Solution Manual for Modern Physics, 2nd Edition
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INSTRUCTOR
SOLUTIONS
MANUAL
MODERN PHYSICS
S E C O N D E D I T I O N
RANDY HARRIS
University of California, Davis
SOLUTIONS
MANUAL
MODERN PHYSICS
S E C O N D E D I T I O N
RANDY HARRIS
University of California, Davis
iii
Contents
2. Special Relativity 1
3. Waves and Particles I: Electromagnetic Radiation Behaving as Particles 31
4. Waves and Particles II: Matter Behaving as Waves 41
5. Bound States: Simple Cases 55
6. Unbound States: Obstacles, Tunneling and Particle-Wave Propagation 76
7. Quantum Mechanics in Three Dimensions and the Hydrogen Atom 92
8. Spin and Atomic Physics 115
9. Statistical Mechanics 132
10. Bonding: Molecules and Solids 151
11. Nuclear Physics 166
12. Fundamental Particles and Interactions 178
Contents
2. Special Relativity 1
3. Waves and Particles I: Electromagnetic Radiation Behaving as Particles 31
4. Waves and Particles II: Matter Behaving as Waves 41
5. Bound States: Simple Cases 55
6. Unbound States: Obstacles, Tunneling and Particle-Wave Propagation 76
7. Quantum Mechanics in Three Dimensions and the Hydrogen Atom 92
8. Spin and Atomic Physics 115
9. Statistical Mechanics 132
10. Bonding: Molecules and Solids 151
11. Nuclear Physics 166
12. Fundamental Particles and Interactions 178
1
CHAPTER 2
Special Relativity
2.1 According to Anna, if clocks at the train’s ends sent light signals when they read noon, the signals would reach
Anna together somewhat after noon. According to Bob outside, the signals must start at different times, so that
both reach the moving Anna together. The clocks read noon at different times according to Bob.
2.2 She decides to turn on the lights simultaneously. According to her, light signals leave her brain together and
reach her hands simultaneously, so her hands act simultaneously. According to Bob outside, the light signal
heading toward her trailing hand will reach that hand first, for the hand moves toward the signal. This hand acts
first. The signal heading toward her front hand has to catch up with that hand, taking more time and causing that
hand to act later.
2.3 By symmetry, if an observer in S sees the origin of frame S´ moving at speed v, an observer in S´ must see the
origin of S moving at the same speed.
2.4 (a) Slower.
(b) Later.
(c) Ground observers must see my clock running slowly, so their clock at Y must be ahead. I don’t see clocks
X and Y as synchronized, so when I pass by X, the clock at Y certainly does not read zero, so even though
it does run slowly according to me, it might (must) nevertheless be ahead of mine when I pass it.
2.5 If it passes through in one frame, it must do so in all others. Moving parallel to the ground is an issue of
simultaneity. If you are “on the disk,” the plate has both x- and y-components of velocity and it will not be equal
distances from the ground at the same time—it will be slanted. If slanted, the disk can pass through it without
collision.
2.6 Physical laws are not the same if you are not in an inertial frame. If you are in an accelerating frame, you know it,
no matter what others may be doing. Objects in your frame would appear to change velocity without cause. The
physical laws are always obeyed for the observer in the inertial frame, but not for the twin who turns around.
There is an asymmetry.
2.7 No, clocks run at different rates no matter how low the relative velocity, though considerable total travel time
might be required before a significant effect is noticed.
2.8 I am inertial, so I must see this moving clock running slowly. Each time the space alien passes in front of me I
will see his clock getting further and further behind. Passing observers always agree on the readings of local
clocks, so he will agree that his clock is getting further and further behind—that my clock is running faster than
his. He is in a frame constantly accelerating toward me, so according to him, my clock is continuously “jumping
ahead,” as in the Twin Paradox.
2.9 Sound involves a medium, and three speeds—wave, source, and observer—relative to that. Light has no medium.
The only speeds are the relative velocity of the source and observer, and of course the speed of light.
CHAPTER 2
Special Relativity
2.1 According to Anna, if clocks at the train’s ends sent light signals when they read noon, the signals would reach
Anna together somewhat after noon. According to Bob outside, the signals must start at different times, so that
both reach the moving Anna together. The clocks read noon at different times according to Bob.
2.2 She decides to turn on the lights simultaneously. According to her, light signals leave her brain together and
reach her hands simultaneously, so her hands act simultaneously. According to Bob outside, the light signal
heading toward her trailing hand will reach that hand first, for the hand moves toward the signal. This hand acts
first. The signal heading toward her front hand has to catch up with that hand, taking more time and causing that
hand to act later.
2.3 By symmetry, if an observer in S sees the origin of frame S´ moving at speed v, an observer in S´ must see the
origin of S moving at the same speed.
2.4 (a) Slower.
(b) Later.
(c) Ground observers must see my clock running slowly, so their clock at Y must be ahead. I don’t see clocks
X and Y as synchronized, so when I pass by X, the clock at Y certainly does not read zero, so even though
it does run slowly according to me, it might (must) nevertheless be ahead of mine when I pass it.
2.5 If it passes through in one frame, it must do so in all others. Moving parallel to the ground is an issue of
simultaneity. If you are “on the disk,” the plate has both x- and y-components of velocity and it will not be equal
distances from the ground at the same time—it will be slanted. If slanted, the disk can pass through it without
collision.
2.6 Physical laws are not the same if you are not in an inertial frame. If you are in an accelerating frame, you know it,
no matter what others may be doing. Objects in your frame would appear to change velocity without cause. The
physical laws are always obeyed for the observer in the inertial frame, but not for the twin who turns around.
There is an asymmetry.
2.7 No, clocks run at different rates no matter how low the relative velocity, though considerable total travel time
might be required before a significant effect is noticed.
2.8 I am inertial, so I must see this moving clock running slowly. Each time the space alien passes in front of me I
will see his clock getting further and further behind. Passing observers always agree on the readings of local
clocks, so he will agree that his clock is getting further and further behind—that my clock is running faster than
his. He is in a frame constantly accelerating toward me, so according to him, my clock is continuously “jumping
ahead,” as in the Twin Paradox.
2.9 Sound involves a medium, and three speeds—wave, source, and observer—relative to that. Light has no medium.
The only speeds are the relative velocity of the source and observer, and of course the speed of light.
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Chapter 2 Special Relativity
2
2.10 First, for a massive object, E = mc2 is incomplete. If something is moving, it is E =
umc2, which would
invalidate the argument. Second, some things have energy and no mass, so if any of these are generated, the
energy and mass of what remains could change.
2.11 No, for if we consider a frame in which the initial object is at rest, kinetic energy clearly increases, so mass
would have to decrease.
2.12 No, for if we consider a frame in which the final object is at rest, kinetic energy clearly decreases, so mass would
have to increase.
2.13 Yes, of course. We could consider a frame in which it is either at rest or moving, giving different values for KE.
No, for mass/internal energy is the same in any frame of reference, so the change in mass is the same, and so the
corresponding opposite change in kinetic energy must be the same.
2.14 Yes, if it loses internal energy—in whatever form it may be taken away—it loses mass.
2.15 No, the light was emitted in a frame that, relative to me, is moving away, so I see a longer wavelength (a very
slight change). The observer at the front waits longer to see the light, so a greater relative velocity is involved,
and thus a larger red shift. Simply turn the bus up on end and let gravity accomplish the same thing.
2.16 No electron travels from one side of the screen to the other. Nothing that can have any information about the left-
hand-side of the screen moves to the right-hand-side. The events (electrons hitting phosphors) are all planned
ahead of time, and have no effect upon one another.
2.17
v ≥ 1. Classical mechanics applies when v << c, and
v = 1. At what speed will
v be 1.01?2 2
1
1− v c
= 1.01 0.14c
2.18 Inserting x´ = –vt´ and x = 0 into (2-4) yields –vt´ = 0 + Bt and t´ = 0 + Dt. Dividing these equations gives –v =
B/D. Inserting x´ = ct´ and x = ct into (2-4) yields ct´ = Act + Bt = At(c – v) or t´ = At(1 – v/c) and t´ = Cct + Dt =
t(Cc + A). Setting these two equations equal gives A (1 – v/c) = Cc + A or C = –v/c2 A.
2.19 If I am in frame S and say that the post in S´ is shorter, then my saw will not slice off any of the post in frame S´,
but the saw in frame S´ will slice off some of the post in my frame S. An observer in frame S´, where the relative
speed (if not direction) is the same, would also have to say that the post in the “other frame,” frame S, is shorter.
Therefore, the saw in frame S will slice off some of the post in frame S´. This is a contradiction. Thus, the posts
cannot be contracted in either frame. The same argument would work if we supposed that the post in the “other
frame” were elongated rather than contracted.
2.20 Your time is longer. tyou =
v tCarl → 60s =2
1
1 (0.5)− tCarl tCarl = 52s
2.21 L = L02 2
1− v c → 35m = L02
1 (0.6)− L0 = 43.75m
2.22 Let Anna be S, Bob S. It is simplest to let t = t = 0 be the time when Anna passes Earth. According to Bob, the
explosion event occurs at x = 5ly, t = 2yr.
2
2.10 First, for a massive object, E = mc2 is incomplete. If something is moving, it is E =
umc2, which would
invalidate the argument. Second, some things have energy and no mass, so if any of these are generated, the
energy and mass of what remains could change.
2.11 No, for if we consider a frame in which the initial object is at rest, kinetic energy clearly increases, so mass
would have to decrease.
2.12 No, for if we consider a frame in which the final object is at rest, kinetic energy clearly decreases, so mass would
have to increase.
2.13 Yes, of course. We could consider a frame in which it is either at rest or moving, giving different values for KE.
No, for mass/internal energy is the same in any frame of reference, so the change in mass is the same, and so the
corresponding opposite change in kinetic energy must be the same.
2.14 Yes, if it loses internal energy—in whatever form it may be taken away—it loses mass.
2.15 No, the light was emitted in a frame that, relative to me, is moving away, so I see a longer wavelength (a very
slight change). The observer at the front waits longer to see the light, so a greater relative velocity is involved,
and thus a larger red shift. Simply turn the bus up on end and let gravity accomplish the same thing.
2.16 No electron travels from one side of the screen to the other. Nothing that can have any information about the left-
hand-side of the screen moves to the right-hand-side. The events (electrons hitting phosphors) are all planned
ahead of time, and have no effect upon one another.
2.17
v ≥ 1. Classical mechanics applies when v << c, and
v = 1. At what speed will
v be 1.01?2 2
1
1− v c
= 1.01 0.14c
2.18 Inserting x´ = –vt´ and x = 0 into (2-4) yields –vt´ = 0 + Bt and t´ = 0 + Dt. Dividing these equations gives –v =
B/D. Inserting x´ = ct´ and x = ct into (2-4) yields ct´ = Act + Bt = At(c – v) or t´ = At(1 – v/c) and t´ = Cct + Dt =
t(Cc + A). Setting these two equations equal gives A (1 – v/c) = Cc + A or C = –v/c2 A.
2.19 If I am in frame S and say that the post in S´ is shorter, then my saw will not slice off any of the post in frame S´,
but the saw in frame S´ will slice off some of the post in my frame S. An observer in frame S´, where the relative
speed (if not direction) is the same, would also have to say that the post in the “other frame,” frame S, is shorter.
Therefore, the saw in frame S will slice off some of the post in frame S´. This is a contradiction. Thus, the posts
cannot be contracted in either frame. The same argument would work if we supposed that the post in the “other
frame” were elongated rather than contracted.
2.20 Your time is longer. tyou =
v tCarl → 60s =2
1
1 (0.5)− tCarl tCarl = 52s
2.21 L = L02 2
1− v c → 35m = L02
1 (0.6)− L0 = 43.75m
2.22 Let Anna be S, Bob S. It is simplest to let t = t = 0 be the time when Anna passes Earth. According to Bob, the
explosion event occurs at x = 5ly, t = 2yr.
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Instructor Solutions Manual for Harris, Modern Physics, Second Edition
3
According to Anna,
x =
v (x − v t) =( )
2
1 (5ly) (0.8 )(2yr)
1 (0.8) −
− c = 5.67ly,
and t =
v2
− +
v x t
c =22
1 0.8 (5ly) 2yr
1 (0.8 )
− +
−
c
cc = −3.33yr.
Negative!? How can this be? To gain some appreciation of the Lorentz transformation, let’s see how involved it
is to solve the problem without it. According to Bob: The planet explodes at t = 2yr. At this time, Anna will have
moved (0.8c) (2yr) = 1.6ly. Bob sees a distance between Anna and Planet Y of 3.4ly and a relative velocity
between Anna and the light from the explosion of 1.8c. So the light from the explosion will reach Anna in
another t = 3.4ly/1.8c = 1.89yr—at the time, according to Bob, of 2 + 1.89 = 3.889yr. The distance he now sees
between Anna and Planet Y (or the center of the debris) is 5ly − (0.8c) (3.889yr) = 1.889ly. Meanwhile he will
have seen less time go by on Anna’s clock:2
1 (0.8)− 3.889yr = 2.33yr. According to Anna: The distance Planet
Y is from herself when she receives the bad news will be shorter:2
1 (0.8)− 1.889ly = 1.133ly. The big
question: If Planet Y moves toward Anna at 0.8c, and the light from the explosion moves toward Anna at c, how
long will it take for the light to get 1.133ly in front of the planet? The relative velocity between Planet Y and the
light from the explosion is 0.2c, and 0.2ct = 1.133ly t = 5.667yr. If Anna’s clock reads 2.33yr when she
gets the bad news, and it took 5.667yr for the news to reach her, the explosion must have occurred at t = −3.33yr.
The problem is a good example of two events that occur in a different order in two frames of reference. The
explosion (one event) occurs before Anna and Bob cross (another event) according to Anna, but after according
to Bob.
2.23 Let Anna be S, Bob S. The flash of the flashbulb is the event in question here. We seek Bob’s time. We know
that it is 100ns 27ns, i.e., either 127ns or 73ns.2v
v
t x t
c
= +
=22
1 0.6 100ns
1 (0.6 )
c x
cc
+
− =0.75
c x + 125ns.
If x is positive (Anna’s arm is stretched out in the positive direction), then it must be 127ns.
Later. We might be tempted to use time dilation to answer this, but care is necessary. There is no doubt that the
wristwatch on Anna’s hand reads 100ns when the flash goes off. But Bob will not agree that it reads zero when
Anna passed him. It is set back a bit according to Bob (like the front clocks in Example 2.4). Thus, Bob sees
slightly more than 100ns go by on the wristwatch. Time dilation would then give the time Bob sees go by on his
own: slightly more than
0.6(100ns) = 125ns.
(b) Now knowing that t must be 127ns, we can solve for x, the location in Anna’s frame where the flash occurs
(i.e., her hand). 12710−9s =8
0.75
3 10 m/s x + 12510−9s x = 0.8m. (A reasonable arm length.)
2.24 Bob, standing in the barn/S, sees a length L =2 2
1 /− v c L. 10ft =2 2
1 / (16ft)− v c v = 0.78c.
(b) The observer stationary in the barn. Pole−vaulter Anna, S, will never agree that the pole fits in all at
once; the barn is shorter than 10ft!
(c) We seek t, knowing that, according to a barn observer, the ends of the pole are at the doors
simultaneously, i.e., t = 0, and that the distance between these events is 10ft = 3.048m.2 1 2 1 2 12 ( ) ( )
− = − − + −
v
v
t t x x t t
c
=22
1 0.78 (3.08m) 0
1 (0.78 )
− +
−
c
cc = −1.2710−8s.
3
According to Anna,
x =
v (x − v t) =( )
2
1 (5ly) (0.8 )(2yr)
1 (0.8) −
− c = 5.67ly,
and t =
v2
− +
v x t
c =22
1 0.8 (5ly) 2yr
1 (0.8 )
− +
−
c
cc = −3.33yr.
Negative!? How can this be? To gain some appreciation of the Lorentz transformation, let’s see how involved it
is to solve the problem without it. According to Bob: The planet explodes at t = 2yr. At this time, Anna will have
moved (0.8c) (2yr) = 1.6ly. Bob sees a distance between Anna and Planet Y of 3.4ly and a relative velocity
between Anna and the light from the explosion of 1.8c. So the light from the explosion will reach Anna in
another t = 3.4ly/1.8c = 1.89yr—at the time, according to Bob, of 2 + 1.89 = 3.889yr. The distance he now sees
between Anna and Planet Y (or the center of the debris) is 5ly − (0.8c) (3.889yr) = 1.889ly. Meanwhile he will
have seen less time go by on Anna’s clock:2
1 (0.8)− 3.889yr = 2.33yr. According to Anna: The distance Planet
Y is from herself when she receives the bad news will be shorter:2
1 (0.8)− 1.889ly = 1.133ly. The big
question: If Planet Y moves toward Anna at 0.8c, and the light from the explosion moves toward Anna at c, how
long will it take for the light to get 1.133ly in front of the planet? The relative velocity between Planet Y and the
light from the explosion is 0.2c, and 0.2ct = 1.133ly t = 5.667yr. If Anna’s clock reads 2.33yr when she
gets the bad news, and it took 5.667yr for the news to reach her, the explosion must have occurred at t = −3.33yr.
The problem is a good example of two events that occur in a different order in two frames of reference. The
explosion (one event) occurs before Anna and Bob cross (another event) according to Anna, but after according
to Bob.
2.23 Let Anna be S, Bob S. The flash of the flashbulb is the event in question here. We seek Bob’s time. We know
that it is 100ns 27ns, i.e., either 127ns or 73ns.2v
v
t x t
c
= +
=22
1 0.6 100ns
1 (0.6 )
c x
cc
+
− =0.75
c x + 125ns.
If x is positive (Anna’s arm is stretched out in the positive direction), then it must be 127ns.
Later. We might be tempted to use time dilation to answer this, but care is necessary. There is no doubt that the
wristwatch on Anna’s hand reads 100ns when the flash goes off. But Bob will not agree that it reads zero when
Anna passed him. It is set back a bit according to Bob (like the front clocks in Example 2.4). Thus, Bob sees
slightly more than 100ns go by on the wristwatch. Time dilation would then give the time Bob sees go by on his
own: slightly more than
0.6(100ns) = 125ns.
(b) Now knowing that t must be 127ns, we can solve for x, the location in Anna’s frame where the flash occurs
(i.e., her hand). 12710−9s =8
0.75
3 10 m/s x + 12510−9s x = 0.8m. (A reasonable arm length.)
2.24 Bob, standing in the barn/S, sees a length L =2 2
1 /− v c L. 10ft =2 2
1 / (16ft)− v c v = 0.78c.
(b) The observer stationary in the barn. Pole−vaulter Anna, S, will never agree that the pole fits in all at
once; the barn is shorter than 10ft!
(c) We seek t, knowing that, according to a barn observer, the ends of the pole are at the doors
simultaneously, i.e., t = 0, and that the distance between these events is 10ft = 3.048m.2 1 2 1 2 12 ( ) ( )
− = − − + −
v
v
t t x x t t
c
=22
1 0.78 (3.08m) 0
1 (0.78 )
− +
−
c
cc = −1.2710−8s.
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Chapter 2 Special Relativity
4
Why negative? Since we chose x2−x1 as positive, it must be that the front of the pole reaching its door is
Event 2, the back reaching its door Event 1. The answer shows that the front time is a smaller time; it
occurs earlier, as it must. According to the pole−vaulter/S, front leaves before back enters.
