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Use a stationary, non-streaming tracer population with spherical symmetry in its full velocity distribution. In particular, the mean velocities vanish, the mixed covariances vanish, and the two tangential components have equal velocity dispersions. WriteHere ${\sigma }_t^2$ is the variance of one tangential component. In the alternative convention using the sum of both tangential variances, the same velocity-anisotropy parameter is $1-({\sigma }_{\theta }^2+{\sigma }_{\varphi }^2)/(2{\sigma }_r^2)$. These definitions agree; they must not be mixed by inserting an extra factor of two.

The time derivative and mean-velocity terms in the supplied Jeans equation vanish, as do its angular mixed-moment terms. Its remaining geometric term is $2\nu ({\sigma }_r^2-{\sigma }_t^2)/r$. Thus the Spherical Jeans equation becomesThe no-streaming assumption excludes a steady radial flow; stationarity of the density alone would not remove its advective terms.

By the shell theorem, the gravitational potential of a spherical mass distribution satisfies $d\Phi /dr=GM(<r)/r^2$. Expanding the derivative above, dividing by $\nu$, and using logarithmic derivatives givesThe tracer density $\nu$ need not be the density of the gravitating matter. The circular speed is $V_c^2=rd\Phi /dr=GM(<r)/r$, whether or not the tracers themselves follow circular orbits.

Now let $\beta$ be constant, $\nu ={\nu }_0(r/r_0{)}^{-\gamma }$, and $V_c^2=V_0^2(r/r_0{)}^{2\alpha }$. With $y={\sigma }_r^2$, the Spherical Jeans equation readsFor a power-law ansatz $y=K{V}_c^2$, $y^{\prime }=2\alpha y/r$, so $(\gamma -2\beta -2\alpha )y=V_c^2$. Therefore the scale-free spherical Jeans solution isThe positivity condition is necessary for this particular solution to be a physical velocity dispersion.

There is a boundary condition implicit in selecting this solution. The integrating factor $r^{2\beta -\gamma }$ gives, when $D=\gamma -2\beta -2\alpha \ne 0$,The homogeneous term represents an additional radial-pressure boundary contribution. It is removed, for example, by requiring $r^{2\beta }\nu {\sigma }_r^2\to 0$ at infinity. Under that condition,and convergence requires $D>0$, giving the boxed result. If $D=0$, the solution instead has the logarithmic formso the quoted constant proportionality does not apply. A finite outer boundary can retain a homogeneous term even when $D\ne 0$.

To derive the heliocentric projection of a spherical velocity dispersion, put the Sun at vector ${\mathbf{R}}_{\odot }$ of length $R_{\odot }$, and a tracer at $\mathbf{r}=r\hat{\mathbf{r}}$. Its heliocentric line of sight isIf $\chi$ is the angle between $\hat{\mathbf{\ell }}$ and the Galactocentric radial direction, thenAfter correction for the Sun's motion, the intrinsic covariance matrix has eigenvalues ${\sigma }_r^2,{\sigma }_t^2,{\sigma }_t^2$ in the local spherical basis. Projection onto the line of sight therefore givesAt fixed $r$, ${\sigma }_r$ and $\beta$ are constant across the spherical shell. Averaging its squared line of sight velocity, with zero mean, givesFor a uniformly sampled Galactocentric shell, the solid angle measure makes $u$ uniform on $[-1,1]$, giving the geometric correction for heliocentric velocity dispersionTo evaluate it, set $a=r^2+R_{\odot }^2$ and $b=2r{R}_{\odot }$, and divide the integrand:The odd term integrates to zero, while ${\int }_{-1}^{1}du/(a-bu)=b^{-1}\ln [(a+b)/(a-b)]$. For $r>0$, $R_{\odot }>0$, and $r\ne R_{\odot }$, this givesThe removable limit at $r=R_{\odot }$ is $A=1/2$. As $r/R_{\odot }\to \infty$, $A\simeq 2R_{\odot }^2/(3r^2)$, while as $r/R_{\odot }\to 0$, $A\to 2/3$. If $R_{\odot }=0$, $A=0$ and the observed velocity dispersion is exactly radial. For isotropic velocities, $\beta =0$, the correction vanishes at every radius.

The averaging measure matters. If directions are instead uniform in heliocentric solid angle at a fixed Galactocentric radius $r>R_{\odot }$, let $\vartheta$ be the angle between the line of sight and ${\mathbf{R}}_{\odot }$. Geometry gives ${\sin }^2\chi =(R_{\odot }/r{)}^2{\sin }^2\vartheta$, so that particular weighting gives $A=2R_{\odot }^2/(3r^2)$ exactly. The logarithmic expression above uses uniform tracer weighting on a Galactocentric shell. For an observational sample, use its actual angular selection weights in $A=⟨{\sin }^2\chi ⟩$; the boxed dispersion relation remains valid.

Write the position of a mass element relative to the center of mass as $\mathbf{r}$, so its inertial-frame velocity is $\dot{\mathbf{X}}+\mathbf{\omega }\times \mathbf{r}$. The center-of-mass identity $\int \mathbf{r}dm=0$ removes all cross terms. Define the inertia tensor about the center of mass byThe total angular momentum about the origin and kinetic energy are then

Let $\rho (r)=C{r}^{-1}\sin (\pi r/R)$. Spherical symmetry makes $I=I_{0}1$. The mass normalization isso $C=M/(4R^2)$. A moment about any diameter isTherefore

In spherical symmetry, the velocity is radial and constant over each sphere. HenceApply the divergence theorem to the fluid shell between radii $R_1$ and $R_2$. Since the flow is incompressible,Thus $Q$ is independent of $R$ and is a function of time alone. This is the flux law for spherically symmetric incompressible radial flow.

Let $u(r)>0$ denote the inward radial velocity. Steady spherical symmetry and mass conservation giveThe radial Euler momentum equation for the globally isothermal equation of state $p=c_s^2\rho$ isThe logarithmic derivative of mass conservation is ${\rho }^{\prime }/\rho =-u^{\prime }/u-2/r$. Substitution gives the Isothermal Bondi equationAt a smooth sonic point, $u=c_s$ makes the coefficient of $u^{\prime }$ vanish, so the right-hand side must vanish too. ThereforeThe solution that crosses this critical point of the isothermal Bondi equation continuously is the transonic accretion solution.