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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.

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.

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 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.

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.

Steady spherical mass conservation gives $4\pi r^2\rho u=\dot{M}$. For the polytropic equation of state, $c_s^2=dp/d\rho$, soSubstitution into the radial Euler equations for an inviscid fluid gives the Parker wind equationAt a sonic point, the coefficient of $du/dr$ vanishes. A smooth transonic branch can pass it only if the right-hand side vanishes simultaneously. Hence

Let $u=-v_r>0$ denote the inward speed. Steady spherical mass conservation and the radial momentum equation for an isothermal gas giveEliminating $d\log \rho /dr$ produces the Isothermal Bondi equationA smooth flow can cross Mach number one only where both sides vanish. Its Bondi sonic point is thereforeThe integrated Bernoulli equation which approaches rest and density ${\rho }_{\infty }$ at infinity isAt the sonic point this gives ${\rho }_s=e^{3/2}{\rho }_{\infty }$, and hence the unique regular transonic branch has