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Geodesic Sphere

  1. Jan 23, 2009 #1
    [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.

    2. Relevant equations

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

    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:
    [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?
  2. jcsd
  3. Jan 23, 2009 #2


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    Gold Member

    It might be more useful to write this as:

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

    I think the point of the hint is that the above relation holds for all allowable values of [itex]\theta[/itex] and that [itex]\theta=0[/itex] is one such allowable value. What does [itex]C^2[/itex] equal when [itex]\theta=0[/itex] assuming that [itex]\phi'[/itex] is bounded (i.e. not infinite)? Since [itex]C[/itex] is a constant, it must have this value for all [itex]\theta[/itex].
  4. Jan 24, 2009 #3
    Thanks for your reply.

    Now I can see why C = 0. If I am doing my math correctly that means [tex]\phi (\theta ) = A[/tex] (another constant).

    If [tex]\phi (\theta )[/tex] is just some constant, does that mean this constant A will be a segment of the great circle between two locations on the sphere?
  5. May 17, 2010 #4
    As the choice of axis is arbitrary, you can choose one such that [tex]\theta[/tex] is zero. Then varying [tex]\phi[/tex] (which as you point out should have no [tex]\theta[/tex] dependence) is just tracing out a great circle.

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