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Write $J_T(x)=|\det DT(x)|$, the absolute Jacobian determinant. Under the usual meaning of a $C^1$ map on domains, the injective area formula gives, for any nonnegative measurable $a$,This version of the change of variables formula does not require $DT$ to be nonsingular everywhere. A $C^1$ map is a locally Lipschitz function, and the formula applies by exhaustion if the domains are unbounded. For non-open $X$, one needs the corresponding extension or regularity interpretation of the stated $C^1$ assumption.

First suppose $f(x)=g(T(x))J_T(x)$ almost everywhere with respect to Lebesgue measure. For each bounded nonnegative measurable $a$,Taking indicator functions proves $T_{\#}\mu =\nu$.

Conversely, the area formula shows that the measure $\tilde{\mu }$ with probability density function $g(T(x))J_T(x)$ has total mass one and satisfies $T_{\#}\tilde{\mu }=\nu$. The assumed equality $T_{\#}\mu =\nu$ then implies $\mu =\tilde{\mu }$, since the measurable bijection $T$ has a measurable inverse on these domains. Equivalently, apply both pushforward measure identities to $T(A)$ for each Borel set $A\subseteq X$. Uniqueness of the Radon-Nikodym derivative yields

Interpret the invertibility assumption on $G$ as continuity and strict increase on the relevant range, so that $\nu$ is an atomless measure. Suppose a non-decreasing transport map $T$ with $T_{\#}\mu =\nu$ exists. Then $\mu$ is also an atomless measure: if $\mu (\{x\})>0$, the pushforward measure would give $\nu (\{T(x)\})\ge \mu (\{x\})>0$. Thus $F$ is continuous.

At any point $x$ where the non-decreasing representative is defined, the definition of a monotone function givesUsing the pushforward measure identity and continuity of the two cumulative distribution functions, we obtainConsequentlyThe monotone rearrangement in part (c) is optimal for the convex difference cost, so $T$ has the same cost and solves the Monge optimal transport problem. The meaningful uniqueness is up to a $\mu$-null set; arbitrary values away from the source do not affect transport or cost. Existence of the non-decreasing transport map supplies the source condition missing in part (c).

The intended monotone rearrangement isHere $G^{-1}$ is the quantile function of $\nu$, agreeing with the ordinary inverse when $G$ is continuous and strictly increasing. With an atomless measure $\mu$, its cumulative distribution function $F$ is continuous, and the probability integral transform makes $F(X)$ uniform on $(0,1)$ for $X\sim \mu$. Thus $G^{-1}(F(X))\sim \nu$. The one-dimensional monotone rearrangement theorem says this transport map minimizes the cost $d(x-y)$ for convex continuous $d$, whenever the cost integrals are well defined. Values at exceptional endpoints may be chosen arbitrarily.

The printed assumptions omit an essential source condition. Invertibility of $G$ alone does not ensure an admissible transport map. For example, $\mu ={\delta }_0$ and $\nu$ a standard normal distribution satisfy the stated condition on $G$, but $T_{\#}{\delta }_0$ is always a Dirac measure. There is no solution to the Monge optimal transport problem in this example. The boxed answer therefore requires the additional assumption that $\mu$ is an atomless measure, or an equivalent condition making the displayed map admissible. For arbitrary sources the always admissible monotone transport plan is $(F^{-1},G^{-1}{)}_{\#}\mathcal{U}(0,1)$, which need not be induced by a map.

Given any admissible transport map $T$, form its graph transport planFor Borel sets $A\subseteq X$ and $B\subseteq Y$, the definition of a pushforward measure givesHence ${\pi }_T\in \Pi (\mu ,\nu )$. Integration against a pushforward measure also givesThe Kantorovich optimal transport problem therefore has at least all the competitors of the Monge optimal transport problem, with exactly the same costs. ConsequentlyIf there is no admissible transport map, the left side is $+\infty$ by the convention $\inf \varnothing =+\infty$, so the conclusion still holds. No existence of an optimizer is needed.

For a cost $c:X\times Y\to [0,+\infty ]$ that is a Borel measurable function, a transport map is a measurable $T:X\to Y$ whose pushforward measure satisfiesThe Monge optimal transport problem moves every source point to one destination:The Kantorovich optimal transport problem permits mass to split. Its admissible transport plans are the probability measures on $X\times Y$ with prescribed marginal distributions:Thus a transport plan is a coupling of probability distributions. The set $\Pi (\mu ,\nu )$ is never empty: it contains the product measure $\mu \otimes \nu$. Signed costs can also be used when their integrals are well defined, for example with an integrable lower bound of the form $a(x)+b(y)$.

