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Christoffel symbols

  1. Oct 11, 2008 #1
    Hi guys, I'm studying C. symbols for my G.R. class and have some doubts I hope you can clear out. First, I just saw this in the wikipedia article for C.s.:
    0 = gik;l= gik;l - gmk [itex]\Gamma[/itex]mil - gim [itex]\Gamma[/itex]mkl

    By permuting the indices, and resumming, one can solve explicitly for the Christoffel symbols:

    [itex]\Gamma[/itex]ikl = [itex]\frac{1}{2}[/itex]gim (THREE TERMS GO HERE)
    I tried hard to write the three terms in Latex but I couldn't so I apologize... the expression I am referring to is here: http://en.wikipedia.org/wiki/Christoffel_symbols
    Go down a little to where it says Definition... the 2nd and 3rd expressions are the ones I am referring to.
    So my question is... how do they get from 0 = gik;l= gik;l - gmk [itex]\Gamma[/itex]mil - gim [itex]\Gamma[/itex]mkl to the next expression, what steps did they follow?

    I also have a general question about C.s. ... As far as I know, C.s. of the second kind are used quite frequently, much more that those of the first kind... What are these C.s. of the first kind for then? What do they mean compared to those of the second kind and why are the latter more frequent?
    Thanks a lot for any help... you can also just comment on C.s. I'd appreciate it.
  2. jcsd
  3. Oct 12, 2008 #2


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    I'm not sure if this is relevant to GR, but in neuroscience they are used to write the metric geodesic equation in coordinates. Say you have a "standard" form of geodesic equation with one sort of CS. If you multiply that equation with the inverse of the metric, you get another form of the geodesic equation, with another sort of CS. The other sort of CS is defined to make this other form also look "standard".

    Biess A, Liebermann DG, Flash T. A computational model for redundant human three-dimensional pointing movements: integration of independent spatial and temporal motor plans simplifies movement dynamics. J Neurosci. 2007 Nov 28;27(48):13045-64.
    The paper should be free from http://www.pubmed.org [Broken] or the Journal's page, because of the recent NIH public access policy. I think there's a typo in their definition of the relation between the two sorts of CS, but their equations make sense if I use the definitions at http://mathworld.wolfram.com/ChristoffelSymbol.html.
    Last edited by a moderator: May 3, 2017
  4. Oct 12, 2008 #3


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    You misquoted a semicolom.

    Using the definition of the covariant derivative:

    \nabla_{c}g_{ab} = \partial_{c}g_{ab} - \Gamma^{d}_{ca}g_{db} - \Gamma^{d}_{cb}g_{ad}

    \nabla_{b}g_{ca} = \partial_{b}g_{ca} - \Gamma^{d}_{bc}g_{da} - \Gamma^{d}_{ba}g_{cd}

    \nabla_{a}g_{bc} = \partial_{a}g_{bc} - \Gamma^{d}_{ab}g_{dc} - \Gamma^{d}_{ac}g_{bd}

    Now subtract the 2 last ones from the first equation, and you get those partial derivative terms and all those connection terms. Now use the symmetry of the connection, and the only two connection terms which survive are

    \Gamma^{d}_{ba}g_{cd} + \Gamma^{d}_{ab}g_{dc} = 2 \Gamma^{d}_{ba}g_{cd}

    So appearantly, all thos partial derivative terms are equal to the expression here above.
  5. Oct 12, 2008 #4
    Thanks guys.
    So CS are simply objects which are not invariant under coordinate change... they don't transform under homogeneous transformations that is. Is there some other way to "picture" these CS? Or are they simply useful mathematical objects?
    Last edited: Oct 12, 2008
  6. Oct 12, 2008 #5


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    Yes, Christoffel Symbols are not tensors- they are not invariant under coordinate change. Essentially, the Christoffel symbols measure the difference between a derivative measured in the coordinate plane and the rate of change along the actual surface. That's basically what Haushofer's
    [tex]\nabla_{c}g_{ab} = \partial_{c}g_{ab} - \Gamma^{d}_{ca}g_{db} - \Gamma^{d}_{cb}g_{ad}[/tex]
    says. The term on the left is a rate of change in the surface, the partial derivative is the derivative in the tangent plane, and the Christoffel symbols give the difference.
  7. Oct 12, 2008 #6
    I see. Thanks HallofIvy, that was helpful.
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