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A quantity conserved to leading order when a parameter varies slowly compared with the natural motion period. The action variable of a regular periodic Hamiltonian orbit is a standard example, as explained by adiabatic invariance of the action.

A periodic canonical coordinate paired with an action variable. In an integrable background Hamiltonian $H_0(\mathbf{J})$, the angles advance uniformly at ${\dot{\theta }}_i=\partial H_0/\partial J_i$, while the action variables remain constant. The periodicity permits an integer-indexed Fourier series on each regular orbital torus.

For a harmonic oscillator with Hamiltonian $H=p^2/(2m)+m{\omega }^2{q}^2/2$, where $m,\omega >0$, defineThese action-angle variables are defined for $I>0$, with a local angle $\theta$ modulo $2\pi$. A proof of canonicity is ${\theta }_q=p/(2I)$ and ${\theta }_p=-q/(2I)$, which give $\{\theta ,I\}=1$ using the Poisson bracket. Substitution gives $H=\omega I$. Hamilton's equations then prove $I(t)=I(0)$ and $\theta (t)=\theta (0)+\omega t$. The action integral around the corresponding phase space ellipse is $\oint pdq=2\pi I$, so $I$ is the usual normalized action variable. The angle is undefined at the zero-energy origin.

As $t\to +\infty$, the displayed limit holds in distribution theory. Indeed $\sin (tx)/x=\frac{1}{2}{\int }_{-t}^t{e}^{iux}du$ and the Fourier representation of the Dirac delta function gives the limit on smooth rapidly decaying test functions. For a regular detuning $\Delta (\mathbf{J})$, the coarea formula reduces its use under an integral over action variables to this one-dimensional identity. The kernel grows like $t$ at exact resonance, so this is not pointwise convergence.

Work on $I>0$, with a local choice of the periodic angle $\theta$. The inverse formulas areIn particular, ${\theta }_q=p/(2I)$, ${\theta }_p=-q/(2I)$, $I_q=\alpha q$ and $I_p=p/\alpha$. Their Poisson bracket isThe other fundamental Poisson brackets vanish by antisymmetry. Thus this is a canonical transformation for every $\alpha >0$, on local charts away from the origin. At $I=0$ the angle is undefined.

In the new coordinates the harmonic oscillator Hamiltonian isFor $I$ to be a conserved quantity throughout phase space, Hamilton's equations require $\dot{I}=-{\partial }_{\theta }{H}_0=0$ identically. The two coefficients must coincide. For the usual $m>0$ and $\omega >0$, positivity of $\alpha$ selectsThese are harmonic oscillator action-angle variables; the action variable is $I=H_0/\omega$.

Define the unperturbed orbital angular frequency vector byIt is the rate of advance of the angle variables along the unperturbed Hamiltonian flow; the action variables are constant. Expand the Collisionless Boltzmann equation to first order in the perturbations. Since $f_0$ and $H_0$ depend only on the action variables, $[f_0,H_0]=0$, andThe omitted Poisson bracket $[\delta f,\delta \Phi ]$ is quadratic in the perturbations. Use the Fourier series conventionFor each harmonic the linear ordinary differential equation isMultiplication by the integrating factor $e^{i\mathbf{k}\cdot \mathbf{\Omega }t}$ and use of $\delta f_{\mathbf{k}}(0)=0$ give the causal linear response in action-angle variables:The zero harmonic stays zero at this order. This formula retains the entire forcing history; replacing it immediately by a resonant Dirac delta function would discard the transient response.

Write $m=k_{\varphi }$, $B_{\mathbf{k}}=\mathbf{k}\cdot {\partial }_{\mathbf{J}}{f}_0$, and let the detuning from the rotating perturbation beSubstitute $\delta {\Phi }_{\mathbf{k}}={\Psi }_{\mathbf{k}}{e}^{-im{\Omega }_{\mathrm{p}}t}$ into the linear response in action-angle variables, then set $\tau =t-t^{\prime }$:At exact resonance the last quotient is interpreted by its continuous limit $it$. Inserting this response in c givesPair $\mathbf{k}$ with $-\mathbf{k}$. Because $B_{-\mathbf{k}}=-B_{\mathbf{k}}$, ${\Delta }_{-\mathbf{k}}=-{\Delta }_{\mathbf{k}}$, and $|{\Psi }_{-\mathbf{k}}{|}^2=|{\Psi }_{\mathbf{k}}{|}^2$, the imaginary cosine terms cancel. The finite-time torque isThis is a leading, second-order response in the perturbation amplitude.

