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A phase-space distribution function describes the mass in a small cell of positions and velocities: $F(\mathbf{x},\mathbf{v},t)d^{3}x{d}^{3}v$ is that mass. Its velocity integral is the spatial mass density, and $F/\rho$ at a fixed position is a normalized velocity probability density. A number-normalized galactic distribution function differs by the stellar mass convention.

Take $a>0$ and $v_0>0$. For the cusped logarithmic spherical galaxy model, differentiate the gravitational potential and use the spherical Poisson equation for Newtonian gravity:ThusA circular orbit requires $v_c^2/r={\Phi }^{\prime }$. Its galaxy rotation curve is thereforeIt rises as $v_0\sqrt{r/a}$ at small radius and approaches $v_0$ from below at large radius.

The outer flat galaxy rotation curve is qualitatively useful for galaxies. However, the density has an inner $1/r$ cusp and an outer $1/r^2$ tail, while $M(r)\sim v_0^{2}r/G$ diverges. The model therefore requires an outer truncation for a finite galaxy, and does not describe a central constant-density core. Its gravitational potential cannot be normalized to zero at infinity.

For a star, the specific energy is $E=v^2/2+\Phi$ and the specific angular momentum is $\mathbf{L}=\mathbf{x}\times \mathbf{v}$. Since the gravitational potential is stationary and spherical,Thus both the specific energy and the entire angular-momentum vector are conserved.

For Jeans theorem, follow a stellar trajectory in phase space. The Collisionless Boltzmann equation is preciselyIn a steady system ${\partial }_{t}F=0$, so the value of the galactic distribution function is constant along every trajectory: it is itself an integral of motion. Conversely, if $I_1,\dots ,I_m$ are time-independent integrals of motion, then $F=h(I_1,\dots ,I_m)$ satisfies $dF/dt=\sum h_{,a}{\dot{I}}_a=0$. Locally, a complete set of invariant orbit labels expresses any steady galactic distribution function this way. This is the weak form of Jeans theorem; a stronger assertion about a small set of isolating integrals needs additional orbital regularity or mixing assumptions. In particular, the theorem does not say that every steady galaxy must have a galactic distribution function depending on only $E$ and $L$.

For the proposed constant-anisotropy distribution function, write $v_T=(v_{\theta }^2+v_{\varphi }^2{)}^{1/2}$ and $L=r{v}_T$. Introduce a polar angle $\psi$ in the tangential velocity plane. Then $d^{3}v=d{v}_r{v}_{T}d{v}_{T}d\psi$, so the apparent $1/L$ singularity cancels against the velocity Jacobian. This inverse-angular-momentum velocity integration makes the density finite for every $r>0$.

Let $t=e^{-\Phi /v_0^2}=a/(a+r)$, $c_1=1$, $c_2=2$, and ${\alpha }_k=k/(2v_0^2)$. Each exponential component contributesHere the two Gaussian integrals and the angular integral have product ${\pi }^2/{\alpha }_k$. Consequently,The galactic distribution function is nonnegative, gives the required density, and depends on conserved $E,L$; by Jeans theorem it is a self-consistent steady solution.

Differentiating the Gaussian integral with respect to ${\alpha }_k$ shows that each component has $⟨v_r^2{⟩}_k=1/(2{\alpha }_k)=v_0^2/k$, and the same value for $⟨v_T^2{⟩}_k$. Angular averaging gives $⟨v_{\theta }^2{⟩}_k=⟨v_{\varphi }^2{⟩}_k=v_0^2/(2k)$. Weighting these by ${\rho }_1,{\rho }_2$ yields the velocity dispersionsThe radial numerator includes $2r$, as in the original PDF. All mean velocities vanish by parity. The same parity, or the zero angular average of $\sin \psi \cos \psi$, gives $⟨v_r{v}_{\theta }⟩=⟨v_r{v}_{\varphi }⟩=⟨v_{\theta }{v}_{\varphi }⟩=0$.

The velocity-anisotropy parameter isThe constant-anisotropy distribution function therefore favors radial motion at every radius: the radial variance is twice either individual tangential variance. It still has a substantial population with nonzero angular momentum, rather than exclusively radial orbits. The $L^{-1}$ factor enhances low-angular-momentum stars. The radial velocity dispersion approaches $3v_0^2/4$ at the centre and $v_0^2$ far out; each individual tangential variance is half as large. There is no mean rotation, despite the well-defined galaxy rotation curve for hypothetical circular orbits.

