# Solving the Geodesic equations for a space

1. Dec 7, 2011

### Lyalpha

I'm in an intro course and my shaky ability to solve differential equations is apparent.

How would you go about solving

$\ddot{r}$-r$\ddot{\theta}$=0

$\ddot{\theta}$+$\frac{1}{r}$$\dot{r}$$\dot{\theta}$=0

It might be obvious. They're the geodesic equations for a 2d polar coordinate system (if i'm correct).

2. Dec 8, 2011

### pervect

Staff Emeritus

The second equation is missing a factor of 2. I'm not sure exactly how you derived the geodesic equations, I'm hoping that if I remark that $$\Gamma^{\theta}{}_{{r}{\theta}} = \Gamma^{\theta}{}_{{\theta}{r}} = \frac{1}{r}$$ you'll see what you omitted. If not, you'll need to write down your derivation.

After this correction, the second equation is equivalent to saying a geodesic conserves angular momentum, i.e

$$\frac{d}{dt} \left( r^2 \, \dot{\theta} \right) = 0$$

If you expand this via the chain rule you'll get

$$r^2 \ddot{\theta} + 2\,r\,\dot{r}\,\dot{\theta} = 0$$

which is equivalent to the corrected version of eq 2 (just divide both sides by r^2). This should allow you to separate the variables and solve the equatinos without too much difficulty by solving for $\dot{\theta}$ as a function of r and the conserved, constant angular momentum - which you can also regard as a constant of integration.