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For a stationary axisymmetric tracer with isotropic velocity dispersion, vanishing tilted stress, density $\nu \propto e^{-R/R_d}$, and ${\sigma }^2\propto e^{-R/R_{\sigma }}$, the radial Jeans equation gives the displayed expression. More generally its right side is $-{\sigma }^{2}d\ln (\nu {\sigma }^2)/d\ln R$. It requires a nonnegative streaming-speed square; the density and dispersion laws must describe the same tracer and the chosen height or consistent vertical average.

Reflection symmetry makes $\overline{v_R{v}_z}$ odd in height, so its value at the midplane vanishes while its derivative need not vanish. The radial Jeans equation includes this derivative in the stellar asymmetric drift. For a velocity ellipsoid locally aligned with spherical coordinates, $\overline{v_R{v}_z}\simeq ({\sigma }_R^2-{\sigma }_z^2)z/R$, and its midplane contribution is $-({\sigma }_R^2-{\sigma }_z^2)$. Dropping a mixed moment at the symmetry plane does not justify dropping its derivative.

For the spherical Abel deprojection, put $A_r=r^2+b^2$ and $s=(R^2+b^2)/A_r$. Then $d{R}^2=A_{r}ds$ and $\sqrt{R^2-r^2}=A_r^{1/2}\sqrt{s-1}$. Consequently the integral appearing in the reconstruction of $n$ isDifferentiate before substituting the numerical value of $I_{5/2}$:For the projected second stellar velocity moment, $N{\sigma }^2=N_0{\sigma }_0^2{b}^6(R^2+b^2{)}^{-3}$. The identical substitution in isotropic stellar pressure deprojection givesFor completeness, setting $t=1/s$ yields $I_{\alpha }=B(1/2,\alpha -1/2)$, where $B$ is the Euler beta function. The Gamma function recurrence gives $I_{5/2}=4/3$ and $I_3=3\pi /8$. This also checks the normalization of the power-law spherical projection kernel.

The radial Jeans equation is $p^{\prime }=n{\psi }^{\prime }$. Since $p^{\prime }=-105N_0{\sigma }_0^2{b}^{6}r/(16A_r^{9/2})$, division by the tracer number density givesWriting $A=315\pi {\sigma }_0^{2}b/128$, the total gravitational acceleration is thereforeTo find all gravitating mass density, apply the Poisson equation for Newtonian gravity, with the same sign convention:Equivalently, the enclosed mass is $M(r)=A{r}^3/[G(r^2+b^2{)}^{3/2}]$, and ${\rho }_{\mathrm{tot}}=M^{\prime }/(4\pi r^2)$. The inferred total mass is $A/G$ and its relative potential is $A/\sqrt{r^2+b^2}$, a Plummer model. The tracer number density has a different radial exponent from this total mass density. One must not multiply $n$ by an arbitrary stellar mass and identify it with all matter: the inferred field can include unobserved stars, gas, and dark matter.

The pressure tensor has Cartesian components $p_{ab}=\int f(v_a-u_a)(v_b-u_b)d^{3}v$. With zero mean velocity and an isotropic galactic distribution function, off-diagonal components vanish by reflection symmetry and diagonal components are equal by rotational symmetry. ThereforeHere $p$ is a tracer number-weighted random-motion stress, rather than thermodynamic pressure per volume. Its tensor divergence is $(\mathrm{div}\mathbf{p}{)}_a={\partial }_b(p{\delta }_{ab})={\partial }_{a}p$. The time-independent, nonstreaming Jeans equation consequently reduces toThe gravitational acceleration convention is $\nabla \psi$: this $\psi$ differs by a sign from the usual Newtonian gravitational potential $\Phi$.

Let $z$ measure distance along the line of sight. Spherical density projection givesBecause isotropy makes the local second stellar velocity moment along every line of sight equal to $p/n$, and because there is no streaming motion, the projected moment isThis is the forward relation for isotropic stellar pressure deprojection.

