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The hierarchy obtained by integrating the $N$-particle Hamiltonian Liouville equation over successively fewer particle coordinates. Each marginal equation involves the next marginal, so an autonomous one-particle equation requires a closure. For a pair potential, its first equation couples the phase-space distribution function to the two-particle distribution function. Number-normalized factorial marginals and unit-normalized probability marginals have different explicit combinatorial factors.

The Boltzmann equation evolves a phase-space distribution by balancing free streaming, external forces, and collisions.

For bound Kepler orbits, the Delaunay variables obey $0<G<L$, $-G<H<G$, and three periodic angles. A canonical transformation preserves phase space volume. At fixed $L$, integrating $H$ supplies the measure $2GdG$. Since $G=L\sqrt{1-e^2}$, the fraction with orbital eccentricity less than $e$ is $L^{-2}{\int }_{L\sqrt{1-e^2}}^{L}2GdG=e^2$. Thus the probability density function is $2e$, independently of any normalizable energy-only phase-space distribution function. A Boltzmann distribution is unnecessary. Even the time-dependent density $F(E(L))[1+\eta \cos (l-n(L)t)]$, $|\eta |<1$, has this same marginal because integrating the mean anomaly removes its oscillatory part; an eccentricity marginal alone does not imply equilibrium.

The collisionless equation in the title transports a gyrophase-independent phase-space distribution function at fixed first adiabatic invariant. Its leading guiding center position has $\dot{\mathbf{R}}={\mathbf{u}}_{\perp }+v_{\parallel }\mathbf{b}$. The parallel acceleration includes $(q/m)E_{\parallel }-\mu \mathbf{b}\cdot \nabla B$ and, when required, terms from motion of the local frame. Both $v_{\parallel }$ and the parallel derivative refer to the exact perturbed magnetic field.

For a homogeneous collisionless stellar model with its mean field subtracted, the response to an imposed Newtonian gravitational potential is ${\tilde{\Phi }}_{\mathbf{k}}^{\mathrm{tot}}={\tilde{\Phi }}_{\mathbf{k}}^{\mathrm{ext}}/{ϵ}_{\mathbf{k}}$, whereThe phase-space distribution function $f_0$ is mass-normalized, and $\mathbf{k}\ne 0$. Causal analytic continuation uses the Landau contour wherever the equilibrium permits it. Zeros describe collective Landau modes. This response function is a gravitational analogue of the electrostatic dielectric function; its susceptibility sign reflects attraction.

Collisionless damping of a collective plasma perturbation by phase mixing and resonant exchange with particles near its phase velocity. In linear electrostatic response, the imaginary part of the dielectric function at real frequency is proportional to the derivative of the equilibrium phase-space distribution function at resonance. A velocity plateau removes this contribution on an interval; compact velocity support also permits undamped waves whose phase velocities lie outside that support.

A longitudinal electron plasma oscillation whose long-wavelength frequency approaches the electron plasma frequency when ions can be treated as stationary. Its dispersion relation and Landau damping depend on the electron phase-space distribution function. Compact velocity support can permit exactly undamped branches outside the support.

A lowered Maxwellian velocity distribution subtracts the value of a Maxwell-Boltzmann velocity distribution at the escape velocity and sets the result to zero beyond escape. This gives a nonnegative phase-space distribution function vanishing continuously at the boundary. If $A=n_0/(2\pi {\sigma }^2{)}^{3/2}$ and $x_e=v_e/(\sqrt{2}\sigma )$, its actual number density is $n=n_0[\mathrm{erf}{x}_e-2x_e{e}^{-x_e^2}/\sqrt{\pi }-4x_e^3{e}^{-x_e^2}/(3\sqrt{\pi })]$. Thus $n_0$ is a normalization parameter, not generally the integrated density. Integrating only speeds below $V<v_e$ gives the same bracket with the first two terms at $X=V/(\sqrt{2}\sigma )$ and the last term $-4X^3{e}^{-x_e^2}/(3\sqrt{\pi })$.

The mass-normalized first velocity moment of a distribution on phase space. For a velocity-only distribution, omit the position integral. Nonzero finite mass and a finite first moment are required. For a probability distribution this is its expected value; its unnormalized counterpart is the momentum.

