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In special relativity, the spacetime interval between the same two events measured by two different inertial observers are numerically the same: $$c^2dt^2-|d\vec{x}|^2=c^2dt^{'2}-|\vec{dx'}|^{2}.$$

In general relativity, does the same thing remain true i.e. if two arbitrary observers measure the same two events, does their spacetime intervals $g_{ab}(t,|\vec{x}|)dx^a dx^b$ and $g_{ab}(t',|\vec{x}'|)dx^{'a} dx^{'b}$ are also numerically equal?

Solidification
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Yes they are the same, but just to be clear, you need to indicate in the notation that the second set of metric components are not the same as the first set of metric components. This can be done by using a prime, as in $$ g_{ab} dx^a dx^b = g'_{ab} dx'^a dx'^b $$ This equality is one of the foundational ideas of General Relativity. It announces that we are dealing with a Riemannian or pseudo-Riemannian manifold. In tensor analysis this expression has the correct form to give an invariant scalar.

Andrew Steane
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  • Yes, your answer is correct, but there seems to be a counter example I can't resolve. Let ds be the proper distance between two LIGO mirrors. Kip Thorne says a GW changes the this ds and that is why LIGO can detect GWs. When a GW wave passes, he says in Local Lorentz coordinates dx changes but not g, while in Transverse Traceless coordinates g changes but not dx. A GW does a transverse space-space squash or parallel-piped strain .... a change of coordinates just like doing a rotation or boost (=space-time parallel-piped strain). Why isn't this ds invariant? – Gary Godfrey Sep 26 '20 at 19:17
  • @GaryGodfrey All frames agree that the ds in this example changes as a g-wave passes by; it is simply that, depending on your choice of frame, you can attribute the change to a movement of mirror relative to the coordinates (so dx changes) or to a mirror staying still in the coordinates but the metric changing. – Andrew Steane Sep 26 '20 at 22:28
  • @AndrewSteane Sorry, I want to come back to your answer once more. When you say, "Yes, they are the same," or "This equality is one of the foundational ideas of General Relativity", do you mean that this equality is a postulate (but other things such as $g_{\alpha\beta}$ behaves as a 2nd rank tensor is derived from it)? – Solidification Dec 30 '23 at 15:33
  • I think that it is a postulate that we cannot derive in GR (and your answer gives me the same impression). Correct me if I misinterpret your answer. In this post, https://physics.stackexchange.com/q/275032/164488 , N0va gave a "proof". But as far as I understand, this is by no means a proof because he assumed that $g_{\alpha\beta}$ transforms like a 2nd rank tensor which itself requires the invaariance of $ds^2$ under general coordinate transformation to start with. – Solidification Dec 30 '23 at 15:39
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    It is a postulate. The postulate can be stated "spacetime is well-modelled by a pseudo-Riemannian manifold". – Andrew Steane Dec 30 '23 at 20:53
  • Thanks for entertaining my question on this very old post. :-) – Solidification Dec 31 '23 at 14:51