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Kinetic decoupling occurs when momentum-exchange collisions cease to maintain the temperature and velocity dispersion of a particle species in equilibrium with a surrounding bath. Afterwards, a freely moving nonrelativistic particle has momentum proportional to $a^{-1}$ and effective kinetic temperature proportional to $a^{-2}$, where $a$ is the scale factor.

The M-sigma relation is an empirical correlation between supermassive black hole mass and the stellar velocity dispersion of its host's spheroidal component. A commonly used rough parametrization has exponent near four, but the normalization, exponent, scatter, and selection effects depend on the observed sample. It is distinct from the black-hole--bulge mass relation.

For a finite stationary axisymmetric self-gravitating system with similar oblate density surfaces, let $q=b/a$, $s=\sqrt{1-q^2}$, $A_1=q\arccos (q)/s^3-q^2/s^2$, and $A_3=2/s^2-2q\arccos (q)/s^3$. Its stellar potential-energy tensor has $Q=A_1/(q^2{A}_3)$; the common radial integral cancels. The spherical limit has $A_1=A_3=2/3$ and $Q=1$. With mass-weighted in-plane velocity dispersion ${\sigma }_x$ and constant azimuthal streaming $v$, the tensor virial theorem gives $v^2/{\sigma }_x^2=2[(1-\beta )Q-1]$, where ${\sigma }_z^2=(1-\beta ){\sigma }_x^2$. Neither $Q=1/q$ nor the omission of the Cartesian streaming factor is exact.

The three probes constrain the same deep gravitational potential by different observations. A large member-galaxy velocity dispersion requires a large binding mass. Applying the virial theorem, with a gravitational scale radius $R$ and approximately isotropic orbits, gives a cluster velocity-dispersion mass estimate of order $M\sim R{\sigma }_v^2/G$, with a profile-dependent numerical coefficient. The inferred mass-to-light ratio is much larger than the stellar value. Membership, orbital anisotropy and equilibrium assumptions affect the precision.

The hot intracluster medium radiates through thermal bremsstrahlung and line emission. Its X-ray spectrum measures temperature, while its surface brightness constrains gas density. Confining this gas requires a potential with characteristic $GM/R\sim k_{B}T/(\mu m_p)$. Combining resolved profiles through the cluster hydrostatic mass estimator gives the total mass, not just the gas mass. The detected gas contributes substantial baryonic mass but does not normally account for the full gravitating mass. Nonthermal pressure or a merging cluster can invalidate a purely thermal hydrostatic equilibrium estimate.

Finally, gravitational lensing measures mass through light deflection. Strong gravitational lensing supplies multiple images and arcs; weak gravitational lensing measures coherent background-image distortion over larger radii. The lensing convergence is projected surface mass density divided by the critical surface density, irrespective of whether the material emits light. Cluster gravitational-lensing mass estimates agree broadly with the high masses implied by galaxy motions and hot gas, while having different equilibrium assumptions. Together these observations support a dominant nonluminous component in galaxy clusters, after accounting for their stars and observed gas.

Work in the centre-of-mass frame and define $I={\sum }_{i}m|{\mathbf{r}}_i{|}^2$, $T=\frac{1}{2}{\sum }_{i}m|{\mathbf{v}}_i{|}^2$. Differentiation givesFor each gravitational pair, the two contributions combine toSumming pairs therefore gives the exact scalar virial theoremIn a stationary equilibrium, or a bounded long-time average with vanishing average $\ddot{I}$, $2T=-W$. With $M=Nm$, the stated gravitational radius gives $W=-G{M}^2/R_g$, henceThe rms speed itself is $[GNm/R_g{]}^{1/2}$. This is a three-dimensional mean-square speed; for an isotropic system the one-component velocity dispersion is smaller by a factor of three in its square. No assumption about a uniform spatial density was used.

