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Steady mass conservation with inward-positive accretion rate givesWrite the specific angular momentum as $l=R^2\Omega$. Multiplying the azimuthal equation by $2\pi R$ and using the mass equation shows that the sum of advected and viscous angular-momentum flux is constant:The right-hand side implements the zero-torque inner boundary condition. For a Keplerian accretion disk, $l\propto R^{1/2}$ and $R^{3}d\Omega /dR=-(3/2)l$. ThereforeandFar outside the inner edge, $\Sigma \simeq \dot{m}/(3\pi \nu )$ and $u_R\simeq -3\nu /(2R)$.

Denote the four terms byso the radial or poloidal momentum equation is $I=P+G+C$.

i) In a thin Keplerian accretion disk, radial inertia and pressure are higher-order in $H/R$, leaving $G+C\simeq 0$.

ii) In a nearly static stellar atmosphere, $I=C=0$ and hydrostatic pressure balance gives $P+G=0$.

iii) In pressureless gravitational collapse, rotation and pressure are negligible, so $I=G$; this is free fall.

iv) A slim accretion disk retains radial inertia and radial pressure together with gravity and centrifugal support, so all four terms generally survive: $I=P+G+C$.

v) A stationary geometrically thick disk or torus has negligible poloidal inertia but order-one pressure support, giving $P+G+C=0$.

vi) Nonrotating Bondi accretion has $C=0$ and $I=P+G$. In a highly supersonic Bondi--Hoyle limit the pressure term is also negligible, reducing this to ballistic $I=G$.

vii) A sub-Keplerian advection-dominated accretion flow has significant pressure support and radial inflow as well as rotation, so again $I=P+G+C$, with $C$ smaller than the Keplerian value and the remaining inward gravity balanced by $P$ and $I$.

Take $\dot{M}>0$ for inward accretion, so the outward radial mass flux is $F=-\dot{M}$. Since $F=-(dh/dr{)}^{-1}d\mathcal{G}/dr$ and $\mathcal{G}(r_{\mathrm{in}})=0$,Using $\mathcal{G}=-2\pi \bar{\nu }\Sigma r^{3}d\Omega /dr$ and $x=r/r_S$ givesThus $\boxed{A=3\sqrt{3}/2}$. Far from the hole this approaches the Keplerian accretion disk result $\bar{\nu }\Sigma =\dot{M}[1-(r_{\mathrm{in}}/r{)}^{1/2}]/(3\pi )$, although the pseudo-Newtonian boundary factor retains a different finite-radius shape.

For inward accretion rate $\dot{M}>0$, the steady conservation of angular momentum equation is $\mathcal{G}=\dot{M}(h-h_{\mathrm{in}})$ when the inner torque vanishes. The viscous torque in an accretion disk is $\mathcal{G}=2\pi qh\bar{\nu }\Sigma$, where $h=r^2\Omega$ and $q$ is the orbital shear parameter. ThusThe familiar Keplerian accretion disk follows when $q=3/2$ and $h\propto r^{1/2}$. The general form also applies to the non-Keplerian Paczyński-Wiita circular orbit rotation law.

For a steady Keplerian accretion disk with inward accretion rate $\dot{M}$ and vanishing inner torque, the outward angular-momentum torque is $\mathcal{G}=\dot{M}(\sqrt{GMr}-\sqrt{GM{r}_{\mathrm{in}}})$. It transports energy outward at rate $\Omega \mathcal{G}$. The binding-energy acquisition rate between radii $b$ and $a$ is $B_{a,b}=GM\dot{M}(1/a-1/b)/2$. Integrating the vertical viscous dissipation and then the annulus givesSince $\Omega \mathcal{G}=GM\dot{M}[r^{-1}-\sqrt{r_{\mathrm{in}}}{r}^{-3/2}]$, subtraction proves the displayed energy balance. Far from the inner edge $D_{a,b}\simeq 3B_{a,b}$, whereas the boundary torque powers vanish for the complete disk and $D=B=GM\dot{M}/(2r_{\mathrm{in}})$. Local dissipation is therefore not supplied solely by local changes in orbital binding energy.

In a steady Keplerian accretion disk with $\Sigma ={\Sigma }_0{R}^{-2}$, $\nu ={\nu }_0{R}^2$, and tracer diffusivity $k=\zeta \nu$, the trace passive scalar concentration obeys the displayed constant-coefficient advection-diffusion equation in logarithmic radius. Contaminant mass is proportional to $CdY$, because $2\pi R\sigma dR=2\pi {\Sigma }_{0}CdY$. A radial impulse $C_0\delta (R-R_0)$ therefore has logarithmic amplitude $C_0/R_0$ and evolves by the Gaussian heat kernel centered at $\ln R_0-vt$. The full-line solution idealizes radii far from the disk boundaries.