37.1:
If O sees simultaneous flashes then O will see the A(A ) flash first since O would
believe that the A  flash must have traveled longer to reach O , and hence started first.
37.2:
a) γ 
1
1  (0.9) 2
 2.29.
t  γ  (2.29) (2.20  10 6 s)  5.05  10 6 s.
b) d  vt  (0.900) (3.00  108 m s) (5.05  10 6 s)  1.36  10 3 m  1.36 km.
37.3:
1  u 2 c 2  (1  u 2 c 2 )1 2  1 
u2

2c 2
 (t  t 0 )  (1  1  u 2 c 2 )( t ) 
(250 m s) 2 (4 hrs)  (3600)
u2

t

2c 2
2(3.00  108 m s) 2
 (t  t 0 )  5.00  10 9 s.
The clock on the plane shows the shorter elapsed time.
37.4:
γ
1
1  (0.978) 2
 4.79.
γt  (4.79) (82.4  10 6 s)  3.95  10 -4 s  0.395 ms.
37.5:
t 0
u 2  t 
a) t 
 1 2   0 
c
 t 
1 u2 c2
2
 t 
 2.6 
 u  c 1  0   c 1 

 42 
 t 
2
37.6:
γ  1.667
a) t 0 
2
t 1.20  108 m

 0.300 s.
γ
γ(0.800c)
b) (0.300 s) (0.800c)  7.20  10 7 m.
c) t 0  0.300 s γ  0.180 s. (This is what the racer measures your clock to read at that
instant.) At your origin you read the original
1.20  10 8 m
 0.5 s.
(0.800) (3  10 8 m s)
Clearly the observes (you and the racer) will not agree on the order of events!
2
37.7:
 4.80  10 6 m s 
 (1 yr)
t 0  1  u c t  1  
8
 3.00  10 m s 
2
2
 (t  t 0 )  (1.28  10 4 ) yr  1.12 hrs.
The least time elapses on the rocket’s clock because it had to be in two inertial frames
whereas the earth was only in one.
37.8:
a) The frame in which the source (the searchlight) is stationary is the spacecraft’s frame,
so 12.0 ms is the proper time. b) To three figures, u  c. Solving Eq. (37.7) for u c in terms of
γ,
u
1
 1  (1 γ) 2  1  2 .
c
2γ
Using 1 γ  t 0  t  12.0 ms 190 ms gives u c  0.998.
37.9:
γ
a) l 
1
1  (u c)
2

1
1  (0.9860) 2
 6.00
l 0 55 km

 9.17 km.
γ
6.00
b) In muon’s frame:
d  ut  (0.9860c)( 2.20  10 6 s)  0.651 km.
%
c) In earth’s frame:
d 0.651

 0.071  7.1%.
h 9.17
4.50  10 4 m
37.10: a) t 
 1.51  10 4 s.
0.99540c
b) γ 
h 
c)
1
1  (0.9954) 2
 10.44
h 45 km

 4.31 km.
γ 10.44
h
t
 1.44  10 5 s, and  1.44  10 5 s; so the results agree but the particle’s
0.99540c
γ
lifetime is dilated in the frame of the earth.
37.11: a) l0  3600 m
 l  l0 1 
u2
(4.00  107 m s) 2

l
(
3600
m
)
1

0
c2
(3.00  108 m s) 2
 (3600 m)(0.991)  3568 m.
b ) t 0 
l0
3600 m

 9.00  10 5 s.
7
u 4.00  10 m s
t 
l
3568 m

 8.92  10 5 s.
7
u 4.00  10 m s
c)
37.12: γ  1 0.3048 , so u  c 1  (1 γ) 2  0.952c  2.86  10 8 m s .
37.13: l  l 0 1 
u2
l
 l0 
2
c
1 u2 c2
 l0 
74.0 m
 0.600c 
1 

 c 
2
 92.5 m.
37.14: Multiplying the last equation of (37.21) by u and adding to the first to eliminate t
gives
 u2  1
x  ut   x 1  2   x,
 c  
and multiplying the first by
u
and adding to the last to eliminate x gives
c2
t 
 u2  1
u

1  2   t ,
x

γ
t
c2
 c  γ
so x  γ( x  ut ) and t  γ(t   ux c 2 ),
which is indeed the same as Eq. (37.21) with the primed coordinates replacing the
unprimed, and a change of sign of u.
37.15: a) v 
v  u
0.400c  0.600c

 0.806c
2
1  (0.400) (0.600)
1  uv  c
b) v 
v  u
0.900c  0.600c

 0.974c
2
1  (0.900)(0.600)
1  uv  c
c) v 
v  u
0.990c  0.600c

 0.997c.
2
1  (0.990)(0.600)
1  uv  c
37.16: γ  1.667(γ  5 3 if u  (4 5)c). a) In Mavis’s frame the event “light on” has spacetime coordinates x  0 and t   5.00 s, so from the result of Exercise 37.14 or