For $S_t(z)=\Phi (t,0,z)$, differentiability with respect to initial data gives the variational equationThe Jacobi determinant derivative formula givesThe mixed derivatives in the Hamiltonian vector field cancel:Consequently $J^{\prime }(t)=0$, and $J(0)=1$ gives preservation of phase space volume:The same argument applies to $\Phi (t,s)$ for every starting time $s$. This is the Liouville theorem in Hamiltonian mechanics. Explicit time dependence of $H$ does not affect the cancellation.

Assume $\omega \ne 0$. With $h=0$, the method of characteristics gives $f(t,\Phi (t,0,z))=f_0(z)$. HenceLet $K=\mathrm{supp}{f}_0$ and $E_{\ast }={\max }_K{H}_{\omega }$; if $f_0=0$, the conclusion is immediate. By the Hamiltonian energy balance, every image point of $K$ remains in the energy sublevelThis is a fixed compact ellipsoid in phase space. Therefore the support bound is uniform in time:This uniform support bound from a coercive conserved energy requires no explicit solution of the curves.

The nonzero-frequency qualification is necessary. At $\omega =0$ the energy does not control $x$, and the equation is free transport equation. Choose a smooth compactly supported $f_0$ that is nonzero at $(x_0,v_0)$ with $v_0\ne 0$. Its transported value at $(x_0+t{v}_0,v_0)$ stays nonzero, so the union of the supports is unbounded. Each individual support is compact, but there is no fixed compact support for all times.

A small-angle gravitational encounter between a test star and a field star of mass $m$ gives a transverse kick $\Delta v_{\perp }\simeq 2Gm/(bu)$, where $b$ is the impact parameter and $u$ the relative speed. For number density $n$, encounters in $b,b+db$ occur at rate $2\pi bdbnu$. Independent kicks add in mean square, giving the random-walk derivation of stellar relaxationThe Coulomb logarithm in stellar dynamics has $b_{\max }$ of order system size $r_s$, and $b_{\min }$ of order the strong-deflection scale $Gm/u^2$. For a virialized $N$-star system, $u^2\sim GNm/r_s$, so $\Lambda$ is of order $N$. The stellar relaxation time is the time for accumulated velocity variance to become comparable with $u^2$:Using $n\sim 3N/(4\pi r_s^3)$ gives $t_{\mathrm{rel}}\sim [N/(6\ln N)]r_s/u$. With a diameter-crossing convention $T_{\mathrm{cross}}\sim 2r_s/u$, this is $[N/(12\ln N)]T_{\mathrm{cross}}$, conventionally rounded toThe coefficient is approximate: spatial profile, velocity averages and crossing-time convention change it by factors of order unity. The robust result is $N/\ln N$ crossing times.

For $N\sim 10^{10}$ and $T_{\mathrm{cross}}\sim 10^8$ years, the estimate is about $4\times 10^{15}$ years, enormously longer than a Hubble time. Most galaxies therefore behave as collisionless stellar systems. Dense nuclei and star clusters can relax more rapidly. Negligible stellar encounters do not mean negligible collective gravitational instabilities.

Define the mass-weighted galactic distribution function by $dM=F(\mathbf{x},\mathbf{v},t)d^{3}x{d}^{3}v$, so $\rho =\int F{d}^{3}v$. Hamiltonian gravitational motion preserves phase space volume by the Liouville theorem in Hamiltonian mechanics. In the collisionless regime no encounter term redistributes stars between neighbouring phase space trajectories, so $dF/dt=0$. With acceleration $\nabla \psi$, the Collisionless Boltzmann equation is

For a steadily rotating galactic bar, write the inertial polar angle as ${\varphi }_{\mathrm{I}}=\varphi +{\Omega }_{b}t$. The inertial radial and angular equations are $\ddot{R}-R{\dot{\varphi }}_{\mathrm{I}}^2={\psi }_R$ and $R{\ddot{\varphi }}_{\mathrm{I}}+2\dot{R}{\dot{\varphi }}_{\mathrm{I}}={\psi }_{\varphi }/R$. Substitution gives the equations of motion in a rotating frame:Here the velocities are measured in the bar frame. Set $v_R=\dot{R}$, $v_{\varphi }=R\dot{\varphi }$ and define the positive-force rotating-frame relative effective potentialThe centrifugal sign is positive in this relative-potential convention. The characteristics areConsequently the rotating-frame collisionless Boltzmann equation isFor these planar equations, integrate over $d{v}_{R}d{v}_{\varphi }$ and interpret $\rho$ as the planar or vertically integrated mass density. Define $\rho ⟨A⟩=\int AFd{v}_{R}d{v}_{\varphi }$. Boundary terms in velocity vanish for a sufficiently decaying distribution.

