Metric Tensor – A Derivation About Metric Tensor and Christoffel Symbol

Question

I am reading Landau & Lifshitz's The Classical Theory of Fields. There is a derivation about metric tensor and Christoffel symbol I cannot get. On page 261, section 86, first the authors introduce the relation between the differential of the metric tensor and the differential of its determinant: $$dg=gg^{ik}dg_{ik}=-gg_{ik}dg^{ik}.\tag{86.4}$$ Later, they compute a term:$$g^{kl}\Gamma^{i}_{kl}=\frac{1}{2}g^{kl}g^{im}(\frac{\partial g_{mk}}{\partial x^l}+\frac{\partial g_{lm}}{\partial x^k}-\frac{\partial g_{kl}}{\partial x^m})=g^{kl}g^{im}(\frac{\partial g_{mk}}{\partial x^l}-\frac{1}{2}\frac{\partial g_{kl}}{\partial x^m})$$and say, with the help of $(86.4)$, this can be transformed to $$g^{kl}\Gamma^{i}_{kl}=-\frac{1}{\sqrt{-g}}\frac{\partial (\sqrt{-g}g^{ik})}{\partial x^k}.\tag{86.6}$$ I cannot get the result $(86.6)$. Can someone help?

Answer 1

The piece you may be missing is that $$ 0 = \frac{\partial \delta^i {}_k}{\partial x^l} = \frac{\partial (g^{im} g_{mk})}{\partial x^l} = g^{im} \frac{\partial g_{mk}}{\partial x^l} + g_{mk}\frac{\partial g^{im} }{\partial x^l} $$ where I have applied the product rule in the last step. So we have $$ g^{kl}g^{im}(\frac{\partial g_{mk}}{\partial x^l}-\frac{1}{2}\frac{\partial g_{kl}}{\partial x^m}) = -g^{kl} g_{mk} \frac{\partial g^{im}}{\partial x^l} - \frac{1}{2} g^{im} g^{kl} \frac{\partial g_{kl}}{\partial x^m}. $$ From there, it is mainly a matter of applying the product rule to (86.6) and using the identity in (86.4).