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A converging-diverging nozzle that can accelerate a compressible flow from subsonic to supersonic speed through a sonic throat. The nozzle area-velocity relation requires acceleration in a converging subsonic section and a diverging supersonic section. Reservoir and downstream pressures determine whether this transonic branch operates; the shape alone does not ensure choked flow.

A steady spherical subsonic isothermal breeze with isothermal sound speed $a$ obeys $(u-a^2/u)u^{\prime }=2a^2/r-GM/r^2$. Since $0<u<a$, its finite-radius speed maximum lies at $r_M=GM/(2a^2)$. Spherical mass conservation then gives the displayed density, where ${\mathcal{M}}_M=u(r_M)/a\in (0,1)$ is the peak Mach number. This free parameter is selected by a boundary condition, not by $T$, $M$ and $\dot{M}$ alone. Therefore ${\rho }_M>\dot{M}{a}^3/(\pi G^2{M}^2)$, while a peak Mach number of order unity gives that expression as a characteristic scale.

A direct demonstration of the free parameter comes from the first integral $u^2/2-a^2\ln (u/a)=2a^2\ln (r/r_M)+GM/r+C$. Its additive constant remains free for a breeze; imposing the regular transonic branch would be a different condition.

For steady Isothermal Bondi accretion from gas at rest at infinity, define $\lambda =\dot{M}{c}_s^3/(4\pi G^2{M}^2{\rho }_{\infty })$, $x=r/r_s$ and $y=u/c_s>0$, where $r_s=GM/(2c_s^2)$. The Bernoulli function and mass conservation give $y^2/2-\ln y=2\ln x+2/x-\ln (4\lambda )$. The left side has minimum $1/2$ at $y=1$, and the radius-dependent part on the right has minimum two at $x=1$. Existence over all positive radii therefore requires the displayed bound. Equality permits the smooth accelerating-inward transonic branch and gives the usual isothermal Bondi accretion rate. Smaller positive rates have an everywhere-subsonic reservoir-connected branch, and zero rate gives a hydrostatic atmosphere. An absorbing inner boundary inside the sonic radius selects the critical inflow; steady isothermal equations and the outer density alone do not specify that inner condition.

With $x=r/r_s$ and $\mathcal{M}=|u_r|/c_s$, the radial isothermal equation integrates to ${\mathcal{M}}^2/2-\ln \mathcal{M}=2/x+2\ln x+C$. A regular transonic branch has $C=-3/2$, and the reservoir-fed accretion branch crosses $(x,\mathcal{M})=(1,1)$ with slope $-1$.

A narrow dipolar flux-tube area $A(z)\propto z^3$ gives $A\rho v=\dot{M}$ for inward speed $v>0$. The isothermal magnetic Bernoulli integral in $\Phi =-GM/z$ then gives $(v-c_s^2/v)v\prime =3c_s^2/z-GM/z^2$. A smooth transonic branch crosses at $z_s=GM/(3c_s^2)$ with $v_s=c_s$, ${\rho }_s=e^{5/2}{\rho }_{\infty }$ and $\dot{M}={\rho }_s{c}_{s}A(z_s)$. The crossing must lie in the domain; subsonic branches need not attain it.

A steady spherical Parker wind with constant isothermal sound speed $c$ and negligible wind self-gravity obeys $4\pi r^2\rho u=\dot{M}$ and $(u-c^2/u)u^{\prime }=2c^2/r-GM/r^2$. A regular sonic point has $u=c$ at $r_s=GM/(2c^2)$. With $x=r/r_s$ and $w=u^2/c^2$, integration gives the displayed equation for the smooth accelerating transonic branch. Its slope at the sonic point is $u^{\prime }(r_s)=c/r_s$. Maintaining constant gas temperature during expansion generally requires heat input, so this model is distinct from an adiabatic wind with $\gamma >1$.

For a barotropic fluid, define $h(\rho )=\int dp/\rho$. Dotting the steady Euler equation with $\mathbf{u}$ givesThus the Bernoulli function $H=u^2/2+h+\psi$ is constant along each streamline. Different streamlines can have different constants; steady flow alone does not imply one spatially uniform $H$.

