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The Hernquist model has relative potential $\psi =GM/(r+a)$ and density $\rho =Ma/[2\pi r(r+a{)}^3]$. It has enclosed mass $M{r}^2/(r+a{)}^2$, a central $r^{-1}$ cusp and an outer $r^{-4}$ tail. Since $r\rho =a{\psi }^3/(2\pi G^3{M}^2)$, the half-anisotropic distribution function is $f(E,L)=3a{E}^2/(4{\pi }^3{G}^3{M}^{2}L)$ for $0<E<GM/a$. Its moments are $⟨v_r^2⟩=\psi /4$ and each tangential moment $\psi /8$. Isotropic and other admissible anisotropic distributions can also support this same density; the gravitational potential does not select a unique velocity anisotropy.

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 $\beta =1/2$.