For the long-time limit, assume a continuum of occupied action variables, sufficiently smooth integrable weights, convergent harmonic sums, and regular resonance surfaces ${\nabla }_{\mathbf{J}}{\Delta }_{\mathbf{k}}\ne 0$ wherever a contributing surface intersects the distribution. Assume also that the perturbation is weak enough to neglect nonlinear trapping and secular changes in $f_0$, and that the prescribed stellar bar pattern and its coefficients stay fixed. Then phase mixing removes nonresonant transients in the integrated response. The relevant long-time resonance kernel is the distributional convergenceIt is not a pointwise limit: at $x=0$ the kernel equals $t$. To justify its use under the integral over action variables, resolve the integral into detuning surfaces; the coarea formula givesSmoothness and decay of this surface weight reduce the limit to the one-dimensional long-time resonance kernel. Equivalently, a causal convergence factor $e^{-\eta \tau }$ gives $\mathrm{Re}{\int }_0^{\infty }{e}^{-\eta \tau -i\Delta \tau }d\tau =\eta /({\eta }^2+{\Delta }^2)\to \pi \delta (\Delta )$ as $\eta \downarrow 0$.

Therefore the resonant torque from a rotating perturbation isOnly orbital resonances contribute to this steady integrated torque. The limit is a formal weak-perturbation limit, or an intermediate-time approximation after phase mixing but before substantial secular redistribution. It does not prove an unchanged halo can absorb angular momentum at this rate forever. A discrete set of orbital frequencies, singular resonance surfaces, or strong trapping can invalidate the stated steady limit.

For an integrable or approximately integrable background Hamiltonian, action-angle variables turn an otherwise complicated stellar orbit into $\dot{\mathbf{J}}=0$ and $\dot{\mathbf{\theta }}=\mathbf{\Omega }(\mathbf{J})$. This has two useful consequences. First, a phase-mixed equilibrium can be represented directly as $f_0(\mathbf{J})$: the action variables label the unperturbed orbits, while angle variables describe position along them. Second, the angles are periodic, so Fourier series reduce unperturbed transport to multiplication by $i\mathbf{n}\cdot \mathbf{\Omega }$ and make orbital resonances explicit. The action variables also separate rapid orbital motion from slow secular changes. These advantages apply on regular orbital regions; a globally chaotic background need not admit global action-angle variables.

The transformation is canonical, preserving the phase space measure and the stated total mass $M=\int f{d}^3\theta d^{3}J$. The collisionless Vlasov equation isHere $H$ is the single-star specific Hamiltonian; the normalization of $f$ by mass changes neither the Poisson bracket nor the transport equation.

Since $H_0$ and $f_0$ depend only on the action variables, $\{f_0,H_0\}=0$. Drop the quadratic term $\{\delta f,\delta {\Phi }^{\mathrm{tot}}\}$ from the Vlasov equation in a. Defining the orbital frequency vector by $\mathbf{\Omega }(\mathbf{J})=\partial H_0/\partial \mathbf{J}$, the linearized equation isThe sign of the driving term follows from the second term in the canonical Poisson bracket, because the background has no angle dependence.

For a regular integrable Hamiltonian system, action-angle variables offer two related simplifications.

These advantages assume regular or approximately integrable mean motion; globally defined action-angle variables need not exist in chaotic regions.

For $k_R=0$, the detuning is ${\omega }_{m0}=m({\Omega }_{\phi }-{\Omega }_{\mathrm{p}})$. The Gaussian resonance broadening factor selects the corotation resonance, where the star's azimuthal frequency matches the pattern speed. The derivative driving the flux is just $m\partial f_0/\partial J_{\phi }$. In fact this contribution alone has the formA single Fourier series harmonic changes the action variables in the direction $(m,k_R)$, because $\dot{\mathbf{J}}=-i\mathbf{k}\delta {\varphi }_{\mathbf{k}}{e}^{i\mathbf{k}\cdot \mathbf{\theta }}$ plus its conjugate. For this harmonic, $\Delta J_R=0$: stars change angular momentum and shift their epicyclic guiding center without a leading-order increase of random radial motion. This is stellar radial migration driven by the corotation resonance. The resonance condition explains the absence of leading heating: $\Delta E\simeq {\Omega }_{\phi }\Delta J_{\phi }={\Omega }_{\mathrm{p}}\Delta J_{\phi }$ at the corotation resonance.

