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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.

For $h=({\sin }^2\theta +q^{-2}{\cos }^2\theta {)}^{-\gamma /2}$, separation in spherical coordinates makes both stellar potential-energy tensor integrals contain $\int r^{3-\gamma }{\Phi }^{\prime }(r)dr$. Their ratio is the displayed positive finite angular expression. If the radial integrals diverge, identical spherical cutoffs give this same ratio of force moments, but there is no finite global pressure-energy ratio without further hypotheses. A finite-aperture tensor virial theorem retains its surface tensor.