# The spherically symmetric metric and the general metric

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 $$\vec{r}$$?

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.

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.

• Why the downvote? Nov 17, 2020 at 17:30

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.