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Assume a time-independent spherical relative potential $\psi (r)$, with acceleration $\nabla \psi$. The relative energy and specific angular momentum areAlong an orbit, $\dot{E}=\nabla \psi \cdot \mathbf{v}-\mathbf{v}\cdot \nabla \psi =0$ and $\dot{\mathbf{\ell }}=\mathbf{r}\times \nabla \psi =0$, since a spherical gradient is radial. Thus both $E$ and $L$ are integrals of motion. Choose the escape level $\psi =0$ and take the bound galactic distribution function to vanish for $E\le 0$.

For an isotropic $f(E)$, integrate over a spherical coordinate system in velocity space:Differentiating gives ${\rho }^{\prime }(\psi )=2\pi \sqrt{2}{\int }_0^{\psi }f(E)/\sqrt{\psi -E}dE$. Convolve once more with the reciprocal-square-root kernel. Reversing the order of integration, under the usual integrability assumptions, usesTherefore ${\int }_0^{E_0}{\rho }^{\prime }(\psi )/\sqrt{E_0-\psi }d\psi =2\sqrt{2}{\pi }^2{\int }_0^{E_0}f(E)dE$. Differentiating proves Eddington inversion:Here ${\pi }^2$ is outside the square root, as in the original PDF; the TeX transcription wrongly places it inside. The assumptions include a locally integrable bound distribution and sufficient density regularity, with no extra unbound or boundary population. An inverted expression must additionally be nonnegative to be physical. Since the velocity measure and $f$ are invariant under all velocity rotations, $⟨v_r^2⟩=⟨v_{\theta }^2⟩=⟨v_{\varphi }^2⟩=⟨v^2⟩/3$ whenever these moments exist; mixed moments vanish.

For the half-anisotropic distribution function, write the polar velocity angle from the radial direction as $\eta$. Then $L=rv\sin \eta$, andThe apparent $1/L$ singularity is integrable because it cancels the factor $v\sin \eta$ in the velocity measure. Integrating $\eta \in [0,\pi ]$ and $\phi \in [0,2\pi ]$ givesThe density relation requires $r\rho \to 0$ at the escape boundary and $f_E\ge 0$ for a physical distribution. Consider a monotonically decreasing $\psi$. The largest allowed radius at fixed binding energy is the radial-orbit limit $r_E$ with $\psi (r_E)=E$; a nonzero-$L$ orbit normally has a smaller apocenter because of its centrifugal term. Changing from $\psi$ to $r$ givesThe numerator in the original PDF instead uses $\rho +d\rho /d\psi$. That is a genuine printed error: the chain rule for $d(r\rho )/d\psi$ requires $\rho +rd\rho /dr$. It is not legitimate to prove the printed expression as written. A concrete counterexample is the unit-parameter Hernquist model at $r=1$: ${\psi }^{\prime }=-1/4$, ${\rho }^{\prime }=-(5/2)\rho$ and $d\rho /d\psi =10\rho$. The printed quotient is $-44\rho$, whereas the correct $g$ is $6\rho >0$. The former would give a negative distribution.

The angular weight after cancellation is uniform in $\eta$, so the averages of ${\cos }^2\eta$ and ${\sin }^2\eta$ are both $1/2$. The azimuthal angle splits the tangential term equally. Hence $⟨v_r^2⟩=⟨v_{\theta }^2+v_{\varphi }^2⟩$ and each individual tangential moment is half the radial one. The velocity-anisotropy parameter is consequentlyThis is a radially biased constant-anisotropy distribution function, not an isotropic one or purely radial motion. It does not by itself prove dynamical stability; it preferentially weights small angular momentum while remaining integrable.

For the Hernquist model, use the Poisson equation for Newtonian gravity with the relative-potential sign, ${\nabla }^2\psi =-4\pi G\rho$. Since ${\psi }^{\prime }=-GM/(r+a{)}^2$,Its enclosed mass is $M{r}^2/(r+a{)}^2$, which tends to $M$. Expressing the augmented density in the relative potential gives $r\rho =a{\psi }^3/(2\pi G^3{M}^2)$. The preceding derivative therefore giveswith zero bound-model population outside the allowed energy domain. It is nonnegative and reproduces the density by direct integration. The Hernquist model has a central density cusp, so the formulas for the local $1/L$ distribution are interpreted at $r>0$. As a further check, the radial second moment obtained from the same integral is $\psi /4$ and each tangential second moment is $\psi /8$, consistent with $\beta =1/2$.

