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For transonic accretion in a dipolar flux tube, the Bernoulli equation at the sonic point contains $c_s^2[1/(\gamma -1)-5/2]$. A positive reservoir energy requires $\gamma <7/5$. Compression in the $r^3$ dipolar flux-tube area raises the specific enthalpy more rapidly than in the $r^2$ geometry of spherical Bondi accretion, whose corresponding upper index is $5/3$.

Let $v=-u_r>0$, $c=v_s$ and $c_0=v_{s0}$. Steady smooth adiabatic flow has $p=K{\rho }^{\gamma }$ with the entropy parameter fixed by infinity, andThe second expression follows by integrating the radial momentum equation with $dp/\rho =d(c^2/(\gamma -1))$ and imposing the outer rest condition. Differentiate continuity to give ${\rho }^{\prime }/\rho =-2/r-v^{\prime }/v$, and substitute into momentum. This yields the Bondi accretion wind equationAt a smooth positive-radius sonic point, $v=c=c_s$ and both sides' vanishing factors require $GM/r_s=2c_s^2$. Evaluate the Bernoulli relation there:Both the squared sound speed and sonic radius must be positive and finite. HenceFor $\gamma =5/3$ no finite-radius regular sonic crossing is possible; the critical solution discussed next approaches sonic speed only at the central limit.

Write the inward radial velocity as $u_R=-v(R)$, with $v>0$. Steady spherical mass conservation and the radial Euler equations for an inviscid fluid giveFor a barotropic fluid, define the sound speed $c_s^2=dp/d\rho >0$ and the barotropic enthalpy function by $dh=dp/\rho$. Choosing $\Phi (\infty )=0$, integration gives the Bernoulli equationThe continuity equation gives ${\rho }^{\prime }/\rho =-2/R-v^{\prime }/v$. Eliminating mass density from the momentum equation produces the barotropic spherical accretion equationThe sonic point is where the inward Mach number $v/c_s$ equals one. A smooth solution cannot cross the vanishing coefficient on the left with a nonzero right-hand side, so a regular sonic radius must obeyConversely a continuous flow that is subsonic at infinity and supersonic inward must cross unit Mach number; smooth passage selects a simultaneous zero of the two sides, hence a regular sonic point. The associated critical accretion branch, rather than an arbitrary subsonic solution with smaller flux, is selected by these outer and inner requirements.

For the polytropic equation of state, set $\Gamma =1+1/m$ to distinguish its exponent from any separately specified perturbation exponent. Assume $K>0$ and first take the usual positive index $m>0$. ThenFor the attractive mixed potential, assume $\lambda ,\mu >0$. The sonic condition and Bernoulli equation giveHence polytropic accretion in a mixed inverse-power potential reduces toFor $0<m\le 1/2$ all nonzero coefficients in the left side are nonnegative for $R_s>0$, with at least one strictly positive: there is no finite positive root. For $m>1/2$ the constant term is negative and the leading coefficient positive, giving exactly one positive root. Define $B=(2m-3)\mu /4$ and $C=(2m-1)\lambda /2$. ThenThis is the finite-radius transonic range for the two positive attractive terms.

The critical point has real distinct crossing slopes. To check this rather than just solve the sonic equalities, put $U=\lambda /R_s^2$, $V=\mu /(2R_s)$ and $x=R_s{v}_s^{\prime }/v_s$. Differentiating the barotropic spherical accretion equation at its simultaneous zero givesIts discriminant is $8m[2(2m-1)U+(2m-3)V]/(U+V)$. The sonic Bernoulli equation says $(2m-1)U+(2m-3)V=2m{c}_{\infty }^2$, so this discriminant is strictly positive for $m>1/2$. The accretion crossing is the slope for which $d\log (v/c_s)/d\log R<0$, namely the smaller root. The two slopes have opposite signs of this Mach-number derivative. This is the regular transonic saddle, not a singular vertical tangent.