Example 37.7, x  γ( x   ut ) and t  γ  t  
ux 
 x  γut   2.00  109 m, t  γt  
2 
c 
8.33 s.
b) The 5.00-s interval in Mavis’s frame is the proper time t 0 in Eq. (37.6), so
t  γt 0  8.33 s, as in part (a).
c) (8.33 s) (0.800c)  2.00  10 9 m, which is the distance x found in part (a).
37.17: Eq. (37.18): x  
x  ut
1 u c
2
2
Eq. (37.19): x   ut   x 1  u 2 c 2
Equate: ( x  ut ) γ  ut  
x
γ
 xγ

x
x1
 t    
 tγ    tγ    γ 
uγ 
uγ
 u

1  u c   1
 u2 c2
1
1
2
 γ  1  u c  


 γu 2 c 2
2
2
2
γ
1  u c 
1  u c 
1  u c 
2
 t   tγ 
t  ux c 2
xuγ

.
c2
1  (u c) 2
37.18: Starting from Eq. (37.22),
v 
vu
1  uv c 2
v (1  uv c 2 )  v  u
v   u  v  v uv c 2
 v(1  uv  c 2 )
from which Eq. (37.23) follows. This is the same as switching the primed and unprimed
coordinates and changing the sign of u .
37.19: Let the unprimed frame be Tatooine and let the primed frame be the pursuit ship. We
want the velocity v  of the cruiser knowing the velocity of the primed frame u and the velocity
of the cruiser v in the unprimed frame (Tatooine).
v 
v u
0.600c  0.800c

 0.385c
uv 1  (0.600) (0.800)
1 2
c
 the cruiser is moving toward the pursuit ship at 0.385c.
37.20: In the frame of one of the particles, u and v are both 0.9520c but with opposite sign.
v 
 v  (u )
 0.9520c  0.9520c

 0.9988c.
2
1  (0.9520)( 0.9520)
1  (u ) (v) c
Thus, one particle moves at a speed 0.9988c toward the other in the other particle’s
frame.
37.21: v 
v  u
 0.950c  0.650c

 0.784c.
uv  1  (0.950)(0.650)
1 2
c
37.22: a) In Eq.(39-24), u  0.400c, v  0.700c  v 
v  u
0.700c  0.400c

2
1  uv c
1  (0.700) (0.400)
 0.859c.
x 8.00  10 9 m
b)

 31.0 s.
v
0.859c
37.23: v  
vu
uvv 
 v  2  v  u
2
1  uv c
c
v  v
 vv  
 u 1  2   v  v   u 
c 
(1  vv  c 2 )

0.360c  0.920c
u
 0.837c
(1  (0.360)(0.920))
 moving opposite the rocket, i.e., away from Arrakis.
24:
Solving Eq. (37.25) for u c , (see solution to Exercise 37.25)
u 1  ( f f0 )2

,
c 1  ( f f0 )2
and so (a) if f f 0  0.98, u c   0.0202, the source and observer are moving away from
each other. b) if f f 0  4, u c   0.882, they are moving toward each other.
37.25: a) f 
cu
f 0  (c  u ) f 2  (c  u ) f 02
cu
c( f 2  f 02 ) c(( f f 0 ) 2  1) c(( λ 0 λ) 2  1)
u 


f 02  f 2
( f f0 )2  1
(( λ 0 λ) 2  1)
((675 575) 2  1)
u  c
 0.159c  4.77  10 7 m s  4.77  10 4 km s  1.72  108 km h.
2
((675 575)  1)
b) (1.72  108 km h  90 km h) ( $1.00)  $172 million dollars!
37.26: Using u  0.600c  3 5c in Eq. (37.25) gives
f 
1  3 5
25
f0 
f 0  f 0 2.
1  3 5
85
2
1
1

dp d 
mv
ma
2 mv  2 ( 2va c )

37.27: a) F 



2
2
dt dt  1  v 2 c 2 
(1  v 2 c 2 ) 3 / 2
1

v
c


 (1  v 2 c 2 )  v 2 c 2 
  ma (1  v 2 c 2 ) 3 2
 ma
2
2 32
(
1

v
c
)


