Solution
= Solution
\b[<photon> orbit and the Newtonian limit.] Write $r_g=GM/c^2$ and $e=E/(mc^2)$. For a <timelike circular Schwarzschild geodesic>,
$$
e(r)=\frac{1-2r_g/r}{\sqrt{1-3r_g/r}}.
$$
The denominator tends to zero at $r=3r_g$, so the <energy> per unit <rest mass> diverges as a massive circular orbit approaches that <radius>. The corresponding local <orbital speed> measured by a static observer satisfies $v_{\rm loc}^2/c^2=r_g/(r-2r_g)$ and tends to one. This suggests a null circular orbit, but one must not simply insert $m=0$ into a massive-particle formula. For a <photon>, the radial <effective potential> is proportional to $(1-2r_g/r)/r^2$; its <derivative> vanishes at $r=3r_g$ and its second <derivative> there is negative. Thus \b[the <photon sphere> is at $3GM/c^2$, and its circular null orbits are unstable].
At $r\gg r_g$, expansion gives
$$
e=1-\frac{r_g}{2r}+O\left(\frac{r_g^2}{r^2}\right),\qquad
\boxed{E=mc^2-\frac{GMm}{2r}+\cdots}.
$$
The second term is the total Newtonian circular-orbit <energy>, kinetic plus potential; the relativistic <energy> also includes <rest energy>. Here $E$ is the conserved <energy> referred to a stationary observer at infinity, not the local <kinetic energy>.
\b[The stable inner edge.] At fixed <angular momentum>, write $\ell=L/(mc)$. The timelike radial <geodesic> equation can be written
$$
\left(\frac{1}{c}\frac{dr}{d\tau}\right)^2+V_{\rm eff}(r)=e^2,\qquad
V_{\rm eff}=\left(1-\frac{2r_g}{r}\right)\left(1+\frac{\ell^2}{r^2}\right).
$$
The circular condition $V_{\rm eff}'=0$ gives $\ell^2=r_g r^2/(r-3r_g)$. Substituting into the second <derivative> taken at fixed $\ell$ gives
$$
V_{\rm eff}''=\frac{2r_g(r-6r_g)}{r^3(r-3r_g)}.
$$
Timelike circular orbits exist for $r>3r_g$; they are stable against radial perturbations for $r>6r_g$ and unstable for $3r_g<r<6r_g$. Therefore
$$
\boxed{r_{\rm ISCO}=6r_g,\qquad e_{\rm ISCO}=\sqrt{\frac89}=0.942809.}
$$
The boundary orbit is marginally stable. Differentiating $e(r)$ also gives a minimum there, consistently with this fixed-angular-momentum stability calculation.
\b[<accretion> efficiency and <black-hole spin>.] Matter supplied slowly from far away initially brings <rest energy> $dm_0c^2$. A radiatively efficient, zero-inner-torque <thin disk> releases the <binding energy> down to the <innermost stable circular orbit>; the remaining <energy> $e_{\rm ISCO}dm_0c^2$ is carried into the hole during the plunge. Thus
$$
\boxed{\eta=1-e_{\rm ISCO}=1-\sqrt{\frac89}=0.05719}.
$$
Here $\eta$ denotes the efficiency called $\epsilon$ in this question. The ideal relation assumes negligible additional dissipation during plunge and neglects <photon> capture and <advection>. A hot radiatively inefficient flow need not have this efficiency.
For a <Kerr black hole>, an equatorial prograde disk has a smaller <innermost stable circular orbit> and greater binding-energy efficiency than the Schwarzschild value; a retrograde disk has the opposite trend. The formal extremal prograde limits are $r_{\rm ISCO}\to r_g$ and $\eta\to1-1/\sqrt3\simeq0.423$. These are ideal limiting values, not a universal realized efficiency. <photon> capture and the <accretion> geometry affect the achievable <black-hole spin> and escaping <radiation>.
\b[<radiation> <force> and the limiting <luminosity>.] For isotropic <luminosity> $L$, the outward <radiation> <momentum flux> at <radius> $r$ is $L/(4\pi r^2c)$. In fully ionized <hydrogen>, <Thomson scattering> acts mainly on the <electron>, while electrostatic coupling makes the <electron> and <proton> move together. Neglecting the <electron> <mass>, one <electron>–<proton> pair experiences
$$
F_{\rm rad}=\frac{\sigma_T L}{4\pi r^2c},\qquad
F_{\rm grav}=\frac{GMm_p}{r^2}.
$$
Equating them yields
$$
\boxed{L_{\rm Edd}=\frac{4\pi GMm_pc}{\sigma_T}}.
$$
This <Eddington luminosity> uses <electron-scattering opacity>, spherical or isotropic <force> balance and negligible relativistic corrections at the force-comparison <radius>. Composition, other <opacity>, anisotropy and <photon> trapping can change the relevant limit.