Both inversions follow from one Abel transform calculation. Write $F(u)=N(\sqrt{u})$ and $a(v)=n(\sqrt{v})$, so $F(u)={\int }_u^{\infty }a(v)(v-u{)}^{-1/2}dv$. Compose the transforms, using Tonelli's theorem for nonnegative densities, or absolute convergence for signed profiles:Assume sufficient decay to make these integrals finite and sufficient regularity for the last derivative. Then $J^{\prime }(x)=-\pi a(x)$. Setting $x=r^2$ gives the spherical Abel deprojectionReplacing $F(u)$ by $N(\sqrt{u}){\sigma }^2(\sqrt{u})$ and $a(v)$ by $p(\sqrt{v})$ provesThese are the requested expressions, with $u=R^2$. The composed-transform argument avoids an invalid separate differentiation of a divergent lower-end kernel.

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.

In an axisymmetric cylindrical coordinate system, the steady Collisionless Boltzmann equation isLet $\nu =\int f{d}^{3}v$ and $\overline{A}={\nu }^{-1}\int Af{d}^{3}v$. Multiply by $v_R$ and integrate over velocities. The spatial terms give ${\partial }_R(\nu \overline{v_R^2})+{\partial }_z(\nu \overline{v_R{v}_z})$. Integration by parts in $v_R$ supplies $\nu {\Phi }_R-\nu \overline{v_{\varphi }^2}/R$, and the $v_{\varphi }$ derivative supplies $\nu \overline{v_R^2}/R$. The $v_z$ force term vanishes because $v_R$ does not depend on $v_z$. This assumes sufficient decay of the galactic distribution function to discard all velocity boundary terms. The radial Jeans equation is consequentlyReflection symmetry about $z=0$ makes $\nu$ even and $\overline{v_R{v}_z}$ odd. At the midplane the latter vanishes, and ${\nu }^{-1}{\partial }_z(\nu \overline{v_R{v}_z})={\partial }_z\overline{v_R{v}_z}$. Multiplying by $R/\nu$ provesAll quadratic velocities here are population moments, not individual stellar velocities. In particular, symmetry does not make the midplane tilt contribution to asymmetric drift disappear: an odd mixed moment can have a nonzero derivative at zero.

For application to the Milky Way, assume negligible mean radial and vertical flows, and define the circular speed by $v_c^2=R{\Phi }_R(R,0)$. Write $\overline{v_R^2}={\sigma }_R^2$ and $\overline{v_{\varphi }^2}={\overline{v}}_{\varphi }^2+{\sigma }_{\varphi }^2$. Rearranging gives the exact stress-support relationThus the stellar asymmetric drift $v_a=v_c-{\overline{v}}_{\varphi }$ is the right-hand side divided by $v_c+{\overline{v}}_{\varphi }$. For $v_a\ll v_c$, this denominator is approximately $2v_c$. If $\nu \propto e^{-R/h_{\nu }}$ and ${\sigma }_R^2\propto e^{-R/h_{\sigma }}$ locally, the bracket becomes $R/h_{\nu }+R/h_{\sigma }-1+{\sigma }_{\varphi }^2/{\sigma }_R^2$, making the density and dispersion-gradient contributions explicit.

A radially declining random-motion stress supplies some of the force needed to support a galactic disk, so its stars need less ordered azimuthal motion than a population on circular orbits. Hotter, often older, stellar populations therefore generally rotate more slowly than cold populations in the same gravitational potential; this is the observed stellar asymmetric drift. It is a stress-support effect, not a frictional slowing of each orbit. For a velocity ellipsoid aligned approximately with spherical coordinates, $\overline{v_R{v}_z}\simeq ({\sigma }_R^2-{\sigma }_z^2)z/R$, and the midplane tilt term contributes $-({\sigma }_R^2-{\sigma }_z^2)$ rather than zero.

Local measurements of mean rotation and all components of the velocity dispersion, combined with tracer-density and dispersion gradients, can therefore constrain $v_c$ and the radial Newtonian gravitational field. A hotter sample's mean rotation must not be identified directly with the circular speed. The tracer number density need not be the gravitating mass density. Nonaxisymmetric streaming, a nonstationary disk, poorly measured radial gradients, or a neglected tilt term can bias this inference; these are assumptions of the derivation rather than extra free corrections to the circular force.

