general relativity - Timelike, spacelike, and null geodesics using Euler-Lagrange equations

I'm attempting to plot geodesics in curved spacetime (e.g. the Schwarzschild metric) starting from the Lagrangian

$$ L = \frac{1}{2} g_{\alpha \beta} \dot{x}^\alpha \dot{x}^\beta $$ using the Euler Lagrange equations:

$$ \frac{\partial L}{\partial x^\alpha} = \frac{d}{d \lambda} \frac{\partial L}{\partial \dot{x}^\alpha} $$ My question is mostly on how to specify what kind of geodesics I wish to get in the resulting differential equations. For timelike, null, and spacelike particles $2L = -1,0,1$, respectively, so I was thinking of potentially working this is as a constraint and using

$$ \frac{\partial L}{\partial x^\alpha} + \kappa \frac{\partial f}{\partial x^\alpha} = \frac{d}{d \lambda} \frac{\partial L}{\partial \dot{x}^\alpha} $$

with $f = 2L$. Or is it a matter of specifying initial conditions such that you get the kind of particle you want, i.e. $c^2 = v_{x0}^2 + v_{y0}^2 + v_{z0}^2$ for null particles?

Answer

The Lagrangian does not depend explicitly on the "time" parameter you use to define $\dot{x}$. So, the Hamiltonian function is conserved along the solutions of the E.L. equation. In this case, the Hamiltonian function coincides with the Lagrangian one which, in turn, is the Lorenzian squared norm of the tangent vector. In other words, the type of geodesic is decided on giving the initial tangent vector: its nature (spacelike, timelike, lightlike) is automatically preserved along the curve.