2.25
0.8c =2
1
1 (0.8)− =5 .
3 Bob sees Anna’s ship contracted to 100m/
v = 100m/5
3 = 60m, so Bob Jr. will have to be
at x = 60m.
(b) We seek t, knowing x, x, and t.2
= − +
v
v
t x t
c =8
5 0.8 (60m) 0
3 3 10 m/s
− +
= −2.67 10−7s.
2.26 Calling the front light Event 2, Anna frame S,( ) ( )
2 1 2 1 2 12
− = − + −
v
v
t t x x t t
c =2 (60m) 0
+
v
v
c . Since
this is positive, the front time is the larger (later), so back light must go on first.
(b) 4010−9s =22 2
1 (60m)
1 /−
v
cv c →( )
2 2 9
1 / (40 10 s)−
− v c 2 c2
= (v2/c2) (60m)2 → (4010−9s)2 (3108m/s)2
=( )
22 9 2 8
(60m) (40 10 s) (3 10 m/s−
+ v2/c2
v/c = 0.196
2.27 Bob and Bob Jr. see a 25m−long plane. L = L0 /
v → 25m = 40m2 2
1 /− v c v
c = 0.781
(b) If zero on Anna’s clock is the front of her spaceplane (where Bob is), then the time at the back occurs at
x = −40m, as well as at x = −25m and t = 0.2
= − +
v
v
t x t
c
=82
1 0.781 ( 25m) 0
3 10 m/s1 (0.781)
− − +
−
= 1.0410−7s.
A positive number. Sensible. Back enters after front leaves.
(c) We know that t = 104ns (“at this time”) and x = 0 (Bob’s location), and seek x. x =
v (x + v t) →( )
8 7
2
1
0 (0.781 3 10 m/s)(1.04 10 s)
1 (0.781)
−
= +
− x
x = −24.375m. The front of her ship is the
origin, so 24.375m sticks out. Note: “at this time”, observers at the back (x = −40m) and at x = −24.375m,
i.e., 15.625m apart in Anna’s frame, see the ends of Bob’s hangar. This is correct: Anna should see a hangar
length of L = 25m/
v = 25m2
1 (0.781)− = 15.625m.
4
Why negative? Since we chose x2−x1 as positive, it must be that the front of the pole reaching its door is
Event 2, the back reaching its door Event 1. The answer shows that the front time is a smaller time; it
occurs earlier, as it must. According to the pole−vaulter/S, front leaves before back enters.
2.25
0.8c =2
1
1 (0.8)− =5 .
3 Bob sees Anna’s ship contracted to 100m/
v = 100m/5
3 = 60m, so Bob Jr. will have to be
at x = 60m.
(b) We seek t, knowing x, x, and t.2
= − +
v
v
t x t
c =8
5 0.8 (60m) 0
3 3 10 m/s
− +
= −2.67 10−7s.
2.26 Calling the front light Event 2, Anna frame S,( ) ( )
2 1 2 1 2 12
− = − + −
v
v
t t x x t t
c =2 (60m) 0
+
v
v
c . Since
this is positive, the front time is the larger (later), so back light must go on first.
(b) 4010−9s =22 2
1 (60m)
1 /−
v
cv c →( )
2 2 9
1 / (40 10 s)−
− v c 2 c2
= (v2/c2) (60m)2 → (4010−9s)2 (3108m/s)2
=( )
22 9 2 8
(60m) (40 10 s) (3 10 m/s−
+ v2/c2
v/c = 0.196
2.27 Bob and Bob Jr. see a 25m−long plane. L = L0 /
v → 25m = 40m2 2
1 /− v c v
c = 0.781
(b) If zero on Anna’s clock is the front of her spaceplane (where Bob is), then the time at the back occurs at
x = −40m, as well as at x = −25m and t = 0.2
= − +
v
v
t x t
c
=82
1 0.781 ( 25m) 0
3 10 m/s1 (0.781)
− − +
−
= 1.0410−7s.
A positive number. Sensible. Back enters after front leaves.
(c) We know that t = 104ns (“at this time”) and x = 0 (Bob’s location), and seek x. x =
v (x + v t) →( )
8 7
2
1
0 (0.781 3 10 m/s)(1.04 10 s)
1 (0.781)
−
= +
− x
x = −24.375m. The front of her ship is the
origin, so 24.375m sticks out. Note: “at this time”, observers at the back (x = −40m) and at x = −24.375m,
i.e., 15.625m apart in Anna’s frame, see the ends of Bob’s hangar. This is correct: Anna should see a hangar
length of L = 25m/
v = 25m2
1 (0.781)− = 15.625m.
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Instructor Solutions Manual for Harris, Modern Physics, Second Edition
5
2.28 Since Bob moves to the right, let’s make his frame S. Thus, we seek times in frame S.
v = 2 →2 2
1
1 /− v c = 2
3
2
=v c . The clock at the back of Anna’s ship is presently at x =1
2 L0. (It is also at x = L0.) Bob looks at this
clock at t = 0 on his own clock. Call this Event a. With xa =1
2 L0, ta = 0, we may find ta, the time in Anna’s
frame.2
= +
a v a a
v
t x t
c0
3 2 1
2 0
2
= +
L
c =3
2c L0. The clock at the center of Anna’s ship is at x =1
4 L0
(rather than1
2 L0) in Bob’s frame. The time on this clock, tb, will therefore simply be half the previous result.
tb =3
4c L0.
2.29 If it takes light 40ns to travel half the length of Bob’s ship, then:1
2 L0 = (3108m/s)(4010−9s) L0 = 24m.
(b) The origins are at the center. We know that the event at x = 12m (i.e., the front of Anna’s ship) and
t = −24ns occurs at t = 0.2
= +
v
v
t x t
c
→9
2
0 (12m) ( 24 10 s)
−
= + −
v
v
c v
c = 0.6c.
(c)
0.8c =2
1 5
41 (0.8) =
− . Logically, at the initial instant, if the front clock says −24ns, the back has to say
+24ns. Now consider Anna’s origin at t = 20ns.2
= +
v
v
t x t
c → 2010−9s =8
5 0.6 0
4 3 10 /
− +
t
m s t
= 1610−9s. Agrees. Another route: If 20ns passes for Bob, he must see less time pass on the clock’s in
Anna’s frame. Since
v =5 1
4 0.8
= , he must see 80% as much time pass, and 80% of 20ns is 16ns. Similarly,
Bob must see 16ns pass on all of Anna’s clocks as diagram shows. At t = 40ns, Bob must see that 80% of
40ns, 32ns, has passed on Anna’s clocks as shown. Finally, 80ns for Bob must show 80% of 80ns, or 64ns,
pass on Anna’s clocks.
2.30 It is tricky to use time dilation to relate readings on two different clocks moving relative to you. You would never
agree that the gas station clocks were synchronized. However, gas station observers (S) watch a single clock
(yours, S) and must observe time dilation. When you get to the second station, observers there must see your
clock as having registered less time than in their frame: the gas station clock must be ahead.
5
2.28 Since Bob moves to the right, let’s make his frame S. Thus, we seek times in frame S.
v = 2 →2 2
1
1 /− v c = 2
3
2
=v c . The clock at the back of Anna’s ship is presently at x =1
2 L0. (It is also at x = L0.) Bob looks at this
clock at t = 0 on his own clock. Call this Event a. With xa =1
2 L0, ta = 0, we may find ta, the time in Anna’s
frame.2
= +
a v a a
v
t x t
c0
3 2 1
2 0
2
= +
L
c =3
2c L0. The clock at the center of Anna’s ship is at x =1
4 L0
(rather than1
2 L0) in Bob’s frame. The time on this clock, tb, will therefore simply be half the previous result.
tb =3
4c L0.
2.29 If it takes light 40ns to travel half the length of Bob’s ship, then:1
2 L0 = (3108m/s)(4010−9s) L0 = 24m.
(b) The origins are at the center. We know that the event at x = 12m (i.e., the front of Anna’s ship) and
t = −24ns occurs at t = 0.2
= +
v
v
t x t
c
→9
2
0 (12m) ( 24 10 s)
−
= + −
v
v
c v
c = 0.6c.
(c)
0.8c =2
1 5
41 (0.8) =
− . Logically, at the initial instant, if the front clock says −24ns, the back has to say
+24ns. Now consider Anna’s origin at t = 20ns.2
= +
v
v
t x t
c → 2010−9s =8
5 0.6 0
4 3 10 /
− +
t
m s t
= 1610−9s. Agrees. Another route: If 20ns passes for Bob, he must see less time pass on the clock’s in
Anna’s frame. Since
v =5 1
4 0.8
= , he must see 80% as much time pass, and 80% of 20ns is 16ns. Similarly,
Bob must see 16ns pass on all of Anna’s clocks as diagram shows. At t = 40ns, Bob must see that 80% of
40ns, 32ns, has passed on Anna’s clocks as shown. Finally, 80ns for Bob must show 80% of 80ns, or 64ns,
pass on Anna’s clocks.
2.30 It is tricky to use time dilation to relate readings on two different clocks moving relative to you. You would never
agree that the gas station clocks were synchronized. However, gas station observers (S) watch a single clock
(yours, S) and must observe time dilation. When you get to the second station, observers there must see your
clock as having registered less time than in their frame: the gas station clock must be ahead.
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Chapter 2 Special Relativity
6
(b) In the frame of the gas stations, t = 900m/20m/s = 45s. t =
v t → 45s =8 2
1 (20 / 3 10 )
−
t t =( )
1
8 2 21 (20 / 3 10 ) (45s)−
8 21
1 (20 / 3 10 ) (45s)
2
−
= 45s − 10−13s. Yours is behind by 10−13s. Another
route: Time zero is you (S) passing first station. What would an observer in your frame and abreast the
second station at t = 0 (it’s a long bus) see on the second station’s clock? x = 900m, t = 0 t = ?2
0
= − +
v
v x t
c
→8 2
20m/s
0 (900m)
(3 10 m/s)
= − +
v t t = 210−13s. According to you, then, the
second station’s clock is ahead of the first’s by 210−13s. Now, you do see less time pass on those clocks
than on your own—10−13s less—but this will still leave the second station’s clock ahead of yours by 10−13s
when you get there. Yet another route: What t will your clock read when you pass the second station, at x =
900m, t = 45s?2
= − +
v
v
t x t
c
=8 28 2
1 20m/s (900m) 45s
(3 10 m/s)1 (20 / 3 10 )
− +
−
( )
8 2 131
1 (20 / 3 10 ) 2 10 s 45s
2
−
+ − +
= 45s − 210−13s + 110−13s − 210−26s
45s − 110−13s, i.e., 10−13s behind.
2.31 According to those on the ground, this is simple classical mechanics: a speed and distance according to a ground
observer are used to find a time according to a ground observer. Let’s call the ground frame S. t =6
8
4 10 m
0.8 3 10 m/s
= 1.6710−2s. There are two ways to find the time on the spaceplane—time dilation, or length
contraction. Time dilation: The ground observer will see more time pass on his clock than on the plane. His is t,
the plane’s t. (Note:
0.8c = 1.67) t =
v t → 1.6710−2s = (1.67) t t = 0.0100s.
Length contraction: The plane observer sees a different coast−to−coast distance. He sees L
=2 6
1 (0.8) (4 10 m)− = 2.4106m. How much time will it take a 2,400m object to pass at 0.8c? Classical
mechanics: t =6
8
2.4 10 m
0.8 3 10 m/s
= 0.0100s. The clock seen on the plane from the ground will be 0.0167s −
0.0100s = 0.0067s behind. But how can it be behind if the people on the plane must see time on ground clocks
passing more slowly? Should they not see the ground clocks behind? Assuming the plane and ground clocks read
zero as the plane started across the country, the plane clock will indeed be behind the ground clock on the other
coast, just as we calculated. The people on the plane will by no means agree that the clock on the far coast was
synchronized, that it read zero when the plane started across country (relative simultaneity). In fact, they will say
that it was significantly ahead, so that, even though they “see” less time pass on this far−off clock, they will agree
that it is ahead when they pass it.
2.32 Passengers see a 5km-long object shrunk to L = LRest/v =2
5km 1 (0.8)− = 3km. If the ends of this object are at
the train’s ends simultaneously (according to observers on the train), the train must be 3km long.
(b) Look at it from the ground’s point of view, the train is not 5km long; it’s not even 3km long, but L = LRest/v
=2
3km 1 (0.8)− = 1.8km. Surely the back end will pass its station before the front end passes its, so the
front station is the more current information. No reason to slow down.
6
(b) In the frame of the gas stations, t = 900m/20m/s = 45s. t =
v t → 45s =8 2
1 (20 / 3 10 )
−
t t =( )
1
8 2 21 (20 / 3 10 ) (45s)−
8 21
1 (20 / 3 10 ) (45s)
2
−
= 45s − 10−13s. Yours is behind by 10−13s. Another
route: Time zero is you (S) passing first station. What would an observer in your frame and abreast the
second station at t = 0 (it’s a long bus) see on the second station’s clock? x = 900m, t = 0 t = ?2
0
= − +
v
v x t
c
→8 2
20m/s
0 (900m)
(3 10 m/s)
= − +
v t t = 210−13s. According to you, then, the
second station’s clock is ahead of the first’s by 210−13s. Now, you do see less time pass on those clocks
than on your own—10−13s less—but this will still leave the second station’s clock ahead of yours by 10−13s
when you get there. Yet another route: What t will your clock read when you pass the second station, at x =
900m, t = 45s?2
= − +
v
v
t x t
c
=8 28 2
1 20m/s (900m) 45s
(3 10 m/s)1 (20 / 3 10 )
− +
−
( )
8 2 131
1 (20 / 3 10 ) 2 10 s 45s
2
−
+ − +
= 45s − 210−13s + 110−13s − 210−26s
45s − 110−13s, i.e., 10−13s behind.
2.31 According to those on the ground, this is simple classical mechanics: a speed and distance according to a ground
observer are used to find a time according to a ground observer. Let’s call the ground frame S. t =6
8
4 10 m
0.8 3 10 m/s
= 1.6710−2s. There are two ways to find the time on the spaceplane—time dilation, or length
contraction. Time dilation: The ground observer will see more time pass on his clock than on the plane. His is t,
the plane’s t. (Note:
0.8c = 1.67) t =
v t → 1.6710−2s = (1.67) t t = 0.0100s.
Length contraction: The plane observer sees a different coast−to−coast distance. He sees L
=2 6
1 (0.8) (4 10 m)− = 2.4106m. How much time will it take a 2,400m object to pass at 0.8c? Classical
mechanics: t =6
8
2.4 10 m
0.8 3 10 m/s
= 0.0100s. The clock seen on the plane from the ground will be 0.0167s −
0.0100s = 0.0067s behind. But how can it be behind if the people on the plane must see time on ground clocks
passing more slowly? Should they not see the ground clocks behind? Assuming the plane and ground clocks read
zero as the plane started across the country, the plane clock will indeed be behind the ground clock on the other
coast, just as we calculated. The people on the plane will by no means agree that the clock on the far coast was
synchronized, that it read zero when the plane started across country (relative simultaneity). In fact, they will say
that it was significantly ahead, so that, even though they “see” less time pass on this far−off clock, they will agree
that it is ahead when they pass it.
2.32 Passengers see a 5km-long object shrunk to L = LRest/v =2
5km 1 (0.8)− = 3km. If the ends of this object are at
the train’s ends simultaneously (according to observers on the train), the train must be 3km long.
(b) Look at it from the ground’s point of view, the train is not 5km long; it’s not even 3km long, but L = LRest/v
=2
3km 1 (0.8)− = 1.8km. Surely the back end will pass its station before the front end passes its, so the
front station is the more current information. No reason to slow down.
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Instructor Solutions Manual for Harris, Modern Physics, Second Edition
7
(c) From the ground point of view, the train travels 5km – 1.8km = 3.2km from the time the employee at the
back sees his sign to the time the conductor sees his: t = dist/speed = 3.2103m / (0.8 3 108m/s) =
13.3μs.
2.33 According to the muons, the distance from the mountaintop to sea level is2
1910m 1 (0.9952) 187m.− =
According to a muon, this distance would pass in7
8
187m 6.26 10 s 0.626 s
0.9952 3 10 m/s
−
= =
.0.626 / 2.2
0
0.75.−
= =
N e
N
And395 0.75.
527 =
2.34 We may “work in either frame”. Muon’s frame:
= 2.2s. Distance to Earth is shorter: 4km2
1 (0.93)−
= 1.47km. t =8
dist 1, 470m
speed 0.93 3 10 m/s
= = 5.2710−6s.0
N
N = e−(5.27/2.2) = e−2.4 or 9.1%.
Earth frame: Distance to Earth is 4km. Lifetime is longer.2
2.2
1 (0.93)
−
s = 5.99s.
t =8
dist 4, 000m
speed 0.93 3 10 m/s
= = 1.4310−5s.0
N
N = e−(14.3/5.99) = e−2.4 or 9.1%
(b)
= 2.2
s. t =8
dist 4, 000m
speed 0.93 3 10 m/s
= = 1.4310−5s.0
N
N = e−(14.3/2.2) = e−6.5 or 0.14%
2.35 According to observers on the plane, these two events occur at the same location. Their times differ by t0.0
2 2
1 /
= →
−
t
t
v c( )
1
28 2
0 1 (420 / 3 10 ) (10s) = − t
( )
8 21
2
1 (420 / 3 10 ) (10s)− = 10s − 9.810−12s.
The plane’s clock will read 9.8ps earlier. May also solve in plane frame, where 4.2 km is contracted.
2.36 L =2 2
1 /− v c L → 50m − 0.110−9m =2 2
1 / (50m)− v c . Must use the approximation, since v/c is apparently
small.( )
( )
1
22 2
1+ −v c
1 −2
2
1
2
v
c .
50m − 0.110−9m =2
2
1
1 (50m)
2
−
v
c
0.110−9m =2
2
1
2
v
c (50m)
v = 210−6c = 600m/s.
2.37 There are two equivalent routes. First: Anna doesn’t see a distance of 20ly, but rather a length−contracted one of
20ly/
, and it passes in twenty years of her life, so its speed is v =20ly /
20yr
v = v
c . Solving,v
c =2
2
1− v
c or2
2
v
c =
1 −2
2
v
c or v =1
2 c.
7
(c) From the ground point of view, the train travels 5km – 1.8km = 3.2km from the time the employee at the
back sees his sign to the time the conductor sees his: t = dist/speed = 3.2103m / (0.8 3 108m/s) =
13.3μs.
2.33 According to the muons, the distance from the mountaintop to sea level is2
1910m 1 (0.9952) 187m.− =
According to a muon, this distance would pass in7
8
187m 6.26 10 s 0.626 s
0.9952 3 10 m/s
−
= =
.0.626 / 2.2
0
0.75.−
= =
N e
N
And395 0.75.
527 =
2.34 We may “work in either frame”. Muon’s frame:
= 2.2s. Distance to Earth is shorter: 4km2
1 (0.93)−
= 1.47km. t =8
dist 1, 470m
speed 0.93 3 10 m/s
= = 5.2710−6s.0
N
N = e−(5.27/2.2) = e−2.4 or 9.1%.
