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Homework Help: Variational calculus

  1. Oct 18, 2007 #1
    1. The problem statement, all variables and given/known data

    I am trying to find a geodesic with Euler-Lagrange equation by varying the function

    [tex] ds/d\tau = \sqrt{\dot{x} + \dot{y}} [/tex]

    EDIT: it should be [tex] ds/d\tau = \sqrt{\dot{x}^2 + \dot{y}^2} [/tex]

    where tau is a parametrization and the dot means a tau derivative.

    However, when I plug that into the EL equation for either x or y, I get:

    [tex] (\dot{x} + \dot{y})^{3/2}(\ddot{x}+\ddot{y})=0 [/tex]

    How do I get a line from that?

    2. Relevant equations

    3. The attempt at a solution
    Last edited: Oct 18, 2007
  2. jcsd
  3. Oct 18, 2007 #2
    I'm don't know much about variational calculus, but as far as geodesics go, why does it have to be a line?
  4. Oct 18, 2007 #3
    Because the closest distance between 2 points in R^n is and the ds I am using is the differential line element in R^n.

    In R^2 the differential line element is: ds = sqrt(dx^2+dy^2)

    I just divided both sides by the parameter dtau to get the first equation in my last post.
  5. Oct 18, 2007 #4


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    If I'm understanding what you're saying, shouldn't you get

    [tex] ds/d\tau = \sqrt{\dot{x}^{2} + \dot{y}^{2}} [/tex] ?

    Otherwise, the expression you had isn't even correct dimensionally...
  6. Oct 18, 2007 #5
    Yes, that was foolish. But i had it correctly in my calculations so my question remains unanswered
    Last edited: Oct 18, 2007
  7. Oct 19, 2007 #6


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    Lemme see if I remember how this goes...

    Your equation

    [tex] (\dot{x} + \dot{y})^{3/2}(\ddot{x}+\ddot{y})=0 [/tex]

    implies that either

    [tex] (\dot{x} + \dot{y})^{3/2}=0 [/tex] or [tex] (\ddot{x}+\ddot{y})=0 [/tex].

    For the first term to be zero, you'd need

    [tex] \dot{x} = - \dot{y}[/tex], which would give you dy/dx = -1 , no?

    For the second term to be zero, you'd have

    [tex] \ddot{x} = -\ddot{y} [/tex] , which takes more antidifferentiation, but I believe also leads to a linear solution. (Not very rigorous, to be sure, but I believe that's basically how the argument runs.)
  8. Oct 19, 2007 #7
    That's rigorous enough for me. Its kind of weird we have that -1 solution though. Did you check that my equation is correct?
    Last edited: Oct 19, 2007
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