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Although the energy in velocities is $m({\dot{r}}^2+r^2{\dot{\varphi }}^2)/2+2{\pi }^{2}k(r-r_0{)}^2$, the Hamiltonian in canonical variables isThe field therefore enters Hamilton's equations through the relation between canonical and mechanical momentum, even though a magnetic field does no work.

For asymptotically Cartesian spatial metrics $h_{ij}={\delta }_{ij}+O(r^{-1})$, unit asymptotic lapse and vanishing asymptotic shift, the curvature term in the bulk gravitational Hamiltonian has variationThe surface variation is finite under the usual differentiable asymptotic falloff and is not removed by simply taking the boundary to infinity. Adding Arnowitt-Deser-Misner energy cancels it and makes Hamilton's equations well defined for allowed variations that need not fix the energy. On the constraint surface, the total Hamiltonian for that time translation equals the boundary energy.

Along Hamilton's equations, spatial derivatives of the Hamiltonian function cancel in its total time derivative. Thus $dH/dt={\partial }_{t}H$. An autonomous Hamiltonian function is conserved; explicit time dependence generally prevents conservation even though phase space is still preserved.

Hamiltonian flow is the phase-space evolution generated by Hamilton's equations for a fixed Hamiltonian.

Transport a phase-space density along Hamilton's equations. The Hamiltonian vector field is $b=({\nabla }_{v}H,-{\nabla }_{x}H)$ and the equation is ${\partial }_{t}f+b\cdot {\nabla }_{x,v}f=h$. Its zero divergence gives volume-preserving transport. For $h=0$, the density is constant along characteristics; a source is integrated by the Duhamel formula for Hamiltonian transport. This is distinct from the elliptic Liouville equation.

A dynamical system on a symplectic manifold whose vector field is generated by a Hamiltonian. In canonical variables it obeys Hamilton's equations.

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.

The Hénon-Heiles system is a Hamiltonian system with two degrees of freedom and Hamiltonian functionIts Hamilton's equations give $\ddot{x}=-x-2xy$ and $\ddot{y}=-y-x^2+y^2$. Its potential energy combines an isotropic quadratic minimum with a cubic interaction.

A Lagrangian is hyperregular when its fiber derivative, sending a velocity to its conjugate momentum, is a global diffeomorphism from the tangent bundle to the cotangent bundle. This gives a globally defined Legendre transform in mechanics and a global equivalence between the Euler-Lagrange equations and Hamilton's equations. With time dependence this requirement is imposed at each time, with a smooth inverse depending on time.

For a charged particle in a uniform magnetic field, the symmetric gauge $\mathbf{A}=\mathbf{B}\times \mathbf{x}/2$ makes the displayed magnetic pseudomomentum a conserved quantity when no other position-dependent potential breaks translation symmetry. It differs from the kinetic momentum $\mathbf{\pi }=\mathbf{p}-e\mathbf{A}=m\dot{\mathbf{x}}$. With $\mathbf{B}=B{\mathbf{e}}_3$, Hamilton's equations for $H=|\mathbf{p}-e\mathbf{A}{|}^2/(2m)$ give ${\dot{p}}_1=(eB/2){\dot{x}}_2$ and ${\dot{p}}_2=-(eB/2){\dot{x}}_1$, proving conservation of $K_1=p_1-eB{x}_2/2$ and $K_2=p_2+eB{x}_1/2$. The charge factor is necessary unless it has already been absorbed into the field convention. In quantum mechanics the transverse components need not commute, even though each commutes with this Hamiltonian operator; magnetic translation algebra records that distinction.

In canonical coordinates on phase space, the Poisson bracket isIt treats $t$ as a parameter. Along a solution of Hamilton's equations, the chain rule givesTherefore the Hamiltonian conservation law isThe total derivative follows the trajectory; the explicit partial derivative holds $(\mathbf{q},\mathbf{p})$ fixed.

Define the canonical momenta byTo obtain an ordinary unconstrained Hamiltonian, assume a regular Lagrangian: the velocity Hessian matrix $({\partial }^{2}L/\partial {\dot{q}}_i\partial {\dot{q}}_j)$ is invertible. The inverse function theorem then lets us solve locally for $\dot{\mathbf{q}}=\mathbf{v}(\mathbf{q},\mathbf{p},t)$. The phase space is the cotangent bundle $T^{\ast }\mathcal{Q}$, with local canonical coordinates $(\mathbf{q},\mathbf{p})$ and dimension $2n$. The Legendre transform in mechanics definesThe velocities in this formula have been eliminated in favour of phase space variables. A hyperregular Lagrangian makes this construction global; regularity alone guarantees it locally. Hamilton's equations are

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 $I=H_0/\omega$.

For a time-dependent type-two generating function for a canonical transformation, the differential identitygives $K=H+{\partial }_t{F}_2$, with the partial derivative taken at fixed $(q,P)$. Here $F_2=\cos q\sin P$ has no explicit time dependence, so the Hamiltonian correction isAs always, adding a function of time alone to the new Hamiltonian leaves Hamilton's equations unchanged. Choosing a time-independent type-two generating function for a canonical transformation fixes that irrelevant freedom and gives the displayed $\Lambda =0$.

The action-angle variables $(\mathbf{\theta },\mathbf{J})$ are canonical coordinates, with $[{\theta }_a,J_b]={\delta }_{ab}$. In the convention $\dot{g}={\partial }_{t}g+[g,H]$, their Poisson bracket isIn particular, Hamilton's equations read $\dot{\mathbf{\theta }}={\partial }_{\mathbf{J}}H$ and $\dot{\mathbf{J}}=-{\partial }_{\mathbf{\theta }}H$, fixing the sign needed below.

