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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.

A singular isothermal sphere has total density and circular speedIf gas traces the gravitating matter, ${\rho }_g=f_{\mathrm{gas}}{\rho }_{\mathrm{tot}}$. A moving black hole gravitationally focuses that gas, so the relevant estimate is the Bondi--Hoyle--Lyttleton accretion rateHydrostatic gas in the isothermal potential has $c_s$ of order $\sigma$, and thereforeup to an order-one coefficient. Taking $c_s=\sigma$ and $v_c=\sqrt{2}\sigma$ gives $2/(3\sqrt{3})$ multiplying the displayed expression. The same $G^2{M}^2\rho /v^3$ scaling that produces Chandrasekhar dynamical friction therefore sets the likely capture rate as the hole spirals inward. Feedback, angular momentum, a gas core, or a multiphase medium can reduce the actual rate.

For a dusty shell, direct ultraviolet absorption, infrared trapping, and incomplete ultraviolet absorption give the radiation forceBalancing it against gravity gives the general critical luminosity of a dusty shell

For a singular isothermal sphere,soandIn the infrared-thick limit $F_{\mathrm{rad}}\simeq {\tau }_{\mathrm{IR}}L/c$; in the single-scattering limit it is $L/c$; and in the ultraviolet-thin limit it is ${\tau }_{\mathrm{UV}}L/c$. Therefore

For a singular isothermal sphere host, ${\rho }_h=A_h{r}_o^{-2}$,Now $d\log M_h/d\log r_o=1$, so the circular tidal formula has a factor two:Using $M_s=2\pi A_s{r}_t^2$ givesThe frictional acceleration is now $a_{\mathrm{df}}=-G{\mu }_i\mathcal{L}{r}_o^2$. Since the specific angular momentum is $V{r}_o$,ThusNeither radius nor mass reaches zero at finite time. Compared with the $r^{-1}$ host, the steeper isothermal cusp shrinks the tidal radius as $r_o^2$, so stripping suppresses the friction rapidly enough to produce dynamical-friction stalling by tidal stripping.

If $c_{s,\infty }\gg \sigma$ and $M_{\mathrm{enc}}\to M$, expansion of the denominator givesthe stated model's pressure-dominated Bondi accretion limit. Its order-unity coefficient differs from the $\pi e^{3/2}$ found for the exactly isothermal critical solution because such coefficients depend on the adopted equation of state and interpolation.

If instead $\sigma \gg c_{s,\infty }$,The gas's thermal motion is then negligible beside the galactic velocity dispersion, and the enclosed galactic mass, rather than the black hole alone, focuses the gas. For a singular isothermal sphere, $M_{\mathrm{enc}}(R)=2{\sigma }^{2}R/G$, so this becomes $\dot{M}\simeq 4\pi {\rho }_{\infty }\sigma R^2$. This second limit describes capture controlled by the host potential and is therefore not a genuinely spherical black-hole Bondi solution.

Chandrasekhar dynamical friction is the drag exerted by the overdense gravitational wake that a massive body creates in a background of lighter particles. It transfers orbital energy and angular momentum to the host, causing satellites, star clusters, and massive black holes to spiral inward and promoting galaxy mergers. In a homogeneous isotropic Maxwellian background its force iswhere $\log \Lambda$ is the Coulomb logarithm in stellar dynamics. Its scaling can be reconstructed from the strong-deflection impact parameter $b_{\mathrm{cr}}\sim GM/v^2$: the encountered mass rate is $\sim \rho v\pi b_{\mathrm{cr}}^2$, and multiplying by momentum change $\sim v$ gives $F_{\mathrm{df}}\sim G^2{M}^2\rho /v^2$.

For a spherical host with a flat galaxy rotation curve,This is a singular isothermal sphere with one-dimensional dispersion $\sigma =v_c/\sqrt{2}$, so $X=1$. DefineFor a constant-mass satellite on a circular orbit, the drag magnitude and its torque areIt follows thatThe quadratic radius dependence and inverse mass dependence explain why massive nearby satellites merge much faster than light or distant ones.

