# Einstein Tensor; What is wrong here?

1. Feb 18, 2012

### nobraner

$\nabla^{\mu}$R$_{\mu\nu}$=$\nabla^{\mu}$R$_{\mu\nu}$

Insert

$\nabla^{\mu}$R$_{\mu\nu}$=$\nabla^{\mu}$$\frac{g_{\mu\nu}g^{\mu\nu}}{4}$R$_{\mu\nu}$

Contract the Ricci Tensor

$\nabla^{\mu}$R$_{\mu\nu}$ = $\nabla^{\mu}$$\frac{g_{\mu\nu}}{4}$R

Thus

$\nabla^{\mu}$R$_{\mu\nu}$=$\frac{1}{4}$$\nabla^{\mu}$${g_{\mu\nu}}$R

But General Relativity says

$\nabla^{\mu}$R$_{\mu\nu}$=$\frac{1}{2}$$\nabla^{\mu}$${g_{\mu\nu}}$R

What is wrong here?

2. Feb 18, 2012

### robphy

$\nabla^{\mu}$R$_{\mu\nu}$=$\nabla^{\mu}$$\frac{g_{\sigma\tau}g^{\sigma\tau}}{4}$R$_{\mu\nu}$ (since $\mu$ and $\nu$ are already "taken").

Note that since your proposed proof makes no use of the unique properties of Ricci, it would seem that your result would work for any symmetric tensor. So, you must look at it with suspicion.

3. Feb 18, 2012

### nobraner

I don't understand your declaration that $\mu \nu$ are already taken. Does that mean we can never assume that such a metric as

g$^{\mu\nu}$

exists with out first proving that it is so for the specific case of

$\nabla^{\mu}$R$_{\mu\nu}$=$\frac{1}{4}$$\nabla^{\mu}$g$_{\mu\nu}$R

4. Feb 18, 2012

### elfmotat

There is a double sum in the term that you inserted. Therefore you cannot contract the Ricci leaving out the g$_{\mu\nu}$