(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

Use the result (6.41) of Problem 6.1 to prove that the geodesic (shortest path) between two given points on a sphere is a great circle. [Hint: The integrand f(ψ,ψ',θ) in (6.41) is independent of ψ, so the Euler-Lagrange equation reduces to ∂f/ψ' = c, a constant. This gives you ψ' as a function of θ. ou 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

L = R \int_{\theta_{1} }^{\theta_{2}}\sqrt{1+\sin{\theta}\phi'^{2}}d\theta (6.41)

\frac{\partial f}{\partial \phi}-\frac{\mathrm{d} }{\mathrm{d} \theta}\frac{\partial f}{\partial \phi'} = 0

3. The attempt at a solution

I am able to get down to \phi' = \frac{c}{\sin{\theta}\sqrt{\sin^{2}{\theta}-c^{2}}} by using the Euler-Lagrange equation and finding that ∂f/∂ψ' = c, but I am confused as to what the problem means by choosing the z axis to pass through point 1. The solution from (6.1), which is the L integral was performed in spherical polar coordinates, so where does this z axis come from? Any help getting past \phi' would be helpful. The integration of \phi' looks very difficult which is why I understand why the question recommends a trick; but I do not understand how to go about performing this 'trick'.

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# Homework Help: Classical Mechanics: Minimization of geodesic on a sphere

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