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Linearize the Collisionless Boltzmann equation and the perturbation Poisson equation about a homogeneous velocity distribution $f_0$. A plane wave gives $f_1={\varphi }_1\mathbf{k}\cdot {\nabla }_v{f}_0/(\mathbf{k}\cdot \mathbf{v}-\omega )$. Integrating this density and using $-k^2{\varphi }_1=4\pi G\int f_1{d}^{3}v$ yields $1+(4\pi G/k^2)\int [\mathbf{k}\cdot {\nabla }_v{f}_0/(\mathbf{k}\cdot \mathbf{v}-\omega )]d^{3}v=0$. Growing modes have $\mathrm{Im}\omega >0$ and no real-velocity pole; real and damped frequencies require the causal continuation prescription. The Newtonian homogeneous background requires the Jeans swindle.

A collisionless stellar system is one for which discrete gravitational encounters change its distribution negligibly over the time under study, usually $T\ll t_{\mathrm{rel}}$. Stars can orbit and respond collectively to a varying smooth gravitational potential even in this regime; collisionless does not mean force-free or stable against collective gravitational instabilities. Its smooth distribution follows the Collisionless Boltzmann equation.

The Jeans equations are velocity moments of the Collisionless Boltzmann equation. They relate density, mean motion, velocity dispersion, and gravitational acceleration without requiring a closed fluid equation of state.

Jeans theorem states that a steady Collisionless Boltzmann equation solution depends on phase-space coordinates only through integrals of motion, and that a nonnegative function of isolating integrals gives a steady collisionless distribution function on its domain.

Linearizing the Collisionless Boltzmann equation around $f_0(\mathbf{J})$ in an integrable background Hamiltonian $H_0(\mathbf{J})$ gives $({\partial }_t+i\mathbf{k}\cdot \mathbf{\Omega })\delta f_{\mathbf{k}}=i\mathbf{k}\cdot {\partial }_{\mathbf{J}}{f}_0{\Phi }_{\mathbf{k}}$, with $\mathbf{\Omega }={\partial }_{\mathbf{J}}{H}_0$. An integrating factor gives the displayed zero-initial-perturbation solution. A nonzero initial perturbation adds the freely evolving term $e^{-i\mathbf{k}\cdot \mathbf{\Omega }t}\delta f_{\mathbf{k}}(0)$.

Let $n(\mathbf{x})=\int f{d}^{3}v$ and $\overline{A}=n^{-1}\int Af{d}^{3}v$. Assume the one-particle distribution function and its weighted tails decay enough that all required velocity integrals converge and boundary terms at infinite velocity vanish. Multiply the steady Collisionless Boltzmann equation by $v_j$ and integrate over velocity. In Cartesian coordinates, spatial differentiation commutes with this integration, while integration by parts givesConsequently the three force-balance equations areExplicitly the $x$ component is ${\partial }_x(n\overline{v_x^2})+{\partial }_y(n\overline{v_x{v}_y})+{\partial }_z(n\overline{v_x{v}_z})=-n{\Phi }_x$, with cyclic counterparts for $y,z$. For equal stellar masses, multiplying by the common mass replaces $n$ by the mass density $\rho$; for multiple stellar masses, use a mass-weighted one-particle distribution function or derive a separate equation for each tracer population.

Define mean streaming velocity $u_i=\overline{v_i}$ and the velocity dispersion tensor ${\sigma }_{ij}^2=\overline{(v_i-u_i)(v_j-u_j)}=\overline{v_i{v}_j}-u_i{u}_j$. The zeroth velocity moment of the steady Collisionless Boltzmann equation gives the continuity equation ${\sum }_i{\partial }_i(n{u}_i)=0$. ThereforeSubstituting in a yields the standard steady Jeans equations:The unsteady version adds ${\partial }_t{u}_j$ on the left. Defining the stellar pressure tensor $P_{ij}=\rho {\sigma }_{ij}^2$ makes the analogy with fluid momentum balance explicit. The off-diagonal terms are correlations between random velocity components; they need not vanish or equal the diagonal terms.

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.

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.