The result is a fluid Eulerian perturbation generated by translating the entire equilibrium phase-space distribution function through ${\xi }_x$:Here $\Delta$ follows the displaced flux tube. Magnetic flux freezing carries the tube and its particles across the equilibrium pressure gradient. To leading order the tube changes neither its field strength nor its length by enough to alter the particle phase-space distribution function: these changes are smaller by plasma beta or $L_p/R$. Parallel streaming remains within the same displaced tube, on which the original equilibrium phase-space distribution function was uniform. It therefore cannot erase the displacement-generated contrast between adjacent tubes. This is why the term $v_{\parallel }\delta b_x{\partial }_x{f}_{0\alpha }$ in c exactly cancels the apparent parallel relaxation of $-{\xi }_x{\partial }_x{f}_{0\alpha }$. No collisional restoration of a Maxwell-Boltzmann distribution is needed.

The step in the source must be at zero. We use the Heaviside step function $H(x)=1$ for $x>0$, zero for $x<0$, with its value at zero immaterial here. The printed threshold $1$ is inconsistent with both $H^{\prime }=\delta (x)$ and the stated number density. For example, measuring velocities in fixed units and taking $v_F=2$, the literal printed cutoff fills only $v<1$ and gives one eighth of the stated number density. Its distributional derivative is $\delta (x-1)$, not $\delta (x)$. The zero-threshold convention restores the stated Fermi sphere and its normalization of a probability distribution.

Let $k=|\mathbf{k}|>0$, choose $u=\mathbf{v}\cdot \mathbf{k}/k$, and abbreviate $V_{\alpha }=v_{\mathrm{F}\alpha }$. The three-dimensional phase-space distribution function is $f_{0\alpha }=C_{\alpha }{1}_{v<V_{\alpha }}$, whereIntegrate over the transverse velocity disk to obtain the parallel marginal distributionThere are no edge Dirac delta functions in $F_{\alpha }^{\prime }$ because $F_{\alpha }$ itself vanishes continuously at the endpoints. In particular, the three-dimensional constant occupation does not imply a constant parallel marginal distribution.

Use time dependence $e^{pt+i\mathbf{k}\cdot \mathbf{x}}$, with $p=\gamma -i\omega$ and $\gamma <0$ for damping. The given dielectric function becomesThe Landau contour supplies causal continuation. For a real phase velocity $a=\omega /k>V_{\alpha }$, there is no resonant pole in the populated interval, andHence this species contributes

For comparable species number densities and fixed charge ratios, $m_e\ll m_i$ implies $V_i/V_e=O(m_e/m_i)$ and ${\omega }_{\mathrm{p}i}^2/{\omega }_{\mathrm{p}e}^2=O(m_e/m_i)$. Since $\omega >k{V}_e$, the ions have $s_i\gg 1$ and ${\chi }_i=-{\omega }_{\mathrm{p}i}^2/{\omega }^2+O(k^2{V}_i^2{\omega }_{\mathrm{p}i}^2/{\omega }^4)$. Their response is negligible to leading order in the mass ratio. The real dispersion relation is thereforeThis degenerate-plasma dielectric response treats a completely degenerate, nonrelativistic Fermi gas with classical particle trajectories. It does not include quantum recoil or a collision term. A finite ion response shifts the root slightly; the boxed equation is the stated leading electron approximation.

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.

To obtain the Boltzmann equation, take a dilute gas with a short-range pair potential. Its interaction range is small compared with the mean free path and the spatial scale on which the phase-space distribution function changes. The binary collision approximation treats encounters as isolated elastic collisions; simultaneous encounters involving three or more particles can be neglected. The encounter duration is short compared with the collision relaxation time and macroscopic evolution time, and the external force changes negligibly during an encounter. This permits local, effectively instantaneous scattering at one position $\mathbf{x}$.

The essential statistical closure is molecular chaos for the Boltzmann equation: the distribution of incoming pairs factorizes as $f{f}_{\star }$. Outgoing pairs need not be independent. Follow each incoming pair through its two-body scattering, and use conservation of momentum and kinetic energy, together with time reversal and the invariant scattering measure, to express both gain and loss in the same variables. This givesHere $d\Omega =\sin \theta d\theta d\phi$; when the measure is written as $d\theta d\phi$, its Jacobian is included in the Boltzmann scattering kernel. In a scattering-angle convention, that kernel contains relative speed times differential scattering cross-section. The hemisphere prevents double counting equivalent outcomes for indistinguishable particles; the precise factor is included in $B$. The primed velocities are connected to the unprimed ones by the reversible elastic collision map. This closure is appropriate for a dilute short-range gas, rather than a generic long-range interacting system with an important collective mean-field theory contribution.

