- #1

roeb

- 107

- 1

## Homework Statement

[tex]L = R \int_{\theta_1}^{\theta_2} \sqrt{1 + sin^2(\theta ) \phi ' ^ 2} d\theta [/tex]

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.

## Homework Equations

[tex]\partial f / \partial x = d/du \partial f / \partial x'[/tex]

## The Attempt at a Solution

I am having a bit of difficulty interpreting this problem. Using the Euler-Lagrange equation I get the following:

[tex]\phi ' ^ 2 = C^2 / ( sin^4 (\theta ) - C^2 sin^2(\theta ) [/tex]

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 [tex]\theta_1[/tex] = 0 if anything.

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