Geodesic on a cone, calculus of variations

Telemachus

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

[tex]x=r \cos\theta[/tex]
[tex]y=r \sin \theta[/tex]
[tex]z=Ar[/tex]

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

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

And using Euler-Lagrange equation:
[tex]\frac{\partial f}{\partial \theta}=0[/tex]
[tex]\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}}[/tex]
[tex]\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
[/tex]

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.

algebrat

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

[itex]d\theta=g(r)dr[/itex]

Telemachus

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