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Black-hole spin is the angular momentum of a black hole, commonly expressed by the dimensionless Kerr parameter $a_{\ast }=cJ/(G{M}^2)$. It changes the innermost stable circular orbit, accretion efficiency, and relativistic frame dragging.

Coherent prograde gas accretion increases the angular momentum of a black hole. For a zero-torque thin disc, the rest mass $d{m}_0$ entering from the innermost stable circular orbit changes the hole by $dM=E_{\mathrm{ISCO}}d{m}_0$ and $dJ={\ell }_{\mathrm{ISCO}}d{m}_0$. Its dimensionless spin obeys $d{a}_{\ast }/d\log M=c{\ell }_{\mathrm{ISCO}}/(GM{E}_{\mathrm{ISCO}})-2a_{\ast }$. Starting from zero spin and neglecting captured radiation, the formal extremal endpoint has $M_f=\sqrt{6}{M}_i$ and requires rest-mass supply approximately $1.8464M_i$. Photon capture leads instead to the Thorne spin limit.

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.

Let $r_S=2GM/c^2$. In the Paczyński-Wiita potential, circular orbit balance and the radial epicyclic frequency formula giveThus the innermost stable circular orbit lies at $r=3r_S$. Its specific orbital energy is $-c^2/16$, giving a zero-torque radiative efficiency of black-hole accretion of $1/16$. The model reproduces the Schwarzschild inner orbit exactly but approximates its relativistic energy and frequency. Abramowicz's derivation explains the potential's relation to the Schwarzschild effective potential.

Let $\mathcal{G}$ be the outward viscous torque in an accretion disk. For a declining angular velocity, the vertically integrated viscous shear stress givesIn a steady accretion disk without radiated angular momentum, the net inward angular-momentum flux $Fh-\mathcal{G}$ is constant and equals the swallowed flux $F{h}_0$. Thus conservation of angular momentum gives $F(h-h_0)=\mathcal{G}$. Substituting $F=2\pi r\Sigma u$ cancels the surface density and provesThis uses the stipulated phenomenological viscous shear law; it does not replace it by a different fully relativistic stress prescription.

For the orbital calculation let a dot denote a derivative with respect to proper time. Substituting the geodesic energy and specific angular momentum first integrals into the timelike normalization in Schwarzschild spacetime givesDifferentiate with $h$ and $ϵ$ constant along the geodesic. Where $\dot{r}\ne 0$, division by $2\dot{r}$ givesThis equation also holds at turning points and on circular orbits. To justify that without dividing by zero, put $A=1-2m/r$ in the equatorial geodesic Lagrangian $L=(A{c}^2{\dot{t}}^2-A^{-1}{\dot{r}}^2-r^2{\dot{\varphi }}^2)/2$. Its radial Euler-Lagrange equation is $\ddot{r}=(A^{\prime }/2A){\dot{r}}^2-A{A}^{\prime }{c}^2{\dot{t}}^2/2+Ar{\dot{\varphi }}^2$. Substitution of the timelike normalization cancels the ${\dot{r}}^2$ terms and yields $\ddot{r}=-A^{\prime }{c}^2/2+(A/r-A^{\prime }/2)h^2/r^2$, exactly the displayed radial equation, including when $\dot{r}=0$.

A circular orbit requires $\ddot{r}=\dot{r}=0$. Thereforechoosing the positive sense of rotation. Differentiating the square is simpler than differentiating $h$:It is negative for $3m<r<6m$ and positive for $r>6m$, so the global minimum over timelike circular orbits isAs a stability check, the second radial derivative of $W=(1-2m/r)(c^2+h^2/r^2)$ at a circular orbit, holding $h$ fixed for the perturbation, is $W^{\prime \prime }=2m{c}^2(r-6m)/[r^3(r-3m)]$. Thus the exterior branch is stable, and $6m$ is the innermost stable circular orbit. It is the natural matching radius for the idealized zero-torque inner boundary condition.

