A twin paradox question

Question

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.

Answer 1

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In a comment you linked an essay by Ron Hatch (seemingly unpublished), and the section "Velocity Effects upon the Clock Rates" in it.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.

Answer 2

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.