Earth frame: Distance to Earth is 4km. Lifetime is longer.2
2.2
1 (0.93)
−
s = 5.99s.
t =8
dist 4, 000m
speed 0.93 3 10 m/s
= = 1.4310−5s.0
N
N = e−(14.3/5.99) = e−2.4 or 9.1%
(b)
= 2.2
s. t =8
dist 4, 000m
speed 0.93 3 10 m/s
= = 1.4310−5s.0
N
N = e−(14.3/2.2) = e−6.5 or 0.14%
2.35 According to observers on the plane, these two events occur at the same location. Their times differ by t0.0
2 2
1 /
= →
−
t
t
v c( )
1
28 2
0 1 (420 / 3 10 ) (10s) = − t
( )
8 21
2
1 (420 / 3 10 ) (10s)− = 10s − 9.810−12s.
The plane’s clock will read 9.8ps earlier. May also solve in plane frame, where 4.2 km is contracted.
2.36 L =2 2
1 /− v c L → 50m − 0.110−9m =2 2
1 / (50m)− v c . Must use the approximation, since v/c is apparently
small.( )
( )
1
22 2
1+ −v c
1 −2
2
1
2
v
c .
50m − 0.110−9m =2
2
1
1 (50m)
2
−
v
c
0.110−9m =2
2
1
2
v
c (50m)
v = 210−6c = 600m/s.
2.37 There are two equivalent routes. First: Anna doesn’t see a distance of 20ly, but rather a length−contracted one of
20ly/
, and it passes in twenty years of her life, so its speed is v =20ly /
20yr
v = v
c . Solving,v
c =2
2
1− v
c or2
2
v
c =
1 −2
2
v
c or v =1
2 c.
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Chapter 2 Special Relativity
8
Second: From Bob’s perspective, Planet Y is 20ly away. But Bob, seeing Anna age twenty years, must, owing to
time−dilation, see more time go by on his own clock:
v 20yr. Anna’s speed is thus v =20ly
20yr
v =
v
c . Same
equation.
2.38 The plank has moved to a different frame, frame S. The length along the x−axis is Lx = Lo cos
0, and along the
y−axis it is Ly = Lo sin
o. An observer in S will see the x−leg contracted, but not the y. Lx = Lx/
v = Lo cos
o2 2
1 / ,− v c
and Ly = Ly = Lo sin
o. L =2 2
+x yL L =( )
2 2 2 2 2 2
0 0cos 1 sin .
− +o oL v c L But sin2
o = 1− cos2
o.
L = Lo( ) ( )
2 2 2 2
cos 1 1 cos
− + −o ov c = Lo2 2 2
1 ( )cos .
− ov c As the angle
o goes from zero to 90º, the
length of the plank to an observer in S goes from Lo2 2
1− v c (simple length contraction) to merely Lo (no
contraction perpendicular to axes of relative motion). Now, tan
=y
x
L
L =2 2
sin
cos 1
−
o o
o o
L
L v c = tan
o2 2
1
1− v c
=
tan
o. As the speed increases, the angle
seen by an observer in S increases. This makes sense in view of
the fact that the faster the relative motion, the shorter the x−component of the plank will be. The y−component
doesn’t change, so the angle is larger.
2.39 If it can pass through according to one observer it must be able to pass through according to the other! There
cannot be a collision in one frame and not in the other! It’s just a matter of when. One argument is that its
dimensions perpendicular to the direction of relative motion don’t change—its height is the same in either frame,
so it cannot hit. Another argument is that while Bob sees a plank and doorway at the same angle
, Anna, at rest
with respect to the plank, sees a plank at a smaller angle,
0. She also sees a wall moving toward her, contracted
along the direction of relative motion and therefore at a larger angle than the
seen by Bob. The top will pass
through first! Let’s say that the bottom passes through at time zero on both clocks. According to Bob, the top will
also (simultaneously) pass through at time zero: t = 0. It can’t according to Anna! The top passing through (an
event) occurs at t = 0 and at x = L0cos
0. Find t.2
= +
v
t x t
c →0 02
0 cos
= +
v
v L t
c t= −2
v
c L0
cos
0. According to Anna, in frame S, the top passes through2
v
c L0 cos
0 earlier. This adds up! We know that
Anna sees a smaller angle than Bob. If it passes through at once according to Bob, the top must pass through first
according to Anna. If it does not go through all at once according to Anna, it can indeed be larger than the “hole”.
8
Second: From Bob’s perspective, Planet Y is 20ly away. But Bob, seeing Anna age twenty years, must, owing to
time−dilation, see more time go by on his own clock:
v 20yr. Anna’s speed is thus v =20ly
20yr
v =
v
c . Same
equation.
2.38 The plank has moved to a different frame, frame S. The length along the x−axis is Lx = Lo cos
0, and along the
y−axis it is Ly = Lo sin
o. An observer in S will see the x−leg contracted, but not the y. Lx = Lx/
v = Lo cos
o2 2
1 / ,− v c
and Ly = Ly = Lo sin
o. L =2 2
+x yL L =( )
2 2 2 2 2 2
0 0cos 1 sin .
− +o oL v c L But sin2
o = 1− cos2
o.
L = Lo( ) ( )
2 2 2 2
cos 1 1 cos
− + −o ov c = Lo2 2 2
1 ( )cos .
− ov c As the angle
o goes from zero to 90º, the
length of the plank to an observer in S goes from Lo2 2
1− v c (simple length contraction) to merely Lo (no
contraction perpendicular to axes of relative motion). Now, tan
=y
x
L
L =2 2
sin
cos 1
−
o o
o o
L
L v c = tan
o2 2
1
1− v c
=
tan
o. As the speed increases, the angle
seen by an observer in S increases. This makes sense in view of
the fact that the faster the relative motion, the shorter the x−component of the plank will be. The y−component
doesn’t change, so the angle is larger.
2.39 If it can pass through according to one observer it must be able to pass through according to the other! There
cannot be a collision in one frame and not in the other! It’s just a matter of when. One argument is that its
dimensions perpendicular to the direction of relative motion don’t change—its height is the same in either frame,
so it cannot hit. Another argument is that while Bob sees a plank and doorway at the same angle
, Anna, at rest
with respect to the plank, sees a plank at a smaller angle,
0. She also sees a wall moving toward her, contracted
along the direction of relative motion and therefore at a larger angle than the
seen by Bob. The top will pass
through first! Let’s say that the bottom passes through at time zero on both clocks. According to Bob, the top will
also (simultaneously) pass through at time zero: t = 0. It can’t according to Anna! The top passing through (an
event) occurs at t = 0 and at x = L0cos
0. Find t.2
= +
v
t x t
c →0 02
0 cos
= +
v
v L t
c t= −2
v
c L0
cos
0. According to Anna, in frame S, the top passes through2
v
c L0 cos
0 earlier. This adds up! We know that
Anna sees a smaller angle than Bob. If it passes through at once according to Bob, the top must pass through first
according to Anna. If it does not go through all at once according to Anna, it can indeed be larger than the “hole”.
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Instructor Solutions Manual for Harris, Modern Physics, Second Edition
9
2.40 We have a speed and time according to the lab and wish to know a distance according to that frame.
distance = (0.94 3108m/s)(0.03210−6s) = 9.02m.
(b) If the experimenter sees 0.032s pass on his own clock, he will see less pass on the clock glued to the
particle.
t =2
1 (0.94)
−
t → 0.032
s =2
1 (0.94)
−
t t = 0.011s.
(c) Length contraction. According to the particle, the lab is2
1 (0.94) (9.02m)− = 3.08m long. Let’s see, if
3.08m of lab passes by in 0.011s, how fast is the lab moving?6
3.08m
0.011 10 s−
= .94c. It all fits!
2.41 According to Earth observer, muon “lives” longer.
t =0
2 2
1 /
−
t
v c =2
1
1 (0.92)− 2.2s = 2.55(2.2s) = 5.61s.
Dist = speed time = (0.92 3108m/s)(5.6110−6s) = 1,549m.
2.42 According to an observer in the lab, the pion survives t =2 2
26ns .
1− v c But moving a distance of 13m at speed v,
this time must also be t =13m
v . Thus:2 2
26ns
1 /− v c =13m
v → v8
2.6 10 s
13m
−
=2 2
1 /− v c →8
5 10 m/s
v
=2 2
1 /− v c →2
8 2
(5 10 )
v = 1 −2
8 2
(3 10 )
v v = 2.57108m/s.
2.43 Calling to the right positive and Anna frame S, we know that this tick of the clock/event occurs at t = 0 and
x =1
2 according to Bob. We seek t. t =2
− +
v
v x t
c .
v =2
1
1
1 2
−
=2 .
Thus t =1 2
2 (30m) 0
− +
c = −8
30m
3 10 m/s = −10−7s = −100ns.
(b) We have a distance traveled according to Bob, 30m, and a speed, and so find a time according to Bob via
classical mechanics. t =30m
2c =8
30m 2
3 10 m/s = 1.4110−7s = 141ns.
(c) If Anna’s back clock has moved to the middle of Bob’s ship, Anna’s front clock, formerly at Bob’s middle,
will now be at Bob’s front, x = 60m. Thus the locations, according to Bob, are x = 30 and x = 60m. The
times are both 141ns.
For Anna’s back clock: t =91 2
2 (30m) 141 10 s−
− +
c = −8
30m
3 10 m/s +2 14110−9s = 100ns.
For Anna’s front: t =91/ 2
2 (60m) 141 10 s
c
−
− +
= −8
60 m
3 10 m / s +2 14110−9s = 0.
9
2.40 We have a speed and time according to the lab and wish to know a distance according to that frame.
distance = (0.94 3108m/s)(0.03210−6s) = 9.02m.
(b) If the experimenter sees 0.032s pass on his own clock, he will see less pass on the clock glued to the
particle.
t =2
1 (0.94)
−
t → 0.032
s =2
1 (0.94)
−
t t = 0.011s.
(c) Length contraction. According to the particle, the lab is2
1 (0.94) (9.02m)− = 3.08m long. Let’s see, if
3.08m of lab passes by in 0.011s, how fast is the lab moving?6
3.08m
0.011 10 s−
= .94c. It all fits!
2.41 According to Earth observer, muon “lives” longer.
t =0
2 2
1 /
−
t
v c =2
1
1 (0.92)− 2.2s = 2.55(2.2s) = 5.61s.
Dist = speed time = (0.92 3108m/s)(5.6110−6s) = 1,549m.
2.42 According to an observer in the lab, the pion survives t =2 2
26ns .
1− v c But moving a distance of 13m at speed v,
this time must also be t =13m
v . Thus:2 2
26ns
1 /− v c =13m
v → v8
2.6 10 s
13m
−
=2 2
1 /− v c →8
5 10 m/s
v
=2 2
1 /− v c →2
8 2
(5 10 )
v = 1 −2
8 2
(3 10 )
v v = 2.57108m/s.
2.43 Calling to the right positive and Anna frame S, we know that this tick of the clock/event occurs at t = 0 and
x =1
2 according to Bob. We seek t. t =2
− +
v
v x t
c .
v =2
1
1
1 2
−
=2 .
Thus t =1 2
2 (30m) 0
− +
c = −8
30m
3 10 m/s = −10−7s = −100ns.
(b) We have a distance traveled according to Bob, 30m, and a speed, and so find a time according to Bob via
classical mechanics. t =30m
2c =8
30m 2
3 10 m/s = 1.4110−7s = 141ns.
(c) If Anna’s back clock has moved to the middle of Bob’s ship, Anna’s front clock, formerly at Bob’s middle,
will now be at Bob’s front, x = 60m. Thus the locations, according to Bob, are x = 30 and x = 60m. The
times are both 141ns.
For Anna’s back clock: t =91 2
2 (30m) 141 10 s−
− +
c = −8
30m
3 10 m/s +2 14110−9s = 100ns.
For Anna’s front: t =91/ 2
2 (60m) 141 10 s
c
−
− +
= −8
60 m
3 10 m / s +2 14110−9s = 0.
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Chapter 2 Special Relativity
10
(d) In the diagrams, Event 1 shows an observer in Anna’s ship, at t = −100ns by her wristwatch, looking at a
clock at the middle of Bob’s, which reads zero. Event 2 shows another observer in Anna’s ship, at t = +
100ns by her wristwatch, looking at the same clock in Bob’s, which reads 141ns. The time elapsing between
these events in Anna’s frame is +100−(−100) = 200ns, but they see this single clock at a fixed location in
Bob’s ship mark off only 141ns, which is less than 200ns by the usual factor:
v. That is, 200ns =2
141ns. Events 3 and 4 show observers in Anna’s frame, who are less than 60m apart (not both being at the
very ends of their own ship) viewing at the same time (t = 0) the very ends of Bob’s ship. It is less than
60m long according to them. How much less? According to Bob, the two clocks on Anna’s ship are 30m
apart, but this is a length−contracted observation, so they must be more than 30m apart in their own frame:2
30m = 42.4m. As noted, this is how long Bob’s ship appears from Anna’s frame. This fits! Anna
should see the length of Bob’s ship as60m
v =60m
2 =2 30m = 42.4m.
2.44 When v c,
v approaches 1, so equations (2-12) become x´ = x − vt and t´ = t −2
v
c x, while equations (2-13)
become x = x´ + vt´ and t = t´ +2
v
c x´. So long as neither x nor x´ is large, the two time equations become
equivalent, t´ = t, and then the two position equations are also equivalent, x´ = x – vt. These are equations (2-1).
If, however, we are talking about events at such large x that2
v
c x is not negligible, then the correspondence fails.
2.45 Bob will wait t =12ly
0.6c = 20yr for Anna to get there and 20yr for her to return. Bob ages 40yr. Bob, always in an
inertial frame, will observe Anna’s aging slowly the whole way.v
40yr
=2
40yr
1/ 1 (0.6)− = 32yr.
Anna “sees” Bob age slower than herself on the way out and on the way back, but “sees” Bob age horribly during
the interval in which she accelerates. We need not calculate from Anna’s perspective. Both must agree on their
respective ages when they reunite. Bob is 20 + 40 = 60yr. Anna is 20 + 32 = 52yr.
2.46 A distance of 30ly is traveled at 0.9c.
Time in Bob’s frame =distance in Bob's frame
speed =30ly
0.9c = 33.3yr.
(b) Anna will see Bob age less than herself, but how much does Anna age? She sees a distance of2
1 (0.9)−
30ly = 13.08ly pass by her window at 0.9c.
Time in Anna’s frame =distance in Anna's frame
speed =13.08ly
0.9c = 14.5yr.
This is Anna’s age according to herself, reckoned via length contraction. It is also Anna’s age according to
Bob, for Bob ages 33.3yr, but will determine that Anna ages less (events transpire less rapidly):2
1 (0.9)−
33.3yr = 14.5yr.
10
(d) In the diagrams, Event 1 shows an observer in Anna’s ship, at t = −100ns by her wristwatch, looking at a
clock at the middle of Bob’s, which reads zero. Event 2 shows another observer in Anna’s ship, at t = +
100ns by her wristwatch, looking at the same clock in Bob’s, which reads 141ns. The time elapsing between
these events in Anna’s frame is +100−(−100) = 200ns, but they see this single clock at a fixed location in
Bob’s ship mark off only 141ns, which is less than 200ns by the usual factor:
v. That is, 200ns =2
141ns. Events 3 and 4 show observers in Anna’s frame, who are less than 60m apart (not both being at the
very ends of their own ship) viewing at the same time (t = 0) the very ends of Bob’s ship. It is less than
60m long according to them. How much less? According to Bob, the two clocks on Anna’s ship are 30m
apart, but this is a length−contracted observation, so they must be more than 30m apart in their own frame:2
30m = 42.4m. As noted, this is how long Bob’s ship appears from Anna’s frame. This fits! Anna
should see the length of Bob’s ship as60m
v =60m
2 =2 30m = 42.4m.
2.44 When v c,
v approaches 1, so equations (2-12) become x´ = x − vt and t´ = t −2
v
c x, while equations (2-13)
become x = x´ + vt´ and t = t´ +2
v
c x´. So long as neither x nor x´ is large, the two time equations become
equivalent, t´ = t, and then the two position equations are also equivalent, x´ = x – vt. These are equations (2-1).
If, however, we are talking about events at such large x that2
v
c x is not negligible, then the correspondence fails.
2.45 Bob will wait t =12ly
0.6c = 20yr for Anna to get there and 20yr for her to return. Bob ages 40yr. Bob, always in an
inertial frame, will observe Anna’s aging slowly the whole way.v
40yr
=2
40yr
1/ 1 (0.6)− = 32yr.
Anna “sees” Bob age slower than herself on the way out and on the way back, but “sees” Bob age horribly during
the interval in which she accelerates. We need not calculate from Anna’s perspective. Both must agree on their
respective ages when they reunite. Bob is 20 + 40 = 60yr. Anna is 20 + 32 = 52yr.
2.46 A distance of 30ly is traveled at 0.9c.
Time in Bob’s frame =distance in Bob's frame
speed =30ly
0.9c = 33.3yr.
(b) Anna will see Bob age less than herself, but how much does Anna age? She sees a distance of2
1 (0.9)−
30ly = 13.08ly pass by her window at 0.9c.
Time in Anna’s frame =distance in Anna's frame
speed =13.08ly
0.9c = 14.5yr.
This is Anna’s age according to herself, reckoned via length contraction. It is also Anna’s age according to
Bob, for Bob ages 33.3yr, but will determine that Anna ages less (events transpire less rapidly):2
1 (0.9)−
33.3yr = 14.5yr.
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Instructor Solutions Manual for Harris, Modern Physics, Second Edition
11
Anna, aging 14.5yr, will determine that Bob ages less than herself, by the same factor.2
1 (0.9)− 14.5yr
= 6.33yr. (c) and (d) already answered: 14.5yr. Yes, each says, correctly, that the other is younger.
2.47 For Anna in S´, where the planets are 24ly apart, (2-12b) evaluated for two events is2 1 2 1 22 12 ( ) ( )
− = + − + −
v
v
t t x x t t
c
=5 0.8 (24ly) (0)
3
+ + =
c 32yr .
(b) Relative to Earth, Carl simply has a v that is the opposite of Anna’s, so in this case the time interval is
–32yr.
(c) Bob waits 50yr, so the Planet X clock (synchronized in his frame at least) reads 50yr. All observers agree
that the clock on Planet X reads 50yr when they pass. According to Anna, the clock on Planet X (event 2) is
32 yr ahead of that on Earth, so she says that the Earth clock is chiming 18yr now. Carl says that the Planet
X clock is 32yr behind, so the Earth clock is right now chiming 82yr.
2.48 Calling to the right positive and Anna frame S, we know that this tick of the clock/event occurs at t = 0 and x =
−24m according to this observer in Bob’s frame. We seek t.
t =2
− +
v
v x t
c and
v =2
1
1 (0.6)− = 1.25.
Thus,
t =0.6
1.25 ( 24m) 0
− − +
c =8
18m
3 10 m/s = 610−8s = 60ns.
(b) Before your friend steps on, Anna, at the center of her ship, is seen from Bob’s frame to be age zero.
Afterward, your friend is in Anna’s frame, where the clock reads 60ns. But once in Anna’s frame, all clocks
are synchronized in that frame. Specifically, the clock right at Anna’s location now also must read 60ns.
Anna’s age jumps forward by 60ns.
(c) x would be +24m rather than −24m, so the new time would be −60ns and Anna’s age, starting at zero,
jumps backward by 60ns.
