Modern Physics for Scientists and Engineers 4th Edition Thornton Solutions Manual
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Modern Physics for Scientists and Engineers 4th Edition Thornton Solutions Manual Download: https://goo.gl/zpsgvf mode...
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Chapter 2 1
Special Theory of Relativity
Chapter 2
Special Theory of Relativity 1
Modern Physics for Scientists and Engineers 4th Edition Thornton Solutions Manual Download: http://testbanklive.com/download/modern-physics-for-scientistsand-engineers-4th-edition-thornton-solutions-manual/
Chapter 2 1. For a particle Newton’s second law says F ma
2
md dt
d2 y ˆ j
xˆ i 2
dt
2
d 2 zˆ dt
2
k .
Take the second derivative of each of the expressions in Equation (2.1):
d2x
d2x
d2 y
d2y
d 2z
dt 2
dt 2
dt 2
dt 2
dt 2
d 2z . Substitution into the previous equation gives
F ma m
d x
2
i
dt 2
2
2
dt 2
dt 2
dˆ y d zˆ j k F.ˆ
dt 2
2. From Equation (2.1)
m
dx
In a Galilean transformation
dy ˆ dz ˆ iˆ j k. dt dt dt dx dx dt
dy dy
v
dt
Substitution into Equation (2.1) gives
m However, because p m
dx ˆ dy ˆ dz ˆ i j k dt
dt
dz dz
dt
dt
dx
v i dt
.
dt ˆ
dt
dy j dt
ˆ
dz k p . ˆ dt
the same form is clearly retained, given
dt
dx dx the velocity transformation v. dt dt c2 v 2
3. Using the vector triangle shown, the speed of light coming toward the mirror is and the same on the return trip. Therefore the total time is t 2
distance speed
Notice that sin
v
, so sin
1
v
.
2
2
c2 v2
.
Chapter 2 2
Special Theory of Relativity
c
Chapter 2
c
Special Theory of Relativity 2
Chapter 2 3
Special Theory of Relativity
Chapter 2
1
4. As in Problem 3, sin v / v1 , so sin 1 (v / v ) 1sin 2 2
Special Theory of Relativity 3
0.350 m/s
16.3
and
1.25 m/s 2
v
2
v2 v1
2
(1.25 m/s) (0.35 m/s)
2
1.20 m/s .
5. When the apparatus is rotated by 90°, the situation is equivalent, except that we have effectively interchanged
1
and
2
. Interchanging
1
and
in Equation (2.3) leads to
2
Equation (2.4). 6. Let n = the number of fringes shifted; then n
v2
ct t n
1
n
. Solving for v and noting that
3.0 0 108 m/s
1
. Because d c t t , we have
2
c2
v c
d
0.005 589 10
m
1
+
2
= 22 m,
3.47 km/s.
2 9
7. Letting
1
1
2
1
(where
22 m v / c ) the text equation (not currently numbered) for
t1 becomes
t1
2
1
c1
1
2
2
c 1
1 2
which is identical to t 2 when
1
2
so t 0 as required.
8. Since the Lorentz transformations depend on c (and the fact that c is the same constant for all inertial frames), different values of c would necessarily lead two observers to different conclusions about the order or positions of two spacetime events, in violation of postulate 1. 9. Let an observer in K send a light signal along the + x-axis with speed c. According to the Galilean transformations, an observer in K measures the speed of the signal to be dx dx v c v . Therefore the speed of light cannot be constant under the Galilean dt dt transformations. 10. From the Principle of Relativity, we know the correct transformation must be of the form (assuming y y and z z ):
Chapter 2 4
Special Theory of Relativity
x ax bt;
Chapter 2
Special Theory of Relativity 4
x ax bt .
The spherical wave front equations (2.9a) and (2.9b) give us: ct (ac b)t ; ct (ac b )t . Solve the second wave front equation for t and substitute into the first:
Chapter 2 5
Special Theory of Relativity
(ac b)(ac b)t
ct
Chapter 2
Special Theory of Relativity 5
or c 2 (ac b)(ac b) a 2c 2 b 2 .
c Now v is the speed of the srcin of the
x -axis. We can find that speed by setting x 0
which gives 0 ax bt , or v x / t b / a , or equivalently b = av. Substituting this into the equation above for c 2 yields c 2 a 2c2 a 2v 2 a 2 c 2 v2 . Solving for a: 1
a 1 v
2
.
2
/c
This expression, along with b = av, can be substituted into the srcinal expressions for x and x to obtain:
x x vt;
x x vt
which in turn can be solved for t and t to complete the transformation. 11. When v c we find 1
x ct
x
1
t
t x/c 1
2
2
1, so:
x ct x vt ;
t x/c t ;
2
x x ct x ct x vt ; 1 2 t
tx / c
t x/ c t .
1
2
12. (a) First we convert to SI units: 95 km/h = 26.39 m/s, so 8
v / c (26.39 m/s) / 3.00 10 m/s 8.8108 (b)
v / c (240 m/s) / 3.00108 m/s 8.0 107
(c) v 2.3 vsound 2.3330 m/s so
v / c 2.3 330 m/s / 3.00108 m/s 2.5106 (d) Converting to SI units, 27,000 km/h = 7500 m/s, so
v / c (7500 m/s) / 3.00108 m/s 2.5105 8
(e) (25 cm)/(2 ns) = 1.2510 m/s so 14
22
8
v / c (1.25108 m/s) / 3.00108 m/s 0.42
(f) 110 m / 0.3510 s 2.857 10 m/s
, so
Chapter 2 6
Special Theory of Relativity
8
Chapter 2
8
v / c (2.857 10 m/s) / 3.0010 m/s 0.95
Special Theory of Relativity 6
Chapter 2 7
Special Theory of Relativity
Chapter 2
13. From the Lorentz transformations t
Special Theory of Relativity 7
t vx / c 2 . But t 0 in this case, so
solving for v we find v c2 t / x . Inserting the values t t
2
a / 2c and
t1
2
c a / 2c x x2 x1 a , we find v
c / 2 . We conclude that the frame K travels a at a speed c/2 in the x -direction. Note that there is no motion in the transverse direction. 14. Try setting x 0 x vt . Thus 0 x vt a va / 2c . Solving for v we find
v 2c , which is impossible. There is no such frame K . 15. For the smaller values of β we use the binomial expansion 1
(b)
1
(c)
1
(d) (e) (f)
/ 2 1 3.2 1013 / 2 1 3.11012
2
/ 2 1 3.11010
2
1
2
1
2
1/ 2
/ 2 1 3.8710 2
1
2
15
2
(a)
1
1/ 2
1/ 2
1/ 2
2
1 0.42
1.10
1/ 2
2
1 0.95
3.20
16. There is no motion in the transverse direction, so y z 3.5 m. 1
1 2
1
x x vt
5
5/3 2
1 0.8
2m 0.8c 0 10 / 3 m 3
t t vx / c 2
5
2
9
0 0.8c 2 m / c 8.9 10 s 3 2
3 m (5 m)2 (10 m)2
x2 y 2 z 2
8
17. (a) t
3.86 10 s 8
c 0.8 we find
(b) With
x x vt
5
8
3.00 10 m/s 5 / 3 . Then y y 5 m, -8
3 m 2.4010 m/s (3.8610 s) 10.4 m 3
z z 10 m,
1
2
/ 2.
