Viscous torque in an accretion disk ($\mathcal{G}$)

For vertically integrated kinematic viscosity $\bar{\nu }$, the viscous torque across a cylinder of radius $r$ is

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.

A steady Keplerian disk can transport angular momentum by viscous stress with zero mass flux. The radial drift vanishes, but the torque is constant and nonzero. An inner source and outer sink must supply its boundary angular-momentum exchange. This is distinct from a decretion flow, which transports mass outward.

Integrating the Keplerian accretion disk diffusion equation gives $dJ/dt=6\pi \sqrt{G{M}_{\star }}[r^{3/2}{\partial }_r(\bar{\nu }\Sigma ){]}_{\mathrm{inner}}^{\mathrm{outer}}$ when the angular-momentum integral and boundary limits are legitimate. An inner zero-torque inner boundary condition alone does not eliminate the outer angular-momentum flux. For divergent total $J$, use finite-radius cutoffs before making a conservation claim.

The zero-torque inner boundary condition sets the viscous torque to zero where nearly circular disk flow joins a plunging or stress-free region.

For constant kinematic viscosity $\nu$, use the phenomenological viscous torque in an accretion disk, specific angular momentum $h=rc\sqrt{m/(r-3m)}$, and $h_0=\sqrt{12}mc$ at the innermost stable circular orbit. Neglect radial kinetic motion in the clock relation and neglect differential light travel time. The Schwarzschild time to drift formally from $r_1\ge 6m$ to $6m$ is the displayed expression, withIndeed $\mathcal{T}(2)=0$ and ${\mathcal{T}}^{\prime }(X)=6X[X-2\sqrt{X-1}]/(X-1)\ge 0$ for $X\ge 2$. The zero-torque inner boundary condition makes the formal drift speed diverge as $r\downarrow 6m$, so the nearly circular approximation must give way to the plunging region of a black-hole accretion disk sufficiently close to that orbit. This formula describes the stipulated approximation, not an exact relativistic flow solution.

The plunging region begins near the innermost stable circular orbit, where nearly circular motion becomes unstable and radial inflow becomes rapid. If the inflow time is shorter than the stress-transport time, the gas approximately conserves its specific angular momentum there.

A thin disk with alpha disk viscosity $\nu \sim \alpha H^2\Omega$ has a viscous timescale much longer than its dynamical infall time in the plunging region of a black-hole accretion disk. At fixed nonzero accretion rate, rapid inflow reduces its surface density of a disk. The resulting viscous torque in an accretion disk is smaller by the displayed factor than in the slowly drifting circular disk. This estimate concerns radii outside the narrow matching layers.