Let the satellite also have a flat internal rotation curve of speed $v_s$. Equating its edge density to the host density gives the tidal radiusBecause $M_s(<R_t)=v_s^2{R}_t/G$, the bound mass decreases linearly:Under the question's literal closure that the derivative of the remaining satellite's total orbital angular momentum equals the frictional torque,and henceIf stripped material is explicitly assigned the satellite's instantaneous specific orbital angular momentum, the balance for the bound remnant is instead $M_{s}d(v_{c}r)/dt=-r{F}_{\mathrm{df}}$; that convention gives $t_{\mathrm{df},\mathrm{strip}}=2t_{\mathrm{df},0}$. Both treatments show the robust point: tidal stripping weakens the drag as the orbit shrinks and substantially delays coalescence. In less idealized profiles the mass can fall faster than linearly, producing dynamical-friction stalling by tidal stripping.

For a swept shell, momentum conservation may be writtenThe first term is direct ultraviolet absorption plus trapped-infrared radiation pressure, the second is gravity from the black hole and host, and the last is external pressure. The derivative also accounts for the inertia of newly swept-up gas.

For a singular isothermal sphere,Outside the black hole's sphere of influence and with external pressure neglected, gravity is the constant force $4f_{\mathrm{gas}}{\sigma }^4/G$. The shell optical depths areThus the dust transparency radius and dimensionless optical depths areWith $R^{\prime }=R/R_{\tau }$, $t^{\prime }=t/t_{\tau }$, $t_{\tau }=R_{\tau }/(\sqrt{2}\sigma )$, andthe shell equation becomes

In the optically thick single-scattering regime, $1-e^{-{\tau }_{\mathrm{UV}}}\simeq 1$ and ${\tau }_{\mathrm{IR}}$ is neglected. Integrating the constant right-hand side givesThe zero-radius member has $R^{\prime }=\sqrt{\xi -1}{t}^{\prime }$ and hence the physical constant speed

As the shell becomes ultraviolet-thin,The net force becomes negative beyond the force-balance radiusbut inertia carries the shell farther before it stalls. Match the thick solution at $R^{\prime }=1$, where ${\dot{R}}^{\prime 2}=\xi -1$. In the thin approximation,Using $d/d{t}^{\prime }={\dot{R}}^{\prime }d/d{R}^{\prime }$ and integrating givesThe outer zero is the stalling radiusIncluding the optically thick travel time from a negligible launch radius, the dimensionless stalling time isSolving $d{t}_{\mathrm{stall}}^{\prime }/d\xi =0$ gives $\xi \simeq 1.248$, soFor $1<\xi <1+ϵ$, a stronger mildly supercritical source reaches transparency so much sooner that its total stalling time decreases even though it travels farther. Above this range, the increasing coasting distance dominates. Stalling means that loss of ultraviolet optical depth reduces radiation coupling below gravity; without renewed driving, the swept gas falls back or remains bound rather than escaping the halo.

In the infrared multi-scattering regime, the dominant force is $\xi {\tau }_{\mathrm{IR}}=a/R^{\prime }$ with $a=\xi {\kappa }_{\mathrm{IR}}/{\kappa }_{\mathrm{UV}}$. Seeking $R^{\prime }=A{t}^{\prime 2/3}$ ingives $A^3=9a/2$ and thereforeInfrared trapping supplies more than the single-scattering momentum while the shell is compact and optically thick. The speed nevertheless decreases as $t^{\prime -1/3}$ because the shell sweeps up mass and its infrared optical depth falls. Such driving can launch a powerful dusty radiation-pressure-driven shell, but propagation to halo scales requires enough integrated momentum before the shell becomes transparent; otherwise gravity eventually stalls it.

For a self-consistent Gaussian angular-momentum suppression in a stellar distribution, put $y=[\psi (r)-\psi (0)]/{\sigma }^2$ and $s=\log (r/r_a)$. At large radius the Poisson equation for Newtonian gravity reduces to $y_{ss}+y_s\sim -\lambda e^y$, where $\lambda =4\pi G{\rho }_c{r}_a^2/{\sigma }^2$. The asymptotic balance $y_s\sim -1/s$ gives $e^y\sim 1/(\lambda s)$, the displayed density, and circular speed $v_c^2\sim {\sigma }^2/\log (r/r_{\ast })$. A sufficiently small regular core can have an intermediate isotropic singular isothermal sphere envelope with $\rho \simeq {\sigma }^2/(2\pi G{r}^2)$ and $v_c^2\simeq 2{\sigma }^2$. At the actual regular centre, however, $\rho \to {\rho }_c$ and $v_c\propto r$; the intermediate power law must not be mistaken for a central limit.