Use cylindrical coordinates and tracer density $\nu (R,z)$. A steady axisymmetric Jeans equation, with no mean radial or vertical flow, isWrite $\overline{v_R^2}={\sigma }_R^2$, $\overline{v_{\varphi }^2}={\bar{v}}_{\varphi }^2+{\sigma }_{\varphi }^2$, and $V_c^2=R{\partial }_R\Phi$. Rearrangement gives the stellar asymmetric drift equationFor an isotropic velocity ellipsoid, all three dispersions equal $\sigma$ and the cross stress vanishes. Therefore the required general result isThe positive root is for a prograde galactic disk. A declining radial stress gives $A>0$: random motions supply some support, so mean stellar rotation is slower than cold-gas circular rotation. For $A{\sigma }^2\ll V_c^2$, $V_c-{\bar{v}}_{\varphi }\simeq A{\sigma }^2/(2V_c)$. The condition $\sigma <V_c$ alone is insufficient; one also needs $A{\sigma }^2\le V_c^2$.

There is no universal numerical coefficient expressible from $V_c$ and $\sigma$ alone. For constant velocity dispersion and $\nu \propto R^{-p}$, $A=p$. Thus $p=2$ gives the often-used estimate ${\bar{v}}_{\varphi }=\sqrt{V_c^2-2{\sigma }^2}$, while $p=1$ gives a different answer with the same $V_c$ and $\sigma$. For an exponential galactic disk tracer with $\nu \propto e^{-R/R_d}$ and ${\sigma }^2\propto e^{-R/R_{\sigma }}$, the exponential-disk asymmetric drift has $A=R(R_d^{-1}+R_{\sigma }^{-1})$.

Real galactic disks generally have anisotropic velocity dispersions, often ${\sigma }_R>{\sigma }_{\varphi }>{\sigma }_z$. In the small-random-motion epicyclic motion limit for a flat galaxy rotation curve, ${\sigma }_{\varphi }^2/{\sigma }_R^2\simeq 1/2$, so even the non-gradient term matters. The velocity ellipsoid can tilt, and bars or spiral structure can violate steady axisymmetry. Age-dependent galactic disk heating makes the lag population dependent. Gas has finite turbulent and thermal stresses too, though a cold gas disk often approximates $V_c$ much better than a hot stellar population.

Write $u_i=⟨v_i⟩$ and use summation over repeated Cartesian indices. A mass-normalized galactic distribution function obeys the Collisionless Boltzmann equationAssume $f$ decays in velocity sufficiently fast that the necessary boundary terms vanish. Its zeroth velocity moment gives the continuity equationFor the first velocity moment, multiply by $v_j$ and integrate. In the gravitational term, integration by parts gives $\int v_j{\partial }_{v_i}f{d}^{3}v=-{\delta }_{ij}\rho$. ThereforeThe velocity-dispersion tensor is the centered second moment, so $⟨v_i{v}_j⟩=u_i{u}_j+{\sigma }_{ij}^2$. Expanding the streaming terms, the factor multiplying $u_j$ cancels by the continuity equation:Hence the Jeans equation isIts fluid analogue is the momentum equation with pressure tensor $P_{ij}=\rho {\sigma }_{ij}^2$. For isotropic random motion, $P_{ij}=P{\delta }_{ij}$, and this becomes the Euler equations for an inviscid fluid:The tensor form allows anisotropic stellar stresses. In a fluid an equation of state can provide a pressure closure; the Jeans equation alone does not determine the velocity-dispersion tensor of a collisionless stellar system.

The stellar population has appreciable random velocity dispersion, and its declining radial density creates an outward stress gradient that supplies some support against gravity; cold gas has much less random-motion support and follows nearly the local circular speed. Neglecting velocity-ellipsoid tilt, the radial Jeans equation gives $v_c^2-{\overline{v}}_{\varphi }^2={\sigma }_R^2[-d\ln (n_{\ast }{\sigma }_R^2)/d\ln R-1+{\sigma }_{\varphi }^2/{\sigma }_R^2]$, generally positive for the stellar disk. The slower mean stellar rotation is asymmetric drift.

Older disc stellar populations generally have larger velocity dispersions, partly through accumulated galactic disk heating, so random-motion stresses supply more of their support against gravity. The radial Jeans equation then requires a smaller mean rotation than the circular speed, producing larger stellar asymmetric drift than in dynamically colder young stars.