The corresponding mass accretion rate follows without an additional arbitrary mass density normalization:One may equivalently replace ${\rho }_s$ by $[c_{s,s}^2/(\Gamma K){]}^m$. With a nonzero $\lambda$ term, inner free fall has $v\sim \sqrt{2\lambda }/R$ and continuity gives $\rho \propto R^{-1}$. Thus $v^2/c_s^2\propto R^{-2+1/m}$ increases without bound precisely in the positive-index range $m>1/2$, consistent with the required inner supersonic branch.

The assumptions on the parameters matter. If $\lambda =0$ and $\mu >0$, the finite sonic radius instead requires $m>3/2$, the usual Bondi accretion range; $m=3/2$ is the zero-radius limiting critical case. If $\mu =0$ and $\lambda >0$, the condition remains $m>1/2$. With arbitrary signs of $\lambda ,\mu$, one must use the quadratic and retain only positive roots with $c_{s,s}^2>0$ and real admissible crossing slopes; attraction was not explicitly given as a sign condition in the source.

If positive polytropic index was not intended as an implicit physical convention, there is a further mathematical range. For $K>0$, positive compressibility also permits $m<-1$, giving $0<\Gamma <1$. Both potential terms attractive then give a positive sonic root as well:The same mass accretion rate formula applies; its crossing-slope discriminant is positive. The range $-1<m<0$ has $c_s^2<0$ and no physical acoustic sonic point, $m=-1$ has constant pressure and zero sound speed, and $m=0$ does not define the printed equation of state. Thus the usual answer $m>1/2$ assumes $m>0$ and positive attractive coefficients; these conventions should not be confused with additional printed restrictions.

Use $v=|u_r|>0$ for the radial speed and distinguish the central mass $M$ from the Mach number $\mathcal{M}=v/c$. In each smooth nonzero-flow region, mass conservation, the adiabatic equation of state, and the Bernoulli equation giveHere $K>0$ is constant on a smooth isentropic flow branch, $c$ is the adiabatic sound speed, and $\mathcal{E}$ is the Bernoulli function. Either sign of $u_r$ is possible. The positive-radius nondimensionalization in the question requires $\mathcal{E}>0$; flows with $\mathcal{E}\le 0$ do not have these positive $r_s,x,\lambda$ and require another scaling.

Indeed, the supplied adiabatic process equations give $D(p{\rho }^{-3/2})/Dt=0$, so $K$ is spatially constant along a steady nonzero radial fluid flow. The adiabatic sound speed is $c^2=(\partial p/\partial \rho {)}_s$, and the specific enthalpy is $h=c^2/(\gamma -1)=2c^2$. The radial Euler momentum equation, $vdv/dr=-{\rho }^{-1}dp/dr-GM/r^2$, integrates to $\mathcal{E}$ because $dh=dp/\rho$ on this isentropic flow branch.

Put $D=[j(3K/2{)}^2{]}^{2/5}$. Eliminating the mass density from the continuity equation givesWith $y={\mathcal{M}}^{2/5}$, the Bernoulli equation becomesTherefore the physical constants of the spherical polytropic flow with adiabatic exponent three halves are

For $t>0$, $F^{\prime }(t)=4(t^5-1)/t^2$, so $F$ decreases from infinity to its unique minimum $F(1)=5$, then increases to infinity. This completely determines the algebraic curves. For $0<\lambda <1$, there are two disconnected branches at every $x>0$, one subsonic flow with $y<1$ and one supersonic flow with $y>1$. For $\lambda =1$, the line $y=x$ and a decreasing branch meet at $(1,1)$. Their slopes are $+1$ and $-1$, since the leading expansion is $F(1+\eta )=5+10{\eta }^2+O({\eta }^3)$. For $\lambda >1$, solutions exist only where $F(x)\ge 5\lambda$: if $x_{-}<1<x_{+}$ solve $F(x_{\pm })=5\lambda$, there is a forbidden interval $x_{-}<x<x_{+}$. At either endpoint the two branches turn at $y=1$ and have an infinite slope.

A smooth sonic point needs $F^{\prime }(x)=F^{\prime }(y)=0$; hence a regular sonic transition requires $\lambda =1$ and $r=r_s$. The critical speed of a polytropic flow is then $c_s^2=GM/(2r_s)=2\mathcal{E}$, and the allowed mass flux is $j=(2\mathcal{E}{)}^{5/2}{r}_s^2/(3K/2{)}^2$. Here the subscript on $c_s$ denotes evaluation at the sonic point, rather than the isothermal sound speed convention used in the next solution.