At rest with all separations initially infinite, both initial kinetic and gravitational energies vanish, so the conserved total energy is zero. During an isolated collapse, $T+W=0$, rather than the equilibrium relation $2T+W=0$. Hence at the stated instantaneous gravitational potential energy,The rms collapse speed is $\sqrt{2}$ times the rms equilibrium speed for the same instantaneous $W$ and mass. For a perfectly radial cold infall this mean-square motion includes coherent inward streaming; it need not equal a local random velocity dispersion after subtracting that streaming. The quoted comparison follows the question's global mean-square convention.

Integrate the reduced vertical phase-space distribution function over $v_z$. The Gaussian velocity integral cancels its normalization, givingHere the constant vertical velocity dispersion is ${\sigma }_z$, and $\Phi (0)=0$ fixes the potential reference. Neglecting the radial derivatives in the Poisson equation for Newtonian gravity leavesWith $\varphi =\Phi /{\sigma }_z^2$, $\zeta =z/z_0$ and $z_0^2={\sigma }_z^2/(8\pi G{\rho }_0)$, this becomesThus the stated dimensionless equation follows with its factor of two.

For a direct solution, multiply by $2{\varphi }^{\prime }$ and integrate from the midplane. The initial conditions giveFor $\zeta >0$ take the increasing branch and put $u=e^{\varphi /2}$. Then $4(u^{\prime }{)}^2=u^2-1$, whose smoothly increasing solution consistent with the second-order equation is $u=\cosh (\zeta /2)$. Reflection about the midplane supplies the negative-$\zeta$ branch. ThereforeSubstitution verifies $2{\varphi }^{\prime \prime }={\mathrm{sech}}^2(\zeta /2)=e^{-\varphi }$ and both boundary conditions. The smooth second-order equation has a unique solution for those initial data, so no extra constant branch from the first-integral square root is admissible.

This is the self-gravitating isothermal slab. Its scale convention matters: if its profile is instead written $\rho ={\rho }_0{\mathrm{sech}}^2(z/H)$, then $H=2z_0={\sigma }_z/\sqrt{2\pi G{\rho }_0}$. Its surface density is $\Sigma =4{\rho }_0{z}_0$. The force approaches $|{\Phi }^{\prime }|={\sigma }_z^2/z_0=2\pi G\Sigma$, consistent with the field of a plane sheet. These checks fix both the scale-height and Poisson factors.

Write ${\sigma }_0^2=⟨v_z^2⟩=v_{\mathrm{rms}}^2/3$ for the one-dimensional thermal velocity dispersion of an isotropic monatomic ideal gas. An isothermal process keeps ${\sigma }_0$ fixed, and the vertical pressure is $P=\rho {\sigma }_0^2$. Hydrostatic equilibrium in the Newtonian gravitational potential $\Phi (z)$ givesPlanar symmetry reduces the Poisson equation for Newtonian gravity to ${\Phi }^{\prime \prime }=4\pi G\rho$. Combining the two equations yieldsThe velocity factor here is a velocity second moment, not the square of a particular particle's instantaneous vertical velocity.

Set $H^2={\sigma }_0^2/(2\pi G{\rho }_0)$ and try the symmetric self-gravitating isothermal slab profile $\rho ={\rho }_0{\mathrm{sech}}^2(z/H)$. Its logarithmic derivatives areThe chosen $H$ makes the second expression satisfy the required equation. The central conditions $\rho (0)={\rho }_0$ and ${\rho }^{\prime }(0)=0$ fix the smooth even solution uniquely. In the scale-height convention used here,The undefined constant printed as $a$ in the scale-height denominator must be $G$; this follows both from substitution and dimensions.

Integrating hydrostatic equilibrium, with the additive potential constant chosen so that $\Phi (0)=0$, givesThe orbits in a self-gravitating isothermal slab separate into horizontal and vertical motions. There is no horizontal force, so $x=x_i+v_{x}t$ and $y=y_i+v_{y}t$, with constant horizontal velocities. For collisionless test particles, or between gas collisions, conservation of energy fixes the vertical energyEvery finite $E_z>0$ has two turning points $\pm A$, whereThus the general orbit drifts uniformly in the plane while oscillating periodically across it. A particle with $E_z=0$ stays in the midplane; an orbit with nonzero horizontal velocity is generally not closed.

