There are several ways to derive the geodesic equation. One of which is the variational method which I seemed to understand it because it was written in great details. Then it was mentioned that the geodesic equation can be derived from the Euler-Lagrange equations only. I tried to plug in the Lagrangian $$ L= \frac{1}{2} g_{\mu\nu}\frac{dx^\mu}{d\lambda}\frac{dx^\nu}{d\lambda}$$
in $$\frac{d}{d\lambda}\frac{\partial L}{\partial(dx^\mu/d\lambda)} = \frac{\partial L}{\partial x^\mu} $$
but I am running into derivation problems and the corresponding chain rule. May you please help me out here, I have to understand how the geodesic equation is derived from the Euler-Lagrange equations. Thank you very much in advance!!
Best Answer
Let us do the RHS first. This just gives us a derivative on the metric: $$\frac{\partial L}{\partial x^\lambda}=\frac{1}{2}\partial_\lambda g_{\mu\nu}\dot x^\mu\dot x^\nu$$ The first derivative on the LHS is essentially a derivative of a square, thus $$\frac{\partial L}{\partial \dot x^\lambda}=g_{\mu\lambda}(x(\lambda))\dot x^\mu$$ where we have made the dependence of $g$ on $\lambda$ clear for the next step. Now we differentiate with respect to the curve parameter: $$\frac{\mathrm{d}}{\mathrm{d}\lambda}[g_{\mu\lambda}(x(\lambda))\dot x^\mu]=\partial_\nu g_{\mu\lambda}\dot x^\mu\dot x^\nu+g_{\mu\lambda}\ddot x^\mu=\frac{1}{2}\partial_\nu g_{\mu\lambda}\dot x^\mu\dot x^\nu+\frac{1}{2}\partial_\mu g_{\nu\lambda}\dot x^\mu\dot x^\nu+g_{\mu\lambda}\ddot x^\mu$$ where in the last step we split the first term apart and rearranged indices. Putting it all together, we obtain $$g_{\mu\lambda}\ddot x^\mu=-\frac{1}{2}\left(\partial_\nu g_{\mu\lambda}+\partial_\mu g_{\nu\lambda}-\partial_\lambda g_{\mu\nu}\right)\dot x^\mu\dot x^\nu=-\Gamma_{\lambda\mu\nu}\dot x^\mu\dot x^\nu$$ where in the last step we used the definition of the Christoffel symbols with three lower indices. Now contract with the inverse metric to raise the first index and cancel the metric on the LHS. We obtain $$\ddot x^\lambda=-\Gamma^\lambda{}_{\mu\nu}\dot x^\mu\dot x^\nu$$ as was to be shown.