The simple transonic branch is $y=x$. Direct substitution givesThus the radial velocity has constant magnitude, while the mass density and pressure scale as $r^{-2}$ and $r^{-3}$. For an outward fluid flow, this is a subsonic flow becoming a supersonic flow as its adiabatic sound speed decreases.

On the other transonic branch, $y$ decreases as $x$ increases. At large $r$, the terms $4/y$ and $x^4$ dominate, giving $y\sim 4x^{-4}$. At small $r$, the terms $y^4$ and $4/x$ dominate, giving $y\sim 4^{1/4}{x}^{-1/4}$. The resulting asymptotic expansions areFor inward fluid flow, these are the nearly static reservoir at infinity and the inner free fall region of Bondi accretion. The specific enthalpy is smaller than $GM/r$ in the inner limit, which explains free fall. Reversing the radial velocity gives a decelerating outflow with the same profiles. The formal small-$r$ limit describes the point-mass model: if the spherical gravitating body has a finite surface, the exterior solution stops there.

A stationary, nonradiative normal shock wave conserves the mass flux and the sum $v^2/2+h$, while the Newtonian gravitational potential is continuous across its thin layer. Hence $\mathcal{E}$ and $r_s$ are unchanged. The entropy production in a perfect-gas shock increases $K=p/{\rho }^{3/2}$, soMore explicitly, the Rankine-Hugoniot condition for $\gamma =3/2$ gives, with upstream $m={\mathcal{M}}_1^2>1$,Thus the shock wave jumps vertically, at the same $x$, from the supersonic flow to a subsonic flow on a larger-$\lambda$ curve. The figure illustrates an outward flow on $y=x$ with a shock wave at $x=1.4$; the downstream subsonic flow has $\lambda \simeq 1.224$ and continues to larger $x$. The shock wave radius is an additional boundary-data choice, rather than being determined by the smooth equations alone.

Take $u(r)>0$ to be the inward radial speed and select the regular critical solution for a gas reservoir of density ${\rho }_{\infty }$ and adiabatic sound speed $c_{\infty }$. Other mathematically subsonic steady solutions need a different inner boundary condition. For the polytropic equation of state, let $\gamma =1+1/n$, soMass conservation and the Bernoulli equation giveThe gas is at rest at infinity, so its kinetic and gravitational terms vanish there.

Differentiate these two conserved quantities, using $d\log \rho /dr=-2/r-u^{\prime }/u$. Eliminating the density derivative yields the polytropic Bondi accretion equationAt a regular sonic point, both sides vanish:The Bernoulli equation at that point reduces to $n{c}_{\infty }^2=(n-3/2)c_c^2$. For $n>3/2$, equivalently $1<\gamma <5/3$,Since $c_s^2\propto {\rho }^{1/n}$, the critical density is ${\rho }_c={\rho }_{\infty }[2n/(2n-3){]}^n$. Evaluate the conserved mass accretion rate at the sonic point:whereThe isothermal limit $\gamma \to 1$ gives $\lambda =e^{3/2}/4$, agreeing with Isothermal Bondi accretion.

For $\gamma =5/3$, $n=3/2$ and the Newtonian critical radius tends to zero while $c_c$ diverges. The limiting critical solution approaches Mach number one only as $r\to 0$; it has no finite-radius Newtonian sonic transition. This is a failure of the Newtonian inner approximation, not an obstruction to accretion onto a black hole. A physical Schwarzschild black hole has a finite event horizon, and the regular relativistic inflow becomes supersonic outside it.

The singular-looking factor in the rate has a finite limit. Writing $ϵ=n-3/2$, its logarithm is $ϵ\log [n/ϵ]\to 0$, so the critical Bondi accretion rate for gamma equals five thirds isThis is also the leading cold-reservoir limit of relativistic spherical accretion; an exact rate for relativistically hot gas requires the relativistic equations, not this Newtonian expression.