Use the same specified homogeneous Newtonian background and Jeans swindle, but describe stars by a mass-normalized galactic distribution function. The Collisionless Boltzmann equation is ${\partial }_{t}f+\mathbf{v}\cdot {\nabla }_{x}f-\nabla \varphi \cdot {\nabla }_{v}f=0$. Let $f=f_0(\mathbf{v})+f_1{e}^{i(\mathbf{k}\cdot \mathbf{x}-\omega t)}$ and similarly perturb the gravitational potential. To first order,Eliminating $f_1$ and the nonzero gravitational potential amplitude gives the collisionless Jeans dispersion relationFor growing modes $\mathrm{Im}\omega >0$ there is no real-velocity pole. Real and damped frequencies require a causal contour prescription or analytic continuation; one must not simply integrate through a pole without specifying it.

Take $b>0$ in the Cauchy velocity distribution. Its normalization is $\int f_0{d}^{3}v={\rho }_0$. Orient $\mathbf{k}$ along the $u=v_z$ axis and integrate over the perpendicular velocities:At the marginal mode $\omega =0$, $F^{\prime }(u)/u=-2{\rho }_{0}b/[\pi (u^2+b^2{)}^2]$, with its finite limiting value at zero. The provided Gamma function integral at power two givesConsequentlyTo check that this really separates growing modes, put $c=\omega /k$ in the upper half-plane. Residue integration, or integration by parts followed by the Cauchy resolvent, gives $\int F^{\prime }(u)/(u-c)du={\rho }_0/(c+ib{)}^2$. The upper-half-plane solution has $\omega =i(\sqrt{4\pi G{\rho }_0}-kb)$, which grows exactly for $0<k<k_J$. At the threshold it approaches the marginal mode continuously. Although $b$ plays the threshold role of a velocity scale, this distribution's second moment diverges: the radial integrand for $\int v^2{f}_0{d}^{3}v$ tends to a nonzero constant at large speed. It is therefore incorrect to identify $b^2$ with a finite velocity dispersion or to infer this result by substituting an rms speed into a fluid formula.

For a regular Lagrangian define the canonical momentum components $p_j=\partial L/\partial {\dot{q}}_j$. Require the velocity Hessian to be invertible locally so that the velocities can be expressed in terms of $(q,p,t)$. The Legendre transform in mechanics definesThe explicit time argument cannot generally be omitted when the given Lagrangian depends on time. Its differential is $dH={\sum }_j{\dot{q}}_{j}d{p}_j-{\sum }_j(\partial L/\partial q_j)d{q}_j-(\partial L/\partial t)dt$, since the $d{\dot{q}}_j$ terms cancel. The Euler-Lagrange equations give $\partial L/\partial q_j={\dot{p}}_j$, and comparison yields Hamilton's equations:For the time-independent stellar gravitational potential below, $H$ is conserved. Its sign convention differs from the binding energy in question 2: here $E=H=v^2/2-\psi$ is the ordinary specific orbital energy.

The prescribed nondegenerate prolate spheroidal coordinates require $0<\alpha <\beta$; equality collapses the coordinate family. Their physical velocity components are $v_j=h_j{\dot{q}}_j$, and $h_{\varphi }=R$. Per unit stellar mass the Lagrangian is $L=\frac{1}{2}{\sum }_j{h}_j^2{\dot{q}}_j^2+\psi$, so $p_j=h_j^2{\dot{q}}_j=h_j{v}_j$. Its Hamiltonian isThis uses acceleration $+\nabla \psi$, as specified, rather than importing the opposite gravitational potential convention.

For the Stäckel potential, use the Hamilton-Jacobi equation with principal function $S=W_{\lambda }(\lambda )+W_{\mu }(\mu )+l\varphi -Et$. Axisymmetry makes $p_{\varphi }=l=R{v}_{\varphi }$ constant. Write $d=\lambda -\mu$ and $A(\tau )=(\tau +\alpha )(\tau +\beta )$. Multiplying $H=E$ by $d$ and usingseparates it as $F(\lambda )-E\lambda =F(\mu )-E\mu$, where $F(\tau )=2A(\tau )(W_{\tau }^{\prime }{)}^2-(\alpha -\beta )l^2/[2(\tau +\alpha )]-G(\tau )$. Each side depends on its own coordinate only and therefore equals a constant, chosen as $-I$. Here $G(\tau )$ is the generating function of the gravitational potential, not the gravitational constant. HenceThis proves the conserved axisymmetric Stäckel third integral. Eliminating $E$ between the two expressions and using $p_j=h_j{v}_j$ givesAt regular generic phase-space points the three separation constants $E,l,I$ are independent; special circular or degenerate orbits can reduce their local independence. Thus the model is an integrable Hamiltonian system.

