Frequency of Rotation

4. The evolution of a particle in the periodic electromagnetic field of a travelling plane wave is mathematically equivalent to four rotation processes. Together these processes, Eq.()-(51), describe a point moving uniformly on a four dimensional torus. The four radii of the torus are proportional to the action of four degrees of freedom and their respective frequencies are

$$\nu_u \equiv\frac{d\phi^u}{d\tau}= \frac{\partial H(\mathbf{I})}{\partial I_u}= \left( -2\frac{\omega_2 I_v}{2\pi m}\right) \frac{\omega_1}{2\pi}$$ (55)

$$\nu_v \equiv\frac{d\phi^v}{d\tau}= \frac{\partial H(\mathbf{I})}{\partial I_v}= \left( -2\frac{\omega_1 I_u}{2\pi m}\right) \frac{\omega_2}{2\pi}$$ (56)

$$\nu_x \equiv\frac{d\phi^x}{d\tau}= \frac{\partial H(\mathbf{I})}{\partial I_x}= \left( \frac{ I_x}{m L_x}\right) \frac{1}{L_x}$$ (57)

$$\nu_y \equiv\frac{d\phi^y}{d\tau}= \frac{\partial H(\mathbf{I})}{\partial I_y}= \left( \frac{ I_y}{m L_y}\right) \frac{1}{L_y}~.$$ (58)

These equations will be recognized as the first half of Hamilton's equations relative to action-angle variables. The second half are simply

$$\frac{dI_u}{d\tau} =-\frac{\partial H((\mathbf{I})}{\partial \phi^u}~(=0)$$ (59)

$$\frac{dI_v}{d\tau} =-\frac{\partial H((\mathbf{I})}{\partial \phi^v}~(=0)$$ (60)

$$\frac{dI_x}{d\tau} =-\frac{\partial H((\mathbf{I})}{\partial \phi^x}~(=0)$$ (61)

$$\frac{dI_y}{d\tau} =-\frac{\partial H((\mathbf{I})}{\partial \phi^y}~(=0)~,$$ (62)

and they express what one knew all along, namely that the action variables are constant along the particle path.