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

Question

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 1

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.

Answer 2

I am not sure to have understood what you asked... But the geodesics equation is only one, given the levi-Civita connection. It changes just a bit if you do not use proper time (or an affine parameter) by which you have that $\dot{x}^{\mu}$ is the four velocity s.t. $(\dot{x},\dot{x})=1$ (signature $(+,-,-,-)$). Here you can find more details: https://en.wikipedia.org/wiki/Geodesics_in_general_relativity#Mathematical_expression.