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I understand how to derive the spacetime interval being invariant for Minkowski space, but I've never seen any derivation of it in general curved spacetime. Is the invariance just derived for Minkowski space and then postulated that it holds for all metric tensors in general relativity, or is there a proof to show it is invariant in general relativity?

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  • $\begingroup$ There seems to be a confusion between the two answers below. The first shows that the number $ds^2$ is always invariant. The second shows that the form of $ds^2$, i.e. $dt^2 - dx^2 - dy^2 - dz^2$ is invariant, which is equivalent to saying that the metric tensor components $g_{\mu\nu}$ are invariant, under Lorentz transformations. $\endgroup$ – knzhou Aug 18 '16 at 18:44
  • $\begingroup$ Then the respective answers are yes, always; and no, only in SR. $\endgroup$ – knzhou Aug 18 '16 at 18:45
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Lets look at an arbitrary invertable coordinate transformation: $$ x^\mu \rightarrow x'^{\mu}=x'^{\mu}(x^\nu). $$ The corresponding Jacobian $\Lambda$ $$ \Lambda^\mu_{~~\rho}=\frac{\partial x'^{\mu}}{\partial x^{\rho}}$$ is invertable $$ \Lambda_{\sigma}^{~~\nu}=\frac{\partial x^{\nu}}{\partial x'^{\sigma}}.$$ A vector tansforms like $$x'^\mu=\frac{\partial x'^{\mu}}{\partial x^{\sigma}}x^\sigma=\Lambda^\mu_{~~\sigma}x^\sigma.$$ The defining property of a tensor of second rank (the metric tensor is such a tensor) is that it transforms like $$g'_{\rho\sigma}=\frac{\partial x^{\mu}}{\partial x'^{\rho}}\frac{\partial x^{\nu}}{\partial x'^{\sigma}}g_{\mu\nu}=\Lambda_{\rho}^{~~\mu}\Lambda_{\sigma}^{~~\nu}g_{\mu\nu}.$$

With that in mind lets give this tensor calculus a go on our line element:

\begin{align} ds'^2 & = g'_{\mu\nu}dx'^\mu dx'^\nu \\\\ & =g'_{\mu\nu}\Lambda^\mu_{~~\rho}\Lambda^\nu_{~~\sigma}dx^\rho dx^\sigma\\\\ & = g_{\mu\nu}dx^\rho dx^\sigma \\\\ & = ds^2.\end{align} That would be the textbook calculation for the invariance of the line element using tensor calculus. To prove the transformation property of a second rank tensor one would express everything via base vectors and use the relations of those base vectors.

So the invariance of the line element is more a feature of tensor calcus. A scalar is invariant under coordinate transformations that has nothing to with special or general relativity.

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  • $\begingroup$ Thanks very much, that's a lot simpler than I'd thought it would be. $\endgroup$ – Jack Aug 18 '16 at 15:19
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You cannot derive the invariance of the line element because it is one of the assumptions on which relativity (both flavours) is based. When you say:

I understand how to derive the spacetime interval being invariant for minkowski space

I would guess you mean that you can show the Lorentz transformations preserve the line element. However most of us would take the view that the invariance of the line element was more fundamental, then derive the Lorentz transformations from the requirement that the line element be preserved.

There isn't a simple equivalent to the Lorentz transformations in general relativity. The Lorentz transformations are a coordinate transformation but a very simple one where the transformation is between inertial frames in flat spacetime. While we use coordinate transformations extensively in GR they are usually far more involved than the Lorentz transformations.

However in GR, just as in SR, the invariance of the line element:

$$ ds^2 = g_{\alpha\beta}x^\alpha x^\beta $$

always applies though the metric $g_{\alpha\beta}$ is generally more complicated.

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  • $\begingroup$ +1 because it answers the question nicely and because you really need the rep $\endgroup$ – Jim Aug 18 '16 at 13:58
  • $\begingroup$ Thanks for the answer, but I have to disagree that you can't derive the spacetime interval in special relativity. From the postulates of homogeneity of spacetime, isotropy of space and the invariance of the speed of light it is trivial to derive the invariance of the spacetime interval in special relativity, as is done on page 4 of Landau's classical theory of fields. There is no need to introduce lorentz transformations to show it is invariant. $\endgroup$ – Jack Aug 18 '16 at 14:19
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    $\begingroup$ @Jack: again I would look at it the other way round. Starting from the invariance of the line element it's trivial to show that the speed of light is a constant for all observers. It just depends on what quantities you regard as fundamental and what quantities are derived. My view is that the invariance of the line element is the most fundamental principle in relativity (both flavours). $\endgroup$ – John Rennie Aug 18 '16 at 14:22
  • $\begingroup$ You're completely correct of course, though it is much more common to consider the invariance of the speed of light as a fundamental postulate of special relativity than the line element (and historically that is how special relativity was first arrived at). $\endgroup$ – Jack Aug 18 '16 at 14:26
  • $\begingroup$ @Jack: I know a fair few people working in relativity, and they all consider the invariance of the line element more fundamental. Undergraduate textbooks don't do this, which I think is silly because it makes SR easier to understand. I confidently predict that if you go on to a PhD in relativity you'll end up agreeing with me :-) $\endgroup$ – John Rennie Aug 18 '16 at 14:37
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The spacetime interval is a concept of Minkovski spacetime. It does also appear in general relativity in its infinitesimal form $ds$, as the principles of special relativity apply locally within curved spacetime of general relativity. In general relativity, the distances between two points in curved spacetime are described by geodesics or by a path integral over $ds$.

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