Chapter 2 8
Special Theory of Relativity
t t vx / c 2
5
Chapter 2
2
8 8 8 3.8610 s 2.4010 m/s 3 m / 3.0010 m/s 51.0 ns 3
Special Theory of Relativity 8
Chapter 2 9
Special Theory of Relativity
(c)
Chapter 2
2
x2 y2 z 2 t
2
10.4 m 5 m 10 m
2
2.994 108 m/s which equals c
9
51.0 10 s
to within rounding errors
Special Theory of Relativity 9
.
18. At the point of reflection the light has traveled a distance L vt1 ct1 . On the return
trip it travels L vt2 ct2 . Then the total time is t t1 t2 2
2
2
2 Lc . 2 c v
2L / c 1v /c
2L / c
2 L0 / c
But from time dilation we know (with t proper time 2L0 / c ) that
t t
2 L0 / c
. Comparing these two results for t
we get
1 v2 / c 2
2
1 1 v2 / c 2
which reduces to L L0
L0
2
v
2
1v /c
c2
. This is Equation (2.21).
19. (a) With a contraction of 1%, L / L0 0.99
2
2
1 v / c . Thus 1
2
(0.99)2 0.9801.
0.14 or v 0.14c.
Solving for , we find
(b) The time for the trip in the Earth-based frame is 6 d 5.00 10 m 1 t 1.19 10 s . With the relativistic factor 1.01 v 0.14 3.00 108 m/s (corresponding to a 1% shortening of the ship’s length), the elapsed time on the rocket ship is 1% less than the Earth-based time, or a difference of 1 3 0.011.210 s = 1.2 10 s. 20. The round-trip distance is d = 40 ly. Assume the same constant speed v c for the entire round trip. In the rocket’s reference frame the distance is only d d
in the rocket’s frame of reference v
distance time
Rearranging
v
1
2
40ly 12
d 40 y
. Solving for we find
c
1 2.
Then
1 2.
40 y 0.5 , or v
0.5 c 0.71c .
c To find the elapsed time t on Earth, we know t 40 y, so t t
1
1 40y
2
Chapter 2 10
Special Theory of Relativity
Chapter 2
Special Theory of Relativity 10
56.6 y.
21. In the muon’s frame T0 2.2 μs. In the lab frame the time is longer; see Equation (2.19): T T0 . In the lab the distance traveled is 9.5cm vT vT0 c T0 , since v c .
Chapter 2 11
Special Theory of Relativity
Chapter 2
9.5 cm 1 2 Therefore
9.5 cm 1
v , so
cT0
c
2
. Now all quantities are
c 2.2μs
4 known except β. Solving for β we find 1.410
Special Theory of Relativity 11
or v 1.4 104 c .
22. Converting the speed to m/s we find 25,000 mi/h = 11,176 m/s. From tables the distance is 3.84108 m . In the earth’s frame of reference the time is the distance divided by
t t/ t
3.84 108 m 34, 359 s . In the astronauts’ frame the time elapsed is 11,176 m/s
d v
speed, or t
1
2
. The time difference is t t t t t
1
2
t1
1
2
.
2
Evaluating numerically t 34, 359 s 1
23. T T0 , so we know that 24. L L0 / so clearly
v
3c 2
1
5
1
5/3
2.4 10 s.
11,176 m/s 3.00 108 m/s
. Solving for v we find v 4c / 5.
1 v2 / c 2
1
2 in this case. Thus 2
2
2
and solving for v we find
1v /c . 2
2
25. The clocks’ rates differ by a factor of 1 / 1 v / c . Because the binomial theorem approximation 1
t t t t t t 1 . Using 1
2
2
is very small we will use
/ 2 . Then the time difference is
/ 2 and the fact that the time for the trip
equals distance divided by speed, 375 m/s 2
t t /2
6 8 10 m 3.00 10 m/s
375 m/s 8
t 1.67 10 s 26. (a) L L / L
2
8
2
16.7 ns. 1 v 2 / c 2 3.58104 km
1 0.942 1.22104 km
(b) Earth’s frame:
Chapter 2 12
Special Theory of Relativity
Chapter 2 7
3 . 5
t L/v
8 10 m 0.94 3.00 108 m/s
Golf ball’s frame: t t / 0.127s 1 0.942 0.0433 s
0.127 s
Special Theory of Relativity 12
Chapter 2 13
Special Theory of Relativity
Chapter 2
Special Theory of Relativity 13
27. Spacetime invariant (see Section 2.9): c2 t 2 x2 c 2 t 2 x2 . We know x 4 km,
t 0 , and x 5 km. Thus t 2
2
2 5000 m 4000 m2
xx
c
2
2
1.0 10 3.00 108 m/s
10
s
2
2
and t 1.0 105 s. 28. (a) Converting v = 120 km/h = 33.3 m/s. Now with c = 100 m/s, we have 1
v / c 0.333 and
1 2
1
1.061. We conclude that the moving 1 0.3332
person ages 6.1% slower. (b) L L / (1 m) /(1.061) 0.942 m. 29. Converting v = 300 km/h = 83.3 m/s. Now with c =100 m/s, we have
and
1
1 1
2
1.81. So the length is
v / c 0.833
L L0 / 40 / 1.81 22.1 m.
