# Deriving Einstein tensor, Riemmann tensor step

1. Mar 13, 2013

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?

2. Mar 13, 2013

### George Jones

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

3. Mar 13, 2013

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

4. Mar 13, 2013

### George Jones

Staff Emeritus
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.

5. Mar 13, 2013

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

6. Mar 13, 2013

It is on page 540

7. Mar 13, 2013

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?

8. Mar 13, 2013

### Bill_K

(gασgαρ)(gασgαρ) = (δσρ)(δσρ) = δσσ = 4

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

9. Mar 13, 2013

Is my last post about Riemann tensors correct?

10. Mar 13, 2013

### Mentz114

Like this ?

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

11. Mar 13, 2013

### Fredrik

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

12. Mar 13, 2013

### Fredrik

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

13. Mar 13, 2013

### Bill_K

Or using UTF, like this: Rραβσ.

14. Mar 13, 2013

### George Jones

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

$$g^{\alpha\sigma}g_{\alpha \rho}g_{\alpha\sigma}g^{\alpha\rho} = (g^{\alpha\sigma} g_{\alpha\sigma}) (g_{\alpha \rho} g^{\alpha\rho} ) = 4 \times 4 = 16$$
Sorry, I still don't know what you are asking. I think you need to practise index gymnastics.

Last edited: Mar 13, 2013
15. Mar 13, 2013

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.

16. Mar 13, 2013

### Fredrik

Staff Emeritus
Wikipedia defines $R_{\mu\nu}$ by $R_{\mu\nu} = {R^\rho}_{\mu\rho\nu}$. You understand that repeated indices imply summation, right?

17. Mar 13, 2013

### George Jones

Staff Emeritus
The text that pleasehelpmeno is using, Hobson et al, uses $R_{\mu\nu} = {R^\rho}_{\mu\nu\rho}$.

18. Mar 14, 2013

### jfy4

$$\delta s = \delta \int d^4 x \sqrt{g} R = \int d^4 x \delta (\sqrt{g} R_{\mu\nu}g^{\mu\nu}) = \int d^4 x \left[ (\delta\sqrt{g})R + \sqrt{g}(\delta R_{\mu\nu})g^{\mu\nu} + \sqrt{g}R_{\mu\nu}(\delta g^{\mu\nu}) \right]$$