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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.

The thin-sheet field used above requires the gas to sample heights beyond most of the stellar mass, while the stellar galactic disk still dominates vertical gravity and is approximately locally plane-parallel. In inner regions a cold gas layer can lie inside the thicker stellar layer, where reflection symmetry gives $K_z\simeq {\nu }_z^{2}z$ rather than a constant field. The vertical Jeans equation, or gas hydrostatic equilibrium, then givesHere $H_g$ is the Gaussian width and ${\nu }_z$ is the vertical epicyclic frequency; in a locally flat galaxy rotation curve, ${\nu }_z^2\simeq 4\pi G{\rho }_{\mathrm{total}}(R,0)$. A stellar layer of approximately fixed thickness then gives $H_g\propto {\Sigma }_{\ast }^{-1/2}$, rather than the inverse surface density law.

At large radii, the dark matter halo or the gas's own gravity can exceed the dwindling stellar restoring force, so replacing the total force by $2\pi G{\Sigma }_{\ast }$ fails. A nearly spherical dark matter halo instead gives ${\nu }_z\simeq v_c/R$ near its plane, and $H_g\simeq {\sigma }_{g}R/v_c$ for a flat galaxy rotation curve, much slower than exponential galactic disk flaring. Warps, radial gradients across a thick gas layer, non-equilibrium disturbances, and height-dependent velocity dispersion also invalidate the locally plane-parallel steady isothermal approximation. The inverse stellar surface-density law applies only while its force and geometric assumptions apply.

In the thin-sheet regime, the gas scale height is $h_g(R)={\sigma }_g^2(R)/(2\pi G{\Sigma }_{\ast }(R))$. Hence radius-independent gas velocity dispersion givesFor an exponential galactic disk, ${\Sigma }_{\ast }(R)={\Sigma }_{\ast ,0}{e}^{-R/R_d}$, soThe gas therefore exhibits galactic disk flaring as stellar surface density falls outward. Stellar galactic disks, by contrast, commonly have roughly constant thickness over their main optical region; applying the same scale height relation to the stars requires ${\sigma }_{z,\ast }^2(R)\propto {\Sigma }_{\ast }(R)$, or ${\sigma }_{z,\ast }\propto e^{-R/(2R_d)}$ for an exponential galactic disk. This comparison does not follow by imposing a radially constant stellar velocity dispersion: that extra assumption would predict stellar galactic disk flaring too. The constancy used in the vertical calculation is constancy with height locally, not a mandatory constancy with radius.

Assume a steady, locally plane-symmetric galactic disk, with negligible radial tilt term in the vertical Jeans equation. Write $\rho (z)$ for the stellar mass density, and $K_z={\partial }_z\Phi$ for the upward gradient of the Newtonian gravitational potential; the actual vertical gravitational acceleration is $-K_z$. Constant vertical velocity dispersion givesIntegrating the plane-parallel Poisson equation for Newtonian gravity from the symmetry plane gives $K_z(z)=4\pi G{\int }_0^z\rho (z^{\prime })d{z}^{\prime }$. Outside most of the mass, this tends to $2\pi G\Sigma$ for $z>0$, where the surface density includes both sides of the galactic disk. Treating this field as constant and integrating the vertical Jeans equation yieldsThis exponential is the thin-sheet approximation or the outer tail, not the exact profile at every height. To see the distinction without an additional assumption, set $\rho ={\rho }_0{e}^{-u}$ with $u=(\Phi -\Phi (0))/{\sigma }^2$. The Poisson equation for Newtonian gravity becomes $u^{\prime \prime }=(4\pi G{\rho }_0/{\sigma }^2)e^{-u}$, with $u(0)=u^{\prime }(0)=0$, and is solved bySince $\Sigma =2{\rho }_0{z}_0$, the exact self-gravitating isothermal slab has $z_0={\sigma }^2/(\pi G\Sigma )$. At large $|z|/z_0$, ${\mathrm{sech}}^2(z/z_0)\sim 4e^{-2|z|/z_0}$, recovering precisely the boxed scale height; near the plane its slope vanishes, whereas an exponential has a cusp.

