I need to find the variation of the purely covariant Riemann tensor with respect to the metric $g^{\mu \nu}$, i.e. $\delta R_{\rho \sigma \mu \nu}$.

I know that, $R_{\rho \sigma \mu \nu} = g_{\rho \eta} R^\eta_{\sigma \mu \nu}$

and $\delta R^\eta_{\sigma \mu \nu} = \nabla_\mu (\delta \Gamma^\eta_{\nu \sigma})- \nabla_\nu (\delta \Gamma^{\eta}_{\mu \sigma}).$

So do I need to use the product rule like so, $\delta R_{\rho \sigma \mu \nu} = \delta (g_{\rho \eta} R^\eta_{\sigma \mu \nu}) = \delta (g_{\rho \eta}) R^\eta_{\sigma \mu \nu} + g_{\rho \eta} \delta( R^\eta_{\sigma \mu \nu})$,

which gives me a relatively messy answer. Is there another way to approach this?

  • 6
    $\begingroup$ Thats a pretty compact answer (and it looks correct) if you ask me, and thats the nicest answer you are going to get. The variation of a complicated object like the Riemann tensor is messy. $\endgroup$
    – Winther
    Commented Jul 11, 2014 at 11:37
  • 4
    $\begingroup$ In general, general relativity is relatively messy. $\endgroup$ Commented Jul 11, 2014 at 11:52


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