Write $s=(r^2+z^2{)}^{1/2}$. The Paczyński-Wiita potential is a spherical Newtonian gravitational potential with ${\Phi }^{\prime }(s)=GM/(s-a{)}^2$ and ${\Phi }^{\prime \prime }(s)=-2GM/(s-a{)}^3$. At $z=0$, $s=r$, soSubstituting into the orbital formulas givesChoosing positive orbital and vertical frequencies, $\Omega ={\Omega }_z=\sqrt{GM}/[\sqrt{r}(r-a)]$. For $r>3a$, $\kappa =\Omega \sqrt{(r-3a)/(r-a)}$. For $a<r<3a$, ${\kappa }^2<0$ and radial disturbances grow at rate $\sqrt{-{\kappa }^2}$, even though the vertical motion remains stable. Thus circular orbits are radially unstable for $a<r<3a$, and $r=3a$ is the marginal orbit. This recovers the innermost stable circular orbit of the Paczyński-Wiita circular orbit model.

For an equatorial circular orbit in the Paczyński-Wiita potential, radial force balance givesIts specific angular momentum is $j=R^2\Omega =\sqrt{GM}{R}^{3/2}/(R-R_S)$. Perturbing the circular radius at fixed $j$ gives the radial epicyclic frequency squared ${\kappa }^2=R^{-3}d(j^2)/dR$. ThereforeCircular motion is radially stable for $R>3R_S$, marginal at $3R_S$, and unstable between $R_S$ and $3R_S$. Thus the inner edge of the nearly circular thin accretion disk is $R_{\mathrm{in}}=3R_S$, joining a plunging region of a black-hole accretion disk inside the innermost stable circular orbit. This conclusion concerns the circular thin-disk model, not an assertion that all gas disappears inside this radius.

Take the inward mass accretion rate $\dot{M}$ as positive, and write the positive outward viscous torque in an accretion disk asSteady conservation of angular momentum makes the net inward flux $C=\dot{M}j-\mathcal{G}$ independent of radius. Impose a zero-torque inner boundary condition at $3R_S$, appropriate to an approximately stress-free transition to plunging motion. ThenThis is the steady accretion disk with arbitrary rotation law angular-momentum balance.

The local viscous dissipation integrated through the full disk thickness is $D=\nu \Sigma (R{\Omega }^{\prime }{)}^2=-\mathcal{G}{\Omega }^{\prime }/(2\pi R)$. It is the power per unit planar area summed over both radiating faces. Sincesubstitution gives the two-face dissipation of a Paczyński-Wiita disk:In particular $D$ vanishes at the stress-free inner edge and is nonnegative outside it because $j$ increases there. A one-face radiative flux is $D/2$; the total accretion luminosity is therefore $L={\int }_{3R_S}^{\infty }2\pi RDdR$, with no additional factor of two.

For a direct evaluation put $x=R/R_S$. ThenThe first integral is $7/4$, using $(3x-1)/(x-1{)}^3=3/(x-1{)}^2+2/(x-1{)}^3$. For the second,so it equals $2/[\sqrt{3}(3-1)]=1/\sqrt{3}$. HenceThis is the radiative efficiency of black-hole accretion within this pseudo-Newtonian model.

Independently, the specific orbital energy of a circular particle isThus $E(\infty )=0$ and $E(3R_S)=-GM/(8R_S)=-c^2/16$. The energy released per unit accreted mass is exactly $c^2/16$, confirming the integrated accretion luminosity. The Paczyński-Wiita circular orbit reproduces the Schwarzschild inner radius but gives an approximate binding energy; the efficiency derived here is for the stated potential.

At a circular orbit, $V^{\prime }=0$. Differentiating the Schwarzschild effective potential at fixed $h$ givesA timelike circular orbit requires $r>3r_s/2$. Evaluating the second derivative after this substitution givesA stable nondegenerate minimum therefore requires $r>3r_s$. The boundary $r=3r_s$ is the marginal innermost stable circular orbit, not a strict stable minimum.

The specific orbital energy released in a black-hole accretion flow reaching an inner radius $r_{\mathrm{in}}$ is of order $GM/r_{\mathrm{in}}$. For a circular Keplerian orbit, the binding energy is more precisely $GM/(2r_{\mathrm{in}})$, with the other half of the potential-energy decrease retained as orbital kinetic energy. Multiplying by the mass accretion rate gives the accretion luminosity scaleFor an inner radius of order the Schwarzschild radius, $r_{\mathrm{in}}\sim 2GM/c^2$, this becomesThis is an energy-release scaling, not unit radiative efficiency of black-hole accretion. The actual inner orbit, stresses and advected energy determine the coefficient and the emitted luminosity; the innermost stable circular orbit of a Schwarzschild black hole lies outside its horizon.

