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Oscillatory solution for a given Lagrangian

  1. Mar 29, 2013 #1
    1. The problem statement, all variables and given/known data

    Consider the following Lagrangian:


    [tex] L = \frac{m}{2}(x'^2+y'^2+z'^2) + \frac{q}{2}(xy'-yx') [/tex]

    Where q denotes a charged particle.

    a) Find the equations of motion
    b) Find the solution for z
    c) Find the solution in the x-y plane, and prove that it corresponds to an oscillatory motion along both axes.


    3. The attempt at a solution

    Considering the Euler-Lagrange equations, the solution for coordinate z is given by:

    d/dt (mz') = 0
    Thus, a solution for z corresponds to a uniform linear motion along z axis.

    For x and y, I get the following systems of dif. equations:

    mx'' - qy' = 0
    my'' + qx' = 0

    I tried solving this system by integrating and substituting,

    i.e. mx'' = qy' /integrate
    mx' = qy + C
    x' = qy/m + C

    Substituting, we get
    y'' + y(q^2)/(m^2) = C

    The inverse process could be done to get an equation for x.

    Which I do not think is the right answer, and I don't know how many laws of mathematics I violated while doing that. Could anyone please show me what is the right procedure to solve this system of equations?

    Also, I have the impression this Lagrangian corresponds to a charge q interacting with an electromagnetic field. It would be great if anyone could explain me a little bit further about that.


    Any help would be greatly appreciated.
    Thanks a lot.
     
    Last edited: Mar 29, 2013
  2. jcsd
  3. Mar 29, 2013 #2

    TSny

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    Should both equations have a negative sign in the left side?
     
  4. Mar 29, 2013 #3
    Sorry, my bad, I missed a minus sign in the second equation, then the system becomes:

    mx'' - qy' = 0
    my'' + qx' = 0

    Following the same process I mentioned before, equation for y becomes:

    y'' + y(q^2)/(m^2) = C
     
  5. Mar 29, 2013 #4

    TSny

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    I'm not sure of your math background. If you've had a course in differential equations, then you know that the general solution to this type of equation is y = yh + yp where yh is the general solution of the homogeneous equation y'' + (q/m)2y = 0 and yp is any particular solution of y'' + (q/m)2y = C.
     
  6. Mar 29, 2013 #5
    I'm sorry if this becomes too obvious. I've taken Dif. Equations, but my math background is a bit blurry at the moment, because I've just re-taken classes. Once I recall the methods things go easier.

    I can see that the solution for the homogeneus equation is indeed an oscillatory motion, both for x and y. I'm having trouble about what to do with the particular solution since I don't have any initial condition.

    Would you say that the method I applied to solve the problem is correct?

    Thanks for your help.
     
  7. Mar 29, 2013 #6

    TSny

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    Yes, your method is a correct way to do it. You've found the general solution to y'' + (q/m)2y = 0. Now you just need any particular solution to y'' + (q/m)2y = C. You should be able to this by inspection. You do not need initial conditions here.
     
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