32
F  v2 
 a  1  2  .
m c 
b) If the force is perpendicular to velocity then denominator is constant  F 
m dv dt
1 v c
2
2
a
dp

dt
F
1  v2 c2 .
m
37.28: The force is found from Eq. (37.32) or Eq. (37.33). (a) Indistinguishable from
F  ma  0.145 N. b) γ 3 ma  1.75 N. c) γ 3 ma  51.7 N. d) γma  0.145 N,
0.333 N, 1.03 N.
37.29: a) p 
mv
1 v2 c2
 2mv
 1  2 1 v2 c2 
 v2 
1
v2
 1 2
4
c
3 2
3
c v
c  0.866c.
4
2
b) F  γ 3 ma  2ma  γ 3  2  γ  (2)1 / 3 so
1
v
 22 / 3 
2
c
v
1 2
c
 1  2 2 / 3  0.608.
37.30: a) γ  1.01, so v c   0.140 and v  4.21  108 m s.
b) The relativistic expression
is always larger in magnitude than the non-relativistic expression.
37.31: a) K 

mc 2
1 v c
2
 mc 2  mc 2
2
1
1 v2 c2
b) K  5mc 2 
2
1
v2
 1 2  v 
4
c
1
1 v2 c2
6
3
 0.866c.
4
1
v2
35
 1 2  v 
c  0.986c.
36
36
c
37.32: E  2mc 2  2(1.67  10 27 kg) (3.00  108 m s) 2  3.01 10 10 J  1.88  10 9 eV.
37.33: K  ( γ  1)mc 2 
1 2 3 mv 4
mv 

2
8 c2
1 2
3 v 4 0.02 2
if K  K 0  1.02 mv 

v
2
8 c2
2

150 2
4
v  c2  v 
c  0.163c  4.89  10 7 m s.
4
150
37.35: a) Your total energy E increases because your potential energy increases;
E  mgy
E  (m)c 2 so m   E c 2  mg (y ) c 2
 m m  ( gy ) c 2  (9.80 m s 2 )(30 m) (2.998  10 8 m s) 2  3.3  10 13 %
This increase is much, much too small to be noticed.
b) E  U  12 kx 2  12 (2.00  10 2 N m)(0.060 m) 2  36.0 J
m  (E ) c 2  4.0  10 16 kg
Energy increases so mass increases. The mass increase is much, much too small to be
noticed.
37.36: a) E 0  m0 c 2
2 E  mc 2  2m0 c 2
 m  2 m0 
m0
1 v2 / c2
 2 m0
1
v2
v2 3
 1 2  2   v  c 3 4
4
4
c
c
v  0.866 c  2.60  10 8 m s
b) 10 m0 c 2  mc 2 
1
m0
1 v c
2
2
c2
v2
1
v2
99



2
2
100
100
c
c
99
vc
 0.995 c  2.98  10 8 m s
100
37.37: a) E  mc 2  K , so E  4.00mc 2 means K  3.00mc 2  4.50  10 10 J
b) E 2  (mc 2 ) 2  ( pc) 2 ;
E  4.00mc 2 , so 15.0(mc 2 ) 2  ( pc) 2
p  15mc  1.94  10 18 kg  m s
c) E  mc 2
1 v2 c2
E  4.00mc 2 gives 1  v 2 c 2  1 16 and v  15 16c  0.968c
37.38: The work that must be done is the kinetic energy of the proton.


 1

a) K  (   1)m0 c  
 1 m0 c 2
2
 1  v2

c


2

1

 (1.67  10 27 kg)(3.00  108 m s) 2 
 1  0. 1 c
c

1


 (1.50  10 10 J)
 1
 1  0.01 
 
2


 1


 7.56  10 13 J
b) K  (1.50  10 10 J)

1 (0.5)2
 1

1
1(0.9)2
 1

1
 2.32  10 11 J
c) K  (1.50  10 10 J)

 1.94  10 10 J
37.39: (m  6.64  10 27 kg,
a) E 
p  2.10  10 18 kg  m s)
(mc 2 ) 2  ( pc) 2
 8.68  10 10 J.
b) K  E  mc 2  8.68  10 10  (6.64  10 27 kg) c 2  2.70  10 10 J.
2.70  10 10 J
K

 0.452.
c)
mc 2 (6.64  10 27 kg) c 2
37.40: E  (m c  p c )
2
4
2
2 12
  p 2 

 mc 1  
  mc  


12
2
 1 p2 
p2
1
2

 mc 2 1 

mc

 mc 2  mv 2
2 2 
2m
2
 2m c 
the sum of the rest mass energy and the classical kinetic energy.
37.41: a) v  8  10 7 m s  γ 
m  mp
1
1  v2 c2
K0 
 1.0376
1 2
mv  5.34  10 12 J
2
K  ( γ  1)mc 2  5.65  10 12 J 
K
 1.06.
K0
b) v  2.85  108 m s  γ  3.203
1 2
mv  6.78  10 11 J
2
K  (γ  1)mc 2  3.31  10 10 J
K0 
K K 0  4.88.
37.42: (5.52  10 27 kg)(3.00  108 m s) 2  4.97  10 10 J  3105 MeV.
37.43: a) K  qV  eV


