# Action-angle coordinates for the spherical pendulum

• luisgml_2000
In summary, the student is trying to find the limits of ##\theta## and they are not from -theta0 to theta0. They are from some minimum positive value to some maximum value. The limits of ##\theta## can be determined by considering what the value of ##p_{\theta}## must be at the limits of ##\theta##.
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

Last edited:
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...

Zeroxt said:
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##.

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.

Zeroxt said:
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.

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

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##?

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...

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?

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.

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

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.

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}##?

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?

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.

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...

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.

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

## 1. What are action-angle coordinates for the spherical pendulum?

Action-angle coordinates are a set of coordinates used to describe the motion of a spherical pendulum. They are a combination of the pendulum's angle of rotation and its corresponding action, which is a measure of the pendulum's momentum. This coordinate system is useful for analyzing the motion of a pendulum, as it allows for a more intuitive understanding of its behavior.

## 2. How are action-angle coordinates calculated for a spherical pendulum?

The action-angle coordinates for a spherical pendulum can be calculated using the Hamiltonian formalism, which is a mathematical framework for describing the dynamics of a system. The Hamiltonian equations of motion can be used to derive the action and angle variables for the pendulum, which can then be used to describe its motion.

## 3. What is the significance of using action-angle coordinates for a spherical pendulum?

Action-angle coordinates have several advantages when studying the motion of a spherical pendulum. They allow for a more intuitive understanding of the pendulum's behavior, and they can simplify the equations of motion, making them easier to solve. Additionally, these coordinates can reveal important information about the pendulum's energy and stability.

## 4. How do action-angle coordinates compare to other coordinate systems for a spherical pendulum?

There are several other coordinate systems that can be used to describe the motion of a spherical pendulum, such as Cartesian coordinates or spherical coordinates. However, action-angle coordinates have the advantage of simplifying the equations of motion and providing a more intuitive understanding of the pendulum's behavior. They are also particularly useful for studying the stability of the pendulum's motion.

## 5. Can action-angle coordinates be used for other systems besides the spherical pendulum?

Yes, action-angle coordinates can be used for a variety of physical systems, not just the spherical pendulum. They are particularly useful for systems with periodic motion, such as a pendulum or a harmonic oscillator. They are also commonly used in celestial mechanics and other areas of physics where periodic motion is present.

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