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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.

Interstellar dust both hides and reveals galaxies. Absorption and scattering remove short-wavelength photons from a line of sight, causing interstellar extinction, reddening, anisotropic scattering, and polarization. These effects obscure embedded star formation, bias luminosities, colours, stellar masses, and star-formation rates, and can make an edge-on or dusty galaxy appear older and fainter. Dust also converts absorbed ultraviolet and optical power into far-infrared and submillimetre thermal radiation. That reradiation exposes otherwise hidden star formation, constrains dust temperature and mass, and helps map cold molecular material; polarization traces magnetic-field orientation. Grain surfaces also catalyse molecular-hydrogen formation, while photoelectric emission from grains heats interstellar gas.

One spherical grain has emitting area $4\pi a^2$. Since isotropic blackbody intensity $B_{\nu }$ gives surface flux $\pi B_{\nu }$, its spectral luminosity isFor $N=M_d/m_d$ optically thin grains at distance $D$, the inverse-square law givesDefining the grain mass absorption coefficientthe optically thin dust-mass estimator is

For a constant source function and no incident background, the radiative transfer equation integrates toIf the cloud subtends solid angle $\Omega$, uniform intensity givesThe physical projected area is $A=\Omega D^2$. Its grain column density is $N/A=M_d/(m_d\Omega D^2)$, so the dust optical depth isEliminating ${\tau }_{\nu }$ gives the finite-optical-depth resultIt requires $F_{\nu }<\Omega B_{\nu }$, as demanded by the blackbody brightness limit. When ${\tau }_{\nu }\ll 1$, $1-e^{-{\tau }_{\nu }}={\tau }_{\nu }+O({\tau }_{\nu }^2)$, so $F_{\nu }\simeq \Omega B_{\nu }{\tau }_{\nu }={\kappa }_{\nu }{M}_d{B}_{\nu }/D^2$ and the result reduces to the optically thin estimator.

Let the source's hydrogen-ionizing photon production rate bewhere $h{\nu }_0$ is the hydrogen ionization energy. The idealized Strömgren sphere has an almost fully ionized interior and a thin ionization front. In photoionization equilibrium, every ionization is balanced by a Case B recombination, so spherical symmetry givesFor pure hydrogen of constant number density $n$, the interior has $n_e=n_p=n$, and the Strömgren radius is therefore

Now let the effective number of dust grains per hydrogen nucleus be $f_d$, so the dust absorption coefficient is $a(r)=f_d{\sigma }_{d}n(r)$. If $Q(r)$ is the ionizing-photon rate crossing the sphere of radius $r$, recombinations and dust absorption give the linear ordinary differential equationIf $f_d$ denotes a dust mass fraction instead, the grain mass and gas mean particle mass are simply absorbed into the effective product $f_d{\sigma }_d$. Define the dust optical depthMultiplication by the integrating factor $e^{{\tau }_d(r)}$ and integration to the dusty front gives its governing equationThe exponential function weights recombinations at large optical depth by the extra source photons that dust must remove before those photons reach that radius.

For constant $n$, put $a=f_d{\sigma }_{d}n$ and $\tau =a{R}_d$. The elementary integralreduces the equation toEquivalently, with $y=R_d/R_S$ and ${\tau }_S=a{R}_S$,Its dust-free limit is $y\to 1$, while absorption makes $R_d<R_S$ for nonzero dust abundance.