In a steady state, mass conservation makes $\mathcal{F}$ independent of radius. Taking the accretion rate $\dot{M}>0$ for inward flow gives $\mathcal{F}=-\dot{M}$. The conservation of angular momentum equation then makes $-\dot{M}h+\mathcal{G}$ constant. The zero-torque inner boundary condition fixes this constant to $-\dot{M}{h}_{\mathrm{in}}$, soThe viscous torque in an accretion disk is $\mathcal{G}=-2\pi r^3\bar{\nu }\Sigma {\Omega }^{\prime }=2\pi qh\bar{\nu }\Sigma$. Combining these expressions gives the steady accretion disk with arbitrary rotation lawFor $x=r/R_g$, the Paczyński-Wiita circular orbit formulas giveConsequentlyIn particular, $f(6)=0$ and $f\to 1$ far from the black hole, recovering the outer Keplerian accretion disk.

Inside the innermost stable circular orbit, the gas enters the plunging region of a black-hole accretion disk. Its rapid inward motion leaves little time for stresses to exchange angular momentum, motivating the zero-torque inner boundary condition. This is a thin-disc approximation; a strong magnetic stress could change it.

Matter supplied from very large radius has negligible specific orbital energy, while matter crossing the inner edge carries ${\epsilon }_{\mathrm{in}}=-\eta c^2$. With zero inner torque, no energy is supplied by a stress at that edge. If the heat released outside it escapes by radiative transfer, rather than being lost through inward advection, the integrated conservation of energy balance gives

The usual Pressure-opacity zones of a Shakura--Sunyaev thin disk, in order of increasing cylindrical radius, are an inner region dominated by radiation pressure and electron-scattering opacity, a middle region dominated by gas pressure and electron-scattering opacity, and an outer region dominated by gas pressure and the Kramers' opacity law, often approximated by free-free opacity. The boundaries depend on mass, mass accretion rate, and the alpha disk viscosity parameter. At sufficiently large radius, accretion-disk self-gravity or changes in ionization can invalidate this three-zone model.

For the inner region writeHere $F$ is the standard thin-disk dissipation flux through one face, and $R_{\ast }$ is the innermost stable circular orbit treated as a zero-torque inner boundary. The given factor $1/2$ in the dissipation rate is appropriate to one face. For Keplerian rotation, it is $(9/8)\nu \Sigma {\Omega }_K^2$.

Let $H$ denote the disk scale height or vertical half-thickness, not the full thickness. In vertical hydrostatic equilibrium, the leading vertical acceleration is ${\Omega }_K^{2}z$, so a one-zone estimate givesIn the inner zone, $P_c\simeq P_{\mathrm{rad},\mathrm{c}}=a_{\mathrm{rad}}{T}_c^4/3$. For an optically thick medium with constant electron-scattering opacity, radiative diffusion gives the emergent one-face fluxEquivalently, combine $F=-c({\kappa }_{\mathrm{es}}\rho {)}^{-1}d{P}_{\mathrm{rad}}/dz$ with vertical pressure balance. Equating the radiative flux to the dissipated flux yields the radiation-supported inner-disk height:Order-unity vertical-profile factors depend on the precise height convention. The central mass cancels at fixed physical mass accretion rate, because both vertical gravity and local dissipation are proportional to $G{M}_{\mathrm{BH}}/R^3$.

If ${\dot{m}}_{\mathrm{Edd}}=L_{\mathrm{Edd},\mathrm{es}}/({\eta }_0{c}^2)$ and $r_g=G{M}_{\mathrm{BH}}/c^2$, this can also be writtenThe radiation-supported height rises from the formal zero at $R_{\ast }$ toward the approximately constant value $H_0=3{\kappa }_{\mathrm{es}}\dot{m}/(8\pi c)$. That zero is not a reliable description of the physical plunging region: the radiation-dominated approximation and the Newtonian zero-torque formula fail very close to the inner edge.