1
K  mc 2 
 1  4.025mc 2  3.295  10 13 J  2.06 MeV
 1 v2 c2



6
V  K e  2.06  10 V
b) From part (a), K  3.30  10 13 J  2.06 MeV
37.43: a) K  qV  eV


1
K  mc 2 
 1  4.025mc 2  3.295  10 13 J  2.06 MeV
 1 v2 c2



6
V  K e  2.06  10 V
b) From part (a), K  3.30  10 13 J  2.06 MeV
37.44: a) According to Eq. 37.38 and conservation of mass-energy
2Mc 2  mc 2   2Mc 2    1 
m
9.75
 1
 1.292.
2M
2(16.7)
Note that since  
1
1v 2 c 2
, we have that
v
1
1
 1 2  1
 0.6331.
c

(1.292) 2
b) According to Eq. 37.36, the kinetic energy of each proton is
 1.00 MeV 
 
K  (  1) Mc 2  (1.292  1)(1.67  10 27 kg)(3.00  108 m s) 2 
13
 1.60  10 J 
274 MeV.
c) The rest energy of  0 is mc 2  (9.75  10 28 kg)(3.00  108 m s) 2

1.00 MeV
1.601013 J

548 MeV.
d) The kinetic energy lost by the protons is the energy that produces the  0 ,
548 MeV  2(274 MeV).
37.45: a) E  0.420 MeV  4.20  10 5 eV.
b) E  K  mc 2  4.20  10 5 eV 
(9.11  10 31 kg) (3.00  108 m s) 2
1.6  10 19 J eV
 4.20  10 5 eV  5.11 105 eV  9.32  10 5 eV.
 mc 2 

 v  c 1  
c) E 
1 v2 c2
 E 
mc 2
2
2
 5.11  10 5 eV 
  0.836c  2.51  108 m s .
 c 1  
5
9.32

10
eV


d) Nonrel: K 
1 2
mv  v 
2
2K

m
 3.84  108 m s .
2(4.20  105 eV)(1.6  10 19 J eV)
9.11  10 31 kg
37.46: a) The fraction of the initial mass that becomes energy is
1
(4.0015 u)
 6.382  10 3 , and so the energy released per kilogram is
2(2.0136 u)
(6.382  10 3 )(1.00 kg)(3.00  108 m s) 2  5.74  1014 J.
1.0  1019 J
b)
 1.7  10 4 kg.
14
5.74  10 J kg
37.47: a) E  mc 2 , m  E c 2  (3.8  10 26 J) (2.998  108 m s) 2  4.2  10 9 kg
1 kg is equivalent to 2.2 lbs, so m  4.6  10 6 tons
b) The current mass of the sun is 1.99  10 30 kg, so it would take it
(1.99  10 30 kg) (4.2  10 9 kg s)  4.7  10 20 s  1.5  1013 y to use up all its mass..
37.48: a) Using the classical work-energy theorem we obtain
x 
m(v 2  v 2 0 ) (2.00  10 12 kg)[(0.920 )(3.00  108 m s)] 2

 1.81 m.
2F
2(4.20  10 4 N)
b) Using the relativistic work-energy theorem for a constant force (Eq. 37.35) we obtain
x 
For the given speed,  
x 
1
1 0.9202
(   1)mc 2
.
F
 2.55, thus
(2.55  1)( 2.00  10 12 kg)(3.00  10 8 m s) 2
 6.65 m.
(4.20  10 4 N)
c) According to Eq. 37.30,
3
3
3
(4.20  10 4 N)  v 2  2
F  v2  2
v2  2
2 
16



a  1  2  
1


(
2
.
10

10
m
s
)
1

 c 2  ,
m c 
(2.00  10 12 kg)  c 2 


which yields, i ) a  2.07  1016 m s ( β  0.100)
2
ii) a  1.46  1016 m s ( β  0.462)
2
iii) a  0.126  1016 m s ( β  0.920).
2
37.49: a) d  ct  t 
1200 m
d

 4.00  10 6 s (v is very close to c).
8
c 3  10 m s
t
u2
 t 
u
But 0  1  2     1   0 
t
c
c
 t 
2
 t 
 (1  ) 2  1   0 
 t 
2
2
  t 0  2 
   1  1  
 
  t  
12
  2.6  10 8  2 
 
 1  1  
6 
  4.00  10  
12
 2.11  10 5.
 t 
 4.00  10 6 
2


b) E  mc  
 mc   2.6  10 8  139.6 MeV

t


 0
2
 E  2.15  10 4 MeV.
37.50: One dimension of the cube appears contracted by a factor of
1
, so the volume in S  is
γ
a 3 γ  a 3 1  (u c) 2 .
37.51: Need a  b  l 0  a, l  b