2
1 0.833
30. Let subscript 1 refer to firing and subscript 2 to striking the target. Therefore we can see that x1 1 m, x2 121m, and t1 3 ns.
t2 t1
distance speed
3 ns +
120 m
3 ns + 408 ns = 411 ns. 0.98c
To find the four primed quantities we can use the Lorentz transformations with the known values of x1 , x2 , t1 , and t 2 . Note that with v 0.8c , /c
t t vx 1
1
2
2
/c
2
5/3.
0.56 ns
1 2
t t vx 2
2
1v
/ c 147 ns 2
x1 x1 vt1 0.47 m x2 x2 vt2 37.3 m 31. Start from the formula for velocity addition, Equation (2.23a):
ux
ux v 2
1 vu x/ c (a) u
0.62c 0.84c
1.46c
0.96 c
.
Chapter 2 14
Special Theory of Relativity x
1 (0.62c)(0.84c) / c 0.62c 0.84c
(b) u x
Chapter 2 2
0.22c
1 (0.62c)(0.84c) / c
1.52
0.46 c
2
0.48
ux v
32. Velocity addition, Equation (2.24): u x
ux
0.8c (0.8c) 1 (0.8c)(0.8c) / c
1.6c 2
Special Theory of Relativity 14
0.976 c 1.64
1 vux / c
with v 0.8 c and u 2
0.8 c. x
Chapter 2 15
Special Theory of Relativity
Chapter 2
Special Theory of Relativity 15
33. Conversion: 110 km/h = 30.556 m/s and 140 km/h = 38.889 m/s. Let ux 30.556 m/s and v 38.889 m/s. Our premise is that c 100 m/s. Then by velocity addition, 30.556 m/s 38.889 m/s
u v x
ux
1 vu
x
/ c2
1 38.889 m/s 30.556 m/s / 100 m/s
62.1 m/s. By symmetry
2
each observer sees the other one traveling at the same speed.
34. From Example 2.5 we have u
u
c 1 nv / c . For light traveling in opposite directions n 1 v / nc
c 1 nv / c 1 nv / c . Because v / c is very small, use the binomial expansion: 1 v / nc n 1 v / nc
1 nv / c 1 v / nc
1
1 nv / c 1 v / nc 1 nv / c 1 v / nc 1 nv / c v / nc , where we
have dropped terms of order v 2 / c 2 . Similarly
u
c
1 nv / c v / nc 1 nv / c v / nc
n
1 nv / c
1 nv / c v / nc. Thus 1 v / nc 2v 2 11 / n 2v 11/ n . Evaluating
n
numerically we find u 2(5 m/s) 1
1
4.35 m/s. 1.332 35. Clearly the speed of B is just 0.60 c . To find the speed of C use ux 0.60 c and
v 0.60 c :
ux v
u x
0.60c (0.60c)
1 vux / c
2
0.88 c.
1 (0.60c)(0.60c) / c
2
36. We can ignore the 400 km, which is small compared with the Earth-to-moon distance 3.84 108 m. The rotation rate is 2
rad 100 s1 2 102
rad/s. Then the speed
across the moon’s surface is v R 2 102 rad/s3.84108 m 2.411011 m/s. 37. Classical: t
4205 m
5
1.4310 s . Then
ln 2 t exp
14.6 5
0.98c or about 15 muons. 5
Relativistic: t t /
N N
1.4310 s 2.86 106 s
Chapter 2 16
0
Special Theory of Relativity
t1/ 2
Chapter 2
Special Theory of Relativity 16
so ln 2 t N N 0 exp t1/ 2
2710 muons. Because of the exponential nature of the decay
curve, a factor of five (shorter) in time results in many more muons surviving. 38. The circumference of the fixed point’s rotational path is 2 R cos(39 ) , where E
RE
Earth’s radius = 6378 km. Thus the circumference of the path is 31,143 km. The
Chapter 2 17
Special Theory of Relativity
Chapter 2
Special Theory of Relativity 17
rotational speed of that point is v 31,143 km / 24 h 1298 km/h 360.5 m/s . The observatory clock runs slow by a factor of
1 1
2
1
2
/ 2 1 7.22 1013 . In
41.2 h the observatory clock is slow by 41.2 h 7.221013 2.9746 1011 h = 107 ns. In 48.6 h it is slow by 48.6 h 7.221013 3.5089 1011 h = 126 ns. The Eastward- moving clock has a ground speed of 31,143 km/41.2 h = 755.9 km/h = 210.0 m/s and thus has a net speed of 210.0 m/s + 360.5 m/s = 570.5 m/s. For this clock 1 1
1
2
/ 2 1 1.811012 and in 41.2 hours it runs slow by
2
12
41.2 h 1.8110
11
7.4572 10
h = 268 ns. The Westward-moving clock has a
ground speed of 31,143 km/48.6 h = 640.8 km/h = 178.0 m/s and thus has a net speed of 360.5 m/s − 178.0 m/s = 182.5 m/s. For this clock 1 and in 48.6 hours it runs slow by 1 2 / 2 1 1.85 1013 1 2 13
48.6 h 1.8510
12
8.99110
h = 32 ns. So our prediction is that the Eastward-
moving clock is off by 107 ns – 269 ns 162 ns, while the Westward-moving clock is off by 126 ns − 32 ns = 94 ns. These results are correct for special relativity but do not reconcile with those in the table in the text, because general relativistic effects are of the same order of magnitude. 39. The derivations of Equations (2.31) and (2.32) in the beginning of Section 2.10 will suffice. Mary receives signals at a rate f for t1and a rate f for t 2. Frank receives signals at a rate f for t1 and a rate f for t 2 .
L L L L v c v c f T 2 f L / v signals. 40. 1T t2 t
2L
T tt 1
41. s
2
s2
2L ; Frank sends signals at rate f, so Mary receives v
; Mary sends signals at rate f , so Frank receives
2
v
x 2 y 2 z 2 c 2t 2 ; Using the Lorentz transformation 2
2
2
2
22
2 2
( x vt ) y z c (t vx / c ) x 1 v 2 / c 2 y2 z 2 c 2t 2 2 1 v 2 / c 2 22
fT 2 f L / v
signals.