For completeness, the finite relativistic sonic point can be seen without assigning it an arbitrary inner radius. Let $b=c_s^2/c^2$ denote the relativistic squared sound speed and $g=\gamma -1$. For an isentropic relativistic gas with $p=K{\rho }_{\mathrm{rest}}^{\gamma }$, its dimensionless specific enthalpy is $h=g/(g-b)$. Conservation of rest-mass flux and relativistic Bernoulli energy gives the critical relationsFor $g=2/3$ and $c_{\infty }\ll c$, expansion gives $b_{\infty }\simeq (9/4)b_c^2$, henceHere the asymptotic relativistic and Newtonian sound speeds agree to leading order. This finite-radius transition explains why the gas can reach the horizon despite the Newtonian $r_c=0$ result.

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.

Well inside the Bondi accretion radius, pressure is negligible and the inflow approaches the free-fall speedSteady spherical mass conservation givesEstimating the outward optical depth across a radial scale $r$ with electron-scattering opacity $\kappa ={\sigma }_T/m_p$,The integral optical depth to infinity differs only by a factor of two in this free-fall approximation.

At the Schwarzschild radius, $\tau =1$ whenIf the conventional Eddington accretion rate is ${\dot{M}}_{\mathrm{Edd}}=L_{\mathrm{Edd}}/({\eta }_0{c}^2)$, then ${\dot{M}}_{\mathrm{crit}}=2{\eta }_0{\dot{M}}_{\mathrm{Edd}}$. This comparison depends on the reference efficiency ${\eta }_0$, while ${\dot{M}}_{\mathrm{crit}}$ itself follows directly from optical depth.

Writing $\dot{m}=\dot{M}/{\dot{M}}_{\mathrm{crit}}$ givesFor $\dot{m}>1$, the inner flow is optically thick and its photosphere lies at

The diffusion speed is $v_{\mathrm{diff}}\sim c/\tau$. Henceand equality defines the photon-trapping radiusInside $r_t$, inward advection outruns outward diffusion, so photon trapping in an accretion flow carries most released radiation into the black hole.

Only the binding energy released outside $r_t$ escapes efficiently. Its order of magnitude isThe radiative efficiency of black-hole accretion therefore decreases inversely with supply rate:Order-one coefficients depend on the optical-depth and inner-boundary conventions, but the luminosity saturation and $eta\propto {\dot{M}}^{-1}$ are robust.

Combining the two expressions for the sonic sound speed givesFor a polytropic gas, $c_s^2\propto {\rho }^{\gamma -1}$, and henceThe transonic accretion in a steep dark-matter cusp has mass accretion rateThus $\dot{M}\propto c_0^{-7}$, whereas classical Bondi accretion onto a point mass has ${\dot{M}}_{\mathrm{B}}\propto c_0^{-3}$. The steeper dependence comes from the cusp potential, whose sonic radius scales as $c_0^{-4}$ instead of $c_0^{-2}$.

This mode is most plausible for a nearly stationary supermassive black hole immersed in the hot, pressure-supported, X-ray-emitting atmosphere of a massive elliptical galaxy. Such gas is comparatively smooth, its random and rotational speeds can be smaller than its sound speed, and radiative cooling can be slow on the inflow scale.

The ideal Bondi accretion assumptions can fail in several independent ways. The gas may carry enough specific angular momentum to circularize into an accretion disk; a galaxy's stars and dark matter can dominate the gravitational potential outside the black hole's sphere of influence; the hole can move relative to the gas; cooling, conduction, or external heating can destroy adiabaticity or isothermality; a multiphase or clumpy medium is not a uniform reservoir; magnetic fields, turbulence, and viscosity add stresses absent from the spherical model; and AGN feedback, winds, or jets can expel or recirculate the inflow. Time dependence and self-gravity provide further failures. Thus a Bondi estimate is best interpreted as an idealized supply rate, not automatically as the rate crossing the event horizon.

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.

