I'm studying General Relativity using Ray D'Inverno's book "Introducing Einstein's relativity". I don't understand what the author writes in paragraph 14.3 ("Static solutions") where he proves that for a static spacetime there are no cross-terms $dx^0dx^\alpha$ in $(ds)^2$, where $x^0$ is a timelike coordinate and $x^\alpha$ is a spacelike coordinate. He says that

"the assumption that the solution is static means that $(ds)^2$ is invariant under a time reversal about any origin of time"

but a time reversal is a coordinate transformation and $(ds)^2$ is a scalar, so it is automatically invariant under a time reversal (for it is a coordinate transformation), and not only specifically for a static metric. Is there anyone who can resolve my doubt?


2 Answers 2


Indeed, what the author should have said is that the components of the metric have to remain invariant; that's what an isometry is. The scalar $ds^2$ is always invariant If there is a term

$$ds^2 = \cdots + 2 g_{0i}\, dt\, dx^i + \cdots$$

in the metric, then under a transformation $t \to -t$ the component $g_{0i}$ also has to change sign to keep $ds^2$ invariant. So if we demand that the components of the metric don't change, the cross term has to be zero.

  • $\begingroup$ Thank you very much! $\endgroup$
    – Al01
    Sep 14, 2023 at 16:20

I think I can provide an answer, so I will try... Let the infinitesimal line element in a spacetime be


where I assume that $g_{\mu\nu}$ is a function of $(x^0,x^i),\ i=1,...,d-1$ with $(d-1)$ being the number of spatial dimensions of your spacetime.

If the spacetime is static, then the various components of the metric does not depend on the time-like coordinate, i.e. $x^0$ $$d\tilde{s}^2\rightarrow ds^2=g_{00}(x^i)(dx^0)^2+2g_{0i}(x^i)(dx^0)(dx^i)+...$$ Furthermore, as you correctly state, static also implies that the line element

is invariant under a time reversal about any origin of time

This simply means that if I pefrom the transformation $x^0\rightarrow-x^0\Rightarrow dx^0\rightarrow-dx^0$, then the cross terms in the infinitesimal line element $$g_{0i}(dx^0)d(x^i)\rightarrow -g_{0i}(dx^0)d(x^i)$$ with the remaining terms remaining unaffected by this coordinate transformation. The only way the line element remaining the same under such "time-reversal" transformation, is $g_{0i}=0$.

I hope this helps.

EDIT: So, after reading your comments, I think I can add something to clarify some concerns of yours (if I understand them correctly). Assume a spacetime that is not static, i.e. $g_{\mu\nu}=g_{\mu\nu}(x^0,x^i)$. Then, yes, you are right, performing time reversal transformations would not alter the (scalar) line element in any way and $$g_{0i}(x^0,x^i)dx^0dx^i\rightarrow -g_{0i}(-x^0,x^i)dx^0dx^i$$ This is not really helpful, since you don't know the dependence of $g_{\mu\nu}$ on the temporal coordinate. So, the invariance of the metric in this (a bit more complicated case) can only imply that $$g_{0i}(x^0,x^i)= -G_{0i}(-x^0,x^i)$$ where I have used another letter (capital G) to denote the new metric in the transformed coordinates. So, saying that the spacetime is static means that we force the metric elements to be time-independent, such that when we impose "time-reversal" symmetry on the scalar infinitesimal line element, we obtain this vanishing of the metric cross terms.

  • $\begingroup$ This is precisely the way D'Inverno proves that cross terms vanish... the only thing I don't understand in his proof (which is identical to yours) is when he says: "since the metric is static the line element is invariant under a time reversal". But, I think, a time reversal is a coordinate transformation, and the line element is a scalar, so it must be invariant under coordinate transformation (= definition of a scalar), aside from the metric being static or not... I'm sure that my error is in a wrong consideration of time reversal as a coordinate transformation, but I can't find the error... $\endgroup$
    – Al01
    Sep 14, 2023 at 10:16
  • 1
    $\begingroup$ So my problem is not in the understanding the mathematics $\endgroup$
    – Al01
    Sep 14, 2023 at 10:19
  • $\begingroup$ Okay, I see... I will try to edit and if this covers you, then ok. If not, please leave another comment $\endgroup$
    – schris38
    Sep 14, 2023 at 10:21
  • $\begingroup$ There is a problem. We have to distinguish between static and stationary metrics. Stationary metric is a metric which admits a coordinate system where the metric is time-independent. If we assume that $ds^2$ is always invariant since it is a scalar, we end up concluding that a stationary metric is always static (it is sufficient to repeat your proof in the coordinate system in which the metric is time independent). But D'Inverno adds that there are metrics which are stationary but not static... $\endgroup$
    – Al01
    Sep 14, 2023 at 10:49
  • $\begingroup$ You can read this about static and stationary metrics: physicsforums.com/threads/… . It turns out that a static metric is also stationary, so a metric being static is stronger than it being stationary $\endgroup$
    – schris38
    Sep 14, 2023 at 11:35

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