1
$\begingroup$

I am a little bit confused about the convention of the curvature tensor. The books of Wald and Misner/Deser/Wheeler seem to have the same conventions, i.e. the indices of the Riemann curvature tensor are defined by

$${R_{\alpha\beta\gamma}}^{\delta}v_{\delta}=(\nabla_{\alpha}\nabla_{\beta}-\nabla_{\beta}\nabla_{\alpha})v_{\gamma}$$

and the Ricci tensor is defined by the contraction

$$R_{\alpha\beta}:={R_{\alpha\gamma\beta}}^{\gamma}={R^{\gamma}}_{\alpha\gamma\beta}.$$

However, the expression of the Ricci tensor in coordinates seems to be different:

$$R_{\alpha\beta}=\partial_{\lambda}\Gamma^{\lambda}_{\alpha\beta}-\partial_{\alpha}\Gamma^{\lambda}_{\beta\lambda}+\Gamma^{\lambda}_{\alpha\beta}\Gamma^{\mu}_{\lambda\mu}-\Gamma^{\lambda}_{\alpha\mu}\Gamma^{\mu}_{\beta\lambda} $$ $$R_{\alpha\beta}=\partial_{\lambda}\Gamma^{\lambda}_{\alpha\beta}-\partial_{\beta}\Gamma^{\lambda}_{\alpha\lambda}+\Gamma^{\lambda}_{\alpha\beta}\Gamma^{\mu}_{\lambda\mu}-\Gamma^{\lambda}_{\alpha\mu}\Gamma^{\mu}_{\beta\lambda}.$$

The first one is taken from Wald and the second one from Misner/Deser/Wheeler. The only difference is in the ordering of $\alpha$ and $\beta$ in the second term. Does anyone know why? Am I missing something?

$\endgroup$
1
  • 1
    $\begingroup$ The second term is actually symmetric in $\alpha$ and $\beta$ so the ordering is irrelevant. $\endgroup$
    – Prahar
    Nov 29, 2022 at 8:49

1 Answer 1

2
$\begingroup$

The expressions are same, since $\Gamma^{\lambda}_{\alpha\lambda}=\partial_{\alpha}\log\sqrt{|g|}$ and that the ordering of $\partial_{\alpha}\partial_{\beta}$ acting on something doesn't matter as long as they are continuous.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.