(d) In part (b) your friend is accelerating toward Anna, and Anna’s age jumps forward. In part (c) your friend
accelerates away from Anna (before the ship-changing, Anna was approaching your friend, but afterward is
no longer doing so), and Anna’s age jumps backward. If you think about this with relative simultaneity in
mind, it makes sense. If your two friends jump off your ship at the same time (t1 = t2 = 0), they cannot
possibly arrive on Anna’s ship at the same time (t1 ≠ t2). They must necessarily arrive there at two different
times (ages) of Anna.
2.49 The whole idea here is that you are jumping into a new frame, and the clocks in the now moving
Earth−Centaurus A frame are unsynchronized. How out of synch is the one on Centaurus A? Let us call the
Earth−Centaurus A system frame S, with their separation given the symbol W. You are initially in frame S, but at
the passing of the origins you instantly jump into frame S, moving at 5m/s toward Centaurus A. We know that
your time is t = 0, and we seek t, the time on a clock in Centaurus A at x = + W in frame S.
t =2
−
v
v
t x
c → 0 =2 W
−
v
v
t c , or t =2
v
c W.
This being positive, the clocks in front of you will all be ahead.2
v
c
=16 2 2
5m/s
9 10 m /s = 5.5610−17s/m.
Thus, t = W 5.5610−17s/m
11
Anna, aging 14.5yr, will determine that Bob ages less than herself, by the same factor.2
1 (0.9)− 14.5yr
= 6.33yr. (c) and (d) already answered: 14.5yr. Yes, each says, correctly, that the other is younger.
2.47 For Anna in S´, where the planets are 24ly apart, (2-12b) evaluated for two events is2 1 2 1 22 12 ( ) ( )
− = + − + −
v
v
t t x x t t
c
=5 0.8 (24ly) (0)
3
+ + =
c 32yr .
(b) Relative to Earth, Carl simply has a v that is the opposite of Anna’s, so in this case the time interval is
–32yr.
(c) Bob waits 50yr, so the Planet X clock (synchronized in his frame at least) reads 50yr. All observers agree
that the clock on Planet X reads 50yr when they pass. According to Anna, the clock on Planet X (event 2) is
32 yr ahead of that on Earth, so she says that the Earth clock is chiming 18yr now. Carl says that the Planet
X clock is 32yr behind, so the Earth clock is right now chiming 82yr.
2.48 Calling to the right positive and Anna frame S, we know that this tick of the clock/event occurs at t = 0 and x =
−24m according to this observer in Bob’s frame. We seek t.
t =2
− +
v
v x t
c and
v =2
1
1 (0.6)− = 1.25.
Thus,
t =0.6
1.25 ( 24m) 0
− − +
c =8
18m
3 10 m/s = 610−8s = 60ns.
(b) Before your friend steps on, Anna, at the center of her ship, is seen from Bob’s frame to be age zero.
Afterward, your friend is in Anna’s frame, where the clock reads 60ns. But once in Anna’s frame, all clocks
are synchronized in that frame. Specifically, the clock right at Anna’s location now also must read 60ns.
Anna’s age jumps forward by 60ns.
(c) x would be +24m rather than −24m, so the new time would be −60ns and Anna’s age, starting at zero,
jumps backward by 60ns.
(d) In part (b) your friend is accelerating toward Anna, and Anna’s age jumps forward. In part (c) your friend
accelerates away from Anna (before the ship-changing, Anna was approaching your friend, but afterward is
no longer doing so), and Anna’s age jumps backward. If you think about this with relative simultaneity in
mind, it makes sense. If your two friends jump off your ship at the same time (t1 = t2 = 0), they cannot
possibly arrive on Anna’s ship at the same time (t1 ≠ t2). They must necessarily arrive there at two different
times (ages) of Anna.
2.49 The whole idea here is that you are jumping into a new frame, and the clocks in the now moving
Earth−Centaurus A frame are unsynchronized. How out of synch is the one on Centaurus A? Let us call the
Earth−Centaurus A system frame S, with their separation given the symbol W. You are initially in frame S, but at
the passing of the origins you instantly jump into frame S, moving at 5m/s toward Centaurus A. We know that
your time is t = 0, and we seek t, the time on a clock in Centaurus A at x = + W in frame S.
t =2
−
v
v
t x
c → 0 =2 W
−
v
v
t c , or t =2
v
c W.
This being positive, the clocks in front of you will all be ahead.2
v
c
=16 2 2
5m/s
9 10 m /s = 5.5610−17s/m.
Thus, t = W 5.5610−17s/m
Loading page 14...
Chapter 2 Special Relativity
12
(a) If W = 21023m then clock jumps ahead by t = 1.11107s = 128days.
(b) If W = 4.5109m then t = 250ns.
(c) Need only reverse the sign of x. Clocks will be behind by same amounts.
(d) If the traveler is moving away from Earth, then decelerates to a stop (stopping being the reverse of starting
to jog away from a planet), he moves to a frame in which clocks back on Earth are immediately advanced.
If he furthermore turns around and jumps to a frame moving back toward Earth, he moves to a frame in
which clocks back on Earth are again immediately advanced. Acceleration toward Earth causes clocks there
to advance. The effect depends not only on the speed involved, but also on the distance away; as we see, by
merely jogging this way and that at ordinary speeds, we move through frames in which clocks on heavenly
bodies very far away change by a great deal. Readings on local clocks, those around the solar system, don’t
vary much.
2.50 I am S´. At t´ = 0, I seek t on a clock at the 100cm mark, i.e., x = 50cm (with origin at meterstick center).
t =2
− +
v
v x t
c → =2 50cm
− +
v
v t
c or t =v
c2 50cm .
The clock at 0cm, or x = –50cm, would readv
c2
– 50cm .
(b) Changing direction would simply change the sign on each value calculated.
(c) When I turn around, the clock at the 0cm mark jumps forward byv
c2 1m , as I accelerate toward it, just as
Bob’s clock on Earth jumps forward as Anna accelerates toward it.
2.51 Toward is
= 180º, cos
= − 1. fobs = fsource2 2
1 /
1 /
−
−
v c
v c = fsource(1 / )(1 / )
(1 / )(1 / )
− +
− −
v c v c
v c v c = fsource1 /
1 /
+
−
v c
v c
2.52 fobs = fsource1 /
1 /
−
+
v c
v c →obs
c =source
1 /
1 /
−
+
c v c
v c
obs =1 /
1 /
+
−
v c
v c
source =1 0.9
1 0.9
+
−
source = 4.36
source
2.53 If we see a shorter wavelength, a higher frequency, it must be moving toward.obs
source
f
f
=1 /
1 /
+
−
v c
v c →obs
source
/
/
c
c =1 /
1 /
+
−
v c
v c →532
412 =1 /
1 /
+
−
v c
v c → 1.667 =1 /
1 /
+
−
v c
v c v
c = 0.25.
(b) Away:obs
c =source
1 /
1 /
−
+
c v c
v c →obs
1
=1 1 0.25
532 1 0.25
−
+
obs = 687nm.
(c) If
= 90º, fobs = fsource2 2
1 /− v c →obs
c =2
1 (0.25)
532nm −
c
obs = 549nm.
A period in the observer’s frame is longer than for 532nm light, due solely to time dilation.
12
(a) If W = 21023m then clock jumps ahead by t = 1.11107s = 128days.
(b) If W = 4.5109m then t = 250ns.
(c) Need only reverse the sign of x. Clocks will be behind by same amounts.
(d) If the traveler is moving away from Earth, then decelerates to a stop (stopping being the reverse of starting
to jog away from a planet), he moves to a frame in which clocks back on Earth are immediately advanced.
If he furthermore turns around and jumps to a frame moving back toward Earth, he moves to a frame in
which clocks back on Earth are again immediately advanced. Acceleration toward Earth causes clocks there
to advance. The effect depends not only on the speed involved, but also on the distance away; as we see, by
merely jogging this way and that at ordinary speeds, we move through frames in which clocks on heavenly
bodies very far away change by a great deal. Readings on local clocks, those around the solar system, don’t
vary much.
2.50 I am S´. At t´ = 0, I seek t on a clock at the 100cm mark, i.e., x = 50cm (with origin at meterstick center).
t =2
− +
v
v x t
c → =2 50cm
− +
v
v t
c or t =v
c2 50cm .
The clock at 0cm, or x = –50cm, would readv
c2
– 50cm .
(b) Changing direction would simply change the sign on each value calculated.
(c) When I turn around, the clock at the 0cm mark jumps forward byv
c2 1m , as I accelerate toward it, just as
Bob’s clock on Earth jumps forward as Anna accelerates toward it.
2.51 Toward is
= 180º, cos
= − 1. fobs = fsource2 2
1 /
1 /
−
−
v c
v c = fsource(1 / )(1 / )
(1 / )(1 / )
− +
− −
v c v c
v c v c = fsource1 /
1 /
+
−
v c
v c
2.52 fobs = fsource1 /
1 /
−
+
v c
v c →obs
c =source
1 /
1 /
−
+
c v c
v c
obs =1 /
1 /
+
−
v c
v c
source =1 0.9
1 0.9
+
−
source = 4.36
source
2.53 If we see a shorter wavelength, a higher frequency, it must be moving toward.obs
source
f
f
=1 /
1 /
+
−
v c
v c →obs
source
/
/
c
c =1 /
1 /
+
−
v c
v c →532
412 =1 /
1 /
+
−
v c
v c → 1.667 =1 /
1 /
+
−
v c
v c v
c = 0.25.
(b) Away:obs
c =source
1 /
1 /
−
+
c v c
v c →obs
1
=1 1 0.25
532 1 0.25
−
+
obs = 687nm.
(c) If
= 90º, fobs = fsource2 2
1 /− v c →obs
c =2
1 (0.25)
532nm −
c
obs = 549nm.
A period in the observer’s frame is longer than for 532nm light, due solely to time dilation.
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Instructor Solutions Manual for Harris, Modern Physics, Second Edition
13
2.54 Since the observed frequency is larger (wavelength smaller) than the source, the galaxy must be moving toward
Earth. fobs = fsource1 /
1 /
+
−
v c
v c →obs
source
f
f =1 /
1 /
+
−
v c
v c →obs
source
/
/
c
c
=1 /
1 /
+
−
v c
v c →396.85
396.58 =1 /
1 /
+
−
v c
v c → 1.00136 =1 /
1 /
+
−
v c
v c
v
c = 0.000681. The wavelength shift is about one part in a thousand, and the speed is roughly
one−thousandth that of light.
2.55 Yes—movement partially toward must be compensating for time dilation effect.
fobs = fsource2 2
1 /
1 ( / ) cos
−
+
v c
v c but fobs = fsource 1 + (v/c) cos
=2 2
1 /− v c
→ 1+ 0.8 cos
=2
1 (0.8)−
= 120º. This fits, for the movement is partially toward.
2.56 Toward:obs
c =source
1 /
1 /
+
−
c v c
v c →540nm
c =source
1 /
1 /
+
−
c v c
v c
Away:obs
c =source
1 /
1 /
−
+
c v c
v c →650nm
c =source
1 /
1 /
−
+
c v c
v c
Divide first equation by second:650
540 =1
1
+
−
v c
v c v
c = 0.0924
Plug back in:540nm
c =source
1 0.0924
1 0.0924
+
−
c
source = 592nm (Yellow.)
2.57 When moving away, fobs = fsource1 /
1 /
−
+
v c
v c orobs
c =source
c1 /
1 /
−
+
v c
v c or
obs =
source (1 + v/c)1/2 (1 − v/c)−1/2
source
(1 +1
2 v/c)(1 +1
2 v/c)
source +
source (v/c). Moving toward would just change the sign. The total range is thus 2source v/c =/B3k T m
c
2
.
(b) vrms =23 4 27
3(1.38 10 J/K)(5 10 K) /1.67 10 kg− −
= 3.5104m/s.
Thus,
=4
8
3.5 10
2 656 nm
3 10
= 0.15nm.
2.58 The car moving away would detect a frequency fobs = fsource1 /
1 /
−
+
v c
v c = fsource (1 − v/c)1/2(1 + v/c)−1/2 fsource
(1−1
2 v/c) (1−1
2 v/c) fsource (1− v/c). It then becomes a source of this frequency, moving away from the final
observer (i.e., the radar gun). The final observed frequency would be f'obs = f'source1 /
1 /
−
+
v c
v c
=
1 /
(1 / ) 1 /
−
− +
source
v c
f v c v c
(1 / ) (1 / )− −sourcef v c v c fsource (1−2 v/c) = 900MHz (1−28
30
3 10 ) = 900MHz − 180Hz.
The beat frequency (the difference) is 180Hz.
13
2.54 Since the observed frequency is larger (wavelength smaller) than the source, the galaxy must be moving toward
Earth. fobs = fsource1 /
1 /
+
−
v c
v c →obs
source
f
f =1 /
1 /
+
−
v c
v c →obs
source
/
/
c
c
=1 /
1 /
+
−
v c
v c →396.85
396.58 =1 /
1 /
+
−
v c
v c → 1.00136 =1 /
1 /
+
−
v c
v c
v
c = 0.000681. The wavelength shift is about one part in a thousand, and the speed is roughly
one−thousandth that of light.
2.55 Yes—movement partially toward must be compensating for time dilation effect.
fobs = fsource2 2
1 /
1 ( / ) cos
−
+
v c
v c but fobs = fsource 1 + (v/c) cos
=2 2
1 /− v c
→ 1+ 0.8 cos
=2
1 (0.8)−
= 120º. This fits, for the movement is partially toward.
2.56 Toward:obs
c =source
1 /
1 /
+
−
c v c
v c →540nm
c =source
1 /
1 /
+
−
c v c
v c
Away:obs
c =source
1 /
1 /
−
+
c v c
v c →650nm
c =source
1 /
1 /
−
+
c v c
v c
Divide first equation by second:650
540 =1
1
+
−
v c
v c v
c = 0.0924
Plug back in:540nm
c =source
1 0.0924
1 0.0924
+
−
c
source = 592nm (Yellow.)
2.57 When moving away, fobs = fsource1 /
1 /
−
+
v c
v c orobs
c =source
c1 /
1 /
−
+
v c
v c or
obs =
source (1 + v/c)1/2 (1 − v/c)−1/2
source
(1 +1
2 v/c)(1 +1
2 v/c)
source +
source (v/c). Moving toward would just change the sign. The total range is thus 2source v/c =/B3k T m
c
2
.
(b) vrms =23 4 27
3(1.38 10 J/K)(5 10 K) /1.67 10 kg− −
= 3.5104m/s.
Thus,
=4
8
3.5 10
2 656 nm
3 10
= 0.15nm.
2.58 The car moving away would detect a frequency fobs = fsource1 /
1 /
−
+
v c
v c = fsource (1 − v/c)1/2(1 + v/c)−1/2 fsource
(1−1
2 v/c) (1−1
2 v/c) fsource (1− v/c). It then becomes a source of this frequency, moving away from the final
observer (i.e., the radar gun). The final observed frequency would be f'obs = f'source1 /
1 /
−
+
v c
v c
=
1 /
(1 / ) 1 /
−
− +
source
v c
f v c v c
(1 / ) (1 / )− −sourcef v c v c fsource (1−2 v/c) = 900MHz (1−28
30
3 10 ) = 900MHz − 180Hz.
The beat frequency (the difference) is 180Hz.
Loading page 16...
Chapter 2 Special Relativity
14
2.59 The “object” here is the projectile. Let’s choose away from Earth as positive. Therefore Bob, on Earth, is frame
S, while Anna (moving in the positive direction) is frame S. We are given that the velocity is v = 0.6c between
the frames, and that the object moves in the positive (away from Earth) direction at u = 0.8c relative to frame
S/Bob. u =2
1
−
−
u v
uv
c =0.8 0.6
1 (0.8)(0.6)
−
−
c c = 0.385c.
(b) The relativistic velocity transformation works for light just as for a massive object. Thus, u = c, and u =2
1
−
−
u v
uv
c
=0.6
1 (1)(0.6)
−
−
c c = c. This had better be the case! Light traveling at c in both frames is built in to the
Lorentz Transformation equations, from which the velocity transformation is derived.
2.60 The “object” here is Carl. Let’s choose Anna, on Earth, as frame S and Bob as frame S. (Since the S frame by
definition moves in the positive direction relative to S, we’ve chosen toward Earth as positive.) We are given that
the velocity is v = 0.8c between the frames, and that the object Carl moves in the positive (toward Earth)
direction at u = 0.9c relative to frame S/Anna. We seek the velocity u of the object/Carl relative to frame S/Bob.
u =2
1
−
−
u v
uv
c =0.9 0.8
1 (0.8)(0.9)
−
−
c c = 0.357c. Positive means toward Earth.
(b) Bob sees Carl moving at 0.357c in one direction and Earth moving at 0.8c in the other (i.e., toward Bob).
These are both velocities according to Bob. They may be added using the classical expression to find a
relative velocity according to Bob. 1.157c. Note: This is a velocity of Carl relative to Earth according to
Bob. It is not said that an observer sees something else moving at greater than c relative to him or herself.
2.61 (c−u)(c−v) > 0 c2− (u+ v)c + uv > 0 → c2+ uv > (u+ v)c → c >2
1
+
+
u v
u v
c = u
2.62 The lab is S; Particle 2 is S, moving at v = + 0.99c relative to the lab; and Particle 1 is the object, which moves at
u = − 0.99c through the lab. u =2
1
−
−
u v
uv
c =0.99 0.99
1 ( 0.99)(0.99)
− −
− −
c c = −0.9995c
2.63 uy =.2 2
xc u−
(b)xu =2
1
−
−
x
x
u v
u v
c =2
2
1
−
−
x x
x
u u
u
c = 0 andyu =2
1
−
y
x
v
u
u v
c =2 2
2
2
1
−
−
x
x
v
c u
u
c =2 2
2
22 2
1 1
1
−
−
−
x
x
x
c u
u
cu c = c.
The light beam has no x-component, and its speed overall must be c.
2.64 In frame S, the velocity components of the light beam are ux = c cos
and uy = c sin
. Equations (2-20) apply.xu
=2
1
−
−
x
x
u v
u v
c andyu =2
1
−
y
x
v
u
u v
c .
14
2.59 The “object” here is the projectile. Let’s choose away from Earth as positive. Therefore Bob, on Earth, is frame
S, while Anna (moving in the positive direction) is frame S. We are given that the velocity is v = 0.6c between
the frames, and that the object moves in the positive (away from Earth) direction at u = 0.8c relative to frame
S/Bob. u =2
1
−
−
u v
uv
c =0.8 0.6
1 (0.8)(0.6)
−
−
c c = 0.385c.
(b) The relativistic velocity transformation works for light just as for a massive object. Thus, u = c, and u =2
1
−
−
u v
uv
c
=0.6
1 (1)(0.6)
−
−
c c = c. This had better be the case! Light traveling at c in both frames is built in to the
Lorentz Transformation equations, from which the velocity transformation is derived.
2.60 The “object” here is Carl. Let’s choose Anna, on Earth, as frame S and Bob as frame S. (Since the S frame by
definition moves in the positive direction relative to S, we’ve chosen toward Earth as positive.) We are given that
the velocity is v = 0.8c between the frames, and that the object Carl moves in the positive (toward Earth)
direction at u = 0.9c relative to frame S/Anna. We seek the velocity u of the object/Carl relative to frame S/Bob.
u =2
1
−
−
u v
uv
c =0.9 0.8
1 (0.8)(0.9)
−
−
c c = 0.357c. Positive means toward Earth.
(b) Bob sees Carl moving at 0.357c in one direction and Earth moving at 0.8c in the other (i.e., toward Bob).