The further replacement by the BGK model is a phenomenological relaxation closure. Replace the full nonlinear Boltzmann collision operator by relaxation towards a local Maxwellian, with a single, velocity-independent collision relaxation time $\tau >0$:It models relaxation towards a local Maxwellian without resolving the detailed scattering. Treating $\tau$ as constant is an additional simplification; it is of the order of a mean time between effective collisions. Near a local Maxwellian, departures in the nonconserved velocity moments decay on this timescale.

Choose $f^{(0)}$ to match the collision invariants $1$, $\mathbf{v}$ and $|\mathbf{v}{|}^2$ of $f$. For unit particle mass, this meansHere $\rho$ is the mass density, $\mathbf{u}$ is the mean velocity, and $\Theta$ is the temperature in squared-velocity units, $\Theta =k_{B}T/m$ before setting $m=1$. In particular,Thus relaxation preserves local mass, momentum and kinetic energy rather than relaxing them to externally prescribed values.

For the two outer branches, $|u|\gg v_0$ at small $k$. An integration by parts and a large-$u$ expansion giveThere is no boundary contribution because the phase-space distribution function vanishes at both endpoints. Direct area integration givesThus the exact small-$k$ cutoff, retaining the paper's definition $n_0=v_0{A}_0$, is $\omega \to \pm {\omega }_{\mathrm{pe}}\sqrt{n_{\ast }/n_0}$. To leading order in the narrow-velocity plateau approximation this isThese are Langmuir waves. The difference between $n_{\ast }$ and the reference $n_0$ is only first order in $\delta /v_0$, so it does not alter the requested leading-order result.

For the finite-phase velocity branch put $u=v_1+\delta a$, with $0<a<1$. To leading order in $\delta /v_0$ and $\delta /v_1$, the real dispersion relation becomesAt long wavelength set $K=0$ to leading order. Then $a=v_1^2/v_0^2$, andFor comparison, the exact $k\to 0$ phase velocity from the cubic equation in b is $B/A$; expanding it gives precisely $v_1+\delta v_1^2/v_0^2+O({\delta }^2/v_0)$ at fixed $v_1/v_0$.

A single Maxwell-Boltzmann velocity distribution has a nonzero slope at every finite nonzero $u$ and has no velocity plateau. The would-be electron acoustic wave is therefore subject to strong Landau damping, so there is no corresponding completely undamped acoustic branch in a single Maxwellian electron population with immobile ions. At its only stationary point $u=0$, a Maxwellian has $\int f_0^{\prime }(v)/vdv<0$, making the static dielectric function positive rather than zero. This statement does not exclude acoustic modes of suitably distinct electron populations or an ion-supported mode when ion dynamics is retained.

Use the usual spherical circular-shell adiabatic contraction approximation: initial dark-matter orbits are circular, slow accumulation keeps them circular, and dark shells do not cross. Angular momentum is conserved in a spherical potential, and the vanishing radial action stays zero as an adiabatic invariant. For a circular shell,Here $r_i$ is the initial shell radius and $r$ its final radius. The final flat circular speed $V=2v_{\mathrm{NFW},\mathrm{c},\max }$ gives $M_f(r)=V^{2}r/G$.

For an initial shell in the inner NFW profile, write $A=2\pi {\rho }_1{r}_1$, so $M_i(r_i)\simeq A{r}_i^2$. The invariant givesConservation of dark shell mass yieldsand differentiation givesTo express the requested ratios, set $K=F(x_{\max })\simeq 0.2162166$ and $B=8K\simeq 1.72973$, so $V^2/(GA)=B{r}_1$. Since the original cusp has ${\rho }_D(r)\simeq A/(2\pi r)$,The total density of the final spherical flat-rotation model is ${\rho }_{\mathrm{tot}}=V^2/(4\pi G{r}^2)$. Its dark fraction isSubtracting dark matter from the total gives the added-matter density, ${\rho }_B={\rho }_{\mathrm{tot}}(1-h)$, and thusThese expressions describe the leading inner approximation, not an extension to radii where the denominator becomes negative or the initial shell is outside the NFW cusp. As $r/r_1\to 0$, dark matter is enhanced relative to its original density while its fraction of the even steeper total density tends to zero. This is inner NFW contraction to a flat rotation curve.

The orbital assumptions matter. An initial density profile does not specify the phase-space distribution function. For eccentric orbits, slow evolution conserves their actions, but $rM(r)$ at an instantaneous radius is not generally their invariant. The above ratios are the intended circular-shell estimate; their numerical coefficients cannot be uniquely inferred from the density profile and slow accumulation alone.

