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NtE Coursework 7 Time Dilation Answers

PHY 116 From Newton to Einstein
Answers to Problem Sheet 7: Time Dilation
Marked out of 20 +10
A spaceship flies past Mars with a speed of 0.978c relative to the surface of
the planet. When the spaceship is directly overhead at an altitude of 1200km, a very
bright signal light on the Martian surface blinks on and then off. An observer on Mars
measures that the signal was on for 82.4 µs. What is the duration of the light pulse
measured by the pilot of the spaceship?
Answer: The events, the light going on and off, are in the same place in the Martian [1]
frame of reference. It is in this frame of reference that the time interval is the proper
time T0. In any other frame at speed v, it is the dilated time that is observed. So, in
the spaceship, it is T = T0 that is observed. [1]
 
 4.79
1  0.978
T  T0  4.79  82.4  395 s
The negative pion (π-) is an unstable particle with an average lifetime of 2.60 x
10 s (measured in the rest frame of the pion). a). If the pion is made to travel at very
high speed relative to a laboratory, its average lifetime is measured in the laboratory
to be 4.20 x 10-7 s. Calculate the speed of the pion expressed as a fraction of c. b).
What distance, measured in the laboratory, does the pion travel during its average
Answer: (a) The creation and destruction of the pion occur at the same place in the
rest frame of the pion, so its decay time is the proper
[1] time in this frame. The decay
time in the lab frame is therefore the dilated time. So
4.2  10 7
 
 16.15 
2.6  10
1 
   1  2  0.996
(b) s = vt, so the pion travels 0.996 c  4.20  10-7 s = 126 m
As you pilot your space utility vehicle at a constant speed towards the moon, a
racing pilot flies past you in her spaceracer with a constant speed of 0.8c relative to
you. At the instant the spaceracer passes you, both of you start timers at zero. a) At
the instant when you measure that the spaceracer has travelled 1.20 x 108 m past you,
what does the racing pilot read on her timer? b) When the racing pilot reads the value
calculated in part (a) on her timer, what does she measure to be her distance from
you? c). At the instant when the racing pilot reads the value calculated in part (a) on
her timer, what do you read on yours?
Answers: (a) The two events on your timer take place at the same place, so the
1.2  10 8
 0.5 sec . [1]
interval is a proper time in your frame, of TE  
v 0.8  3  10 8
Consequently, it is a dilated time in the spaceracer’s frame, to be compared with her
proper time TS . So
TE  TS 
 S
1  0.8 2 0.6
TS  0.6TE  0.3 s
(b) She measures you to be receding at 0.8 c, so when she reads 0.3 sec she knows
that you are at 0. 3 × 0.8 light-seconds away = 0.24 × 3 × 108 = 7.2 × 107 m
(c) Her interval of 0.3 s is between two events that happened in the same place in the
spaceship. But you reading your timer occurs in your frame,[1]so you are reading a
proper time corresponding to her dilated time. Consequently, you read 0.6 times her
reading, or 0.16 s.
A spacecraft flies away from the Earth with a speed of 4.80 x 106 m/s relative
to the Earth and the returns at the same speed. The spacecraft carries an atomic clock
that had been carefully synchronised with an identical clock which remains at rest on
the Earth. The spacecraft returns to its starting point 365 days later, as measured by
clocks that remained on the Earth. What is the difference in elapsed times on the two
clocks, measured in hours? Which clock, the one on the spacecraft or the one on
Earth, shows the smallest elapsed time?
 0.016 , so  
 1.00013 . The atomic clock measures the
proper time TS on the spaceship, so time on Earth is dilated, i.e. 365 days =  TS. So
TE  TS  1.12hours
The spaceship clock therefore records the smaller elapsed time.
Answer:  
An alien spacecraft is flying overhead at a great distance from you. You see its
searchlight blink on for 0.19 s. On the spacecraft they measure the searchlight as
being on for 12 ms. a). Which of these two times is the proper time? b). What is the
speed of the spacecraft relative to the Earth expressed as a fraction of the speed of
Answer: (a) The searchlight switched on (Event 1) and switched off (Event 2) in the
same place in the spacecraft frame of reference. The time measured on the spacecraft
is therefore the proper time. You are travelling at some velocity with respect to this
frame, your measured time is therefore the dilated time.
Answer: (b) The dilated time is T=190 ms, the proper time T0 is 12 ms. So
T  T0
 
T 190
 15.8
 1  2  0.996