# Null Geodesics in flat 2+1 dimensional Minkowski space

For a given line element in flat 2+1 dimensional Minkowski space

$$g = ds^{2} = − dz \otimes dz + dx \otimes dx + dy \otimes dy .$$

The null geodesics are supposedly given by:

$$x = lu + l'$$

$$y = mu + m'$$

$$z = nu + n'$$

where $l, m, n, l', n'$, and $m'$ are constants (of integration, presumably), and $u$ is a parameter. I'm not sure how to arrive at this. The Geodesic Equation seems to just evaluate to zero when I plug these in (which makes sense), but that doesn't constitute a proof, and I'm struggling to work backwards from there.

Can anyone give me a push in the right direction or show me what I'm missing?

You're already there. The method would just be to solve the geodesic equation. To me it seems you have a completely flat manifold. The affine connection vanishes on a flat manifold, reducing the geodesic equation to $$\frac{d^{2}x^{\mu}} {du^{2}} = 0$$ where $\mu \; \epsilon \; 1,2,3$ and $x^{1} = x$, $x^{2}=y,$ and $x^{3}=z$ Trivial result, but the result nonetheless. Your equations satisfy this one. They can thus have been arrived at by solving the initial value problem for each variable to calculate the constants.

• Thanks a lot for the guidance. It was hard for me to believe that the example being given was really so trivial, I suppose. – user42745 Mar 18 '14 at 3:11
• Don't feel bad. I tend to never take trivial results at face value. C: – Doryan Miller Mar 18 '14 at 3:13