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This question The "Satellite Paradox": Twin paradox in orbiting satellites asks something similar, but none of the answers quite get at the meat of what I'm trying to understand. On the accuracy of " Freefalling reference frames are equivalent to an inertial reference frame." also asks something similar, but gets into "fictitious forces" and I'm hoping for a simpler answer more fit for a layman (i.e. I didn't understand the answer given there).

Suppose A is on a very tall tower on a non-rotating planet, some hundreds of miles tall. B is orbiting at that same height. As B passes A, the two of them synchronize their clocks.

From the perspective of the planet, which I believe we can assume is inertial, A is clearly in an inertial reference frame. However B, in freefall and feeling no forces pushing or pulling on him, is also in an inertial reference frame as the orbit simply follows the curvature of space-time (or so I've been told, but it sure doesn't look inertial to me).

When B passes A a second time, which twin's clock will have had more time pass? Why?

Readin
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A is clearly in an inertial reference frame.

Not in General Relativity. B, who is in free-fall is in a GR inertial frame, but A is not.

Math Keeps Me Busy
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    "Not in General Relativity. B, who is in free-fall is in a GR inertial frame, but A is not." Is that because the force of the planet pushing up on A (to counter the gravity and prevent A from going to the center of the planet) is treated as an acceleration that prevents A from being in an inertial reference frame? – Readin Sep 03 '21 at 05:15
  • @Readin Yes. Exactly. – Chiral Anomaly Sep 03 '21 at 13:12
  • @Readin General Relativity (GR) asserts that a object traversing a geodesic path through space-time (i.e. in free-fall) is locally in an inertial frame per Special Relativity (SR). This local frame is expressed in the technical language of GR as a "tangent space". You can think of it, as you have done, that "A" is "accelerating" (equivalence of gravity and acceleration) if that helps your intuition. – Math Keeps Me Busy Sep 03 '21 at 13:13
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Math Keeps Me Busy correctly pointed out that A is accelerating (not inertial) in your problem, but it's worth saying a bit more.

The usual twin paradox is seen as paradoxical because there's a symmetry argument: A sees B move off into the distance and then come back, and B also sees A move off into the distance and then come back, so how could there be a measurable difference between them at the end? The paradox is usually resolved by pointing out that one of them accelerates and the other doesn't, and they can detect this (acceleration can be felt, unlike velocity).

This resolution may have left some people with the idea that acceleration is somehow the cause of difference in aging, but that's not the case. The only reason to mention acceleration is to refute the incorrect symmetry argument. Without the symmetry argument, there's no reason to expect the twins to end up the same age, but that doesn't tell you that they end up different ages, or which one ends up older if they do. To determine that, you measure the length of their worldlines, because the length of a worldline is the elapsed proper time.

In your problem, you could make a similar symmetry argument: A sees B moving in a circle, and B also sees A moving in a circle. You could refute the symmetry argument in the same way, by pointing out that A is accelerating (feels their weight) while B isn't (feels weightless). To determine which (if either) ends up younger, you still have to measure their worldlines. When you do that calculation, you find that the orbiting twin ends up younger, even though the orbiting twin is the one not accelerating, and in the original twin paradox the twin who doesn't accelerate ends up older.

You can set up problems where neither twin accelerates, but one still ends up older – by putting one in a circular orbit and the other in an elliptical orbit, for example. You would have to find some other way to refute the symmetry argument in that case, but it is still invalid. If the argument is irrefutable because there really is a symmetry between the twins, then a direct calculation will always find that they end up the same age.

benrg
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  • Would it be possible to calculate the inverse time dilations from Special Relativity, without having to go into GR tensor calculus? That is, how to calculate the A's time dilation from B's perspective, and B's time dilation from A's perspective? – fishinear Feb 08 '22 at 14:16
  • @fishinear You can't do it in SR, but the calculation in GR (for a circular orbit) is easy. Use the Schwarzschild metric with constant $r$ and $θ$, which is $ds^2=(1-2GM/rc^2)dt^2-(r\sin θ)^2 d\phi^2$. Rewrite that as $ds=\sqrt{(1-2GM/rc^2)-(r\sin θ)^2 \phi'(t)^2}dt$. Plug in $r$ and $θ$ and $\phi(t)=ωt$, and compute $\int ds$. For the orbit you can use $θ=π/2$ and approximate $ω$ from $r$ using Newtonian gravity. For the twin on the surface, $ω=1\text{ sidereal day}^{-1}$ and $θ$ is their latitude. The ratio of the two integrals is the average Doppler shift / relative time dilation. – benrg Feb 08 '22 at 18:25
  • Thanks, but I don't think I understand fully. The $ds/dt$ gives the time dilation ratio wrt. a far away observer, correct? Why would the two time dilation ratios wrt. a far away observer give information about their respective ratios? It doesn't necessarily do that in Special Relativity. Is that because they both travel between the same two events? That is, with $\int ds$, you mean to integrate over a full circle of the orbit? How does the planet-bound observer (A) know how long a full orbit is on a non-rotating planet in deep-space (when there is no "sidereal day")? – fishinear Feb 08 '22 at 20:16