Action Variables

The four periods give rise to the four action variables

$$I_u\equiv \int_0^{2\pi/\omega_1}p_u du$$ (34)

$$I_v\equiv \int_0^{2\pi/\omega_2}p_v dv$$ (35)

$$I_x\equiv \int_0^{L_x}p_x dx$$ (36)

$$I_y\equiv \int_0^{L_y}p_y dy~.$$ (37)

The evaluation of these integrals is done with the help of Eqs.(16)-(18) and (31)-(32). The result is the transformation

$$(P_v,P_x,P_y,P_\tau)\leadsto \left\{ \begin{array}{rl} I_u=&\disp... ..._v\frac{2\pi}{\omega_2}\\ ~&~\\ I_x=& P_xL_x\\ I_y=& P_yL_y \end{array} \right.$$ (38)

and its inverse

$$(I_u,I_v,I_x,I_y)\leadsto \left\{ \begin{array}{rl} -P_\tau=&\dis... ...\frac{\omega_2}{2\pi}\\ ~&~\\ P_x=&I_x /L_x\\ P_y=&I_y /L_y \end{array} \right.$$ (39)

The effect of this transformation is that it decomposes the dynamical system into its fundamental physical - and hence mathematical - components, each having its own frequency.

Indeed, introducing the action variables into the dynamical phase, Eq.(22), one finds that, with the help of Eqs.(31) and (32), its form (a.k.a. ``Hamilton's principal function'') is

$$\overbrace{ W_0(u;\mathbf{I})+W_1(v;\mathbf{I})+W_2(x;\mathbf{I})... ...lpha,\mathbf{I})} -W(x_0^\alpha,\mathbf{I})-(\tau-\tau_0) H(\mathbf{I})}~~~~~~~$$ (40)

where the four contributions to Hamilton's characteristic function

$$W(x^\alpha;\mathbf{I})=\int_0^u p_udu+\int_0^v p_vdv+ \int_0^x p_xdx+\int_0^y p_ydy$$ (41)

are

$$W_0(u,\mathbf{I}) =u \frac{\omega_1}{2\pi}I_u$$

$$%+v\frac{\omega_2}{2\pi}I_v + +\frac{1}{4I_v}\frac{2\pi}{\omega_2}\int_{0}^u 2m\, du \left\{ \f... ...x} \eta_x \sin\omega_1 u+ \frac{I_y}{L_y} \eta_y \sin(\omega_1 u+\delta)\right.$$

$$~~~~~~~~~~~~~~~-\frac{\eta^2_x}{4} m\cos 2\omega_1 u \left. -\frac{\eta^2_y}{4} m\cos (2\omega_1 u+2\delta) \right\}$$ (42)

$$W_1(v,\mathbf{I}) =v\frac{\,\omega_2}{2\pi}I_v$$ (43)

$$W_2(x,\mathbf{I}) =\frac{x}{L_x}I_x$$ (44)

$$W_3(y,\mathbf{I}) =\frac{y}{L_y}I_y$$ (45)
with

$$\eta_x \equiv \frac{qE_x}{m\omega_1}~~~\textrm{and}~~~ \eta_y\equiv \frac{qE_y}{m\omega_1}$$ (46)

as the dimensionless relativistic impulse factors, which express the interaction between the laser and the charge, while

$$H(\mathbf{I})=\displaystyle\frac{1}{2m}\left\{ \frac{I_x^2}{L_x^2... ...i}\frac{\omega_2}{2\pi}+ \frac{m^2}{2}\left( \eta_x^2+\eta_y^2\right) \right\}$$

is the conserved superhamiltonian, Eq.(40), expressed in terms of the four action variables.

The introduction of these variables into the dynamical phase, and their application to the principle of constructive interference,

$$\frac{\partial \tilde{S'}(x;\mathbf{I};\tau)}{\partial I_\beta}=0~~~~~~\beta=u,v,x,y$$

again leads to a straightening out of the phase space trajectories, just like Eq.(23) led to Figure 8b. The only difference is that now the straight lines are tilted relative to the coordinate plane $\tau=\tau_0$. This means that the straight lines have nonzero projections onto this plane. These projected paths are

$$\frac{\partial W(x^\alpha;I)}{\partial I_\beta} = \frac{\partial W(x_0^\alpha;I)}{\partial I_\beta}+(\tau-\tau_0) \frac{\partial H(I)}{\partial I_\beta}~~~~~~\beta=u,v,x,y~.$$

They lead to a number of conclusions.

Problem6: a) Point out why these four equations express the same spacetime trajectory as Eqs.(23).

b) show that their explicit form is

$$(u-u_0)\frac{\omega_1}{2\pi} = -(\tau-\tau_0)\frac{2}{m} \frac{\omega_1}{2\pi}\frac{\omega_2}{2\pi}I_v$$

$$(v-v_0)\frac{\omega_2}{2\pi} + \frac{1}{4I_v^2}\frac{2\pi}{\omega_2}\frac{2m}{\omega_1} \left[... ...L_x}I_x \cos \omega_1 u + \frac{\eta_y}{L_y}I_y \cos (\omega_1 u+\delta)\right.$$

$$~~~~~~~~~~~~~~~+\left.\frac{m}{8}\left( \eta_x^2 \sin 2\omega_1 u+ \eta_y^2 \sin (2\omega_1 u+\delta)\right) \right]^u_{u_0}$$

$$= -(\tau-\tau_0)\frac{2}{m} \frac{\omega_1}{2\pi}\frac{\omega_2}{2\pi}I_u$$

$$\frac{x-x_0}{L_x} -\frac{1}{4I_v}\frac{2\pi}{\omega_2}\frac{2m}{\omega_1} \left. \f... ...\eta_x}{L_x}\cos \omega_1 u\right\vert^u_{u_0} =(\tau-\tau_0)\frac{I_x}{mL^2_x}$$

$$\frac{y-x_0}{L_y} -\frac{1}{4I_v}\frac{2\pi}{\omega_2}\frac{2m}{\omega_1} \left.\fr... ...}\cos (\omega_1 u+\delta) \right\vert^u_{u_0} = (\tau-\tau_0)\frac{I_y}{mL^2_y}$$