Solution Manual for Modern Physics 5th Edition by Stephen Thornton, Andrew Rex

Solution Manual for
Modern Physics for
Scientists and Engineers
5th Edition by Stephen
Thornton & Andrew Rex


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,Chapter 2 Special Theory of Relativity 1




Chapter 2
d 2x d 2 y d 2z 
1. For a particle Newton’s second law says F = ma = m  2 iˆ + 2 ˆj + 2 kˆ  .
 dt dt dt 
Take the second derivative of each of the expressions in Equation (2.1):
d 2 x d 2 x d 2 y  d 2 y d 2 z  d 2 z
= 2 = 2 = 2 . Substitution into the previous equation gives
dt 2 dt dt 2 dt dt 2 dt
 d 2 x ˆ d 2 y  ˆ d 2 z  ˆ 
F = ma = m  2 i + 2 j + 2 k  = F .
 dt dt dt 
 dx dy ˆ dz ˆ 
2. From Equation (2.1) p = m  iˆ + j + k .
 dt dt dt 
dx dx dy dy dz  dz
In a Galilean transformation = −v = = .
dt dt dt dt dt dt
 dx  ˆ dy ˆ dz  ˆ
Substitution into Equation (2.1) gives p = m  + v i + j+ k  p .
 dt  dt dt
 dx dy ˆ dz  ˆ 
However, because p = m  iˆ + j + k  the same form is clearly retained, given
 dt dt dt 
dx dx
the velocity transformation = −v.
dt dt
3. Using the vector triangle shown, the speed of light coming toward the mirror is c 2 − v 2
distance 2 2
and the same on the return trip. Therefore the total time is t2 = = .
speed c2 − v2
v v
Notice that sin  = , so  = sin −1   .
c c

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2 Chapter 2 Special Theory of Relativity



 0.350 m/s 
4. As in Problem 3, sin  = v1 / v2 , so  = sin −1 (v1 / v2 ) = sin −1   = 16.3 and
 1.25 m/s 
v = v22 − v12 = (1.25 m/s)2 − (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 2 in Equation (2.3) leads to
Equation (2.4).
d
6. Let n = the number of fringes shifted; then n = . Because d = c ( t  − t ) , we have

c ( t  − t ) v2 ( + ) . Solving for v and noting that
n= = 1 2
1 + 2 = 22 m,
 c 
2



n ( 0.005) ( 589 10−9 m )
v=c = ( 3.00 10 m/s ) 8
= 3.47 km/s.
1+ 2 22 m

7. Letting 1 → 1 1 −  2 (where  = v / c ) the text equation (not currently numbered) for
t1 becomes
2 1−  2 2 1
t1 = 1
=
c (1 −  ) c 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  ):
x = ax + bt ; 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:
 (ac + b)(ac − b)t 
ct =   or c2 = (ac + b)(ac − b) = a2c2 − b2 .
 c 