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Which coordinate is cyclic in this case

  1. Oct 12, 2012 #1
    Consider a simple two particle system with two point masses of mass m at x1 and x2 with a potential energy relative to each other which depends on the difference in their coordinates V = V(x1-x2)

    The lagrangian is:

    L = ½m(x1')2 + ½m(x2')2 + V(x1-x2)

    Obviously their total momentum is conserved d/dt(mx1' + mx2') = 0, which can be verified by plugging in to the lagrangian. But there is no cyclic coordinates in the lagrangian. Is it possible to put it in a form where this hidden cyclic coordinate is shown?
     
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  3. Oct 12, 2012 #2

    mfb

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    Hint: Try to express your lagrangian in terms of
    y1=x1+x2
    y2=x1-x2
     
  4. Oct 12, 2012 #3
    ahh nice.
    So you get:

    1/4my12 + 1/4my22 - V(y2) = 0

    Is it possible to transform to a situation with all coordinates cyclic?
     
  5. Oct 13, 2012 #4
    It seems to be the motion in an inertial frame of reference. Well in that case x1 - x2 must be constant I think, so potential energy can be omitted as a constant. What is clear is that potential energy is a function of generalised coordinates, so it has to be Cartesian coordinates as well.
     
  6. Oct 13, 2012 #5

    vanhees71

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    First of all the Lagrangian in the new coordinates
    [tex]L=\frac{\mu}{2}(\dot{y}_1^2+\dot{y}_2^2)-V(y_2)[/tex]
    with [itex]\mu=m/2[/itex].

    To answer the question, whether you can find a set of coordinates, which all are cyclic, you should read about the Hamilton-Jacobi partial differential equation and action-angle variables.

    In your case there is for sure another conserved quantity! Think which that might be!
     
  7. Oct 13, 2012 #6
    well that's the energy but that has nothing to do with cyclic coordinates.
    Also I did read Hamilton-Jacobi theory but that takes its basis in the hamiltonian formulation. So I guess it's not really possible to transform to a frame with all coordinates cyclic UNLESS you use the hamiltonian formulation with more freedom to vary your conjugate variables?
     
  8. Oct 13, 2012 #7

    mfb

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    To get a second cyclic variable, you would need some parameter which describes the time-evolution of your (y2-)system with fixed energy. If V is quadratic (or at least gives oscillations in some way), this would be the phase of the oscillation, for example.
    I doubt that you can do this transformation in an explicit way with a general V.
     
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