Let the velocity accelerations above be $A_R,A_{\varphi }$. Their velocity divergence is ${\partial }_{v_R}{A}_R+{\partial }_{v_{\varphi }}{A}_{\varphi }=-v_R/R$. Thus integration by parts in the zeroth moment contributes $\rho ⟨v_R⟩/R$, producing the cylindrical continuity equation for a rotating stellar systemThis is conservation of mass: change of density balances flux through the radial and azimuthal sides of a small cylindrical element. The factor $R$ is geometric; the physical phase space measure is $RdRd\varphi d{v}_{R}d{v}_{\varphi }$.

For the radial first moment, multiply the rotating-frame collisionless Boltzmann equation by $v_R$. Its velocity integrations are $\int v_R{A}_R{F}_{v_R}{d}^{2}v=-\rho ⟨A_R⟩$ and $\int v_R{A}_{\varphi }{F}_{v_{\varphi }}{d}^{2}v=\rho ⟨v_R^2⟩/R$. Hence the radial cylindrical Jeans equations in a rotating frame areFor the azimuthal first moment, the $A_R$ velocity integral vanishes, while ${\partial }_{v_{\varphi }}(v_{\varphi }{A}_{\varphi })={\psi }_{\mathrm{eff},\varphi }/R-2{\Omega }_b{v}_R-2v_R{v}_{\varphi }/R$. This givesThe Jeans equations are local momentum-balance equations. The second moments include streaming momentum flux and random-velocity stress, while gravity, centrifugal acceleration and Coriolis acceleration supply the frame-dependent forces. They do not by themselves close the full distribution: a stress prescription or a galactic distribution function is also needed.

Use the Minkowski metric ${\eta }_{mn}=\mathrm{diag}(-1,1,\dots ,1)$ and units with speed of light one. The contractions are $P^2={\eta }^{mn}{P}_m{P}_n$ and $(T{X}^{\prime }{)}^2=T^2{\eta }_{mn}(X^m{)}^{\prime }(X^n{)}^{\prime }$; $(X^m{)}^{\prime }{P}_m$ and ${\dot{X}}^m{P}_m$ pair a vector with a covector without another metric. The Lagrange multipliers impose the two first-class constraintsThey generate worldsheet diffeomorphisms, so the phase space contains both constrained directions and gauge redundancy. Two first-class constraints remove two canonical pairs, leaving $D-2$ physical degrees of freedom per point.

In Monge gauge, $X^0=t$ and $X^1=\sigma$. Write the transverse canonical variables as $\mathbf{X}=(X^2,\dots ,X^{D-1})$ and $\mathbf{P}=(P_2,\dots ,P_{D-1})$. Solving the first-class constraints givesThe negative root selects positive energy. Substitution into the phase-space action gives the Hamiltonian reductionFor a static segment, $\mathbf{P}=0$, its proper length element is $d\ell =\sqrt{1+|{\mathbf{X}}^{\prime }{|}^2}d\sigma$, and $\mathcal{E}d\sigma =Td\ell$. Thus the string tension is the rest energy per unit proper length. In particular a straight resting segment has $\mathcal{E}=T$. Monge gauge is a local choice on a string embedding map for which $X^1$ is a valid coordinate; it need not cover folded strings or all endpoint configurations.

For an unperturbed Kepler orbit, successive revolutions advance the planet-relative orientation by $-n_p{T}_c$ modulo $2\pi$, where $T_c$ is the comet's orbital period. An irrational period ratio makes these orientations dense; sufficiently distant mean-motion resonance and adequate observation time justify the corresponding phase mixing approximation.

At each radius between its pericentre distance and apocentre distance, the comet eventually visits every azimuth in the rotating reference frame. The spatial projection is thereforeIt is an annulus, with a nonuniform radial residence probability, rather than a uniformly filled area. Strictly, the full position-velocity phase space is not this annulus: at each radius the energy and angular momentum constrain the velocity, with inward and outward branches. The annulus is its position-space projection. This description assumes the orbital elements have not yet been substantially changed by planetary scattering.

