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Help! Geodesics of a sphere

  1. Jul 25, 2011 #1
    1. The problem statement, all variables and given/known data
    Hi, just want to get a couple of things straight regarding finding the geodesics of a sphere not using polar coordinates, but rather, Lagrange multipliers...

    I want to minimize I = int (|x-dot|2 dt)
    subject to the constraint |x|=1 (sphere)
    which gives an Euler equation of [itex]\lambda[/itex]x - x-doubledot = 0

    I have to show that the Euler equation is actually |x-dot|2x - x-doubledot = 0
    Is it right to assume that [itex]\lambda[/itex]=|x-dot|2 simply by the fact that it minimizes I* = int [|x-dot|2 - [itex]\lambda[/itex](|x|2-1)dt] which is [itex]\geq0[/itex], so the [itex]\lambda[/itex] that minimizes I* is |x-dot|2?

    If I then try to integrate the Euler equation, then I get a SHM equation:

    x1= A1 cos(|x-dot| t - C1) where A, C are constants
    and similarly for x2, x3

    But how do I combine them to give the equation of a great circle, since I don't know the Ci's?

    Thank you for any enlightenment!


    2. Relevant equations
    See above


    3. The attempt at a solution
    See above
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Jul 25, 2011 #2

    hunt_mat

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    Homework Helper

    So you are trying to minimise the following:
    [tex]
    \int |\dot{x}|^{2}dt
    [/tex]
    One way to think about this to say that [itex]\dot{x}=0[/itex] which integrates up to [itex]x=\textrm{constant}[/itex] which implies that [itex]|x|^{2}=\textrm{constant}[/itex]. Does that help?
     
  4. Jul 25, 2011 #3
    Thanks, hunt_mat,
    Why can you take x-dot = 0?
     
  5. Jul 25, 2011 #4

    Ray Vickson

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    Science Advisor
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    Your formulation is incorrect: it should be [itex] \min \int |\dot{x}(t)|^2 dt [/itex], subject to [itex] |x(t)|^2 = 1 \; \forall t [/itex]. So, basically, you have infinitely many constraints, one for each t.

    RGV
     
  6. Jul 26, 2011 #5

    hunt_mat

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    Homework Helper

    Because
    [tex]
    \int |\dot{x}(t)|^{2}dt\geqslant 0
    [/tex]
    for all t and so it must be smallest when the integrand is identically zero.
     
  7. Jul 26, 2011 #6
    I guess I meant how does one know that 0 is attained? Also, I believe I was given that |x|^2 =1 (constant)
    However does your suggestion mean that I can set the augmented integrand |x-dot|2-lamda*|x|2 to 0? then I will have the desired lamda = |x-dot|2 ?
     
    Last edited: Jul 26, 2011
  8. Jul 26, 2011 #7
    Problem resolved! Thanks everyone.
     
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