In a quasi-one-dimensional flow through area $A(z)$ with negligible gravity, the axial momentum equation and $dp=c_s^{2}d\rho$ giveMeanwhile the mass flow rate $\dot{M}=\rho uA$ is constant, so $d\ln \rho =-d\ln u-d\ln A$. Substitution proves the nozzle area-velocity relationA subsonic flow accelerates where $A$ decreases, whereas a supersonic flow accelerates where $A$ increases. A smooth accelerating transonic branch in a de Laval nozzle therefore crosses $u=c_s$ at a minimum-area throat. Suitable reservoir and downstream pressures are also required: a converging-diverging shape by itself does not exclude a wholly subsonic solution or a downstream shock.

For the stipulated constant-$K$ polytropic flow, $c_s^2=\gamma K{\rho }^{\gamma -1}$. At the throat $u_m=c_{s,m}$, soThe specific enthalpy is $h=\gamma K{\rho }^{\gamma -1}/(\gamma -1)=c_s^2/(\gamma -1)$. Evaluating $H$ at the throat and substituting $\rho =\dot{M}/(Au)$ elsewhere givesOn the outward supersonic branch, $A\sim C{r}^2$ lets the gas expand and its specific enthalpy tend to zero for $\gamma >1$. ConsequentlyFor $\gamma =5/3$, $u_{\infty }=2c_{s,m}$. Also $T\propto {\rho }^{\gamma -1}\propto r^{-2(\gamma -1)}$. These are the adiabatic nozzle terminal speed and expansion laws for a lossless escaping branch; a subsonic branch could instead approach finite pressure with $u\propto A^{-1}$. A pressure-shaped astrophysical jet is being approximated by a nozzle here: the assumed geometry does not by itself explain the confinement or account for magnetic forces or relativistic motion in an actual active galactic nucleus.

A stellar wind carries matter away from a star, altering its mass, angular momentum and surroundings. Hot coronal gas can escape through pressure-driven expansion; radiation transfers momentum to gas through continuum or spectral line absorption, or to dust that then drags gas. A magnetic field can transfer rotational energy and angular momentum to an outflow. These mechanisms need not give the same terminal speed or mass loss rate.

For a steady spherical pressure-driven model, conservation of mass gives $\dot{M}=4\pi r^2\rho u$. Combining the radial Euler equation with an isothermal equation of state $p=\rho a^2$ gives the Parker wind equationA smooth transonic branch passes through $u=a$ at $r_s=GM/(2a^2)$, connecting a subsonic inner atmosphere to an escaping supersonic wind. The isothermal Parker wind requires continuing heat input as the gas expands; a polytropic stellar wind instead tracks a changing temperature and must satisfy its own terminal-energy condition. A subsonic isothermal breeze generally tends to a nonzero external pressure, so outward velocity alone does not select the escaping solution.

The solar wind supplies a nearby example of an outward plasma flow carrying a magnetic field. Profiles of spectral lines, especially absorption shifted by the Doppler effect together with emission, diagnose more distant stellar winds. The critical-point condition and boundary heating constrain $\dot{M}$; it is not determined by an escape-speed estimate alone.

Assume a steady spherical outflow of fixed composition. Let $\bar{m}$ be its mean particle mass, and define the constant isothermal sound speed by $a^2=k_B{T}_0/\bar{m}$. Subsonic here refers to this isothermal closure, so $0<u<a$. Mass conservation and radial momentum balance giveLogarithmically differentiating the first equation yields $d\ln \rho /dr=-2/r-u^{\prime }/u$. Substitution eliminates the density gradient and gives the isothermal Parker wind equationAt a smooth speed maximum, $u^{\prime }=0$, while the left prefactor is finite and negative. The right side must vanish, henceFor $r<r_M$ both the right side and the prefactor are negative, so $u^{\prime }>0$; for $r>r_M$, $u^{\prime }<0$. This is a true maximum on a subsonic isothermal breeze, not the accelerating transonic branch.

There is an underdetermination in the requested density estimate. Introduce the peak Mach number ${\mathcal{M}}_M=u_M/a$, where $0<{\mathcal{M}}_M<1$. The density at the maximum speed of an isothermal breeze is exactlyThe hypotheses do not fix ${\mathcal{M}}_M$. A boundary density or pressure, or an equivalent wind integration constant, is needed. The definite statement following from subsonicity is the lower boundIf a maximum speed of order $a$ is intended as an extra estimate, then ${\rho }_M\sim {\rho }_{\mathrm{char}}\propto \dot{M}{T}_0^{3/2}{M}^{-2}$. A very slow breeze can have a much larger density, so setting $u_M=a$ as an exact equality would contradict the stated always-subsonic assumption. The mean particle mass is another implicit fixed-composition input needed to express a temperature as a squared sound speed.

