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

<br /> 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}<br />

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

 

[mjsd]

<br /> 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}<br />

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,&amp;\mbox{ if }<br /> \alpha=\beta\\0, &amp; \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,&amp;\mbox{ if }\alpha=\beta\\0, &amp; \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,
<br /> 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}<br />

 

[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]

<br /> 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}<br />

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.