For a homework question, we are given the metric $$ds^2=dt^2-\frac{2m}{F}dr^2-F^2d\Omega^2\ ,$$ where F is some nasty function of $r$ and $t$. We're asked to then show that this satisfies the Field Equations.

I know exactly how to proceed with this computation, but the algebra is so annoying and I'm so likely to make mistakes that I wondered if there was a way to obtain the Ricci tensor from a spherically symmetric spacetime without doing a whole bunch of pointless computation? For instance, our prof has posted such formulae for a Riemann Tensor: $$R_{2323}=\frac{\sin^2{\theta}}{4ab}(c^2_{r}a-c^2_{t}b-4abc)$$ where the subscripts denote partial differentiation and the metric has the form $$ds^2=a(r,t)dt^2-b(r,t)dr^2-c(r,t)d\Omega^2\ .$$ Is there a resource that shows how to now go from the Riemann to the Ricci tensor? Or still better, formulae that give the Ricci tensor directly for a general spherically symmetric metric?

  • $\begingroup$ This link might be of help - arxiv.org/abs/gr-qc/9602015 $\endgroup$
    – Prahar
    Nov 27, 2014 at 4:29
  • $\begingroup$ I haven't got time to do it manually, but I've computed the Ricci tensor and scalar using Mathematica, and both are non-zero for a generic $F(r,t)$, so in general it's not a solution to the vacuum Einstein field equations. $\endgroup$
    – JamalS
    Nov 27, 2014 at 12:28
  • $\begingroup$ If you are familiar with differential forms from differential geometry, I highly recommend the Cartan method - it's much faster than doing the whole computation directly. $\endgroup$
    – JamalS
    Nov 27, 2014 at 12:30
  • $\begingroup$ @Prahar Thank you for the link! This is exactly the kind of thing I was looking for. $\endgroup$
    – Anthony
    Nov 28, 2014 at 2:49
  • $\begingroup$ @JamalS I haven't seen the Cartan method before, but I looked it up and I think it's definitely worth learning. $\endgroup$
    – Anthony
    Nov 28, 2014 at 2:50

1 Answer 1


I take the metric in the question to mean explicitly,

$$ds^2 = dt^2 - \frac{2m}{F(t,r)}dr^2-F(t,r)^2 \left( d\theta^2 + \sin^2 \theta\, d\phi^2\right).$$

Computing the curvature tediously, one finds the Einstein tensor $G_{ab}=R_{ab}-\frac12 g_{ab}R$ has eight non-vanishing components. For example, from the field equations, one finds,

$$T_{00} = \frac{m - F(t,r)F_r(t,r)^2 - F(t,r)^2 F_{rr}(t,r)}{8\pi G m \, F(t,r)^2}.$$

If one wants the metric to satisfy the vacuum field equations, one can explicitly solve $T_{00}=0$, but there are further constraints on $F(t,r)$ from the rest of the components of $T_{ab}$ - I have not been able to explicitly prove yet whether this system can be solved to yield a $F(t,r)$ for which the metric is a vacuum solution.

However, demoting $F(t,r)$ to $F(r)$ to make the system somewhat simpler, there certainly is a choice of $F(r)$ for which the metric is a vacuum solution, e.g.

$$F(r) = \left[ \frac32 \left( \pm \sqrt{2m}r + c\right)\right]^{2/3}$$

for $c\in\mathbb R$, which certainly makes me suspect there are time-dependent $F(t,r)$ for which the metric is a vacuum solution as well.


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