l b
b
 
 1  u 2 c2
l0 a 1.40b
2
2
b
 1 
 u  c 1    c 1 
  0.700c.
a
 1.40 
 2.10  108 m s .
37.52: The change in the astronaut’s biological age is t 0 in Eq. (37.6), and t is the distance
to the star as measured from earth, divided by the speed. Combining, the astronaut’s biological
age is
19 yr 
42.2 yr c
42.2 yr
 19 yr 
 24.7 yr.
γu
γ(0.9910)
37.53: a) E  mc 2 and   10 
1
1  (v c) 2

v

c
γ2  1
v
99
 c
2
γ
c
100
 0.995.
  v 2 2 
b) ( pc)  m v γ c , E  m c    γ  1
 c 



2
2 2

2 2
2
E 2  ( pc) 2

E2
2 4
1
v
1 2 
c
2

1
 0.01  1%.
1  (10 (0.995)) 2
37.54: a) Note that the initial velocity is parallel to the x-axis. Thus, according to Eqn. 37.30,
3
F
ax  x
m
3
 v 2  2 (3.00  10 12 N)
2
2 2
1  2  
(
1

0
.
900
)
 1.49  1014 m s .
 27
(1.67  10 kg)
 c 
Now note that the initial velocity is perpendicular to the y-axis. Thus, according to Eqn. 37.33,
1
1
Fy  v 2  2 (5.00  10 12 N)
2
2 2


ay 
1 2  
(1  0.900 )  1.31  1015 m s .
 27

m c 
(1.67  10 kg)
b) The angle between the force and acceleration is given by
cos θ 
Fx a x  Fy a y
Fa

( 3.001012 N)( 1.491014 m s 2 )  ( 5.001012 N)(1.311015 m s 2 )
( 3.001012 N)2  ( 5.001012 N)2 ( 1.491014 m s 2 ) 2  (1.311015 m s 2 ) 2

θ  24.5.
37.55: a) K  20  1012 eV  3.204  10 6 J


1
K  mc 2 
 1, so
 1 v 2 c 2



1
1
1 v c
2
2
 2.131  10 4
v2
1
v 2 (c  v)(c  v) 2(c  v)

;
1



since c  v  2c
c
c 2 (2.131  10 4 ) 2
c2
c2
v  (1  )c so 1  v 2 c 2  2 and   1.1  10 9
b) mrel 
m
1 v c
2
2

m
2
 (2.1  10 4 )m
37.56: a) (8.00 kg)(1.00  10 4 )(3.00  108 m s) 2  7.20  1013 J.
b) (E t )  (7.20  1013 J) (4.00  10 6 s)  1.80  1019 W.
c) M 
(7.20  1013 J)
E

 7.35  10 9 kg.
gh (9.80 m s 2 )(1.00  10 3 m)
37.57: Heat in Q  mL f  (4.00 kg)(3.34 105 J kg)  1.34 106 J
 m 
(1.34  10 6 J)
Q

 1.49  10 11 kg.
2
8
2
c
(3.00  10 m s)
p ( E c)
E


, where the atom and the photon have the same magnitude
m
m
mc
E
  c, so E  mc 2 .
of momentum, E c . b) v 
mc
37.58: a) v 
37.59: Speed in glass v 

1
1 v2 c2
c
c

 1.97  10 8 m s
n 1.52
 1.326
37.60:
 K  (   1)mc 2  (0.326)(0.511 MeV)  0.167 MeV  1.67  10 5 eV.
80.0 m s is non-relativistic, and K 
1 2
mv  186 J.
2
b) (   1)mc 2  1.31 1015 J.
c) In Eq. (37.23),
v  2.20  108 m s , u  1.80  108 m s , and so v  7.14  10 7 m s .
a)
d)
20.0 m
 13.6 m.

e)
20.0 m
 9.09  10 8 s.
8
2.20  10 m s
f) t  
13.6 m
t
 6.18  10 8 s, or t  
 6.18  10 8 s.

2.20  10 8 m s
37.60: a) 80.0 m s is non-relativistic, and K 
1 2
mv  186 J.
2
b) (   1)mc 2  1.31 1015 J.
c) In Eq. (37.23),
v  2.20  108 m s , u  1.80  108 m s , and so v  7.14  10 7 m s .
d)
20.0 m
 13.6 m.

e)
20.0 m
 9.09  10 8 s.
8
2.20  10 m s
f) t  
13.6 m
t
 6.18  10 8 s, or t  
 6.18  10 8 s.
8

