# Geodesic Sphere

1. Jan 23, 2009

### roeb

1. The problem statement, all variables and given/known data
$$L = R \int_{\theta_1}^{\theta_2} \sqrt{1 + sin^2(\theta ) \phi ' ^ 2} d\theta$$

Use the result to prove that the geodesic (shortest path) between two given points on a sphere is a great circle. [Hint: The integrand f(phi,phi_prime,theta) in the result is independent of phi so the Euler-Lagrange equation reduces to partial_f/partial_phi_prime = c, a constant. This gives you phi_prime as a function of theta. You can avoid doing the final integral by the following trick: There is no loss of generality in choosing your z axis to pass through the point 1. Show that with this choice the constant c is necessarily zero and describe the corresponding geodesics.

2. Relevant equations

$$\partial f / \partial x = d/du \partial f / \partial x'$$

3. The attempt at a solution

I am having a bit of difficulty interpreting this problem. Using the Euler-Lagrange equation I get the following:
$$\phi ' ^ 2 = C^2 / ( sin^4 (\theta ) - C^2 sin^2(\theta )$$
The hint kind of confusing me. How can 'c' be zero? If I were to have the z-axis at point 1, I would think that $$\theta_1$$ = 0 if anything.

Can anyone help guide me on the right path for this problem?

2. Jan 23, 2009

### gabbagabbahey

It might be more useful to write this as:

$$\left( sin^4 (\theta ) - C^2 sin^2(\theta ) \right) \phi ' ^ 2 = C^2$$

I think the point of the hint is that the above relation holds for all allowable values of $\theta$ and that $\theta=0$ is one such allowable value. What does $C^2$ equal when $\theta=0$ assuming that $\phi'$ is bounded (i.e. not infinite)? Since $C$ is a constant, it must have this value for all $\theta$.

3. Jan 24, 2009

### roeb

Now I can see why C = 0. If I am doing my math correctly that means $$\phi (\theta ) = A$$ (another constant).
If $$\phi (\theta )$$ is just some constant, does that mean this constant A will be a segment of the great circle between two locations on the sphere?
As the choice of axis is arbitrary, you can choose one such that $$\theta$$ is zero. Then varying $$\phi$$ (which as you point out should have no $$\theta$$ dependence) is just tracing out a great circle.