Why Lagrangian has this form in general relativity?

Question

One can derive the geodesic equation by Euler-Lagrangian equation, \begin{equation} \dfrac{\partial \mathcal{L}}{\partial x^\gamma} - \dfrac{d}{ds}\bigg(\dfrac{\partial \mathcal{L}}{\partial (dx^\gamma/ds)}\bigg) =0 \end{equation} where $$\mathcal{L}= \bigg[ -g_{\mu \nu}\dfrac{dx^\mu}{ds}\dfrac{dx^\nu}{ds}\bigg]^{1/2}.$$ And by searching in some references, i didn't find any clear reason for this form of Lagrangian $\mathcal{L}$. Can you give me the reason for this form of Lagrangian?

Answer 1

The integral of the Lagrangian is the proper time elapsed along the particle's worldline $\gamma$: $$ \tau = \int_\gamma d \tau = \int_\gamma \sqrt{-g_{\mu \nu} dx^\mu dx^\nu} = \int_\gamma \sqrt{-g_{\mu \nu} \frac{dx^\mu}{ds} \frac{dx^\nu}{ds}}ds = \int_\gamma \mathcal{L} \, ds. $$ So any solution to the Euler-Lagrange equations for this "Lagrangian" will extremize the proper time elapsed between the endpoints of $\gamma$. But geodesics extremize the proper time between their endpoints, and so a solution to the Euler-Lagrange equations will be a geodesic.