Measure the number-density profile $n(z)$ and vertical velocity dispersion ${\sigma }_z(z)$ of a tracer stellar population. The vertical Jeans equation estimates the restoring field fromThe second term accounts for the tilt of the velocity ellipsoid; an exponential tracer with constant velocity dispersion and negligible tilt gives $K_z={\sigma }_z^2/h$ above the plane. The tracer need not supply all the gravitating mass. To infer the dynamical surface density from vertical stellar motions, integrate the axisymmetric Poisson equation for Newtonian gravity:For a nearly flat galaxy rotation curve with a small radial correction, this reduces to $K_z/(2\pi G)$. Estimate stellar surface density independently from counts, the initial mass function and faint objects, and add the gas surface density and stellar remnants. The residuallimits the allowed unseen material within the same height interval. A fitted dark matter halo contributes too: measurements at several heights help distinguish its extended density from a thin dark component in the galactic disk. Under a specified decomposition, the disc dark-matter fraction is the inferred dark-disc surface density divided by the total disc surface density, with uncertainty or an upper limit if the residual is consistent with zero. Selection effects, disequilibrium, the tilt term and uncertain baryonic counts must enter the error budget; an excess alone does not establish that all of it is in the galactic disk.

Using the given constants, the mass is $2\times 10^{41}\mathrm{kg}$ and the velocity dispersion is $2\times 10^5\mathrm{m}{\mathrm{s}}^{-1}$. The nominal galactic dynamical time isThus the collapse clock is a few tens of millions of years, of order $10^8$ years, with $R/\sigma \simeq 3\times 10^7$ years in the interior singular isothermal sphere convention. The nominal radius $GM/{\sigma }^2\simeq 3.5\times 10^{20}\mathrm{m}\simeq 12\mathrm{kpc}$ is a plausible galactic scale. A full stellar orbit takes several crossing times, giving order $10^8$ years near a galactic disc radius, whereas the age of the Milky Way is of order $10^{10}$ years. Many orbital passages therefore fit into the lifetime of the galaxy, as expected; a galactic dynamical time need not equal either its full orbital period or its age.

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.

For an isolated final system with isotropic one-dimensional velocity dispersion $\sigma$, the random kinetic energy is $K=\frac{1}{2}M⟨v^2⟩=\frac{3}{2}M{\sigma }^2$. The virial theorem gives $W=-2K$ and total energy $E_f=K+W=-K$. Starting from a diffuse, almost unbound gas reservoir with $E_i\simeq 0$, conservation of energy therefore requires the radiative energy of virialized gravitational collapseIt is the decrease in total energy, not the whole final magnitude $|W|$, that is lost; half of the binding-energy magnitude remains as kinetic energy. Assuming radiation carries most of this loss, the average luminosity during the nominal galactic dynamical time isusing $L_{\odot }\simeq 4\times 10^{26}\mathrm{W}$. This is comparable in order of magnitude to the stellar luminosity of a characteristic $L_{\ast }$ galaxy in the Schechter function, often of order $10^{10}{L}_{\odot }$ with a band-dependent definition. The equality is an energy-scale comparison: ordinary present starlight is powered mainly by stellar nucleosynthesis, whereas this estimate concerns gravitational energy lost during formation. Appreciable initial binding energy, mass loss or external stresses alter the coefficient, and the finite-boundary singular isothermal sphere cannot be used as an isolated exact virial equilibrium without its boundary term.

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.

Many lower-luminosity elliptical galaxies have dense central stellar cusps, giving far more photons and kinematic tracers near a central black hole than the shallow depleted cores common in massive elliptical galaxies, as measured in Central galaxy profiles. Thus central surface brightness can make the smaller black hole easier to measure, despite its smaller mass, whereas sparse central tracers and uncertain stellar orbital anisotropy weaken measurements in a large core. This is not a universal mass ordering: the black-hole sphere of influence $r_{\inf }\simeq G{M}_{\bullet }/{\sigma }^2$ must still be sufficiently resolved in angle, so distance, velocity dispersion and angular resolution also matter.

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.

Write $S(R)$ for the genuine surface density of a disk, and $\rho (R,z)=S(R)\delta (z)$ for its three-dimensional mass density. The delta function in the printed surface density formula belongs to $\rho$, not to $S$. Introduce a reference length $R_{\ast }$ to make logarithms dimensionless. The intended model is the infinite, self-gravitating, scale-free Mestel disk, with no extra source or imposed gravitational field.

