a. They will measure their speed to be that contracted
distance divided by the years of travel time (as measured on the ship). Use that speed to find the work
done (the kinetic energy of the ship).
υ=
A 1 − υ 2 /c 2
A
= 0
Δt0
Δt 0
→
⎛
W = ΔKE = (γ − 1)mc 2 = ⎜
⎜
⎝
υ
1
2
⎛
⎜ 1 + ⎛ cΔt0 ⎞
⎜
⎟
⎜⎜
⎝ A0 ⎠
⎝
⎞
1
− 1⎟ mc 2
⎟
1 − υ 2 /c 2
⎠
=
c
⎞
⎟
⎟⎟
⎠
=
1
= 0.9887c
2⎞
⎛
⎜ 1 + ⎛ 1.0 ly ⎞ ⎟
⎜
⎟
⎜⎜
⎝ 6.6 ly ⎠ ⎟⎟
⎝
⎠
⎛
⎞
1
=⎜
− 1⎟ (3.6 × 104 kg)(3.00 × 108 m/s) 2 = 1.8 × 1022 J
⎜
⎟
2
⎝ 1 − 0.9887
⎠
Note that, according to Problem 73, this is about 100 × the annual energy consumption of the United
States.
73.
The kinetic energy is given by Eq. 26–5b.
KE
⎛
⎞
⎛
⎞
1
1
= (γ − 1)mc 2 = ⎜
− 1⎟ mc 2 = ⎜
− 1⎟ (14,500 kg)(3.00 × 108 m/s) 2
⎜
⎟
2 2
⎜ 1 − (0.90) 2
⎟
⎝ 1 − υ /c
⎠
⎝
⎠
= 1.7 × 1021 J
The spaceship’s kinetic energy is approximately 20 times as great as the annual U.S. energy use.
74.
The pi meson decays at rest, so the momentum of the muon and the neutrino must each have the same
magnitude (and opposite directions). The neutrino has no mass, and the total energy must be
conserved. We combine these relationships using Eq. 26–9.
Ev = ( pv2 c 2 + mv2 c 4 )1/2 = pv c;
Eπ = Eμ + Ev
pμ = pv = p
→ mπ c 2 = ( pμ2 c 2 + mμ2 c 4 )1/2 + pv c = ( p 2 c 2 + mμ2 c 4 )1/2 + pc →
mπ c 2 − pc = ( p 2 c 2 + mμ2 c 4 )1/2
→ (mπ c 2 − pc) 2 = ( p 2 c 2 + mμ2 c 4 )
Solve for the momentum.
mπ2 c 4 − 2mπ c 2 pc + p 2 c 2 = p 2 c 2 + mμ2 c 4
→
pc =
mπ2 c 2 − mμ2 c 2
2mπ
Write the kinetic energy of the muon using Eqs. 26–6b and 26–9.
K μ = Eμ − mμ c 2 ;
Eμ = Eπ − Ev = mμ c 2 − pc →
K μ = (mπ c 2 − pc) − mμ c 2 = mπ c 2 − mμ c 2 −
=
=
2mπ (mπ c 2 − mμ c 2 )
2mπ
−
2mπ
(mπ2 c 2 − mμ2 c 2 )
2mπ
(2mπ2 − 2mμ mπ − mπ2 + mμ2 )c 2
2mπ
(mπ2 c 2 − mμ2 c 2 )
=
(mπ2 − 2mμ mπ + mμ2 )c 2
2mπ
=
(mπ − mμ ) 2 c 2
2mπ
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The Special Theory of Relativity
75.
(a)
26-21
Earth observers see the ship as contracted, as given by Eq. 26–3a.
A = A 0 1 − υ 2 /c 2 = (23 m) 1 − (0.75) 2 = 15.2 m ≈ 15 m
(b)
Earth observers see the launch as dilated (lengthened) in time, as given by Eq. 26–1a.
Δt =
76.
(a)
Δt0
2
2
1 − υ /c
=
28 min
1 − (0.75) 2
= 42.3 min ≈ 42 min
The relative speed can be calculated in either frame and will be the same in both frames. The
time measured on Earth will be longer than that measured on the spaceship, by Eq. 26–1a.
υ=
Δtspaceship
ΔxEarth
; ΔtEarth =
=
ΔtEarth
1 − υ 2 /c 2
Δtspaceship
⎛ Δx
⎞
1 − ⎜ Earth ⎟
⎝ cΔtEarth ⎠
2
2
→
2
⎛ Δx
⎞
(ΔtEarth ) 2 − ⎜ Earth ⎟ = (Δtspaceship ) 2
c
⎝
⎠
⎛ Δx
⎞
→ (ΔtEarth ) 2 − ⎜ Earth ⎟ = (Δtspaceship ) 2
c
⎝
⎠
→
2
⎛ Δx
⎞
ΔtEarth = ⎜ Earth ⎟ + (Δtspaceship ) 2 = (6.0 yr) 2 + (3.50 yr) 2 = 6.946 yr ≈ 6.9 yr
⎝ c ⎠
(b)
The distance as measured by the spaceship will be contracted.