One can see the free parameter directly by integrating the wind equation:Different $C$ give different subsonic maxima at the same radius. The original plot illustrates this nonuniqueness. These breeze branches require nonzero ambient pressure at large radius; they are not the usual transonic escape into a low-pressure exterior.

Maintaining an isothermal expanding wind generally requires heat exchange; isothermal motion is not adiabatic motion.^[ref]

Following a parcel. Let $\mathbf{X}(t)$ be a fluid pathline, with $\dot{\mathbf{X}}=\mathbf{u}(\mathbf{X},t)$. The multivariable chain rule gives the material derivative,The first term measures change at fixed position; the advection term accounts for the parcel moving through spatial gradients. The derivative is also called the Lagrangian derivative.

The forces on a parcel. A material volume $dV$ has mass $\rho dV$. The pressure forces on its boundary sum to $-\nabla pdV$, while the gravitational acceleration is $-\nabla \psi$. Newton's second law therefore gives the Euler equations for an inviscid fluid,This neglects viscosity and other body forces. The left side is the parcel's acceleration, not just the time derivative at a fixed spatial location.

Bernoulli conservation. The claimed streamline result needs steady flow and a barotropic fluid, $p=p(\rho )$, or an equivalent enthalpy relation along each isentropic streamline. These hypotheses are necessary; the unsteady Euler equation alone does not imply the stated conservation law. Define the barotropic enthalpy functionDot the steady Euler equation with $\mathbf{u}$, using $\mathbf{u}\cdot [(\mathbf{u}\cdot \nabla )\mathbf{u}]=\mathbf{u}\cdot \nabla (u^2/2)$. This givesThus the Bernoulli function isEvery term has units of energy per mass, ${\mathrm{m}}^2{\mathrm{s}}^{-2}$. Different streamlines can have different constants.

The spherical accretion equation. Write the inward speed as $u(r)>0$, so the velocity vector is $-u(r)\hat{\mathbf{r}}$ and the positive mass accretion rate isAssume a prescribed central potential $\psi =-GM/r$ and negligible gas self-gravity. The steady radial Euler equation isFor a barotropic equation of state, $dp/dr=c_s^{2}d\rho /dr$, with $c_s^2=dp/d\rho$. Eliminate the density derivative to obtainA smooth sonic point requires the right side to vanish where $u=c_s$, henceFor a nonisothermal gas this is an implicit relation involving the local sound speed. In the isothermal case $c_s$ is constant and this is the Bondi sonic point.

The isothermal integral. For $p=c_s^2\rho$, choose the enthalpy reference so that $h_b=c_s^2\ln (\rho /{\rho }_{\infty })$. The reservoir is at rest at infinity and $\psi (\infty )=0$, making its Bernoulli constant zero. ThusorThe logarithm has a dimensionless argument. Constant temperature during compression generally requires heat removal, so this is Isothermal Bondi accretion, rather than adiabatic accretion with $\gamma >1$.

The critical accretion rate. Select the smooth reservoir-connected transonic branch. Evaluate the last relation at $u_s=c_s$ and $GM/r_s=2c_s^2$:Mass conservation then gives the isothermal Bondi accretion rate,The units are $\mathrm{kg}{\mathrm{s}}^{-1}$. Equivalently $\dot{M}=4\pi \lambda G^2{M}^2{\rho }_{\infty }/c_s^3$, with $\lambda =e^{3/2}/4\simeq 1.12$.

The inner-boundary selection of isothermal Bondi flow is important: a standard absorbing accretor inside the sonic radius selects this critical solution. The steady equations and ${\rho }_{\infty }$ alone also admit subcritical subsonic branches supported by a different inner boundary; they do not by themselves fix a unique rate. The usual answer presumes transonic capture.

Hydrostatic equilibrium with self-gravity. A resting fluid obeys $\nabla p=-\rho \nabla \Psi$. Spherical symmetry and the Poisson equation for Newtonian gravity giveEliminate the Newtonian gravitational potential derivative to obtainEquivalently, the shell theorem gives ${\Psi }^{\prime }=GM(r)/r^2$ and $M^{\prime }(r)=4\pi r^2\rho$, which leads to the same hydrostatic equilibrium equation.

