The Lagrangain 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?

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?