Deriving Einstein tensor, Riemmann tensor step

  • #1

Main Question or Discussion Point

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?
 

Answers and Replies

  • #2
George Jones
Staff Emeritus
Science Advisor
Gold Member
7,261
790
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]
 
  • #3
mmm, how can one expand R_{\mu\nu} then? Does one have to multiply by eta's?
 
  • #4
George Jones
Staff Emeritus
Science Advisor
Gold Member
7,261
790
First, why does

[tex](g^{\alpha\sigma}g_{\alpha \rho})(g_{\alpha\sigma}g^{\alpha\rho} ) = 16?[/tex]

Can you be more specific? I have Hobson et al.
 
  • #5
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
It is on page 540
 
  • #7
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?
 
  • #8
Bill_K
Science Advisor
Insights Author
4,155
194
(gασgαρ)(gασgαρ) = (δσρ)(δσρ) = δσσ = 4

(Of course the original expression is invalid since it contains four α's, but I'm assuming...)
 
  • #9
Is my last post about Riemann tensors correct?
 
  • #10
5,428
290
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]
 
  • #11
Fredrik
Staff Emeritus
Science Advisor
Gold Member
10,851
406
Like this ?

[itex] R_{\alpha \beta}\ne{R^{\rho}}_{\alpha \beta \sigma}[/itex]
[itex] R_{\alpha \beta}={R^{\sigma}}_{\alpha\sigma \beta }[/itex]
Or like this ##R^\rho{}_{\alpha\beta\sigma}##.
 
  • #12
Fredrik
Staff Emeritus
Science Advisor
Gold Member
10,851
406
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?
The numbers could be equal, but the tensors can't be.
 
  • #13
Bill_K
Science Advisor
Insights Author
4,155
194
Or like this ##R^\rho{}_{\alpha\beta\sigma}##.
Or using UTF, like this: Rραβσ.
 
  • #14
George Jones
Staff Emeritus
Science Advisor
Gold Member
7,261
790
(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]
It is on page 540
Sorry, I still don't know what you are asking. I think you need to practise index gymnastics.
 
Last edited:
  • #15
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.
 
  • #16
Fredrik
Staff Emeritus
Science Advisor
Gold Member
10,851
406
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?
 
  • #17
George Jones
Staff Emeritus
Science Advisor
Gold Member
7,261
790
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}##.
 
  • #18
649
3
[tex]
\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]
[/tex]
does this look what you have to start with?
 

Related Threads for: Deriving Einstein tensor, Riemmann tensor step

  • Last Post
Replies
23
Views
3K
  • Last Post
Replies
5
Views
3K
  • Last Post
Replies
15
Views
4K
Replies
2
Views
678
  • Last Post
Replies
3
Views
3K
  • Last Post
Replies
1
Views
763
  • Last Post
Replies
1
Views
3K
  • Last Post
Replies
6
Views
2K
Top