# Second derivative of a metric and the Riemann curvature tensor

1. Apr 30, 2013

### MarkovMarakov

I can't see how to get the following result. Help would be appreciated!

This question has to do with the Riemann curvature tensor in inertial coordinates.

Such that, if I'm not wrong, (in inertial coordinates) $$R_{abcd}=\frac{1}{2} (g_{ad,bc}+g_{bc,ad}-g_{bd,ac}-g_{ac,bd})$$
where $",_i"$ denotes $\partial \over \partial x^i$.

How does $$g_{ab,cd}=-\frac{1}{3}(R_{acbd}+R_{adbc})$$?
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So
$$-\frac{1}{3}(R_{acbd}+R_{adbc})=-\frac{1}{6}(g_{bc,ad}+g_{ad,bc}-g_{ba,cd}-g_{cd,ab}+g_{bd,ac}+g_{ac,bd}-g_{ab,cd}-g_{cd,ab})$$
$$=-\frac{1}{6}(g_{bc,ad}+g_{ad,bc}+g_{bd,ac}+g_{ac,bd})+\frac{1}{3}g_{cd,ab}+\frac{1}{3}g_{ab,cd}$$
But how does $$\frac{1}{6}(g_{bc,ad}+g_{ad,bc}+g_{bd,ac}+g_{ac,bd})+\frac{1}{3}g_{cd,ab}=\frac{2}{3}g_{ab,cd}$$
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