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For an isotropic self-gravitating collisionless stellar system with $k<3$ and no central point mass, $M(r)=4\pi A{r}^{3-k}/(3-k)$. Integrating the Spherical Jeans equation to an outer radius $b$ gives the displayed pressure, with $P_b=\rho (b){\sigma }_r^2(b)$. For $k<1$ the inner pressure is set by outer radii and ${\sigma }_r^2\sim C{r}^k$, with $C$ depending on the boundary. An unbounded pure power law has divergent pressure in this case. For $1<k<3$, inner radii dominate and ${\sigma }_r^2\sim 2\pi GA{r}^{2-k}/[(3-k)(k-1)]$. At $k=1$ a logarithm appears. For $k\ge 3$ the enclosed central mass diverges unless the density law is modified at small radii.

For the $r^{-2}$ singular isothermal sphere, the isotropic Spherical Jeans equation gives ${\sigma }_r^2={\sigma }_{\theta }^2={\sigma }_{\varphi }^2=v_0^2/2$, so $⟨v^2⟩=3v_0^2/2$. Randomly oriented circular orbits instead have ${\sigma }_r^2=0$, each transverse second moment $v_0^2/2$, and total mean square speed $v_0^2$.

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.

For a spherical, nonstreaming stellar population with isotropic velocity dispersion, each component of the random pressure tensor equals $p=n⟨v_r^2⟩$. The observed second stellar velocity moment satisfies $N(R){\sigma }_{\mathrm{los}}^2(R)=2{\int }_R^{\infty }p(r)r/\sqrt{r^2-R^2}dr$. Composing this Abel transform with itself gives the displayed inverse under sufficient decay and regularity. The Spherical Jeans equation then gives ${\psi }^{\prime }=p^{\prime }/n$, with gravitational acceleration $\nabla \psi$. The Poisson equation for Newtonian gravity recovers total gravitating mass density, not just the tracer density. Without velocity isotropy the same projected moment also contains anisotropic stresses.

With radial pressure zero at $\psi =0$, an augmented density supplies $\rho ⟨v_r^2⟩={\int }_0^{\psi }\rho (r,q)dq$ and $\rho ⟨v_{\theta }^2⟩=(2r{)}^{-1}{\int }_0^{\psi }{\partial }_r[r^2\rho (r,q)]dq$. The derivative holds $q$ fixed. Their chain-rule combination satisfies the Spherical Jeans equation, but the physical density alone does not choose an augmented extension.

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.

Interpret the barred velocities as second moments of a steady nonstreaming collisionless stellar system, and write ${\sigma }_r^2=\overline{v_r^2}$. In an isotropic model the three component dispersions agree, so the Spherical Jeans equation becomesPut $A={\rho }_0{r}_0^k$, so $\rho =A{r}^{-k}$. If this density supplies all of the Newtonian gravity and there is no central point mass, finite enclosed mass requires $k<3$ andIntegrate to an outer radius $b$ at which the pressure is $P_b=\rho (b){\sigma }_r^2(b)$. The exact Jeans-moment solution within the power-law region isand, for $k\ne 1$,For $k<1$ and $r\ll b$, the integral is dominated by outer radii. ThereforeThe coefficient cannot be determined from the local density cusp alone. Extending the power law to infinity gives a divergent pressure integral when $k<1$, so an outer scale or boundary condition is essential. For the usual shallow cusp $0<k<1$, the central velocity dispersion tends to zero even though the density diverges.

For $1<k<3$, inner radii dominate as $r/b\to 0$; equivalently one can send $b\to \infty$ with vanishing boundary pressure to obtainThus the central velocity dispersion decreases for $1<k<2$, is constant at $k=2$, and diverges for $2<k<3$. At the omitted threshold $k=1$, ${\sigma }_r^2=(P_b/A)r+2\pi GAr\ln (b/r)$. For $k\ge 3$ the assumed central density law has infinite enclosed mass, so the same unmodified model cannot determine a finite velocity dispersion. These boundary and domain restrictions are the boundary dependence of an isotropic self-gravitating cusp; specifying density without them does not specify a physical finite galaxy.

Let $p(r)=\rho (r){\sigma }_r^2(r)$ and use the velocity-anisotropy parameter $\beta =1-{\sigma }_{\theta }^2/{\sigma }_r^2$, with ${\sigma }_{\theta }^2={\sigma }_{\varphi }^2$. Here the tangential variance is per component; a definition using the sum of both tangential variances instead has an extra factor of two. The original PDF has this correct per-component ratio, which the TeX corrupts.

At a point along the line of sight, $r^2=R^2+z^2$. The radial direction makes an angle with the sightline satisfying ${\sin }^2\chi =R^2/r^2$. Spherical symmetry eliminates mixed moments, and projection givesUsing both sides of the galaxy and $dz=rdr/\sqrt{r^2-R^2}$,For this formula use $I$ as the projected density of the same tracer represented by $\rho$. If $\rho$ is a mass density and $I$ an observed surface brightness, a constant mass-to-light ratio rescales $I$ by that factor; otherwise replace $\rho$ by the luminosity density throughout the tracer equation. The gravitating $M(r)$ remains the total mass, not necessarily the tracer's mass.

The Spherical Jeans equation is $p^{\prime }+2\beta p/r=-\rho GM/r^2$. Eliminate $\beta p=-r[p^{\prime }+\rho GM/r^2]/2$ from the projected integrand:Rearranging provesNo constancy of $\beta$ was needed. Convergence of the projected moments and the steady spherical assumptions are required. This is the integral form of the projected spherical Jeans residual.

Although $\beta$ has disappeared explicitly, the unknown radial pressure $p$ still carries the orbital information. Thus the identity does not remove the mass--anisotropy--density degeneracy: projected brightness and one projected second moment do not uniquely determine both $M(r)$ and the radial/tangential partition. A central rise in velocity dispersion can reflect a compact central mass, radial orbital bias, or a changing mass-to-light ratio; flattening and departures from equilibrium add further ambiguities.

For example, within a point-mass-dominated region with tracer $\rho \propto r^{-k}$ and constant $\beta$, direct solution gives ${\sigma }_r^2=G{M}_{\bullet }/[(k+1-2\beta )r]$. For $k>1$ the projected velocity dispersion of a Keplerian cusp isThe beta function factors come from setting $r=R\sec \chi$ in the integrals along the line of sight. Different allowed $\beta$ generally change the mass needed to fit the same normalization; $k=2$ is a special cancellation, not the general case. Resolving the central black-hole sphere of influence and constraining orbital structure with higher velocity moments or additional kinematic directions strengthens a compact-mass inference. Brightness and velocity dispersion profiles alone do not establish a unique central black-hole mass, still less an event horizon.

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.

Integrating the luminosity density along a line of sight gives its surface brightness:The second identity is Spherical Abel deprojection. For the second stellar velocity moment, the angle between the radial direction and the line of sight has ${\sin }^2\theta =R^2/r^2$. The line-of-sight velocity dispersion is therefore ${\sigma }_r^2{\cos }^2\theta +{\sigma }_t^2{\sin }^2\theta ={\sigma }_r^2(1-\beta R^2/r^2)$, andFrom the Spherical Jeans equation, $\beta q=-r{q}^{\prime }/2-\gamma \lambda L/(2r)$. Substitution producesThe primary PDF has a genuine symbol error in its projected gravity term: it prints a second $\gamma$ multiplying $L$ inside the integral, where $\lambda$ is required. The corrected projected spherical Jeans residual isTo justify the derivative at the singular-looking lower limit, write $A(R)={\int }_0^{\infty }q(\sqrt{R^2+z^2})dz$. Differentiation then gives $A^{\prime }(R)=R{\int }_R^{\infty }{q}^{\prime }(r)(r^2-R^2{)}^{-1/2}dr$. In particular, $Rp=(R^{2}A{)}^{\prime }$. If $R^{2}A$ tends to zero both at zero and infinity, as for a finite isolated equilibrium with sufficiently decaying stresses,If a boundary stress survives, its contribution must be retained; the zero-integral statement is not a finite-aperture identity.