Write $w=1/r$ and $h=r^2\dot{\theta }=b{v}_{\infty }$. Since $\dot{r}=-hdw/d\theta$, the radial equation becomes the Binet equationwhose solution isChoose the incoming asymptote at $\theta =\pi$ and the downstream axis at $\theta =0$. Then $w\to 0$ as $\theta \to \pi$, while $r\sin \theta \to b$. These two conditions giveThe mirror-image streamlines meet on the downstream axis atAt that point each streamline has radial velocity $-v_{\infty }$ and equal and opposite azimuthal velocity. An inelastic collision cancels the latter, so the specific energy afterwards isThe gas is bound when $E_{\mathrm{after}}<0$, orSweeping the corresponding capture cylinder through gas of density ${\rho }_{\infty }$ gives the Bondi--Hoyle--Lyttleton accretion rateUnlike stationary spherical Bondi accretion, this is a directed, supersonic flow with a downstream focusing wake; bulk speed replaces sound speed as the main resistance to capture.

For an adiabatic Bondi accretion flow, $\dot{M}\propto M^2{\rho }_{\infty }{c}_{s,\infty }^{-3}$. Inside the Bondi radius the speed and ion temperature are approximately virial, so $v\propto (M/r{)}^{1/2}$ and $T\propto M/r$, while mass conservation gives $n\propto \dot{M}{M}^{-1/2}{r}^{-3/2}$. The frequency-integrated thermal bremsstrahlung emissivity is proportional to $n^2{T}^{1/2}$. Its volume integral is dominated by the inner flow and consequently scales asSince the Eddington luminosity is proportional to $M$, this may be written $L/L_{\mathrm{Edd}}=C(\dot{M}{c}^2/L_{\mathrm{Edd}}{)}^2$. With the standard fully ionized-plasma constants and the Bondi profiles, $\sqrt{C}=9\times 10^{-3}$. Eliminating $\dot{M}$ from $\eta =L/(\dot{M}{c}^2)$ then givesThe cancellation of $M$, ${\rho }_{\infty }$, and $c_{s,\infty }$ expresses the scale-free character of the ideal flow. More physically, two-body emission scales as density squared, so an increasingly dilute flow radiates a progressively smaller fraction of its available accretion power.

Denote the four terms byso the radial or poloidal momentum equation is $I=P+G+C$.

i) In a thin Keplerian accretion disk, radial inertia and pressure are higher-order in $H/R$, leaving $G+C\simeq 0$.

ii) In a nearly static stellar atmosphere, $I=C=0$ and hydrostatic pressure balance gives $P+G=0$.

iii) In pressureless gravitational collapse, rotation and pressure are negligible, so $I=G$; this is free fall.

iv) A slim accretion disk retains radial inertia and radial pressure together with gravity and centrifugal support, so all four terms generally survive: $I=P+G+C$.

v) A stationary geometrically thick disk or torus has negligible poloidal inertia but order-one pressure support, giving $P+G+C=0$.

vi) Nonrotating Bondi accretion has $C=0$ and $I=P+G$. In a highly supersonic Bondi--Hoyle limit the pressure term is also negligible, reducing this to ballistic $I=G$.

vii) A sub-Keplerian advection-dominated accretion flow has significant pressure support and radial inflow as well as rotation, so again $I=P+G+C$, with $C$ smaller than the Keplerian value and the remaining inward gravity balanced by $P$ and $I$.

Let $u>0$ be the inward radial speed. Steady spherical Bondi accretion conserves mass:For $\gamma =5/3$, $p=K{\rho }^{5/3}$ and the adiabatic sound speed satisfies $v_s^2=\gamma p/\rho \propto {\rho }^{2/3}$. Matching the reservoir therefore givesThe Bernoulli integral isWriting the Mach number as $\mathcal{M}=u/v_s$ and eliminating $v_s$ with mass conservation givesMultiplication of the Bernoulli equation by $r/2$ now yieldswhere

The radiative efficiency of black-hole accretion isFor a steady thin disk with negligible stress at the innermost stable circular orbit, matter radiates the binding energy lost before plunging, so $\eta =1-E_{\mathrm{ISCO}}$. Black-hole spin changes both the ISCO radius and its specific orbital energy. A prograde disk around a rapidly rotating Kerr black hole reaches deeper into the potential and is more efficient than a retrograde disk; representative ideal values run from $\eta \simeq 0.057$ for a Schwarzschild hole toward $0.42$ for an extremal prograde Kerr hole, reduced to about $0.3$ when photon capture limits astrophysical spin-up.

