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Action-angle coordinates for the spherical pendulum

  1. Nov 22, 2009 #1
    1. The problem statement, all variables and given/known data

    To obtain the action angle coordinates por the spherical pendulum.

    2. Relevant equations

    [tex]
    H=\frac{1}{2mh^2}\left(p_\theta^2+\frac{p_\phi^2}{\sin^2\theta}\right)+mgh\cos\theta
    [/tex]

    [tex]
    \frac{1}{2mh^2}\left(\left(\frac{\partial G}{\partial \theta}\right)^2+\frac{\left(\frac{\partial G}{\partial\phi}\right)^2}{\sin^2\theta}\right)+mgh\cos\theta=-\frac{\partial G}{\partial t}
    [/tex]

    [tex]
    G=-\alpha_1 t+ \alpha_2\phi \pm\int{\sqrt{2mh^2(\alpha_1-mgh\cos\theta)-\frac{\alpha_2^2}{\sin^2\theta}}d\theta}
    [/tex]

    [tex]
    p_\phi&=&\alpha_2\\
    p_\theta&=& \pm\sqrt{2mh^2(\alpha_1-mgh\cos\theta)-\frac{\alpha_2^2}{\sin^2\theta}}
    [/tex]

    [tex]
    J_\phi=\frac{1}{2\pi}\oint p_\phi d\phi=\alpha_2
    [/tex]

    [tex]
    J_\theta=\frac{1}{2\pi}\oint p_\theta d\theta=\frac{1}{2\pi}\oint \pm\sqrt{2mh^2(\alpha_1-mgh\cos\theta)-\frac{\alpha_2^2}{\sin^2\theta}} d\theta

    [/tex]

    3. The attempt at a solution

    The Hamilton-Jacobi equation is separable in these coordinates and hence the calculation of the action coordinates gets simpler. In fact, the action corresponding to [tex] \phi [/tex] is trivial but I cannot evaluate the integral for the other action. Can someone help me out?

    By the way somewhere on the Internet I saw that one cannot define action-angle coordinates for this problem but I find that assertion puzzling since this system is an integrable one and because the phase space for it is bounded.

    Thanks
     
    Last edited: Nov 22, 2009
  2. jcsd
  3. Jul 25, 2015 #2
    I have the same problem, but in my case the integral I express in other way and for small angles, so in my case the integral was from -theta0 to theta0
    ∫±(1/θ)√(-α+Eθ^2 -mglθ^4)dθ.
    So I did a sustitution with y=θ^2, and the integral was the form ∫(1/y)√(-α+Ey-mgly^2)dy, and that integral is in some tables, but I have problem with the limits of these integral, because when I evaluate the integral goes to zero....
    I don't know where is the mistake...
     
  4. Jul 26, 2015 #3

    TSny

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    The limits will not be from ##-\theta_0## to ##\theta_0##. In fact, ##\theta## is always non-negative in spherical coordinates. ##\theta## will vary between some minimum positive value ##\theta_{min}## to some maximum value ##\theta_{max}##. These limits of ##\theta## can be determined by considering what the value of ##p_{\theta}## must be at the limits of ##\theta##.
     
  5. Jul 26, 2015 #4
    Yes, I did that, I integrate with the substitution and then I come back to the original variable and evaluated the result of integral, but in the result I only have squares variables, so the result of the evaluation is zero.
     
  6. Jul 26, 2015 #5

    TSny

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    You stated that your integration range for ##\theta## is from ##-\theta_{0}## to ##\theta_{0}##. But ##\theta## cannot take on negative values.
     
  7. Jul 26, 2015 #6
    But so, how can I define the limits? Because for action of theta, you have to do a line integral, so you have to integrate in a period... For this I suppose that the limits are -theta0 to theta0... In other case I don't know which are the possible limits
     
  8. Jul 26, 2015 #7

    TSny

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    When ##\theta## reaches one of its extreme values (limits), what is the value of ##\dot{\theta}##? What does this tell you about the value of ##p_{\theta}## at the limiting values of ##\theta##?
     
  9. Jul 26, 2015 #8
    I found this value of theta assuming that the velocity in theta (theta dot) was zero and I use the condition of the energy to find these theta. So in this angle the momentum will be zero...
     
  10. Jul 26, 2015 #9

    TSny

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    OK. The limiting values of ##\theta## will correspond to ##\dot{\theta} = 0##. Since ##p_{\theta}## is proportional to ##\dot{\theta}##, you can find the limiting values of ##\theta## by finding where ##p_{\theta} = 0##. Your integrand for the action variable ##J_{\theta}## represents ##p_{\theta}##. For what values of ##\theta## is the integrand equal to zero?
     
  11. Jul 26, 2015 #10
    In the problem appears that the condition initial is theta0 <<1. So the value of theta that I found is a constant times theta0, this constant can take two possible values depending of the large of the pendulum, g and the velocity initial (w0) in fhi (which is constant too). In case that lw0>g the constant take the value of lw0/g in the other case the value of the constant is 1.
     
  12. Jul 26, 2015 #11

    TSny

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    What are all the specific initial conditions that you are assuming for the problem?
     
  13. Jul 26, 2015 #12
    I write you the problem:
    A spherical pendulum it's build with a rod of length, massless and inextensible, which in the extreme of the rod put a punctual mass m under the gravity acceleration. In a initial moment the particle it's shifted an angle theta0 respect to the vertical and it's impulse with a rotational velocity azimutal w0. Using the spherical coordinates determined the action-angles variables and obtain the period of oscillation/libration characteristic of the system in the limit theta0<<1.
     
  14. Jul 26, 2015 #13

    TSny

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    OK. So, the initial conditions may be taken to be ##\theta = \theta_{0}##, ##\dot{\theta} = 0##, ##\phi = 0##, and ##\dot{\phi} = \omega_0##.

    If you follow my suggestions earlier, you should be able to find the limits of integration for ##J_{\theta}##. As expected, you should find ##\theta_{min} = \theta_0##. Did you find a value for ##\theta_{max}##?
     
  15. Jul 26, 2015 #14
    I think that I found the angle, because I compute the Energy with the initial condition and then I compute the Energy in the moment when the velocity in theta is zero, so the energies have the same value therefore I found an angle with these condition...
    Now I have a doubt, the minimum angle is theta0?
     
  16. Jul 26, 2015 #15

    TSny

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    I shouldn't have said that ##\theta_{min}## is necessarily ##\theta_0##.

    ##\theta_0## could be either ##\theta_{min}## or ##\theta_{max}## depending on whether or not ##\omega_0## is greater than or less than ##\sqrt{g/l} \equiv \Omega_0##, where ##l## is the pendulum length.

    If ##\omega_0 > \Omega_0##, then the pendulum will swing out to larger values of ##\theta##, and ##\theta_0## will be the minimum value of ##\theta## for the motion.
    If ##\omega_0 < \Omega_0##, then the pendulum will "fall" to smaller values of ##\theta##, and ##\theta_0## will be the maximum value of ##\theta## for the motion.
    If ##\omega_0 = \Omega_0##, then the pendulum mass will move in a horizontal circle so that ##\theta## doesn't change during the motion.
     
  17. Jul 26, 2015 #16
    Yes I found angles in term of these condition that you say, so I have two possible values for theta one of them greater than theta0 and the other equal to theta0...
     
  18. Jul 26, 2015 #17

    TSny

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    OK. One of the limits of integration will be ##\theta_0## and the other limit will be greater than, less than, or equal to ##\theta_0## depending on the initial conditions.
     
  19. Jul 26, 2015 #18
    Okay, thanks you so much for your help, you help me a lot.
     
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