- #1

luisgml_2000

- 49

- 0

## Homework Statement

To obtain the action angle coordinates por the spherical pendulum.

## Homework 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]

## 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: