Christofle symbol and determinant of metric tensor

[off-diagonal]

Hi, every one I'm newbie here. I have a few problem with my study about GR.
Here's a problem

$$\partial_a(g^{ad})g_{cd}-\partial_d(g^{ad})g_{ac}=\\0$$


Could I prove these relation by change index (in 1st term ) from a -> d and also d -> a?

and let's defined $${g}=det{\\g_{ab}\\}$$



$${g^{ab}}\partial_c(\\g_{ab})=\frac{1}{g} \partial_c(\\g)$$

How I prove these equation ? Any one got an idea?

 

[Fredrik]

Could I prove these relation by change index (in 1st term ) from a -> d and also d -> a?

Yes. (You also have to use that the metric is a symmetric tensor: $g_{ab}=g_{ba}$).

$${g^{ab}}\partial_c(\\g_{ab})=\frac{1}{g} \partial_c(\\g)$$

How I prove these equation ? Any one got an idea?

I haven't given much thought to this specific identity, but the only time I proved an identity involving the determinant of the metric, I had to use this crap.

 

[shoehorn]

and let's defined $${g}=det{\\g_{ab}\\}$$



$${g^{ab}}\partial_c(\\g_{ab})=\frac{1}{g} \partial_c(\\g)$$

How I prove these equation ? Any one got an idea?

By definition $\sqrt{-g}$ is a tensor density of weight one. As a consequence of this and the fact that the covariant derivative of the metric is zero, one has the result

$$\nabla_a\sqrt{-g} = \partial_a\sqrt{-g} - \Gamma^b_{\phantom{b}ab}\sqrt{-g} = 0$$

It's trivial to go from here to the result you want to prove.

 

[haushofer]

You can check Ray d'Inverno's book on GR; in chapter 7 I believe he derives these things quite extensively :)