The circular speed satisfies $v_c^2=R{\partial }_R\Phi (R,0)$. Thus the flat galaxy rotation curve givesA reflection-symmetric harmonic function with this midplane boundary value isVerify this continuation using the Poisson equation for Newtonian gravity. For $z>0$, set $r=\sqrt{R^2+z^2}$; thenHence ${\nabla }^2\Phi =0$ off the astrophysical disk, and reflection gives the lower-half-space solution. Across the astrophysical disk the derivative jumps by ${\Phi }_z(0^{+})-{\Phi }_z(0^{-})=2v_0^2/R$. The distributional Poisson equation for Newtonian gravity therefore givesAt the origin the enclosed disk mass tends to zero linearly with radius, so there is no additional central point mass. This verifies the Mestel disk potential-density pair. The astrophysical disk has infinite total mass and a logarithmic Newtonian gravitational potential, so it is not an isolated finite-mass model with Newtonian gravitational potential zero at infinity. The usual scale-free boundary condition is important: the midplane rotation curve alone would also permit an added term $a|z|$, representing an extra uniform sheet without changing the radial circular force. That contribution is excluded in the intended Mestel disk model.

Use a mass-weighted planar galactic distribution function, so $F(\mathbf{x},\mathbf{v})d^{2}x{d}^{2}v$ is the stellar mass in a small planar phase space element. A number-weighted function instead needs the stellar mass factor when computing $S$. Assume a steady collisionless stellar system and isotropy in the two in-plane velocity components. Put $u=|\mathbf{v}{|}^2/2$ and write $F=f(\mathbf{x},u)$. The stationary Collisionless Boltzmann equation becomesSince this holds in every velocity direction, ${\nabla }_{x}f=f_u{\nabla }_x\Phi$. In coordinates $(\mathbf{x},E)$ with $E=u+\Phi$, this is exactly ${\nabla }_{x}f{|}_E=0$. Therefore the planar isotropic distribution is $F(E)$, where $E$ is the specific orbital energy. Stationarity is essential; instantaneous isotropy alone would not imply this result.

Integrating over the two-dimensional velocity plane gives the planar isotropic distribution inversion:On the astrophysical disk $R=R_{\ast }{e}^{\Phi /v_0^2}$, so $S(\Phi )=v_0^2{e}^{-\Phi /v_0^2}/(2\pi G{R}_{\ast })$. Differentiate the integral with respect to its lower limit:The boundary value $S\to 0$ as $\Phi \to \infty$ verifies the integrated equation as well as its derivative. Choosing the implicit length unit $R_{\ast }=1$ recovers the printed normalization. Changing the additive energy zero changes this prefactor accordingly.

At a fixed radius, the normalized velocity density isIt is a product of centered Gaussian distributions. Differentiating the supplied Gaussian integral with respect to its coefficient gives the second moments, and odd moments vanish. Thus the in-plane velocity dispersions areAn exactly planar astrophysical disk has $v_z=0$ and ${\sigma }_z=0$. The nonzero circular speed is a property of the force field, not a statement that this hot stellar distribution has net rotation.

Reversing every retrograde star folds the azimuthal Gaussian to a half-normal distribution. Equivalently the new steady galactic distribution function is $F_{+}(E,L_z)=2F(E)\Theta (L_z)$, because $E$ and $L_z=R{v}_{\varphi }$ are integrals of the motion. Its density and even velocity moments are unchanged. Its streaming velocity isThis maximally prograde stellar distribution still has radial motion and a spread of azimuthal speeds; it does not place every star on a circular orbit. In particular $⟨v_{\varphi }^2{⟩}_{+}=v_0^2$ while ${\sigma }_{\varphi ,+}^2=v_0^2(1-2/\pi )$. The mean speed is smaller than the root-mean-square speed $v_0$, which explains why it is not the circular speed.

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 phase-space distribution function describes the mass in a small cell of positions and velocities: $F(\mathbf{x},\mathbf{v},t)d^{3}x{d}^{3}v$ is that mass. Its velocity integral is the spatial mass density, and $F/\rho$ at a fixed position is a normalized velocity probability density. A number-normalized galactic distribution function differs by the stellar mass convention.