On the Polish space $\mathbb{R}$, take the Dirac measuresEvery measurable map satisfies $T_{\#}{\delta }_0={\delta }_{T(0)}$, which cannot equal $\nu$. A transport map cannot split an atom of a measure, whereas the transport plan $\frac{1}{2}{\delta }_{(0,-1)}+\frac{1}{2}{\delta }_{(0,1)}$ can.

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.

Analyze a local temperature perturbation at fixed radius and fixed surface density of a disk $\Sigma$. The thermal timescale of an accretion disk is shorter than its viscous timescale, so mass redistribution is negligible during the perturbation. Let $Q^{+}$ be viscous heating and $Q^{-}$ radiative cooling, measured consistently through one face or both faces, and define net cooling by $\dot{Q}=Q^{-}-Q^{+}$. A temperature increase is unstable if it decreases net cooling.

With gas pressure dominant, vertical hydrostatic equilibrium and the ideal gas law giveFor the alpha disk prescription with temperature-independent $\alpha$, $\nu =\alpha c_{s}H\propto T_c$, and the Keplerian rotation heating law isOne must not hold the steady-state $\dot{m}$ fixed while perturbing the temperature: the stress and heating respond through $\nu$, even though $\Sigma$ is temporarily fixed.

At an equilibrium temperature $T_0$ where $Q^{+}=Q^{-}=Q_0$, if $Q^{-}\propto T_c^q$ and $Q^{+}\propto T_c$, thenThe cooling exponent therefore decides the sign.

For hot, ionized, optically thin gas, the usual intended cooling model is thermal bremsstrahlung, whose volume emissivity scales as ${\rho }^2{T}_c^{1/2}$. Integrated vertically,Thus $q=0$, andA hotter annulus expands vertically, lowering its density enough to offset the direct increase in bremsstrahlung emission; viscous heating still rises.

For an optically thick medium with the Kramers' opacity law, $\kappa \propto {\rho }_c{T}_c^{-7/2}\propto T_c^{-4}$ at fixed $\Sigma$. Thermalized radiative diffusion givesIts falling opacity lets a hotter annulus radiate much more efficiently.

For an optically thick medium with dominant electron-scattering opacity, $\kappa$ is nearly independent of temperature. Provided enough true absorption is present to thermalize the radiation,This conclusion differs from the thermal instability of a radiation-pressure-dominated alpha disk, where the heating has a different temperature dependence. The question explicitly excludes that pressure regime.

The first conclusion needs its cooling assumption stated. $T>10^4\mathrm{K}$ alone does not guarantee that thermal bremsstrahlung dominates; atomic-line cooling, ionization changes, external illumination, or energy advection can change the answer. For a general optically thin emissivity ${\rho }^2\Lambda (T_c)$, the corresponding exponent isAccordingly, under the same fixed-$\Sigma$ alpha prescription,The unstable/stable/stable classification is the standard free-free/diffusion answer, rather than a universal result specified by temperature and optical depth alone.

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.

Work in the rest frame of the hole and neglect stellar velocities compared with $v$. In addition to $v\gg {\sigma }_{\ast }$, assume the galactic encounter speed is nonrelativistic, $v\ll c$. Treat a star as a test particle with specific angular momentum $\ell =bv$. At its pericentre $r_{\min }$, its radial velocity vanishes. Conservation of energy and conservation of angular momentum giveThe second term is gravitational focusing. For the rough Newtonian estimate in the question, assume capture when the pericentre reaches the Schwarzschild radius, $r_S=2G{M}_{\mathrm{BH}}/c^2$. It follows thatThe approximation keeps the focusing term because $v\ll c$.