Integrate the reduced vertical phase-space distribution function over $v_z$. The Gaussian velocity integral cancels its normalization, givingHere the constant vertical velocity dispersion is ${\sigma }_z$, and $\Phi (0)=0$ fixes the potential reference. Neglecting the radial derivatives in the Poisson equation for Newtonian gravity leavesWith $\varphi =\Phi /{\sigma }_z^2$, $\zeta =z/z_0$ and $z_0^2={\sigma }_z^2/(8\pi G{\rho }_0)$, this becomesThus the stated dimensionless equation follows with its factor of two.

For a direct solution, multiply by $2{\varphi }^{\prime }$ and integrate from the midplane. The initial conditions giveFor $\zeta >0$ take the increasing branch and put $u=e^{\varphi /2}$. Then $4(u^{\prime }{)}^2=u^2-1$, whose smoothly increasing solution consistent with the second-order equation is $u=\cosh (\zeta /2)$. Reflection about the midplane supplies the negative-$\zeta$ branch. ThereforeSubstitution verifies $2{\varphi }^{\prime \prime }={\mathrm{sech}}^2(\zeta /2)=e^{-\varphi }$ and both boundary conditions. The smooth second-order equation has a unique solution for those initial data, so no extra constant branch from the first-integral square root is admissible.

This is the self-gravitating isothermal slab. Its scale convention matters: if its profile is instead written $\rho ={\rho }_0{\mathrm{sech}}^2(z/H)$, then $H=2z_0={\sigma }_z/\sqrt{2\pi G{\rho }_0}$. Its surface density is $\Sigma =4{\rho }_0{z}_0$. The force approaches $|{\Phi }^{\prime }|={\sigma }_z^2/z_0=2\pi G\Sigma$, consistent with the field of a plane sheet. These checks fix both the scale-height and Poisson factors.

Use physical momentum $p_{\mathrm{phys}}$ and conserved-background comoving momentum $q=a{p}_{\mathrm{phys}}$. At fixed $q$, the homogeneous collisionless phase-space distribution function is $f_0(q)$, with no spatial or angular dependence and no explicit background time derivative. The printed relation $p_i=aq{n}_i$ cannot have this meaning: freely propagating physical momentum redshifts as $a^{-1}$, so it should read $p_i=q{n}_i/a$. Keeping the printed relation literally would give $q=p_{\mathrm{phys}}/a$ and a nonzero zeroth-order redshift, inconsistent with the target equation. The standard definition is positively confirmed by the phase-space convention in Ma and Bertschinger's perturbation treatment.

With this conserved $q$, the orders of the factors are:

Thus the linear Collisionless Boltzmann equation isFor photons $v_0^i=n^i/a$. For a particle of mass $m$, $v_0^i=q{n}^i/[a\sqrt{q^2+a^2{m}^2}]$. The order counting does not assume that the first-order angular deflection itself vanishes; only its multiplication by an already perturbed angular distribution makes it higher order.

Use $\mu =\gamma m$ for the gravitational parameter, to distinguish it from the Delaunay action called $G$. The displayed actions are per unit reduced mass. Accordingly use specific canonical momenta; with physical relative momenta, all three actions acquire a common reduced-mass factor, which leaves the eccentricity fraction below unchanged.

The canonical transformation. In spherical coordinates, separation of the Kepler Hamilton-Jacobi equation givesA local type-two generating function, continued across turning points with the appropriate momentum branches, isThe old momenta are $\partial W/\partial (r,\theta ,\varphi )$ and the new angles are $l={\partial }_{L}W$, $g={\partial }_{G}W$, $h={\partial }_{H}W$. This defines a canonical transformation, since its generating differential is $dW=p_{r}dr+p_{\theta }d\theta +p_{\varphi }d\varphi +ldL+gdG+hdH$. With standard choices of angle origins, the resulting Delaunay variables arewith $l$ the mean anomaly, $g$ the argument of periapsis, and $h$ the longitude of ascending node. In particular, the signed $H$ includes retrograde orbits; the squared equation in the PDF must not be read as $H\ge 0$.

For bound noncollision Kepler orbits, the independent ranges areThe limiting circular, radial and coplanar cases sit on boundaries where some angle charts degenerate. They have zero measure in the generic phase-volume calculation. The ranges are constrained as above, rather than a Cartesian box with three unconstrained actions.