2.20  10 m s
37.61: x  2  c 2 t  2

 ( x  ut ) 2  2  c 2  2 t  ux c 2

2
 x  ut  c(t  ux c 2 )
 u 1
 x1    x(u  c)  t (u  c)  x  ct
 c c
 x 2  c 2t 2 .
37.62: a) From Eq. (37.37),
(3.00  10 4 m s) 4
1 2 3 v4 3
K  mv  m 2  (90.0 kg)
 304 J.
2
8 c
8
(3.00  108 m s) 2
2
(3 8) mv4 c 2 3  v 
b)
    7.50  10  9.
2
(1 2)mv
4c
37.63:
dv F
 (1  v 2 c 2 ) 3 2
dt m
a

v
0
dv
F t
F
  dt  t
2
2 32
0
(1  (v c ))
m
m
 c
v c
0
dx
x
c
2 32
(1  x )
1  x2
 Ft 
 v2   
m
v
2
vc

0
v
1  (v c )
2

F
t
m
2
  v  2   Ft  2
1         v 2  Ft 
 c   m 
 mc 


Ft m
1  ( Ft mc) 2
.
So as t  , v  c.
37.64: Setting x  0 in Eq. (37.21), the first equation becomes x    ut and the last, upon
multiplication by c, becomes ct   ct. Squaring and subtracting gives
c 2t 2  x2  γ 2 (c 2t 2  u 2t 2 )  c 2t 2 ,
or x   c t  2  t 2  4.53  10 8 m.
37.65: a) Want t   t 2  t1
x1  ( x1  ut1 ) γ  x2  ( x2  ut 2 ) γ  u 
x2  x1 x

t 2  t1
t

( x  x1 ) 
 . Since u  x t ,
And t   γ t 2  t1  u 2
c2



ux 
x 2 
x 2

2
t   γ t  2   γ t  2   t   t 1   xt  c 2  t 2  2
c 
c t 
c


There’s no physical solution for t 
x
 x  ct.
c
b) Simultaneously  t   0  t1 
ux1
ux
 t 2  22
2
c
c
 t 
u
c 2 t

x

u

.
t
c2
Also
x   x 2  x1 
 x 1 
1
1  (u c) 2
(x  ut ) 

t 2 
 x  c 2

x 
1  c 2 t 2 x 2 
1
c 2 t 2
 x  x 2  c 2 t 2 .
x 2
c) Part (b): t 
1
1
(x) 2  (x ) 2 
(5.00 m) 2  (2.50 m) 2  1.44  10 8 s.
c
c
37.66: a) (100 s)(0.600)( 3.00  108 m s)  1.80  1010 m.
b) In Sebulbas frame, the
relative speed of the tachyons and the ship is 3.40c, and so the time t2  100 s 
1.80  1010 m
 118 s. At t 2 Sebulba measures that Watto is a distance from him of
3.4c
(118 s)(0.600)( 3.00  108 m s)  2.12  1010 m. c) From Eq. (37.23), with
v   4.00c and u  0.600c, v  2.43c, with the plus sign indicating a direction in the same
direction as Watto’s motion (that is, away from Sebulba). d) As the result of part (c) suggests,
Sebulba would see the tachyons moving toward Watto and hence t 3 is the time they would
have left Sebulba in order to reach Watto at the distance found in part
2.12  1010 m
(b), or 118 s 
 89 s, and so Sebulba receives Watto’s message before even
2.43c
sending it! Tachyons seem to violate causality.
37.67: Longer wavelength (redshift) implies recession. (The emitting atoms are moving away.)
Using the result of Ex. 37.26: u  c
(λ 0 λ) 2  1
(λ 0 λ)  1
 (656.3 953.4) 2  1
8
 u  c
  0.3570c  1.071  10 m s
2
 (656.3 953.4)  1
37.68: The baseball had better be moving non-relativistically, so the Doppler shift formula (Eq.
(37.25)) becomes f  f 0 (1  (u c)). In the baseball’s frame, this is the frequency with which
the radar waves strike the baseball, and the baseball reradiates at f. But in the coach’s frame,
the reflected waves are Doppler shifted again, so the detected frequency is
f (1  (u c))  f 0 (1  (u c)) 2  f 0 (1  2(u c)), so f  2 f 0 (u c) and the fractional
frequency shift is
f
 2(u c). In this case,
f0
u
f
(2.86  10 7 )
c
(3.00  10 8 m)
2 f0
2
 42.9 m s  154 km h
 92.5 mi h.
37.69: a) Since the two triangles are similar:
H  A  mc 2   E.
b) O 
H 2  A 2  E 2  (mc 2 ) 2  pc.
c) K  E  mc 2 .
The kinetic energy can be obtained by the difference between the hypoteneuse and adjacent
edge lengths.
37.70: a) As in the hint, both the sender and the receiver measure the same distance.
However, in our frame, the ship has moved between emission of successive wavefronts, and we
can use the time T  1 f as the proper time, with the result that f  f 0  f 0 .
b) Toward: f  f 0
cu
 1  0.758 
 345 MHz 

cu
 1  0.758 
12
 930 MHz
f  f 0  930 MHz  345 MHz  585 MHz.
Away:
cu
 1  0.758 
f  f0
 345 MHz 

