# Homework Help: Geodesic on a cone, calculus of variations

1. Jun 22, 2012

### Telemachus

I have to find the geodesics over a cone. I've used cylindrical coordinates. So, I've defined:

$$x=r \cos\theta$$
$$y=r \sin \theta$$
$$z=Ar$$

Then I've defined the arc lenght:
$$ds^2=dr^2+r^2d\theta^2+A^2dr^2$$

So, the arclenght:
$$ds=\int_{r_1}^{r_2}\sqrt { 1+A^2+r^2 \left ( \frac{d\theta}{dr}\right )^2 }dr$$

And using Euler-Lagrange equation:
$$\frac{\partial f}{\partial \theta}=0$$
$$\frac{\partial f}{\partial \dot \theta}=-\frac{r^2\frac{d\theta}{dr}}{\sqrt{1+A^2+r^2\left ( \frac{d\theta}{dr}\right )^2}}$$
$$\frac{d}{dr}\frac{\partial f}{\partial \dot \theta}=0 \rightarrow \frac{\partial f}{\partial \dot \theta}=-\frac{r^2\frac{d\theta}{dr}}{\sqrt{1+A^2+r^2\left ( \frac{d\theta}{dr}\right )^2}}=K$$

The differential equation which I've arrived is non linear. I don't know if what I've done is fine. I know the problem can also be done using spherical coordinates instead of cylindrical, but I've choossen to do it this way.

2. Jun 23, 2012

### algebrat

Is it separable? I didn't simplify your expression, but it looks like you might have

$d\theta=g(r)dr$

3. Jun 23, 2012

### Telemachus

Yes, you're right:
$$r^2\frac{d\theta}{dr}=K \sqrt{ 1+A^2+r^2\left ( \frac{d\theta}{dr}\right )^2 }$$
$$r^4 \left ( \frac{d\theta}{dr} \right )^2=\gamma \left [ 1+A^2+r^2\left ( \frac{d\theta}{dr}\right )^2 \right ]$$
$$r^4 \left ( \frac{d\theta}{dr} \right )^2=\gamma \left [ 1+A^2+r^2\left ( \frac{d \theta}{dr}\right )^2 \right ]$$
$$(r^4-r^2\gamma) \left ( \frac{d\theta}{dr} \right )^2=\gamma \left [ 1+A^2 \right ]$$
$$d\theta=\sqrt{ \frac { \eta }{r^2(r^2-\gamma) }}dr$$
eta and gamma are constants.

Anyway, I wanted to know if what I did was ok. I don't care that much about the integral :P

Thank you algebrat.