Question about "odometer" analogy and time dilation

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Mister T said:
By symmetrical do you mean simultaneous? Recall that simultaneity is relative. Two events that are simultaneous in the rest frame of one of the ships will not be simultaneous in the rest frame of the other ship.

Edit: By this I mean that if the two launches are simultaneous in the rest frame of one of the ships, they will not be simultaneous in the rest frame of the other ship.
Later in the thread I amended the scenario so that the probe launches occur when the two ships measure an agreed upon distance between themselves after they pass.

After seeing the spacetime diagram Ibix provided I understand it doesn't matter, if motion is regarded as relative the proper time each probe accumulates will always be identical. I was supposing some version of absolute motion without entirely realizing it, I thought it was more of a case of "objectively, there is a true velocity each ship carries with respect to the other, but there's no way to detect it" as opposed to it truly being relative.
 
Ibix said:
both at 0.4c. Five minutes later they launch probes, represented by the thinner lines moving at 0.95c. They catch up with the opposite spaceship at 12.27 minutes after the first meeting.
Why were these values chosen?
Of course, they are correct,
and a computer wouldn't have any problem with them,
but the arithmetic seems (in my opinion) unnecessarily complicated to make your point.
A return leg of 0.8c would have been better.
(An outgoing leg of 0.6 would also help simplify the arithmetic.
I have a fondness for 3/5, 4/5, 15/17... https://en.wikipedia.org/wiki/Pythagorean_triple .)
 
robphy said:
Why were these values chosen?
0.6c and 0.8c made for a very long chase phase (5 mins to launch, then 30 mins to overtake) and I wanted the diagram to be reasonably wide and short so you can see the ticks clearly. 0.4 and 0.95 were pretty arbitrary, I admit.