= Solution
Write $A=\lambda\beta\ell_E$, where $\ell_E=L_{\rm Edd}/M_{\rm BH}$. For ionized <hydrogen> with <Thomson scattering> opacity, radiation force on one electron-proton pair balances gravity when
$$
\frac{\sigma_T L}{4\pi r^2c}=\frac{GM_{\rm BH}m_p}{r^2},
\qquad
\ell_E=\frac{4\pi Gm_pc}{\sigma_T}.
$$
The mass-loading coefficient $A$ has units of inverse time. With the stated neglect of returned and directly accreted gas, <mass conservation> gives
$$
\dot M_g=\Phi-\epsilon M_g-A M_s,\qquad
\dot M_s=\epsilon M_g.
$$
At a specified stellar mass, the <gas-balance nullcline> is
$$
M_g^{\rm eq}(M_s)=\frac{\Phi-A M_s}{\epsilon}.
$$
A positive reservoir exists only below the limiting mass
$$
\boxed{M_{s,\max}=\frac{\Phi}{A}
=\frac{\Phi\sigma_T}{4\pi Gm_pc\,\lambda\beta}.}
$$
Equality has zero gas, so this is the supremum of masses with a positive instantaneous <equilibrium>. For $\Phi,\epsilon,A>0$, the corresponding gas fraction is
$$
\boxed{f_{\rm gas}(M_s)=
\frac{\Phi-A M_s}{\Phi+(\epsilon-A)M_s}
=\left[1+\frac{\epsilon M_s}{\Phi-A M_s}\right]^{-1},
\quad0\leq M_s<M_{s,\max}.}
$$
It decreases from one to zero: the derivative is $-\epsilon\Phi/[\Phi+(\epsilon-A)M_s]^2<0$. This is a <gas regulator> with an outflow increasing with the stellar mass through the assumed black-hole scaling.
There is an important time-dependence qualification. A positive gas mass cannot remain exactly constant forever while the stellar mass continues to increase and $M_{\rm BH}=\beta M_s$ is maintained: differentiating $\dot M_g=0$ would require $A\epsilon M_g=0$. The expression above is the instantaneous <equilibrium>, or a quasisteady approximation when gas relaxation is fast relative to stellar-mass evolution. The only exact <fixed point> of the coupled idealized equations is $(M_g,M_s)=(0,\Phi/A)$. It is not a sustained star-forming state.
Above the limiting mass, $\dot M_g<0$ for every nonnegative reservoir: the assumed Eddington-driven wind already demands more than the incoming supply, even before <star formation> consumes gas. The reservoir is depleted and <star formation> is quenched. The equations must cease to apply when gas runs out; one cannot continue to negative gas mass or power an indefinitely Eddington-luminous nucleus without fuel. At fixed $M_s>M_{s,\max}$ the depletion time from $M_{g,0}$ is
$$
t_{\rm dep}=\epsilon^{-1}\ln\left(1+
\frac{\epsilon M_{g,0}}{A M_s-\Phi}\right).
$$
Allowing $M_s$ to grow only strengthens the initial depletion tendency. If $\lambda\beta=0$, this particular feedback threshold is absent.
\Image[/past-exam-of-the-formation-of-structure-in-the-universe-course-of-the-university-of-cambridge/2016/paper-1-gas-regulator.png]
{title=Original dimensionless gas-regulator nullcline and its decreasing equilibrium gas fraction}
Back to article page
