# Einstein Tensor; super simple derivation; where did I go wrong?

1. Feb 5, 2012

### nobraner

$\nabla_{μ}R^{\mu\nu}=\nabla_{μ}R^{\mu\nu}$

insert the multiplicative identity, expressed as the product of the covariant and contravariant metric

$\nabla_{μ}R^{\mu\nu}=\nabla_{μ}(g^{\mu \nu}g_{\mu\nu})R^{\mu\nu}$

contract the indices of the Ricci Tensor, to get

$\nabla_{μ}R^{\mu\nu}=\nabla_{μ}g^{\mu\nu}R$

but the general theory tells us that

$\nabla_{μ}R^{\mu\nu}=\frac{1}{2} \nabla_{μ}g^{\mu\nu}R$

Where have I gone wrong?

Last edited: Feb 5, 2012
2. Feb 5, 2012

### Matterwave

Er, $g^{\mu\nu}g_{\mu\nu}=4$, so it appears you inserted 4 into the right hand side on the second line. You are also seriously overloading your indices.

The divergence free nature of the Einstein tensor arises from the Bianchi Identity.

Last edited: Feb 5, 2012
3. Feb 6, 2012

### nobraner

Professor Lenard Susskind explicitly states in his YouTube videos that

$g^{\mu\nu}g_{\mu\nu}=\delta^{\mu}_{\nu}$

The product of the covariant and contravariant metric is the kroniker delta (the multiplicative identity matrix). Although, he does say somewhere that if you have

$\delta^{a}_{\nu}$ and $\delta^{\nu}_{b}$

and you identify a with b as

$\delta^{a}_{\nu}\delta^{\nu}_{a}$=$\delta^{\nu}_{\nu}$

Then you are summing over a which is simply the sum of the 1's in the diagonal of the identity matrix and that gives you 4.

Last edited: Feb 6, 2012
4. Feb 6, 2012

### Simon_G

This relation is wrong.

$g^{\mu\nu} g_{\mu\rho} = \delta^\nu_\rho$

then, if we take $$\nu = \rho$$ we obtain:

$$g^{\mu\nu} g_{\mu\nu} = \delta^\nu_\nu$$

but the last term is the trace of Kronecker delta which is four if $$dim(M) = 4$$

Last edited: Feb 6, 2012
5. Feb 6, 2012

### nobraner

If as you say, Professor Susskind is wrong, then I feel betrayed that a physicist of his stature would teach error.

6. Feb 6, 2012

### Simon_G

Why? Isn't he human? :D

Anyway, relation above hasn't free indices:

$$g^{\mu\nu}g_{\mu\nu}$$

so it has to a scalar. Indeed it is dimension of manifold.

Sorry for my poor english :D

Last edited: Feb 6, 2012
7. Feb 6, 2012

### haushofer

In that case he made a notational error. An equation with no free indices on the right, but free indices on the left could never be right.

8. Feb 6, 2012

### Matterwave

It's very easy to make such a simple error, especially if you are teaching a class.

Also, don't overload your indices, that's another very important point. If you already see mu's and nu's and then you introduce another set of mu's and nu's and you sum over them, for example, you are bound to make errors.

9. Feb 6, 2012

### MiljenkoM

I think he meant $g^{\mu\nu}g_{\mu\nu}=\delta^{\mu}_{\mu}$ and this is just trace of Identity matrix, and equals n, where n is dimension of Manifold (usually 4).