A uniform-density liquid. With constant $\rho$, $M(r)=4\pi \rho r^3/3$. There is no central point mass, so the interior gravitational acceleration magnitude is $4\pi G\rho r/3$. ThusUse $p(R)=0$ at a vacuum boundary, neglecting surface tension. The uniform-density planetary pressure profile isThe central value can also be writtenIt has units pascals and decreases smoothly from its maximum at the centre to zero at the surface.

An isothermal gas. Here $c$ is the constant isothermal sound speed, with $p=c^2\rho$. Try $p=A/r^2$, so $\rho =A/(c^2{r}^2)$. The two sides of the equilibrium equation becomeThus $A=c^4/(2\pi G)$, giving the singular isothermal sphere,This does not describe a finite gaseous sphere in vacuum. Both pressure and mass density diverge at the centre, and $p(R)>0$ for every finite outer radius $R$, so it cannot meet $p=0$ there. Extending it to infinity removes the finite-radius pressure jump but makes the total mass infinite. It is useful as an idealized profile over an intermediate radial range, or as a truncated model with nonzero external pressure. The vacuum boundary obstruction for an isothermal sphere also applies to the regular, nonsingular isothermal solution; removing the central singularity does not produce a finite vacuum surface.

Steady outflow. In the wind calculation, take the central mass $M$ to be fixed and neglect the wind's own self-gravity. Constant mass flow rate and the isothermal equation of state giveThe steady radial Euler equation isSubstituting the density derivative givesThis is the isothermal Parker wind equation. A smooth sonic point requires both sides to vanish when $u=c$, soMaintaining constant temperature during expansion generally requires heat input; it is an isothermal model, not the adiabatic blast-wave model of Question 5.

The escaping branch at large radius. Integrate the wind equation, using dimensionless logarithm arguments:Set $x=r/r_s$ and $w=u^2/c^2$. ThenFor the accelerating transonic branch, $w(1)=1$, which fixes $C_0=-3$. At $x\gg 1$ on its supersonic branch, $w\to \infty$ and $\ln w$ is smaller than $w$. ConsequentlyA first correction makes the ordering explicit:This is the large-radius asymptotics of an isothermal Parker wind. The leading expression also holds for other supersonic branches with finite $C_0$.

The source implicitly selects an escaping wind. Positive steady outward flow alone does not determine that branch: a subsonic isothermal breeze also satisfies the differential equation. Its $w\ll 1$ limit gives $w\sim e^{-C_0}{x}^{-4}$, hence $u\propto r^{-2}$ and $\rho \to {\rho }_{\infty }>0$. Such a breeze needs nonzero pressure at infinity. Selecting a regular outflow from a subsonic base into vanishing ambient pressure instead gives the accelerating transonic wind used above.

The asymptotic speed has the same units as $c$; the logarithm is dimensionless. Its unlimited growth is a formal feature of indefinite isothermal heating and a prescribed point-mass potential. Real heating, surrounding pressure, and eventually the neglected wind mass limit that extrapolation.

Take $u<0$ for inward radial velocity, and assume a constant specific-heat ratio $\gamma >1$. The isentropic flow has $p=K{\rho }^{\gamma }$ and adiabatic sound speed $c^2=\gamma K{\rho }^{\gamma -1}$. Steady spherical mass conservation and the radial Euler equation for fluid motion giveEliminating the mass density derivative gives $(u-c^2/u)u^{\prime }=2c^2/r-GM/r^2$, orDifferentiating $c^2\propto {\rho }^{\gamma -1}$ gives the corresponding sound speed equation:At a smooth sonic point, $u_s^2=c_s^2$. A finite derivative requires the numerator of the flow equation to vanish too. ThusThis regularity requirement concerns a smooth transonic branch, rather than an arbitrary singular solution reaching unit Mach number.

Integrating the momentum equation using the specific enthalpy $c^2/(\gamma -1)$ yields the Bernoulli equationFor a finite positive mass accretion rate, its large-radius solution hasThese expressions satisfy both the flow equations asymptotically and show $u\to 0$ and $\rho \to {\rho }_{\infty }$. In particular $u^{\prime }/u\sim -2/r$ agrees with the first boxed equation. Combining the Bernoulli equation with sonic regularity additionally gives $c_s^2=2c_{\infty }^2/(5-3\gamma )$, so a finite positive sonic point occurs in the range identified below.

