# Brachistochrone on the surface of a sphere

1. Apr 5, 2009

### TimJ

1. The problem statement

I'm trying to solve the brachistochrone problem between two points on the surface of a sphere.

2. The attempt at a solution

The Lagrangian for this problem in spherical coordinates is
$$L=\frac{1}{2}m \left(r^2 \left (\frac{d \theta}{dt}\right)^2+r^2 \sin^2(\theta) \left(\frac{d \phi}{dt}\right)^2\right)-mgr\cos(\theta)$$

After applying the Euler-Lagrange equation we get:

For $$\theta$$:

$$\frac{\partial L}{\partial (\frac{d \theta}{dt})} = m r^2\frac{d \theta}{dt}$$

$$\frac{\partial L}{\partial \theta} = m r^2 \left(\frac{d \phi}{dt}\right)^2 \sin(\theta) \cos(\theta) + mg r \sin(\theta)$$

$$\frac{\partial L}{\partial \theta} - \frac{d}{dt} \left ( \frac{\partial L}{\partial (\frac{d \theta}{dt})} \right ) = 0 \;\; \Rightarrow \;\; m r^2 \left(\frac{d \phi}{dt}\right)^2 \sin(\theta) \cos(\theta) + mg r \sin(\theta) - m r^2 \frac{d^2 \theta}{dt^2}=0$$

For $$\phi$$:

$$\frac{\partial L}{\partial \phi} = 0 \;\; \Rightarrow \;\; \frac{\partial L}{\partial (\frac{d \phi}{dt})}=m r^2 \sin^2(\theta) \frac{d \phi}{dt}= const.$$

Now we have two equations:

$$\frac{d^2 \theta}{dt^2}-\frac{g}{r}\sin(\theta)-\sin(\theta) \cos(\theta) \left(\frac{d \phi}{dt}\right)^2=0$$

and

$$m r^2 \sin^2(\theta) \frac{d \phi}{dt}= const. = A$$

And here is where it stopped. I don't know how to solve this two equations.

2. Apr 5, 2009

### Hurkyl

Staff Emeritus
The two things I first observe about that pair of equations is:
(1) It's easy to eliminate one of the variables
(2) A change of variable could help simplify things

3. Apr 5, 2009

### TimJ

Then I get this equation:

$$\frac{d^2 \theta}{dt^2}-\frac{g}{r}\sin(\theta)- \frac{A^2 \cos(\theta)}{(m r^2)^2 \sin^3(\theta) }=0$$

I tried to solve it with Mathematica, but i didn't get a result.