Action-angle coordinates for the spherical pendulum

[luisgml_2000]

Homework Statement

To obtain the action angle coordinates por the spherical pendulum.

Homework Equations

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

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

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

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

$$J_\phi=\frac{1}{2\pi}\oint p_\phi d\phi=\alpha_2$$

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

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 $$\phi$$ 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

 

[Zeroxt]

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...

 

[TSny]

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...

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 $\theta$.

 

[Zeroxt]

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.

 

[TSny]

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.

You stated that your integration range for $\theta$ is from $-\theta_{0}$ to $\theta_{0}$. But $\theta$ cannot take on negative values.

 

[Zeroxt]

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

 

[TSny]

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 $\theta$?

 

[Zeroxt]

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...

 

[TSny]

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?

 

[Zeroxt]

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.

 

[TSny]

What are all the specific initial conditions that you are assuming for the problem?

 

[Zeroxt]

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.

 

[TSny]

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 $\theta_{max}$?

 

[Zeroxt]

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?

 

[TSny]

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.

 

[Zeroxt]

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...

 

[TSny]

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.

 

[Zeroxt]

Okay, thanks you so much for your help, you help me a lot.