1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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
    [tex]
    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)
    [/tex]

    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]

    [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
    [/tex]

    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:

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

    and

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

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

    Hurkyl

    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:

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

    I tried to solve it with Mathematica, but i didn't get a result.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Brachistochrone on the surface of a sphere
  1. The brachistochrone (Replies: 1)

Loading...