Near the plane, $\Phi (z)={\sigma }_0^2{z}^2/H^2+O(z^4)$, so the vertical motion is a harmonic oscillator with ${\nu }^2=2{\sigma }_0^2/H^2=4\pi G{\rho }_0$. Far from the plane, $\Phi \sim 2{\sigma }_0^2|z|/H$ and the force tends to a constant magnitude. Indeed the surface density is $\Sigma =\int \rho dz=2{\rho }_{0}H$, and that magnitude is $2\pi G\Sigma$. Large-amplitude vertical motion is therefore approximately uniformly accelerated on each side, rather than sinusoidal. Because this ideal slab extends infinitely in the horizontal directions and its potential grows without bound as $|z|\to \infty$, no finite-energy particle escapes vertically.

For an isolated smooth mass distribution with spherical symmetry, assemble each shell $dM$ at radius $r$ against the Newtonian gravitational potential of its interior mass. The spherical shell theorem givesAssume finite total mass, finite binding energy and no central point-mass self-energy, so $M(r{)}^2/r$ vanishes at both endpoints. Integration by parts then givesFor example, $M(r)\propto r^3$ near a regular center guarantees the central boundary condition, while finite total mass guarantees the one at infinity. A divergent central binding energy would fall outside this formula's finite-energy hypotheses.

For the discrete system use the centre of mass frame, let $N$ be the particle number, and write $M_{\mathrm{tot}}=Nm$. The printed $M$ in the particle-count factor and the subsequent $n$ must be understood consistently as $N$. DefineBy Newton's second law, differentiation givesEach pair contributesTherefore the exact dynamical virial theorem is $\ddot{I}/2=2K+W$. In a stationary statistical state, or on a bounded-motion time average for which the mean $\ddot{I}$ vanishes, it becomesThere is no external pressure, tidal field or boundary work in this isolated point-particle derivation. For a time-averaged system, the following $W$ and gravitational radius refer to that same averaged binding energy.

Using $W=-G(Nm{)}^2/R_g$ and $2K=Nm⟨v_{\mathrm{eq}}^2⟩$, virial equilibrium givesThis is the three-dimensional mean squared speed about the centre of mass. For an isotropic system each one-dimensional velocity dispersion squared is one third of it.

Set the Newtonian gravitational potential to zero at infinity and exclude each particle's own field. The local escape velocity in the instantaneous frozen potential obeys $v_{\mathrm{esc},i}^2=-2{\Phi }_i$, where ${\Phi }_i=-Gm{\sum }_{j\ne i}{r}_{ij}^{-1}$. Since every pair occurs twice, ${\sum }_{i}m{\Phi }_i=2W$. Averaging over the identical particles proves the gravitational rms escape and collapse speeds relationAn actual encounter in a changing potential can exchange energy, so this is the usual local escape-speed definition, not a claim about every many-body trajectory.

Finally, dissipation-free collapse from rest at infinite separation has total energy zero. Conservation of energy at the specified binding energy therefore gives $K_{\mathrm{coll}}=-W$, twice the equilibrium kinetic energy at the same $W$. HenceThis mean squared speed includes coherent infall; a local random velocity dispersion after subtracting the infall velocity need not obey that ratio.

For a relaxed cluster, galaxy motions act as tracers of its gravitational potential. The virial theorem gives a dynamical mass of orderwith a profile-dependent coefficient and a projection correction because the observed velocity dispersion is usually only the line-of-sight component. For an isotropic uniform-sphere estimate using its one-dimensional dispersion, $T=3M{\sigma }_v^2/2$ and $W=-3G{M}^2/(5R)$ give $M=5R{\sigma }_v^2/G$. Measured dispersions require much more gravitating mass than the luminous galaxies provide, hence large cluster mass-to-light ratios. Interlopers, orbital anisotropy and departures from equilibrium are systematic uncertainties, so the independent gas and lensing measurements matter.

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.

