Einstein Tensor; What is wrong here?

[nobraner]

Start with

\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?

 

[robphy]

Start with

\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}

[snip]

What is wrong here?

Your inserted factor should be
\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.

 

[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 without 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

 

[elfmotat]

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 without 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

You're overloading your indices. An index shouldn't appear more than twice in any term.

 

[juanrga]

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 without 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

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