ch37 - Chapter 37 1. From the time dilation equation t = t0...

1417
Chapter 37
1. From the time dilation equation  t =   t
0
(where  t
0
is the proper time interval,
    1 1 2 / , and  = v / c ), we obtain
  
F
H
G
I
K
J
1 0
2


t
t
.
The proper time interval is measured by a clock at rest relative to the muon. Specifically,
 t
0
= 2.2000  s. We are also told that Earth observers (measuring the decays of moving
muons) find  t = 16.000  s. Therefore,
2
2.2000 s
1 0.99050.
16.000 s



 
  
 
 
2. (a) We find  from     1 1 2 / :
  2 2
1 1
1 1 0.14037076.
1.0100000


    
(b) Similarly,
 
2
1 10.000000 0.99498744. 

  
(c) In this case,
  2
1 100.00000 0.99995000. 

  
(d) The result is
 
2
1 1000.0000 0.99999950. 

  
3. (a) The round-trip (discounting the time needed to “turn around”) should be one year
according to the clock you are carrying (this is your proper time interval  t
0
) and 1000
years according to the clocks on Earth, which measure  t . We solve Eq. 37-7 for  :
2
2
0
1y
1 1 0.99999950.
1000y
t
t

  
 
    
 
 

 
 

1418 CHAPTER 37
(b) The equations do not show a dependence on acceleration (or on the direction of the
velocity vector), which suggests that a circular journey (with its constant magnitude
centripetal acceleration) would give the same result (if the speed is the same) as the one
described in the problem. A more careful argument can be given to support this, but it
should be admitted that this is a fairly subtle question that has occasionally precipitated
debates among professional physicists.
4. Due to the time-dilation effect, the time between initial and final ages for the daughter
is longer than the four years experienced by her father:
t
f daughter
– t
i daughter
=  (
4.000 y
)
where  is the Lorentz factor (Eq. 37-8). Letting T denote the age of the father, then the
conditions of the problem require
T
i
= t
i daughter
+ 25.00 y , T
f
= t
f daughter
– 25.00 y .
Since T
f
 T
i
= 4.000 y, then these three equations combine to give a single condition
from which  can be determined (and consequently v ):
54 = 4    = 13.5   = 0.9973.
5. In the laboratory, it travels a distance d = 0.00105 m = vt , where v = 0.992 c and t is the
time measured on the laboratory clocks. We can use Eq. 37-7 to relate t to the proper
lifetime of the particle t
0
:
 
2
2
0
0
2
1 1 0.992
0.992
1 /
t v d
t t t
c c
v c
 
     
 
 

which yields t
0
= 4.46  10
–13
s = 0.446 ps.
6. From the value of  t in the graph when  = 0, we infer than  t
o
in Eq. 37-9 is 8.0 s.
Thus, that equation (which describes the curve in Fig. 37-22) becomes
0
2 2
8.0 s
1 ( / ) 1
t
t
v c 

  
 
.
If we set  = 0.96 in this expression, we obtain approximately 29 s for  t .
7. We solve the time dilation equation for the time elapsed (as measured by Earth
observers):


t
t


0
2 1 0 9990 ( . )

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where  t
0
= 120 y. This yields  t = 2684 y 3 2.68 10 y.  
8. The contracted length of the tube would be
L  L
0
1   2  5.00m
 
1  (0.999987) 2  0.0255m.