A local stellar relaxation time depends on the number density $n$ of scatterers. Calling $N$ a total number, as the PDF does, leaves the displayed expression dimensionally wrong: ${\sigma }^3/(G^2{m}^{2}N)$ has units of time per volume. The intended local expression must use $n$, or $N/V$ for a homogeneous sample of volume $V$. We shall also make the normalization of the stellar relaxation time explicit, since an unspecified representative one-dimensional velocity dispersion does not fix an exact numerical coefficient.

In a small-angle gravitational encounter, approximate the relative trajectory by a straight line of speed $u$ and impact parameter $b$. The perpendicular acceleration of one star due to the other is $Gmb/(b^2+u^2{t}^2{)}^{3/2}$. Its impulse is thereforeA star encounters $2\pi bdbnu$ scatterers per unit time in this interval of impact parameters. Independent deflections have random directions, so their variances add as in a random walk. Hence the accumulated squared velocity change has rateThe Coulomb logarithm in stellar dynamics arises because each logarithmic interval of impact parameter contributes equally. The lower cutoff is approximately the strong-deflection scale, $b_{\min }\simeq G(2m)/u^2$ for equal masses, or the physical collision scale if larger. The upper cutoff is a scale on which the local density and straight-line encounter approximation cease to apply, such as the local disk thickness. Thus $\Lambda$ is dimensionless; it is not a number density or an arbitrary dimensional argument of a logarithm.

To display the normalization issue, use the representative relative speed $u=\sqrt{2}\sigma$ and define $T(\kappa )$ by $D_{v}T=\kappa {\sigma }^2$. This givesChoosing $\kappa =3/4$ recovers precisely the printed coefficient after repairing $N$ to $n$:This is an explicitly defined convention, not a uniquely derivable normalization from the supplied data. If instead relaxation means a mean-square change equal to the full initial random-speed variance, $3{\sigma }^2$, the same encounter approximation gives $T_R=4T_R^{\mathrm{printed}\mathrm{convention}}$. Averaging the encounter speed distribution instead of using one representative $u$ changes the coefficient again. The robust result is the scaling $T_R\propto {\sigma }^3/(n{G}^2{m}^2\ln \Lambda )$.

For a transparent solar-neighbourhood estimate, take $n=0.1{\mathrm{pc}}^{-3}$, $m=M_{\odot }$, $\sigma =30{\mathrm{km}\mathrm{s}}^{-1}$ and $b_{\max }=300\mathrm{pc}$. With $G=4.3009\times 10^{-3}\mathrm{pc}{M}_{\odot }^{-1}({\mathrm{km}\mathrm{s}}^{-1}{)}^2$, one obtains $b_{\min }=4.78\times 10^{-6}\mathrm{pc}$, $\ln \Lambda =17.96$, andThese illustrative parameters give an order $10^{13}$--$10^{14}$ year stellar relaxation time, much longer than a galactic lifetime of order $10^{10}$ years. The solar neighbourhood is consequently a collisionless stellar system with respect to encounters between individual stars. Scattering by massive clouds and collective structure is a different process; the estimate does not assert that every mechanism of disk heating is negligible.

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 the stellar position second-moment tensor $I_{ij}=\int \rho x_i{x}_j{d}^{3}x$. This is the $I$ entering the tensor virial theorem, rather than the engineering moment-of-inertia tensor $\mathrm{tr}(I)1-I$. For a density pattern rotating steadily about its third principal axis, choose $t=0$ when its in-plane axes are principal axes. Then $I(0)=\mathrm{diag}(I_1,I_2,I_3)$ andIn particular, $I_{11}=I_1{\cos }^2\Omega t+I_2{\sin }^2\Omega t$, $I_{22}=I_1{\sin }^2\Omega t+I_2{\cos }^2\Omega t$, and $I_{12}=(I_1-I_2)\sin \Omega t\cos \Omega t$, while $I_{33}=I_3$ and $I_{13}=I_{23}=0$. Two differentiations at $t=0$ prove the inertia-tensor acceleration of a steadily rotating stellar pattern:Both the off-diagonal acceleration and the in-plane trace acceleration vanish at this instant.

