Two spaceships are orbiting a massive body

[CarlB]

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.

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 traveling 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 \phi angle set to 90 degrees (i.e. equatorial motion):

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

Carl

 

[thepatientmental]

: According to the Schwarzschild metric provided, the proper time interval along the path of spaceship "A" can be calculated as follows:

Δt_A = ∫√(1 - r_s/r) dt

And the proper time interval along the path of spaceship "B" can be calculated as:

Δt_B = ∫√(1 - r_s/r) dt

Since both spaceships are on the same circular orbit, their radial distance (r) from the massive body will be the same. Therefore, the only difference between the two equations is the direction of motion, which is accounted for by the negative sign in the dr term.

In the coordinate system centered on spaceship "A", the position of spaceship "B" can be defined as r = constant and θ = π (180 degrees), since they are on the same circular orbit but moving in opposite directions. This means that the radial distance (r) and the angular distance (θ) are fixed for both spaceships.

Plugging in these values, we get:

Δt_A = ∫√(1 - r_s/r) dt = ∫√(1 - r_s/r) dr = √(1 - r_s/r) Δr = √(1 - r_s/r) (r_B - r_A)

And:

Δt_B = ∫√(1 - r_s/r) dt = -∫√(1 - r_s/r) dr = -√(1 - r_s/r) Δr = -√(1 - r_s/r) (r_B - r_A)

Since r_B and r_A are the same for both equations, the only difference is the negative sign in front of Δr for spaceship "B". This negative sign accounts for the opposite direction of motion and ensures that the proper time interval for spaceship "B" is equal to that of spaceship "A".

Therefore, in this coordinate system centered on spaceship "A", the proper time intervals for both spaceships are equal, as expected by symmetry. This also confirms the intuition that in special relativity, the traveling twin experiences time dilation while the stationary twin does not. Thanks to the Schwarzschild metric, we can see that this holds true even in the presence of a massive gravitating body.