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Covariant derivative of riemann tensor

  1. Aug 3, 2011 #1
    what would Rabcd;e look like in terms of it's christoffels?

    or Rab;c
     
  2. jcsd
  3. Aug 3, 2011 #2

    WannabeNewton

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    Well [tex]\bigtriangledown \gamma R^{\alpha }_{\beta \mu \nu } = \frac{\partial R^{\alpha }_{\beta \mu \nu }}{\partial x^{\gamma }} + \Gamma ^{\alpha }_{\sigma \gamma }R^{\sigma }_{\beta \mu \nu } - \Gamma^{\sigma }_{\gamma \beta }R^{\alpha }_{\sigma \mu \nu } - \Gamma ^{\sigma }_{\gamma \mu }R^{\alpha }_{\beta \sigma \nu } - \Gamma ^{\sigma }_{\gamma \nu }R^{\alpha }_{\beta \mu \sigma }[/tex] by applying the general formula for the covariant derivative. If you want to see what it looks like in terms of the christoffel symbols use the respective formula for the Riemann tensor [tex]R^{\alpha }_{\beta \mu \nu } = \frac{\partial \Gamma ^{\alpha }_{\beta \nu } }{\partial x^{\mu }} - \frac{\partial \Gamma ^{\alpha }_{\beta \mu } }{\partial x^{\nu }} + \Gamma ^{\alpha }_{\sigma \mu }\Gamma ^{\sigma }_{\beta \nu} - \Gamma ^{\alpha }_{\sigma \nu }\Gamma ^{\sigma }_{\beta \mu }[/tex] but it will look pretty messy.
     
  4. Aug 4, 2011 #3
    OK thank you. I saw in a youtube lecture though that Rab;a=1/2gabR,a. I cannot find the proof for this. can anyone help me with this problem?
     
    Last edited: Aug 4, 2011
  5. Aug 4, 2011 #4

    dextercioby

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    That's standard textbook material. But you don't need the textbook itself, just know what the tensors look like in terms of the torsion-less connection. So pick up your pencil and a sheet of paper & do it.
     
  6. Aug 4, 2011 #5
    I've tried over and over again. If someone could post a solution I'd really appreciate it.
     
  7. Aug 4, 2011 #6

    WannabeNewton

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    Could you give a link to the video =D?
     
  8. Aug 4, 2011 #7
    about 1:05:44 in
     
    Last edited by a moderator: Sep 25, 2014
  9. Aug 4, 2011 #8

    WannabeNewton

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    It essentially boils down to a really long but not hard calculation but its easy to make mistakes with all the indeces. If you're intent on doing it then try [tex]\bigtriangledown _{\mu }R^{\mu \nu } = \bigtriangledown _{\mu }(g^{\mu \alpha }g^{\nu \beta }R_{\alpha \beta }) = g^{\mu \alpha }g^{\nu \beta }\bigtriangledown _{\mu }R^{\sigma }_{\alpha \sigma \beta }[/tex] and work down to the christoffel symbols and try to get the expression that equals [itex]\frac{1}{2}\bigtriangledown _{\mu }(g^{\mu \nu }R)[/itex] with R expanded, formula - wise,up to the christoffel symbols. You can try googling or looking through your texts but most texts (at least the ones I own) simply skip the tedious calculations and also just state the bianchi identities which are related to what Susskind is talking about in the video.
     
  10. Aug 4, 2011 #9
    yeah ive seen the method for the contraction of bianchi identities to arrive at the einstein tensor but I was determined to use this method too. It's really tedious and i just can't find the expression. I guess I'm just forced to believe the equation but thanks so much for the help.
     
  11. Aug 5, 2011 #10

    dextercioby

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    The second Bianchi identity for the Riemann tensor (torsion-less manifold, so that the curvature 2-form is closed) by double contraction with the covariantly constant metric tensor immediately yields

    [tex] \nabla^{\mu}\left(R_{\mu\nu} - \frac{1}{2} g_{\mu\nu}R\right) = 0 [/tex]
     
  12. Aug 5, 2011 #11

    Ben Niehoff

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    The curvature 2-form is not closed, generally speaking. Everything else you say is correct, though. The second Bianchi identity

    [tex]\nabla_{[\lambda} R_{\mu\nu]}{}^\rho{}_\sigma = 0[/tex]
    is not the exterior derivative of the curvature 2-form. (I used symmetries [itex]R^\rho{}_{\sigma\mu\nu}[/itex] to make the formula more legible). Here is a case where index notation can be deceptive.
     
  13. Aug 6, 2011 #12

    dextercioby

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    Yes, you are right. I stand corrected.
     
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