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Homework Help: Brachistochrone on the surface of a sphere

  1. Apr 5, 2009 #1
    1. The problem statement

    I'm trying to solve the brachistochrone problem between two points on the surface of a sphere.

    2. The attempt at a solution

    The Lagrangian for this problem in spherical coordinates is
    L=\frac{1}{2}m \left(r^2 \left (\frac{d \theta}{dt}\right)^2+r^2 \sin^2(\theta) \left(\frac{d \phi}{dt}\right)^2\right)-mgr\cos(\theta)

    After applying the Euler-Lagrange equation we get:

    For [tex] \theta [/tex]:

    [tex] \frac{\partial L}{\partial (\frac{d \theta}{dt})} = m r^2\frac{d \theta}{dt}[/tex]

    [tex] \frac{\partial L}{\partial \theta} = m r^2 \left(\frac{d \phi}{dt}\right)^2 \sin(\theta) \cos(\theta) + mg r \sin(\theta) [/tex]

    \frac{\partial L}{\partial \theta} - \frac{d}{dt} \left ( \frac{\partial L}{\partial (\frac{d \theta}{dt})} \right ) = 0 \;\; \Rightarrow \;\;
    m r^2 \left(\frac{d \phi}{dt}\right)^2 \sin(\theta) \cos(\theta) + mg r \sin(\theta) -
    m r^2 \frac{d^2 \theta}{dt^2}=0

    For [tex] \phi[/tex]:

    [tex] \frac{\partial L}{\partial \phi} = 0 \;\; \Rightarrow \;\; \frac{\partial L}{\partial (\frac{d \phi}{dt})}=m r^2 \sin^2(\theta) \frac{d \phi}{dt}= const.[/tex]

    Now we have two equations:

    \frac{d^2 \theta}{dt^2}-\frac{g}{r}\sin(\theta)-\sin(\theta) \cos(\theta) \left(\frac{d \phi}{dt}\right)^2=0


    m r^2 \sin^2(\theta) \frac{d \phi}{dt}= const. = A

    And here is where it stopped. I don't know how to solve this two equations.
  2. jcsd
  3. Apr 5, 2009 #2


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    The two things I first observe about that pair of equations is:
    (1) It's easy to eliminate one of the variables
    (2) A change of variable could help simplify things
  4. Apr 5, 2009 #3
    Then I get this equation:

    \frac{d^2 \theta}{dt^2}-\frac{g}{r}\sin(\theta)-
    \frac{A^2 \cos(\theta)}{(m r^2)^2 \sin^3(\theta) }=0

    I tried to solve it with Mathematica, but i didn't get a result.
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