Let $u_i=⟨v_i⟩$, ${\sigma }_{ij}=⟨(v_i-u_i)(v_j-u_j)⟩$, and $P_{ij}=\rho {\sigma }_{ij}$. The Jeans equation obtained by taking the first velocity moment of the Collisionless Boltzmann equation isFor an equilibrium population multiply by $x_i$ and integrate. Integration by parts givesAssume the integrals exist and the boundary flux vanishes, as for an isolated finite system with adequate decay. With the stellar kinetic-energy tensor, stellar pressure-energy tensor and stellar potential-energy tensor defined bywe obtain the equilibrium tensor virial theoremThe stellar kinetic-energy tensor describes ordered motion, whereas the stellar pressure-energy tensor describes random motion. In particular, $T_{ij}=0$ does not mean the stars have no kinetic energy. If a surface flux is retained, the right side of the displayed equilibrium identity is that surface tensor; it cannot be discarded for an arbitrary finite aperture. For an isolated self-gravitating population, $W_{ij}$ is symmetric by pairwise interchange in the Newtonian interaction integral, and the time-dependent identity is $\frac{1}{2}{\ddot{I}}_{ij}=2T_{ij}+{\Pi }_{ij}+W_{ij}$ with $I_{ij}=\int \rho x_i{x}_j{d}^{3}x$.

For the spherical Newtonian gravitational potential, ${\partial }_j\Phi ={\Phi }^{\prime }(r)x_j/r$. In the pressure-supported case the tensor virial theorem gives ${\Pi }_{ii}=\int \rho {\Phi }^{\prime }(r)x_i^2/r{d}^{3}x$. The notation ${\Pi }_{RR}$ used in this question is the planar virial trace ${\Pi }_{xx}+{\Pi }_{yy}$, not just the integral of the local cylindrical radial velocity dispersion. In cylindrical coordinates this trace includes both $v_R^2$ and $v_{\varphi }^2$. Summing the two Cartesian components therefore givesThe distinction explains the factor of two in the spherical limit.

Use ordinary spherical coordinates, $R=r\sin \theta$, $z=r\cos \theta$. The mass density separates as ${\rho }_0{r}^{-\gamma }h(\theta )$, whereBoth potential-energy integrals have the same radial factor $2\pi {\rho }_0{\int }_0^{\infty }{r}^{3-\gamma }{\Phi }^{\prime }(r)dr$. When this factor is finite and positive it cancels, leaving the scale-free tracer virial ratioThe angular factor in the printed intermediate formula has its sine and cosine interchanged. Keeping its sine/cosine weights while using that printed factor would instead produce the opposite sign in the final small-flattening correction. Merely measuring $\theta$ from the equatorial plane cannot fix this: the measure and both moment weights would have to change as well.

The angular ratio is finite and positive for every fixed $q>0$ and finite real $\gamma$, because $h$ is positive and bounded above and below. However, the claim that the unrestricted global virial integrals are always well-defined needs a hypothesis. In a point-mass Newtonian gravitational potential, the common radial factor is proportional to ${\int }_0^{\infty }{r}^{1-\gamma }dr$: convergence at zero requires $\gamma <2$, while convergence at infinity requires $\gamma >2$. There is no such $\gamma$. If the global energies diverge, the boxed angular ratio is still the ratio of the two potential-energy integrals with identical spherical inner and outer cutoffs, independent of the cutoffs. It is not a quotient of two finite global stellar pressure-energy tensors; applying a finite-aperture tensor virial theorem additionally requires its surface terms. These convergence and boundary qualifications are separate from the printed angular typo.

Set $ϵ=q^{-2}-1$. For modest flattening, the Taylor expansion is $h=1-(\gamma /2)ϵ{\cos }^2\theta +O({ϵ}^2)$. The required elementary angular integrals areThese follow directly by $t=\cos \theta$, or from the supplied Gamma function identity. Hence the numerator is $(4/3)[1-\gamma ϵ/10+O({ϵ}^2)]$ and the denominator is $(2/3)[1-3\gamma ϵ/10+O({ϵ}^2)]$. The flattening–anisotropy virial relation becomesFor a decreasing tracer mass density ($\gamma >0$), oblate flattening ($q<1$) requires more random kinetic support per in-plane direction than vertically. In a spherical Newtonian gravitational field, directional orbital anisotropy supplies the flattening rather than an anisotropic force law. A steeper radial mass density amplifies the required velocity dispersion anisotropy. A prolate tracer reverses the leading inequality. The relation constrains integrated velocity dispersions, not a unique local galactic distribution function.

