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A steady galactic distribution function depending only on relative energy $E=\psi -v^2/2$ is isotropic and has zero mean velocity. If its one-dimensional velocity dispersion is constant, the Spherical Jeans equation reduces to ${\sigma }_0^2{\rho }^{\prime }=\rho {\psi }^{\prime }$, yielding the displayed density. This is a tracer relation in a specified gravitational potential, not a requirement that the tracer generate that potential. With no escape-energy cutoff its local velocity distribution is Maxwell-Boltzmann velocity distribution.

Use the relative potential $\Psi ={\Phi }_{\ast }-\Phi$ and integrate the stated, untruncated Maxwell-Boltzmann velocity distribution over ${\mathbb{R}}^3$. The normalization cancels its three Gaussian integrals:The gravitational potential sign reverses under the relative-potential convention, so the spherical Poisson equation for Newtonian gravity isAlso ${\Psi }^{\prime }=-Gm(r)/r^2$, with $m^{\prime }=4\pi r^2\rho$. Differentiating the density relation yieldsAn ideal gas with constant temperature has $P=\rho k_{B}T/m_0$. Its hydrostatic equilibrium equation is precisely the same whenTogether with the same enclosed-mass equation, central density and boundary conditions, this proves the isothermal stellar and gas density equivalence. In dimensionless terms, $y=-[\Psi -\Psi (0)]/{\sigma }^2$ and $x=r\sqrt{4\pi G\rho (0)}/\sigma$ obey $x^{-2}(x^2{y}^{\prime }{)}^{\prime }=e^{-y}$ with $y(0)=y^{\prime }(0)=0$ in both systems.

This calculation uses the distribution as printed for all velocities. Imposing an additional cutoff at positive relative energy changes the integral to ${\rho }_1[e^W\mathrm{erf}\sqrt{W}-2\sqrt{W}/\sqrt{\pi }]$, where $W=\Psi /{\sigma }^2\ge 0$, and no longer gives an exactly isothermal gas profile. An untruncated self-gravitating isothermal sphere also needs an outer modification for finite total mass. The function $\mathrm{erf}$ here is the error function. Finally, the half-line Gaussian integral printed in the note lacks a factor: the correct value is ${\int }_0^{\infty }{e}^{-\alpha x^2}dx=\frac{1}{2}\sqrt{\pi /\alpha }$.

Use relative energy $E=\psi -v^2/2$, with $\psi =-\varphi$. If the galactic distribution function depends on $E$ alone, rotations and sign reversals in velocity space leave it unchanged. Thus all mean velocities vanish, and the three component second moments are equal. The anisotropy term in the Spherical Jeans equation cancels. With constant radial velocity dispersion ${\sigma }_0$, it reduces toIntegration gives the isothermal collisionless tracer densityThe constant ${\rho }_0$ changes if a constant is added to the potential; this has no effect on the velocity distribution.

Apply the given untruncated Eddington inversion, retaining its lower limit $-\infty$. Define its integral before differentiation by $I(E)$. With $\psi =E-{\sigma }_0^2{s}^2$, the Gaussian integral givesDifferentiating and multiplying by $1/(\sqrt{8}{\pi }^2)$ yieldsIndeed, at a fixed position this iswhose integral over all velocities is $\rho$ and whose component second moments are ${\sigma }_0^2$. This checks both the normalization and the velocity dispersion.

Let $w$ be the line of sight component of velocity. Integrate this Maxwell-Boltzmann velocity distribution over the two transverse components:At projected radius $R$, integrating along the line of sight simply replaces $\rho$ by the projected surface density $\Sigma (R)$. Dividing by $\Sigma (R)$ gives the spectroscopist's normalized Gaussian line-of-sight velocity distribution:It is a zero-mean normal distribution with width ${\sigma }_0$, independent of projected radius. This is the untruncated model specified by the lower limit in the inversion formula: no escape-energy cutoff has been imposed. A population restricted to positive binding energy would have truncated velocity tails and would not obey this exact Gaussian density law in a finite escape potential.

A spherically symmetric stellar model has no preferred direction in the tangent plane at a fixed radius. Rotations about the radial axis exchange its two transverse velocity directions, so $⟨v_{\theta }^2⟩=⟨v_{\varphi }^2⟩$. This uses spherical symmetry of the velocity distribution, not just a spherical mass density.

