# Einstein's tensor divergence

1. Aug 19, 2011

### teddd

Hi everyone!
I'm having a lillle problem proving that the einstein tensor is divergence free!!
I don't know how to begin, i start with
$$\nabla_\mu G^{\mu\nu}=\nabla_{\mu}(R^ {\mu\nu} -\frac{1}{2}g^{\mu\nu}R)$$
i tried to do $$\nabla_\mu G^{\mu\nu}=\nabla_{\mu}g^{\mu\nu}(g_{\mu\nu}R^{\mu\nu}-\frac{1}{2}R)$$
(by the way, is that right? I guess no becaouse then I get to $$g^{mu\nu}(\nabla_{\mu}(R-\frac{1}{2}R))=\frac{1}{2}g^{\mu\nu}\frac{\partial R}{\partial \mu}$$ which i guess never goes to zero!)

can you help me out?

Last edited: Aug 19, 2011
2. Aug 19, 2011

### Ben Niehoff

You index manipulations are incorrect. You can't have an index repeated more than twice.

The way to get the divergence of the Einstein tensor is to start with the second Bianchi identity of the Riemann tensor, and then contract the upper index with one of the lower ones.

3. Aug 20, 2011

### teddd

Thanks pal, I had already figured the doubly contracted Bianchi identity thing out, but I wanted to know where was the mistake in that!!!

But why I can't repeat an index more than twice? I guess becaouse it clashes against the summation convention isnt'it?

4. Aug 20, 2011

### haushofer

Yes. In that case, it's not clear over which indices you should sum.