Beyond the inner zone, the height grows slowly with radius rather than staying on that plateau forever. Far from the inner boundary, the standard gas-pressure scalings give $H\propto R^{21/20}$ in the electron-scattering zone and $H\propto R^{9/8}$ in the Kramers zone. These follow from $H\propto T_c^{1/2}{R}^{3/2}$ with $T_c\propto R^{-9/10}$ and $R^{-3/4}$, respectively. A schematic profile is

vertical half-thickness H
  ^
  |                                               / outer gas/Kramers
  |                                          ____/
  |                                     ____/ middle gas/scattering
  |              ______________________/
  |          ___/ inner radiation/scattering plateau
  |       __/
  |     _/
  +----|----------------------------------------------> radius R
      R_*              zone boundaries are schematic

The height formula also tests the thin disk assumption. For a nonrotating hole with $R_{\ast }=6r_g$, the largest $H/R$ in this inner approximation occurs at $R=(9/4)R_{\ast }$. At $\dot{m}={\dot{m}}_{\mathrm{Edd}}$ and ${\eta }_0=0.1$, it is about $0.37$, so a disc this close to the conventional Eddington accretion rate is only marginally thin. Slim accretion disk effects and advected heat can then matter.

Interpret $\dot{m}$ as the rest-mass supply rate through a coherent disc; the hole's gravitating mass grows more slowly because radiation carries away binding energy. Initially the hole is a Schwarzschild black hole, whose innermost stable circular orbit is at $R_{\ast }=6G{M}_0/c^2$. The relativistic orbital constants there areThey follow by extremizing the timelike geodesic effective potential for a circular orbit and imposing marginal stability. Assume negligible torque inside the inner edge, so the gas carries this specific angular momentum into the hole. Its initial spin-up torque isfor $M_0=10^8{M}_{\odot }$ and $\dot{m}=0.1M_{\odot }\mathrm{y}{\mathrm{r}}^{-1}$. In SI units this is $9.7\times 10^{41}\mathrm{kg}{\mathrm{m}}^2{\mathrm{s}}^{-2}$. A purely Newtonian estimate $\ell \sim \sqrt{G{M}_0{R}_{\ast }}$ gives a similar order of magnitude, but the relativistic value is appropriate at the innermost stable circular orbit.

The maximal Kerr black hole angular momentum at the initial mass is $J_{\max ,0}=G{M}_0^2/c\simeq 8.8\times 10^{64}\mathrm{g}\mathrm{c}{\mathrm{m}}^2{\mathrm{s}}^{-1}$. Holding both that mass and the initial torque fixed would giveInitially ${\dot{M}}_{\mathrm{BH}}=E_{\mathrm{ISCO}}\dot{m}\simeq 0.094M_{\odot }\mathrm{y}{\mathrm{r}}^{-1}$, so the corresponding frozen-coefficient estimate of gravitating mass gain is about $2.7\times 10^7{M}_{\odot }$. These are useful initial spin-growth scales, not a self-consistent time and mass increase for reaching extremality: the target angular momentum itself grows as $M^2$, and the inner orbit changes as the spin increases.

A consistent endpoint estimate is obtained from black-hole spin-up by coherent accretion. Let $a_{\ast }=cJ/(G{M}^2)$ and $r=R_{\mathrm{ISCO}}{c}^2/(GM)$, where $M$ is the current mass. On the prograde Kerr black hole inner-orbit branch, $1\le r\le 6$, the orbital constants may be writtenIgnoring photon capture, an accreted rest mass $d{m}_0$ changes the hole by $dM=Ed{m}_0$ and $dJ=\ell d{m}_0$. Differentiating the spin definition givesSubstitution of the orbital constants gives $dr/d\log M=-2r$, henceAt the formal extremal limit $r=1$, the result isFor constant rest-mass supply, use $r=6(M_0/M{)}^2$ to integrate the elapsed time:Thus the idealized coherent-disc endpoint isIf the quoted $0.1M_{\odot }\mathrm{y}{\mathrm{r}}^{-1}$ were instead the actual gravitating mass-growth rate, the corresponding endpoint time would be $\Delta M_{\mathrm{BH}}/{\dot{M}}_{\mathrm{BH}}\simeq 1.45\times 10^9\mathrm{yr}$. The two rate conventions cannot be interchanged.

The exact extremal endpoint is an idealization. Capture of disc photons exerts a counteracting torque and gives the Thorne spin limit, $a_{\ast }\simeq 0.998$, for the standard radiatively efficient thin-disc model. Therefore physical disc accretion approaches a large spin below unity rather than producing an exactly extremal hole. Thorne's original spin-evolution calculation describes that correction. The independently integrated expressions above explain why the frozen-mass estimate is too short for the ideal endpoint.