Chapter 2 18
x2 y2 z 2 c2t 2 s2
Special Theory of Relativity
Chapter 2
Special Theory of Relativity 18
Chapter 2 19
Special Theory of Relativity
Chapter 2
Special Theory of Relativity 19
42. For a timelike interval s 2 0 so x 2 c 2t 2 . We will prove by contradiction. Suppose that there is a frame K is which the two events were simultaneous, so that t 0 . Then by the spacetime invariant x2 c 2t 2 x2 c 2 t 2 x2 . But because x 2 c 2t 2 , this implies x2 0 which is impossible because x is real. 43. As in Problem 42, we know that for a spacelike interval s
2
a frame K in which the two events occur in the same place,
0 so x
2
2 2 c t . Then in
x 0 and
x2 c2t 2 x2 c 2 t 2 c 2 t 2 . But because x 2 c2 t 2 we have c2 t 2 0 , which is impossible because t is real. 44. In order for two events to be simultaneous in K , the two events must lie along the x axis, or along a line parallel to the x axis. The slope of the x axis is v / c , so
ct . Solving for v, we find v c 2 t / x . Since the slope of the x axis x 2 2 2 2 so s x c t 0 is required. must be less than one, we see that x ct v / c slope =
45. parts (a) and (b) To find the equation of the line use the Lorentz transformation. With t 0 we have t 0 t vx / c 2 or, rearranging,
ct vx / c c . Thus the graph of ct vs. x is a straight line with a slope . (c) Now with t constant, the Lorentz transformation 2 gives t t vx / c . Again we solve for ct :
ct x ct / x constant . This line is parallel to the t 0 line we found earlier but shifted by the constant. (d) Here both the x and ct axes are shifted
ned, copied or duplicated, or posted to a or in part.
Chapter 2 20
Special Theory of Relativity
from their normal (x, ct) orientation and they are not perpendicular.
© 2013 Cengage Learning. All Rights Reserved. May not be scan
Chapter 2
Special Theory of Relativity 20
Chapter 2 21
Special Theory of Relativity
Chapter 2
Special Theory of Relativity 21
46. The diagram is shown here. Note that there is only one worldline for light, and it bisects both the x, ct axes and the x, ct axes. The x and ct axes are not perpendicular. This can be seen as a result of the Lorentz transformations, since x 0 defines the ct axis and t 0 defines the x axis.
47. The diagram shows that the events A and B that occur at the same time in K occur at different times in K .
48. The Doppler shift gives
1 0
1
. With numerical values
540 nm, solving this equation for gives
0
650 nm and
0.183 . The astronaut’s speed is
v c 5.50107 m/s. In addition to a red light violation, the astronaut gets a speeding ticket. 49. According to the fixed source (K) the signal and receiver move at speeds c and v, respectively, in opposite directions, so their relative speed is c v . The time interval t between receipt of signals is t / (c v) 1 / f . By time dilation t . 0 (c v ) 1
c 1 v2 / c2
1
2
Chapter 2 22
Special Theory of Relativity
Using c / v0
f
and
f 0 (1 )
1
t
1
2
1v /c 1
f 2
Chapter 2
0
1
.
2
we find t
f 0 (c v )
Special Theory of Relativity 22
f 0 (1 )
and
Chapter 2 23
Special Theory of Relativity
Chapter 2
Special Theory of Relativity 23
50. For a fixed source and moving receiver, the length of the wave train is cT vT. Since n cn cT vT . waves are emitted during time T, and the frequency f c / is f cT vT n
cf T / 0
As in the text n f 0T0 and T0 T / . Therefore f 1
51. f f 0
1400 kHz
cT
0
2
1
f
.
vT
1
1
224 kHz
1 0.95
52. The Doppler shift function f f 0
1
0
1 0.95
1
f
1 1
is the rate at which #1 and #2 receive signals from each other and the rate at which #2 and #3 receive signals from each other. But for signals between #1 and #3 the rate is
f
f
1 1
f
1 . 1
0
53. The Doppler shift function f f 0
1 1
is the rate at which #1 and #2 receive signals from each other and the rate at which #2 and #3 receive signals from each other. As for #1 and #3 we will assume that these plumbing vans are non-relativistic v c . Otherwise it would be necessary to use the velocity addition law and apply the transverse Doppler shift. From the figure we see 1 f 0 1/ t0 and that f . Now t0 (t 2 t1 )
t2 t1
f
2 x 2vt0 cos
c
c
. With an angle of 45 , cos(45) 1/ 2
1 1/ f 0 (2v cos ) / cf0
f0 1 2v cos / c
and
f0 . 1 2v / c 1
54. The Doppler shift to higher wavelengths is (with
0
589 nm) 700 nm
0
1
.
Chapter 2 24
Special Theory of Relativity
Solving for
we find
0.171. Then t
Chapter 2
v
Special Theory of Relativity 24
0.171 3.00 108 m/s 2
6
1.75 10 s
a 29.4 m/s which is 20.25 days. One problem with this analysis is that we have only computed the
Chapter 2 25
Special Theory of Relativity
Chapter 2
Special Theory of Relativity 25
time as measured by Earth. We are not prepared to handle the non-inertial frame of the spaceship. 55. Let the instantaneous momentum be in the x-direction and the force be in the y-direction. v m . Then d Fdt md v and dv is also in the y-direction. So we haveF m d dt
v2 56. The magnitude of the centripetal force is ma m v2 particle F qvB , so qvB m
r
r
r for circular motion. For a charged
or, rearranging qBr mv p . Therefore
p . qB
When the speed increases the momentum increases, and thus for a given value of B the radius must increase. d mv mv and F 57. . The momentum is the product of two factors that 2 2 dt 1v /c contain the velocity, so we apply the product rule for derivatives: mv d
F m
dt 1 v 2 / c 2 d v / dt
m
v
2
1v /c
m a mv
2
a 3ma
2
v
c2 v2
2 3
a 1
m
v 2
c a
1
dt
1v /c
2v
1 2
m
d
2
c
c2
2
3
dv dt
2
Chapter 2 26
Special Theory of Relativity
Chapter 2
Special Theory of Relativity 26
3
58. From the preceding problem F 1 (a)
3
ma . We have a 1019 m/s2 and m 1.67 1027
1 2
1v /c
2
1.00005 2 1 0.01
3
F 1.00005 1.67 1027 kg 1019 m/s2 1.67 108 N (b) As in (a) 1.005 and
8 F 1.70 10 N
kg.
a
Chapter 2 27
Special Theory of Relativity
Chapter 2
F 2.02 107 N
(c) As in (a)
2.294 and
(d) As in (a)
7.0888 and
6 F 5.9510 N
1
59. p mv with
Special Theory of Relativity 27
1 2
1v /c
2.5516 ; 2 1 0.92
2
16
m
p v
10
kg·m / s
1.42 1025 kg
2.5516 0.92 3.00 108 m/s
1
60. The initial momentum is p0 mv
1 0.5
m 0.5c 0.57735mc . 2
(a) p / p0 1.01 1.01
mv 0.57735mc
v 1.01 0.57735c 0.58312c 1/ 2
1 1 Substituting for and solving for v, v 2 2 c .58312c
0.504c.
