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SR tell us that inside ISS time is running slower than on the earth. But velocity is relative. We can choose to consider either ISS or earth to be at rest. An observer inside ISS sees the earth revolve around the ISS. By his frame,time is running slower on the earth. What is really happening?

Qmechanic
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elias2010
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    Look up the twin paradox. The ISS is accelerating so the setup is not symmetric. – Connor Behan Dec 20 '21 at 10:41
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    You've described what's happening and then asked what's happening. ISS clocks run slow in the earth's frame. Earth clocks run slow in the (instantaneous) ISS frame. That's, as you say, what is happening. What is your question? – WillO Dec 20 '21 at 14:06
  • @ConnorBehan: GR tells us that the ISS is closer to being "non-accelerating" than the observers on Earth are. – Michael Seifert Dec 20 '21 at 22:01

2 Answers2

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While velocity is relative, spacetime curvature is not. The Earth and the ISS are not symmetric because the Earth is lower in the gravitational potential well and the gravitational effects are non-negligible. This problem must be treated using general relativity, not special relativity.

For an example of the ISS’ reference frame please see my answer here: The "Satellite Paradox": Twin paradox in orbiting satellites

In particular, equation (6) says: $$ds^2 =-c^2 dt'^2 + r^2 d\phi'^2 + \frac{2 r v}{\sqrt{1-\frac{R}{r}-\frac{v^2}{c^2}}} dt' d\phi'$$ Note that both the $d\phi'^2$ term and the $dt' d\phi'$ term depend on $r$. We can write $$d\tau^2 = -ds^2/c^2 =dt'^2-\frac{r^2}{c^2} d\phi'^2 - \frac{2rv}{c^2 \sqrt{1-\frac{R}{r}-\frac{v^2}{c^2}}} dt' d\phi'$$ $$\frac{d\tau^2}{dt'^2} =1 -\frac{r^2}{c^2} \frac{d\phi'^2}{dt'^2} - \frac{2rv}{c^2 \sqrt{1-\frac{R}{r}-\frac{v^2}{c^2}}}\frac{d\phi'}{dt'}$$ and since for a clock at rest on the ground $d\phi'/dt'=-v/r$ we have $$\frac{1}{\gamma}=\frac{d\tau}{dt'} = \sqrt{ 1 - \frac{v^2}{c^2} + \frac{2v^2}{c^2\sqrt{1-\frac{R}{r}-\frac{v^2}{c^2}}}}$$ The time dilation in the ISS reference frame does not have the familiar form from SR, even in the limit where $\frac{R}{r} \rightarrow 0$

Dale
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None of the answers are correct above. Special relativity and general relativity occur for satellites, at the altitude of the ISS special relativity is the main causal factor. See

https://demonstrations.wolfram.com/RelativisticEffectsOnSatelliteClockAsSeenFromEarth/

or

https://link.springer.com/article/10.12942/lrr-2003-1

The increased relative velocity of the clocks on any satellite needs to be taken into account. Even ones stationary above the earth or the observer such as geostationary !

For the ISS special relativty effect is more than gravitational, so time is slowed and then astronauts are fractionally younger.

"By his frame,time is running slower on the earth"

A rash claim. If they compared clocks after a time I'm sure they'd find his clock slower.

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    Unfortunately your answer is very incomplete, especially compared to other answers to this question. You should expand to included more details so your answer is self-contained rather than expect people to read two papers. – ZeroTheHero Feb 02 '23 at 13:35
  • As it’s currently written, your answer is unclear. Please [edit] to add additional details that will help others understand how this addresses the question asked. You can find more information on how to write good answers in the help center. – Community Feb 02 '23 at 16:50