- #1
miew
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Homework Statement
L = R [tex]\int \sqrt{1+ sin^2 \theta \phi ' ^ 2} d\theta
[/tex] from theta 1 to theta 2
Using this result, prove that the geodesic (shortest path) between two given points on a sphere is a great circle. [Hint: The integrand f([tex]\phi,\phi',\theta[/tex]) in the result is independent of phi so the Euler-Lagrange equation reduces to [tex]\partial f [/tex]/[tex]\partial \phi ' [/tex] = c, a constant. This gives you [tex]\phi ' [/tex] 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
The Attempt at a Solution
Doing some calculations, I ended up getting
[tex]\phi ' ^ 2[/tex] - A sin^2 [tex]\theta[/tex] [tex]\phi ' ^ 2 [/tex]= A^2. The hint tells me that I can take theta as 0, and so I get that phi_prime = A.
I integrate that, but I don't get anything that looks like a circle.