For the intended regular, phase-mixed steady model, Jeans theorem permits a galactic distribution function $f=F(E,l,I)$. At any fixed point, $E$ and $I$ depend on $v_{\lambda }$ and $v_{\mu }$ only through their squares, and $l$ depends only on $v_{\varphi }$. Thus $f$ is even separately in $v_{\lambda }$ and $v_{\mu }$, whether or not it is even in $v_{\varphi }$. Odd integration proves velocity-ellipsoid alignment in a Stäckel potential:The argument allows azimuthal streaming: it does not require zero mean $v_{\varphi }$. The means in the two meridional directions vanish, so the off-diagonal velocity covariance also vanishes. Second moments must exist for this conclusion.

There is a hypothesis behind the printed universal wording. On a generic nonresonant regular invariant torus, a continuous steady distribution is constant along a dense orbit and hence depends only on the torus integrals; this justifies the phase-mixed form. Bare stationarity alone is not sufficient in a degenerate resonant gravitational potential with additional integrals. For example $H=\frac{1}{2}(v^2+{\omega }^2{r}^2)$ is a Stäckel Hamiltonian: choose $G(\tau )=-\frac{1}{2}{\omega }^2[{\tau }^2+(\alpha +\beta )\tau ]$, up to a constant vacuum shift. The additional conserved quantity $K_{xy}=v_x{v}_y+{\omega }^{2}xy$ gives a positive normalizable steady tracer $f\propto \exp (-aH+ϵK_{xy})$ for $a>|ϵ|>0$. Its local $⟨v_x{v}_y⟩=ϵ/(a^2-{ϵ}^2)\ne 0$. At azimuth zero and a regular point where the $\lambda$ direction has a radial component, this gives nonzero $⟨v_{\lambda }{v}_{\varphi }⟩$. This tracer need not itself source the harmonic gravitational potential. It exhibits why the stronger, ordinary phase-mixed three-integral assumption is necessary instead of claiming that every imaginable steady distribution has the asserted alignment.

The geometry follows by multiplying the coordinate-defining equation and evaluating at $\tau =-\alpha ,-\beta$:Constant $\lambda$ surfaces are prolate spheroids, with equatorial semiaxis $\sqrt{\lambda +\alpha }$ and polar semiaxis $\sqrt{\lambda +\beta }$. Constant interior $\mu$ surfaces are two-sheeted hyperboloids. Their common foci lie at $z=\pm \sqrt{\beta -\alpha }$. The orbit volume in an axisymmetric Stäckel potential is determined byIts allowed coordinate intervals have turning boundaries or coordinate endpoints. A typical bound orbit with $l\ne 0$ oscillates between inner and outer spheroids, oscillates vertically between hyperboloidal limits, and circulates about the symmetry axis with $\dot{\varphi }=l/R^2$. The equatorial fold $\mu =-\beta$ joins the positive and negative $z$ branches; it is not necessarily a physical turning point in $z$.

For three incommensurate frequencies, the regular trajectory densely samples its projected allowed volume, with nonuniform occupation determined by its speeds. It does not follow one coordinate surface, and it need not close. Resonant orbits may be closed or lie in a smaller projected set; $l=0$ confines azimuthal motion to a meridional plane (with the usual axis coordinate qualification), and circular or equatorial limits are also lower-dimensional. Smooth Stäckel separation removes the generic chaotic behavior of nonintegrable potentials; it does not assert that every orbit fills a three-dimensional volume.

At fixed radius, use spherical coordinates in velocity space with polar angle $\alpha$ measured from the radial direction:The galactic distribution function and volume element give angular weightAll dependence on $v$, $r$, and $g(E)$ cancels from ratios of second moments. Symmetry in the tangential plane givesWriting the angular integrals as beta function integrals and using the Gamma function recurrence,Therefore, for every admissible energy factor $g$,Equivalently, the velocity-anisotropy parameter is the constant $\beta =1-p/2$.

At fixed energy, the circular orbit has $L_c^2=GMa$, whileAn isotropic galactic distribution function has conditional angular-momentum density $n(L)dL\propto LdL$. ThereforeNormalization on $0\le e<1$ gives the thermal eccentricity distribution

With the convention in the question, the constant velocity-anisotropy parameter is $\beta =1-{\sigma }_t^2/{\sigma }_r^2$. Since $V_c^2=rd\Phi /dr$, the Spherical Jeans equation becomesFor $\nu ={\nu }_0{r}^{-\alpha }$ and $V_c^2=V_0^2{r}^{2\gamma }$, its general integrating factor solution isThe homogeneous term represents a boundary pressure. For an extended scale-free system the physical boundary condition normally removes it, leavingPositivity requires $\alpha -2\beta -2\gamma >0$. A complete physical model must also have a nonnegative galactic distribution function and sensible inner and outer boundary behaviour; for example, strong radial anisotropy is restricted by density-slope--anisotropy inequalities.

