Derivation of the Schwarzschild metric: why are $g_{22}$ and $g_{33}$ the same as for flat spacetime?

Question

I'm trying to understand the derivation of the Schwarzschild metric from Wikipedia, but I simply do not understand why, therein, $g_{22}$ and $g_{33}$ must be those of the flat spacetime.

Couldn't $g_{22}$ and $g_{33}$ have any other radial dependence than that of the flat space? If $g_{22}$ and $g_{33}$ were only dependent on $r$ (arbitrary, how exactly), that would be spherical symmetric as well, I suppose.

Why are they set to the coefficients of the flat space time?

Addendum: I especially don't see why they can't be another function of r. For example how about $g_{22}=A(r)r^2 d\theta^2$ and $g_{33} = A(r)r^2sin^2\theta$? That would be also spherically symmetric as $g_{22}$ and $g_{33}$ only depend on $r$. Could those second A(r)s simply be transformed to the flat spacetime coefficients? Please, show how, in detail.

Addendum 2: Meanwhile, I came across a coordinate change in Eddingtons Mathematics of Relativity. They start with U(r), V(r) , W(r) as prefactors for radial, tangential, and temporal component, respectively (as this is indeed the most general sperically symmetric metric). Then, they do the coordinate transformation r1^2=r^2V(r) and end up with only U and W and simply r1 instead of r as radial coordinate. However, now I don't see why the r1 should still be the radial coordinate of normal spherical coordinates. It's totally messed up if V(r) is complicated, isn't it? However, in the derivation of the Schwarzschild metric, it's treated as the normal spherical symmetric radial coordinate.

Answer 1

I simply do not understand why, there, $g_{22}$ and $g_{33}$ must be those of the flat spacetime.

In fact, it is not necessary that $g_{22}$ and $g_{33}$ take on the form of the angular part of the metric for flat space in spherical coordinates, even for the Schwarzschild solution. The reason is that you can choose to describe a geometry in any coordinates you want to, so you always can make the metric look arbitrarily complicated by doing a coordinate transformation. As a result, you should not be looking for a logical reason why the metric must take a particular form.

Instead, you should ask whether the metric can take on a particular form. The idea of the usual derivation of the Schwarzschild solution in Schwarzschild coordinates, is to take advantage of the spherical symmetry of the problem to guarantee that you can write the angular part of the metric as a function of $r$ times the metric induced on a 2-sphere. That function can then be fixed by an additional radial coordinate transformations.

Answer 2

The Schwarzschild solution is a spherically symmetric solution produced by a central source. This means that at $t = \mathrm{const}$ the metric should be invariant under rotations.

\begin{equation} ds^{2} = -A_{1} (r,t) c^{2} dt^{2} + A_{2}(r,t) dr^{2}+A_{3}(r,t) dr dt +A_{4}(r,t)(d\theta^{2} + \sin^{2}\theta \,d\phi^{2}) \end{equation}

From gauge invariance we can choose $t = f_{1} (\tilde{t}, \tilde{r})$, $r = f_{2} (\tilde{t}, \tilde{r})$ so that $\tilde{A}_{3} = 0$ and $\tilde{A}_{4} = \tilde{r}^{2}$. Then if we forget about the new notation and refer at the new coordinates simply by $t$ and $r$, at constant $t$ and $r$ we have \begin{equation} d\sigma^{2} = r^{2} (d\theta^{2} + \sin^{2} \theta \,d\phi^{2}) \end{equation}

This argument is the one presented by M. Gasperini in https://link.springer.com/book/10.1007/978-3-319-49682-5

Answer to comment:

Suppose that $d\sigma^{2} = C(r)r^{2} (d\theta^{2} + \sin^{2} \theta \,d\phi^{2})$, then you can choose new coordinates \begin{equation} \tilde{r}^{2} = C(r)r^{2} \end{equation} so that $dr^{2} = F(\tilde{r}) d\tilde{r}^{2}$, where $F(\tilde{r})$ include the function $C(\tilde{r})$ and its derivative expressed as functions of $\tilde{r}$. The metric then become

\begin{equation} ds^{2} = -A_{1} (\tilde{r},t) c^{2} dt^{2} + A_{2}(\tilde{r},t)F(\tilde{r})d\tilde{r}^{2}+\tilde{r}^{2}(d\theta^{2} + \sin^{2} \theta \,d \phi^{2}) \end{equation}

At this point we can simply drop the tilde for the notation and call $A_{2}(\tilde{r},t)F(\tilde{r}) \equiv A_{2}(r,t)$.