Christoffel Symbols: Intuitive Proof for Covariant Derivative of Metric Tensor

[zwoodrow]

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.

 

[nicksauce]

Check out chapter 3 of Wald's GR book.

 

[Fredrik]

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.

 

[HallsofIvy]

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.

 

[dx]

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.