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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.

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.

The planar trace of the stellar pressure-energy tensor is the sum of its two Cartesian in-plane components. For an axisymmetric nonstreaming population it equals $\int \rho (⟨v_R^2⟩+⟨v_{\varphi }^2⟩)d^{3}x$, not merely the cylindrical radial second moment. A spherical isotropic population has ${\Pi }_{\perp }=2{\Pi }_{zz}$.

This tensor records the ordered kinetic energy of the streaming velocity $u_i=⟨v_i⟩$. It excludes random motion; its vanishing is compatible with a nonzero stellar pressure-energy tensor.