υ=
ΔxEarth Δxspaceship
=
ΔtEarth Δtspaceship
Δxspaceship =
Δtspaceship
ΔtEarth
→
ΔxEarth =
3.50 yr
(6.0 ly) = 3.023 ly ≈ 3.0 ly
6.946 yr
This distance is the same as that found using the length contraction relationship.
77.
(a)
The kinetic energy is 5.00 times greater than the rest energy. Use Eq. 26–5b.
KE
(b)
= (γ −1)mc 2 = 5.00mc 2
1
1 − υ 2 /c 2
= 6.00 → υ = c 1 −
1
(6.00) 2
= 0.986c
The kinetic energy is 999 times greater than the rest energy. We use the binomial expansion.
KE
= (γ − 1)mc 2 = 999mc 2
υ = c 1−
78.
→
→
1
1 − υ 2 /c 2
= 1000 →
2⎞
⎛
1 ⎛ 1 ⎞ ⎟ = c (1 − 5.0 × 10−7 )
⎜
≈
−
1
c
⎜ 2 ⎜⎝ 1000 ⎟⎠ ⎟
(1000) 2
⎝
⎠
1
Every observer will measure the speed of a beam of light to be c. Check it with Eq. 26–10.
u=
υ + u′ (−c) + 0.70c
=
= −c
1 + υ u′ 1 + (0.70)(−1)
The beam’s speed as a positive value, relative to Earth, is c.
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26-22
79.
Chapter 26
From the boy’s frame of reference, the pole remains at rest with respect to him. As such, the pole will
always remain 13.0 m long. As the boy runs toward the barn, relativity requires that the (relatively
moving) barn contract in size, making the barn even shorter than its rest length of 13.0 m. Thus it is
impossible, in the boy’s frame of reference, for the barn to be longer than the pole. So according to the
boy, the pole will never completely fit within the barn.
In the frame of reference at rest with respect to the barn, it is possible for the pole to be shorter than the
barn. We use Eq. 26–3a to calculate the speed that the boy would have to run for the contracted length
of the pole, A, to equal the length of the barn.
A = A 0 1 − υ 2 /c 2
80.
→ υ = c 1 − A 2 /A 20 = c 1 − (10.0 m) 2 /(13.0 m) 2 = 0.6390c
At the North Pole the clock is at rest, while the clock on the Equator travels the circumference of the
Earth each day. We divide the circumference of the Earth by the length of the day to determine the
speed of the equatorial clock. We set the dilated time equal to 2.0 years and solve for the change in rest
times for the two clocks.
υ=
2π R 2π (6.38 × 106 m)
=
= 464 m/s
T
(24 h)(3600 s/h)
Δt =
Δt =
Δt0,eq
⎛
υ2 ⎞
→ Δt0,eq = Δt 1 − υ 2 /c 2 ≈ Δt ⎜ 1 + 2 ⎟
⎜ 2c ⎟
⎝
⎠
1 − υ 2 /c 2
Δt0,pole
1− 0
→ Δt0,pole = Δt
⎛
υ2
Δt0,eq − Δt0,pole = Δt ⎜ 1 + 2
⎜ 2c
⎝
= Δt
81.
υ2
2c 2
=
⎞
⎟⎟ − Δt
⎠
(2.0 yr)(464 m/s) 2 (3.156 × 107 s/yr)
2(3.00 × 108 m/s) 2
= 75 μ s
We treat the Earth as the stationary frame and the airplane as the moving frame. The elapsed time in
the airplane will be dilated to the observers on the Earth. Use Eq. 26–1a.
tEarth =
2π rEarth
υ
;
tplane = tEarth 1 − υ 2 /c 2 =
Δt = tEarth − tplane =
2π rEarth
υ
⎡
(1 −
)
1 − υ 2 /c 2 ≈
⎛ 1 m/s ⎞ ⎤
⎟⎥
⎝ 3.6 km/h ⎠ ⎦
π (6.38 ×106 m) ⎢1300 km/h ⎜
=
2π rEarth
⎣
(3.00 × 108 m/s) 2
υ
1 − υ 2 /c 2
2π rEarth ⎡ ⎛ 1 υ 2 ⎞ ⎤ π rEarthυ
⎢1 − ⎜⎜ 1 − 2 2 ⎟⎟ ⎥ =
υ
c ⎠ ⎦⎥
c2
⎣⎢ ⎝
= 8.0 × 10−8 s
Solutions to Search and Learn Problems
1.