cu
 1  0.758 
f  f 0  217 MHz.
12
 128 MHz
c) γf 0  1.53 f 0  528 MHz , f  f 0  183 MHz. The shift is still bigger than f 0 , but
not as large as the approaching frequency.
37.71: The crux of this problem is the question of simultaneity. To be “in the barn at one time”
for the runner is different than for a stationary observer in the barn.
The diagram below, at left, shows the rod fitting into the barn at time t  0 , according
to the stationary observer.
The diagram below, at right, is in the runner’s frame of reference.
The front of the rod enters the barn at time t1 and leaves the back of the barn at time
t2 .
However, the back of the rod does not enter the front of the barn until the later
time t 3 .
37.72: In Eq. (37.23), u  V , v  (c n), and so
v
(c n)  V (c n)  V

.
cV
1  (V nc)
1 2
nc
For V non-relativistic, this is
v  ((c n)  V )(1  (V nc))
 (c n)  V  (V n 2 )  (V 2 nc)



so k  1 
c 
1 
 1  2  V
n  n 
1 
. For water, n  1.333 and k  0.437.
n2 
37.73: a) a  
dv 
dv
dt 
dt   (dt  udx c 2 )
dv
vu
u

dv
2
2 2
(1  uv c ) (1  uv c ) c 2
dv
1
vu
u


.
2 2  2 
2
dv 1  uv c
(1  uv c )  c 

 1 u2 c2 
(v  u ) u c 2 
1



 dv  dv


dv
2
2 2 
(1  uv c 2 ) 2 
 1  uv c
 (1  uv c ) 
(1  u 2 c 2 )
1
(1  uv c 2 ) 2 dv (1  u 2 c 2 )
 a 
 .
.
2
2 2
γdt  udx c
dt (1  uv c ) γ(1  uv c 2 )
dv
 a(1  u 2 c 2 ) 3 2 (1  uv c 2 ) 3 .
b) Changing frames from S   S just involves changing
3
 uv  
a  a , v   v   a  a (1  u 2 c 2 ) 3 2 1  2  .
c 

37.74: a) The speed v is measured relative to the rocket, and so for the rocket and its
occupant, v  0. The acceleration as seen in the rocket is given to be a   g , and so the
acceleration as measured on the earth is
32
 u2 
du
a
 g 1  2  .
dt
 c 
b) With v1  0 when t  0 ,
1
du
g (1  u 2 c 2 ) 3 2
t1
1 v1
du
dt

0

0
g (1  u 2 c 2 ) 3 2
v1
t1 
.
g 1  v12 c 2
dt 
c) dt   dt  dt / 1  u 2 c 2 , so the relation in part (b) between dt and du, expressed
in terms of dt  and du, is
dt   dt 
1
1 u c
2
2
du
1
du

.
2
2 32
g (1  u 2 c 2 ) 2
g (1  u c )
Integrating as above (perhaps using the substitution z  u c ) gives
t1 
c
v 
arctanh  1 .
g
c
For those who wish to avoid inverse hyperbolic functions, the above integral may be done by
the method of partial fractions;
gdt  
du
1  du
du 
 

,
(1  u c)(1  u c) 2 1  u c 1  uc 
which integrates to
t1 
c  c  v1 
.
1n
2 g  c  v1 
d) Solving the expression from part (c) for v1 in terms of t1 , (v1 c)  tanh( gt1 c), so that
1  (v1 c) 2  1 cosh ( gt1 c), using the appropriate indentities for hyperbolic functions. Using
this in the expression found in part (b),
t1 
c tanh( gt1 c)
c
 sinh( gt1 c),
g 1 cosh( gt1 c) g
which may be rearranged slightly as
gt1
 gt  
 sinh  1 .
c
 c 
If hyperbolic functions are not used, v1 in terms of t1 is found to be
 gt  c
v1 e gt1 c  e 1 1

c e gt1 c  e  gt1 c
which is the same as tanh( gt1 c ). Inserting this expression into the result of part (b) gives, after
much algebra,
t1 
c gt1 c
(e
 e  gt1 c ),
2g
which is equivalent to the expression found using hyperbolic functions.
e) After the first acceleration period (of 5 years by Stella’s clock), the elapsed time on
earth is
t1 
c
sinh( gt1 c)  2.65  10 9 s  84.0 yr.
g
The elapsed time will be the same for each of the four parts of the voyage, so when Stella has
returned, Terra has aged 336 yr and the year is 2436. (Keeping more precision than is given in
the problem gives February 7 of that year.)
37.75: a) f 0  4.568110  1014 Hz f   4.568910  1014 Hz f   4.567710  1014 Hz
f 
f 
c  (u  v) 
 f0 
c  (u  v) 
f 2 (c  (u  v))  f 02 (c  (u  v))


f 2 (c  (u  v))  f 02 (c  (u  v))
c  (u  v) 
 f0
c  (u  v) 
where u is the velocity of the center of mass and v is the orbital velocity.
 (u  v) 
( f 2 f 02 )  1
( f f0 )2  1
c
(
u

v
)

c
and
( f f0 )2  1
( f 2 f 02 )  1
 u  v  5.25  10 4 m s u  v  2.63  10 4 m s
 u  1.31 10 4 m s  moving toward at 13.1 km s.
b) v  3.94  10 4 m s T  11.0 days.
(3.94  10 4 m s) (11.0 days) (24 hrs day) (3600 sec hr)

