1. Limited time only! Sign up for a free 30min personal 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!

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


    User Avatar
    Homework Helper

    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


    User Avatar
    Homework Helper

    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
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook