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In Ta-Pei Cheng's Relativity, Gravitation and Cosmology book, pg. 88, it was stated that the infinitesimal invariant interval $ds^2=g_{\mu\nu}dx^\mu dx^\nu$ for a spherically symmetric metric $g_{\mu\nu}$ is: $$ds^2=Ad\vec{r}\cdot d\vec{r}+B(\vec{r}\cdot d\vec{r})^2+Cdt(\vec{r}\cdot d\vec{r})+Ddt^2$$ where $A$, $B$, $C$ and $D$ are scalar functions of $t$ and $\vec{r}\cdot\vec{r}$.

Is it correct to say that for a general metric $g_{\mu\nu}$ (not necessarily spherically symmetric), the invariant interval $ds^2$ will be the similar as above, i.e.

$$ds^2=Ad\vec{r}\cdot d\vec{r}+B(\vec{r}\cdot d\vec{r})^2+Cdt(\vec{r}\cdot d\vec{r})+Ddt^2$$ but now with $A$, $B$, $C$ and $D$ being scalar functions of $t$ and $\vec{r}$?

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3 Answers 3

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To describe the most general spacetime metric you need 10 functions (the metric is a symmetric 4x4 matrix). However, using the transformation law for the metric, you can always choose a new coordinate system in which you eliminate 4 of the above 10 functions. Thus, the most general metric (regardless of whether you represent it in spherical coordinates) has 6 free functions. In your example you have only 4 functions: $A, B, C$ and $D$, so it cannot describe the most general spacetime.

By the way, note that the very meaning of the coordinate system is not given a priory. So, before you get to know more about the details of your spacetime, it is meaningless to say, for example, that $\vec{r}$ is the usual radial vector.

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Not quite. There are more possible terms one can add that are forbidden by spherical symmetry. For example, rotations in some direction $\theta$ are characterised by a term $~dtd\theta$ in the metric. Expanding your metric using $d \vec{r} \cdot d \vec{r}=d r^{2}+r^{2}\left(d \theta^{2}+\sin ^{2} \theta d \phi^{2}\right)$, we see that

$$ d s^{2}=A\left[d r^{2}+r^{2}\left(d \theta^{2}+\sin ^{2} \theta d \phi^{2}\right)\right]+B r^{2} d r^{2}+C r d r d t+D d t^{2} $$

Since there are no terms that go like $dtd\theta$ or $dtd\phi$, we know that this can't be the most general metric.

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  • $\begingroup$ Why the downvote? $\endgroup$
    – Akoben
    Nov 17, 2020 at 17:30
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Spherical symmetry implies that the metric is independent of angles, that is, where you are standing on a spherical surface. Basically, entire surface of the sphere (at every r) looks the same everywhere on it. Mathematically, this translates to the metric coefficients, $A$, $B$, $C$ and $D$ being independent of $\theta$ and $\phi$ coordinates.

If you forgo spherical symmetry, in the most general metric, the coefficients $A$, $B$, $C$ and $D$ will be functions of $r$, $\theta$, and $\phi$.

About the dependence on $t$, if a metric is time-independent people generally add a qualifier of steady and/or stationary. So when you are talking about just a spherically symmetric metric, the metric coefficients depend on $t$ anyway in general.

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