Use a mass-normalized galactic distribution function $F$, so that $\rho =\int F{d}^{3}v$ and $\rho ⟨v_i{v}_j⟩=\int v_i{v}_{j}F{d}^{3}v$. Multiplying the stationary Collisionless Boltzmann equation by $v_j$ and integrating over velocity space givesAssume the galactic distribution function decays sufficiently rapidly for the velocity boundary term to vanish. Integration by parts then gives $\int v_j{\partial }_{v_i}F{d}^{3}v=-{\delta }_{ij}\rho$, and therefore the Jeans equations areThe second moment includes both ordered motion and velocity dispersion; no assumption of zero mean velocity was needed.

To obtain the tensor virial theorem, multiply the $j$th Jeans equation by $x_i$ and integrate over position. For an isolated, finite system with a vanishing spatial surface term,Since the left side is $-2K_{ij}$, $2K_{ij}+W_{ij}=0$. This convention puts all stellar second moments into $K_{ij}$; it does not separate ordered and random kinetic energies.

For a self-gravitating system, the trace $W$ is the Newtonian gravitational potential energy. Indeed, symmetrizing the pair integral givesThus the trace of the tensor virial theorem is $2K+W=0$, and the total energy obeysThese identities require self-gravity without an additional external potential or an omitted confining boundary pressure.

Let $⟨v_I^2⟩$ now denote a mass-weighted average over the entire initial system. Then $K_I=M_I⟨v_I^2⟩/2$. The gravitational radius is defined by $R_I=-G{M}_I^2/W_I$, so the virial theorem immediately yieldsThe gravitational radius measures total binding energy, rather than a particular geometric edge or half-mass radius.

For the accreted systems define the mass-weighted internal mean-square speed by $M_A⟨v_A^2⟩={\sum }_s{M}_s⟨v_s^2⟩$. If each satellite initially satisfies the virial theorem, its internal energy contributes to $E_A=-M_A⟨v_A^2⟩/2$. In parabolic dry-merger energy accounting, the orbital energy at large separation is zero. Assume a dry galaxy merger, no loss of mass or energy through escaping stars, no external work, and a final relaxed system satisfying the virial theorem. Then conservation of energy gives $E_F=E_I+E_A$ and $M_F=M_I+M_A$. Energy transferred by Chandrasekhar dynamical friction remains part of the total energy under these assumptions. Consequently,Applying the final virial theorem and dividing by the initial relation givesFor a mass doubling, $\eta =1$. A merger of two identical equilibrated galaxies has $ϵ=1$, hence $R_F/R_I=2$. For accretion through many minor galaxy mergers whose satellites have much smaller internal mean-square speeds, $ϵ\ll 1$, hence $R_F/R_I\simeq 4$, tending to $4$ in the cold-satellite limit. Equivalently, nearly fixed total energy makes the gravitational radius scale as $M^2$ during cold minor-merger size growth. Small satellite mass alone does not logically imply $ϵ\ll 1$: the factor four also requires this weak-binding assumption. Likewise, equal masses require comparable internal binding to give the factor two. These are the physical limits behind the two stated merger comparisons.

A small-angle gravitational encounter between a test star and a field star of mass $m$ gives a transverse kick $\Delta v_{\perp }\simeq 2Gm/(bu)$, where $b$ is the impact parameter and $u$ the relative speed. For number density $n$, encounters in $b,b+db$ occur at rate $2\pi bdbnu$. Independent kicks add in mean square, giving the random-walk derivation of stellar relaxationThe Coulomb logarithm in stellar dynamics has $b_{\max }$ of order system size $r_s$, and $b_{\min }$ of order the strong-deflection scale $Gm/u^2$. For a virialized $N$-star system, $u^2\sim GNm/r_s$, so $\Lambda$ is of order $N$. The stellar relaxation time is the time for accumulated velocity variance to become comparable with $u^2$:Using $n\sim 3N/(4\pi r_s^3)$ gives $t_{\mathrm{rel}}\sim [N/(6\ln N)]r_s/u$. With a diameter-crossing convention $T_{\mathrm{cross}}\sim 2r_s/u$, this is $[N/(12\ln N)]T_{\mathrm{cross}}$, conventionally rounded toThe coefficient is approximate: spatial profile, velocity averages and crossing-time convention change it by factors of order unity. The robust result is $N/\ln N$ crossing times.