# Contracting the Riemann tensor issues, p540 hobson

I am stuck trying to work through something on p540 in Hobson (General Relativity: An Introduction for Physicists), one is supposed to use the variation of the full Riemann tensor and then contract it to get $\delta R_{\mbox{ }\mu \nu }$: $$\delta R^{\sigma}_{\mbox{ }\mu \nu \rho} = \delta\partial_{\nu}\Gamma^{\sigma}_{\mbox{ }\mu\rho}-\delta\partial_{\rho}\Gamma^{\sigma}_{\mbox{ }\mu\nu} +\delta\Gamma^{\sigma}_{\mbox{ }\tau\nu}\Gamma^{\tau}_{\mbox{ }\rho\mu}-\delta\Gamma^{\sigma}_{\mbox{ }\tau\rho}\Gamma^{\tau}_{\mbox{ }\nu\mu}=\delta\partial_{\nu}\Gamma^{\sigma}_{\mbox{ }\mu\rho}-\delta\partial_{\rho}\Gamma^{\sigma}_{\mbox{ }\mu\nu}$$ and then contract it to show $$\delta R_{\mbox{ }\mu \nu }=[ \partial_{\nu}(\delta \Gamma^{\sigma}_{\mbox{ }\mu \sigma}) - \partial_{\sigma}(\delta \Gamma^{\sigma}_{\mbox{ }\mu \nu}) ]$$

MY ATTEMPT:

I thought to contract it by multiplying by $g_{\sigma \alpha}g^{\alpha \rho}$ $$g_{\sigma \alpha}g^{\alpha \rho}\delta R^{\sigma}_{\mbox{ }\mu \nu \rho}=\delta R_{\mbox{ }\mu \nu }$$ but then I get $$g_{\sigma \alpha}g^{\alpha \rho}(\delta\partial_{\nu}\Gamma^{\sigma}_{\mbox{ }\mu\rho}-\delta\partial_{\rho}\Gamma^{\sigma}_{\mbox{ }\mu\nu})=(\delta \partial_{\nu}\Gamma_{\mu\rho}-\delta \partial_{\rho}\Gamma^{\rho}_{\mbox{ }\mu\nu})$$ Can anyone help me get the correct answer (4th line), I may have made a mistake or could be completely wrong I am not sure.

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It seems like your penultimate Christoffel is missing an index, i.e. $\Gamma_{\mu \rho}$ needs to be $\Gamma^{\rho}{}_{\mu \rho}.$ In that case you're done, you have the correct answer! – Vibert Apr 7 '13 at 22:59
Looks like en.wikipedia.org/wiki/… is just what you want. – user21433 Apr 8 '13 at 1:05
If this is homework, please use the homework tag. – Ben Crowell May 8 '13 at 4:52
it is not but i have done it now anyway – user21119 May 9 '13 at 8:01

Apart from the missing index, which was already mentioned, you are fine. The partial derivative commutes with the variation, i.e. $$\partial_\rho(\delta{\Gamma^{\rho}}_{\mu\nu})=\delta\partial_\rho{\Gamma^{\rho}}_{\mu\nu}.$$
surely though the $g^{\alpha \rho}$ takes away this index? If not thewn is $g_{\sigma\alpha}g^{\alpha rho} \Gamma^{\sigma}_{\mbox{ }\mu \rho}=\Gamma^{\rho}_{\mbox{ }\mu \rho}$ – user21119 Apr 8 '13 at 10:53