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Imagine twin A sits on top of Mount Everest. Twin B is in a jet that flys westward at the speed of the rotating earth, so is always directly under the sun, at the same altitude as Mount Everest so there is no gravitational time dilation issue. They wave at each other every day as they pass. They do this for one year, so we have the situation where A has travelled a greater distance than B, with the difference being the circumference of the earth x 365.

After one year, who is younger and how do we reconcile two conflicting opinions? Under normal twin paradox thoughts, I would think that since A has travelled much further, and is the one who has gone "out and back" 365 times, A should be the younger twin. But based on Ron Hatch's paper on time dilation with ICBMs referencing GPS satellites for their positioning, and GPS satellites reference an earth based clock for their adjustment, I would think that B is the younger twin.

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    $\begingroup$ Where are you stuck? $\endgroup$
    – WillO
    Sep 15, 2022 at 21:46
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    $\begingroup$ Who is younger after one year? The second paragraph lays out the problem. $\endgroup$ Sep 15, 2022 at 21:56
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    $\begingroup$ And where are you stuck? $\endgroup$
    – WillO
    Sep 15, 2022 at 22:26
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    $\begingroup$ I don't think I can explain the second paragraph any clearer. Most answers to the twin paradox would say that A is younger because he has travelled further and has gone "out and back". But Hatch's use of GPS satellites shows (to him at least) that there is some sort of ether and B would be younger. How do we reconcile these conflicting opinions? $\endgroup$ Sep 15, 2022 at 22:30
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    $\begingroup$ The basis of his theory is that missiles that rely on GPS satellites for positioning do not set their internal clocks relative to the GPS satellite clock, as you would think necessary with special relativity. Rather they set their clocks relative to an Earthbound clock. He worked for the GPS system prior to his death so he knows exactly what he was talking about. His theory of ether is controversial, but he was right on with his knowledge of clocks in GPS satellites and missiles. This goes right to the heart of my question. $\endgroup$ Sep 15, 2022 at 23:04

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Your analysis of your experiment is correct. The gravitational field can be approximated reasonably well by the Schwarzschild metric. (Frame dragging is negligible since the Earth's rotation speed is so small, and I'll neglect the mass of Everest.) Everything happens at a fixed $r$, and you can orient the coordinates so $θ$ is also constant, and the remaining metric is just

$$dτ^2 = \left(1-\frac{2M}{r}\right)\,dt^2 - r^2 \sin^2 θ\, d\phi^2 \quad \text{or} \quad \left(\frac{dτ}{dt}\right)^2 = 1-\frac{2M}{r} - r^2 \sin^2 θ\, \left(\frac{d\phi}{dt}\right)^2$$

which shows that your aging rate ($dτ/dt$) is maximal when your velocity relative to the fixed stars ($r\sin θ\,d\phi/dt$) is zero.

In a comment you linked an essay by Ron Hatch (seemingly unpublished), and the section "Velocity Effects upon the Clock Rates" in it. Hatch argues that a magazine article by Neil Ashby is wrong. Perhaps it is; I think that's too far removed from your question to discuss. But fundamentally, the way in which GPS satellites are synchronized doesn't have any connection to the aging of your twins. The purpose of GPS is to broadcast signals that can be used to derive one's spacetime coordinates in an agreed-upon coordinate system. The clocks on the satellites need to keep the time of that coordinate system so that they can broadcast it. The satellites don't need to know their own elapsed proper time, nor the coordinate time of some random inertial system in which they're instantaneously at rest. That doesn't mean there is anything wrong about those quantities. They're just not relevant to the satellites' purpose.

If both of your twins had wristwatches that got the current time from the GPS system, then each time they met, their wristwatches would show the same time. Nevertheless, the earthbound twin would age less between meetings. If it were humanly possible to notice such small differences, then the earthbound twin would see their own wristwatch ticking slightly faster than the twin on the plane would see their own wristwatch ticking.

Hatch seems to believe that the Earth-centered nonrotating coordinate system used by GPS is the rest frame of the luminiferous ether, and something goes wrong if you try to do physics in other coordinate systems. He's wrong about that. But he's right about another thing: it's perfectly fine to use that coordinate system to solve any problem about satellites, twins, etc. You are not obliged to pick a coordinate system in which certain objects are at rest, as some other people seem to think.

