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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.

The swept shell has $M_s=Ar$, where $A=2f_{\mathrm{gas}}{\sigma }^2/G$. Its steady radial thin-shell momentum equation isLet $r_0$ be the launch radius and impose $v(r_0)=0$.

In the single-scattering regime, defineThe force difference is constant, and integration givesThe shell accelerates monotonically, but continuous sweeping of $M_s\propto r$ makes the speed approach the finite maximum

For the optically thin ultraviolet regime, define the launch Eddington factorSince $F_{\mathrm{rad}}={\tau }_{\mathrm{UV}}L/c\propto r^{-1}$ while the shell's gravitational force is constant, integration givesIt initially accelerates, reaches its maximum atand then decelerates, formally stalling at $r=r_0(2{\Gamma }_{\mathrm{UV}}-1)$. The contrast is physical: single-scattering transfers the same $L/c$ to the shell at every radius, whereas an ultraviolet-thin shell intercepts a fraction proportional to its declining optical depth, so the isothermal host's gravity eventually wins.