A question that led to an interesting answer was posed in a newsgroup recently that I thought I'd share here. Someone proposed a scenario where there were two rocket ships that would travel at constant speed along a circle in opposite directions. They would pass each other twice per revolution. The times of the onboard crew's watches are easy to express as reckoned from an inertial frame still with respect to the circle. According to the inertial frame observer both accelerated frame watch times are simply given by special relativistic time dilation(adsbygoogle = window.adsbygoogle || []).push({});

[tex]t = \frac{t'}{\sqrt{1 - \frac{v^2}{c^2}}} [/tex]

[tex]t = \frac{t''}{\sqrt{1 - \frac{v^2}{c^2}}}[/tex]

This means that the accelerated frame observers must agree that their watches should read the same time each time they pass nearby. However, while they are nearby eachother each will observe the others local motion yielding special relativistic time dilation as they pass. This means that while they are far apart each must find that the other is "time contracted". This has been understood qualitatively in terms of the curvalinear nature of accelerated frame coordinates which are composed of a continuum of inertial frames with different standards of simultaneity. In other words at each point in the trip the rocket is instantaneously at rest with respect to some inertial frame, but in the next moment it is at rest with respect to another inertial frame and so the crews standards of simultaneity are continuously adjusted. The question that was posed was at what time does one crew reckon that the other transitions from a state of time dilation to time contraction with respect to their own accelerated frame. For the simplest choice of accelerated frame coordinates the A' frame observers find that the A" frame observers watch runs according to

[tex]\gamma t' - r_{0}\frac{v}{c^2}sin[\frac{v}{r_{0}}\gamma (t' + t'')] - \gamma t'' = 0[/tex]

where

[tex]\gamma = \frac{1}{\sqrt{1 - \frac{v^{2}}{c^{2}}}}[/tex]

and v is the speed of one with respect to the inertial frame and [tex]r_0[/tex] is the radius of the circle according to the inertial frame.

This equation implies that the watches do synch up according to the accelerated frames not only upon passing, but also at opposite ends of the circle. Differentiation of this does also yield that "upon passing" eachother the A' frame observers do observe the A" frame observer's watch as time dilated according to

[tex]dt' = \frac{dt''}{\sqrt{1 - \frac{u^2}{c^2}}}[/tex]

with

[tex]u = \frac{2v}{1 + \frac{v^{2}}{c^{2}}}[/tex]

and at opposite ends where the clocks resynch as well to be time contracted maximally by

[tex]dt'' = \frac{dt'}{\sqrt{1 - \frac{u^2}{c^2}}}[/tex]

These were derived as the answer to problem 5.4.5 from equation 5.4.4 at

http://www.geocities.com/zcphysicsms/chap5.htm

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# Counter rotating frames

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