- #1
Physicist97
- 31
- 4
In Sean Carroll's Lecture Notes on General Relativity (Chapter 3, Page 60), in the chapter on Curvature, he derives the definition of the Christoffels Symbols by assuming the connection is metric compatible and torsion free. He then takes the covariant derivative of the metric and cycles through the indices to arrive at the usual definition of the Christoffel Symbols, that is ##{\Gamma}^{\sigma}_{\mu\nu}=1/2g^{\sigma\rho}({\partial}_{\mu}g_{\nu\rho}+{\partial}_{\nu}g_{\rho\mu}-{\partial}_{\rho}g_{\mu\nu})## , but why is it not possible to derive the definition of the Christoffel Symbols this way. Assuming metric compatibility , ##{\nabla}_{\mu}g_{\nu\sigma}={\partial}_{\mu}g_{\nu\sigma}-{\Gamma}^{\lambda}_{\mu\nu}g_{\lambda\sigma}-{\Gamma}^{\lambda}_{\mu\sigma}g_{\nu\lambda}=0## . From here you can subtract the partial derivative from both sides and multiply by a negative to get you, ##{\Gamma}^{\lambda}_{\mu\nu}g_{\lambda\sigma}+{\Gamma}^{\lambda}_{\mu\sigma}g_{\nu\lambda}={\partial}_{\mu}g_{\nu\sigma}## . Now multiplying both sides by ##g^{\lambda\sigma}## leaves ##{\Gamma}^{\lambda}_{\mu\nu}+{\delta}^{\sigma}_{\nu}{\Gamma}^{\lambda}_{\mu\sigma}=g^{\lambda\sigma}{\partial}_{\mu}g_{\nu\sigma}## . The delta is the Kronecker Delta, ##{\delta}^{\sigma}_{\nu}=g^{\lambda\sigma}g_{\lambda\nu}## and it is 1 when ##{\sigma}={\nu}## and 0 otherwise. The Kronecker Delta will simply change the ##{\sigma}## of the Christoffel Symbol to a ##{\mu}## , thus getting you ##{\Gamma}^{\lambda}_{\mu\nu}=1/2g^{\lambda\sigma}{\partial}_{\mu}g_{\nu\sigma}## . How is any of what I did wrong, other than arriving at what I assume is a wrong definition of the Christoffel Symbol? All I did was assume metric compatibility, and then solved for the Christoffel Symbol. I would appreciate some clarification to whether this is wrong, and if so why :) .