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Proper time problem

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1. Consider two observers moving in the x direction with respect to a stationary reference frame. The worldline of one observer is given by x1 = (1/3)t while that of the other is given by x2 = (1/18)t^2.
(a) Determine the event at which the two observers meet after leaving the origin.
(b) Determine the proper time measured by both observers between the origin and the point found in part (a).

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A detailed solution is given. The expert determines the event at which the two observers meet after leaving the origin. The proper time measured by both observers between the origin and the points found are determined.

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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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