These are both velocities according to Bob. They may be added using the classical expression to find a
relative velocity according to Bob. 1.157c. Note: This is a velocity of Carl relative to Earth according to
Bob. It is not said that an observer sees something else moving at greater than c relative to him or herself.
2.61 (c−u)(c−v) > 0 c2− (u+ v)c + uv > 0 → c2+ uv > (u+ v)c → c >2
1
+
+
u v
u v
c = u
2.62 The lab is S; Particle 2 is S, moving at v = + 0.99c relative to the lab; and Particle 1 is the object, which moves at
u = − 0.99c through the lab. u =2
1
−
−
u v
uv
c =0.99 0.99
1 ( 0.99)(0.99)
− −
− −
c c = −0.9995c
2.63 uy =.2 2
xc u−
(b)xu =2
1
−
−
x
x
u v
u v
c =2
2
1
−
−
x x
x
u u
u
c = 0 andyu =2
1
−
y
x
v
u
u v
c =2 2
2
2
1
−
−
x
x
v
c u
u
c =2 2
2
22 2
1 1
1
−
−
−
x
x
x
c u
u
cu c = c.
The light beam has no x-component, and its speed overall must be c.
2.64 In frame S, the velocity components of the light beam are ux = c cos
and uy = c sin
. Equations (2-20) apply.xu
=2
1
−
−
x
x
u v
u v
c andyu =2
1
−
y
x
v
u
u v
c .
Loading page 17...
Instructor Solutions Manual for Harris, Modern Physics, Second Edition
15
Plugging in:xu =
−
−
cos
cos
1
c v
v
c ,yu =
−
v
sin
cos
1
c
v
c
2 2 +x yu u
=2
2
( cos )
cos
1
−
−
c v
v
c +2 2
2 2
( sin )
1 cos
1
− −
v c
c v
c =( )
2 2 2 2
2
( cos ) 1 / ( sin )
cos
1
− + −
−
c v v c c
v
c .
Multiplying out the numerator, c2 cos2
− 2 c v cos
+ v2 + c2 sin2
− v2 sin2
= c2 cos2
− 2 c v cos
+ v2 +
c2 (1−cos2
) − v2 (1−cos2
) = − 2 c v cos
+ c2 + v2 cos2
. Multiplying out the denominator, 1 − 2 (v/c) cos
+
(v2/c2) cos2
. The numerator is c2 times the denominator. Conclusion: Though the components may be different,
the light beam moves at c in both frames of reference.
2.65 ux = – c cos 60° = – c/2. uy = c sin 60° =3 2.cxu =1 1
2 2
1 1
2 2
2
1
4
5
− − = −
−
−
c c
c c
c
c andyu
=2
1
−
y
x
v
u
u v
c =1 1
2 2
221
2
3 2
1 1
1 ( )
−
−
−
c
c c
cc =.c
3
5
´ = tan-1 (–3/4) = 36.9° north of west. u´ =2 2
(0.8 ) (0.6 )+c c = c, as it must be.
(b) Would change only the sign of v. Thus,xu =1 1
2 2
1 1
2 2
2
( )
1
− +
− −
−
c c
c c
c = 0, andyu
=1 1
2 2
221
2
3 2
( )1 1
1 ( )
− −
−
−
c
c c
cc = c. Again, speed is c, but direction is along y´.
2.66 Whenxu =2
1
−
−
x
x
u v
u v
c is zero, it divides positive x´−components from negative ones according to an observer in
S´. This occurs when ux = v or u cos
= v. But the “object” moving in frame S here is light, for which u = c. Thus
c cos
= v or
= cos−1(v/c).
(b) At v = 0, this is 90°, which makes sense. The beacon and Anna are in the same frame, and light emitted on
the +x side would be on the +x side according to both. At v = c, the angle is 0°. Only the light emitted by the
beacon directly along the horizontal axis would appear to Anna to be moving in the positive direction. All
the rest would appear to have a negative component according to Anna.
(c) According to Anna, the beacon shines essentially all of its light in front of it, in the direction it is moving
relative to Anna.
2.67
u 2
2
1
1
− u
c =2 2 2
2
1
1 + +
− x y zu u u
c =( ) ( ) ( )
2 2 2
2 2 2 2
1
1
1 1 1 1
− − + +
− − −
yx z
x v x v x
uu v u
c u v c u v c u v c
15
Plugging in:xu =
−
−
cos
cos
1
c v
v
c ,yu =
−
v
sin
cos
1
c
v
c
2 2 +x yu u
=2
2
( cos )
cos
1
−
−
c v
v
c +2 2
2 2
( sin )
1 cos
1
− −
v c
c v
c =( )
2 2 2 2
2
( cos ) 1 / ( sin )
cos
1
− + −
−
c v v c c
v
c .
Multiplying out the numerator, c2 cos2
− 2 c v cos
+ v2 + c2 sin2
− v2 sin2
= c2 cos2
− 2 c v cos
+ v2 +
c2 (1−cos2
) − v2 (1−cos2
) = − 2 c v cos
+ c2 + v2 cos2
. Multiplying out the denominator, 1 − 2 (v/c) cos
+
(v2/c2) cos2
. The numerator is c2 times the denominator. Conclusion: Though the components may be different,
the light beam moves at c in both frames of reference.
2.65 ux = – c cos 60° = – c/2. uy = c sin 60° =3 2.cxu =1 1
2 2
1 1
2 2
2
1
4
5
− − = −
−
−
c c
c c
c
c andyu
=2
1
−
y
x
v
u
u v
c =1 1
2 2
221
2
3 2
1 1
1 ( )
−
−
−
c
c c
cc =.c
3
5
´ = tan-1 (–3/4) = 36.9° north of west. u´ =2 2
(0.8 ) (0.6 )+c c = c, as it must be.
(b) Would change only the sign of v. Thus,xu =1 1
2 2
1 1
2 2
2
( )
1
− +
− −
−
c c
c c
c = 0, andyu
=1 1
2 2
221
2
3 2
( )1 1
1 ( )
− −
−
−
c
c c
cc = c. Again, speed is c, but direction is along y´.
2.66 Whenxu =2
1
−
−
x
x
u v
u v
c is zero, it divides positive x´−components from negative ones according to an observer in
S´. This occurs when ux = v or u cos
= v. But the “object” moving in frame S here is light, for which u = c. Thus
c cos
= v or
= cos−1(v/c).
(b) At v = 0, this is 90°, which makes sense. The beacon and Anna are in the same frame, and light emitted on
the +x side would be on the +x side according to both. At v = c, the angle is 0°. Only the light emitted by the
beacon directly along the horizontal axis would appear to Anna to be moving in the positive direction. All
the rest would appear to have a negative component according to Anna.
(c) According to Anna, the beacon shines essentially all of its light in front of it, in the direction it is moving
relative to Anna.
2.67
u 2
2
1
1
− u
c =2 2 2
2
1
1 + +
− x y zu u u
c =( ) ( ) ( )
2 2 2
2 2 2 2
1
1
1 1 1 1
− − + +
− − −
yx z
x v x v x
uu v u
c u v c u v c u v c
Loading page 18...
Chapter 2 Special Relativity
16
=( )
2
2 2 2
2 2 2
2 2 2 2
1
1
1 1 1
−
− − − + − + −
x
x
x y z
u v
c
u v v v
u v u u
c c c c
=2
2 2 2 2 2 2
2 2
2 4 2 2 2 2 2 2
1
1
1 2 2 1 1
−
− + − + − − − + −
x
x x x x
y z
u v
c
u v u v u u v v v v
u u
c c c c c c c c
=2
2 2 2 2
2 2 2
2 2 2 2 2 2
1
1 1
1 1 1 1
−
− − − − − + −
x
x y z
u v
c
v v v v
u u u
c c c c c c =2
2 2 2 2
2 2
1
1
1 1
−
+ +
− −
x
x y z
u v
c
v u u u
c c =2
1
−
xu v
c
v
u
2.68 We suppress the index i for clarity.2
1
= −
x
u
u v
mu c
v
u m2
1
−
−
x
x
u v
u v
c =
v
umux – v
v
um. Summing this over all
particles of the system would reproduce equation (2-23).
2.69
u mu
mu =
u. This becomes significant when u nears c.
2.70 p =
umu =27 8
2
1 (1.67 10 kg)(0.8 3 10 m/s)
1 (0.8)
−
− = 6.6810−19kgm/s.
(b) E =
umc2 =27 8 2
2
1 (1.67 10 kg)(3 10 m/s)
1 (0.8)
−
− = 2.5110−10J.
(c) KE =( )
1
−u mc2 =27 8 2
2
1 1 (1.67 10 kg)(3 10 m/s)
1 (0.8)
−
−
− = 1.0010−10J.
2.71 Einternal = mc2 = (1kg) (3108m/s)2 = 91016J (Huge!);
KE = (
u−1)mc2 = (5
3 −1)(1kg) (3108m/s)2 = 61016J;
Etotal =
umc2 = Einternal+KE = 1.51017J.
2.72 To melt ice, energy (heat) must be added. This increases the internal thermal energy, hence the mass. It takes
3.33105J/kg to change ice at 0ºC to water at 0ºC. Water is 18g/mol, so, with one mole, we have 18g.
3.33105J
kg 0.018kg = 6103J. E = m c2 m =2
E
c =3
16 2 2
6 10 J
9 10 m /s
= 6.710−14kg. The mass of the ice
is less, by 67pg. Not much!
2.731
2 k x2 =1
2 (18N/m) (0.5m)2 = 2.25J. If it gains this much internal energy, its mass increases correspondingly:
Einternal = m c2 m = (2.25J)/91016m2/s2 = 2.510−17kg.
16
=( )
2
2 2 2
2 2 2
2 2 2 2
1
1
1 1 1
−
− − − + − + −
x
x
x y z
u v
c
u v v v
u v u u
c c c c
=2
2 2 2 2 2 2
2 2
2 4 2 2 2 2 2 2
1
1
1 2 2 1 1
−
− + − + − − − + −
x
x x x x
y z
u v
c
u v u v u u v v v v
u u
c c c c c c c c
=2
2 2 2 2
2 2 2
2 2 2 2 2 2
1
1 1
1 1 1 1
−
− − − − − + −
x
x y z
u v
c
v v v v
u u u
c c c c c c =2
2 2 2 2
2 2
1
1
1 1
−
+ +
− −
x
x y z
u v
c
v u u u
c c =2
1
−
xu v
c
v
u
2.68 We suppress the index i for clarity.2
1
= −
x
u
u v
mu c
v
u m2
1
−
−
x
x
u v
u v
c =
v
umux – v
v
um. Summing this over all
particles of the system would reproduce equation (2-23).
2.69
u mu
mu =
u. This becomes significant when u nears c.
2.70 p =
umu =27 8
2
1 (1.67 10 kg)(0.8 3 10 m/s)
1 (0.8)
−
− = 6.6810−19kgm/s.
(b) E =
umc2 =27 8 2
2
1 (1.67 10 kg)(3 10 m/s)
1 (0.8)
−
− = 2.5110−10J.
(c) KE =( )
1
−u mc2 =27 8 2
2
1 1 (1.67 10 kg)(3 10 m/s)
1 (0.8)
−
−
− = 1.0010−10J.
2.71 Einternal = mc2 = (1kg) (3108m/s)2 = 91016J (Huge!);
KE = (
u−1)mc2 = (5
3 −1)(1kg) (3108m/s)2 = 61016J;
Etotal =
umc2 = Einternal+KE = 1.51017J.
2.72 To melt ice, energy (heat) must be added. This increases the internal thermal energy, hence the mass. It takes
3.33105J/kg to change ice at 0ºC to water at 0ºC. Water is 18g/mol, so, with one mole, we have 18g.
3.33105J
kg 0.018kg = 6103J. E = m c2 m =2
E
c =3
16 2 2
6 10 J
9 10 m /s
= 6.710−14kg. The mass of the ice
is less, by 67pg. Not much!
2.731
2 k x2 =1
2 (18N/m) (0.5m)2 = 2.25J. If it gains this much internal energy, its mass increases correspondingly:
Einternal = m c2 m = (2.25J)/91016m2/s2 = 2.510−17kg.
Loading page 19...
Instructor Solutions Manual for Harris, Modern Physics, Second Edition
17
2.74 500103W3,600s = 1.8109J. E = mc2 → m =9
16 2 2
1.8 10 J
9 10 m /s
= 210−8kg or 20g.
2.75 p =
umu =31 4
24
8
1 (9.11 10 kg)(2.4 10 m/s)
2.4 10 m/s
1 3 10 m/s
−
−
= (1.000000003) (9.1110−31kg) (2.4104m/s)
= 2.1910−26kgm/s.
(b) p =
umu =31 6
28
8
1 (9.11 10 kg)(2.4 10 m/s)
2.4 10 m/s
1 3 10 m/s
−
−
= (1.667) (9.1110−31kg)(2.4108m/s) = 3.6410−22kgm/s.
(c) % error =−classical correct
correct
p p
p 100%
=
−
u
u
mu mu
mu 100% =1 1
−
u 100%. In the first case,1 1
1.000000003
−
100%
= (0.999999997 − 1) 100% = − 310−7% or 310−7% low.
In the second case,1 1
1.667
−
100% = (0.6 − 1) 100% = − 40% or 40% low.
Simply put, the classical expression is good so long as
u is not significantly different from 1.
2.76 Before: ptotal =
0.6(16)(0.6c) +
0.8(9)(−0.8c) =5
4 (16)(0.6c) +5
3 (9)(0.8c) = 0
(b) v = 0.6c and we seek u, given various u. u =2
1
−
−
u v
uv
c .
Given u = +0.6c: u =0.6 0.6
1 (0.6)(0.6)
−
−
c c = 0. Given u = −0.8c: u =0.8 0.6
1 ( 0.8)(0.6)
− −
− −
c c = −0.946c.
Given u = −0.6c: u =0.6 0.6
1 ( 0.6)(0.6)
− −
− −
c c = −0.882c. Given u = +0.8c: u =0.8 0.6
1 (0.8)(0.6)
−
−
c c = 0.385c.
(c) Before: ptotal =1 (16)(0)
1 0− +2
1 (9)( .946 )
1 (0.946) −
− c = −26.25c.
After: ptotal =2
1 (16)( 0.882 )
1 (0.882) −
− c +2
1 (9)(0.385)
1 (0.385)− = −26.25c.
2.77 Momentum can be arbitrarily large. p =
umu = m c
uu = c u = c2 2
1 /− u c u = c/2
2.78 (
u−1) mc2 =( )
1
22 2
(1 / ) 1
−
− −u c mc2 ( )
2 21
2
[1 ( ) / ] 1− − −u c mc2 =1
2 mu2
17
2.74 500103W3,600s = 1.8109J. E = mc2 → m =9
16 2 2
1.8 10 J
9 10 m /s
= 210−8kg or 20g.
2.75 p =
umu =31 4
24
8
1 (9.11 10 kg)(2.4 10 m/s)
2.4 10 m/s
1 3 10 m/s
−
−
= (1.000000003) (9.1110−31kg) (2.4104m/s)
= 2.1910−26kgm/s.
(b) p =
umu =31 6
28
8
1 (9.11 10 kg)(2.4 10 m/s)
2.4 10 m/s
1 3 10 m/s
−
−
= (1.667) (9.1110−31kg)(2.4108m/s) = 3.6410−22kgm/s.
(c) % error =−classical correct
correct
p p
p 100%
=
−
u
u
mu mu
mu 100% =1 1
−
u 100%. In the first case,1 1
1.000000003
−
100%
= (0.999999997 − 1) 100% = − 310−7% or 310−7% low.
In the second case,1 1
1.667
−
100% = (0.6 − 1) 100% = − 40% or 40% low.
Simply put, the classical expression is good so long as
u is not significantly different from 1.
2.76 Before: ptotal =
0.6(16)(0.6c) +
0.8(9)(−0.8c) =5
4 (16)(0.6c) +5
3 (9)(0.8c) = 0
(b) v = 0.6c and we seek u, given various u. u =2
1
−
−
u v
uv
c .
Given u = +0.6c: u =0.6 0.6
1 (0.6)(0.6)
−
−
c c = 0. Given u = −0.8c: u =0.8 0.6
1 ( 0.8)(0.6)
− −
− −
c c = −0.946c.
Given u = −0.6c: u =0.6 0.6
1 ( 0.6)(0.6)
− −
− −
c c = −0.882c. Given u = +0.8c: u =0.8 0.6
1 (0.8)(0.6)
−
−
c c = 0.385c.
(c) Before: ptotal =1 (16)(0)
1 0− +2
1 (9)( .946 )
1 (0.946) −
− c = −26.25c.
After: ptotal =2
1 (16)( 0.882 )
1 (0.882) −
− c +2
1 (9)(0.385)
1 (0.385)− = −26.25c.
2.77 Momentum can be arbitrarily large. p =
umu = m c
uu = c u = c2 2
1 /− u c u = c/2
2.78 (
u−1) mc2 =( )
1
22 2
(1 / ) 1
−
− −u c mc2 ( )
2 21
2
[1 ( ) / ] 1− − −u c mc2 =1
2 mu2
Loading page 20...
Chapter 2 Special Relativity
18
2.79 The area that Earth presents to (and that absorbs) the incoming sunlight is simply a circle whose radius is that of
Earth. area =
RE
2 =
(6.37106m)2 = 1.271014m2. The power absorbed is thus power =power
area area =
(1.53W/m2)(1.271014m2) = 1.911017W.
m
t =int
2
/ E t
c =17
16 2 2
1.91 10 W
9 10 m / s
= 2.1kg/s or 1.835kg/day.
2.80 In orbit: F = ma →earth
2
GM m
r = m2
v
r or v2 =earthGM
r . If r = Rearth (i.e., 6.37106m) then
v =11 2 2 24
6
(6.67 10 N m /kg )(5.98 10 kg)
6.37 10 m
−
= 7913m/s. The Empire State Building is of mass 365103ton
8890N
ton 2
1
9.8m/s = 3.31108kg. Thus its KE needs to be1
2 (3.31108kg) (7913)2 = 11016J. But 1kg converts
to (1kg)(3108m/s)2 = 91016J. More than enough.
2.81 Intensity Power
Area =2
Power
4
R Power = Intensity 4
R2 = (1.53W/m2)11 2
4 (1.5 10 m)
= 4.241026W.
So every second, 4.241026J of energy is put out by the sun. m =int
2
E
c =26
16 2 2
4.24 10 J/s
9 10 m /s
= 4.71109kg per
second.
2.82 If one kilogram explodes, 106J is released. But how much mass must actually be converted to produce such
energy?
m =int
2
E
c =6
16 2 2
10 J
9 10 m /s = 1.1110−11kg.11
1.11 10 kg
1kg
−
= 1.1110−11
(b) Suppose we have one kilogram. If one part in ten−thousand is converted,1kg
10,000 = 0.0001kg is converted.
How much energy is released? Eint = m c2 = (0.0001kg) (91016m2/s2) = 91012J. Explosive yield:
91012J/kg. A much greater percent is converted, so it is much more powerful.
2.83 (
u−1)mc2 = mc2
u = 2 →2
1
1 ( / )− u c = 2 u = c3 /2. Fast! Internal energy is large.