In adiabatic Bondi accretion, spherical compression raises the gas's internal energy reversibly, but much of that energy is advected through the horizon. There is no sustained shear stress that converts orbital binding energy into heat at a sequence of radii. A Shakura--Sunyaev thin disk, by contrast, must transport angular momentum outward. Its differential rotation stores free energy, local stress dissipates that energy as heat, and the short cooling time lets an optically thick disk radiate it before accretion. High efficiency therefore requires irreversible heating beyond adiabatic compression.

The likely source is MRI-driven magnetohydrodynamic turbulence. A weak magnetic field couples neighboring annuli; when angular velocity decreases outward, magnetic tension transfers angular momentum outward and amplifies the displacement. The alpha disk prescription replaces the unresolved turbulent stress by $T_{r\varphi }=\alpha P$, or equivalently $\nu \sim \alpha c_{s}H$. It captures the correct dimensional scale because subsonic turbulent motions are bounded by $c_s$ and their largest local eddies by $H$.

Its limitations include the following.

For steady spherical isentropic flow with $p=K{\rho }^{3/2}$ around a point mass, put $j=r^2\rho v>0$ and $\mathcal{E}=v^2/2+2c^2-GM/r>0$. Eliminating the mass density gives $c^2=[j(3K/2{)}^2{]}^{2/5}{r}^{-4/5}{\mathcal{M}}^{-2/5}$. With $r_s=GM/(4\mathcal{E})$, $x=(r/r_s{)}^{1/5}$, $y={\mathcal{M}}^{2/5}$ and $\lambda =[j(3K/2{)}^2{]}^{2/5}/(2\mathcal{E}{r}_s^{4/5})$, the Bernoulli equation reduces to $\lambda F(y)=F(x)$, where $F(t)=t^4+4/t$. Since $F$ has its unique minimum at $t=1$, a regular sonic point requires $\lambda =1$ and $r=r_s$. One transonic branch is $y=x$, which has $v=\sqrt{2\mathcal{E}}$, $\rho \propto r^{-2}$ and $p\propto r^{-3}$. The decreasing branch has inner free fall and an outer nearly static reservoir, giving Bondi accretion for inward fluid flow. A nonradiative stationary shock wave keeps $j,\mathcal{E},r_s$ fixed but raises $K$, so $\lambda$ increases by the factor $(K_2/K_1{)}^{4/5}$. The inward transonic solution is the spherical $n=2$, $\gamma =3/2$ specialization of transonic accretion in a power-law tube.

Heating a gas reservoir reduces its Bondi accretion rate because ${\dot{M}}_B\propto {\rho }_{\infty }{c}_s^{-3}$. At fixed density, increasing the sound speed by ten reduces the rate by about a thousand and the gravitational capture radius by a hundred. Expansion and density loss can strengthen this negative feedback. Neither the density nor the cold pre-heating accretion rate is determined by the temperature change alone.

Consider steady isentropic flow toward a Newtonian gravitational potential $-GM/r$ through a tube of cross-sectional area $A(r)=C{r}^n$. Write $v>0$ for the inward speed and use a polytropic equation of state with specific-heat ratio $\gamma >1$. Mass conservation and the Euler equations for an inviscid fluid givewhere $c_s$ is the adiabatic sound speed. A regular sonic point therefore has $v_s=c_s$ and $r_s=GM/(n{c}_s^2)$. The Bernoulli equation, matched to a nearly stationary reservoir with sound speed $c_0$, givesFor $n>1/2$, a finite positive sonic point requires $1<\gamma <(2n+1)/(2n-1)$. Differentiating the flow equation at that point, with $x=r_s{v}_s^{\prime }/c_s$, givesIts discriminant is $4n[(2n+1)-(2n-1)\gamma ]$, so the same bound permits real regular slopes. The branch on which the Mach number rises inward selects the negative sign inFor a spherical tube $n=2$, this recovers the $5/3$ threshold of Bondi accretion; for a dipolar flux-tube area, $n=3$ gives $7/5$. The tube approximation must remain valid between the reservoir matching region and the accretor.