How to show the spacetime interval is invariant in general?

Question

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?

Answer 1

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:

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

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

Answer 2

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.

Answer 3

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 $ds$.

Answer 4

The line element $ds$ respectively its integral $s=\int_{s_1}^{s_2} ds$ measures the length of (a segment of) a (world)line in space(time), i. e. it has very simple geometrical meaning. The space(time) can be 2-dimensional, for instance the 2-dimensional surface of a sphere, or 3-dimensional as in 3D Euclidean space, or 4-dimensional in flat Minkowski-space or even 4-dimensional in curved spacetime. There is no apparent difference.

Obviously, the line element is coordinate-independent, it makes no sense that the length of a (segment of a) (world)line would depend on the coordinate system. And the answer of "NOva" demonstrates that this is indeed the case if the coefficients $g_{\mu\nu}$ transform like a tensor. Then we have the right to call the ensemble of these coefficients "metric tensor". The independence of the line element from the chosen coordinate system is called "invariance of the line element".

For instance one can take a line element (let's say a segment that corresponds to half a circle) on a circle -- in order to reduce the algebra of the computation we only consider a very simple 2D case -- and express it in polar coordinates and cartesian coordinates. In cartesian coordinates one has of course to take into account that the line considered lies on a circle, i.e. it fulfills the contraint $r^2 = x^2 + y^2$.

The line element in cartesian coordinates in 2D Euclidean space is:

$$ds^2 = dx^2 + dy^2 \quad\text{with the contraint}\quad r^2 = x^2 + y^2$$

We solve for $x$:

$$ x = \sqrt{r^2 - y^2} \quad\text{consequently}\quad dx = - \frac{ydy}{\sqrt{r^2 - y^2}}$$

We plug this into the cartesian line element:

$$ds^2 = \frac{y^2 dy^2}{r^2 -y^2} + dy^2 = \frac{r^2}{r^2 - y^2} dy^2$$

We want to know the length of the line segment of half a circle, in the $y$-coordinate it is $y_1 =r$ to $y_2=-r$ assuming that the origin of the cartesian coordinate system is in the middle of the circle. Then (integral substitution $u = y/r$):

$$ s = \int_{-r}^r \frac{r}{\sqrt{r^2 -y^2}} dy = r \int_{-1}^{1} \frac{u}{\sqrt{1-u^2}} du = r \arcsin(u)|^1_{-1} = r(\frac{\pi}{2} -(-\frac{\pi}{2}))=r \pi$$

Now we compute it in polar coordinates. Actually this problem is only one-dimensional, i.e. our line element is only

$$ds^2 =r^2 d\varphi^2$$

Taking the integral over $[0,\pi]$ yields:

$$ds = r \int_0^{\pi} d\varphi = r\pi$$

So we see that the result is the same. A demonstration of the fact that the line-element is independent of the coordinates. As little detail note that in the given example the line element on a circle is curved.

Answer 5

The answer by N0va, proves that $ds^2$ is invariant in GR. I doubt that this really "proves" the invariance of $ds^2$ in GR because he used the transformation properties of $g_{αβ}$ to "prove" $ds^2=ds′^2$. But the transformation rule of $g_{αβ}$ itself rests on the invariance of $ds^2$ under general coordinate transformation.

The short answer is that in Special Relativity the invariance of $ds^2$ can be shown to imply the Lorentz transformations or equivalently the form of of the Lorentz transformation can be shown to imply the invariance of $ds^2$. Then, when we go to General Relativity we demand that the equations in the absence of gravity reduce to Special Relativity and that equations preserve their form under general coordinates transformations. In other words, N0va's "proof" seems to be "fine" in the sense that N0va just chooses one of the directions of implication for the "Special Relativity component" (and it seems to be just a matter of opinion as to which direction is "more fundamental") and then the rest of the proof is just rote transformation from locally Euclidean coordinates to general coordinates.

Following Weinberg's "Gravitation and Cosmology: Principles and Applications of the General Theory of Relativity," a number of relevant equations are provided below to further elucidate my short answer above.

