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In special relativity, we know that if an inertial observer $A$, having a relative motion WRT an inertial observer $B$, attributes, say, time dilation and length contraction to respectively $B$'s clocks and rulers, $B$ can do the same to $A$'s too.

However, I want to know why this is not the case in general relativity. Assume $A$ is located on a massive planet, and $B$ is a Schwarzschild observer both at rest WRT the planet. If $B$ measures $A$'s clocks run slower and his ruler contracted, $A$ detects vice versa, i.e., $B$'s clocks run faster and his ruler is possibly expanded. Why is this the case? What happens to symmetry? Is it because, in the latter example, one observer ($A$) is non-inertial and the other one ($B$) is inertial?

Moreover, please explain which ruler (radial or tangential) is contracted or expanded, and in which directions the speed of light is measured smaller or greater than $c$ from the viewpoint of both $A$ and $B$.

As far as I remember, radial rulers are left unchanged from the viewpoint of both of these observers. Is it correct?

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  • $\begingroup$ By a "Schwarzschild observer", do you mean "an observer at rest at a great distance from the planet"? (I'm guessing so, since otherwise observer $B$ isn't inertial any more than observer $A$ is.) $\endgroup$ Dec 10, 2019 at 22:40
  • $\begingroup$ @MichaelSeifert Yes, I do. $\endgroup$ Dec 11, 2019 at 7:39

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In the special relativity case, neither frame is privileged. You can take either as at rest and the other moving. Or neither at rest, choosing some other frame as the origin. You can move between them with a simple change of velocity.

You can't do that with the observer on the surface of the planet. The planet breaks the symmetry. That observer is in a gravity well. Space-time is distorted there. Both observers will agree on this. To the observer outside the well, things appear slower inside the well. To the observer inside the well, things appear faster outside the well.

Or to put it another way, there's only one planet. You can't transform it away by a change of coordinates.

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  • $\begingroup$ I added a sentence in my question that somehow verifies your perspective, just before you send your answer. Thanks. $\endgroup$ Dec 10, 2019 at 21:42

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