Christoffel Symbols: Intuitive Proof for Covariant Derivative of Metric Tensor

zwoodrow
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I am learning about christoffel symbols and there is a pretty standard representation of christoffel symbols as a linear combination of products of the metric tensor and the metric tensors derivative. However when this is derived it is always done in a hoakey manner. Something along the lines of ... do these permutations add this subtract that and walllaaa. I am trying to make a more physically intuitive proof based off the covariant derivative of the metric tensor being equal to zero. Has anyone seen this proof somewhere i haven't got it to work out and i am looking go help.
 
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Check out chapter 3 of Wald's GR book.
 
I think the best place to read about connections is "Riemannian manifolds: an introduction to curvature", by John Lee. But I don't remember how he did this particular thing.
 
The ordinary derivative of a tensor is NOT a tensor. In order to make it one, the "covariant derivative", you have to subtract off the Christoffel symbols- or, to put it another way, the Chrisoffel symbols are the covariant derivative minus the ordinary derivative.
 
zwoodrow said:
I am trying to make a more physically intuitive proof based off the covariant derivative of the metric tensor being equal to zero. Has anyone seen this proof somewhere i haven't got it to work out and i am looking go help.

Yes, you can find it in MTW exercise 8.15. It has an outline solution too.
 
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From $$0 = \delta(g^{\alpha\mu}g_{\mu\nu}) = g^{\alpha\mu} \delta g_{\mu\nu} + g_{\mu\nu} \delta g^{\alpha\mu}$$ we have $$g^{\alpha\mu} \delta g_{\mu\nu} = -g_{\mu\nu} \delta g^{\alpha\mu} \,\, . $$ Multiply both sides by ##g_{\alpha\beta}## to get $$\delta g_{\beta\nu} = -g_{\alpha\beta} g_{\mu\nu} \delta g^{\alpha\mu} \qquad(*)$$ (This is Dirac's eq. (26.9) in "GTR".) On the other hand, the variation ##\delta g^{\alpha\mu} = \bar{g}^{\alpha\mu} - g^{\alpha\mu}## should be a tensor...

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