The augmented density $\rho (r,\psi )$ treats radius and relative potential as independent variables, agreeing with the physical density only along $\psi =\psi (r)$. For any regular bound spherical galactic distribution function $F(E,L)$, write $P_r=\int v_r^{2}F{d}^{3}v$ and $P_t=\int v_{\theta }^{2}F{d}^{3}v$. At fixed $r$, $F_{\psi }=F_E$ and ${\partial }_{v_r}F=-v_r{F}_E$. Integration by parts therefore givesFor $v_t^2=v_{\theta }^2+v_{\varphi }^2$, the velocity measure is $2\pi v_{t}d{v}_{t}d{v}_r$, and $r{F}_L={\partial }_{v_t}F+v_t{F}_E$. Integrating the transverse derivative gives $-2P_r$, while integrating the radial derivative in $\int v_r^2{v}_t^2{F}_E{d}^{3}v$ gives $2P_t$. Consequentlywhere $\psi$ is fixed. These identities show that the augmented representation holds for any such spherical distribution, not only a monomial ansatz. With radial pressure vanishing at the zero-binding boundary, set Along the physical curve, the chain rule gives $dA/dr={\int }_0^{\psi }{\partial }_r\rho (r,q)dq+\rho (r,\psi ){\psi }^{\prime }(r)$. The Spherical Jeans equation then requiresThus the Jeans moments from an augmented density areThe partial derivative holds $q$ fixed. Substitution cancels the integral of ${\partial }_r\rho$ and leaves precisely $\rho {\psi }^{\prime }$ in the Spherical Jeans equation. This is a representation by a chosen augmented density, not a unique solution determined by the one-variable density: different extensions off the physical curve encode different anisotropies, and not every formal extension necessarily gives a nonnegative galactic distribution function.

For the hypervirial density-potential family, differentiate the potential and use the spherical shell theorem:Differentiating $M(<r)$ and dividing by $4\pi r^2$ givesThe enclosed mass tends to $M$ at infinity and to zero at the centre for every $p>0$. The same result follows from ${\nabla }^2\psi =-4\pi G\rho$, with the stated sign convention.

From now on set $G=M=a=1$. The separable augmented density is $\rho (r,q)=(p+1)r^{p-2}{q}^{2p+1}/(4\pi )$. Integrating it givesand henceWith $\psi =(1+r^p{)}^{-1/p}$ these are the requested radial expressions. The velocity-anisotropy parameter is $\beta =1-p/2$, and the total mean square speed is $⟨v^2⟩=\psi /2$.

The local kinetic-energy density is $t=\rho ⟨v^2⟩/2=\rho \psi /4$. With the potential zero at infinity, the gravitational energy density is $w=\rho \Phi /2=-\rho \psi /2$, where $\Phi =-\psi$. ThereforeThis is the local virial relation of the hypervirial model, a special property stronger than the global virial theorem. To compute the global energies as well, set $I=\int \rho \psi d^{3}x$. The substitution $u=r^p$ yields the global binding integral of the hypervirial modelThe beta function integral converges for all $p>0$. ThusRestoring dimensions multiplies $I,T,W$ by $G{M}^2/a$.

To derive the hypervirial distribution function, let $E=\psi -v^2/2$ be the positive relative energy and $L=rv\sin \vartheta$ the magnitude of the specific angular momentum. Seek $F=C{L}^{p-2}{E}^s$ for $E>0$ and zero otherwise. Its density isThe angular integral is $\sqrt{\pi }\Gamma (p/2)/\Gamma ((p+1)/2)$. Substitution $z=v^2/(2\psi )$ makes the radial integralMatching the augmented-density power ${\psi }^{2p+1}$ gives $s=(3p+1)/2$. Matching its coefficient gives the normalization of the hypervirial distribution functionAll integrals are finite in velocity for $p>0$, and $C>0$. Because $E$ and $L$ are orbital integrals, this is a steady solution by the Jeans theorem. As checks, $p=1$ gives $C=3/(4{\pi }^3)$ for the Hernquist model, and $p=2$ gives $C=24\sqrt{2}/(7{\pi }^3)$ for the isotropic Plummer model.

The density and potential do not uniquely fix the stellar distribution. The preceding coefficient is fixed within the chosen separable power-law model, but nonuniqueness of spherical galactic distribution functions remains without that extra restriction. An explicit positive counterexample uses the same $p=2$ density. Define the Osipkov-Merritt distribution function variable $Q=E-L^2/2$ and takeRescale the transverse velocity by $\sqrt{1+r^2}$. Then $Q=\psi -[v_r^2+(1+r^2)v_t^2]/2$ and the velocity Jacobian contributes $(1+r^2{)}^{-1}$. Direct integration givesexactly the same Plummer model density. This anisotropic Plummer distribution with unit anisotropy radius has $\beta =r^2/(1+r^2)$, so it is distinct from the isotropic $E^{7/2}$ distribution while remaining positive and spherical. Its second moments differ, so it does not share the extra augmented-density and local-virial restrictions of the chosen hypervirial model.

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 relative energy $\mathcal{E}=-E=\Psi -v^2/2$. At fixed $r$, write velocity-space spherical coordinates with polar angle $\alpha$ from the radial direction, so $L=rv\sin \alpha$. ThenThe beta function integrals givewhereThus $\gamma =2\beta$ and $p=n+3/2-\beta$, with convergence for $\beta <1$ and $n>-1$.

Comparison with part a givesHence the model has the constant-anisotropy distribution functionwhere matching the two density coefficients gives

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.

The spherical Plummer model has $\psi =GM/\sqrt{a^2+r^2}$ and $\rho =3M{a}^2/[4\pi (a^2+r^2{)}^{5/2}]$. It is the $p=2$ hypervirial model. In unit scales its isotropic distribution is $24\sqrt{2}{E}^{7/2}/(7{\pi }^3)$ for positive relative energy.

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.