For the small-radius inward free fall branch, neglecting enthalpy against $GM/r$ in the Bernoulli equation predicts $u^2\sim 2GM/r$. The continuity equation then fixes $\rho \propto r^{-3/2}$, and consequently $c^2\propto r^{-3(\gamma -1)/2}$. The neglected-to-leading ratio isIt tends to zero precisely whenThis makes the assumed supersonic flow and leading free fall self-consistent. At equality, the ratio is of constant order rather than vanishing; above it, the proposed asymptotic balance fails. The isothermal limit $\gamma =1$ also permits free fall, with logarithmic rather than power-law enthalpy.

For the stellar funnel, write the magnetic scalar potential as $\psi =\mu \cos \theta /r^2$, where $R=r\sin \theta$ and $z=r\cos \theta$. Its magnetic dipole field has $B_r=-2\mu \cos \theta /r^3$ and $B_{\theta }=-\mu \sin \theta /r^3$, so a magnetic-field-line equation isThe edge of the polar flux tube meets the stellar surface at ${\theta }_{\ast }\simeq a/R_{\ast }$. Thus, near the pole, ${\theta }^2\simeq {\theta }_{\ast }^{2}r/R_{\ast }$, and its transverse radius is $R\simeq r\theta$. This proves the dipolar flux-tube area lawEquivalently magnetic flux conservation gives $A|\mathbf{B}|=\mathrm{constant}$ and $|\mathbf{B}|\propto r^{-3}$. The stipulated small cross section keeps the tube close to the polar axis, where its direction is approximately radial; the approximation is not valid all the way to arbitrarily large $r$.

Outside the star the prescribed field is a gradient, hence $\nabla \times \mathbf{B}=0$ and $\mathbf{j}=0$. Its Lorentz force density is therefore $\mathbf{j}\times \mathbf{B}=0$. The magnetic-pressure and magnetic-tension parts cancel in this exact current-free dipole. Field-aligned velocity also has $\mathbf{u}\times \mathbf{B}=0$, consistent with a steady prescribed field. The following calculation uses the narrow polar funnel approximation.

Now steady mass conservation reads $\rho uA=\mathrm{constant}$. With $A\propto r^3$, replace the spherical factor 2 by 3:The Bernoulli equation is unchanged because the magnetic field supplies no work or force. On a gravity-dominated free fall branch, $u^2\sim 2GM/r$, but $\rho \propto r^{-5/2}$ and $c^2\propto r^{-5(\gamma -1)/2}$. HenceThe same bound follows from regular transonic accretion in a power-law tube: $u_s^2=c_s^2=GM/(3r_s)$ and the reservoir-matched Bernoulli equation gives $c_s^2=2c_{\infty }^2/(7-5\gamma )$. The stellar surface cuts off the mathematical small-$r$ limit; the criterion refers to a funnel that becomes gravity dominated and crosses its sonic point before reaching that surface.

In a steady flow the ideal magnetohydrodynamic induction equation gives $\nabla \times (\mathbf{u}\times \mathbf{B})=0$. Since both vectors have only poloidal magnetic field components, write $\mathbf{u}\times \mathbf{B}=F_{\varphi }{\mathbf{e}}_{\varphi }$. Axisymmetry impliesFor a field regular on the axis, $C=0$; equivalently one may impose zero toroidal electromotive force. Under this physical condition, $\mathbf{u}$ is parallel to $\mathbf{B}$. Write $\rho \mathbf{u}=k\mathbf{B}$. Continuity and $\nabla \cdot \mathbf{B}=0$ then give $\mathbf{B}\cdot \nabla k=0$. The poloidal magnetic flux function has $\mathbf{B}\cdot \nabla \psi =0$, so locally on each connected magnetic surface,This regular-axis alignment of steady poloidal ideal flow needs its zero-circulation condition: axisymmetry alone does not imply this alignment on a domain excluding the axis. An explicit counterexample is available in a cylindrical annulus. Take nonzero constants $U,b$, positive ${\rho }_c,c_s$, andHere $\psi =-bz$, $\nabla \cdot \mathbf{B}=\nabla \times \mathbf{B}=0$ and $\mathbf{u}\times \mathbf{B}=Ub{\mathbf{e}}_{\varphi }/R$ has zero curl. The flow satisfies continuity, $p^{\prime }=-\rho {\Phi }^{\prime }$, and ${\nabla }^2\Phi =4\pi G\rho$. Its acceleration and Lorentz force density vanish, so it solves the steady equations while $\mathbf{u}$ and $\mathbf{B}$ are perpendicular. The nonzero toroidal circulation and the axial singularity explain why this counterexample is excluded from a regular polar accretion column. The remaining derivation uses the regular, aligned branch.