1/ 2
1
1
(b) Similarly v .63509c
2
0.536c
c2 1/ 2
(c) Similarly v
1
1
1.1547 c
2
0.756c
c2 3
61. 6.3 GeV protons have K 6.310 MeV and
p
E K E0
7238 MeV. Then
E 2 E0 2 7177 MeV/c . Converting to SI units c 13
1.60 10
c
J
3.8310 18 kg·m/s
p 7177 MeV/c MeV
3.00 108 m/s
p
3.8310
18
kg m/s
Chapter 2 28
Special Theory of Relativity
Chapter 2
From Problem 56 we have B
Special Theory of Relativity 28
1.57 T .
qr
1.60 10
19
C15.2 m
62. Initially Mary throws her ball with velocity (primes showing the measurements are in the elastic collision, the signs on the above Mary’s frame): u x 0 u y u . After 0 expressions are reversed, so the change in momentum as measured by Mary is mu0 mu0 2mu0 . pM 1 u2 / c2 0
1 u2 / c2 0
1 u 2 / c2 0
Chapter 2 29
Special Theory of Relativity
Chapter 2
Now for Frank’s ball, we know u Fx 0 and u for Frank’s ball as measured by Mary: u x To find for Frank’s ball, note that uF
2 x
1
u0 . The velocity transformations give
Fy
Fy
2
2
1 v /c .
v 2 u 2 1 v02 / c 2 . Then
1
1
1 v 2 / c 2 u 2 1 0v 2 / c 2 / c 2
1 2uF/ 2c
2
u y u0
v u
Special Theory of Relativity 29
. Using
1 u 2 /0c 2 1 v 2 / c 2
p mu along with the reversal of velocities in an elastic collision, we find 1 v 2 / c 2 mu
p mu F
1 v 2 / c 2 2 mu
0
0 2
2mu
1 v /c
2
0 2
2
1 u /0c 1 v / c Finally p p
2
2mu
0 2
2
1 v /c 0
F
p
2
M
2
1 u /c
2
2mu0 2mu0
0
0 as required.
1 u 20 / c 2 63. To prove by contradiction, suppose that K
1
2
mv . Then 2
mc2 mc2 1 mc2 1 mv2 . This implies 1 v 2 / 2c2 , or
K E E0
2
1 v 2 / 2c2 , which is clearly false. 64. The source of the energy is the internal energy associated with the change of state, commonly called that latent heat of fusion L f . Let m be the mass equivalent of 2 grams
and M be the mass of ice required.
M
mc L
2
8
0.002 kg 3.00 10 m/s 334 103 J/kg
65. In general K 1 mc2 , so 1
K
m
E c2
Lf M c2
Rearranging
2
5.39 108 kg
. For 9 GeV electrons: mc 2
Chapter 2 30
Special Theory of Relativity
1
9000 MeV
1.76 10 0.511 MeV
1
1 2
9
1
Chapter 2
4
Special Theory of Relativity 30
Then from the definition of we have 1
1.76 104
9
2
v 11.6 10 c 0.9999999984c .
1 1.6 10 . Thus
Chapter 2 31
Special Theory of Relativity
For 3.1 GeV positrons: 1
Chapter 2
3100 MeV 6068 and 0.511 MeV
Special Theory of Relativity 31
1
1
6068
2
8 1 1.4 10 .
Thus v 11.4 108 c 0.999999986c .
66. Note that the proton’s mass is 938 MeV/c 2. In general K 1 mc , so 1 2
K
1
Then from the definition of we have
MeV, and 1
1
2
0.750 MeV 1.00080 with 938 MeV
1
1
2
mc 2
K 0.750
. For the first section
1
.
1 1.00080
2
0.040 .
Thus v = 0.04c at the end of the first stage. For the other stages the computations are similar, and we tabulate the results:
K (GeV) 0.400
1.43
0.71
8
9.53
0.994
150
160.9
0.99998
1000
1067
0.9999996 2
67. (a) p mu
511 keV/c 0.020c 10.22 keV/c ; 2 1 0.020 2
2
511 keV/c (c ) 511.102 keV; 2 1 0.02 K E E0 511.102 keV 511.00 keV 102eV
E mc2
The results for (b) and (c) follow with similar computations and are tabulated:
p (keV/c )
E (keV)
K (keV)
0.20
104.3
521.5
10.5
0.90
1055
1172
661
68. E 2 E0 E0 so
2 . Then
1
1
3
2
23 c
Chapter 2 32
and v .
Special Theory of Relativity
2
Chapter 2
Special Theory of Relativity 32
Chapter 2 33
Special Theory of Relativity
Chapter 2
Special Theory of Relativity 33
69. For a constant force, work = change in kinetic energy = Fd mc2 / 4, because
25% 1 / 4 .
8
80 kg 3.00 10 m/s
2
mc
d
4F
1
17
2.25 10 m = 23.8 ly.
48 N
70. E K E0 2E0 E0 3E0 E0 so
3 . Then
2 2 1 1 32 3 0.943 . Thus v = 0.943c.
2
1
2
71. (a) E K E0 0.1E0 E0 1.1E0 E0 , so 1.1 . Then
1
1
(b) As in (a)
1
1
2
2
1.1
0.417 and v = 0.417c.
2 and v
3 c / 2.
(c) As in (a) 11 and v = 0.996c. 72. K E E
E0 1
E0
so K E
E
0
0
2
. Rearranging 1
0
1
2
2
E0
which
K E
2
0
2
2
2
can be written as
E0 or 1 K E0
1
73. Using E mc2 along with p mv we see that find
0E . K E0
E / mc2 p / mv . Solving for v/c we
v / c pc / E .
