Clearing Up Confusion: G_{\mu\nu} Explained

[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

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

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

 

[mjsd]

$G_{\mu\nu}=R_{\mu\nu}-\frac{1}{2} g_{\mu\nu}g^{\alpha\beta}R_{\alpha\beta}<br /> R_{\mu\nu}-\frac{1}{2} g_{\mu\nu}g^{\mu\beta}g_{\mu\beta}g^{\alpha\beta}R_{\alpha\beta}$

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.

 

[John_Doe]

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

 

[Dick]

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

 

[mjsd]

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

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.

 

[John_Doe]

Thank you very much, except now I seem to have

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

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.

 

[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.

 

[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,
$G_{\mu\nu}<br /> =R_{\mu\nu}-\frac{1}{2} g_{\mu\nu}g^{\alpha\beta}R_{\alpha\beta}<br /> =R_{\mu\nu}-\frac{1}{8} g_{\mu\nu}g^{\mu\beta}g_{\mu\beta}g^{\alpha\beta}R _{\alpha\beta}<br /> =R_{\mu\nu}-\frac{1}{8} \delta^{\beta}_{\nu}\delta^{\alpha}_{\mu}R_{\alpha \beta}<br /> =R_{\mu\nu}-\frac{1}{8} R_{\mu\nu}<br /> =\frac{7}{8} R_{\mu\nu}$

 

[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".

 

[John_Doe]

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

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.

 

[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?

 

[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 $\beta=\delta$?

 

[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.