Recall, per Weinberg that

[i]t was Gauss's great contribution to pick out one particular class of metric spaces... Gauss assumed that in any sufficiently small regions of space it would be possible to find a locally Euclidean coordinate system $(\xi_1,\xi_2)$ so that the distance... $$ds^2 = d\xi_1^2 + d\xi_2^2$$

The above locally (but not necessarily global) Euclidean requirement further implies that (and as it turns out is also implied by) for some other coordinates $(x_1, x_2)$ that may cover the whole space we have: $$ ds^2 = g_{11}dx_1^2 + 2g_{12}dx_1dx_2 + g_22 dx_2^2 $$

This is important in General Relativity since the Equivalence Principle asserts that it is always possible "to choose a locally inertial coordinate system, where the laws of nature take the same form as in unaccelerated Cartesian coordinate systems in the absence of gravitation." (Weinberg Chapter 3.)

However, before running straight at this, let's recall a little bit about Special Relativity.

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In Special Relativity, a Lorentz transformation can be defined as a linear transformation with constants $\Lambda^{\alpha}_\beta$ that are required to satisfy: $$ \Lambda^{\alpha}_\beta \Lambda^{\gamma}_\delta\eta_{\alpha\gamma}=\eta_{\beta\delta}\;,$$ where $\eta = diag(-1,1,1,1)$ is a constant diagonal matrix.

And, given this definition, it can be shown that Lorentz transformations leave the proper time $$ d\tau^2 = -\eta_{\alpha\beta}dx^\alpha dx^\beta $$ invariant.

On the other hand, it can also be shown that Lorentz transformations are the only non-singular coordinate transformation that leave $d\tau$ invariant.

(N.b., Weinberg proves both directions of the implications in his Chapter 2.)

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Now, back to General Relativity.

The Principle of Equivalence tells us that there is some coordinate system $\xi^\alpha$ that can describe a particle moving under gravitational forces by: $$ \frac{d^2\xi^\alpha}{d\tau^2}=0\;, $$ where $d\tau^2 = -\eta_{\alpha\beta}d\xi^\alpha d\xi^\beta$.

In any other coordinate system $x^\mu$ we have: $$ \frac{d^2 x^\alpha}{d\tau^2}+\Gamma^{\alpha}_{\mu\nu}\frac{dx^\mu}{d\tau}\frac{dx^\nu}{d\tau}=0\;, $$ where $\Gamma^{\alpha}_{\mu\nu}$ is the affine connection.

And in this arbitrary coordinate system we have, by definition: $$ d\tau^2 = -g_{\mu\nu}dx^\mu dx^\nu\;, $$ where $$ g_{\mu\nu} \equiv \frac{\partial \xi^\alpha}{\partial x^\mu}\frac{\partial \xi^\beta}{\partial x^\nu}\eta_{\alpha\beta}\tag{A} $$ and where the above also ensures that $g_{\mu\nu}$ has an inverse: $$ g^{\sigma\nu} = \eta^{\alpha\beta}\frac{\partial x^\nu}{\partial \xi^\alpha}\frac{\partial x^\sigma}{\partial \xi^\beta}\;. $$

The rules of partial differentiation show that the coordinate differentials $dx^\mu$ are generally contravariant vectors and that the partial derivatives are generally covariant vectors.

Similarly, by Eq. (A), we can show (see Weinberg Eq. (4.2.6)) that $g_{\mu\nu}$ is a generally covariant tensor, $$ \tilde g_{\mu\nu} = g_{\rho\sigma}\frac{\partial x^\rho}{\partial \tilde x^\mu}\frac{\partial x^\sigma}{\partial \tilde x^\nu}\;.\tag{B} $$

So, we have, in the tilde frame: $$ d\tilde \tau^2 = -\tilde g_{\mu\nu}d\tilde x^\mu d\tilde x^\nu $$ $$ =g_{\rho\sigma}\frac{\partial x^\rho}{\partial \tilde x^\mu}\frac{\partial x^\sigma}{\partial \tilde x^\nu}d\tilde x^\mu d\tilde x^\nu $$ $$ = - g_{\mu\nu}d x^\mu d x^\nu $$ $$ =d\tau^2 $$