Take $a>0$ and $v_0>0$. For the cusped logarithmic spherical galaxy model, differentiate the gravitational potential and use the spherical Poisson equation for Newtonian gravity:ThusA circular orbit requires $v_c^2/r={\Phi }^{\prime }$. Its galaxy rotation curve is thereforeIt rises as $v_0\sqrt{r/a}$ at small radius and approaches $v_0$ from below at large radius.

The outer flat galaxy rotation curve is qualitatively useful for galaxies. However, the density has an inner $1/r$ cusp and an outer $1/r^2$ tail, while $M(r)\sim v_0^{2}r/G$ diverges. The model therefore requires an outer truncation for a finite galaxy, and does not describe a central constant-density core. Its gravitational potential cannot be normalized to zero at infinity.

For a star, the specific energy is $E=v^2/2+\Phi$ and the specific angular momentum is $\mathbf{L}=\mathbf{x}\times \mathbf{v}$. Since the gravitational potential is stationary and spherical,Thus both the specific energy and the entire angular-momentum vector are conserved.

For Jeans theorem, follow a stellar trajectory in phase space. The Collisionless Boltzmann equation is preciselyIn a steady system ${\partial }_{t}F=0$, so the value of the galactic distribution function is constant along every trajectory: it is itself an integral of motion. Conversely, if $I_1,\dots ,I_m$ are time-independent integrals of motion, then $F=h(I_1,\dots ,I_m)$ satisfies $dF/dt=\sum h_{,a}{\dot{I}}_a=0$. Locally, a complete set of invariant orbit labels expresses any steady galactic distribution function this way. This is the weak form of Jeans theorem; a stronger assertion about a small set of isolating integrals needs additional orbital regularity or mixing assumptions. In particular, the theorem does not say that every steady galaxy must have a galactic distribution function depending on only $E$ and $L$.

For the proposed constant-anisotropy distribution function, write $v_T=(v_{\theta }^2+v_{\varphi }^2{)}^{1/2}$ and $L=r{v}_T$. Introduce a polar angle $\psi$ in the tangential velocity plane. Then $d^{3}v=d{v}_r{v}_{T}d{v}_{T}d\psi$, so the apparent $1/L$ singularity cancels against the velocity Jacobian. This inverse-angular-momentum velocity integration makes the density finite for every $r>0$.

Let $t=e^{-\Phi /v_0^2}=a/(a+r)$, $c_1=1$, $c_2=2$, and ${\alpha }_k=k/(2v_0^2)$. Each exponential component contributesHere the two Gaussian integrals and the angular integral have product ${\pi }^2/{\alpha }_k$. Consequently,The galactic distribution function is nonnegative, gives the required density, and depends on conserved $E,L$; by Jeans theorem it is a self-consistent steady solution.

Differentiating the Gaussian integral with respect to ${\alpha }_k$ shows that each component has $⟨v_r^2{⟩}_k=1/(2{\alpha }_k)=v_0^2/k$, and the same value for $⟨v_T^2{⟩}_k$. Angular averaging gives $⟨v_{\theta }^2{⟩}_k=⟨v_{\varphi }^2{⟩}_k=v_0^2/(2k)$. Weighting these by ${\rho }_1,{\rho }_2$ yields the velocity dispersionsThe radial numerator includes $2r$, as in the original PDF. All mean velocities vanish by parity. The same parity, or the zero angular average of $\sin \psi \cos \psi$, gives $⟨v_r{v}_{\theta }⟩=⟨v_r{v}_{\varphi }⟩=⟨v_{\theta }{v}_{\varphi }⟩=0$.

The velocity-anisotropy parameter isThe constant-anisotropy distribution function therefore favors radial motion at every radius: the radial variance is twice either individual tangential variance. It still has a substantial population with nonzero angular momentum, rather than exclusively radial orbits. The $L^{-1}$ factor enhances low-angular-momentum stars. The radial velocity dispersion approaches $3v_0^2/4$ at the centre and $v_0^2$ far out; each individual tangential variance is half as large. There is no mean rotation, despite the well-defined galaxy rotation curve for hypothetical circular orbits.