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

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.

• In raindrop coordinates not only gθθ and gφφ but also grr are euclidean, but for the price of a gtr crossterm. Aug 7 at 8:14

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.

• "... 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." Could you please explain in detail, how it can be "fixed"? For example, $ds^2=A(r)dr^2+A(r)r^2(d\theta^2+sin^2\theta d\phi) + B(r)dt^2$ would be also spherically symmetric. How can the second $A(r)$ be "fixed" then to make it the original again? I don't see, how that can be changed to $ds^2=A(r)dr^2+r^2(d\theta^2+sin^2\theta d\phi) + B(r)dt^2$ again. Aug 7 at 7:34
• @Scibo It's been a while since I've worked through the derivation, but I believe that you use a radial coordinate transformation to set the coefficient of $r^2(d\theta^2+\sin^2\theta d\phi^2)$ to $1$. Then the coefficients of $dr^2$ and $dt^2$ are fixed by solving Einstein's equations. The trick here is that part of the metric is fixed by solving Einstein's equations, but not all of it. There is also a part of the metric that is fixed by choosing coordinates carefully. The coordinate choices are justified given the symmetry of the problem. The details of this are explained in textbooks. Aug 8 at 1:12
• Thank you! Meanwhile, I came across exactly this coordinate change in Eddingtons Mathematics of Relativity. They start with U, V, W as prefactors for radial, tangential, and temporal component, respectively. Then they do the coordinate transformation r1=rV(t) and end up with only U and W. 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? I will write an Addendum to this question. Aug 8 at 5:47

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.

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

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 $$$$d\sigma^{2} = r^{2} (d\theta^{2} + \sin^{2} \theta \,d\phi^{2})$$$$

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

Suppose that $$d\sigma^{2} = C(r)r^{2} (d\theta^{2} + \sin^{2} \theta \,d\phi^{2})$$, then you can choose new coordinates $$$$\tilde{r}^{2} = C(r)r^{2}$$$$ so that $$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}$$. The metric then become
$$$$ds^{2} = -A_{1} (\tilde{r},t) c^{2} dt^{2} + A_{2}(\tilde{r},t)F(\tilde{r})d\tilde{r}^{2}+\tilde{r}^{2}(d\theta^{2} + \sin^{2} \theta \,d \phi^{2})$$$$
At this point we can simply drop the tilde for the notation and call $$A_{2}(\tilde{r},t)F(\tilde{r}) \equiv A_{2}(r,t)$$.
• I see that $A_3=0$. However, I don't see why $d\sigma$ can't be another function of r? simply for example $d\sigma^2=C(r) r^2(d\theta^2+sin^2\theta)$? Could you please explain? What you write in your answer is exactly that what I don't understand. It's written everywhere where the Schwarzschild solution is derived. Aug 7 at 7:03