74. The speed is the same for protons, electrons, or any particle.
K 1 mc2 1.01
1
mv2 0.505mc2
2
2 Rearranging and solving for , we find
so 1
1
1 0.505 2 . 12
0.114 or v = 0.114c.
Chapter 2 34
Special Theory of Relativity
Chapter 2
Special Theory of Relativity 34
75. Converting 0.1 ounce = 2.835103 kg. 2
E mc2 2.835103 kg 3.00108 m/s 2.551014 J . Eating 10 ounces results in a factor of 100 greater mass-energy increase, or 2.551016 J. This is a small increase compared with your srcinal mass-energy, but it will tend to increase your weight; depending on how they are prepared, peanuts generally contain about 100 kcal of food energy per ounce.
Chapter 2 35
Special Theory of Relativity
Chapter 2
Special Theory of Relativity 35
76. The energy needed equals the kinetic energy of the spaceship. 1 2 1 mc 1 2
K 1 mc 2 1
2
2
1 10 4 kg 3.00 108 m/s 4.35 1019 J
1 0.3
21
or 4.35% of 10 J. 77. Up to Equation (2.57) the derivation in the text is complete. Then using the integration by parts formula, xdy xy ydx and noting that in this case x u and y u , we 2 have ud ( u ) u udu . Thus
u
K m u d ( u ) mu 2 m u du 0
u du 2 2 1u /c
mu 2 m
Using integral tables or simple substitution: u
K
mu
2
mc 2
1u /c
2
2
mu
2
mc 2
1 u / c mc
2
2
0
mc 2 mc 2 1 u 2 / c 2 2
1u /c
2
2
mc 2
mc 2 mc 2 mc 2 ( 1) 78. Converting 0.11 cal ·g 1 ∙ C
1
= 460 J ·kg 1 ∙ C
1
c
V
(specific heat at constant volume). From thermodynamics the energy E used to change the temperature by T is mcV T . Thus
E 1000 kg 460 J/(kg C 0.5C 2.30105 J m
and
5
E
2.30 10 J 2
c
8
2.56 10 2
3.00 10 m/s
12
kg . The source of this energy is the internal
Chapter 2 36
Special Theory of Relativity
Chapter 2
Special Theory of Relativity 36
energy of the arrangement of atoms and molecules prior to the collision.
Chapter 2 37
Special Theory of Relativity
Chapter 2
Special Theory of Relativity 37
79.
Eb
2m p 2mn
m He c 2
4.001505 u c
2 1.007276 u 2(1.008665 u)
931.494 MeV
2
28.3 MeV
c 2·u 80.
E
mn m p me c
1.008665 u
2
0.000549 u c
1.007276 u
2
931.494 MeV
0.782 MeV
c 2·u 81.
E K E0 1 TeV 938 MeV 1 TeV ; E2 E2
1 TeV
2
938MeV 938 MeV
2
0
p
1.000938 TeV/c c
c E E0 1.000938 TeV E0
2
1067 ;
1
2 1 8.7810
0.000938 TeV 7
1 8.7810 1 4.39 107 ; and
p 2c2 E 2 0
82. (a) E
1
v c 0.999999561c 2
40GeV 511 keV
2
40.0 GeV
K E E0 40.0 GeV p 2c2 E 2 0
(b) E
2
40 GeV 0.938 GeV
K E E0 40.011 GeV 0.938 GeV
2
40.011GeV
39.07 GeV
83. E K E0 200 MeV 106 MeV 306 MeV
E2 E2
2
2
306 MeV 106 MeV
0
p E 306 MeV
287.05 MeV/c c c 2.887
7
Chapter 2 38
Special Theory of Relativity
E0 1
106 MeV 1 2
0.938
so v 0.938c
Chapter 2
Special Theory of Relativity 38
Chapter 2 39
Special Theory of Relativity
Chapter 2
Special Theory of Relativity 39
3
84. (a) The mass-energy imbalance occurs because the helium-3 ( He) nucleus is more 2
tightly bound than the two separate deuterium nuclei ( H) . (Masses from Appendix 8.)
E [2 m( 2 H)] [ m
n
m(3 He)]c 2
2(2.014102 u) (1.008665 u 3.016029 u) c
2
931.494 MeV 3.27 MeV c2 u
2
(b) The initial rest energy is 2m( H) = 2(2.014102 u)
931.494 MeV 2
c u Thus the answer in (a) is about 0.09% of the initial rest energy.
= 3752 MeV.
85. (a) The mass-energy imbalance occurs because the helium-4 (4 He) is more tightly bound than the deuterium ( 2 H) and tritium nuclei (3 H) .
E [ m( 2 H)+m(3 H)] [ m
n
m( 4 He)] c 2
(2.014102 u 3.016029 u) (1.008665 u 4.002603 u) c 2
931.494 MeV
c2 u
17.6 MeV (b) The initial rest energy is m( 2 H)+ m(3 H) c 2 (5.030131 u) c 2
931.494 MeV
=
2
c u 4686 MeV. Thus the answer in (a) is about 0.37% of the initial rest energy.
86. (a) In the inertial frame moving with the negative charges in wire 1, the negative charges in wire 2 are stationary, but the positive charges are moving. The density of the positive charges in wire 2 is thus greater than the density of negative charges, and there is a net attraction between the wires. (b) By the same reasoning as in (a), note that the positive charges in wire 2 will be stationary and have a normal density, but the negative charges are moving and have an increased density, causing a net attraction between the wires. (c) There are two facts to be considered. First, (a) and (b) are consistent with the physical result being independent of inertial frame. Second, we know from classical physics that
Chapter 2 40
Special Theory of Relativity
Chapter 2
Special Theory of Relativity 40
two parallel wires carrying current in the same direction attract each other. That is, the same result is achieved in the “lab” frame. As in the solution to Problem 21 we have
v d
1
2
where d is the length of the
c ct 87. particle track and t the particle’s lifetime in its rest frame. In this problem t 8.2 1011 s
Chapter 2 41
Special Theory of Relativity
Chapter 2
and d = 24 mm. Solving the above equation we find
E0
E
1672 MeV 1
2
Special Theory of Relativity 41
0.698 . Then
2330 MeV 2
1 0.698
88.
dp
d
dt
dt
F
d mv
m
mv 1
2
1/ 2
2
c
dt
dv mv 3 2v dv 2 dt c 2 dt dv
m
v
1
dv
2 2
dv
m c2
dt
m
v
1
dt 1 v 2 / c 2
2
dv
m
2
1 v2 / c2
dt
m3
v /c
1
dv dt
1
dt 1 v 2 / c 2
3/ 2
89. (a) The number n received by Frank at f is half the number sent by Mary at that rate, or
fL / v . The detected time of turnaround is
n t
L1
L 1
fL / v f 1 /1
L v
v 1
L .
v
c
(b) Similarly, the number n received by Mary at f is
n f
T
f
1
L
f L1
.