Use the Collisionless Boltzmann equation in action-angle variables,Define the orbital frequency vector byThe slow evolution of the angle average $f_0$ starts at second order in the perturbation. At first order, the Poisson bracket therefore givesFor the $\mathbf{k}$th term in the Fourier series, this readsThe lowercase $\delta {\varphi }_{\mathbf{k}}$ here denotes the Fourier coefficient of the uppercase $\delta \Phi$ in the Hamiltonian; these are the same perturbing potential. Multiply by the integrating factor $e^{i\mathbf{k}\cdot \mathbf{\Omega }t}$ and set the initial homogeneous response to zero. One obtainsRetaining the slow time dependence of $f_0$ inside this integral is compatible with the subsequent second-order calculation; changes in $\mathbf{\Omega }$ caused by that slow evolution affect only higher orders in this approximation.

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.

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.

Let $p^0=d\tau /d\lambda$. A comoving observer measures photon energy $p=a{p}^0$, so $q=ap=a^2{p}^0$. The null condition is $({\delta }_{ij}+h_{ij})p^i{p}^j=(p^0{)}^2$. Using the stated Christoffel symbols in the time geodesic equation givesDivide by $p^0$ to differentiate with respect to conformal time. The background redshift term cancels when differentiating $a^2{p}^0$:In the perturbation term use the background relation $p^i/p^0={\hat{n}}^i$, givingWithout metric tensor perturbations, both $q$ and the propagation direction are constant along an unperturbed ray. Thus $q^{\prime }$ and $d{\hat{n}}^i/d\tau$ are first order. Multiplying them by derivatives of $f_1$, which is also first order, gives second-order terms. The term $q^{\prime }d{f}_0/dq$ is first order and must be retained. At linear order the Collisionless Boltzmann equation therefore readsTaking the Fourier transform in the spatial coordinates provesA first-order correction to the propagation velocity multiplying a spatial derivative of $f_1$ is likewise second order.

First velocity moment. Assume a stationary, spherical, nonstreaming collisionless stellar system with constant mass-to-light ratio ${\rm Y}$. Use a luminosity-weighted galactic distribution function, with luminosity density $\lambda$. Multiplying the Collisionless Boltzmann equation by $v_i$ and integrating over velocities, with its velocity-space boundary terms vanishing, givesSpherical symmetry makes the stress tensorHere ${\sigma }_t^2$ is one tangential component, as in the paper; the total tangential variance is $2{\sigma }_t^2$. Its radial divergence is $q^{\prime }+2(q-q_t)/r$, since the divergence of the radial unit vector is $2/r$. With the velocity-anisotropy parameter $\beta =1-{\sigma }_t^2/{\sigma }_r^2$, and enclosed mass $M(r)={\rm Y}L(r)$, this gives the Spherical Jeans equationThis uses a constant ratio of total gravitating mass to luminosity, rather than an arbitrary independent dark halo.

Projection and elimination of anisotropy. Write $R$ for projected radius and $r$ for spherical radius. The Abel transform of the luminosity density giveswhose inverse is the displayed density inversion in the paper. At a point on this line of sight, ${\sin }^2\theta =R^2/r^2$. The line-of-sight second stellar velocity moment is ${\sigma }_r^2{\cos }^2\theta +{\sigma }_t^2{\sin }^2\theta ={\sigma }_r^2(1-\beta R^2/r^2)$. Therefore the measured quantity $J(R)=\mu (R){\sigma }_p^2(R)$ satisfiesThe Spherical Jeans equation gives $\beta q=-r{q}^{\prime }/2-\gamma \lambda L/(2r)$. Substitution yieldsThe inner $\gamma L$ printed in the PDF's definition of $p$ must therefore be $\lambda L$: the literal expression has an extra gravitational factor and lacks the density. Both this derivation and dimensional consistency require the correction. Define the projected spherical Jeans residual byTo express it entirely in terms of the radial stress, putDifferentiating the nonsingular $z$ representation gives $A^{\prime }(R)=R{\int }_R^{\infty }{q}^{\prime }(r)/\sqrt{r^2-R^2}dr$. HenceIf $R^{2}A$ vanishes at zero and infinity, then ${\int }_0^{\infty }6\pi Rp(R)dR=6\pi [R^{2}A{]}_0^{\infty }=0$. These are the absence-of-boundary-stress conditions needed for the global result.

