# Geodesics and their linear equations

1. Feb 16, 2008

### Biest

Hi,

So I am going over on how to find a geodesic from any metric, esp. on a 2-sphere. I have been looking at my lecture notes and am confused as to how my professor solves for the equation in terms of the variables, i.e. $$(\theta , \phi)$$.

If i use the 2-sphere as an example here where the geodesic is given by:

$$\ddot{\phi} = -2\cot \theta \dot{\phi} \dot{\theta}$$

$$\ddot{\theta} = \sin \theta \cos \theta \dot{\phi}^2$$

So from the metric we get a first dervative

$$1 = \dot{\theta}^2 + \sin^2 \theta \dot{\phi}^2$$

My prof then goes on and simply states that

$$\frac{1}{\sin^2 \theta} \frac{d}{d\tau} (\sin^2 \theta \dot{\phi}) = 0$$

Where does the last equation come from... in lecture he simply stated it and moved on from there with

$$(\sin^2 \theta \dot{\phi}) = l = constant$$

I really thinking i got a bit rusty on this since mechanics lies two years in the past.

Thank you very much any help in advance.

Cheers,

Biest

Last edited: Feb 16, 2008
2. Feb 16, 2008

### Rainbow Child

Multiply the first equation with $\sin^2\theta$ and you will arrive to the wanted result.

3. Feb 16, 2008

### Biest

Thanks.... I forgot to add one more question... What do we do when we have three geodesics? Just choose one and solve from there?

4. Feb 16, 2008

### Rainbow Child

Finding the geodesics is a really hard problem. In general you can not find a closed form solution for them. But if you have a killing field, i.e. $\nabla_\alpha\,\xi_\beta+\nabla_\beta\,\xi_\alpha=0$ then you have a constant along the geodesic, i.e. $\xi_\alpha\,u^\alpha=C$ where $u^\alpha$ is tangent to the geodesic.

5. Feb 16, 2008

### Biest

I know it is hard... i had to derive the conditions for homework and was stuck on the derivative of $g_{\mu \nu}$ anyway. At the moment we have just done polar coordinates and the 2-sphere. and now we just moved into particle orbits, which i have to work through as well cause i am trying how the Killing vector works there. I am starting to get it, but it is taking me a while already.