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Derivation of the value of christoffel symbol

  1. Apr 19, 2012 #1
    Hi,
    I am new to general relativity and as I would like to find out how we could derive the value of christoffel symbol in terms of the metric tensor.
    I have also heard that it was given as a definition for the christoffel symbol and would like a clarification on that.

    Regards
    Bltzmn2012
     
  2. jcsd
  3. Apr 19, 2012 #2

    jtbell

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    Staff: Mentor

  4. Apr 19, 2012 #3
    By permuting the indices, and resumming, one can solve explicitly for the Christoffel symbols as a function of the metric tensor:

    I didnt quite get that(although I was given the formula underneath).could you please explain what is permuting the indices. I am really new to this stuff.

    Regards
    Bltzmnn2012
     
  5. Apr 19, 2012 #4
    [tex]\Gamma^\rho_{~\mu \nu}=\frac{1}{2} g^{\rho \lambda} (\partial_\mu g_{\nu \lambda}+\partial_\nu g_{\mu \lambda}-\partial_\lambda g_{\mu \nu})[/tex]

    It's relatively straightforward to calculate the components of the Christoffel symbols from the components of the metric. It can get pretty tedious though.
     
    Last edited: Apr 19, 2012
  6. Apr 19, 2012 #5

    pervect

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    Yes - a few comments on notation for the OP
    [tex]\partial_\mu = \frac{\partial}{\partial_\mu}[/tex]

    Summation over lambda is implied by the Einstein convention, (which is that you sum over repeated indices), i.e: [correction to index]

    [tex]\Gamma^\rho_{~\mu \nu}=\sum_{\lambda=0}^{\lambda=3} \frac{1}{2} g^{\rho \lambda} (\partial_\mu g_{\nu \lambda}+\partial_\nu g_{\mu \lambda}-\partial_\lambda g_{\mu \nu})[/tex]
     
    Last edited: Apr 19, 2012
  7. Apr 19, 2012 #6

    Nabeshin

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    The sum should extend from 0-3, not 0-4 :)
     
  8. Apr 19, 2012 #7

    pervect

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    Ooops - yes, good point, space-time has 4 dimensions, not 5. I corrected the original, for whatever it's worth.
     
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