Deriving Einstein tensor, Riemmann tensor step

[pleasehelpmeno]

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 \delta(g^{\mu\nu}) terms.

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

 

[George Jones]

simply multiply by (g^{\alpha\sigma}g_{\alpha \rho})(g_{\alpha\sigma}g^{\alpha\rho} )

(g^{\alpha\sigma}g_{\alpha \rho})(g_{\alpha\sigma}g^{\alpha\rho} ) = 16

 

[pleasehelpmeno]

mmm, how can one expand R_{\mu\nu} then? Does one have to multiply by eta's?

 

[George Jones]

First, why does

(g^{\alpha\sigma}g_{\alpha \rho})(g_{\alpha\sigma}g^{\alpha\rho} ) = 16?

Can you be more specific? I have Hobson et al.

 

[pleasehelpmeno]

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

 

[pleasehelpmeno]

It is on page 540

 

[pleasehelpmeno]

perhaps i am being thick is R_{\alpha \beta}=R^{\rho}_{\alpha \beta \sigma} 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?

 

[Bill_K]

(g^ασg_αρ)(g_ασg^αρ) = (δ^σ_ρ)(δ_σ^ρ) = δ^σ_σ = 4

(Of course the original expression is invalid since it contains four α's, but I'm assuming...)

 

[pleasehelpmeno]

Is my last post about Riemann tensors correct?

 

[Mentz114]

perhaps i am being thick is R_{\alpha \beta}=R^{\rho}_{\alpha \beta \sigma} 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 ?

R_{\alpha \beta}\ne{R^{\rho}}_{\alpha \beta \sigma}
R_{\alpha \beta}={R^{\sigma}}_{\alpha\sigma \beta }

 

[Fredrik]

Like this ?

R_{\alpha \beta}\ne{R^{\rho}}_{\alpha \beta \sigma}
R_{\alpha \beta}={R^{\sigma}}_{\alpha\sigma \beta }

Or like this $R^\rho{}_{\alpha\beta\sigma}$.

 

[Fredrik]

perhaps i am being thick is R_{\alpha \beta}=R^{\rho}_{\alpha \beta \sigma} so that then one wouldn't need to contract it?

The numbers could be equal, but the tensors can't be.

 

[Bill_K]

Or like this $R^\rho{}_{\alpha\beta\sigma}$.

Or using UTF, like this: R^ρ_αβσ.

 

[George Jones]

(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. I immediately read it as

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

It is on page 540

Sorry, I still don't know what you are asking. I think you need to practise index gymnastics.

 

[pleasehelpmeno]

I am confused can you explain how to turn R^{\sigma}_{\mu\nu\rho} \mbox{ to } R_{\mu\nu} 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.

 

[Fredrik]

I am confused can you explain how to turn R^{\sigma}_{\mu\nu\rho} \mbox{ to } R_{\mu\nu} 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?

 

[George Jones]

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 $R_{\mu\nu} = {R^\rho}_{\mu\nu\rho}$.

 

[jfy4]

<br /> \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]<br />
does this look what you have to start with?