2.84 KE =( )1
−fu mec2 −( )1
−iu mec2 =( )
−f iu u mec2
=31 16 2 2
2 2
1 1 (9.11 10 kg)(9 10 m /s )
1 (0.6) 1 (0.3)
−
−
− − = 1.6510−14J
(b)31 16 2 2
2 2
1 1 (9.11 10 kg)(9 10 m /s )
1 (0.9) 1 (0.6)
−
−
− − = 8.5610−14J
If1
2 mu2 were correct, the second one would require only 67% more energy. Here we see it is more than
five times as much. This is due to the steep rise in KE near c.
18
2.79 The area that Earth presents to (and that absorbs) the incoming sunlight is simply a circle whose radius is that of
Earth. area =
RE
2 =
(6.37106m)2 = 1.271014m2. The power absorbed is thus power =power
area area =
(1.53W/m2)(1.271014m2) = 1.911017W.
m
t =int
2
/ E t
c =17
16 2 2
1.91 10 W
9 10 m / s
= 2.1kg/s or 1.835kg/day.
2.80 In orbit: F = ma →earth
2
GM m
r = m2
v
r or v2 =earthGM
r . If r = Rearth (i.e., 6.37106m) then
v =11 2 2 24
6
(6.67 10 N m /kg )(5.98 10 kg)
6.37 10 m
−
= 7913m/s. The Empire State Building is of mass 365103ton
8890N
ton 2
1
9.8m/s = 3.31108kg. Thus its KE needs to be1
2 (3.31108kg) (7913)2 = 11016J. But 1kg converts
to (1kg)(3108m/s)2 = 91016J. More than enough.
2.81 Intensity Power
Area =2
Power
4
R Power = Intensity 4
R2 = (1.53W/m2)11 2
4 (1.5 10 m)
= 4.241026W.
So every second, 4.241026J of energy is put out by the sun. m =int
2
E
c =26
16 2 2
4.24 10 J/s
9 10 m /s
= 4.71109kg per
second.
2.82 If one kilogram explodes, 106J is released. But how much mass must actually be converted to produce such
energy?
m =int
2
E
c =6
16 2 2
10 J
9 10 m /s = 1.1110−11kg.11
1.11 10 kg
1kg
−
= 1.1110−11
(b) Suppose we have one kilogram. If one part in ten−thousand is converted,1kg
10,000 = 0.0001kg is converted.
How much energy is released? Eint = m c2 = (0.0001kg) (91016m2/s2) = 91012J. Explosive yield:
91012J/kg. A much greater percent is converted, so it is much more powerful.
2.83 (
u−1)mc2 = mc2
u = 2 →2
1
1 ( / )− u c = 2 u = c3 /2. Fast! Internal energy is large.
2.84 KE =( )1
−fu mec2 −( )1
−iu mec2 =( )
−f iu u mec2
=31 16 2 2
2 2
1 1 (9.11 10 kg)(9 10 m /s )
1 (0.6) 1 (0.3)
−
−
− − = 1.6510−14J
(b)31 16 2 2
2 2
1 1 (9.11 10 kg)(9 10 m /s )
1 (0.9) 1 (0.6)
−
−
− − = 8.5610−14J
If1
2 mu2 were correct, the second one would require only 67% more energy. Here we see it is more than
five times as much. This is due to the steep rise in KE near c.
Loading page 21...
Instructor Solutions Manual for Harris, Modern Physics, Second Edition
19
2.85 KE = KEf =31 16 2 2
2
1 1 (9.11 10 kg)(9 10 m /s )
1 (0.9998)
−
−
− = 4.0210−12J = 25.1MeV, requiring an
accelerating potential of 25.1MV.
2.86 First find the kinetic energy fromqV =KE ; then either (1) solve for u, then find p, or (2) calculate E from
KE, then use (2−28).
Let’s do (1).qV =KE → qV = (
u−1)mpc2 → (1.610−19C)(109V)
=27 16 2 2
2
1 1 (1.67 10 kg)(9 10 m /s )
1 ( / )
−
−
− u c u = 0.875c.
p =
u mu =27 8
2
1 (1.67 10 kg)(0.875 3 10 m/s)
1 (0.875)
−
− = 9.0510−19kgm/s.
2.87 It acquires a KE of 500MeV = 810−11J. KE = (
u−1)mc2 →
810−11J =27 16 2 2
2 2
1 1 (1.66 10 kg)(9 10 m /s )
1 /
−
−
− u c u = 0.759c
(b) 4810−11J =27 21 (1.66 10 kg)
2
−
u u = 6.21108m/s = 2.07c
(c) 4 ( )
11
8 10 J−
=27 16 2 2
2 2
1 1 (1.66 10 kg)(9 10 m /s )
1 /
−
−
− u c u = 0.948c
2.88 Momentum conserved:
0.6 m0 (0.6c) =
0.8 (0.66m0)(0.8c) +
u m u
Energy conserved:
0.6 m0 c2 =
0.8 (0.66m0)c2 +
u m c2. The physics is done; the rest is math. We wish to solve
for u. Divide energy equation by c2, and rearrange the equations:
(1)
0.6 m0 (0.6c) −
0.8 (0.66m0)(0.8c) =
u m u (2)
0.6 m0 −
0.8 (0.66m0) =
u m
Divide (1) by (2): u =0.6 0 0.8 0
0.6 0 0.8 0
(0.6 ) (0.66 )(0.8 )
(0.66 )
−
−
m c m c
m m =(5 / 4)(0.6 ) (5 / 3)(0.66)(0.8 )
(5 / 4) (5 / 3)(0.66)
−
−
c c = − 0.867c.
It moves at 0.867c in the opposite direction. Plug back into either (1) or (2) to find m.
Using (2):
0.6 m0 −
0.8 (0.66m0) =
u m → (5/4)m0 − (5/3)(0.66m0) =2
1
1 (0.867)− m
m = 0.0748m0. Much mass is lost, because there is a significant increase in kinetic energy.
2.89 Since carbon−14 is “slow”,
1. Pf = Pi → mC uC +
e me ue = 0. Ef = Ei → mC c2 +
e me c2 = mBc2
→ (13.99995) +2 2
e
1 (0.00055)
1− u c = (14.02266) ue = 0.99971c. Plug back in to momentum equation:
(13.99995)uC +2
1 (0.00055)(0.99971 )
1 (0.99971)− c = 0 uC = −1.6210−3c (~1
600 the electron’s speed).
19
2.85 KE = KEf =31 16 2 2
2
1 1 (9.11 10 kg)(9 10 m /s )
1 (0.9998)
−
−
− = 4.0210−12J = 25.1MeV, requiring an
accelerating potential of 25.1MV.
2.86 First find the kinetic energy fromqV =KE ; then either (1) solve for u, then find p, or (2) calculate E from
KE, then use (2−28).
Let’s do (1).qV =KE → qV = (
u−1)mpc2 → (1.610−19C)(109V)
=27 16 2 2
2
1 1 (1.67 10 kg)(9 10 m /s )
1 ( / )
−
−
− u c u = 0.875c.
p =
u mu =27 8
2
1 (1.67 10 kg)(0.875 3 10 m/s)
1 (0.875)
−
− = 9.0510−19kgm/s.
2.87 It acquires a KE of 500MeV = 810−11J. KE = (
u−1)mc2 →
810−11J =27 16 2 2
2 2
1 1 (1.66 10 kg)(9 10 m /s )
1 /
−
−
− u c u = 0.759c
(b) 4810−11J =27 21 (1.66 10 kg)
2
−
u u = 6.21108m/s = 2.07c
(c) 4 ( )
11
8 10 J−
=27 16 2 2
2 2
1 1 (1.66 10 kg)(9 10 m /s )
1 /
−
−
− u c u = 0.948c
2.88 Momentum conserved:
0.6 m0 (0.6c) =
0.8 (0.66m0)(0.8c) +
u m u
Energy conserved:
0.6 m0 c2 =
0.8 (0.66m0)c2 +
u m c2. The physics is done; the rest is math. We wish to solve
for u. Divide energy equation by c2, and rearrange the equations:
(1)
0.6 m0 (0.6c) −
0.8 (0.66m0)(0.8c) =
u m u (2)
0.6 m0 −
0.8 (0.66m0) =
u m
Divide (1) by (2): u =0.6 0 0.8 0
0.6 0 0.8 0
(0.6 ) (0.66 )(0.8 )
(0.66 )
−
−
m c m c
m m =(5 / 4)(0.6 ) (5 / 3)(0.66)(0.8 )
(5 / 4) (5 / 3)(0.66)
−
−
c c = − 0.867c.
It moves at 0.867c in the opposite direction. Plug back into either (1) or (2) to find m.
Using (2):
0.6 m0 −
0.8 (0.66m0) =
u m → (5/4)m0 − (5/3)(0.66m0) =2
1
1 (0.867)− m
m = 0.0748m0. Much mass is lost, because there is a significant increase in kinetic energy.
2.89 Since carbon−14 is “slow”,
1. Pf = Pi → mC uC +
e me ue = 0. Ef = Ei → mC c2 +
e me c2 = mBc2
→ (13.99995) +2 2
e
1 (0.00055)
1− u c = (14.02266) ue = 0.99971c. Plug back in to momentum equation:
(13.99995)uC +2
1 (0.00055)(0.99971 )
1 (0.99971)− c = 0 uC = −1.6210−3c (~1
600 the electron’s speed).
Loading page 22...
Chapter 2 Special Relativity
20
KEe = (
− 1)mc2 =31 16 2 2
2
1 1 (9.11 10 kg)(9 10 m /c )
1 (0.99971)
−
−
− = 3.310−12J = 20.6MeV.
KEC=1
2 mu2 =1
2 (13.99995u1.6610−27kg/u) (1.6210−33108m/s)2 = 2.7510−15J = 17.1keV
(about1
1, 000 of KEe).
2.90 Let’s use m0 for an “atomic mass unit”, rather than u (for obvious reasons).
Momentum:
0.8 (3m0) (+0.8c) +
0.6 (4m0)(−0.6c) =
0 (6m0)(0) +
u m u → 1m0 c =
u m u
Energy:
0.8 (3m0)c2 +
0.6 (4m0)c2 = +10m0c2 =
0 (6m0)c2 +
u mc2 → +4m0 =
u m
Divide the two: u = c/4. Plug back in: 4m0 =
0.25 m m = 3.87u
(b) Since mass/internal energy increases, KE must decrease. The long way: KEfinal = (
0.25−1) (3.87m0) c2 =
0.127m0 c2. KEinitial = (
0.8−1) (3m0) c2 + (
0.6−1) (4m0)c2 = 3m0 c2
KE = −2.87m0c2. The short way: KE = − m c2 = − ((3.87+6) − (3+4))m0 c2 = −2.87m0 c2
= −2.87u1.6610−27kg/u (3108m/s)2 = − 4.2910−10J
2.91 Momentum:
0.6(10kg)(0.6c) =
0.6m1(−0.6c) +
0.8m2(0.8c) → 7.5kg = −0.75m1+1.33m2
Energy:
0.6 (10kg) c2 =
0.6 m1c2 +
0.8 m2c2 → 12.5kg = 1.25m1 + 1.67m2
Solve: Multiply E−equation by 0.6, then add: 15kg = 2.33 m2 m2 = 6.43kg. Reinsert: m1 = 1.43kg
(b) KE = − mc2 = − (6.43kg + 1.43kg − 10kg) (91016m2/c2) = 1.931017J
2.92 Momentum:
0.8 m1(0.8c) +
0.6 m2 (−0.6c) = 0 1.33 m1 = 0.75 m2 m2 = 1.78 m1
(b) Energy:
0.8 m1c2 +
0.6 m2c2 =
0 mfc2 → 1.67 m1 + 1.25 m2 = mf →
1.67 m1 + 1.25(1.78m1) = mf mf = 3.89m1.
(c) KE = − mc2 = − (3.89m1 − m1 − 1.78m1)c2 = −1.11m1c2
2.93 (
0.9−1) m0c2 =2
1 1
1 (0.9)
−
− m0c2 = 1.29m0c2
(b) 2 (
u−1) m0c2 = 1.29m0c2
u =1.29
2 + 1 = 1.647.2
1
1 ( / )− u c = 1.647 u = 0.795c.
(c) Experiment A: Momentum:
0.9 m0 (0.9c) =
fu m uf . Energy:
0.9 m0c2 + m0c2 =
fu m c2
Divide equations: uf =0.9 0
0.9 0 0
(0.9 )
+
m c
m m =2
2
1 (0.9 )
1 (0.9)
1 1
1 (0.9)
−
+
−
c = 0.627c. Plug back in to momentum equation:2
1
1 (0.9)−
m0 (0.9c) =2
1
1 (0.627)− m (0.627c) m = 2.57m0
20
KEe = (
− 1)mc2 =31 16 2 2
2
1 1 (9.11 10 kg)(9 10 m /c )
1 (0.99971)
−
−
− = 3.310−12J = 20.6MeV.
KEC=1
2 mu2 =1
2 (13.99995u1.6610−27kg/u) (1.6210−33108m/s)2 = 2.7510−15J = 17.1keV
(about1
1, 000 of KEe).
2.90 Let’s use m0 for an “atomic mass unit”, rather than u (for obvious reasons).
Momentum:
0.8 (3m0) (+0.8c) +
0.6 (4m0)(−0.6c) =
0 (6m0)(0) +
u m u → 1m0 c =
u m u
Energy:
0.8 (3m0)c2 +
0.6 (4m0)c2 = +10m0c2 =
0 (6m0)c2 +
u mc2 → +4m0 =
u m
Divide the two: u = c/4. Plug back in: 4m0 =
0.25 m m = 3.87u
(b) Since mass/internal energy increases, KE must decrease. The long way: KEfinal = (
0.25−1) (3.87m0) c2 =
0.127m0 c2. KEinitial = (
0.8−1) (3m0) c2 + (
0.6−1) (4m0)c2 = 3m0 c2
KE = −2.87m0c2. The short way: KE = − m c2 = − ((3.87+6) − (3+4))m0 c2 = −2.87m0 c2
= −2.87u1.6610−27kg/u (3108m/s)2 = − 4.2910−10J
2.91 Momentum:
0.6(10kg)(0.6c) =
0.6m1(−0.6c) +
0.8m2(0.8c) → 7.5kg = −0.75m1+1.33m2
Energy:
0.6 (10kg) c2 =
0.6 m1c2 +
0.8 m2c2 → 12.5kg = 1.25m1 + 1.67m2
Solve: Multiply E−equation by 0.6, then add: 15kg = 2.33 m2 m2 = 6.43kg. Reinsert: m1 = 1.43kg
(b) KE = − mc2 = − (6.43kg + 1.43kg − 10kg) (91016m2/c2) = 1.931017J
2.92 Momentum:
0.8 m1(0.8c) +
0.6 m2 (−0.6c) = 0 1.33 m1 = 0.75 m2 m2 = 1.78 m1
(b) Energy:
0.8 m1c2 +
0.6 m2c2 =
0 mfc2 → 1.67 m1 + 1.25 m2 = mf →
1.67 m1 + 1.25(1.78m1) = mf mf = 3.89m1.
(c) KE = − mc2 = − (3.89m1 − m1 − 1.78m1)c2 = −1.11m1c2
2.93 (
0.9−1) m0c2 =2
1 1
1 (0.9)
−
− m0c2 = 1.29m0c2
(b) 2 (
u−1) m0c2 = 1.29m0c2
u =1.29
2 + 1 = 1.647.2
1
1 ( / )− u c = 1.647 u = 0.795c.
(c) Experiment A: Momentum:
0.9 m0 (0.9c) =
fu m uf . Energy:
0.9 m0c2 + m0c2 =
fu m c2
Divide equations: uf =0.9 0
0.9 0 0
(0.9 )
+
m c
m m =2
2
1 (0.9 )
1 (0.9)
1 1
1 (0.9)
−
+
−
c = 0.627c. Plug back in to momentum equation:2
1
1 (0.9)−
m0 (0.9c) =2
1
1 (0.627)− m (0.627c) m = 2.57m0
Loading page 23...
Instructor Solutions Manual for Harris, Modern Physics, Second Edition
21
Experiment B: Momentum:
0.795 m0 (−0.795c) +
0.795 m0 (+0.795c) =
fu m uf .
Energy: 2
0.795 m0c2 =
fu mc2. From p-equation, uf = 0. E-equation becomes: 2
0.795 m0 = 1 m
m = 22
1
1 (0.795)− m0 = 3.29m0. Though mass increases in the both completely inelastic collisions,
Experiment B, the collider, with the same initial kinetic energy input, yields more mass, simply because uf
= 0. There is no final kinetic energy.
2.94 Energy:
u (8.8710−28kg) c2 =
0.9 (2.4910−28kg)c2+
0.8 (2.4910−28kg)c22
1
1 ( / )− u c
=(2.294)(2.49) (1.67)(2.49)
8.87
+ u = 0.437c.
Momentumx:
0.437 (8.87)(0.437c) =
0.9 (2.49)(0.9c cos
1) +
0.8 (2.49)(0.8c cos
2)
Momentumy: 0 =
0.9 (2.49)(0.9c sin
1) −
0.8 (2.49)(0.8c sin
2)
Or: 4.312 − 5.141 cos
1 = 3.32 cos
2 and 5.141 sin
1 = 3.32 sin
2. Square both:
18.591 − 44.335 cos
1 + 26.432 cos2
1 = 11.022 cos2
2 and 26.432 sin2
1 = 11.022 sin2
2.
Add: 18.591 − 44.335 cos
1 + 26.432 = 11.022
1 = 39.9º (the 0.9c
+)
Plug back in: 5.141 sin 39.9º = 3.32 sin
2
2 = 83.6º (the 0.8c
−)
Mass decreases, so KE must increase. As is often the case, this can be seen another way: There is a frame of
reference moving with the kaon in which the process is simply a stationary object (i.e., the kaon) exploding into
two parts. KEi = 0, KEf > 0.
2.95 In the new frame, the initial particle moves left at 0.6c and the right-hand fragment is stationary. The left-hand
fragment moves at u =2
1
−
−
u v
uv
c =2
0.6 0.6
( 0.6 )(0.6 )
1
− −
−
−
c c
c c
c = −0.882c.
In the new frame, Pinitial =
0.6 m0(–0.6c) = –1.25m00.6c = –0.75 m0c
Pfinal =
0.882 m (–0.882c) = –2.125 m 0.882c. Can relate m0 and m via energy conservation, and using original
frame is easiest: m0c2 = 2(1.25)mc2 m = 0.4 m0. Plug back in: Pfinal = –2.125 (0.4 m0) 0.882c = –0.75 m0c
2.96( )
2 2 2
1
= − + i iu i u i im c m c m c . E = 0 →( )
2 2
1 0
− + =iu i im c m c or( )
2 2
1
− = − iu i im c m c
2.97 E =
u m c2 and p =
umu. Squaring both: E2 =2
1
1 ( / )− u c m2 c4 and p2 =2
1
1 ( / )− u c m2 u2.
E2 − p2 c2 =2
1
1 ( / )− u c m2 c4 −2
1
1 ( / )− u c m2 u2 c2 = m2 c22 2
2
( )
1 ( / )
−
−
c u
u c = m2 c4
2.98 In the product2 2 1 1
2 2 1 1
( ) ( ) ( 2 ) ( )
. . .
( ) ( 2 ) ( ) ( )
− + +
− − +
f r f r dr f r dr f r dr
f r dr f r dr f r dr f r , all terms but2
1
( )
( )
f r
f r cancel.
