Action-Angle Variables

We want to find out the periods of periodic trajectories, even for complex trajectories. Especially, if we don’t need to solve the trajectory.

Suppose we have $H (q, p) = E = α$ is a constant for one dimension. Then, we can get $p = p (q, α)$ . Plotting this orbit in the phase space (q,p), we can get a couple different types of trajectories,

Liberation/ Oscillation motion: ’circular’ phase space trajectories (as q changes [and repeats itself], p repeats itself).
Rotation: ’sinusoidal’ phase space trajectories. (as q continues to increase, p repeats itself)

Let’s introduce the action variable $J$ . Let $J = \oint p d q = \oint p (q, α) d q$ be the integration for one period. Note that $J$ will no longer be dependent upon $q$ so the integration remove the dependence of $q$ . So, $J = J (α)$ . Then, we can write $H = H (J)$ . HPF: $S (q, α, t) = W (q, α) - α t$ . So, $S (q, J, t) = W (q, J) - α (J) t$ .

Let $Q = β = \frac{\partial S}{\partial α}$ . So, $S = S (q, α, t) = W (q, α) - α t = W (q, J) - α (J) t$ hence $Q = β = \frac{\partial W}{\partial α} - t$ . Also, $Q = β = \frac{\partial W}{\partial J} \frac{\partial J}{\partial α} - \frac{\partial J}{\partial α} t$ . Let $\overset{―}{Q} = \overset{―}{β} = \frac{\partial W}{\partial J} - \frac{\partial α}{\partial J} t$ . So, $\overset{―}{Q} = \overset{―}{β} = ω (q, J) - ν (J) t$ . Hence, $ω (q, J) = \overset{―}{β} + ν (J) t$ .

Under a single period of motion, $Δ ω (q, J) = \oint d q \frac{\partial ω}{\partial q} = \oint d q \frac{\partial}{\partial q} (\frac{\partial W}{\partial J}) d q = \frac{\partial}{\partial J} \oint d q \frac{\partial W}{\partial q} = \frac{\partial}{\partial J} \oint d q p = \frac{\partial}{\partial J} J = 1 = ν (J) T$ , where $T$ is the time of one period. Thus, $ν (J) = \frac{1}{T} = \frac{\partial α}{\partial J} = \frac{\partial H}{\partial J}$ . Then, for the particular coordinate (for a seperable system), the frequency for the $σ$ coordinate is $\frac{\partial H (J_{1}, \dots, J_{n})}{\partial J_{σ}}$ .

Example

Suppose we have a one dimensional harmonic oscillator. Then,

\begin{array}{r} H = \frac{1}{2 m} (p^{2} + m^{2} ω^{2} q^{2}) = α, ω = \sqrt{\frac{k}{m}} \end{array}

Then,

\begin{aligned} J & = \oint d q p \\ = \pm \oint d q \sqrt{2 m α - m^{2} ω^{2} q^{2}} \\ = \pm \sqrt{2 m α} \oint d q \sqrt{1 - \frac{m^{2} ω^{2} q^{2}}{2 m α}} \\ = \pm \sqrt{2 m α} \oint d q \sqrt{1 - \frac{m^{2} ω^{2} q^{2}}{2 m α}} \\ = \pm \sqrt{2 m α} \sqrt{\frac{2 α}{m ω^{2}}} \oint d θ \cos^{2} θ \\ = \frac{2 α}{ω} \int_{0}^{2 π} d θ \cos^{2} θ \\ = \frac{2 α}{ω} π . \\ α = \frac{J ω}{2 π} . \\ \frac{\partial H}{\partial J} & = \frac{ω}{2 π} . \\ T & = \frac{2 π}{ω} . \end{aligned}

$\frac{m ω^{2}}{2 α} q^{2} = \sin^{2} θ$ . $q = \sqrt{\frac{2 α}{m ω^{2}}} \sin θ$ . $d q = \sqrt{\frac{2 α}{m ω^{2}}} \cos θ d θ$ .