Let ${\rho }_{\ast }=m{n}_{\ast }$ be the stellar mass density. Multiplying the geometric collision cross-section by the incoming mass flux gives the collisionless black-hole capture rateFor a supersonic gas with negligible sound speed, Bondi--Hoyle--Lyttleton accretion has capture radius $r_a=2G{M}_{\mathrm{BH}}/v^2$ and rateConsequentlyFor comparable densities and $v=200\mathrm{km}{\mathrm{s}}^{-1}$, the suppression is approximately $4\times 10^{-7}$. Gas can collide, dissipate orbital energy, and be captured from a region of size $GM/v^2$; collisionless particles generally leave on an unbound orbit unless their angular momentum is small enough for direct capture. Chandrasekhar dynamical friction can slow the hole by transferring energy to a stellar wake, but does not mean that the stars producing that wake are swallowed.

There is an important accuracy limitation in the requested numerical coefficient. Newtonian orbital mechanics evaluated at a relativistic event horizon cannot give the exact capture boundary. In Schwarzschild spacetime, a slowly incoming massive particle has dimensionless energy $\mathcal{E}\simeq 1$ and timelike geodesic effective potentialThe double turning point satisfying $V=1$ and $dV/dr=0$ occurs at $r=4r_g$ and $\ell =4G{M}_{\mathrm{BH}}/c$. The exact low-speed boundary is therefore $b_{\mathrm{cap}}\simeq 4G{M}_{\mathrm{BH}}/(cv)$, twice the rough value; it makes the rate ratio $4({\rho }_{\ast }/{\rho }_g)(v/c{)}^2$. The key suppression scaling is unchanged.

Direct capture of ordinary stars or dark matter is therefore an inefficient route to rapid growth of a supermassive black hole under these conditions. Tidal disruption events can capture stellar debris without swallowing an intact point-like star, and special dense environments can modify the supply. These qualifications do not make collisionless capture equivalent to dissipative gas accretion.

The singular isothermal sphere has one-dimensional velocity dispersion $\sigma$, density ${\rho }_{\mathrm{tot}}(r)={\sigma }^2/(2\pi G{r}^2)$, and enclosed mass $M(<R)=2{\sigma }^{2}R/G$. The gas mass inside $R$ is thereforeTaking the dynamical crossing time to be $t_{\mathrm{dyn}}\simeq R/\sigma$, the dynamical upper bound on black-hole fuelling isThe cancellation of $R$ is special to the singular isothermal sphere. At $\sigma =200\mathrm{km}{\mathrm{s}}^{-1}$, this is approximately

Use the stated M-sigma relation to set $M_{\mathrm{BH}}\simeq 3\times 10^8{M}_{\odot }$. Defining the Eddington accretion rate with a reference radiative efficiency of black-hole accretion ${\eta }_0$,Combining this with $M_{\mathrm{BH}}\propto {\sigma }^4$ givesThus a gas fraction of order $0.1$--$0.16$ and the usual ${\eta }_0=0.1$ give roughly $60$--$90$ times the Eddington accretion rate, below $100$ at the requested dispersion. Using a crossing time $R/(\sqrt{2}\sigma )$ instead changes the coefficient by $\sqrt{2}$, so this is an order-of-magnitude estimate rather than a sharp universal bound. The problem does not specify $f_{\mathrm{gas}}$ or the rate convention: without these choices, the numerical claim cannot be proved for every gas fraction. Defining the rate as $L_{\mathrm{Edd}}/c^2$ would make this ratio ten times larger.

This is extremely optimistic because it gives every gas element a direct path to the hole on a crossing time. In a real galaxy, gas must shed angular momentum, avoid conversion into stars through star formation, cool sufficiently to flow inward, and survive active-galactic-nucleus feedback. A rotating accretion disk generally transports material on a viscous timescale, much longer than the orbital crossing time. Most available gas need not reach the central hole. If the hole is supplied at approximately the Eddington accretion rate, only about one percent of this optimistic dynamical supply reaches it; the actual percentage depends on its luminosity and duty cycle, so the bound alone is not a measurement of the fuelling efficiency.

For the temperature jump, use Bondi accretion as the local hot-gas estimate:Assuming the initial gas had $c_s\sim \sigma$, its density has not yet changed, and its polytropic coefficient remains comparable, heating to $c_{s,\mathrm{feed}}\sim 10\sigma$ givesThe capture radius $G{M}_{\mathrm{BH}}/c_s^2$ also shrinks by about $100$. Subsequent expansion can lower the density and suppress the mass accretion rate further. The result illustrates thermal suppression of black-hole accretion and negative active-galactic-nucleus feedback: vigorous activity can shut off its own gas supply. The factor $10^{-3}$ is conditional on the fixed-density comparison; the previous dynamical upper bound is not itself an actual pre-heating Bondi rate.

