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Help! Covariant Derivative of Ricci Tensor the hard way.

  1. Mar 3, 2012 #1
    I am trying to calculate the covariant derivative of the Ricci Tensor the way Einstein did it, but I keep coming up with

    [itex]\nabla_{μ}[/itex]R[itex]_{αβ}[/itex]=[itex]\frac{∂}{∂x^{μ}}[/itex]R[itex]_{αβ}[/itex]-2[itex]\Gamma^{α}_{μ\gamma}[/itex]R[itex]_{αβ}[/itex]

    or


    [itex]\nabla_{μ}[/itex]R[itex]_{αβ}[/itex]=[itex]\frac{∂}{∂x^{μ}}[/itex]R[itex]_{αβ}[/itex]-[itex]\Gamma^{α}_{μ\gamma}[/itex]R[itex]_{αβ}[/itex]-[itex]\Gamma^{β}_{μ\gamma}[/itex]R[itex]_{αβ}[/itex]

    Any help will be much appreciated.
     
  2. jcsd
  3. Mar 3, 2012 #2

    Matterwave

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    Your second expression is almost right, except you have some trouble with indices. If alpha and beta are free indices on the left, you should not use them as summation indices on the right. Use gamma as your summation index, and then alpha and beta should remain free indices.
     
  4. Mar 4, 2012 #3
    Then, how did Einstein get

    [itex]\nabla_{μ}[/itex]R[itex]_{αβ}[/itex]=[itex]\frac{1}{2}[/itex]g[itex]_{αβ}[/itex][itex]\nabla_{μ}[/itex]R
     
  5. Mar 4, 2012 #4
    Try the second Bianchi identity, contract it twice and you get the result.
     
  6. Mar 4, 2012 #5
    Thanks for the advice, but anyone can do it that way. I'm trying to do it the way Einstein did it; the hard way. Einstein didn't know about the Bianchi Identities.
     
  7. Mar 4, 2012 #6
    You can put the definition of the Ricci and scalar curvature tensor in term of metric...
     
  8. Mar 4, 2012 #7

    pervect

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    Einstein didn't have symbolic algebra programs, I would guess. But that's what I'd use.
     
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