Einstein Tensor

1. Sep 8, 2007

John_Doe

Dumb question, but...

$G_{\mu\nu}=R_{\mu\nu}-\frac{1}{2} g_{\mu\nu}R$

Since

$R=g^{\mu\nu}R_{\mu\nu}$

and

$g^{\mu\nu}g_{\mu\nu}=1$

it would appear that

$G_{\mu\nu} =R_{\mu\nu}-\frac{1}{2} g_{\mu\nu}g^{\alpha\beta}R_{\alpha\beta} =R_{\mu\nu}-\frac{1}{2} g_{\mu\nu}g^{\mu\beta}g_{\mu\beta}g^{\alpha\beta}R_{\alpha\beta} =R_{\mu\nu}-\frac{1}{2} \delta^{\beta}_{\nu}\delta^{\alpha}_{\mu}R_{\alpha\beta} =R_{\mu\nu}-\frac{1}{2} R_{\mu\nu} =\frac{1}{2} R_{\mu\nu}$

which cannot be correct. I would be very grateful if someone could clear this up for me. Thank you in advance.

Last edited: Sep 8, 2007
2. Sep 8, 2007

mjsd

your $$\mu \nu$$ indices don't match on both side of equation.. check that first and foremost
btw, don't mix fixed indices $$\mu \nu$$ with the summation indices.

Last edited: Sep 8, 2007
3. Sep 8, 2007

John_Doe

Yeah - there should be an = there so that it's
$G_{\mu\nu} =R_{\mu\nu}-\frac{1}{2} g_{\mu\nu}g^{\alpha\beta}R_{\alpha\beta} =R_{\mu\nu}-\frac{1}{2} g_{\mu\nu}g^{\mu\beta}g_{\mu\beta}g^{\alpha\beta}R _{\alpha\beta}$
Then the metric tensors contract, yielding the kronecker deltas. I'm not sure where the mistake is.

Last edited: Sep 8, 2007
4. Sep 9, 2007

Dick

$g^{\mu\nu}g_{\mu\nu}=1$ No. It equals 4. Assuming you are in dimension 4. It's a trace.

5. Sep 9, 2007

mjsd

in the second step you are contracting $$\mu$$ (confusing choice of symbol) with the wrong metric tensor, note as I pointed out before $$\mu, \nu$$ are not summation indices. If in doubt always put back in the summation signs and write things out in components.

6. Sep 9, 2007

John_Doe

Thank you very much, except now I seem to have

$$g_{\mu\alpha}g^{\mu\beta}=\delta_{\alpha}^{\beta}=\left\{\begin{array}{cc}1,&\mbox{ if } \alpha=\beta\\0, & \mbox{ if } \alpha\neq\beta\end{array}\right.$$

but also

$$g_{\mu\nu}g^{\mu\nu}=4$$

Marginal confusion there... and I also don't quite understand why it matters which tensors the contraction is done on, other than you get the wrong answer.

All help is appreciated, thank you.

7. Sep 9, 2007

HallsofIvy

$$g_{\mu\alpha}g^{\mu\beta}=\delta_{\alpha}^{\beta}= \left\{\begin{array}{cc}1,&\mbox{ if }\alpha=\beta\\0, & \mbox{ if } \alpha\neq\beta\end{array}\right.$$
is wrong. As Dick pointed out, for any tensor $A^{ij}$, $A_{ij}A^{ij}$ is the trace of A.

8. Sep 9, 2007

John_Doe

I'm sorry, but you've lost me there. It's not equal to $$\delta_{\alpha}^{\beta}$$?

I thought that $$g^{\mu\nu}$$ was defined as the inverse of the $$g_{\mu\nu}$$. After all, $$g^{\mu\nu}=\frac{G(\mu,\nu)}{g}$$ if $$G(\mu,\nu)$$ denotes the cofactors of $$g_{\mu\nu}$$ and $$g = |g_{\mu\nu}|$$.

Edit:
Taking into account corrections,
$G_{\mu\nu} =R_{\mu\nu}-\frac{1}{2} g_{\mu\nu}g^{\alpha\beta}R_{\alpha\beta} =R_{\mu\nu}-\frac{1}{8} g_{\mu\nu}g^{\mu\beta}g_{\mu\beta}g^{\alpha\beta}R _{\alpha\beta} =R_{\mu\nu}-\frac{1}{8} \delta^{\beta}_{\nu}\delta^{\alpha}_{\mu}R_{\alpha \beta} =R_{\mu\nu}-\frac{1}{8} R_{\mu\nu} =\frac{7}{8} R_{\mu\nu}$

Last edited: Sep 9, 2007
9. Sep 9, 2007

dextercioby

It's still not okay. You must not use the same index both as "dummy index" and as summation index, and here i mean "\mu" and must not use the same index twice as a summation index, and here i mean "\beta".

10. Sep 9, 2007

John_Doe

$G_{\mu\nu} =R_{\mu\nu}-\frac{1}{2} g_{\mu\nu}g^{\alpha\beta}R_{\alpha\beta} =R_{\mu\nu}-\frac{1}{8} g_{\mu\nu}g^{\sigma\tau}g_{\sigma\tau}g^{\alpha\beta}R _{\alpha\beta}$

So all the working here is now correct, even if it doesn't help in the slightest? That's a good thing. Thank you all very much.

11. Sep 9, 2007

Dick

Yes, it's now correct. But it doesn't say very much, just 4*(1/8)=1/2. What were you trying to prove to begin with?

12. Sep 9, 2007

John_Doe

No, it just seemed to me at first, by eye, that the equation should simpllify, which is obviously wrong. It's not supposed to say much now - I'm just insterested in the fact that all the working is now correct.

Except, one last thing: what if I have

$$g_{\alpha\beta}g^{\gamma\delta}$$

and each of the indices does not appear anywhere else in the equation? Shouldn't $\alpha=\gamma$ and $\beta=\delta$?

13. Sep 10, 2007

dextercioby

No, why would they they have to be like that ? Notice it's just a tensor product. 2 2-nd rank tensors multiplied yeilding a (2,2) 4-th rank tensor.