Take $u(r)>0$ to be the inward radial speed and select the regular critical solution for a gas reservoir of density ${\rho }_{\infty }$ and adiabatic sound speed $c_{\infty }$. Other mathematically subsonic steady solutions need a different inner boundary condition. For the polytropic equation of state, let $\gamma =1+1/n$, soMass conservation and the Bernoulli equation giveThe gas is at rest at infinity, so its kinetic and gravitational terms vanish there.

Differentiate these two conserved quantities, using $d\log \rho /dr=-2/r-u^{\prime }/u$. Eliminating the density derivative yields the polytropic Bondi accretion equationAt a regular sonic point, both sides vanish:The Bernoulli equation at that point reduces to $n{c}_{\infty }^2=(n-3/2)c_c^2$. For $n>3/2$, equivalently $1<\gamma <5/3$,Since $c_s^2\propto {\rho }^{1/n}$, the critical density is ${\rho }_c={\rho }_{\infty }[2n/(2n-3){]}^n$. Evaluate the conserved mass accretion rate at the sonic point:whereThe isothermal limit $\gamma \to 1$ gives $\lambda =e^{3/2}/4$, agreeing with Isothermal Bondi accretion.

For $\gamma =5/3$, $n=3/2$ and the Newtonian critical radius tends to zero while $c_c$ diverges. The limiting critical solution approaches Mach number one only as $r\to 0$; it has no finite-radius Newtonian sonic transition. This is a failure of the Newtonian inner approximation, not an obstruction to accretion onto a black hole. A physical Schwarzschild black hole has a finite event horizon, and the regular relativistic inflow becomes supersonic outside it.

The singular-looking factor in the rate has a finite limit. Writing $ϵ=n-3/2$, its logarithm is $ϵ\log [n/ϵ]\to 0$, so the critical Bondi accretion rate for gamma equals five thirds isThis is also the leading cold-reservoir limit of relativistic spherical accretion; an exact rate for relativistically hot gas requires the relativistic equations, not this Newtonian expression.

For completeness, the finite relativistic sonic point can be seen without assigning it an arbitrary inner radius. Let $b=c_s^2/c^2$ denote the relativistic squared sound speed and $g=\gamma -1$. For an isentropic relativistic gas with $p=K{\rho }_{\mathrm{rest}}^{\gamma }$, its dimensionless specific enthalpy is $h=g/(g-b)$. Conservation of rest-mass flux and relativistic Bernoulli energy gives the critical relationsFor $g=2/3$ and $c_{\infty }\ll c$, expansion gives $b_{\infty }\simeq (9/4)b_c^2$, henceHere the asymptotic relativistic and Newtonian sound speeds agree to leading order. This finite-radius transition explains why the gas can reach the horizon despite the Newtonian $r_c=0$ result.

Use a geometrically thin spherical shell with constant mass $M_{\mathrm{sh}}$ and assume that its dust transfers momentum efficiently to its gas. Define the ultraviolet and infrared opacities per unit total shell mass by ${\kappa }_{\mathrm{UV}}$ and ${\kappa }_{\mathrm{IR}}$. Its dust optical depths areFor a central ultraviolet luminosity $L$, the usual dusty radiation-pressure-driven shell model absorbs a fraction $1-e^{-{\tau }_{\mathrm{UV}}}$ of the incident momentum and adds an infrared trapping contribution ${\tau }_{\mathrm{IR}}L/c$. Neglecting swept-up mass, shell self-gravity, and other gravitating matter, its thin-shell momentum equation isBalancing these terms gives the critical luminosity of a dusty shell:This is a local force-balance luminosity, so it can depend on radius even when $L$ is constant.

The three useful limiting Eddington ratios areThe infrared limit assumes ${\tau }_{\mathrm{IR}}\gg 1$ and ultraviolet absorption; the single-scattering limit assumes ${\tau }_{\mathrm{UV}}\gg 1\gg {\tau }_{\mathrm{IR}}$; the ultraviolet-thin limit assumes ${\tau }_{\mathrm{UV}}\ll 1$ with infrared opacity negligible compared with ultraviolet opacity. In the ultraviolet-opaque region the two radiative contributions add, and the effective ratio is ${\Gamma }_{\mathrm{IR}}+{\Gamma }_{\mathrm{ss}}(R)$.

