Deriving Einstein tensor, Riemmann tensor step

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pleasehelpmeno
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Hi i have been following Hobson in their attempt to derive the einstein tensor, I have split the varied action into three terms and want to factor out [itex]\delta(g^{\mu\nu})[/itex] terms.

The Riemann tensor [itex]R_{\mu \nu}[/itex] must be expanded to [itex]R^{\rho}_{\mu \nu p}[/itex] and then contracted back to the original form. To do this should one simply multiply by [itex](g^{\alpha\sigma}g_{\alpha \rho})(g_{\alpha\sigma}g^{\alpha\rho} )[/itex] and expand with the first bracket and then contract down with the second?
 
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pleasehelpmeno said:
simply multiply by [itex](g^{\alpha\sigma}g_{\alpha \rho})(g_{\alpha\sigma}g^{\alpha\rho} )[/itex]

[tex](g^{\alpha\sigma}g_{\alpha \rho})(g_{\alpha\sigma}g^{\alpha\rho} ) = 16[/tex]
 
mmm, how can one expand R_{\mu\nu} then? Does one have to multiply by eta's?
 
Ok well because it is 4x4 isn't it,, each g^{}g_{} will be a kronecker delta = 4 or 0, I ma just finding the page number
 
It is on page 540
 
perhaps i am being thick is [itex]R_{\alpha \beta}=R^{\rho}_{\alpha \beta \sigma}[/itex] so that then one wouldn't need to contract it?

Do you know how to write tensors in latex so that there is say a spacing between top and bottom?
 
Is my last post about Riemann tensors correct?
 
pleasehelpmeno said:
perhaps i am being thick is [itex]R_{\alpha \beta}=R^{\rho}_{\alpha \beta \sigma}[/itex] so that then one wouldn't need to contract it?

Do you know how to write tensors in latex so that there is say a spacing between top and bottom?
Like this ?

[itex]R_{\alpha \beta}\ne{R^{\rho}}_{\alpha \beta \sigma}[/itex]
[itex]R_{\alpha \beta}={R^{\sigma}}_{\alpha\sigma \beta }[/itex]
 
Bill_K said:
(Of course the original expression is invalid since it contains four α's, but I'm assuming...)

Yikes, my eyes skimmed right over the illegal expression. :redface: I immediately read it as

[tex]g^{\alpha\sigma}g_{\alpha \rho}g_{\alpha\sigma}g^{\alpha\rho} = (g^{\alpha\sigma} g_{\alpha\sigma}) (g_{\alpha \rho} g^{\alpha\rho} ) = 4 \times 4 = 16[/tex]
pleasehelpmeno said:
It is on page 540

Sorry, I still don't know what you are asking. I think you need to practise index gymnastics.
 
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I am confused can you explain how to turn [itex]R^{\sigma}_{\mu\nu\rho} \mbox{ to } R_{\mu\nu}[/itex] by contraction, I thought that my method was correct but clearly not, I am self taught so there are some things that i have missed.
 
pleasehelpmeno said:
I am confused can you explain how to turn [itex]R^{\sigma}_{\mu\nu\rho} \mbox{ to } R_{\mu\nu}[/itex] by contraction, I thought that my method was correct but clearly not, I am self taught so there are some things that i have missed.
Wikipedia defines ##R_{\mu\nu}## by ##R_{\mu\nu} = {R^\rho}_{\mu\rho\nu}##. You understand that repeated indices imply summation, right?
 
Fredrik said:
Wikipedia defines ##R_{\mu\nu}## by ##R_{\mu\nu} = {R^\rho}_{\mu\rho\nu}##. You understand that repeated indices imply summation, right?

The text that pleasehelpmeno is using, Hobson et al, uses ##R_{\mu\nu} = {R^\rho}_{\mu\nu\rho}##.
 
[tex] \delta s = \delta \int d^4 x \sqrt{g} R = \int d^4 x \delta (\sqrt{g} R_{\mu\nu}g^{\mu\nu})<br /> = \int d^4 x \left[ (\delta\sqrt{g})R + \sqrt{g}(\delta R_{\mu\nu})g^{\mu\nu} + \sqrt{g}R_{\mu\nu}(\delta g^{\mu\nu}) \right][/tex]
does this look what you have to start with?