21
Experiment B: Momentum:
0.795 m0 (−0.795c) +
0.795 m0 (+0.795c) =
fu m uf .
Energy: 2
0.795 m0c2 =
fu mc2. From p-equation, uf = 0. E-equation becomes: 2
0.795 m0 = 1 m
m = 22
1
1 (0.795)− m0 = 3.29m0. Though mass increases in the both completely inelastic collisions,
Experiment B, the collider, with the same initial kinetic energy input, yields more mass, simply because uf
= 0. There is no final kinetic energy.
2.94 Energy:
u (8.8710−28kg) c2 =
0.9 (2.4910−28kg)c2+
0.8 (2.4910−28kg)c22
1
1 ( / )− u c
=(2.294)(2.49) (1.67)(2.49)
8.87
+ u = 0.437c.
Momentumx:
0.437 (8.87)(0.437c) =
0.9 (2.49)(0.9c cos
1) +
0.8 (2.49)(0.8c cos
2)
Momentumy: 0 =
0.9 (2.49)(0.9c sin
1) −
0.8 (2.49)(0.8c sin
2)
Or: 4.312 − 5.141 cos
1 = 3.32 cos
2 and 5.141 sin
1 = 3.32 sin
2. Square both:
18.591 − 44.335 cos
1 + 26.432 cos2
1 = 11.022 cos2
2 and 26.432 sin2
1 = 11.022 sin2
2.
Add: 18.591 − 44.335 cos
1 + 26.432 = 11.022
1 = 39.9º (the 0.9c
+)
Plug back in: 5.141 sin 39.9º = 3.32 sin
2
2 = 83.6º (the 0.8c
−)
Mass decreases, so KE must increase. As is often the case, this can be seen another way: There is a frame of
reference moving with the kaon in which the process is simply a stationary object (i.e., the kaon) exploding into
two parts. KEi = 0, KEf > 0.
2.95 In the new frame, the initial particle moves left at 0.6c and the right-hand fragment is stationary. The left-hand
fragment moves at u =2
1
−
−
u v
uv
c =2
0.6 0.6
( 0.6 )(0.6 )
1
− −
−
−
c c
c c
c = −0.882c.
In the new frame, Pinitial =
0.6 m0(–0.6c) = –1.25m00.6c = –0.75 m0c
Pfinal =
0.882 m (–0.882c) = –2.125 m 0.882c. Can relate m0 and m via energy conservation, and using original
frame is easiest: m0c2 = 2(1.25)mc2 m = 0.4 m0. Plug back in: Pfinal = –2.125 (0.4 m0) 0.882c = –0.75 m0c
2.96( )
2 2 2
1
= − + i iu i u i im c m c m c . E = 0 →( )
2 2
1 0
− + =iu i im c m c or( )
2 2
1
− = − iu i im c m c
2.97 E =
u m c2 and p =
umu. Squaring both: E2 =2
1
1 ( / )− u c m2 c4 and p2 =2
1
1 ( / )− u c m2 u2.
E2 − p2 c2 =2
1
1 ( / )− u c m2 c4 −2
1
1 ( / )− u c m2 u2 c2 = m2 c22 2
2
( )
1 ( / )
−
−
c u
u c = m2 c4
2.98 In the product2 2 1 1
2 2 1 1
( ) ( ) ( 2 ) ( )
. . .
( ) ( 2 ) ( ) ( )
− + +
− − +
f r f r dr f r dr f r dr
f r dr f r dr f r dr f r , all terms but2
1
( )
( )
f r
f r cancel.
Loading page 24...
Chapter 2 Special Relativity
22
(b) In multiplying out the product1 1 1 2
2 2 2 2
( ) ( ) ( 2 ) ( )
1 1 1 . . . 1
+ + −
− − − −
g r dr g r dr dr g r dr dr g r dr dr
c c c c , any
term with a fraction squared or higher order will be negligible. Thus, only the leading 1 and terms with a
fraction to the first power survive.1 1 1 2
2 2 2 2
( ) ( ) ( 2 ) ( )
1 . . .
+ + −
− − − − −
g r dr g r dr dr g r dr dr g r dr dr
c c c c . The
terms beyond the 1 define the integral of g(r) from r1 to r2.
(c) mg(r) =2
GMm
r g(r) =2
GM
r . So,2
1
2
2 2 2
1 1 2
( ) 1 1 1 1
1 1
( )
= − = − −
r
r
f r GM dr GM
f r r rc r c
or2 1 2
1 2
1 1 1
( ) ( ) 1
= − −
f r f r GM r rc . This is analogous to equation (2-29), where the satellite is the higher
point r2 and Earth is the lower point r1. Thus, analogous to (2-30), we have1 2 2
1 2
1 1 1
( ) ( ) 1
= − −
t r t r GM r rc
orEarth satellite 2
Earth satellite
1 1 1
1
= − −
t t GM r rc
2.99 Although the satellite appears not to move in the sky, it is moving, and the point on Earth’s equator is also
moving. For geosynchronous orbit,2
2
= → =
v GMm GM
m v
r rr and T =2 86, 400s.
=
r
v11 2 2 24
(6.67 10 N m /kg )(5.98 10 kg)
86, 400s / 2
−
=v v
. Solving this for v gives v = 3.07103m/s. Subtracting the tangential
speed of 460m/s at Earth’s surface gives a relative speed of 2.61103m/s. The orbit radius is given by3
2 86, 400s
3.07 10 m/s
=
r
, or r = 4.23107m/s. Now, as in the GPS example, for the speed-dependent part,( )
1/ 211Satellite
Earth Satellite
23
8
1 7.6 10
2.61 10
1 3 10
−−
= = −
−
t
t t( )
11
Satellite1 3.8 10−
+ t
and for the gravitational part,14 3 2
Earth Satellite 8 2 6 7
4.0 10 m /s 1 1
1 (3 10 m/s) 6.4 10 m 4.23 10 m
− −
t t( )
10
Satellite 1 5.89 10−
= − t
. Accounting for
both (5.89 – 0.38 5.5),( )
10
Earth Satellite 1 5.5 10−
− t t . The difference in a day is (86,400s) (5.510−10)
= 4.810−5s = 48μs.
2.100 Radius is of dimensions [L], mass [M], c[L]
[T] , and G (Nm2/kg2)3
2
[L]
[T] [M]
r = Ma Gb cd → [L] = [M]a3b d
2b b d
[L] [L]
[T] [M] [T] . Considering mass gives 0 = a − b, or a = b
Considering time gives 0 = −2b − d, or d = −2b. Considering length gives 1 = 3b + d. But since d = −2b this
becomes 1 = 3b − 2b, b = 1. This in turns gives d = −2 and a = 1. Thus r = MG/c2
2.101 KE + PE = E +2
( / )
−
GM E c
r = 0. The pulse’s energy cancels, leaving r =2
GM
c .
22
(b) In multiplying out the product1 1 1 2
2 2 2 2
( ) ( ) ( 2 ) ( )
1 1 1 . . . 1
+ + −
− − − −
g r dr g r dr dr g r dr dr g r dr dr
c c c c , any
term with a fraction squared or higher order will be negligible. Thus, only the leading 1 and terms with a
fraction to the first power survive.1 1 1 2
2 2 2 2
( ) ( ) ( 2 ) ( )
1 . . .
+ + −
− − − − −
g r dr g r dr dr g r dr dr g r dr dr
c c c c . The
terms beyond the 1 define the integral of g(r) from r1 to r2.
(c) mg(r) =2
GMm
r g(r) =2
GM
r . So,2
1
2
2 2 2
1 1 2
( ) 1 1 1 1
1 1
( )
= − = − −
r
r
f r GM dr GM
f r r rc r c
or2 1 2
1 2
1 1 1
( ) ( ) 1
= − −
f r f r GM r rc . This is analogous to equation (2-29), where the satellite is the higher
point r2 and Earth is the lower point r1. Thus, analogous to (2-30), we have1 2 2
1 2
1 1 1
( ) ( ) 1
= − −
t r t r GM r rc
orEarth satellite 2
Earth satellite
1 1 1
1
= − −
t t GM r rc
2.99 Although the satellite appears not to move in the sky, it is moving, and the point on Earth’s equator is also
moving. For geosynchronous orbit,2
2
= → =
v GMm GM
m v
r rr and T =2 86, 400s.
=
r
v11 2 2 24
(6.67 10 N m /kg )(5.98 10 kg)
86, 400s / 2
−
=v v
. Solving this for v gives v = 3.07103m/s. Subtracting the tangential
speed of 460m/s at Earth’s surface gives a relative speed of 2.61103m/s. The orbit radius is given by3
2 86, 400s
3.07 10 m/s
=
r
, or r = 4.23107m/s. Now, as in the GPS example, for the speed-dependent part,( )
1/ 211Satellite
Earth Satellite
23
8
1 7.6 10
2.61 10
1 3 10
−−
= = −
−
t
t t( )
11
Satellite1 3.8 10−
+ t
and for the gravitational part,14 3 2
Earth Satellite 8 2 6 7
4.0 10 m /s 1 1
1 (3 10 m/s) 6.4 10 m 4.23 10 m
− −
t t( )
10
Satellite 1 5.89 10−
= − t
. Accounting for
both (5.89 – 0.38 5.5),( )
10
Earth Satellite 1 5.5 10−
− t t . The difference in a day is (86,400s) (5.510−10)
= 4.810−5s = 48μs.
2.100 Radius is of dimensions [L], mass [M], c[L]
[T] , and G (Nm2/kg2)3
2
[L]
[T] [M]
r = Ma Gb cd → [L] = [M]a3b d
2b b d
[L] [L]
[T] [M] [T] . Considering mass gives 0 = a − b, or a = b
Considering time gives 0 = −2b − d, or d = −2b. Considering length gives 1 = 3b + d. But since d = −2b this
becomes 1 = 3b − 2b, b = 1. This in turns gives d = −2 and a = 1. Thus r = MG/c2
2.101 KE + PE = E +2
( / )
−
GM E c
r = 0. The pulse’s energy cancels, leaving r =2
GM
c .
Loading page 25...
Instructor Solutions Manual for Harris, Modern Physics, Second Edition
23
2.102 Those of velocity +1m/s will be at 1m, of velocity +2m/s at 2m, of velocity −1m/s at −1m, etc.
(b) If the observer jumps to v meters from the origin, on a particle moving at v meters per second, he will find
particles moving at v+1 meters per second one meter further away, and particles moving at v−1 meters per
second one meter closer to the origin. But if moving at v meters per second himself, he will see relative
velocities, respectively, of +1m/s and −1m/s, just as does the person at the origin.
2.1030 0 0
3 2
0 0 0 0
2 1 0 1 0 1
2 2 2 2 2 2
= − + − + +
v
u u uv c v c v c
t x u c c u c c u c c u or( )
0
2 2
0 0
0
1 2
1
2 2
=
+
+
uv
c cu u cc
c u
.
(b) If u0 < c, then( )
2 2
0
2
1+ u c > 1. Thus, for3
t to be negative means that0
uv
c c , but, as noted, v is not
allowed to exceed u0.
(c)2 2 2
0 0 0 0 0
2 2
1 0 2 1 0 1 2
− → − + → +
u u u u u
c c cc c or0
2 2
0
2
1 1
+
u c
u c
2.1042
= − +
v
v
t x t
c . Divide both sides by t:2 1
= − +
v
t v x
t tc . If the time intervals are of opposite
sign, then2 1 0
− +
v x
tc or2
.
x c c
t v The speed needed to travel the x in time t is greater than c. Using
the complementary Lorentz transformation equation gives2 1 0
+ +
v x
tc or2
−
x c
t v . This too implies a
speed greater than c.
2.105 Asv
c → 0,
v → 1, so the matrix in (1−15) becomes:−
−
1 0 0
0 1 0 0
0 0 1 0
0 0 1
v
c
v
c
(b)0
0
x
ct =1 0 0
0 1 0 0 0
0 0 1 0 0
0 0 1
−
−
v
c
v
c
x
ct =−
−
x vt
0
0
ct xv c .
In the limitv
c → 0, the term xv/c can be ignored, leaving x = x−vt and t = t.
23
2.102 Those of velocity +1m/s will be at 1m, of velocity +2m/s at 2m, of velocity −1m/s at −1m, etc.
(b) If the observer jumps to v meters from the origin, on a particle moving at v meters per second, he will find
particles moving at v+1 meters per second one meter further away, and particles moving at v−1 meters per
second one meter closer to the origin. But if moving at v meters per second himself, he will see relative
velocities, respectively, of +1m/s and −1m/s, just as does the person at the origin.
2.1030 0 0
3 2
0 0 0 0
2 1 0 1 0 1
2 2 2 2 2 2
= − + − + +
v
u u uv c v c v c
t x u c c u c c u c c u or( )
0
2 2
0 0
0
1 2
1
2 2
=
+
+
uv
c cu u cc
c u
.
(b) If u0 < c, then( )
2 2
0
2
1+ u c > 1. Thus, for3
t to be negative means that0
uv
c c , but, as noted, v is not
allowed to exceed u0.
(c)2 2 2
0 0 0 0 0
2 2
1 0 2 1 0 1 2
− → − + → +
u u u u u
c c cc c or0
2 2
0
2
1 1
+
u c
u c
2.1042
= − +
v
v
t x t
c . Divide both sides by t:2 1
= − +
v
t v x
t tc . If the time intervals are of opposite
sign, then2 1 0
− +
v x
tc or2
.
x c c
t v The speed needed to travel the x in time t is greater than c. Using
the complementary Lorentz transformation equation gives2 1 0
+ +
v x
tc or2
−
x c
t v . This too implies a
speed greater than c.
2.105 Asv
c → 0,
v → 1, so the matrix in (1−15) becomes:−
−
1 0 0
0 1 0 0
0 0 1 0
0 0 1
v
c
v
c
(b)0
0
x
ct =1 0 0
0 1 0 0 0
0 0 1 0 0
0 0 1
−
−
v
c
v
c
x
ct =−
−
x vt
0
0
ct xv c .
In the limitv
c → 0, the term xv/c can be ignored, leaving x = x−vt and t = t.
Loading page 26...
Chapter 2 Special Relativity
24
2.106 The matrix to find (x, t) values from (x, t) is of the form:
x
y
x
ct =0 0
0 1 0 0
0 0 1 0
0 0
−
−
v
c
v
c
v v
v v
x
y
x
ct
0.8c =5
3 and5
3 0.8 =4
3 , so this matrix is−
−
0 0
0 1 0 0
0 0 1 0
0 0
5 4
3 3
54
3 3 . The space−time point is5ly
0
0
2yr
c in Bob’s frame S.
Thus, in Anna’s frame S:5 4
3 3
54
3 3
0 0 5ly
0 1 0 0 0
0 0 1 0 0
0 0 2yr
−
− c =(17 / 3)ly
0
0
(10 / 3)yr
− c . The position is 5.67ly and the time −3.33yr,
in agreement with Exercise 22.
2.107
= −
x v x t
v
A A A
c =y yA A =z zA A
= −
t v t x
v
A A A
c
Squaring,2
2 2 2 2
2
2
+
= −
x v x x t t
v v
A A A A A
c c and2
2 2 2 2
2
2
+
= −
t v t x t x
v v
A A A A A
c c
Subtracting, the middle terms cancel:2 2
2 2 2 2 2
2 2
1 1
− = − − −
t x v t x
v v
A A A A
c c .
But
v
2 =2
2
1
1− v
c , so this becomes2 2
−t xA A , and since =y yA A and =z zA A , equation (2-36) follows.
2.108 p =
0.8 m(0.8c) =5
3 (1kg)(0.8c) =4
3 (1kg) c = 4108kgm/s. E =5
3 (1kg)c2 = 1.51017J
The matrix to find( , ) xp E from( , )xp E is of the form
x
y
z
p
p
p
E c =0 0
0 1 0 0
0 0 1 0
0 0
−
−
v
c
v
c
v v x
y
z
v v
p
p
p
E c
Since
0.8c =5
3 and5
3 0.8 =4
3 , the matrix is−
−
0 0
0 1 0 0
0 0 1 0
0 0
5 4
3 3
54
3 3 . Momentum−energy in S is4
3
25
3
(1kg)
0
0
(1kg)
c
c c and
matrix multiplication gives5 54 4
3 3 3 3
5 54 4
3 3 3 3
(1kg) (1kg) 0
0 0
0 0
(1kg) (1kg) (1kg)
−
=
− +
c c
c c c for momentum−energy in S. In its rest frame,
its momentum is zero; and E/c = (1kg)c E = (1kg)c2. Its energy is internal only; no KE.
24
2.106 The matrix to find (x, t) values from (x, t) is of the form:
x
y
x
ct =0 0
0 1 0 0
0 0 1 0
0 0
−
−
v
c
v
c
v v
v v
x
y
x
ct
0.8c =5
3 and5
3 0.8 =4
3 , so this matrix is−
−
0 0
0 1 0 0
0 0 1 0
0 0
5 4
3 3
54
3 3 . The space−time point is5ly
0
0
2yr
c in Bob’s frame S.
Thus, in Anna’s frame S:5 4
3 3
54
3 3
0 0 5ly
0 1 0 0 0
0 0 1 0 0
0 0 2yr
−
− c =(17 / 3)ly
0
0
(10 / 3)yr
− c . The position is 5.67ly and the time −3.33yr,
in agreement with Exercise 22.
2.107
= −
x v x t
v
A A A
c =y yA A =z zA A
= −
t v t x
v
A A A
c
Squaring,2
2 2 2 2
2
2
+
= −
x v x x t t
v v
A A A A A
c c and2
2 2 2 2
2
2
+
= −
t v t x t x
v v
A A A A A
c c
Subtracting, the middle terms cancel:2 2
2 2 2 2 2
2 2
1 1
− = − − −
t x v t x
v v
A A A A
c c .
But
v
2 =2
2
1
1− v
c , so this becomes2 2
−t xA A , and since =y yA A and =z zA A , equation (2-36) follows.
2.108 p =
0.8 m(0.8c) =5
3 (1kg)(0.8c) =4
3 (1kg) c = 4108kgm/s. E =5
3 (1kg)c2 = 1.51017J
The matrix to find( , ) xp E from( , )xp E is of the form
x
y
z
p
p
p
E c =0 0
0 1 0 0
0 0 1 0
0 0
−
−
v
c
v
c
v v x
y
z
v v
p
p
p
E c
Since
0.8c =5
3 and5
3 0.8 =4
3 , the matrix is−
−
0 0
0 1 0 0
0 0 1 0
0 0
5 4
3 3
54
3 3 . Momentum−energy in S is4
3
25
3
(1kg)
0
0
(1kg)
c
c c and
matrix multiplication gives5 54 4
3 3 3 3
5 54 4
3 3 3 3
(1kg) (1kg) 0
0 0
0 0
(1kg) (1kg) (1kg)
−
=
− +
c c
c c c for momentum−energy in S. In its rest frame,
its momentum is zero; and E/c = (1kg)c E = (1kg)c2. Its energy is internal only; no KE.
Loading page 27...
Instructor Solutions Manual for Harris, Modern Physics, Second Edition
25
2.109yp =2
1
−
x
v u
u v
c m2
1
−
y
x
v
u
u v
c =
u m uy = py. The z−component follows similarly.
2.110
= −
x v x
v E
p p c c . If1,
v
c then
v 1 and E mc2. Thus,xp becomes px – mv, essentially equation
(2-21).
= −
v x
E E v p
c c c . In the same limit, this becomes m´c = mc, simply confirming that the mass is the
same in both frames.