Her turnaround time is T / 2 L / v .
2 1 v v (c) For Frank, the time t2 for the remainder of the trip is t2 = T − t1 = L/v − L/c. 1 fL Number of signals = f t2 f . L/v L/c
Total number received =
Chapter 2 42
1
Special Theory of Relativity
Chapter 2
v
fL fL 2 fL . v v v Total number received 2L Mary’s age = . f v (d) For Mary, t2 T t1 L / v . Number of signals = f t2 f
1
L 1 v
fL
Total number received =
v
1 1
fL v 2 fL
1 .
.
v
Special Theory of Relativity 42
Chapter 2 43
Special Theory of Relativity
Frank’s age =
Chapter 2
Special Theory of Relativity 43
Total number received 2L . f v
90. In the fixed frame, the distance is Δx = 8 ly and the elapsed time is Δt = 10 y, so the 2
2
2 2
interval is s x c t
2
64 ly2
100 ly =
2
36 ly . In the moving frame, Mary’s
clock is at rest, so x 0 , and the time interval is t 6 y. Thus the interval is
s2 x2 c2 t 2 0 ly2
36 ly2 =
36 ly2 . The results are the same, as they should
be, because the spacetime interval is the same for all inertial frames. 91. (a) From Table 2.1 number
fL
1
1 0.8 55.9 56 52y 1
v L
L
4.3 ly
(b) t1
c
0.8c
0.8c
4.3 ly 9.68 y ;
v
4.3 ly
fL number
f t1
c
2
1
167.7 168
v
f L
1 2 168 so the total is 168 + 168 = 336. v Mary: number 2 f L / v 559
(c) Frank: f t
2
(d) Frank: T 2L / v 10.75 y Mary: T 2L / v 6.45 y (e) From part (c), 559 weeks = 10.75 years and 336 weeks = 6.46 years, which agrees with part (d). 92. Notice that the radar is shifted twice, once upon receipt by the speeding car and again upon reemission (reflection) of the beam. For this double shift, the received frequency f is 1 1 1 f 1 1 f0 1
For speeds much less than c, a Taylor series approximation gives an excellent result: 1 f 1 1 1 1 2. f0 1 −7
Converting 80 mph to 35.8 m/s gives β = 1.19 × 10 , so the received frequency is approximately
Chapter 2 44
Special Theory of Relativity
Chapter 2
Special Theory of Relativity 44
f f 0 1 2 10.4999975 GHz , which is 2.5 kHz less than the srcinal frequency. 93. For this transformation v = 0.8c (so γ = 5/3), ux 0 and uy 0.8c. Applying the transformations,
Chapter 2 45
Special Theory of Relativity
u x v 0 0.8c
u x
1
vux c2
u
0.8c
y
u
0.8c ; 10
The speed is u
1
vux c2
Special Theory of Relativity 45
0.48c
5 10 3
u x2 u 2 y 0.93c , safely under the speed of light.
94. (a) K E E0 250E0 . Thus (b)
Chapter 2
K 249E0 249511keV 127.2 MeV
250
1
1
0.999992 2
v 0.999992 c E2 E2
2
250 511 keV 511 keV
0
(c) p
128 MeV/c c
95. (a) For the proton: p mu
For the electron: E
2
1
c
938 MeV / c 2 0.9c 1940 MeV / c . 2 1 0.9
p 2c2 E 2 0
2
2
1940 MeV 0.511 MeV 1940 MeV .
E E0 1940 MeV 3797 0.511 MeV 1
1 2
1 1 2 3797
8
1 6.94 10 1 3.97 10
8
v 1 3.97108 c K 1 E0
(b) For the proton:
For the electron:
1 1 0.9
2
1 938 MeV 1214 MeV .
K E0 1214 MeV 0.511 MeV
2377 .
2
1
1
E0 1
1
Chapter 2 46
Special Theory of Relativity
2377
2
Chapter 2
Special Theory of Relativity 46
0 . 5 1 1 M e V 1 1 . 7 7 1 0 7
1 8 . 8 5 1 0 8
v 1 8.85108 c 96. In the frame of the decaying K 0 meson, the pi mesons must recoil with equal speeds in opposite directions in order to conserve momentum. In that reference frame the available kinetic energy is 498 MeV 2 140 MeV 218 MeV . The pi mesons share this equally,
Chapter 2 47
Special Theory of Relativity
Chapter 2
Special Theory of Relativity 47
so each one has a kinetic energy of 109 MeV in that frame. The speed of each pi meson can be found:
K E0 109 MeV 140 MeV
1.779 so u 140 MeV
E0
1
1
c 0.827c .
2
The greatest and least speeds in the lab frame are obtained when the pi mesons are released in the forward and backward directions. Then by the velocity addition laws: 0.9c 0.827c
vmax
1 0.9 0.827 0.9c 0.827c
vmin
1 0.9 0.827
0.990c 0.285c
97. (a) The round-trip distance is L0 8.60 ly . Assume the same constant speed v c for the entire trip. In the rocket’s frame, the distance is only L L
1 0
L
0
1
2
.
Mary will age in the rocket’s reference frame a total of 22 y, and in that frame
v
L 22y
distance time
8.60 ly 1 2 0.39c 22y
1
2
. Therefore,
v
0.39 1 c
2
.
2
0.39 Solving for we find
2
, or
0.36 so Mary’s speed is v 0.36c .