The action variable for a regular periodic one-degree-of-freedom Hamiltonian orbit is $I(E)=(2\pi {)}^{-1}\oint pdq$, with orientation along the motion. Along its energy curve, implicit differentiation gives $\partial p/\partial E=1/H_p=1/\dot{q}$. Differentiating the loop integral branchwise givesAt simple turning points the two momentum branches join and their endpoint contributions cancel, so the integrable turning-point singularities do not spoil this differentiation. The formula concerns regular periodic levels, not a separatrix of infinite period.

The Hamilton equations are $\dot{q}=p/m$ and $\dot{p}=-V_q$. Applying the chain rule to $H(q,p,\lambda (t))$, their two phase-space contributions cancel, giving $\dot{H}=V_{\lambda }\dot{\lambda }=H_{\lambda }\dot{\lambda }$.

For fixed $\lambda$, a closed periodic orbit of energy $E$ has $p=\pm \sqrt{2m(E-V)}$. Thus its action variable isThe sign matches the direction on each half of the orbit, making the enclosed phase-plane area positive. Since $dq=\dot{q}dt=pdt/m$, differentiating under the integral givesFor slowly varying $\lambda$, the exact energy equation averaged over a nearly frozen cycle gives $d⟨H⟩/dt=⟨V_{\lambda }⟩\dot{\lambda }$ to the retained adiabatic order. Using $E=⟨H⟩$, the two chain-rule terms therefore cancel:Here the angle brackets differentiate the explicit parameter dependence along the frozen cycle, not an independently chosen family of energies. The action is an adiabatic invariant, with errors beyond the slow-cycle approximation.

For $V=\lambda q^{2n}$, scale $q=(E/\lambda {)}^{1/(2n)}s$. ThenThe integral is a positive constant depending only on $n$. Constancy of $I$ gives $E^{n+1}/\lambda$ constant, hence

For a regular periodic energy orbit, differentiate the action integral with respect to its energy. On a branch $H(p,q)=E$, implicit differentiation gives $\partial p/\partial E=1/H_p=1/\dot{q}$. ThusThe integral follows the entire orbit, with its orientation; at turning points this formula is interpreted by the limiting integral from each branch. It is valid for a smooth family of nonseparatrix periodic orbits.

An adiabatic invariant means that the action variable remains constant to leading order when the external parameter changes on a timescale much longer than the orbital period. Denote that parameter by $\lambda$. With $H(0,\lambda )=0$, the displayed identity integrates toIf the parameter change decreases $T(J,\lambda )$ for every intervening action $J$, this integrand increases pointwise. At the adiabatically fixed positive $I$, the energy increases. This uses the phrase “with any $I$” as a statement about the family of orbits, not merely a period decrease at a single selected action; that weaker premise alone would not determine the integral.

Hamilton's equations are ${\dot{q}}_j=\partial H/\partial p_j$ and ${\dot{p}}_j=-\partial H/\partial q_j$. The Arnold-Liouville theorem states that $n$ functionally independent first integrals in involution on a $2n$-dimensional symplectic manifold give local integrability; a compact connected regular common level is an $n$-torus. Near such a torus there are action-angle variables $(I,\vartheta )$ with $H=H(I)$, $\dot{I}=0$ and $\dot{\vartheta }=\partial H/\partial I$.

For the displayed diagonal oscillator Hamiltonian set $E_k=(p_k^2+{\omega }_k^2{q}_k^2)/2$. Each is constant, since ${\dot{q}}_k=p_k$ and ${\dot{p}}_k=-{\omega }_k^2{q}_k$. Different $E_k$ depend on disjoint canonical pairs, so their Poisson brackets vanish. Their differentials are independent wherever every $d{E}_k$ is nonzero, an open dense set; this proves complete integrability. For nonzero frequencies put ${\Omega }_k=|{\omega }_k|$. A positive-energy level in each pair is an ellipse with semiaxes $\sqrt{2E_k}$ and $\sqrt{2E_k}/{\Omega }_k$. Its enclosed area is $2\pi E_k/{\Omega }_k$, so the action variables areFor the usual positive oscillator frequencies this is $I_k=E_k/{\omega }_k$. If a printed “constant” frequency is allowed to be zero, that pair is a free particle: $p_k$ is a commuting integral, but its generic level is noncompact and has no closed-cycle oscillator action. Complete integrability persists, while the torus/action formula requires nonzero frequencies.

