# Geodesic equations

1. Dec 16, 2008

### mooshasta

1. The problem statement, all variables and given/known data
I'm given the surface of revolution parametrized by $\psi (t, \theta ) = (x(t), y(t)cos \theta, y(t)sin \theta )$ where the curve $\alpha (t) = (x(t),y(t))$ has unit speed. Also given is that $\gamma (s) = \psi (t(s), \theta (s))$ is a geodesic which implies the following equations hold:

$$\ddot{\theta} = -2 \frac{y'}{y} \dot{\theta} \dot{t}$$ and $$\ddot{t} = y y' \dot{\theta}^2$$

where

$$y' = \frac{dy}{dt}, \dot{\theta} = \frac{d \theta}{ds}, \ddot{\theta} = \frac{d^2 \theta}{ds^2}, \dot{t} = \frac{dt}{ds}, \ddot{t} = \frac{d^2 t}{ds^2}$$

I have to show that the following quantities are independent of $s$:

$$\dot{t}^2 + y^2 \dot{\theta}^2 = E$$
$$y^2 \dot{\theta} = A$$

2. Relevant equations

All that I can think may be of relevance that isn't already listed is that for a unit speed curve, $$y'^2 + x'^2 = 1$$. Not sure that this matters here, though.

3. The attempt at a solution

I've tried rearranging the equations to try to resemble the desired equations, but it's been pretty unfruitful. I was thinking about maybe differentiating one of the equations w.r.t. $s$, but I'm not sure how one would deal with the third derivative of $t$ or $\theta$.

Any help (or hints) are is appreciated! Thanks!

2. Dec 16, 2008

### gabbagabbahey

In this case, I think the easiest way to tackle this is to work backwards from the two equations you are trying to prove:

What is $$\frac{d}{ds}(y^2 \dot{\theta})$$?

3. Dec 16, 2008

### mooshasta

Thank you! Your hint was just what I needed. For some reason I was forgetting that $\frac{dy}{ds} = y' \dot{t}$, but now it makes perfect sense.

Thanks again :)