2.111 p =
0.8 3m0(0.8c) =5
3 3m0 (0.8c) = 4m0c. E =5
3 3m0c2 = 5m0c2.
(b)xu =2
1
−
−
x
x
u v
u v
c =2
0.8 0.5 0.5 .
(0.8 )(0.5 )
1
− =
−
c c c
c c
c
p´ =
0.5 3m0(0.5c) =2
3 3m0 (0.5c) =3 m0c. E´ =2
3 3m0c2 = 23 m0c2.
= −
x v x
v E
p p c c
=
0 0 0
2 4 (0.5)5 3 .
3 − =m c m c m c
= −
v x
E E v p
c c c
0 0 0
2
= 5 (0.5) 4 2 3 .
3 m c m c m c− =
2.112 The frame in which the final single object is at rest is simplest, in which case the invariant is just (Mc2/c)2= M2c2.
(b) In the lab frame there are two terms in the total energy. Inserting both and using the relationship suggested
to eliminate Ptotal, gives2 2 2 2 2 2 2
total total( / ) ( / ) ( / )i iE c P E c mc E c m c− = + − −
=2 2 2 2 2 2 2 2
/ 2 /i i iE c E m m c E c m c+ + − + = 2(mEi + m2c2). The invariant is the same no matter which frame is
considered, so M2c2 = 2(mEi + m2c2) M =2 2
2 2imE c m+ .
(c) Momentum and energy conservation arei fu i u fmu Mu
= and2 2 2
i fu umc mc Mc
+ = . If we square both
sides of both we have2 2 2 2 2 2
i fu i u fm u M u
= and2 2 4 2 4 2 4 2 2 4
2i i fu u um c m c m c M c
+ + = . If we now subtract c2
times the former from the latter we have2 2 4 2 2 2 2 2 4 2 4 2 2 4 2 2 2 2
2i i i f fu u i u u u fm c m u c m c m c M c M u c
− + + = − .
The identify can now be used on each side, yielding2 4 2 4 2 4 2 4
2 ium c m c m c M c
+ + = , or2 2 2 2 2 2
2 2iu m c m c M c
+ =
. Noting that2
iu mc
is Ei, this is the same result as before.
25
2.109yp =2
1
−
x
v u
u v
c m2
1
−
y
x
v
u
u v
c =
u m uy = py. The z−component follows similarly.
2.110
= −
x v x
v E
p p c c . If1,
v
c then
v 1 and E mc2. Thus,xp becomes px – mv, essentially equation
(2-21).
= −
v x
E E v p
c c c . In the same limit, this becomes m´c = mc, simply confirming that the mass is the
same in both frames.
2.111 p =
0.8 3m0(0.8c) =5
3 3m0 (0.8c) = 4m0c. E =5
3 3m0c2 = 5m0c2.
(b)xu =2
1
−
−
x
x
u v
u v
c =2
0.8 0.5 0.5 .
(0.8 )(0.5 )
1
− =
−
c c c
c c
c
p´ =
0.5 3m0(0.5c) =2
3 3m0 (0.5c) =3 m0c. E´ =2
3 3m0c2 = 23 m0c2.
= −
x v x
v E
p p c c
=
0 0 0
2 4 (0.5)5 3 .
3 − =m c m c m c
= −
v x
E E v p
c c c
0 0 0
2
= 5 (0.5) 4 2 3 .
3 m c m c m c− =
2.112 The frame in which the final single object is at rest is simplest, in which case the invariant is just (Mc2/c)2= M2c2.
(b) In the lab frame there are two terms in the total energy. Inserting both and using the relationship suggested
to eliminate Ptotal, gives2 2 2 2 2 2 2
total total( / ) ( / ) ( / )i iE c P E c mc E c m c− = + − −
=2 2 2 2 2 2 2 2
/ 2 /i i iE c E m m c E c m c+ + − + = 2(mEi + m2c2). The invariant is the same no matter which frame is
considered, so M2c2 = 2(mEi + m2c2) M =2 2
2 2imE c m+ .
(c) Momentum and energy conservation arei fu i u fmu Mu
= and2 2 2
i fu umc mc Mc
+ = . If we square both
sides of both we have2 2 2 2 2 2
i fu i u fm u M u
= and2 2 4 2 4 2 4 2 2 4
2i i fu u um c m c m c M c
+ + = . If we now subtract c2
times the former from the latter we have2 2 4 2 2 2 2 2 4 2 4 2 2 4 2 2 2 2
2i i i f fu u i u u u fm c m u c m c m c M c M u c
− + + = − .
The identify can now be used on each side, yielding2 4 2 4 2 4 2 4
2 ium c m c m c M c
+ + = , or2 2 2 2 2 2
2 2iu m c m c M c
+ =
. Noting that2
iu mc
is Ei, this is the same result as before.
Loading page 28...
Chapter 2 Special Relativity
26
2.113
=
x
x
du
a dt =2
2
( ) 1
− −
− +
x
x
v
u v
d u v c
v dx dt
c =2
2
2 2
2
( )
1 1
−
+
− −
− +
x x
x
x x
v
v
u v du
du c
u v u v
c c
v dx dt
c =2 2
2 2
2 2
2
1 ( )
1 1
− −
+
− −
− +
x
x
x
x x
v
u v v
u v
c c
du u v u v
c c
v dx dt
c
=22
2 2
2
1 1
( )
− −
− +
x
x
v
u vv
du c c
v dx dt
c . Dividing top and bottom by dt yields22
2 2
2
1 1
1
− −
=
− +
x x
x
v
du u vv
dt c c
a v dx
dtc
But=x
x
du a
dt , and 1−2
2
v
c =2
1
v and −2
v dx
dtc +1 = 1 −2x
v
u c , so3
3
2
1
=
−
x
x
x
v
a
a u v
c
=
y
y
du
a dt
=2
2
1
−
− +
x
y v
v
u v
d u c
v dx dt
c =2
2
2 2
2
2
1 1
+ − −
− +
y x
y
x x
v
v
u du
du c
u v u v
c c
v dx dt
c .
Dividing top and bottom by dt,2
2
2 2
2
2
1 1
1
+ − −
=
− +
xy
y
x x
y
v
duvdu u dtcdt
u v u v
c c
a v dx
dtc
But=x
x
du a
dt ,=
y
y
du a
dt and −2
v dx
dtc +1 = 1 −2x
v
u c , so2
2 3
2 2
2 2
1 1
= +
− −
y
x
y
y
x x
v v
u v
aa ca u v u v
c c .
2.114
0.5c =2
1
1 (0.5)− =2
3 andv
c
v is therefore1
3 . Thus, the matrix is−
−
0 0
0 1 0 0
0 0 1 0
0 0
2 1
3 3
1 2
3 3 .
26
2.113
=
x
x
du
a dt =2
2
( ) 1
− −
− +
x
x
v
u v
d u v c
v dx dt
c =2
2
2 2
2
( )
1 1
−
+
− −
− +
x x
x
x x
v
v
u v du
du c
u v u v
c c
v dx dt
c =2 2
2 2
2 2
2
1 ( )
1 1
− −
+
− −
− +
x
x
x
x x
v
u v v
u v
c c
du u v u v
c c
v dx dt
c
=22
2 2
2
1 1
( )
− −
− +
x
x
v
u vv
du c c
v dx dt
c . Dividing top and bottom by dt yields22
2 2
2
1 1
1
− −
=
− +
x x
x
v
du u vv
dt c c
a v dx
dtc
But=x
x
du a
dt , and 1−2
2
v
c =2
1
v and −2
v dx
dtc +1 = 1 −2x
v
u c , so3
3
2
1
=
−
x
x
x
v
a
a u v
c
=
y
y
du
a dt
=2
2
1
−
− +
x
y v
v
u v
d u c
v dx dt
c =2
2
2 2
2
2
1 1
+ − −
− +
y x
y
x x
v
v
u du
du c
u v u v
c c
v dx dt
c .
Dividing top and bottom by dt,2
2
2 2
2
2
1 1
1
+ − −
=
− +
xy
y
x x
y
v
duvdu u dtcdt
u v u v
c c
a v dx
dtc
But=x
x
du a
dt ,=
y
y
du a
dt and −2
v dx
dtc +1 = 1 −2x
v
u c , so2
2 3
2 2
2 2
1 1
= +
− −
y
x
y
y
x x
v v
u v
aa ca u v u v
c c .
2.114
0.5c =2
1
1 (0.5)− =2
3 andv
c
v is therefore1
3 . Thus, the matrix is−
−
0 0
0 1 0 0
0 0 1 0
0 0
2 1
3 3
1 2
3 3 .
Loading page 29...
Instructor Solutions Manual for Harris, Modern Physics, Second Edition
27
Now multiplying,2 1 2 1
3 3 3 3
1 2 1 2
3 3 3 3
0 0 0 0
0 1 0 0 0 1 0 0
0 0 1 0 0 0 1 0
0 0 0 0
− −
− −
=−
−
0 0
0 1 0 0
0 0 1 0
0 0
5 4
3 3
54
3 3 . The upper−left element is
v, which in this case is5
3
.2 2
1
1 /− v c =5
3 v = 0.8c. This is as it should be: If S moves at 0.5c relative to S, which moves at 0.5c
relative to S, the velocity addition formula gives a velocity of2
0.5 0.5
(0.5 )(0.5 )
1
+
+
c c
c c
c = 0.8c for S relative to S.
2.115 In the new frame, the right-moving stream is stationary, so the distance between the charges is larger by
v and
the charge density thus smaller by
v.
right =
/
v =2
1 (1 3)
− = 8 3
. This is the charge density in the rest
frame of the charges in the stream. The charges in the left-moving stream move at c/3 relative to the “lab,” and
the lab moves at c/3 relative to the stationary stream, so the speed of the left-moving stream relative to the
stationary stream, by the relativistic velocity transformation, is2
3 3
( 3)( 3)
1
+
+
c c
c c
c = 0.6c. Relative to the stationary
stream, the left-moving charges are close together, so the density is higher by
0.6.
Thus,
left =
0.6
right =5
4 8 3
=5 8 12
.
(b) The streams push in opposite directions on the point charge and are the same distance away, so the net
electric force will depend only on the difference between the charge densities.( )
05 8 12 8 3 2
= = −EF qE q r.08 24
= q r
For the magnetic force, only the left-moving
stream is indeed moving, thus producing a magnetic field. The current is the charge per distance times the
distance per unit time.( )
5 8 12 0.6 8 4.
= =I c c In the new frame, the point charge is moving at speed
c/3, so it experiences a magnetic force( )
2
point charge 0 0( 3) 8 4 2 ( ) 8 24 .
= = =
BF qv B q c c r c q r
Noting that2
0 0
1 ,
=c we see that this is the same as the electric force. Because the point charge would
not experience a net force in the “lab” frame, it must not experience one in the new frame. The electric and
magnetic forces must be equal and opposite.
2.116 W =0 fu
u dp =0 2 2
1
1 /
−
fu
u d mu
u c =2
2 2 3/ 2 2 2 1/ 20
( / ) 1
(1 / ) (1 / )
+ − −
fu u c du
u m u du
u c u c2 2 2
2 2 3/ 2 2 2 3/ 20
/ 1 /
(1 / ) (1 / )
−
+ − −
fu u c u c
u m u du
u c u c
=2 2 3 / 20
1
(1 / )− fu
u m du
u c =2
2 2
01 /−
fu
mc
u c =2
0
fu
u mc
=2
( 1)
−fu mc
2.117 F =dp
dt =2 2 1/ 2
1
(1 / )
−
d mu
dt u c =2
2 2 3 / 2 2 2 1/ 2
/ 1
(1 / ) (1 / )
+
− −
u c du du
m u
dt dtu c u c
=2 2 2
2 2 3/ 2 2 2 3/ 2
( / ) 1 /
(1 / ) (1 / )
−
+
− −
u c u u c du
m dtu c u c =2 2 3 / 2
1
(1 / )−
du
m dtu c =3
u
du
m dt .
27
Now multiplying,2 1 2 1
3 3 3 3
1 2 1 2
3 3 3 3
0 0 0 0
0 1 0 0 0 1 0 0
0 0 1 0 0 0 1 0
0 0 0 0
− −
− −
=−
−
0 0
0 1 0 0
0 0 1 0
0 0
5 4
3 3
54
3 3 . The upper−left element is
v, which in this case is5
3
.2 2
1
1 /− v c =5
3 v = 0.8c. This is as it should be: If S moves at 0.5c relative to S, which moves at 0.5c
relative to S, the velocity addition formula gives a velocity of2
0.5 0.5
(0.5 )(0.5 )
1
+
+
c c
c c
c = 0.8c for S relative to S.
2.115 In the new frame, the right-moving stream is stationary, so the distance between the charges is larger by
v and
the charge density thus smaller by
v.
right =
/
v =2
1 (1 3)
− = 8 3
. This is the charge density in the rest
frame of the charges in the stream. The charges in the left-moving stream move at c/3 relative to the “lab,” and
the lab moves at c/3 relative to the stationary stream, so the speed of the left-moving stream relative to the
stationary stream, by the relativistic velocity transformation, is2
3 3
( 3)( 3)
1
+
+
c c
c c
c = 0.6c. Relative to the stationary
stream, the left-moving charges are close together, so the density is higher by
0.6.
Thus,
left =
0.6
right =5
4 8 3
=5 8 12
.
(b) The streams push in opposite directions on the point charge and are the same distance away, so the net
electric force will depend only on the difference between the charge densities.( )
05 8 12 8 3 2
= = −EF qE q r.08 24
= q r
For the magnetic force, only the left-moving
stream is indeed moving, thus producing a magnetic field. The current is the charge per distance times the
distance per unit time.( )
5 8 12 0.6 8 4.
= =I c c In the new frame, the point charge is moving at speed
c/3, so it experiences a magnetic force( )
2
point charge 0 0( 3) 8 4 2 ( ) 8 24 .
= = =
BF qv B q c c r c q r
Noting that2
0 0
1 ,
=c we see that this is the same as the electric force. Because the point charge would
not experience a net force in the “lab” frame, it must not experience one in the new frame. The electric and
magnetic forces must be equal and opposite.
2.116 W =0 fu
u dp =0 2 2
1
1 /
−
fu
u d mu
u c =2
2 2 3/ 2 2 2 1/ 20
( / ) 1
(1 / ) (1 / )
+ − −
fu u c du
u m u du
u c u c2 2 2
2 2 3/ 2 2 2 3/ 20
/ 1 /
(1 / ) (1 / )
−
+ − −
fu u c u c
u m u du
u c u c
=2 2 3 / 20
1
(1 / )− fu
u m du
u c =2
2 2
01 /−
fu
mc
u c =2
0
fu
u mc
=2
( 1)
−fu mc
2.117 F =dp
dt =2 2 1/ 2
1
(1 / )
−
d mu
dt u c =2
2 2 3 / 2 2 2 1/ 2
/ 1
(1 / ) (1 / )
+
− −
u c du du
m u
dt dtu c u c
=2 2 2
2 2 3/ 2 2 2 3/ 2
( / ) 1 /
(1 / ) (1 / )
−
+
− −
u c u u c du
m dtu c u c =2 2 3 / 2
1
(1 / )−
du
m dtu c =3
u
du
m dt .
Loading page 30...
Chapter 2 Special Relativity
28
(b) F =du
m dt only when
u is essentially unity, at speeds much less than c.
(c) F =2 2 3 / 2
1
(1 / )−
du
m dtu c → (F/m) dt =2 2 3/ 2
(1 / )−
du
u c → (F/m)2 2 3 / 2
(1 / )
= − du
dt u c → (F/m) t
=2 2 1/ 2
(1 / )−
u
u c →2 2 1/ 2
( ) (1 / )− =F m t u c u →2 2 2 2
( ) (1 / )− =Ft m u c u
( )
2 2 2
( ) 1 ( )= +Ft m Ft mc u u =2
1
1 ( )+
F t
mFt mc .
(d) u → c.
2.118 t =8
2
0.99 3 10 m/s
9.8m/s
= 3.03107s = 0.96yr.
(b) u =2
1
1 ( / )+
F
mFt mc t → 0.99 3108m/s =2
22
8
1 9.8 /
9.8m/s
1 + 3 10 m/s
m s t
m t
m
→ (0.99 3108m/s)222
8
9.8m/s
1 3 10 m/s
+
t = (9.8m/s2)2 t2
t =8
2 2
0.99 3 10 m/s
9.8m/s 1 (0.99)
− = 2.15108s = 6.8yr. Because when u approaches c the momentum begins to
grow much more rapidly with speed than classically, force must be applied for a much greater time.
2.119 x =0 2
1
1 ( / )
= +
ft F
u dt t dt
mFt mc =2 2
0 2
( / )
1 ( / )+
ftmc F mc t dt
F Ft mc =2
2
0
1 ( / )+
ft
mc Ft mc
F
=( )
2
2
1 ( / ) 1+ −
mc Ft mc
F
2.120 dt =2 2
1 /− u c dt =2
2 2
1 1
1 1 ( / )
− +
gt
c gt c dt =2 2
2
1 ( / ) ( / )
1 ( / )
+ −
+
gt c gt c
gt c dt =2
1 ( / )+
dt
gt c .
Integrating both sides, t =2
0 1 ( / )+
t dt
gt c =1
sinh−c gt
g c or t =sinh c gt
g c
28
(b) F =du
m dt only when
u is essentially unity, at speeds much less than c.
(c) F =2 2 3 / 2
1
(1 / )−
du
m dtu c → (F/m) dt =2 2 3/ 2
(1 / )−
du
u c → (F/m)2 2 3 / 2
(1 / )
= − du
dt u c → (F/m) t
=2 2 1/ 2
(1 / )−
u
u c →2 2 1/ 2
( ) (1 / )− =F m t u c u →2 2 2 2
( ) (1 / )− =Ft m u c u
( )
2 2 2
( ) 1 ( )= +Ft m Ft mc u u =2
1
1 ( )+
F t
mFt mc .
(d) u → c.
2.118 t =8
2
0.99 3 10 m/s
9.8m/s
= 3.03107s = 0.96yr.
(b) u =2
1
1 ( / )+
F
mFt mc t → 0.99 3108m/s =2
22
8
1 9.8 /
9.8m/s
1 + 3 10 m/s
m s t
m t
m
→ (0.99 3108m/s)222
8
9.8m/s
1 3 10 m/s
+
t = (9.8m/s2)2 t2
t =8
2 2
0.99 3 10 m/s
9.8m/s 1 (0.99)
− = 2.15108s = 6.8yr. Because when u approaches c the momentum begins to
grow much more rapidly with speed than classically, force must be applied for a much greater time.
2.119 x =0 2
1
1 ( / )
= +
ft F
u dt t dt
mFt mc =2 2
0 2
( / )
1 ( / )+
ftmc F mc t dt
F Ft mc =2
2
0
1 ( / )+
ft
mc Ft mc
F
=( )
2
2
1 ( / ) 1+ −
mc Ft mc
F
2.120 dt =2 2
1 /− u c dt =2
2 2
1 1
1 1 ( / )
− +
gt
c gt c dt =2 2
2
1 ( / ) ( / )
1 ( / )
+ −
+
gt c gt c
gt c dt =2
1 ( / )+
dt
gt c .
Integrating both sides, t =2
0 1 ( / )+
t dt
gt c =1
sinh−c gt
g c or t =sinh c gt
g c
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