The eccentricity count. A canonical transformation preserves phase space volume, soIntegrating the three angles contributes $(2\pi {)}^3$, and integrating $H$ gives $2G$. At fixed $L$, the condition that the orbital eccentricity be less than $e_0$ is $G>L\sqrt{1-e_0^2}$. ThusAny energy weight cancels between these integrals. If the total population is finite, or an explicit finite action band is used,This is the Delaunay phase-volume proof of the thermal eccentricity distribution. The source's $e^2$ law is a cumulative count, not a differential eccentricity density. Also, pointwise finite $F(E)$ does not alone imply finite $N_{\mathrm{tot}}$: $F=1$ gives ${\int }_0^{\infty }{L}^{2}dL=\infty$. The finite normalization of an energy-only Kepler distribution is the additional condition needed for a global probability statement; the fraction at every fixed energy is unaffected.

Why the observation does not imply relaxation or equilibrium. No Boltzmann distribution was used: every normalizable energy-only phase-space distribution function has the same eccentricity marginal, irrespective of collisional thermalization. An even stronger counterexample is a genuinely time-dependent collisionless distributionKepler motion has $\dot{l}=\nu (L)$ and constant actions and other angles, so ${\partial }_{t}f+\nu {\partial }_{l}f=0$: this is an exact nonstationary solution of the Collisionless Boltzmann equation. Integrating $l$ removes the cosine and leaves precisely the same $e^2$ cumulative count at every time. An eccentricity marginal alone cannot establish thermal relaxation, a relaxed age, or even a stationary distribution.

Write ${\mu }_g=\gamma m$ for the gravitational parameter of the relative Kepler orbit, and use specific orbital energy and momenta per unit reduced mass. The Hamiltonian isIf physical momenta rather than these specific momenta are used, multiply the actions by the reduced mass. This only introduces a constant factor into the phase space measure and leaves the eccentricity fractions unchanged. The formulas in the question adopt the specific-action convention.

Here is an explicit local canonical transformation. A separated complete integral of the Hamilton-Jacobi equation isChoose branches along an orbit and take $p_j=\partial S/\partial q_j$, $l=\partial S/\partial L$, $g=\partial S/\partial G$, $h=\partial S/\partial H$. Thenwhose exterior derivative proves preservation of the canonical symplectic form. Thus these variables are canonical, and the absolute Jacobian determinant is one. The angular separation gives $G=|\mathbf{r}\times \mathbf{v}|$ and $H=G\cos i$, including the sign of the angular momentum along the reference axis.

For clarity, the radial action can be evaluated rather than assumed. Let $r=a(1-e\cos u)$, with $u$ the eccentric anomaly. Then $p_r=\sqrt{{\mu }_g/a}e\sin u/(1-e\cos u)$, andIn the integral, use $e^2{\sin }^{2}u/(1-e\cos u)=1+e\cos u-(1-e^2)/(1-e\cos u)$ and ${\int }_0^{2\pi }(1-e\cos u{)}^{-1}du=2\pi /\sqrt{1-e^2}$. The polar action is $J_{\theta }=G-|H|$, while the signed azimuthal action is $J_{\varphi }=H$. The separated angular coordinates have the usual Delaunay variables interpretation: $g$ is the argument of periapsis, and $h$ is the longitude of ascending node. In particular,with a suitable origin at periapsis, so $l$ is the mean anomaly. We have therefore recoverednot merely assigned names to the new coordinates.

For nondegenerate bound Kepler orbits, the ranges areThe boundaries describe circular, coplanar or radial degeneracies, where some angles cease to be defined; they have zero measure in the present calculation. The actions are canonical coordinates with nested bounds, rather than three independently unrestricted Cartesian coordinates. The square of $H$ alone would miss the distinction between prograde and retrograde inclinations.

By the canonical transformation, the population measure becomesAt fixed $L$, the condition $e^{\prime }<e$ is $L\sqrt{1-e^2}<G<L$. Integrating all three angles and then $H$ givesConsequently the Delaunay phase-volume proof of the thermal eccentricity distribution yieldsThe PDF's $e^2$ dependence must be interpreted as a cumulative distribution function, not an eccentricity probability density function or the value of the original six-dimensional phase-space distribution function. A finite value of $F(E)$ at every energy also does not suffice to make the total population finite: $F=1$ on all bound energies gives the divergent integral ${\int }_0^{\infty }{L}^{2}dL$. The fractions are well defined at each fixed energy, over a finite allowed energy range, or for an energy distribution with a finite total integral.

A Boltzmann distribution is therefore unnecessary. Any normalizable energy-only distribution has the same eccentricity marginal, although such a distribution is a collisionless steady state by Jeans theorem. The marginal alone does not even prove stationarity: for $0<|\eta |<1$,is a positive time-dependent solution of the Collisionless Boltzmann equation and has exactly the same eccentricity marginal after integration over $l$. An observed $e^2$ cumulative law establishes neither thermal relaxation nor the age of the system or Universe.