Derivative of line element in general relativity is zero?

Question

The Lagrangian for a point particle in general relativity is

$$ L= -m \sqrt{-g_{\mu\nu}\dot{x}^\mu \dot{x}^\nu} $$

where $x^\mu(\lambda)$ is the world line of a particle with mass $m$. The world line is parameterized with an arbitrary parameter $\lambda$. The derivative is denoted with $\dot{x}^\mu=\frac{\partial x^\mu}{\partial\lambda}$,

I can derive the equations of motions with the Euler-Lagrange equations:

$$ \frac{\partial L}{\partial x^\mu}- \frac{d}{d\lambda}\frac{\partial L}{\partial \dot{x}^\mu} =0.$$

However I only get the desired result (the geodesic equation $\ddot{x}^\mu+\Gamma^\mu_{\alpha\beta}\dot{x}^\alpha\dot{x}^\beta=0$) when I assume that:

$$ \frac{d}{d\lambda} \sqrt{-g_{\mu\nu}\dot{x}^\mu \dot{x}^\nu} =0 $$

why should this hold?

Answer 1

Actually you don't need $\frac{d}{d\lambda} \sqrt{-g_{\mu\nu}\dot{x}^\mu \dot{x}^\nu} =0$.

Since $$L=\sqrt{-g_{\mu\nu}\dot{x}^ \mu \dot {x}^\nu} $$, $L^2$ also satisfies Euler-Lagrange equations. So take $L={-g_{\mu\nu}\dot{x}^ \mu \dot {x}^\nu} $. Now $$\frac{\partial L}{\partial x^\mu}- \frac{d}{d\lambda}\frac{\partial L}{\partial \dot{x}^\mu} =0$$

This will give $$\ddot{x}^\mu+ \dot g_{\alpha \nu} g^{\mu \alpha} \dot x^\nu- \frac{1}{2}\frac{\partial g_{\rho \nu}}{\partial x^\alpha}g^{\mu\alpha}\dot x^\rho \dot x^\nu =0$$ Since $\dot g_{\alpha\nu} =\frac{\partial g_{\alpha \nu}}{\partial x^\beta}\dot x^\beta $. Substitute this equation in previous one and use definition of Christoffel symbols as $$\Gamma^{\mu}_{\rho\nu} = \frac{1}{2} g^{\mu\alpha} \left( \frac{\partial g_{\alpha\rho}}{\partial x^{\nu}} + \frac{\partial g_{\alpha\nu}}{\partial x^{\rho}} - \frac{\partial g_{\rho\nu}}{\partial x^{\alpha}} \right)$$ You'll get, $$\ddot{x}^\mu+\Gamma^\mu_{\alpha\beta}\dot{x}^\alpha\dot{x}^\beta=0$$