The qualitative conclusion is robust: a hole that gains most of its mass through a persistent, coherently corotating Shakura--Sunyaev thin disk is expected to have substantial black-hole spin. Very low spins require a different angular-momentum history, such as short episodes with changing orientations, counterrotating accretion, or mergers. Coherent feeding, not merely the existence of a disc during each episode, is the decisive assumption.

The Soltan argument offers a population-level observational test. From the bolometric luminosity function of active galactic nuclei, infer the total energy radiated per comoving volume over cosmic history,Compare it with the increase $\Delta {\rho }_{\mathrm{BH}}$ in supermassive black hole mass density attributed to radiative accretion. For an effective radiative efficiency of black-hole accretion ${\eta }_{\mathrm{eff}}$,In a zero-torque relativistic thin disc, $\eta =1-E_{\mathrm{ISCO}}$ rises from about $0.057$ at zero spin to $0.423$ in the photon-free prograde extremal limit. The measured effective efficiency therefore constrains population spin models under the thin-disc assumption. Soltan's original population argument explains how integrated quasar light constrains accumulated black-hole mass.

This is an accreted-rest-mass-weighted efficiency constraint, not a direct measurement of the arithmetic mean spin of every black hole. Obscured emission, bolometric corrections, seed masses, nonradiative growth, and disc orientation introduce uncertainty, and the efficiency-spin relation is nonlinear. These qualifications are essential when interpreting an inferred average spin for the whole population.

Inside the innermost stable circular orbit, nearly circular motion is unstable and gas enters the plunging region of a black-hole accretion disk. Its inflow time becomes shorter than the time on which internal stress can communicate angular momentum back to the disk, motivating the zero-torque inner boundary condition at $r_{\mathrm{ISCO}}$.

In a steady state, the given diffusion equation impliesThe constant mass supply fixes $C=\dot{M}/(6\pi )$, and a second integration givesZero torque means $\bar{\nu }\Sigma =0$ at $r=r_{\mathrm{ISCO}}$, so $C_0=r_{\mathrm{ISCO}}^{1/2}$. Thus the Keplerian accretion disk relation is

At radius $r$, radiation of luminosity $L$ gives matter of flux-mean opacity $\kappa$ the outward accelerationEquating this with the gravitational acceleration $G{M}_{\mathrm{BH}}/r^2$ gives the Eddington luminosity

Let $L=f{L}_{\mathrm{Edd}}$ and let $\mathcal{M}(r)$ be the radiative force multiplier relative to the Compton or electron-scattering acceleration. A steady spherical wind obeys mass conservation ${\dot{M}}_w=4\pi r^2\rho v$ and the radial Euler momentum equationWhen gas pressure is negligible, radiation accelerates the wind locally if $f\mathcal{M}>1$. Electron scattering alone has $\mathcal{M}\simeq 1$, so it launches a wind only near or above the Eddington ratio $f=1$. Bound-free absorption, spectral lines, or dust can give $\mathcal{M}\gg 1$ and allow a sub-Eddington source to launch a wind, provided the gas is not so highly ionized that those extra opacities disappear. A wind that has already acquired sufficient kinetic energy may coast through a region where $f\mathcal{M}<1$, but such a region cannot launch a cold wind from rest.

Write the innermost stable circular orbit asEstimating the emitting area as $4\pi R_{\mathrm{ISCO}}^2$, the Stefan–Boltzmann law givesand henceThus $T_{\mathrm{BB}}\propto M_{\mathrm{BH}}^{-1/4}$ at fixed $f$ and opacity: a larger black hole radiates from an area growing as $M_{\mathrm{BH}}^2$, faster than its Eddington luminosity grows. Prograde black-hole spin moves the ISCO inward and raises the characteristic temperature, whereas retrograde spin moves it outward and lowers the temperature. Relativistic transfer and the radial temperature profile change the numerical coefficient, not these leading scalings.

The standard Shakura--Sunyaev thin disk assumes: a steady state; axial symmetry; a geometrically thin disk $H/R\ll 1$; Newtonian, nearly Keplerian circular motion outside the inner edge; subsonic radial drift $|u_R|\ll u_{\varphi }$; vertical hydrostatic equilibrium; negligible disk self-gravity; an optically thick, locally thermal spectrum; local radiative balance between viscous heating and cooling; and a local alpha-viscosity stress with a zero-torque inner boundary near the innermost stable circular orbit. These assumptions also exclude dynamically dominant winds and large-scale external torques from the standard solution.

