# Solving Geodesics

I'm wondering how to solve the geodesic equation to get a null geodesic. I know the two equations

$$\frac{d^2x^\mu}{ds^2}+{\Gamma^\mu}_{\nu\lambda}\frac{dx^\nu}{ds}\frac{dx^\lambda}{ds}=0$$

and the null one

$$g_{\mu\nu}\frac{dx^\mu}{ds}\frac{dx^\nu}{ds}=0,$$

but how do you do this, given a metric? Consider, for example,

$$ds^2=\cos\sigma\,d\tau^2+d\sigma^2.$$

I don't understand how to plug this in and actually find the null geodesics once I've calculated the Christoffel symbols. thanks!

• Please use MathJax for all math. Mar 4 at 22:40
• How much experience do you have solving coupled nonlinear differential equations? Mar 4 at 22:43
• i have done some in the past, and i'm sure I can figure it out in general, i'm just generally confused on what this 's' is denoting when taking the derivatives, and overall what the equation should look like plugged in. Mar 4 at 22:44
• Do you understand what a parameterized curve is? $s$ parameterizes the geodesic. When the geodesic is timelike instead of null, it’s the proper time along the geodesic. Mar 4 at 22:47
• your example metric is positive definite and has no null geodesics. Mar 5 at 4:19

Following Ray D'Inverno. Introducing Einstein's Relativity. Page 101.

Assuming u is an affine parameter.

If $$2K\equiv g_{ab}\dot{x}^a\dot{x}^b =\dot{x}^a\dot{x}_a$$ then you can quickly read off $$\Gamma^a _{bc}$$ from the geodesic equation in the form

$$\frac{\partial K}{\partial x^a } - \frac{d}{du} (\frac{\partial K}{\partial \dot{x}^a})=0$$ where $$2K=\text{C}=0,+1,-1$$.

Example:

$$ds^2 = \eta^2 d\tau^2 - d\eta^2 \to K=\frac{1}{2}(\eta^2 \dot\tau^2 - \dot{\eta}^2 )$$

$$\frac{\partial K}{\partial x^a } - \frac{d}{ds} (\frac{\partial K}{\partial \dot{x}^a})=0 \to \frac{\partial K}{\partial \tau } - \frac{d}{ds} (\frac{\partial K}{\partial \dot{\tau}})=0\to2\eta \dot{\eta}\dot{\tau}+\eta^2 \ddot{\tau}=0\to$$

$$\ddot{\tau}+\frac{2}{\eta} \dot{\eta} \dot{\tau}=0\implies\Gamma^\tau _{\eta \tau}=\frac{2}{\eta}=\Gamma^\tau _{\tau\eta}$$

$$ds^2=\cos σ dτ^2+ dσ^2$$

$$\to$$

$$K=\cos σ {\dot{τ}}^2+ {\dot{σ}}^2 =0$$

$$\to$$

$$\cos(\sigma){\dot{τ}}^2 = - {\dot{σ}}^2$$

That's one differential equation.

Next calculate,

$$\frac{\partial K}{\partial x^a } - \frac{d}{du} (\frac{\partial K}{\partial \dot{x}^a})=0$$

$$\to$$

$$\frac{d}{du} (\frac{\partial K}{\partial \dot{\tau}})=0$$

$$\frac{d}{du} (\frac{\partial K}{\partial \dot{\sigma}})=0$$

$$\to$$

$$\frac{\partial (\cos(σ) {\dot{τ}})}{\partial u}=0$$

$$\to$$

$$-\sin(σ) \dot{\sigma} \dot{τ} + cos(\sigma) \ddot{\tau}=0$$

and

$$\ddot{\sigma}=0$$.

$$\cos(\sigma){\dot{τ}}^2 = - {\dot{σ}}^2$$

$$-\sin(σ) \dot{\sigma} \dot{τ} + cos(\sigma) \ddot{\tau}=0$$

$$\ddot{\sigma}=0$$.

$$\to$$

$$\sigma(u)=\frac{a}{2} u^2 +bu +c$$

$$\to$$

$$\dot{\sigma}(u)= au +b$$

$$\to$$

$$\dot{\tau}^2 = \frac{-(u+b)^2}{cos(\frac{1}{2} au^2 +ub +c)}$$

set $$b=c=0$$ and $$a=2$$

$$\to$$

$$\dot{\tau}^2 = \frac{-(u)^2}{cos(u^2)}$$

etc. etc... ?