Use the first-order actionVarying $q$ with fixed endpoints and varying $p$ freely givesThe principle of stationary action therefore yields Hamilton's equations

For $H=p^2+q^{-2}$ these become $\dot{q}=2p$ and $\dot{p}=2q^{-3}$. Along their Hamiltonian flow, $H=E$ is conserved andThus $F=pq-ctH$ is an integral of motion exactly when $c=2$.

Write its constant value as $F_0$. The surfaces $pq-2tE=F_0$ in extended $(q,p,t)$-space giveUsing $E{q}^2=p^2{q}^2+1$,The sign of $q$ is fixed. In phase space, each energy orbit is one of the two branchesEach branch approaches $p=-\sqrt{E}$ and $p=+\sqrt{E}$ as $|q|\to \infty$ and turns at $(p,q)=(0,\pm E^{-1/2})$.

Let $F_1,\dots ,F_n$ be commuting independent integrals for $(M,H)$ and let $G_1,\dots ,G_m$ be such integrals for $(\tilde{M},\tilde{H})$. Regard each as a function on the product phase space. ThenThe $F_i$ commute among themselves, the $G_a$ commute among themselves, and every cross bracket $\{F_i,G_a\}$ vanishes because the functions depend on disjoint canonical coordinates. Their differentials form two independent coordinate blocks, so all $n+m$ are independent on the product of the two regular sets. Thus the combined system is integrable, an instance of the direct product of integrable Hamiltonian systems.

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

A transformation of phase space is canonical when it preserves the symplectic form, or equivalently all Poisson brackets. For one degree of freedom this isequivalently $\{Q,P{\}}_{q,p}=1$.

Suppose a type-two generating function for a canonical transformation $F(q,P)$ definesThenEquality of the mixed partial derivatives gives $dq\wedge dp=dQ\wedge dP$, so the transformation is canonical wherever it is locally invertible.

Forthe fundamental theorem of calculus and differentiation under the integral sign giveThusand in particular $P=(p^2+q^2)/2$.

For the unit-frequency simple harmonic motion, the transformed Hamiltonian is simply $H=P$. Hamilton's equations becomeThereforeand transformation back gives the familiar solution

Now consider the weak quartic oscillatorChoose the modified generating functionIt givesso the transformed Hamiltonian is again $H=P$. Hence $P=E$ and $Q=t-t_0$. Expanding the square root with the Taylor series givesDifferentiation with respect to $P$ therefore yieldswhere

Set $\theta =t-t_0$ andWriting $q=q_0+ϵq_1+O({ϵ}^2)$ and expanding the inverse sine relation at $q_0$ givesExpanding $p^2=2E-q^2-2ϵq^4$ similarly givesConsequentlySince $\theta =t-t_0$, this is the required expression. The formula is understood away from turning points where $p_0=0$ and the ratio $q/p$ itself is singular.

A galactic distribution function $f(\mathbf{x},\mathbf{v},t)$ is the stellar mass or number per six-dimensional phase space volume,Its velocity moments give the spatial density, mean velocity, and velocity dispersion; integrating those quantities along the line of sight and weighting by luminosity produces surface-brightness and line-of-sight-velocity observables. A model is compared with data only after the same projection, selection function, and instrumental convolution have been applied.

Jeans theorem states that every steady solution of the Collisionless Boltzmann equation depends on phase-space coordinates only through integrals of motion. Conversely, every nonnegative function of isolating integrals is a steady collisionless distribution function on the region where those integrals are defined.

For the stated power law in relative energy, isotropy givesWith the requested change of variables $v=\sqrt{2\Psi }\cos \theta$, the density becomeswhere the last equality uses the Beta function and Gamma function. Thusfor $n>1/2$.

The normalized second velocity moment isThe same substitution and the Beta-function recurrence giveConsequently the one-dimensional isotropic velocity dispersion is ${\sigma }^2=\overline{v^2}/3=\Psi /(n+1)$, proving the required linear dependence on the relative potential.

A non-coplanar encounter may change $I$ while preserving $T_p$, so the projection of the accessible phase space onto the $(a,e)$ plane is a region rather than one curve. A pair $(x,e)$ is accessible for some inclination exactly whenFor $T_p=5/2$, the boundary curves are obtained by taking $\cos I=1$ or $-1$; intermediate inclinations fill the region between them. In particular, inclinations at least as large as the value found in part (d) allow the region to meet $e=0$.