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Calculus of Var, Euclidean geodesic

  1. Jul 12, 2012 #1
    1. The problem statement, all variables and given/known data
    Calculate the geodesic for euclidean polar coordinates given ds[itex]^{2}[/itex]=dr[itex]^{2}[/itex]+r[itex]^{2}[/itex]dθ[itex]^{2}[/itex]


    2. Relevant equations
    standard euler-lagrange equation

    3. The attempt at a solution
    I was able to reduce the euler-lagrange equation to [itex]\frac{d^{2}r}{dθ^{2}}[/itex]-rλ=0 where λ=[itex]\sqrt{(\frac{dr}{dθ})^{2}+r^{2}}[/itex] is the Lagrangian itself (namely the linear element)

    My main concern is that I have the correct differential equation, I'm curious because I can't possibly imagine how this author expects me to solve that if it is indeed the correct DE for the lagrangian.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Jul 13, 2012 #2
    Were you asked to solve it? Most geodesic equations cannot be solved analytically and require numerical analysis.
     
  4. Jul 13, 2012 #3
    I would think it would be easy to solve because it is obviously a straight line, that's why I think I have done something wrong to get this non-linear second order DE.
     
  5. Jul 13, 2012 #4
    I do not necessarily think it would be easy. Given an arbitrary straight line that does not cross through the origin, how would you write this in polar coordinates? It is actually a very difficult thing to do.
     
  6. Jul 13, 2012 #5
    so does that mean I have the correct DE? i'm not too worried about actually solving it, just that I have the right one and applied the E-L equations correctly to this problem
     
  7. Jul 13, 2012 #6
    I haven't had to do this stuff in a long time, but I computed my geodesic equations to be

    [tex]\frac{d^2 r}{dt^2} - r \left( \frac{d\theta}{dt}\right)^2 = 0, \qquad r^2 \frac{d^2 \theta}{dt^2} + 2r \frac{dr}{dt} \frac{d\theta}{dt} = 0 [/tex]
     
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