The virial relation, term by term. Interchange the order in the residual integral and useIt follows thatThe first term is $2K/{\rm Y}$: spherical averaging assigns one third of the integrated velocity-square to any fixed viewing direction, giving $K=3\pi {\rm Y}\int RJdR$. The gravitational potential energy isThus the integrated residual is exactly $(2K+W)/{\rm Y}$, and its vanishing is the virial theorem. Solving for $\gamma$ gives the global projected virial estimate of a mass-to-light ratio,It is independent of the radial anisotropy profile when the full aperture and the stated boundary conditions are used. A finite aperture retains a boundary term.

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.

Express the Collisionless Boltzmann equation in variables $(\mathbf{x},q,\mathbf{n},\tau )$ and divide the derivative along the photon ray by $d\tau /d\lambda$. Since $f_0(q)$ has no explicit position, direction or conformal-time dependence,This is the chain rule in momentum magnitude and direction. At linear order, $P^i=n^i+O(h)$, while the last two terms are products of first-order perturbations and can be dropped. The synchronous photon momentum redshift from part i then gives, for spatial Fourier convention $e^{i\mathbf{k}\cdot \mathbf{x}}$,The factor $q$ is essential. It is missing in the PDF's displayed intermediate equation, but follows directly from part i and is needed to obtain its final brightness equation.

Let $I_0={\int }_0^{\infty }{q}^3{f}_0(q)dq$. Absorbing the common phase-space normalization into $f$, the homogeneous photon energy density is ${\rho }_{\gamma }=4\pi I_0/a^4$, so the photon brightness perturbation isA small directional temperature change $\Theta =\Delta T/T$ in a Planck distribution gives $f_1=-q{f}_0^{\prime }(q)\Theta$. Integration by parts yieldsbecause the Planck distribution behaves as $q^{-1}$ at low momentum and decays exponentially at high momentum. Thus $\Delta =4\Theta$. Equivalently, frequency-integrated blackbody radiation energy scales as $T^4$. Identifying brightness with a single temperature perturbation assumes this blackbody spectral form; the brightness integral itself is defined more generally.

Finally, multiply the corrected transport equation by $q^3$, integrate and divide by $I_0$. The spatial streaming factor is independent of $q$, while its gravitational source integrates to $(1/2)(-4)h_{ij}^{\prime }{n}^i{n}^j$. ThereforeThis derives the synchronous photon brightness equation with the normalization and sign fixed, rather than inferring it from the dimensionally inconsistent intermediate formula.

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.

Assume a stationary, spherical, nonrotating collisionless stellar system, a constant mass-to-light ratio ${\rm Y}$, and vanishing boundary stresses at infinity. Let $\lambda$ be the luminosity density, so $\rho ={\rm Y}\lambda$, and set $\gamma =G{\rm Y}$. Distinguish spatial radius $r$ from projected radius $R$. The enclosed luminosity and radial force areHere ${\sigma }_t^2$ is the velocity dispersion of either one of the two tangential components; thus $\beta =1-{\sigma }_t^2/{\sigma }_r^2$. If a combined tangential variance is used instead, its contribution must be divided by two in this definition.

To derive the Jeans equation, start with the steady Collisionless Boltzmann equation $v_j{\partial }_{j}f-({\partial }_j\Phi ){\partial }_{v_j}f=0$. Multiply by $v_i$ and integrate over all velocities. Integration by parts in velocity, with $f$ decaying fast enough to eliminate boundary terms, givesThe same equation holds with luminosity weighting because the tracer's luminosity is conserved along its orbit. Spherical symmetry makes the local stellar velocity moment tensor diagonal, with entries $q=\lambda {\sigma }_r^2$, $q_t=\lambda {\sigma }_t^2$, $q_t$. Its radial divergence is $q^{\prime }+(2q-2q_t)/r$: the second term accounts for the changing directions of the radial and tangential unit vectors. ThereforeThis is the Spherical Jeans equation; it follows from a stellar velocity moment rather than requiring a particular form of the galactic distribution function.