\b[Exponential growth and doubling.] Distinguish rest-mass supply $\dot m_0$ from black-hole <mass> growth $\dot M$:
$$
L=\eta\dot m_0c^2,\qquad
\dot M=(1-\eta)\dot m_0.
$$
At the Eddington limit,
$$
\dot M=\frac{1-\eta}{\eta}\frac{M}{t_E},\qquad
t_E=\frac{\sigma_Tc}{4\pi Gm_p},\qquad
M(t)=M_i\exp\left(\frac{t-t_i}{t_S}\right),
\quad
t_S=\frac{\eta}{1-\eta}t_E.
$$
Using the printed scattering cross section gives $t_E\simeq4.50\times10^8\ \mathrm{yr}$. The <Salpeter time> $t_S$ is an e-folding time; the <Eddington-limited doubling time> is
$$
\boxed{t_2=\frac{\eta}{1-\eta}\frac{\sigma_Tc}{4\pi Gm_p}\log2
\simeq1.89\times10^7\ \mathrm{yr}}
$$
for the Schwarzschild thin-disk efficiency. If one approximates $\dot M\simeq\dot m_0$, the factor $1/(1-\eta)$ is dropped and the estimate is about $1.79\times10^7\ \mathrm{yr}$. The retained-energy version is the mass-conserving expression.
\b[The efficiency bound.] A growth factor $10^8$ requires $N=\log(10^8)=18.4207$ e-folds. Eddington-limited <accretion> is the fastest growth allowed by the adopted <luminosity> bound at fixed efficiency, so growth within a time $\Delta t$ requires
$$
N\frac{\eta}{1-\eta}t_E\leq\Delta t,\qquad
\boxed{\eta\leq\frac{\Delta t}{\Delta t+N t_E}\simeq0.108}.
$$
This <efficiency upper bound from Eddington-limited growth> is an \b[upper bound]: higher efficiency radiates more of each supplied <rest mass>, and the <luminosity> ceiling therefore allows less <mass> to enter per unit time. A duty cycle below unity or sub-Eddington <luminosity> tightens it. The approximate convention $\dot M=\dot m_0$ would instead give $\eta\lesssim0.121$; both figures refer to the same effect, with different treatment of the radiated <mass> deficit. Mergers, super-Eddington growth or a different seed history would change the premises.
Coherent prograde <accretion> <spins> up the hole, moves the inner edge inward and increases $\eta$. For fixed $L/L_{\rm Edd}$ the mass-growth e-folding time then grows, so the logarithmic growth curve becomes less steep; one cannot simply retain the initial Schwarzschild efficiency throughout. Episodic misaligned <accretion> can have a different <black-hole spin> history. In general,
$$
\log\frac{M_f}{M_i}
=\frac1{t_E}\int_{t_i}^{t_f} f_{\rm Edd}(t)\frac{1-\eta(t)}{\eta(t)}\,dt.
$$
\b[<energy> available to affect the <galaxy>.] Let $M_g$ denote <galaxy> <mass>, keeping it distinct from black-hole <mass>. At constant efficiency, negligible seed <mass> and no merger contribution, the emitted <energy> in building $M_{\rm BH}=0.002M_g$ is
$$
E_{\rm rad}=\frac{\eta}{1-\eta}M_{\rm BH}c^2,\qquad
\boxed{\frac{E_{\rm rad}}{E_{\rm bind}}\sim
0.002\frac{\eta}{1-\eta}\left(\frac{c}{v}\right)^2}.
$$
For $v=300\ \mathrm{km\,s^{-1}}$ and the nonspinning efficiency this is \b[approximately $121$]. The common small-efficiency approximation gives $2000\eta\simeq114$, and a fiducial $\eta=0.1$ would give about $222$ with retained <mass> included. These differences do not alter the conclusion: the available <accretion> <energy> substantially exceeds the stated galactic binding-energy estimate.
Only a coupling fraction of order one percent would then be energetically sufficient to affect the gas strongly. <radiation>, winds or jets can heat gas, drive outflows and suppress further cooling and <star formation>, providing <active-galactic-nucleus feedback> and possible self-regulation of black-hole growth. <energy> availability alone does not prove that the whole <galaxy> is expelled: most <radiation> can escape, jets can be collimated, heated gas can cool, and collisionless <stars> couple differently from gas.
\Image[/past-exam-of-the-relativistic-astrophysics-and-cosmology-course-of-the-university-of-cambridge/2015/paper-1-accretion-growth.png]
{title=Original comparison of Eddington-limited growth at fixed efficiencies, including the radiated mass deficit}