I was trying to follow the Cartan method to find the curvature as outlined here, in the case of a static, spherically symmetric spacetime. It all seemed to work fine, and for a metric of

$$ds^2 = -a^2(r)dt^2 + b^2(r)dr^2 + c^2(r)d\theta^2 + c^2(r)\sin^2\theta d\phi^2$$

I got a scalar curvature of

$$R = \frac{2}{ab^2}(\frac{a'b'}{b} - a'') - \frac{4a'c'}{ab^2c} + \frac{4}{b^2c}(\frac{b'c'}{b} - c'') + \frac{2}{c^2}(1 - \frac{c'^2}{b^2}).$$

But I want to use the gauge where $r$ is proper length, i.e. $b(r) = 1$. Of course the problem is that $r$ may in general become timelike, and since the functions we use in this method do not absorb the minus sign, we have to treat each possibility separately. So I have to redo it with $g_{tt}$ positive and $g_{rr}$ negative.

But because the 1-forms we construct do not contain the minus sign, I don't see anything changing in the calculation of the Riemann components. Even in the last step,

${R^\rho}_{\sigma\mu\nu} = e^\rho_a {R^a}_{bcd} e^b_\sigma e^c_{\mu} e^d_{\nu}$

when calculating the inverse frame matrix via

$e^\mu_i = e^j_{\nu} \eta_{ij} g^{\mu\nu}$

since the transform is diagonal anyway, the signs in the frame and coordinate metrics just cancel out.

Well if the (1, 3) coordinate Riemann components are all the same, then so are the coordinate Ricci components. So it's only the Ricci-to-scalar step where anything changes.

$R = g^{\mu\nu}R_{\mu\nu}$

So we just flip the contributions from $R_{tt}$ and $R_{rr}$. But that has an effect that seems completely wrong to me: it wipes out the two middle terms from the expression for $R$ I had above. How could that be at all possible? We just switch which coordinate is timelike and all of a sudden the curvature has a completely different formula? I don't think so... But I've been going over and over the steps and I just don't see what I'm missing. So I'd love to hear your input!

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    $\begingroup$ I didn't follow the steps and am not going to go through the calculation, but just a comment that your original metric is not symmetric under interchanging $g_{tt}$ and $g_{rr}$. For instance, the angular terms depend on $r$ regardless of whether $r$ is spacelike or timelike. So I don't see a reason why the scalar curvature must be symmetric when computed with $r$ spacelike and $r$ timelike. $\endgroup$
    – Andrew
    Sep 27, 2021 at 3:51
  • $\begingroup$ @Andrew Thanks, I think you're right, that is the main point. I think I found the mistake which led to an almost but, as you say, not quite symmetric expression that seems much more likely, so I will post an answer for future reference. $\endgroup$ Sep 27, 2021 at 4:08

1 Answer 1


Well I'll post what I figured out in case anyone ever searches this up. What I was missing is that, for the spin connection,

$\omega^i_j = -\eta^{ii}\eta_{jj} \omega^j_i$

(which I think is because $\eta$ raises and lowers indices on frame objects like $\omega$, and $\eta$ is necessarily diagonal...though I'm not sure why the minus sign). That means that when $i$ and $j$ are both spacelike, $\omega^i_j = -\omega^j_i$, but when one is timelike, $\omega^i_j = \omega^j_i$. So some of their formulae flip when $r$ switches to timelike, and the net effect downstream is to recover those two middle terms in $R$.

But also, there is an even simpler way to reason out $R$ given the symmetry. $R$ comes from the $R_{\mu\nu}$, but due to symmetry we only have $R_{\mu\mu}$, and these in turn only come from Riemman components of the form ${R^\nu}_{\mu\nu\mu}$, and because the frame $e^i_\mu$ is aligned with the coordinate axes (it's merely normalized), we only need to use the ${R^i}_{jij}$. When you take ${R^i}_{jij}$ and convert it back to coordinates, the upper and lower $i$'s cancel, and you're just multiplying by the $j$ coordinate function ($a$, $b$, $c$, or $c\sin\theta$) twice -- that is, by $|g_{jj}|$. But when finding the Ricci scalar, you multiply by the inverse metric, which cancels that out, leaving only the sign, ie. $\eta_{jj}$. Thus

$R = \Sigma \eta_{jj}{R^i}_{jij}$

From this, I found that $R$ when $g_{rr} < 0$ is exactly the negative of what it is when $g_{rr} > 0$, except for the little $\frac{2}{c^2}$ term which retains its sign:

$R = \frac{2}{ab^2}(a'' - \frac{a'b'}{b}) + \frac{4a'c'}{ab^2c} + \frac{4}{b^2c}(c'' - \frac{c'b'}{b}) + \frac{2}{c^2}(1 + \frac{c'^2}{b^2})$

Thus the last term does not cancel even when $b(r) = 1$ and $c(r) = r$, apparently reminding us that in spherical coordinates, switching the timelike and spacelike coordinates does not preserve Minkowski space!


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