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  • $\begingroup$ It is the first sentence in your last paragraph that led to my question in the first place. I'm trying to sketch it out on paper to try to understand it all. Would it be right that if Hatch is correct about the ether, then the formula you are using would not be applicable? (And I'm certainly not saying that he is correct. Just that if he is correct, then the formula would not be applicable?) $\endgroup$ Sep 16, 2022 at 20:17
  • $\begingroup$ @foolishmuse He doesn't seem to disagree with any calculation done in those coordinates (which I think are what he calls the Earth-centered inertial (ECI) frame). I think he would agree with the formula in this answer and the conclusion that the twin on Everest ages less. He just thinks that something goes wrong in other coordinate systems, e.g., that the speed of light isn't isotropic in local inertial frames moving wrt the ECI frame. I think it's the usual confusion about relativity of simultaneity. $\endgroup$
    – benrg
    Sep 16, 2022 at 21:48
  • $\begingroup$ When I look at his paper again, I notice his use of the words "non-rotating frame." Of course this is something I would normally ignore as being unimportant. But referring back to my original question, if the ether frame was rotating with the Earth, then the answer would have been opposite, and the twin on the jet would be younger. $\endgroup$ Sep 19, 2022 at 20:57
  • $\begingroup$ @foolishmuse You're right, for at least some interpretations of a rotating ether. But that would be inconsistent with various experiments, including GPS, and Hatch knew that (as an expert on relativistic effects in the GPS system). For instance, he mentions the Sagnac effect and asserts that it exists; it wouldn't exist if the ether rotated. $\endgroup$
    – benrg
    Sep 19, 2022 at 21:14
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To find out the time dilation we start with the Schwarzschild metric for the non-rotating frame (i.e. the frame where the distant stars are not rotating): $$ds^2=\frac{\mathrm{d}r^2}{1-\frac{2 M}{r}}+\mathrm{d}t^2 \left(\frac{2 M}{r}-1\right)+\mathrm{d}\theta ^2 r^2+\mathrm{d}\phi ^2 r^2 \sin ^2(\theta )$$ for this problem we can simplify with $\mathrm{d}r = \mathrm{d}\theta = 0$ to get $$ds^2=\mathrm{d}t^2 \left(\frac{2 M}{r}-1\right)+\mathrm{d}\phi ^2 r^2 \sin ^2(\theta )$$ from which we can calculate the time dilation as $$\frac{\mathrm{d}\tau}{\mathrm{d}t}=\frac{-\mathrm{d}s}{\mathrm{d}t}=\sqrt{1-\frac{2 M}{r}-r^2 \sin ^2(\theta ) \left(\frac{\mathrm{d}\phi}{\mathrm{d}t}\right)^2}$$ So in this frame A is rotating with $\mathrm{d}\phi/\mathrm{d}t=\omega$ and B is not rotating with $\mathrm{d}\phi/\mathrm{d}t=0$ which we can plug in to get $$\frac{\mathrm{d}\tau_A}{\mathrm{d}t}=\sqrt{1-\frac{2 M}{r}-r^2 \omega ^2 \sin ^2(\theta )} \mathrm{\ \ \ and \ \ \ } \frac{\mathrm{d}\tau_B}{\mathrm{d}t}=\sqrt{1-\frac{2 M}{r}}$$

Now, if we are feeling particularly masochistic then we can check to make sure that everything works out in the rotating coordinate system also. We can do that with the coordinate transform $\theta \rightarrow \theta + \Omega t$. This gives us the "simplified" metric $$ds^2=\mathrm{d}t^2 \left(\frac{2 M}{r}+r^2 \Omega ^2 \sin ^2(\theta )-1\right)+2 \mathrm{d}t \ \mathrm{d}\phi \ r^2 \Omega \sin ^2(\theta )+\mathrm{d}\phi ^2 r^2 \sin ^2(\theta )$$ from which we get the time dilation in the rotating frame $$\frac{\mathrm{d}\tau}{\mathrm{d}t}=\sqrt{1-\frac{2 M}{r}-r^2 \sin ^2(\theta ) \left(\frac{\mathrm{d}\phi}{\mathrm{d}t}+\Omega \right)^2}$$ So in this frame A is rotating with $\mathrm{d}\phi/\mathrm{d}t=\omega-\Omega$ and B is rotating with $\mathrm{d}\phi/\mathrm{d}t=-\Omega$ which we can plug in to get $$\frac{\mathrm{d}\tau_A}{\mathrm{d}t}=\sqrt{1-\frac{2 M}{r}-r^2 \omega ^2 \sin ^2(\theta )} \mathrm{\ \ \ and \ \ \ } \frac{\mathrm{d}\tau_B}{\mathrm{d}t}=\sqrt{1-\frac{2 M}{r}}$$ Note that this result holds for any $\Omega$, not just $\Omega = \omega$. Note also that $\mathrm{d}\tau/\mathrm{d}t$ is not generally frame invariant, instead, $\tau$ is frame invariant and in this specific instance we did not transform our $t$ coordinate so it has the same meaning in both frames. We could have also transformed the $t$ coordinate, and then the time dilation rates would be different but when integrated over one "loop" then they would have been the same as in the non-rotating frame.

Now this means that regardless of whose frame we are using A and B agree on whose clock is ticking more slowly. This can be disconcerting for students of relativity that expect everything to be symmetric. The point is that A and B are not symmetric, and they know that they are not symmetric. This is not a pair of inertial observers in flat spacetime. They are non-inertial (both of them) and in curved spacetime.

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  • $\begingroup$ I appreciate you answer, but there is an old saying "know your audience". Most of it is literally Greek to me. I think the answer comes down to your final sentence in your final paragraph. Could you please expand on that, especially since the Mount Everest twin is swinging around the earth, and we are not concerned with the gravitational time dilation, but only kinetic time dilation. I'd like to understand that better. $\endgroup$ Sep 17, 2022 at 18:25

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