To determine Mr Tompkins’ speed, examine the clock at Lloyd’s Bank. It should be circular, so
measure the height of the clock to find the proper length (260 units) and the width of the clock to find
the contracted length (120 units). Use these lengths in Eq. 26–3a with c = 10 m/s to find his speed.
A = A 0 1 − υ 2 /c 2
→ υ = c 1 − (A /A 0 ) 2 = 10 m/s 1 − (120/260)2 = 8.9 m/s
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The Special Theory of Relativity
26-23
This is not how he would see the world, because there would be additional distortions due to the time it
takes light to travel from objects to him. More distant objects would appear rotated.
2.
In O 2 ’s frame of reference, O 2 observes that both
trains are the same length. But O 2 is observing the
contracted length of O1’s train. Thus the proper
length of O1’s train is longer than O 2 ’s train. In
O1’s frame of reference, O 2 is moving; therefore,
O 2 ’s train is length contracted, making it even
shorter. When the end of O 2 ’s train passes the end of
O1’s train, the first lightning bolt strikes, and both
observers record the strike as occurring at the end of
their train. A short while later, O1 observes that O 2 ’s
train moves down the track until the front ends of the
trains are at the same location. When this occurs, the
second lightning bolt strikes, and both observers note
that it occurs at the front of their train. Since to
observer O 2 the trains are the same length, the strikes
occurred at the same time. To observer O1 , the trains
are not the same length; therefore, train O 2 had to move between
strikes so that the ends were aligned during each strike.
3.
The change in length is the difference between the initial length and the contracted length. The
binomial expansion is used to simplify the square root expression.
(
ΔA = A 0 − A = A 0 − A 0 1 − υ 2 /c 2 = A 0 1 − 1 − υ 2 /c 2
(
)
)
≈ A 0 ⎡1 − 1 − 12 υ 2 /c 2 ⎤ = 12 A 0υ 2 /c 2
⎣
⎦
⎛ m/s ⎞
= 12 (500 m)(100 km/h) 2 ⎜
⎟
⎝ 3.6 km/h ⎠
2
(3.00 × 108 m/s) 2 = 2.14 × 10−12 m
This length contraction is less than the size of an atom.
4.
If the ship travels 0.90c at 30° above the horizontal, then we can consider the motion in two stages:
horizontal motion of 0.90c cos 30° = 0.779c and vertical motion of 0.90c sin 30° = 0.45c. The length
of the painting will be contracted due to the horizontal motion, and the height will be contracted due to
the vertical motion.
A = A 0 1 − υ x2 /c 2 = 1.50 m 1 − 0.7792 = 0.94 m
h = h0 1 − υ y2 /c 2 = 1.00 m 1 − 0.452 = 0.89 m
5.
Using nonrelativistic mechanics, we would find that the muon could only travel a distance
of (3.00 × 108 m/s)(2.20 × 10−6 s) = 660 m before decaying. But from an Earth-based reference
frame, the muon’s “clock” would run slowly, so the time to use is the dilated time. The speed in the
Earth’s reference frame is then the distance of 10 km divided by the dilated time.
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26-24
Chapter 26
υ=
υ
c
=
ΔA 0 ΔA 0 ΔA 0 (1 − υ 2 /c 2 )
=
=
Δt
γΔt0
Δt0
→
(3.0 ×10 m )
4
ΔA 0
c 2 (Δt0 ) 2 + (ΔA 0 ) 2
= 0.9997581c →
=
(3.00 × 108 m/s) 2 (2.20 × 10−6 s)2 + (3.0 × 104 m) 2
υ = 0.99976c
Alternatively, we could find the speed for the muon to travel a contracted length during its “normal”
ΔA
ΔA ⎛ ΔA 0 ⎞ 1
lifetime. The contracted length is ΔA = ΔA 0 /γ , so the speed would be υ =
=⎜
= 0,
⎟
Δt0 ⎝ γ ⎠ Δt0 γΔt0
which is the same as the earlier expression.
The kinetic energy is found from Eq. 26–5b.
KE
⎛
⎞
⎛
⎞
1
1
= (γ − 1)mc 2 = ⎜
− 1⎟ mc 2 = ⎜
− 1⎟ (105.7 MeV)
⎜ (1 − υ 2 /c 2 ) ⎟
⎜ (1 − (0.9997581) 2 ) ⎟
⎝
⎠
⎝
⎠
= 4700 MeV
6.
The electrostatic force provides the radial acceleration, so those two forces are equated. We solve that
relationship for the speed of the electron.
Felectrostatic = Fcentripetal
υ=
1
e2
4πε 0 melectron r
=
→
e2
1
4πε 0 r 2
=
melectronυ 2
r
→
(8.99 × 109 N ⋅ m 2 /C2 ) (1.60 × 10−19 C) 2
9.11× 10
−31
kg
(0.53 × 10
−10
m)
= 2.18 × 106 m/s = 0.0073c
Because this is much less than 0.1c, the electron is not relativistic.
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