1.00 0.39 (b) To find the elapsed time on Earth, we know that T0 22y , so 1
T T0
1
2
22y 23.6 y . Frank will be 53.6 years old when Mary returns at the
age of 52 y. 98. With v 0.995c then
0.995 and
10.01. The distance out to the star is
L0 5.98 ly . In the rocket’s reference frame, the distance is only L L
1 0
5.98 ly
0.597 ly. 10.01
(a) The time out to the star in Mary’s frame is time
distance 0.597 ly 0.6 y. The speed 0.995c
Chapter 2 48
Special Theory of Relativity
Chapter 2
Special Theory of Relativity 48
time for the return journey will be the same. When you include the 3 years she spends at the star, her total journey will take 4.2 y. (b) To find the elapsed time on Earth for the outbound journey, we know that T0 0.6y so
T T0 (10.01)0.6 y 6.006 y. The return journey will take an equal time. The 3 years the spaceship orbits the star will be equivalent for both observers. Therefore Frank will measure a total elapsed time of 15.012 y, which makes him 10.8 years older than her.
Chapter 2 49
Special Theory of Relativity
Chapter 2
Special Theory of Relativity 49
8
99. (a) The Earth to moon distance is 3.84 10 m. The rotation rate is 1
2
1
rad 0.030 s 1.88510 rad/s. Then the speed across the moon’s surface is
v R 1.885101 rad/s3.82108 m 7.24107 m/s . (b) The required speed can be found using c R 2 f R which requires the frequency 8
c
to be f
3.00 10 m/s
2R
0.124 Hz .
8
2 3.84 10 m/s 3.28106 , which is very small. We will use the binomial
100. With the data given,
approximation theorem. From Equation (2.21), we know that
L L0 . 1
1
2
1
1
2
/ 2.
(a) The percentage of length contraction would be: % change
L0 L
100%
L0
1
1
100% 1 (1
0L
L0
1
100%
L0 2
100%
/ 2)
2
100% 5.37 1010 2 (b) The clocks’ rates differ by a factor 1 / 1
2
. The clock on the SR-71 measures
the proper time and Equation 2.19 tells us that T T0
so the time difference is
2
t T T0 T0 T0 T0 ( 1) . Using 1
/ 2 and the fact that the time for
the trip in the SR-71 equals the distance divided by the speed 983 m/s 6
t t 2 /2
3.2 10 m 3.00 10 m/s 983 m/s
1.75 10 8 s
2
8
2
17.5 ns
101. As the spaceship is approaching the observer, we will make use of Equation (2.32),
f
1 1
f 0 with the prime indicating the Doppler shifted frequency (or wavelength).
Chapter 2 50
Special Theory of Relativity
Chapter 2
Special Theory of Relativity 50
This equation indicates that the Doppler shifted frequency will be larger than the frequency measured in a frame where the observer is at rest with respect to the source. Since c f , this means the Doppler shifted wavelengths will be lower. As given in the problem, we see that the difference in wavelengths for an observer at rest with respect to the source is
0
0.5974nm . We want to find a speed so that the Doppler shifted
difference is reduced to 0.55nm . We have
Chapter 2 51
Special Theory of Relativity
c c 2
ff
1
c
1
f2
2
1
f 0,1 c
Special Theory of Relativity 51
f1f 2
f1
1 1
Chapter 2
1
c 1
f 0,2 f 0,1 f 0,2
1
f
f
1
f
f 0,1
1
0,1
c
1
0,2
c
1 0,2
f 0,2
1
0,2
f 0,1
0,1
1
1 1
0
2 2
We can complete the algebra to show that
0 2 2
0.0825 so that
v 2.47107 m/s . 102. As we know that the quasars are moving away at high speeds, we make use of Equation (2.33) and the equation c f . Using a prime to indicate the Doppler shifted frequency (or wavelength), Equation (2.33) indicates that frequency is given by f
f0
1
f
1
z
so
c/ f 1
0 0
0
1 c / f0
1 1
f 0 ,or
Chapter 2 52
Special Theory of Relativity
f0 1 f
1
1
2
Therefore z 1
2
z1 1
Special Theory of Relativity 52
1 1
. We can complete the algebra to show that 1
2
z1 1
Chapter 2
2
and thus v
z 1.9 and v 0.944c
z1 1
c.
2
z1 1
for z
For the values of z given, v 0.787c for
4.9 . 0
103. (a) In the frame of the decaying K meson, the pi mesons must recoil with equal momenta in opposite directions in order to conserve momentum. In that reference frame the available kinetic energy is 498 MeV 2 135 MeV 228 MeV .
Chapter 2 53
Special Theory of Relativity
Chapter 2
Special Theory of Relativity 53
(b) The pi mesons share this equally, so each one has a kinetic energy of 114 MeV in that frame. The energy of each pi meson is E K E0 114 MeV 135 MeV 249 MeV . The momentum of each pi meson can be found: 2
E2 E2
2
249 MeV 135 MeV
0
p 209.2 MeV/c c c 104. (a) For each second, the energy used is 3.9 × 1026 J. The mass used in each second is E 3.9 10 J26 9 2 m c2 4.310 kg 3.0 108 m/s
(b) The time to use this mass is
t
msun
efficiency
2.0 10
kg
30 18
0.007 3.310 9 4.310 kg/s
m/t
1.0 1011 y.
s
That is 20 times longer than the expected lifetime of the sun. 2r 4 105. a) The planet’s orbital speed is vp 1.30 10 m/s . If the system’s center of mass T is at rest, conservation of momentum gives the star ’s speed: mp vp 1.90 1027 4
mp vp ms vs , so
vs
30 1.30 10 m/s = 12.4 m/s 1.99 10 1 (b) The redshifted wavelength is 0 550 nm + 2.3 10 5 nm 1
ms
Similarly, the blueshifted wavelength is
0
1 1
550 nm
5
2.3 10 nm
These small differences can be detected using modern spectroscopic tools. 106. To a good approximation, the shift is the same in each direction, so the redshift for the approaching light is half the difference, or 0.0045 nm. Using this in the equation 1 0 leads to a speed parameter β = 6.9 × 10−6, or v = βc = 2070 m/s. The speed 1 is equal to the circumference 2πR divided by the period, so the period is
2R
Chapter 2 54
Special Theory of Relativity
Chapter 2
8
2 6.96 10 m T
6
v
2070 m/s
2.1110 s = 24.5 days.
Special Theory of Relativity 54
Chapter 2 55
Special Theory of Relativity
Chapter 2
Special Theory of Relativity 55
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