This question is not about the solutions but much rather about the equations we write in GR for a spherically symmetric, static vacuum 4D spacetime.

The Einstein equations are $$G_{\mu\nu}=0\;\;\;\Rightarrow\;\;\; g^{\mu\nu}G_{\mu\nu}=0\;\;\;\Rightarrow \;\;\;R=0$$ so that in conclusion we can solve $$R_{\mu\nu}=0 \;\;\;\text{or}\;\;\;R^{\mu}_{\nu}=0$$ Moreover, we can choose the coordinates so that the metric takes the form $$ds^2=-f(r)\,dt^2+h^{-1}(r)\,dr^2+r^2\!\left(d\theta^2+\sin^2\!\theta \,d\varphi^2\right)$$ In this way when we write explicitly the equations $R^{\mu}_{\nu}=0$ we get three non trivial equations, namely: $$R^{0}_0=0\;\;,\qquad R^r_r=0\;\;,\qquad R^{\theta}_{\theta}\equiv R^{\varphi}_{\varphi}=0$$

Using the first two equations and the formal identity $R^{\theta}_{\theta}\equiv R^{\varphi}_{\varphi}$ , we can write the third simply as $R=0$.

My question is what is the meaning of this third equation, when the variables are just the two functions $f(r)\;$ and $\;h(r)$? Does it act like a constraint or is there any way to get around it? and what does it tell us about GR in spherically symmetric and static space-times?

I would guess that it corresponds to a relation we impose between $f$, $h$ (and the coordinate $r$) when we write the solid angle part as in flat space, but I don't really understand its role.


$\let\a=\alpha \let\b=\beta \let\th=\vartheta$

My question is what is the meaning of this third equation

I think the answer is the following. You know that Einstein's tensor is divergenceless. If you expand $G^\a_{\b;\a}$ you'll find that three of the four equations are trivial identities. The only one surviving is the $r$-component and you'll be able to see that it simply says $$G^r_{r,r} = 0$$ (no sum).

So $G^r_r$ is a constant and if the spacetime is asymptotically flat the constant is zero. Then one of Einstein's equations is automatically satisfied and only the remaining two $$G^t_t = 0 \qquad G^\th_\th = 0$$ are real equations for $f(r)$, $h(r)$.

  • $\begingroup$ Thank you, this is interesting! However it is strange that the radial equation gets eliminated, when the choice of the metric is basically equivalent to renounce to the variations with respect to the angular components of the metric. This intuitively would eliminate the angular equation $G^{\theta}_{\theta}=0$. I would say that the angular equation should be the superfluous one, but maybe an argument like yours can be used to motivate the fact that it is always automatically satisfied anyway. $\endgroup$ – AoZora Jun 21 '19 at 21:31
  • $\begingroup$ I gave it a bit more of a thought and I think you are right. Note that we can shuffle the system so that the redundant equation is that for the angular component of the metric (which is more intuitive maybe since we fixed it). Using $G^r_r\equiv0$ we can say that solving $R^r_r=0$ implies $R=0$. On the other hand if we solve both $R^t_t=0$ and $R^r_r=0$ we can say that solving $R=0$ is equivalent to solve $R^{\theta}_{\theta}=0$. So solving $R^t_t=0$ and $R^r_r=0$ implies also $R^{\theta}_{\theta}=0$. $\endgroup$ – AoZora Jun 21 '19 at 22:06

The metric in a vacuum around a spherically symmetric and static massive body is in principle written as
$ds^2 = -U(r) dt^2 + V(r) dr^2 + W(r) r^2 (d\theta^2 + \sin^2 \theta d\phi^2)$.
The three independent equations $R_{t t} = R_{r r} = R_{\theta \theta} = 0$, ($R_{\theta \theta}$ and $R_{\phi \phi}$ linearly dependent), allow for the determination of the three radial functions $U(r)$, $V(r)$, $W(r)$.

However the three radial functions, being $r$ simply a radial parameter and not an actual distance, can be reduced to two funtions. Put $W(r) r^2 = \hat r^2$, express $U(r) = \hat U(\hat r)$ and $V(r) = \hat V(\hat r)$. Then removing the hats and putting $U(r) = e^{2 \nu(r)}$, $V(r) = e^{2 \lambda(r)}$ the line element becomes
$ds^2 = -e^{2 \nu(r)} dt^2 + e^{2 \lambda(r)} dr^2 + r^2 (d\theta^2 + \sin^2 \theta d\phi^2)$
If you start from this metric assumption, you have three independent equations for two functions $\nu(r)$ and $\lambda(r)$. One equation is redundant, however to determine the expression of the radial functions you may choose which equations are more convenient for an easy computation.

Note: There is not a specific meaning of the third equation, as each of the three equations can be taken as redundant compared to the other two equations.

  • $\begingroup$ Thanks. If this is the case, shouldn't we be able to prove the redundancy explicitly? Is that possible? $\endgroup$ – AoZora Jun 20 '19 at 10:04
  • $\begingroup$ @AoZora. The redundance is because you have three independent equations and two functions. Just note I will edit my post as $R_{\theta \theta}$ and $R_{\phi \phi}$ are not independent, but not the same. $\endgroup$ – Michele Grosso Jun 20 '19 at 15:33
  • $\begingroup$ The fact that you have more equations than unknowns does not mean that the extra equations are trivial. I don't think it is obvious that the third equation, say $R^{\theta}_{\theta}=0$, will be satisfied by solutions of the first two equations. How do you prove this? $\endgroup$ – AoZora Jun 20 '19 at 16:57
  • $\begingroup$ I am thinking that maybe the choice of the function $\hat{r}(r)$ amounts to integrate out the degree of freedom corresponding to $g_{\theta\theta}$ in the path integral, so that is in some way inappropriate to vary the action with respect to this component of the metric, leading to only two equations.. Even if path integral is not the best thing to consider in gravity, maybe there is some argument that eliminates the extra equation once you fix the radial diffeomorphism.. $\endgroup$ – AoZora Jun 20 '19 at 17:06
  • $\begingroup$ @AoZora. If you have a redundance you determine the unknowns with part of the constraints. I agree with you that then you have to check that the solutions comply with the remaining constraints. However in the books I have studied I never found that they were not. $\endgroup$ – Michele Grosso Jun 21 '19 at 14:57

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