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?

  • $\begingroup$ 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
    Commented Aug 18, 2016 at 18:44
  • $\begingroup$ Then the respective answers are yes, always; and no, only in SR. $\endgroup$
    – knzhou
    Commented Aug 18, 2016 at 18:45

5 Answers 5


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.

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    $\begingroup$ +1 because it answers the question nicely and because you really need the rep $\endgroup$
    – Jim
    Commented Aug 18, 2016 at 13:58
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    $\begingroup$ 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
    Commented Aug 18, 2016 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). $\endgroup$ Commented Aug 18, 2016 at 14:22
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    $\begingroup$ 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
    Commented Aug 18, 2016 at 14:26
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    $\begingroup$ As a PhD student in particle physics, I have quite a lot of graduate books in QFT that deals quite heavily in relativity, all of which I remember consider the speed of light being invariant to be a fundamental postulate. You may be correct that PhDs purely in relativity might consider the line element to be more fundamental, but it certainly isn't just an undergraduate thing to think otherwise. And historically the line element was arrived at by Einstein from considering the invariance of the speed of light as a fundamental postulate, not the other way around. $\endgroup$
    – Jack
    Commented Aug 18, 2016 at 14:50

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
    Commented Aug 18, 2016 at 15:19
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    $\begingroup$ I don't think that is a "proof" of the invariance of the spacetime interval. This "proof" assumes that $g_{\alpha\beta}$ behaves as a 2nd rank tensor which itself requires the invariance of $ds^2$ to start with. Experts should correct me. $\endgroup$ Commented Dec 30, 2023 at 15:45
  • $\begingroup$ @Solidification This is a good point. If the tensor nature of $g$ is taken as given, and if it is further taken as a given the coordinates transform like components of a vector, and if it is further taken as a given that the transformation is invertable, then the "proof" can proceed from those axioms very simply. But, usually it is done the other way around... However, this might actually be what OP was asking about... $\endgroup$
    – hft
    Commented Jan 4 at 19:35
  • $\begingroup$ @Solidification I provided an answer that might address your concerns. $\endgroup$
    – hft
    Commented Jan 4 at 21:02
  • $\begingroup$ In my non-expert opinion, the reason why one assumes that $g_{α,β}$ is a tensor is because of the first postulate of special relativity. This reads, "The laws of physics take the same form in all inertial frames of reference." That $g_{α,β}$ is, in particular, a second-rank tensor would need to come from its transformation laws. Epistemologically, if you want to prove true the assumption that $g_{α,β}$ is a tensor, then you first need to prove true the 1st postulate of special relativity. Yet, a postulate is a statement assumed true without proof. $\endgroup$ Commented Jan 5 at 3:34

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 $ds$.


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.

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.)

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: $$ d\tilde \tau^2 = -\tilde g_{\mu\nu}d\tilde x^\mu d\tilde x^\nu $$ $$ =g_{\rho\sigma}\frac{\partial x^\rho}{\partial \tilde x^\mu}\frac{\partial x^\sigma}{\partial \tilde x^\nu}d\tilde x^\mu d\tilde x^\nu $$ $$ = - g_{\mu\nu}d x^\mu d x^\nu $$ $$ =d\tau^2 $$


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.

  • $\begingroup$ "...is indeed the case if the coefficients $g_{\mu\nu}$ transform like a tensor". I agree. But how will you prove/ensure that $g_{\alpha\beta}$ behaves as a second-rank tensor? $g_{\alpha\beta}$ has no other independent expression. I think the only way you can deduce its transformation rule is by starting with the invariance $ds^2$ under general coordinate transformation. I think John Rennie's answer is correct. $\endgroup$ Commented Dec 30, 2023 at 19:23
  • $\begingroup$ @Solidification I only make a reference, I do not proof that the $g_{\mu\nu}$ transform as a tensor. But I think NOva shows it very well in his 5. equation. $\endgroup$ Commented Dec 30, 2023 at 20:18
  • $\begingroup$ In his 5th equation, N0va writes the transformation rule for a general 2nd-rank tensor. But to show that the metric tensor is indeed a 2nd-rank tensor, you have to assume the invariance $ds^{\prime 2}=ds^2$ under general coordinate transformation. There is no way we can show that $g_{\alpha\beta}$ (which makes its appearance through $ds^2$) is a 2nd-rank tensor without assuming $ds^2$ is invariant and therefore, invariance of $ds^2$ is a postulate, at least in GR. The original question was precisely this: whether the invariance of $ds^2$ can be derived in GR or it is a postulate? $\endgroup$ Commented Dec 31, 2023 at 8:18
  • $\begingroup$ @Solidification It is clear for geometrical reasons that the length of a line does not depend on the coordinates. This is what I explain in the first part of my post. If you like you can call it a postulate, but it is just result of differential geometry. $\endgroup$ Commented Dec 31, 2023 at 10:36
  • $\begingroup$ @Solidification To achieve the differential geometrical result the metric tensor is not required. One also see in my example the coordinate independence of $ds$ without using the metric tensor. $\endgroup$ Commented Dec 31, 2023 at 11:04

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