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At a finite smooth sonic point $r_{\ast }$, the equation from (a) requiresSet $\delta =\gamma -1$ and $q=2/\beta -1/2$. Substitution into the Bernoulli function givesAn inflow from a warm reservoir at infinity has $\mathcal{B}=c_{\infty }^2/\delta >0$; an outflow reaching infinity has $\mathcal{B}=u_{\infty }^2/2+c_{\infty }^2/\delta \ge 0$. Therefore a nondegenerate transonic spherical flow in a power-law potential requiresFor the usual range $0<\beta <4$, the critical adiabatic exponent for spherical power-law flow is consequentlyThe printed assumption $\beta >0$ also permits $\beta \ge 4$. In that range the boxed inequality holds for every finite $\gamma >1$: there is no finite upper bound. Equivalently, take $f(\beta )=+\infty$ for $\beta \ge 4$. Extending the rational expression beyond $\beta =4$ would give an incorrect restriction.

The strict inequality also follows from local sonic-point slope discriminant analysis, including the potentially cold zero-energy endpoint. Put $X=(r_{\ast }/u_{\ast })u_{\ast }^{\prime }$. Since mass conservation implies $(c^2{)}^{\prime }/c^2=-\delta (u^{\prime }/u+2/r)$, differentiating the sonic point equation at $r_{\ast }$ yieldsIts discriminant is $8[2\beta -(4-\beta )\delta ]$. Moreover, the logarithmic Mach number derivative at the sonic point isA genuine crossing has two distinct branches and a nonzero derivative. At equality the crossing degenerates. In particular, for $\gamma =f(\beta )$ and $\mathcal{B}=0$, mass conservation and the polytropic equation of state make the Mach number constant along the scale-invariant solution; a solution that is sonic there is sonic everywhere, rather than crossing an isolated sonic point.

Use the positive mass accretion rate $\dot{M}=-\mathcal{J}$, and retain $\delta =\gamma -1$, $q=2/\beta -1/2$. Define $h=1-q\delta$, which is positive under the condition in (b). Matching the Bernoulli function to the reservoir and using the polytropic equation of state givesThe transonic spherical accretion rate in a power-law potential is thereforeThe combination $A/c_{\infty }^2$ has dimensions of length to the power $\beta$, so this expression has dimensions of mass per time. The sonic point selects the flux that connects the subsonic reservoir to the inward supersonic transonic branch.

For the endpoint limits of power-law spherical accretion, hold $A,\beta ,{\rho }_{\infty },c_{\infty }$ fixed. As $\delta \to 0$, $\ln h=-q\delta +O({\delta }^2)$, so $(q-1/\delta )\ln h\to q$. HenceThis is also obtained directly from an isothermal equation of state: the Bernoulli function becomes $u^2/2+c_{\infty }^2\ln (\rho /{\rho }_{\infty })-A/r^{\beta }=0$, so ${\rho }_{\ast }/{\rho }_{\infty }=e^q$. For $\beta =1$, $A=GM$, it recovers the isothermal Bondi accretion rate.

For $0<\beta <4$, $q>0$ and $\gamma \uparrow f(\beta )$ means $h\downarrow 0$. Since $\delta =(1-h)/q$,ThusThe finite limiting flux accompanies $r_{\ast }\to 0$ and $c_{\ast },{\rho }_{\ast }\to \infty$; it does not assert a finite-radius sonic point at the endpoint. For $\beta =1$ this is the familiar $\gamma \uparrow 5/3$ limit $\pi G^2{M}^2{\rho }_{\infty }/c_{\infty }^3$.

For completeness, the printed range $\beta \ge 4$ has the endpoint $f=+\infty$. At $\beta =4$, $q=0$ and the flux is independent of $\gamma$, namely $4\pi {\rho }_{\infty }(2A{)}^{1/2}$. For $\beta >4$, $q<0$ and $h\sim |q|\delta$, so $\dot{M}\to 0$ as $\gamma \to \infty$. These are the corresponding extended endpoint limits.

