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Deriving Einstein tensor, Riemmann tensor step

  1. Mar 13, 2013 #1
    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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  3. Mar 13, 2013 #2

    George Jones

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    [tex](g^{\alpha\sigma}g_{\alpha \rho})(g_{\alpha\sigma}g^{\alpha\rho} ) = 16[/tex]
     
  4. Mar 13, 2013 #3
    mmm, how can one expand R_{\mu\nu} then? Does one have to multiply by eta's?
     
  5. Mar 13, 2013 #4

    George Jones

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    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.
     
  6. Mar 13, 2013 #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
     
  7. Mar 13, 2013 #6
    It is on page 540
     
  8. Mar 13, 2013 #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?
     
  9. Mar 13, 2013 #8

    Bill_K

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

    (Of course the original expression is invalid since it contains four α's, but I'm assuming...)
     
  10. Mar 13, 2013 #9
    Is my last post about Riemann tensors correct?
     
  11. Mar 13, 2013 #10

    Mentz114

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    Like this ?

    [itex] R_{\alpha \beta}\ne{R^{\rho}}_{\alpha \beta \sigma}[/itex]
    [itex] R_{\alpha \beta}={R^{\sigma}}_{\alpha\sigma \beta }[/itex]
     
  12. Mar 13, 2013 #11

    Fredrik

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    Or like this ##R^\rho{}_{\alpha\beta\sigma}##.
     
  13. Mar 13, 2013 #12

    Fredrik

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    The numbers could be equal, but the tensors can't be.
     
  14. Mar 13, 2013 #13

    Bill_K

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    Or using UTF, like this: Rραβσ.
     
  15. Mar 13, 2013 #14

    George Jones

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    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]
    Sorry, I still don't know what you are asking. I think you need to practise index gymnastics.
     
    Last edited: Mar 13, 2013
  16. Mar 13, 2013 #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.
     
  17. Mar 13, 2013 #16

    Fredrik

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    Wikipedia defines ##R_{\mu\nu}## by ##R_{\mu\nu} = {R^\rho}_{\mu\rho\nu}##. You understand that repeated indices imply summation, right?
     
  18. Mar 13, 2013 #17

    George Jones

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    The text that pleasehelpmeno is using, Hobson et al, uses ##R_{\mu\nu} = {R^\rho}_{\mu\nu\rho}##.
     
  19. Mar 14, 2013 #18
    [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?
     
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