Since $L_z=J_{\varphi }$, Hamilton's equations giveFor the spherical unperturbed Hamiltonian, $H_0=H_0(J_r,L)$; in these particular action-angle variables the third component ${\Omega }_{\varphi }=0$ describes the fixed orientation of the orbital plane. It should not be confused with the physical azimuthal angular velocity of a particle around its orbit.

The conserved normalization of a probability distribution $\int f{d}^3\theta d^{3}J=1$ makes the expected value of specific angular momentum equal to $\int L_{z}f{d}^3\theta d^{3}J$. Its time derivative is the torque per unit halo mass:This follows either by averaging the single-particle equation or by integrating the Collisionless Boltzmann equation against $L_z$, with vanishing boundary fluxes.

The term $f_0{\partial }_{{\theta }_{\varphi }}\delta \Phi$ integrates to zero over the periodic angle variable. Thus the first nonzero torque is quadratic in the perturbation. Use the Fourier series expansions of b; the source's reference here to part (a) is a numbering slip. Integrating the product over all three angle variables yieldsReality of the potential energy per unit mass gives $\delta {\Phi }_{-\mathbf{k}}=\delta {\Phi }_{\mathbf{k}}^{\ast }$; likewise $\delta f_{-\mathbf{k}}=\delta f_{\mathbf{k}}^{\ast }$. Conjugate harmonics make the total torque real. The sum is over all integer vectors, so no extra factor of two should be inserted.

Put $z_i=({\mathbf{x}}_i,{\mathbf{v}}_i)$ and $d{z}_i=d^3{x}_i{d}^3{v}_i$. Normalize the symmetric $N$-particle phase-space distribution function $D_N$ to one. Use factorial normalization for its one-particle marginal and two-particle distribution function:Thus $\int fdz=N$ and $\int f_2(z_1,z_2)d{z}_2=(N-1)f(z_1)$. For unit particle mass, $f$ also carries mass density. These conventions explain why no extra factor $N-1$ occurs in the result.

Separate the Hamiltonian as $H=H_1+H_{\mathrm{rest}}+H_{\mathrm{int}}$, whereand $H_{\mathrm{rest}}$ contains all terms involving only particles $2,\dots ,N$. The stated Poisson bracket gives Hamilton's equationsConservation of probability along the Hamiltonian flow gives the Hamiltonian Liouville equationAssume decay of the required velocity moments and either decay or periodic boundary conditions in position. Integrating over $z_2,\dots ,z_N$ eliminates the $H_{\mathrm{rest}}$ terms by integration by parts. The terms in $H_{\mathrm{int}}$ differentiating an integrated velocity also vanish. Indistinguishability makes the remaining $N-1$ terms equal, soRelabeling the two particles gives the first BBGKY hierarchy equation:This is exact for a differentiable pair potential; no independent-particle closure has been made.

Normalize the symmetric $N$-particle phase-space distribution function $D_N$ to one and write $z_i=({\mathbf{x}}_i,{\mathbf{v}}_i)$, $d{z}_i=d^3{x}_i{d}^3{v}_i$. The number-normalized one- and two-particle distribution functions areThus $\int fdz=N$ and $\int f_2(z,z_{\star })d{z}_{\star }=(N-1)f(z)$. These conventions absorb the otherwise explicit factor $N-1$ in the first BBGKY hierarchy equation.

Separate the Hamiltonian into kinetic energy, external potential energy, and pair potential terms. Their Hamilton's equations giveThe sign follows directly from the stated Poisson bracket: ${\dot{v}}_{i\alpha }=\{v_{i\alpha },H\}=-{\partial }_{x_{i\alpha }}H$. The Hamiltonian Liouville equation is consequentlyIntegrate over $z_2,\dots ,z_N$ and multiply by $N$. Assume sufficient decay in velocity space and either decay or periodic boundary conditions in position, so that integration by parts removes all transport terms with $i\ge 2$. The surviving interaction term iswhere indistinguishability makes all $N-1$ terms equal. Relabeling $(z_1,z_2)$ as $(z,z_{\star })$ yields the exact resultNo assumption of molecular chaos for the Boltzmann equation has entered: the dependence on the two-particle distribution function is still exact.

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.

Let $⟨\cdot ⟩$ denote $(2\pi {)}^{-2}\int d^2\theta$. Average the Collisionless Boltzmann equation to second order. All first-order angle derivatives average to zero, leavingAn angle integration by parts shows thatThus the angle-averaged galactic distribution function obeys an action-space continuity equation,This sign also follows from Hamilton's equations, ${\dot{J}}_i=-{\partial }_{{\theta }_i}\delta \Phi$: the flux is $⟨\delta f{\dot{J}}_i⟩$.

Reality gives $\delta {\varphi }_{-\mathbf{k}}=\delta {\varphi }_{\mathbf{k}}^{\ast }$ and $\delta f_{-\mathbf{k}}=\delta f_{\mathbf{k}}^{\ast }$. Orthogonality of the Fourier series impliesInsert b to obtainAlthough individual terms are complex, pairing $\mathbf{k}$ with $-\mathbf{k}$ makes the flux real.

Now take the marginalization specified by $F_0=2\pi {\int }_0^{\infty }{f}_{0}d{J}_R$. Assume vanishing radial-action flux at $J_R=0$ and $J_R\to \infty$: the first boundary does not permit negative radial action, and the second has no escaping flux in this calculation. ThenwithIf radial-action boundary flux were present, its explicit boundary term would have to be retained.