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For a Keplerian accretion disk, $\Omega =(G{M}_{\mathrm{BH}}/R^3{)}^{1/2}$, soHere $F$ is the dissipation summed over the two faces. The total luminosity isThis is the binding energy per unit mass of a circular orbit at the inner edge, multiplied by the accretion rate. The zero-torque inner boundary condition ensures that no additional mechanical work enters from smaller radii.

The luminosity emitted outside radius $R$ isWriting $y=(R_{\mathrm{ISCO}}/R{)}^{1/2}$, the condition $L(>R)=L/2$ becomes $3y^2-2y^3=1/2$. Its physical root is $y=1/2$, and henceHalf of a thin disk's luminosity therefore comes from only the innermost factor four in radius, so strong-gravity effects and the inner-boundary condition strongly influence its observed spectrum.

Let $l(R)=R^2\Omega =\sqrt{GMR}$ be the Keplerian specific angular momentum. In a source-free steady interval, the sum of advected and viscously transported angular momentum is constant:Inside the injection radius, $\dot{m}={\dot{m}}_0$. The zero-torque inner boundary condition at $R_{\mathrm{ISCO}}$ sets $C={\dot{m}}_0{l}_{\mathrm{ISCO}}$, and therefore

Outside $R_0$ there is no net mass flow in the stated steady distribution, but it must carry outward the angular momentum deposited by matter moving from $R_0$ to the ISCO. Its constant viscous torque in an accretion disk is thereforeThusThe two expressions agree at $R_0$; the jump in mass flux there is exactly the injected rate.

An optically thick annulus radiates as a blackbody from both faces, soFor a steady Keplerian accretion disk with a zero-torque inner boundary condition,ConsequentlyandAway from the inner edge, $T\propto R^{-3/4}$.

Ignoring inclination and distance factors, the multitemperature blackbody disk spectrum isSet $x=h\bar{\nu }/(k_{B}T)$. Since $T\propto R^{-3/4}$, $RdR\propto {\bar{\nu }}^{-8/3}{x}^{5/3}dx$, whereas the Planck function contributes ${\bar{\nu }}^3/(e^x-1)$. In the stated intermediate-frequency range the radial endpoints become $0$ and $\infty$, leaving a frequency-independent convergent integral. Therefore

Write $y=\nu \Sigma$. A steady inward accretion rate has $F_M=-\dot{M}$, so the integrated angular-momentum equation isFor $\beta =0$, $\mathcal{T}=0$ and thereforeafter imposing the zero-torque inner boundary condition at $r_{\mathrm{in}}$. In a Keplerian accretion disk, $h\propto r^{1/2}$ and $\mathcal{G}=3\pi hy$, so

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 $u_R\simeq -3\nu /(2R)$.

The steady disk should end at the innermost stable circular orbit, $r_{\mathrm{in}}=3r_S$. Inside it, matter plunges inward on an orbital timescale and cannot maintain the nearly circular shear flow assumed by the viscous-disk equations. If the plunging region communicates negligible stress back across the ISCO, the natural boundary condition is the zero-torque inner boundary condition

For a Keplerian disk with evolution ${\partial }_t\Sigma =(2\pi r{)}^{-1}{\partial }_r\mathcal{F}+S$ and $\mathcal{F}=6\pi r^{1/2}{\partial }_r(r^{1/2}\bar{\nu }\Sigma )$, take zero viscous torque at $r_1>0$ and zero flux outside the source region at $r_2>r_1$. The steady state gives ${\mathcal{F}}^{\prime }=-2\pi rS$, and its outer condition gives the displayed flux. Integrating once more from the zero-torque inner boundary condition yieldsThe equation determines the viscosity-density product, so a thermodynamic viscosity closure can be imposed afterward. For nonnegative integrable $S$, it is nonnegative and the inner accretion rate is exactly the total supplied mass per unit time.

For a steady accretion disk with arbitrary rotation law and a zero-torque inner boundary condition at $R=3R_S$, the outward viscous torque in an accretion disk is $\mathcal{G}=\dot{M}(j-j_{\mathrm{in}})$. The dissipation per unit planar area summed over both disk faces is $D=-\mathcal{G}{\Omega }^{\prime }/(2\pi R)=\nu \Sigma (R{\Omega }^{\prime }{)}^2$. Integrating $2\pi RD$ gives $L=\dot{M}GM/(8R_S)=\dot{M}{c}^2/16$ in the Paczyński-Wiita potential. The radiative flux from one face is $D/2$, preventing a factor-of-two error in the luminosity.