Spaceship "A" is on a circular orbit around the massive body in one direction. "B" is on the same circular orbit in the other direction. As they pass each other, they synchronize their watches. When they meet again, they compare watches.(adsbygoogle = window.adsbygoogle || []).push({});

When the two spaceships separate, but before they've separated very far, they can be assumed to be in such a small region of space that special relativity applies. In this region, in the approximate inertial reference frame of spaceship "A", it is clear that spaceship "B" is moving and should have his time dilated. When "A" and "B" again meet, the same things is true.

Intuitively, this suggests that "A" should be treated as the stationary twin and "B" should be treated as the travelling twin. Yet, by symmetry, we know that the proper time experienced by "A" and "B" must be the same.

So the problem is this. Find a global coordinate system centered on spaceship "A". Define the position of spaceship "B" relative to these coordinates. Let event "1" be the first meeting of the spaceships, and event "2" be the second. Compute the proper time interval along the path of spaceship "A" and the proper time interval along the path of spaceship "B", in your coordinate system centered on "A", and show that these are equal.

To get you started, here's the Schwarzschild metric for a massive gravitating body, with the [tex]\phi[/tex] angle set to 90 degrees (i.e. equatorial motion):

[tex] ds^2 = ( 1 - r_s / r ) dt^2 - ( 1 - r_s / r )^{-1} dr^2 - r^2 d_\theta^2[/tex]

Carl

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# Two spaceships are orbiting a massive body

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