Now identify each integral with a term in the virial theorem. Let $K$ be the total kinetic energy, including all three random components, and $W$ the gravitational potential energy. Averaging over spatial directions in a spherical system gives one third of the total stellar velocity moment along any chosen line of sight, even if the velocities are locally anisotropic. ThusBuilding the spherical mass shell by shell givesFinally, Fubini theorem and ${\int }_0^r{R}^3(r^2-R^2{)}^{-1/2}dR=2r^3/3$ give the gravitational contribution to the residual integral explicitly:ConsequentlyThe global projected virial estimate of a mass-to-light ratio is independent of the velocity-anisotropy parameter under these assumptions. It does not remove the mass--anisotropy--density degeneracy for arbitrary radial mass profiles. In particular, if an independent dark matter halo makes mass fail to follow light, a single constant ${\rm Y}$ is no longer the physical enclosed mass-to-light ratio at every radius.

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.

Use a steady, isotropic, self-gravitating dark matter halo, with $V$ its constant one-dimensional velocity dispersion. The Spherical Jeans equation and spherical mass conservation readFor the scale-free outer region, put $\rho =A{r}^{-p}$. Its enclosed mass scales as $r^{3-p}$, so $GM(r)/r$ is constant only if $p=2$; the Spherical Jeans equation then gives $2V^2=GM(r)/r=4\pi GA$. Thus the singular isothermal sphere hasIts circular speed is $v_c=\sqrt{2}V$. If a speed denoted $V$ were instead the circular speed, the coefficient would be $V^2/(4\pi G)$, but the printed variable is a velocity dispersion.

The assumptions matter. A constant velocity dispersion and spherical shape alone do not specify a unique central mass density or orbital distribution. Even with isotropy, the Spherical Jeans equation integrates to $\rho ={\rho }_0\exp [-(\Phi -{\Phi }_0)/V^2]$, and the self-gravitating Poisson equation for Newtonian gravity givesThere are finite-central-density isothermal solutions as well as the singular one; the $r^{-2}$ expression is the intended outer scale-free halo approximation. A finite halo must also end or steepen at large radius, since the untruncated singular isothermal sphere has infinite total mass.

Let $R$ be a characteristic radius enclosing mass $M$, and let $\sigma$ denote a one-dimensional velocity dispersion. Gravity and random-motion support give a relation $GM/R=\eta {\sigma }^2$, where the dimensionless structure coefficient $\eta$ depends on the chosen radius and density profile. A crossing-based galactic dynamical time is thereforeThe scaling follows by equating acceleration scales ${\sigma }^2/R$ and $GM/R^2$, not by identifying a complete orbital period with a crossing time. An explicit scale-free example checks the coefficient: constant circular speed and spherical symmetry give $M(<r)=v_c^{2}r/G$ and $\rho =v_c^2/(4\pi G{r}^2)$. With isotropic constant velocity dispersion, the Spherical Jeans equation becomesso $v_c^2=2{\sigma }^2$, $M(<R)=2{\sigma }^{2}R/G$, and $R/\sigma =GM/(2{\sigma }^3)$. This is the singular isothermal sphere scaling, valid over an interior radial range; a literal infinite model has no finite total mass, and a hard finite boundary needs a pressure term or an outer transition. A free-fall time of a uniform sphere or orbital period changes the numerical factor. The subsequent estimates use the nominal scale $GM/{\sigma }^3$, keeping these order-unity structural uncertainties explicit.

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.

False. The singular isothermal sphere is infinite in extent and mass. The isolated finite-system virial theorem cannot be applied to its global divergent energies without boundary terms.

A direct Spherical Jeans equation counterexample is enough. For isotropic velocities with constant one-dimensional dispersion ${\sigma }^2$, its density slope givesThus ${\sigma }^2=v_0^2/2$ and the circular and isotropic speed comparison in a singular isothermal sphere isThis is realized by a positive isotropic distribution $F\propto \exp (E/{\sigma }^2)$, which gives a Maxwell-Boltzmann distribution of velocities and $\rho \propto \exp (\psi /{\sigma }^2)\propto r^{-2}$. Circular orbits have instead $⟨v_r^2⟩=0$ and $⟨v_{\theta }^2⟩=⟨v_{\varphi }^2⟩=v_0^2/2$, so their total mean square speed is $v_0^2$. The isotropic model contains eccentric orbits except for a measure-zero subset.

The boundary virial term of a singular isothermal sphere explains the different energy budgets of a finite spherical region. At radius $R$, its radial-pressure surface term is $4\pi R^3\rho (R){\sigma }_r^2$. It is nonzero in the isotropic model and zero in the circular-orbit model. Equality of the density therefore does not enforce equality of the velocity energy through an incorrectly omitted surface term.