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Proper time problem

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I'll assume that we are using units such that c=1. The two world lines are given by:

x1 = t/3

x2 = t^2/18

Here the "18" must carry the units of time used. So, we'll just assume that some units are chosen and the numbers that come out for times/distances are given in terms of that unit.

The time when the observers meet follows from equating x1 to x2:

x1 = x2 -->

t = 0 and t = 6.

At t = 6, x1 = x2 = 2.

The event where they meet again is thus the event t = 6, x = 2.

Now, you know from elementary special relativity that the relation between proper time tau and coordinate time t is:

tau = t/gamma

where gamma = 1/sqrt[1-v^2]

But you must be very careful about applying equations from special relativity derived for inertial observers to accelerating observers. So, let's apply this equation to observer 1 and then I'll show how you deal with the general case. Observer 1 moves ...

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