In Schwarzschild geometry, we often say that when $r<2M$, $r$ becomes timelike and $t$ becomes spacelike. While I understand that this refers to the metric and how it behaves for a radial worldline if r=constant or t=constant, I do not understand what physical consequences this might have. Would that mean that for a time-separation we'd have to measure dr? Would t take values of 1M, 1.5M etc?
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1$\begingroup$ The Schwarzschild Droste coordinates use stationary local observers who have to be lightlike (photons) at the horizon and spacelike (tachyons) behind it, and for tachyons the spacetime axes are flipped. For timelike (human) local observers like you have them in Gullstrand Painlevé or Eddington Finkelstein coordinates that is not the case though, see physics.stackexchange.com/a/565144/24093 $\endgroup$– YukterezCommented Sep 28 at 1:11
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$\begingroup$ @Yukterez “The Schwarzschild Droste coordinates use stationary local observers” - This is not true. The Schwarzschild coordinates are just an abstract standard polar coordinate system with no observers or, if you prefer, with just one asymptotically at infinity. The Droste coordinates differ by defining the timelike radial interval inside as a spacelike reduced circumference, which is absurd. Sadly this absurd is now widely accepted and even called “the Schwarzschild coordinates”. $\endgroup$– safesphereCommented Sep 28 at 4:18
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$\begingroup$ @AgathaHarkness Physical time at a remote location is not necessarily pointed to where your time is pointed. As an object falls to a black hole, the time direction of the object gradually turns toward the black hole. At the horizon this turn becomes 90 degrees. Now the time of the object points inside the black hole, so $r$ becomes a direction in time instead of space. When you ask about the time separation, you need to be more specific by saying whose clock you are using to measure and what specific separation you mean. However just in general $r$ inside is timelike and so is $dr$. $\endgroup$– safesphereCommented Sep 28 at 4:36
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$\begingroup$ You cannot deduce "physical consequences" from coordinate-dependent quantities (like the Schwarzschild metric). You need to find invariants, like this: en.wikipedia.org/wiki/Kretschmann_scalar. Unfortunately there is not a lot of discussion about these outside academic circles. $\endgroup$– m4r35n357Commented Sep 28 at 10:22
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$\begingroup$ @safesphere - if you don't believe me google this. You're also wrong about the 90°, that happens at the singularity, at the horizon it's only 45°. $\endgroup$– YukterezCommented Sep 28 at 12:06
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