Write $\mathbf{u}=\overline{\mathbf{v}}$ for the local mean streaming. The stellar kinetic-energy tensor, stellar pressure-energy tensor, and stellar potential-energy tensor have the conventionsFor an isolated self-gravitating system, $\frac{1}{2}{\ddot{I}}_{ij}=2T_{ij}+{\Pi }_{ij}+W_{ij}$. Subtracting its second diagonal component from the first givessoThis quotient requires $I_{11}\ne I_{22}$. In an axisymmetric density distribution it becomes $0/0$: rotating the pattern about its symmetry axis does not change the density, and this equation cannot determine a pattern speed. Stellar streaming need not be rigid rotation with angular frequency $\Omega$.

Next add the first two diagonal virial equations and use the third. Their acceleration terms vanish, soFor $T_{33}=0$, ${\Pi }_{33}=-W_{33}$. With the planar virial trace ${\Pi }_{\perp }={\Pi }_{11}+{\Pi }_{22}$ and $W_{\perp }=W_{11}+W_{22}$, the prescribed mass-weighted quantities satisfy ${\Pi }_{\perp }=2M{\sigma }_0^2$, $2(T_{11}+T_{22})=M{V}_0^2$, and $1-\delta =2{\Pi }_{33}/{\Pi }_{\perp }$. Therefore $W_{33}=-(1-\delta )M{\sigma }_0^2$ and the planar equation becomesEliminating $M{\sigma }_0^2$ givesThe ${\nu }_0$ in the numerator of the PDF's displayed relation is a notation slip: the velocity subsequently defined is $V_0$. The relation follows from the traces, not from replacing a galaxy by a solid rotating body.

For the axisymmetric oblate spheroid, rotational symmetry gives ${\Pi }_{11}={\Pi }_{22}=M{\sigma }_0^2$, ${\Pi }_{33}=M{\sigma }_0^2(1-\delta )$ and zero integrated off-diagonal random stresses. A line-of-sight unit vector can be chosen as $\mathbf{n}=(\sin i,0,\cos i)$, with $i=0$ face-on. Contracting the random stress tensor with this vector gives the inclination projection of an axisymmetric stellar velocity tensor:This is a mass-weighted random velocity dispersion after subtracting streaming, with consistent aperture weights. Along the projected major axis, azimuthal streaming lies along the direction whose line-of-sight projection is $\sin i$; a rotation amplitude $V_0$ consequently projects asThere is a necessary observational convention here. The globally defined $V_0^2=2(T_{11}+T_{22})/M$ is a streaming mean square. The displayed amplitude relation holds if the rotation curve is constant, or if the measured rotation-amplitude statistic is weighted consistently to represent this intrinsic quantity. Axisymmetry alone does not equate an arbitrary observed peak rotation to the global $V_0$. If $V_r$ instead means the full-aperture line-of-sight streaming root-mean-square, azimuthal averaging yieldsand the corresponding root-mean-square is $V_0\sin i/\sqrt{2}$. Keeping amplitude and root-mean-square distinct prevents a spurious factor of two in the kinetic energy.

Finally, let the intrinsic equatorial and polar semiaxes be $a$ and $c$, and put $q_t=c/a=1-{ϵ}_t$. An intrinsic boundary has $(x^2+y^2)/a^2+z^2/c^2=1$. Its sky minor-axis coordinate is $Y=y\cos i-z\sin i$. Eliminating the line-of-sight coordinate, or maximizing $Y$ on this ellipse, gives the projected minor semiaxis $\sqrt{a^2{\cos }^{2}i+c^2{\sin }^{2}i}$; the major semiaxis remains $a$. Thus the projected axis ratio of an oblate spheroid isSubtracting from unity provesThe face-on limit is round with zero projected rotation amplitude; the edge-on limit reveals the intrinsic flattening and in-plane random velocity dispersion. These limits check both the geometry and the convention for inclination.