2π
5.96  10 9 m  0.040 earth - sun distance.
 2R  vt  R 
Also the gravitational force between them (a distance of 2R) must equal the centripetal
force from the center of mass:
(Gm 2 ) mv 2
4 Rv 2

m

R
G
(2 R) 2
4(5.96  10 9 m)(3.94  10 4 m s) 2
 5.55  10 29 kg  0.279 m sun .
11
2
2
6.672  10 N  m kg
37.76: For any function f  f ( x, t ) and x  x( x , t ), t  t ( x , t ), let F ( x , t ) 
f ( x( x , t ), t ( x , t )) and use the standard (but mathematically improper) notation
F ( x , t )  f ( x , t ). The chain rule is then
f ( x , t ) f ( x, t )

x
x 
f ( x , t ) f ( x, t )

t
x 
x  f ( x , t )

x
t 
x  f ( x , t )

t
t 
t 
,
x
t 
.
t
In this solution, the explicit dependence of the functions on the sets of dependent variables is
suppressed, and the above relations are then
f
f x  f t 


,
x x  x t  x
f
f x  f t 


.
t x  t t  t
x
x
t 
t 
E E
2E 2E
 1,
 v,
 0 and
 1. Then ,

, and

. For the
x
t
x
t
x x
x 2 x 2
f
E yE
 v

. To find the second time derivative, the chain rule must be
time derivative,
t
x  t 
a)
applied to both terms; that is,

t

t
E
2E 2E
 v 2 
,
x 
t x 
x 
E
2E 2E
 v

.
t 
x t  t  2
2E
Using these in 2 , collecting terms and equating the mixed partial derivatives gives
t
2
2E
2E 2E
2  E

v

2
v

x t  t  2
t 2
x 2
and using this and the above expression for
b) For the Lorentz transformation,
2E
gives the result.
x  2
x
x
t 
t 
 γ,
 v,
 v / c 2 and
 γ.
x
t
x
t
The first partials are then
E
E
v E E
E
E
γ
γ 2
,
  γv
γ
x
x
c t  t
x
t 
and the second partials are (again equating the mixed partials)
2
2
2E
2E
2E
2  E
2 v
2 v

γ

γ

2
γ
x 2
x 2
c 4 t  2
c 2 xt 
2
2
2E
2E
2 2  E
2  E
2
γ v
γ
 2γ v
.
t 2
x 2
t  2
xt 
Substituting into the wave equation and combining terms (note that the mixed partials
cancel),
2
2E 1 2E
v2  2E
1  2E
2
2 v





γ
1




 c 2  x 2
 c 4 c 2  t  2
x 2 c 2 t 2




2
2
 E 1  E


x 2 c 2 t  2
 0.
37.77: a) In the center of momentum frame, the two protons approach each other with equal
velocities (since the protons have the same mass). After the collision, the two protons are at
rest─but now there are kaons as well. In this situation the kinetic energy of the protons must
equal the total rest energy of the two kaons  2(γ cm  1)m p c 2  2mk c 2 
γ cm  1 
vcm
mk
 1.526. The velocity of a proton in the center of momentum frame is then
mp
2
γ cm
1
c
 0.7554c.
2
γ cm
To get the velocity of this proton in the lab frame, we must use the Lorentz velocity
transformations. This is the same as “hopping” into the proton that will be our target and asking
what the velocity of the projectile proton is. Taking the lab frame to be the unprimed frame
moving to the left, u  vcm and v  vcm (the velocity of the projectile proton in the center of
momentum frame).
2vcm
v  u

 0.9619c
2
uv
v
1  2 1  cm2
c
c
1
 γ lab 
 3.658
2
vlab
1 2
c
 K lab  ( γ lab  1)m p c 2  2494 MeV.
vlab 
b)
K lab
2494 MeV

 2.526.
2mk 2(493.7 MeV)
c) The center of momentum case considered in part (a) is the same as this situation.
Thus, the kinetic energy required is just twice the rest mass energy of the kaons.
K cm  2(493.7 MeV)  987.4 MeV. This offers a substantial advantage over the fixed target
experiment in part (b). It takes less energy to create two kaons in the proton center of
momentum frame.