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General Relativity: Prove symmetry of Einstein tensor

  1. Apr 19, 2012 #1
    1. The problem statement, all variables and given/known data

    Show that Gij = Gji using the Riemann tensor identity (below)


    2. Relevant equations

    Gij = Rij - 1/2(gijR)

    Rabcd + Rbcad + Rcabd = 0

    R = gmrRmr

    Rmr = Rmnrn

    3. The attempt at a solution

    I have tried to put the Ricci tensor and Ricci scalar (from the Gij equation) into full Riemann tensor form using the metric. For Gij I get

    Rij = Rinjn = gnaRinja

    gijR = gia gjb gab R = gia gjb Rab = gia gjb Ranbn = gia gjb gnd Ranbd

    So Gij = ( gnaRinja - 1/2 gia gjb gnd Ranbd )

    I have done the same for Gji but can't see how to link the two equations using the required Riemann identity. Is this the right approach?
     
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  3. Apr 19, 2012 #2

    George Jones

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    Can you show that the Ricci tensor is symmetric?
     
  4. Apr 19, 2012 #3
    I know that the Riemann tensor is skew-symmetric in the first and second pair of indexes, when it is in fully covariant form. I'm not sure if that holds when it is in mixed form though?

    I know that

    Rabcd = -Rbacd and that Rabcd = -Rabdc

    But I'm not sure if

    Rabcd = -Rbacd and Rabcd = -Rabdc

    Looking at the Ricci tensor, in order for Rmr = Rrm I would need

    Rmnrn = Rrnmn

    And I'm not sure how to swap indices between the first pair and second pair on Riemann. Do I need to go all the way back to the equation for Riemann in terms of Christoffels, and Christoffels from the metric? There must be a quicker way, this is a tiny bit of an exam question that should take 5 minutes!
     
  5. Apr 19, 2012 #4

    George Jones

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    Yes.
    No, indices need to have the same upstairs/downstairs locations on the both sides of an equation, i.e., if you raise the d on the left, then then d on the right should be raised. More explicitly,

    [tex]R_{abcd} g^{de} = -R_{abdc}g^{de}[/tex]

    Raise the d in each term of

    [tex]R_{abcd} + R_{bcad} + R_{cabd} = 0[/tex]

    and contract with one of the other indices.
     
  6. Apr 19, 2012 #5
    Thanks George, that helps a bit but I am still missing something. I tried

    gbd(Rabcd + Rbcad + Rcabd) = 0

    Rabcb + Rbcab + Rcabb = 0

    Rabcb - Rcbab + Rcabb = 0

    Rac - Rca + Rcabb = 0

    So I would be OK if that last term were to vanish somehow! Can someone suggest where I go from here?
     
  7. Apr 19, 2012 #6

    George Jones

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    Put the g back in the last term, and look at it more carefully. g is symmetric in b and d, while the last term is antisymmetric in b and d.
     
  8. Apr 24, 2012 #7
    Sorry for the delay, I had to work on some other stuff. I have tried your suggestion but I'm still not getting this :(

    gbd(Rabcd + Rbcad + Rcabd) = 0

    Rabcb + Rbcab + gbdRcabd = 0

    Rabcb + Rbcab - gdbRcadb = 0

    Rabcb - Rcbab - Rcadd = 0

    Rac - Rca - Rcadd = 0

    I'm not sure how that helps me. Have I done it right?
     
  9. Apr 24, 2012 #8

    George Jones

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    You need to show [itex]R_{cad}{}^d = 0[/itex], i.e, you need to show [itex]g^{db} R_{cadb} = 0[/itex].

    g is symmetric in b and d, while [itex]R_{cadb}[/itex] is antisymmetric in b and d. b and d are both dummy indices, and so can be relabeled. Use these facts to show [itex]g^{db} R_{cadb} = 0[/itex].
     
  10. May 7, 2012 #9
    I forgot to thank you for this George, I figured it out in the end (although it didn't come up in my exam!). Your help is much appreciated.
     
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