Observationally, the tracer density can be estimated from star counts only after correcting distances, extinction, survey selection, and incompleteness. Spectroscopy supplies mainly line-of-sight velocities; proper motions add transverse information but become less precise for distant halo stars. The equation shows the mass--anisotropy--density degeneracy directly: the same measured ${\sigma }_r$ can result from a larger $V_c$, a steeper tracer slope $\alpha$, or a different $\beta$. Even globally constant power laws therefore do not determine the galactic mass profile unless some of these quantities are independently constrained.

If $\alpha$, $\beta$, or $\gamma$ changes near a break radius, the solution at one radius also depends on the outer boundary integral. A break in observed dispersion may be attributed to a mass-profile feature, a tracer-density break, or a change in orbital anisotropy. Separate tracer populations, full three-dimensional velocities, higher velocity moments, and measurements over a wide radial range help break this degeneracy.

More flexible alternatives model a nonnegative solution of the Collisionless Boltzmann equation itself. An action-based galactic distribution function gives an analytic or parametrized $f(\mathbf{J})$; a Schwarzschild orbit-superposition model assigns nonnegative weights to an orbit library; and a made-to-measure stellar-dynamical model adjusts particle weights to reproduce observations. These methods retain more phase-space information than Jeans moments, although their flexibility introduces model choices and regularization.

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.

For a stationary collisionless stellar system confined to a plane, velocity isotropy makes the galactic distribution function depend on speed through $u=v^2/2$. The Collisionless Boltzmann equation implies ${\nabla }_{x}f=f_u{\nabla }_x\Phi$, hence $f=F(E)$ with specific orbital energy $E=u+\Phi$. This conclusion requires stationarity as well as isotropy.

In the midplane Mestel disk potential $\Phi =v_c^2\log (R/R_0)$, take specific energy $E=(v_R^2+v_{\varphi }^2)/2+\Phi$ and $L_z=R{v}_{\varphi }$. A prograde galactic distribution function $f=A{L}_z^q{e}^{-E/s}$ for $L_z>0$, with $s>0$, has surface density proportional to $R^{q-v_c^2/s}$. Matching $\Sigma ={\Sigma }_0{R}_0/R$ gives $q=v_c^2/s-1>-1$ andThe denominator is the product of the full radial Gaussian integral and the positive-azimuthal half-line moment. Here $s$ is a squared-velocity scale, not a velocity dispersion itself. Changing the potential's additive constant changes $A$. With positive exponent $e^{\epsilon /s}$, $\epsilon$ must instead be the negative of the specific total energy.

For a galactic distribution function $f(\mathbf{x},\mathbf{v})$ with tracer number density $\nu =\int f{d}^{3}v$, a velocity moment is the weighted integral of a velocity monomial. The first moment gives the mean velocity ${\overline{v}}_i$; the second moment satisfies $\overline{v_i{v}_j}={\overline{v}}_i{\overline{v}}_j+{\sigma }_{ij}^2$, separating ordered motion from the random velocity dispersion tensor. Multiplying the Collisionless Boltzmann equation by $v_i$ and integrating yields the Jeans equation, with vanishing velocity-space boundary terms when the distribution decays sufficiently rapidly.

The streaming velocity is the local mean stellar velocity obtained by averaging with the galactic distribution function. In an axisymmetric disk it usually denotes $⟨v_{\varphi }⟩$. It differs from the circular speed, which is set by the force field, and from the velocity dispersion, which measures fluctuations about the mean.

A stationary axisymmetric galactic distribution function depending on the specific orbital energy and axial specific angular momentum is a two-integral distribution. These quantities are conserved in a time-independent axisymmetric Newtonian gravitational potential, so the Collisionless Boltzmann equation holds by the chain rule. Even dependence on $L_z$ removes net azimuthal streaming velocity; an odd part can supply rotation without changing the mass density.

If the galactic distribution function is not restricted to positive relative energy, isotropic velocity integration gives $\rho (\psi )=4\pi \sqrt{2}{\int }_{-\infty }^{\psi }f(E)\sqrt{\psi -E}dE$. The same Abel transform argument as Eddington inversion yields the displayed formula, with lower limit $-\infty$. For $\rho ={\rho }_0{e}^{\psi /{\sigma }_0^2}$, substitution $\psi =E-{\sigma }_0^2{s}^2$ gives $f(E)={\rho }_0(2\pi {\sigma }_0^2{)}^{-3/2}{e}^{E/{\sigma }_0^2}$. An escape-energy truncation changes both the density law and the velocity tails.

Velocity space is the space of velocity vectors at a fixed spatial position. Integrating a galactic distribution function over velocity space produces mass density; its normalized moments determine mean velocity and velocity dispersion.