Action-angle variables

For a one-degree-of-freedom closed orbit, action-angle coordinates make the Hamiltonian a function $H(I)$ and give $\dot{I}=0$, $\dot{\varphi }=dH/dI$.

Near a regular Lagrangian torus, choose a one-form $\lambda$ with $d\lambda =-\omega$ and a basis of fibre cycles ${\gamma }_j$. The action $I_j=(2\pi {)}^{-1}{\int }_{{\gamma }_j}\lambda$ has differential of a smooth map $d{I}_j=(2\pi {)}^{-1}{\sum }_i{t}_i^{(j)}d{F}_i$, where $t^{(j)}$ is the corresponding joint-flow period. Nonsingularity of the period matrix makes these actions local coordinates transverse to the tori.

With angle period $2\pi$, the action is the enclosed phase-space area divided by $2\pi$:

When a Hamiltonian parameter varies slowly compared with the orbital period and no separatrix is crossed, the action changes only at higher adiabatic order.

For a one-dimensional particle of energy $E$ between walls separated by $L$, the unnormalized action isAdiabatic conservation gives $E(t)L(t{)}^2=\text{constant}$.

If an action has parameter dependence $I\propto \sqrt{m/|E|}$, adiabatic conservation implies $|E|\propto m$.

When a gravitational potential changes slowly relative to an orbital period, the orbital actions remain constant provided no separatrix is crossed. In a spherical potential, angular momentum is conserved exactly and the radial action is adiabatically conserved.

Adiabatic invariance of an orbital action preserves action values during slow regular changes, but frequencies $\partial H/\partial J_i$ can vary. Closure requires rational frequency ratios, which need not persist. A slowly added harmonic component in a Kepler orbit potential produces generic apsidal precession.

For radial motion between turning radii $r_{-}$ and $r_{+}$ in a central potential, the radial action per unit mass isIt vanishes on a circular orbit and is canonically conjugate to the radial angle.

For a bound Kepler orbit of semimajor axis $a$ around mass $M$,Adiabatic central-mass change therefore preserves $Ma$ and orbital eccentricity.

For $H=p^2/(2m)-1/|q|=-|E|$, collision continuation between the two turning points gives $I=\sqrt{2m/|E|}$.