Start from the continuity equationIntegrate vertically, use axial symmetry and steadiness, and define the surface density of a disk $\Sigma =\int \rho dz$. Thenso the inward-positive accretion rate is

Steady mass conservation gives $\dot{M}=-2\pi R\Sigma u_R$. Substituting this into the angular-momentum equation and integrating from the innermost stable circular orbit with zero torque givesThereforeUsing $\nu =\alpha c_{s}H$, $c_s\simeq H\Omega$, $l=R{u}_{\varphi }$, and $Rd\Omega /dR$ of order $-\Omega$ givesup to the order-unity Keplerian factor $3/2$.

At the sonic transition, $|u_R|\simeq c_s\simeq (H/R)u_{\varphi }$. Hencefor a geometrically thin disk with $\alpha \ll 1$. The specific angular momentum therefore differs only fractionally from $l_{\mathrm{ISCO}}$ before the gas enters the plunging region of a black-hole accretion disk. Its much shorter inflow time then prevents appreciable viscous transport, justifying angular-momentum conservation across the ISCO and the zero-torque boundary condition.

The maximal radiative efficiency of black-hole accretion is the fraction of rest-mass energy available if all binding energy released before capture escapes as radiation. In a Newtonian disk ending at $R_{\mathrm{in}}$,because a circular orbit has specific binding energy $GM/(2R)$. In relativity, ${\eta }_{\max }=1-E_{\mathrm{ISCO}}$. Black-hole spin moves the innermost stable circular orbit inward for prograde flow and outward for retrograde flow, increasing or decreasing this maximum respectively.

Since $L=\eta \dot{M}{c}^2$, a source of fixed luminosity requires $\dot{M}=L/(\eta c^2)$, while black-hole mass grows at approximately $(1-\eta )\dot{M}$. The actual radiative efficiency of black-hole accretion can lie below the maximum when energy is advected through the horizon or carried away mechanically. A low-density, optically thin advection-dominated accretion flow stores dissipated energy in ions, while a high-rate slim accretion disk traps photons and advects their energy inward; both are radiatively inefficient flows.

At fixed radiative efficiency, the Eddington ratio implies $\dot{m}\propto f_{\mathrm{Edd}}M$. Far outside the innermost stable circular orbit, part (a) gives $\Sigma \propto \dot{m}/\nu$. The supplied viscosity law therefore gives, in physical radius,The disk mass follows by radial integration:Since the Schwarzschild radius satisfies $R_S\propto M$, replacing $R$ by $(R/R_S)R_S$ contributes another factor $M^{5/4}$. Thusso

The alpha disk relation and Keplerian angular frequency giveCombining this with $\Omega \propto M^{1/2}{R}^{-3/2}$ and the surface-density scaling from part (d), the Toomre stability criterion becomesIt decreases strictly with radius, so it crosses unity at a unique self-gravitating radius of an accretion disk. Expressing radius in units of $R_S\propto M$ givesSolving $Q(R_{\mathrm{sg}})=1$ therefore yieldsand

Let $x_{\mathrm{ISCO}}=R_{\mathrm{ISCO}}/R_S$. Requiring a non-self-gravitating annulus outside the innermost stable circular orbit gives $R_{\mathrm{sg}}>R_{\mathrm{ISCO}}$, and equality definesFor a nonspinning hole, $x_{\mathrm{ISCO}}=3$, and with $C_2\simeq 10^5$, $\alpha \simeq 0.1$, and $f_{\mathrm{Edd}}\simeq 1$, this is approximately $8\times 10^9{M}_{\odot }$, conventionally quoted as order $10^{10}{M}_{\odot }$. Above this mass the disk would become self-gravitating essentially as soon as stable circular orbits begin, so the assumed smooth Shakura--Sunyaev thin disk cannot provide a broad luminous accretion region.

The corresponding Eddington luminosity is of order $10^{48}\mathrm{erg}{\mathrm{s}}^{-1}$, comparable to the upper envelope of quasar luminosities. This supports self-gravity as one contributor to the observed luminous-mass ceiling. It is not an absolute upper bound on black-hole mass: mergers, radiatively inefficient growth, nonstandard gas supply, spin-dependent inner radii, and fragmented or episodic accretion can all build a more massive hole without maintaining this particular steady thin disk.