Many questions on Physics SE relate to the twin paradox, but I did not find any that ask this specific question. Suppose that object A is in a circular orbit around a spherically symmetric, non-rotating planet (Schwarzschild metric). Suppose that object B is in radial free-fall, with a worldline that meets object A on the way up (passing arbitrarily closely without colliding) and again on the way down, as depicted here:

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Object A executes exactly one period of its orbit between the two meetings, and both objects remain in free-fall between the two meetings.

The question: Which object ages less between the two meetings?

Here's what I already know, expressed using coordinates $t,r,\phi$ where the equation for an object's proper time $\tau$ is $$ d\tau^2=\left(1-\frac{R}{r}\right)\,dt^2-\frac{dr^2}{1-\frac{R}{r}} - r^2d\phi^2, \tag{1} $$ where $R$ is the Schwarzschild radius. (This is in the plane that contains both objects' worldlines, so only $3$ coordinates are needed.) I already know that object A ages less when the coordinate-radius $r$ of object A's orbit is $r=3R/2$, because then A's worldline is lightlike and B's worldline is still timelike.

But does A age less than B for all radii $r>3R/2$? If not, then at what value(s) of $r$ do the two objects age the same amount between meetings?

  • $\begingroup$ So you are just asking what is the proper time along each path? Aren't there duplicates - certainly the object in orbit has been addressed a number of times. $\endgroup$ – Rob Jeffries Dec 15 '18 at 20:32
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    $\begingroup$ This particular problem is discussed at length in the following article: mathpages.com/rr/s6-05/6-05.htm $\endgroup$ – m4r35n357 Dec 15 '18 at 20:41
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    $\begingroup$ For the impatient, the mathpages article doesn't seem to explicitly answer the question, but the answer seems to be implicit in the graph that comes after the words "behaves as shown below." The answer seems to be that in all cases, the radial twin ends up older than the circular twin. $\endgroup$ – Ben Crowell Dec 15 '18 at 21:01
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    $\begingroup$ cool problem. It illustrates nicely how geodesics are local maxima of proper time. I think I will assign it next time I teach GR :). $\endgroup$ – Sam Gralla Dec 15 '18 at 21:40
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    $\begingroup$ @Ben Crowell I think equation (8) answers the OP's question, do you not? $\endgroup$ – m4r35n357 Dec 16 '18 at 12:39

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