How to solve nonlinear diff eq for a general Schwarzschild metric?

Question

So I have a general form of a spherically symmetric metric:

$$ds^2 = -g(r)_t \, dt^2 + g(r)_r \, dr^2 + g(r)_s (d\theta^2 + \sin^2 \theta \, d\Phi^2)$$

$$ R_{\theta\theta} = \frac{-g'_s g'_t g_r + g_t g'_r g'_s - 2 g_r g''_s g_t + 4 g^2_r g_t}{4 g^2_r g_t} $$

$$ R_{\phi\phi} = \frac{\sin^2(\theta)(-g'_s g'_t g_r + g_t g'_r g'_s - 2 g_r g''_s g_t + 4 g^2_r g_t)}{4 g^2_r g_t}$$

$$ R_{rr} = \frac{2g_s g^2_t g'_r g'_s + 2 g_r g^2_t g'^2_s + g^2_s g_t g'_r g'_t + g_r g^2_s g'^2_t - 4 g_r g_s g^2_t g''_s - 2 g_r g^2_s g_t g''_t}{4 g_r g^2_s g^2_t} $$

$$ R_{tt} = \frac{-g_s g_t g'_r g'_t + 2 g_r g_t g'_s g'_t - g_r g_s g'^2_t + 2 g_r g_s g_t g''_t}{4 g^2_r g_s g_t} $$

We can then try and solve these differential equations by setting the Ricci tensor to 0. The primed variables represent a derivative with respect to r. I am unsure how to go about solving for the unknown variables by setting the Ricci tensor to 0 as they are nonlinear equations. I asked for help from someone that was more knowledgable on the topic and they told me that in general I should make my life easier by starting with a simpler ansatz, but that I would also be able to do it with what I currently have. He said to organize the Ricci tensor into a global derivative like $R_{xx}=A(r) \frac{d}{dr}[B(r) \frac{d}{dr}g_y]$ where A and B must be found. He then went on to say that I should use something of the form

$$ \frac{g_s^{a'} g_r^{b'} g_t^{c'}}{r^{d'}} \frac {d}{dr} \frac{r^d}{g_s^a g_r^bg_t^c} \frac{d}{dr}log(g_t) = R_{tt} = 0$$

I'm not really sure why he would choose this form, I am especially confused about the $log(g_t)$ as that should just be $\frac{g_t'}{g_t}$ but wouldn't that denominator just come out when solving for c anyways? To me this should also give the right answer:

$$ {g_s^{a'} g_r^{b'} g_t^{c'}} \frac {d}{dr} {g_s^a g_r^bg_t^c} \frac{d}{dr}g_t = R_{tt} = 0$$

One thing I tried was to rewrite the Ricci tensor as

$$ R_{tt}=0=g_t''+ \frac{-g_s g_t g'_r g'_t + 2 g_r g_t g'_s g'_t - g_r g_s g'^2_t}{8 g^3_r g^2_s g^2_t} $$

After taking the derivatives I have:

$$ g_s^{a'} g_r^{b'} g_t^{c'}(c g_s^a g_r^b g_t^{c-1}g_t'^2 + g_s^a g_r^b g_t^c g_t'' +ag_s^{a-1} g_t^c g_r^b g_s' g_t^{"} +b g_s^a g_r^{b-1} g_t^c g_r' g_t')$$

This would mean that I have $a'+a=b'+b=c'+c=0$ from getting rid of the terms in front of the $ g_t''$ term. Then I said that because there is no r dependence I can just set $d=d'=0$. Now, I tried to match the powers to what I actually have in the equation and I can't find any that fit the form. For example if I choose the $g_r' g_t'$ term and match coefficients I should have $ g_s^{-1}$ term so $a+a'=-1$ but I also have $a+a'=0$ from the $g_t^{''}$ term. So either I made a mistake, or assuming that we have power functions for $B(r)$ is not the way to go. I could go back to the beginning and use symmetry to get $g(r)_s = r^2$, but I had also heard that I would not need to make simplifications a priori. Can anyone offer any insight?

Answer 1

Your ansatz is not the most general form for a metric of a spherically symmetric spacetime. You can only assume that the metric coefficients do not depend on $t$ if you also assume that your spacetime is static.

Anyway, for a general spherically symmetric spacetime, you can always find coordinates such that the metric is written as $$ ds^2=-f(t,r) dt^2 +h(t,r) dr^2 + r^2 (d\theta^2 + \sin(\theta)^2 d\phi^2). $$ You seem to be assuming that you want it to be a vacuum solution of the Einstein equations, so that the Ricci tensor is zero. In this case (spherical symmetry + vacuum), the metric can be shown to be the Schwarzschild metric as follows.

Calculate the Ricci tensor for the above metric and note that $R_{tr}=0$ implies that $h$ can only depend on $r$ (and not on $t$). This in turn implies, together with the equations $R_{tt}=0$ and $R_{rr}=0$ (more specifically, from the equation $h(r) R_{tt} + f(t,r) R_{rr}=0$), that $f(t,r)=A(t)/h(r)$, with $ A(t)>0$. Redefining $t$ in order to incorporate $A(t)$ we then have $ds^2=-h(r)^{-1} dt^2 + h(r) dr^2+r^2 (d\theta^2+ \sin( \theta)^2 d\phi^2)$. Thus, such a metric, besides being spherically static, must also be static for a vacuum solution. Finally, it follows from $R_{\theta\theta}=0$ that $h(r)=\frac{1}{1-k/r}$, for some constant $k$. As a result, we showed that the only spherically symmetric vacuum solution of the Einstein equations is the Schwarzschild metric. This is a notable result, known as Birkhoff theorem.