The ultraviolet dust transparency radius for this constant-mass shell isFor the terminal speed of a constant-mass dusty shell, approximate the ultraviolet absorption factor by $1$ inside $R_{\mathrm{UV}}$ and by ${\tau }_{\mathrm{UV}}$ outside. The infrared contribution outside that radius is neglected; it is small there if ${\kappa }_{\mathrm{IR}}\ll {\kappa }_{\mathrm{UV}}$. Define ${\Gamma }_{\mathrm{ss}}={\Gamma }_{\mathrm{ss}}(R_0)$, so this otherwise radius-dependent ratio in the paper's final formula is an initial value.

Inside $R_{\mathrm{UV}}$ the acceleration isIntegrating from $R_0$ to $R_{\mathrm{UV}}$ givesOutside $R_{\mathrm{UV}}$, $vdv/dR=G{M}_{\mathrm{BH}}({\Gamma }_{\mathrm{UV}}-1)/R^2$. Its integral to infinity supplies $2G{M}_{\mathrm{BH}}({\Gamma }_{\mathrm{UV}}-1)/R_{\mathrm{UV}}$. Combining the two integrals yieldsThis assumes that the outward trajectory actually reaches the transparency radius; a positive formal endpoint energy alone does not rule out an intermediate turning point.

Put $X=R_{\mathrm{UV}}/R_0=\sqrt{{\tau }_{\mathrm{UV}}(R_0)}$ and $v_{\mathrm{esc},0}^2=2G{M}_{\mathrm{BH}}/R_0$. The ratios obey ${\Gamma }_{\mathrm{UV}}={\Gamma }_{\mathrm{ss}}{X}^2$. The preceding expression becomesIf $X\gg 1$ and infrared work is subdominant,For a luminosity of order the initial single-scattering Eddington luminosity, ${\Gamma }_{\mathrm{ss}}\sim 1$, this implies $v_{\infty }/v_{\mathrm{esc},0}\sim \sqrt{2}{\tau }_{\mathrm{UV}}(R_0{)}^{1/4}\gg 1$. The shell continues intercepting approximately $L/c$ of momentum per unit time over a large distance while gravity weakens as $R^{-2}$. The resulting radiative work is not restricted to its initial gravitational binding energy: an Eddington luminosity is a local force threshold, not a universal cap on outflow speed.

If the unqualified $L_{\mathrm{Edd}}$ instead means the electron-scattering value, the statement needs an additional condition. For $L=L_{\mathrm{Edd},\mathrm{es}}$, ${\Gamma }_{\mathrm{UV}}={\kappa }_{\mathrm{UV}}/{\kappa }_{\mathrm{es}}$, andThus speeds greatly above escape velocity require ${\kappa }_{\mathrm{UV}}/{\kappa }_{\mathrm{es}}\gg \sqrt{{\tau }_{\mathrm{UV}}(R_0)}$, together with a viable launch. Large ultraviolet optical depth by itself does not prove the conclusion at fixed electron-scattering luminosity and opacity. The two interpretations must not be conflated.

Treat each grain as an isolated compact sphere and take its wavelength-independent radiation-pressure efficiency to be $Q_{\mathrm{pr}}=1$. At distance $R$, its mass and radiation-force cross-section areThe inward gravitational force and outward radiation pressure force areThus the opacity per unit dust mass and the dust-grain Eddington limit areTo equal the Eddington luminosity based on electron-scattering opacity, ${\kappa }_d$ must equal ${\kappa }_{\mathrm{es}}$. ThereforeFor fully ionized hydrogen, ${\kappa }_{\mathrm{es}}={\sigma }_T/m_p\simeq 0.40\mathrm{c}{\mathrm{m}}^2{\mathrm{g}}^{-1}$, and the stated ${\rho }_d=1\mathrm{g}\mathrm{c}{\mathrm{m}}^{-3}$ givesUsing $0.34\mathrm{c}{\mathrm{m}}^2{\mathrm{g}}^{-1}$ for a roughly solar-composition gas instead gives $2.2\mathrm{cm}$. This composition-dependent normalization should be stated because the problem does not fix it. For a different constant radiation-pressure efficiency, the critical radius is multiplied by $Q_{\mathrm{pr}}$.