Let $e$ be specific internal energy, $Q$ the heat supplied per unit volume per unit time, and $E=\rho (e+|\mathbf{u}{|}^2/2)$. For an inviscid perfect gas in a prescribed Newtonian gravitational potential $\Phi$, the fluid total-energy equation isThere is no gravitational term if no external body is present. For a time-independent $\Phi$, adding potential energy gives the equivalent conservative form ${\partial }_t(E+\rho \Phi )+\nabla \cdot [(E+p+\rho \Phi )\mathbf{u}]=Q$. Subtracting the kinetic-energy equation and using mass conservation givesFor a perfect gas with constant specific-heat ratio $\gamma$, $e=p/[(\gamma -1)\rho ]$. The isothermal equation of state therefore makes $e=c_s^2/(\gamma -1)$ constant. Hence the required heat supply and heat loss areCompression requires cooling; expansion requires heating. The isothermal sound speed $c_s$ differs from the adiabatic sound speed $\sqrt{\gamma }{c}_s$.

Define the isothermal Mach numbers by ${\mathcal{M}}_i=u_i/c_s$. Orient the normal to a stationary isothermal shock along the flow, so the mass flux $j={\rho }_1{u}_1={\rho }_2{u}_2>0$. Tangential velocity is continuous for this planar hydrodynamic shock. Normal momentum conservation givesFor a genuine discontinuity $u_1\ne u_2$, factorization gives $u_1{u}_2=c_s^2$. Thus the isothermal shock jump relations areAcross a thin shock, gravitational potential and tangential kinetic energy are unchanged. The specific enthalpy $h=\gamma c_s^2/(\gamma -1)$ is the same on both sides, so the energy removed per unit area per unit time isPositive net cooling requires $u_1>u_2$, equivalently ${\rho }_2>{\rho }_1$ and ${\mathcal{M}}_1>1>{\mathcal{M}}_2$. Cooling across an isothermal shock permits only compression shocks; the formal expansion discontinuity requires energy supplied by the surroundings. The continuous state $u_1=u_2$ is not a shock.

For steady radial flow, mass conservation gives $r^2\rho u_r=\text{constant}$. Differentiating it and eliminating the mass density gradient from radial momentum balance givesA regular sonic point has $|u_r|=c_s$, so the right-hand side must vanish there as well. ThereforeThis is the Bondi sonic point for accretion; the same critical radius appears in an isothermal Parker wind. A general steady solution need not pass through a sonic point, but a smooth transonic one must obey both critical conditions.

Put $x=r/r_s$ and $\mathcal{M}=|u_r|/c_s$ on a branch whose flow direction is fixed. The radial equation becomesIntegrating yields the integrated isothermal Bondi flow relation:For a transonic branch, $x=\mathcal{M}=1$ gives $C=-3/2$. Expansion about the critical point gives $(\mathcal{M}-1{)}^2=(x-1{)}^2+O((x-1{)}^3)$, with slope $-1$ for inflow supplied by gas at rest at infinity and slope $+1$ for a transonic outflow.

For Isothermal Bondi accretion, integrate radial momentum balance once more, using $u_r\to 0$ and $\rho \to {\rho }_{\infty }$ at infinity:At the Bondi sonic point, $GM/r_s=2c_s^2$ and $u_r^2=c_s^2$, so ${\rho }_s=e^{3/2}{\rho }_{\infty }$. The positive inward isothermal Bondi accretion rate is consequentlyThe critical mass density and rate are fixed by the regular transonic solution and the specified reservoir at infinity; an arbitrary subsonic solution does not share this accretion rate.

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.

Assume a stationary spherical outflow with speed $v>0$, mass-loss rate $\dot{M}>0$, an optically thick thermal radiation field and negligible radiation inertia compared with matter inertia. The stellar wind continuity equation givesUse the printed gas fraction $\beta =P_{\mathrm{gas}}/P$, with $P=P_{\mathrm{gas}}+P_{\mathrm{rad}}$, and assume constant mean molecular weight ${\mu }_{\mathrm{mol}}$. Put $\mathcal{R}=k_B/({\mu }_{\mathrm{mol}}{m}_u)$ and $a_g^2=\mathcal{R}T$ for the isothermal sound speed squared. Then $P_{\mathrm{gas}}=\rho a_g^2$ and $P_{\mathrm{rad}}=a_{\mathrm{rad}}{T}^4/3$. The momentum equation including the radiation-pressure gradient isDo not also add a separate radiative force here: the diffusion radiation-pressure gradient already represents that force.

