How Does Time Dilation Affect Supersonic Jet Pilots?

AI Thread Summary
Supersonic jets traveling at speeds near 3 million times the speed of light experience time dilation, affecting the time perceived by pilots compared to stationary observers. In a year measured by an observer's clock, the pilot's clock would show a negligible time difference of approximately 0.000142 seconds. The proper time is defined as the time experienced by the pilot, which is less than that of the observer due to their relative motion. A mistake was noted in the initial calculations, where it was incorrectly stated that the pilot experiences more time. Overall, the discussion emphasizes the importance of understanding proper time in the context of relativistic speeds.
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Homework Statement


Supersonic jets achieve maximum speeds of about (3 *10^6)*c.

a)During a time of 1 y =3.16*10^7 seconds on your clock, how much time would elapse on the pilot's clock?

b)How many minutes are lost by the pilots clock in 1 year of your time.

Homework Equations



delta(t)=delta(t')/(sqrt(1-v^2/c^2)^.5), where delta(t') is the proper time interval

the proper time would be the time elapse(delta(t')) on my clock and the time dilated (delta (t)) would be the pilot's clock.

The Attempt at a Solution



to solved the first problem, I would have to apply the equations I mentioned and 2 and solved delta(t). delta(t)= (3.16e7)/(sqrt(1-(3e-6)^2)).

delta(t)=31600000.0001422000000095985 seconds . So I say the time dilated on the pilot's clock is very negligble.

The second part of the question just asked the time difference between my clock time and the pilot clock's time and that results to be. delta(t)=.0001422000000095985 seconds =2.37e-5 minutes .
 
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You have one mistake here, other than that your work seems great, assuming you have not made an error while crunching the numbers. You have the wrong time as the proper time. The proper time is time the pilot experiences. This is because the proper time is the time given by an inertial clock present and at both events. Your clock is always with you, but the pilots clock is actually there at the events you are measuring (the beginning and end of the plane's year long trip).

(I like this description of proper time, but I found it on Wikipedia I think, so I am a little suspicious that it may not be a complete description of proper time. The definition seems to be fine to me, but if someone finds that it is not complete, please add on to it.)

Remember that the pilot should experience less time relative to you since he is moving relative to you, yet your answer says the pilot experiences MORE time. This is what gave the mistake away.

Other than that, your solution seems correct. Nice Job!
 
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