Diagonal metric and Ricci tensor

Gregorio Ricci-Curbastro

Just as I was wrapping up section 8.2 on the Robertson-Walker I wondered if the Ricci tensor was always diagonal if the metric was diagonal. The R-W metric and Ricci tensor are both diagonal and Carroll had not said anything about off diagonal components of the Ricci tensor except that they vanished. I was dissatisfied and about to ask on Physics Forums. Then I decided to try to check myself, fumbled for a bit and then found a 1996 paper by K.Z. Win. I suppose in this day and age of Mathematica and the like it has sunk into obscurity but nevertheless it was interesting and useful for me who lacks such tools. 

Win shows that for a diagonal metric the diagonal components of the Ricci tensor are$$
4R_{\mu\mu}=\left(\partial_\mu\ln{\left|g_{\mu\mu}\right|}-2\partial_\mu\right)\partial_\mu\ln{\left|\frac{g}{g_{\mu\mu}}\right|}-\sum_{\sigma\neq\mu}\left[\left(\partial_\mu\ln{\left|g_{\sigma\sigma}\right|}\right)^2+\left(\partial_\sigma\ln{\frac{\left|g\right|}{{g_{\mu\mu}}^2}}+2\partial_\sigma\right)g^{\sigma\sigma}\partial_\sigma g_{\mu\mu}\right]
$$and the off diagonal components are $$
4R_{\mu\nu}=\left(\partial_\mu\ln{\left|g_{\nu\nu}\right|}-\partial_\mu\right)\partial_\nu\ln{\left|\frac{g}{g_{\mu\mu}g_{\nu\nu}}\right|}+\left(\mu\leftrightarrow\nu\right)-\sum_{\sigma\neq\mu,\nu}{\partial_\mu\ln{\left|g_{\sigma\sigma}\right|}\partial_\nu\ln{\left|g_{\sigma\sigma}\right|}}
$$where ##\mu\neq\nu##, there is no summation over ##\mu,\nu## and ##\left(\mu\leftrightarrow\nu\right)##  stands for preceding terms with ##\mu,\nu## interchanged. The partial derivative operator ##\partial_\tau## usually only applies to what immediately follow. ##\partial_\tau\partial_\tau## is a second order derivative. The exception is the final ##2\partial_\sigma## in the first expression which applies to the whole of ##g^{\sigma\sigma}\partial_\sigma g_{\mu\mu}##.

It turns out that in two dimensions a diagonal metric does imply a diagonal Ricci tensor - but we knew that. In more dimensions it's not so simple. It is necessary to calculate the off diagonal components of the R-W Ricci tensor.

To calculate the Ricci tensor more directly one would use $$
R_{\mu\mu}=R_{\ \ \mu\sigma\mu}^\sigma=\partial_\sigma\Gamma_{\mu\mu}^\sigma-\partial_\mu\Gamma_{\sigma\mu}^\sigma+\Gamma_{\sigma\rho}^\sigma\Gamma_{\mu\mu}^\rho-\Gamma_{\mu\rho}^\sigma\Gamma_{\sigma\mu}^\rho
$$ and$$
R_{\mu\nu}=R_{\ \ \mu\sigma\nu}^\sigma=\partial_\sigma\Gamma_{\nu\mu}^\sigma-\partial_\nu\Gamma_{\sigma\mu}^\sigma+\Gamma_{\sigma\rho}^\sigma\Gamma_{\nu\mu}^\rho-\Gamma_{\nu\rho}^\sigma\Gamma_{\sigma\mu}^\rho
$$and once the Christoffel symbols have been calculated, which is quite easy in the R-W case,  the former takes about the same number of lines as Win's formula but the latter is probably much less efficient than Win. However Win's method avoids calculating Christoffel symbols at all.

 

Read the full details at Commentary 8.2 Diagonal metric and Ricci tensor.pdf (11 pages)