Using radiative diffusion in a star,Write $g=G{M}_r/r^2$ and $\Gamma =L_r/L_{\mathrm{crit}}=\kappa L_r/(4\pi cG{M}_r)$, the local ratio to the Eddington luminosity. Substituting continuity into the gas-pressure gradient gives the gas and radiation pressure stellar wind equationThe denominator changes sign at the sonic point $v=a_g$. A smooth transonic branch must have its right-hand numerator vanish there too. In a subsonic flow the denominator is negative, so acceleration requires a negative numerator; for a supersonic flow it requires a positive numerator. A regular subsonic-to-supersonic branch must therefore pass through a compatible common zero, with a finite slope obtained by differentiating the equations. Boundary conditions and that regularity requirement normally select the mass flux. Other branches can remain subsonic, remain supersonic or encounter a singular gradient. In the isothermal nonradiating limit the numerator is $2a_g^2/r-GM/r^2$, recovering the Parker wind equation and critical radius $GM/(2a_g^2)$.

To expose the role of $\beta$, rewrite the diffusion temperature gradient asHenceAt the stipulated local value $\Gamma =1$, with $0<\beta <1$, its numerator isThe gas-pressure contribution is not removed by setting the luminosity equal to the Eddington value. Instead the negative temperature gradient makes the thermal term in the numerator positive as well. A subsonic outflow then has $dv/dr<0$, and no common numerator/denominator zero is available for smooth sonic passage. This is the sonic-point obstruction at the Eddington luminosity.

Thus a subsonically launched wind cannot accelerate smoothly to supersonic speed under these diffusion assumptions at $L_r/L_{\mathrm{crit}}=1$. It is not a claim that every initially supersonic solution decelerates; those have positive denominator and accelerate in this equation. Even in the radiation-dominated limit $\beta \to 0$, the spherical thermal term remains positive for $a_g^2>0$. The singular pressureless limit is not a finite-temperature sonic crossing. Finally $a_g$ is the thermal sound scale of this diffusion-controlled stationary equation, not the full adiabatic sound speed of a tightly trapped gas-plus-radiation perturbation. Optically thin winds or different heating/driving mechanisms require another transport model.

A regular sonic point must make both sides of the flow equation vanish:Substituting into the Bernoulli equation yieldsFor $\gamma >1$, positive finite sound speed therefore requires the critical adiabatic index for dipolar accretionTo check that this gives real regular crossings, differentiate the flow equation at the sonic point and put $x=r_s{v}_s^{\prime }/c_s$. The transonic accretion in a power-law tube calculation givesThe minus sign gives the transonic branch whose Mach number increases inward. At $\gamma =7/5$ the positive-energy reservoir cannot match a finite sonic point; for larger $\gamma$ the required $c_s^2$ is negative. A physical surface-crossing solution also requires $r_s>R_{\star }$ and validity of the narrow flux tube approximation up to the sonic region.

The transonic accretion in a dipolar flux tube values areThe mass density follows from $c^2/c_0^2=(\rho /{\rho }_0{)}^{\gamma -1}$ for the same polytropic equation of state. If $A_{\star }$ is the total loaded area at the surface, mass conservation at the sonic point givesFor two equal polar caps with surface angular radius ${\theta }_{\star }$, $A_{\star }\simeq 2\pi R_{\star }^2{\theta }_{\star }^2$, so the total mass accretion rate isFor only one loaded polar cap, divide this expression by two. Expressing the result first in $A_{\star }$ also covers a bundle whose loading angle is defined at another radius. The smooth isothermal limit $\gamma \to 1$ has $r_s=G{M}_{\star }/(3c_0^2)$, $c_s=c_0$, and ${\rho }_s=e^{5/2}{\rho }_0$.

The dipole magnetic-field line geometry gives the angular radius at the sonic point as ${\theta }_s\simeq {\theta }_{\star }\sqrt{r_s/R_{\star }}$. The sonic region must remain a narrow flux tube for the one-dimensional transonic branch equations to apply. Thus a necessary small-angle constraint isThis also makes the transverse sound-crossing time $r_s{\theta }_s/c_s$ small compared with the radial flow time $r_s/c_s$. The restriction must hold at any larger matching radius as well. To match to a reservoir with negligible gravitational potential within the same narrow dipolar approximation, one needs an overlap regionHence the stronger useful sufficient scaling is ${\theta }_{\star }^2\ll R_{\star }{c}_0^2/(G{M}_{\star })$. The small-angle dipolar model should not be extrapolated to literal infinity, where its boundary magnetic field line turns toward the equatorial plane.

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 continuous transonic branch passes from $dj/du>0$ to $dj/du<0$, so $j(u)$ has a maximum where its derivative vanishes. At this critical speed of a polytropic flow,The maximum mass flux is therefore attained at the sonic point.