# 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?

## 2 Answers

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.

• @user38729 To add to the correct answer of Valter Moretti, you do not need to worry about the type of the geodesic in the equation itself. The type of the geodesic will be defined by a constant of integration in the solution of the differential equation. For example, in a symmetrical 2D case, you would have a system of two equations with a parameter along the geodesics. Eliminating this parameter would give you one differential equation of the geodesics in terms of space and time. One of the constants of integration (+,0,-) in the solution of this equation will define the type of the geodesic. – safesphere Jan 2 '18 at 10:02

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.