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Geodesic equations

  1. Dec 16, 2008 #1
    1. The problem statement, all variables and given/known data
    I'm given the surface of revolution parametrized by [itex]\psi (t, \theta ) = (x(t), y(t)cos \theta, y(t)sin \theta )[/itex] where the curve [itex]\alpha (t) = (x(t),y(t))[/itex] has unit speed. Also given is that [itex]\gamma (s) = \psi (t(s), \theta (s))[/itex] is a geodesic which implies the following equations hold:

    [tex]
    \ddot{\theta} = -2 \frac{y'}{y} \dot{\theta} \dot{t} [/tex] and [tex] \ddot{t} = y y' \dot{\theta}^2
    [/tex]

    where

    [tex]
    y' = \frac{dy}{dt}, \dot{\theta} = \frac{d \theta}{ds}, \ddot{\theta} = \frac{d^2 \theta}{ds^2}, \dot{t} = \frac{dt}{ds}, \ddot{t} = \frac{d^2 t}{ds^2}
    [/tex]

    I have to show that the following quantities are independent of [itex]s[/itex]:

    [tex]\dot{t}^2 + y^2 \dot{\theta}^2 = E[/tex]
    [tex]y^2 \dot{\theta} = A[/tex]

    2. Relevant equations

    All that I can think may be of relevance that isn't already listed is that for a unit speed curve, [tex]y'^2 + x'^2 = 1[/tex]. Not sure that this matters here, though.


    3. The attempt at a solution

    I've tried rearranging the equations to try to resemble the desired equations, but it's been pretty unfruitful. I was thinking about maybe differentiating one of the equations w.r.t. [itex]s[/itex], but I'm not sure how one would deal with the third derivative of [itex]t[/itex] or [itex]\theta[/itex].

    Any help (or hints) are is appreciated! Thanks!
     
  2. jcsd
  3. Dec 16, 2008 #2

    gabbagabbahey

    User Avatar
    Homework Helper
    Gold Member

    In this case, I think the easiest way to tackle this is to work backwards from the two equations you are trying to prove:

    What is [tex]\frac{d}{ds}(y^2 \dot{\theta})[/tex]?:wink:
     
  4. Dec 16, 2008 #3
    Thank you! Your hint was just what I needed. For some reason I was forgetting that [itex]\frac{dy}{ds} = y' \dot{t}[/itex], but now it makes perfect sense.

    Thanks again :)
     
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