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What are the strongest (or most useful) conditions on a metric for it's Ricci tensor to be diagonal? I've read that if the metric is explicitly dependent on only one variable then the Ricci Tensor is diagonal, but this doesn't seem particularly useful to me!

In particular I'd like to be able to tell whether for the standard spherically symmetric Schwarzchild and FLRW metrics I need to be worried about calculating off diagonal terms of the Ricci tensor.

Many thanks!

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If the metric depends on the coordinates $x_{\mu}$ and $x_{\nu}$, then it is possible that $R_{\mu\nu}$ may be nonzero. So when the metric depends on more than one coordinate, we need to put special restrictions on its form in order to guarantee a diagonal Ricci tensor.

One such agreeable metric which is seen in general relativity is the "standard" metric $$ds^2 = B(r) ~dt^2 - A(r) ~dr^2 - r^2(d\theta^2 + \sin^2 \theta ~d\varphi^2).$$ The Ricci tensor for the standard metric is diagonal. To be precise, \begin{align*} R_{tt} & = -\frac{B''(r)}{2A(r)} + \frac{B'(r)}{4A(r)}\left(\frac{A'(r)}{A(r)} + \frac{B'(r)}{B(r)}\right) - \frac{B'(r)}{rA(r)}, \\ R_{rr} & = \frac{B''(r)}{2B(r)} - \frac{B'(r)}{4B(r)}\left( \frac{A'(r)}{A(r)} + \frac{B'(r)}{B(r)}\right) - \frac{A'(r)}{rA(r)}, \\ R_{\theta\theta} & = \frac{r}{2A(r)}\left( \frac{B'(r)}{B(r)} - \frac{A'(r)}{A(r)}\right) + \frac{1}{A(r)} - 1, \\ R_{\varphi\varphi} & = R_{\theta\theta} \sin^2 \theta, \\ R_{\mu\nu} & = 0 \text{ for } \mu \neq \nu. \end{align*} Note that the Schwarzschild metric $$ds^2 = \left(1 - \frac{GM}{r}\right) ~dt^2 - \left(1 - \frac{GM}{r}\right)^{-1} ~dr - r^2(d\theta^2 + \sin^2 \theta ~d\varphi^2)$$ is standard, so indeed you don't need to worry about off-diagonal components of the Ricci tensor when working with the Schwarzschild geometry.

The Friedmann–Lemaître–Robertson–Walker (FLRW) metric for flat spatial sections, $$ds^2 = dt^2 - a(t)^2(dx^2 + dy^2 + dz^2),$$ is diagonal and depends only on $t$, so its Ricci tensor is diagonal.

Cosmological models such as the FLRW metric have certain physical considerations which restrict the metric. In particular, space should be homogeneous and isotropic, or in other words a maximally symmetric $3$-manifold. A Riemannian $n$-manifold is maximally symmetric if and only if its Riemann curvature tensor is of the form $$R_{ijkl} = \frac{R}{n(n-1)}(g_{ik}g_{jl} - g_{il}g_{jk}),$$ where $R$ is the scalar curvature. So the Ricci tensor of a maximally symmetric $3$-manifold satisfies $$R_{ij} = -2Rg_{ij}.$$ If $\tilde{g}_{ij}$ is a maximally symmetric metric on space and we have the metric $$ds^2 = dt^2 - a(t)^2 \tilde{g}_{ij}~dx^i ~dx^j,$$ on spacetime, where Einstein summation notation is being used, then the Ricci curvature of this spacetime metric is given by \begin{align*} R_{tt} & = \frac{3a''}{a}, \\ R_{ij} & = -(aa' + 2(a')^2 + 2k)\tilde{g}_{ij}. \end{align*} So the Ricci tensor is diagonal in such a case if and only if $\tilde{g}_{ij}$ is diagonal. The general FLRW metric $$ds^2 = dt^2 - a(t)^2 \left( \frac{dr^2}{1 -kr^2} + r^2(d\theta^2 + \sin^2 \theta ~d\varphi^2)\right)$$ is of this form, so its Ricci tensor is diagonal as well.


Conclusion: Yes, the Ricci tensor is diagonal for the Schwarzschild metric and the FLRW metric.

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Thank you for the very comprehensive answer! –  Edward Hughes May 23 '12 at 9:51
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