Put $\pi =p-eA/c=m\dot{q}$. The Hamilton equations areDifferentiate the mechanical momentum and substitute:With $B=\nabla \times A$ and $E=-c^{-1}{\partial }_{t}A$ in this zero-scalar-potential gauge, this is $\boxed{m\ddot{q}=eE+(e/c)\dot{q}\times B}$, the Lorentz force in the units of the Hamiltonian.

For $A=(-y{B}_0(z,t),0,0)$, $x$ is cyclic, so $p_x$ is conserved. Choose $p_x=0$, giving $\dot{x}=\Omega y$, where $\Omega =e{B}_0/(mc)$. Since $p_y=m\dot{y}$, its Hamilton equation gives ${\dot{p}}_y=-(e/c)B_0\dot{x}=-m{\Omega }^{2}y$, henceFor nonzero constant $B_0$, the $(y,p_y)$ orbit is an ellipse of semiaxes $\sqrt{2E^{\prime }/(m{\Omega }^2)}$ and $\sqrt{2m{E}^{\prime }}$. Its enclosed phase-space area is $2\pi E^{\prime }/|\Omega |$, so the action variable isThe absolute value keeps the action positive if the signed charge or magnetic field is negative.

When the field varies slowly along the trajectory, with $|d\Omega /dt|\ll {\Omega }^2$ and no zero-frequency crossing, adiabatic invariance of the action makes $I$ approximately constant. Thus $E^{\prime }\simeq I|\Omega (z,t)|$ and the slow motion has effective HamiltonianThe effective potential is the second term. Direct averaging checks its force: the exact equation is $m\ddot{z}=-m\Omega {\partial }_z\Omega y^2$, while the oscillator has $⟨y^2⟩=E^{\prime }/(m{\Omega }^2)$. Hence $m\ddot{z}=-I{\partial }_z|\Omega |$. Spatial field variation along the slow motion must satisfy the same adiabatic condition; slow explicit time variation alone does not control passage through a sharp spatial gradient. A time-dependent effective potential need not conserve total energy.

The Hamiltonian is $\boxed{H(q,p)=p^2/(2m)+V(q)}$. On a regular family of closed phase-space trajectories, action-angle variables are canonical coordinates $(\theta ,I)$ in which $\theta$ is periodic modulo $2\pi$ and $H=H(I)$. Hamilton's equations then give $\dot{I}=0$ and $\dot{\theta }=dH/dI$.

The action variable isThe orientation is that of the motion, so the integral is the positive enclosed phase-space area divided by $2\pi$.

A constant-energy phase-space trajectory of the harmonic oscillator is the ellipseIts area is $\pi \sqrt{2E/(m{\omega }^2)}\sqrt{2mE}=2\pi E/\omega$. Thus the action and slow energy change areThe action variable is an adiabatic invariant when $|\dot{\omega }|\ll {\omega }^2$, with $\omega$ staying positive. Averaging over many short oscillations therefore keeps $I$ approximately fixed; an increase in frequency increases the energy in the same proportion.

Liouville integrability on a $2n$-dimensional symplectic manifold means that there are $n$ functionally independent first integrals, including $H$, whose pairwise Poisson brackets vanish. Near a compact connected regular common level set, the Arnold-Liouville theorem gives action-angle variables $(J_i,{\theta }_i)$ with $J_i=(2\pi {)}^{-1}\oint pdq$, ${\theta }_i$ periodic modulo $2\pi$, $H=H(J)$, ${\dot{J}}_i=0$ and ${\dot{\theta }}_i=\partial H/\partial J_i$.

For a frozen mass, the turning point is $q_{\ast }=(2E{)}^{1/(2k)}$. The action variable isAdiabatic invariance of the action holds to leading order when the mass varies slowly compared with an orbital period, away from degenerating orbits. Keeping $J$ fixed givesFor $k=1$ this agrees with $E=\omega J$ and $\omega \propto m^{-1/2}$.

LetAn orbit of energy $E>0$ is an ellipse in phase space, with intercepts $x=\pm \sqrt{2E/(m{\omega }^2)}$ and $p=\pm \sqrt{2mE}$. Its enclosed area is $2\pi E/\omega$, so the action variable isThe transformationis canonical and gives the action-angle variables. Substitution yields