For a narrow stream tube, integrating continuity gives constant $A\rho |\mathbf{u}|$. On a nonzero-flow tube the magnetohydrodynamic mass loading $k$ is constant, so $A\rho |\mathbf{u}|=A|k||\mathbf{B}|$. Therefore $A|\mathbf{B}|$ is constant and $A\propto |\mathbf{B}{|}^{-1}$. This also follows directly from conserved magnetic flux through a flux tube.

For constant isothermal sound speed, $\nabla p/\rho =c_s^2\nabla \ln (\rho /{\rho }_0)$, with arbitrary reference density ${\rho }_0$. Project momentum along $\mathbf{B}$. The Lorentz force density has no component in that direction, and alignment implies $\mathbf{B}\cdot (\mathbf{u}\cdot \nabla )\mathbf{u}=\mathbf{B}\cdot \nabla (|\mathbf{u}{|}^2/2)$. Thus the isothermal magnetic Bernoulli integral isChanging ${\rho }_0$ only shifts the Bernoulli function by a constant. Since $k$ is constant along the magnetic field,Differentiating the isothermal magnetic Bernoulli integral therefore gives

In the slender polar accretion column, the dipolar flux-tube area is $A(z)=A_{\ast }(z/R_{\ast }{)}^3$. The decreasing axial field is a leading approximation, not an exactly solenoidal field $(0,0,B_z(z))$ throughout a cylinder. Indeed $\nabla \cdot \mathbf{B}={\partial }_z{B}_z\ne 0$ for that literal field. Near the axis a small radial component $B_R\simeq -(R/2){\partial }_z{B}_z=3R{B}_z/(2z)$ supplies the required radial divergence. Its magnitude is smaller by $R/z$, even though its divergence is leading order. Keeping this expanding flux tube while neglecting transverse forces yields the intended one-dimensional model.

Use the positive inward speed $v=-u_z$; the signed sonic velocity is $u_z=-c_s$. Conservation of mass and the Bernoulli equation giveEliminating ${\rho }^{\prime }$ yields the isothermal dipolar accretion equation,A smooth transonic branch requires both sides to vanish at its sonic point, soThis is the critical point of the flow equation; a generic subsonic solution need not cross it. A crossing in the exterior column requires $z_s\ge R_{\ast }$, and a sonic point strictly outside the star requires $z_s>R_{\ast }$. Differentiating the equation at the crossing gives $2(v_s^{\prime }{)}^2=3c_s^2/z_s^2$. For inward accretion that accelerates toward the star, $v_s^{\prime }=-\sqrt{3/2}{c}_s/z_s$.

The reservoir boundary condition fixes $ϵ=c_s^2\ln ({\rho }_{\infty }/{\rho }_0)$. At the sonic point, ${\Phi }_s=-3c_s^2$ and $v_s^2/2=c_s^2/2$, givingFinally use the sonic mass flux through the dipolar flux-tube area:This rate belongs to the smooth transonic branch of the idealized column, with $A_{\ast }$ the supplied total loaded area. The reservoir condition alone does not force every steady solution onto this branch. A real narrow dipolar column must match an outer flow where the slender approximation ceases to apply.

Write the radial velocity as $u(r)$, allowing $u<0$ for accretion flow, and let $c^2=\gamma p/\rho$ be the squared adiabatic sound speed. The steady continuity equation, radial Euler equations for an inviscid fluid and entropy advection equation giveOn a nonzero smooth flow branch, the polytropic equation of state is $p=K{\rho }^{\gamma }$ with constant $K$. Thus the two useful first integrals for spherically symmetric adiabatic flow areHere $\mathcal{J}$ is the signed mass flux and $\mathcal{B}$ is the Bernoulli function. The specific enthalpy is $c^2/(\gamma -1)$; the prescribed Newtonian gravitational potential has no contribution from the gas's own gravity. Eliminating the density derivative using mass conservation gives the differential formThis sonic point equation displays both the singular coefficient and the numerator that must vanish for a smooth transonic branch.

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.