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Riemann-Christoffel Covariant differentation

  1. Jun 16, 2004 #1
    Hi

    Given the Riemann-Christoffel Tensor :

    [tex]R_{i{\text{ }},{\text{ }}jk}^l = \partial _j \Gamma _{ik}^l - \partial _k \Gamma _{ij}^l + \Gamma _{ik}^r \Gamma _{jr}^l - \Gamma _{ij}^r \Gamma _{kr}^l[/tex]

    I'm looking for the proof :

    [tex]\nabla _t R_{i,rs}^l = \partial _{rt} \Gamma _{si}^l - \partial _{st} \Gamma _{ri}^l[/tex]

    Thanks for your help
     
  2. jcsd
  3. Jun 16, 2004 #2
    Sorry I did not find the proof since yesterday evening when I saw your thread for the first time. But one thing is sure: the second equation is true because leading to the Bianchies identities used in general relativity. Blackforest
     
  4. Jun 18, 2004 #3

    robphy

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  5. Jun 18, 2004 #4

    turin

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    You're using two different symbols, [tex]\partial[/tex] and [tex]\nabla[/tex], but I don't quite understand the distinction. Is [tex]\nabla[/tex] the covariant derivative?

    And what's the difference between [tex]\partial _{r}[/tex] and [tex]\partial _{rt}[/tex]? Is the second one just a shorthad for [tex]\partial _{r} \partial _{t}[/tex]?
     
    Last edited: Jun 18, 2004
  6. Jun 22, 2004 #5

    DW

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    I don't think it is generally correct, so good luck.
     
  7. Jun 22, 2004 #6

    robphy

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    The left-hand side is a tensor.
    Is the right-hand side written in a special choice of coordinates?
     
  8. Jun 22, 2004 #7

    DW

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    Right, the right hand side is what one gets for local free fall frames. I don't think the relation will hold in general. When I take the covariant derivative of the Riemann tensor for general coordinates I wind up with many terms on the right hand side a whole lot of which would have to subtract each other off for it to reduce to that and I don't see them doing that. The extra terms are products with Christophel symbols which vanish in local free fall, but otherwise are not zero.
     
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