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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 $5/3$.

At a regular sonic point, differentiating the vanishing coefficient and numerator yields $(2+C/c_s^2)(u^{\prime }{)}^2+4C{u}^{\prime }/(ru)+(4C+2c_s^2)/r^2+{\Phi }^{\prime \prime }=0$. A real slope compatible with the desired branch is needed in addition to the sonic equalities. For constant sound speed this reduces to $(u^{\prime }{)}^2=-c_s^2/r^2-{\Phi }^{\prime \prime }/2$.

The dimensionless accretion factor $(1-q\delta {)}^{q-1/\delta }$ tends to $e^q$ as $\delta \to 0$. For $q>0$, it tends to $1$ as $\delta \uparrow 1/q$, although the sonic point moves to the origin. Writing the latter logarithm as $-qh\ln h/(1-h)$ makes the finite limit transparent. These limits connect transonic spherical accretion rate in a power-law potential to the isothermal equation of state and the limiting critical adiabatic exponent.

In an optically thick steady spherical stellar wind of constant mean molecular weight ${\mu }_{\mathrm{mol}}$, let $v$ be the outward speed and $a_g^2=k_{B}T/({\mu }_{\mathrm{mol}}{m}_u)$ the isothermal sound speed squared. Mass conservation gives ${\rho }^{\prime }/\rho =-2/r-v^{\prime }/v$, where a prime denotes differentiation with respect to radius $r$. Use $P_{\mathrm{gas}}=\rho a_g^2$ and radiative diffusion to write $P_{\mathrm{rad}}^{\prime }/\rho =-g\Gamma$, where $g=G{M}_r/r^2$ and $\Gamma =L_r/L_{\mathrm{Edd}}$ uses the local Eddington luminosity $L_{\mathrm{Edd}}=4\pi cG{M}_r/\kappa$ and opacity $\kappa$. Substituting into the momentum equation yields the displayed wind equation. Smooth sonic point crossing requires both its denominator and numerator to vanish. The radiation-pressure gradient already supplies radiative acceleration; adding that force separately would count it twice.

For $\beta =16c_s^2/c^2$, the exterior sonic point of the barotropic spherical accretion equation lies outside $r_G=2GM/c^2$. Its mass density is ${\rho }_c={\rho }_{\infty }\exp [4/(1+\sqrt{1+\beta })-1/2]$. The selected smooth transonic accretion rate is $4\pi r_c^2{\rho }_c{c}_s$. Subsonic formal solutions are not assigned this rate without transonic selection.

Put $y=\rho r^{4/(\gamma +1)}$ and $\alpha =4(\gamma -1)/(\gamma +1)$. The polytropic stellar wind equations become $F(r)=C{r}^{\alpha }+GM{r}^{\alpha -1}=G(y)=\gamma K{y}^{\gamma -1}/(\gamma -1)+{\dot{M}}^2/(32{\pi }^2{y}^2)$. At a regular sonic point both derivatives vanish. Matching their minima allows the two positive-$y$ branches to join with finite slope; unequal minima either leave a forbidden radial interval or separate the subsonic and supersonic branches. This reduces a singular differential-equation crossing to a geometric minimum comparison.

For isentropic spherical accretion with this adiabatic index, define $\lambda =\dot{M}{c}_{\infty }^3/(4\pi G^2{M}^2{\rho }_{\infty })$ and $x=r{c}_{\infty }^2/(GM)$. The steady-flow equation is $\sqrt{\lambda }[{\mathcal{M}}^{3/2}/2+3{\mathcal{M}}^{-1/2}/2]=1+3x/2$. The bracket has minimum two at Mach one. Reaching arbitrarily small positive radii therefore requires $\lambda \le 1/4$. For smaller rates the outer-rest branch remains subsonic; at the maximal rate it approaches Mach one only as $r\to 0$, not at a finite sonic point. Supersonic outer branches fail the specified resting finite-density condition, and larger rates terminate at a positive minimum radius. An additional inner boundary prescription is needed to select a unique rate among the globally defined subsonic solutions.

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.

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.

A solution reaching arbitrarily small radii must obey $\sqrt{\lambda }h(\mathcal{M})=1+3x/2$ with $h\ge 2$. Taking $x\downarrow 0$ therefore requires $2\sqrt{\lambda }\le 1$, or $\lambda \le 1/4$. This upper bound is attained: at $\lambda =1/4$ the decreasing branch $0<\mathcal{M}<1$ has one solution for every $x>0$ and approaches Mach one as $x\to 0$. HenceThe argument establishes the maximum allowed by a globally continuing steady flow, without incorrectly claiming a finite-radius sonic point at $\gamma =5/3$.

For the isothermal equation of state, $p=c_s^2\rho$, with constant isothermal sound speed $c_s$. A nonzero steady plane-parallel flow has constant mass flux $\rho u=j$. Choose the positive flow direction for convenience. The continuity equation then gives ${\rho }^{\prime }/\rho =-u^{\prime }/u$, and the steady Euler momentum equation becomesIntegrating and absorbing the reference velocity into the constant yields, for the Mach number $\mathcal{M}=u/c_s$,For the opposite flow direction use $\mathcal{M}=|u|/c_s$; the same relation holds.

A smooth sonic point has $u=c_s$, so its momentum equation requires ${\Phi }^{\prime }=0$. Moreover $f(\mathcal{M})={\mathcal{M}}^2/2-\ln \mathcal{M}$ satisfies $f^{\prime }(1)=0$, $f^{\prime \prime }(1)=2$ and has its minimum $f(1)=1/2$. Thus, if the flow reaches $\mathcal{M}=1$ at $x_{\ast }$,A sonic crossing is therefore possible at a maximum of the potential, withAt a nondegenerate maximum, the expansions give $(\mathcal{M}-1{)}^2=-{\Phi }^{\prime \prime }(x_{\ast })(x-x_{\ast }{)}^2/(2c_s^2)+\cdots$. Choosing either smooth sign gives a crossing slope ${\mathcal{M}}^{\prime }(x_{\ast })=\pm \sqrt{-{\Phi }^{\prime \prime }(x_{\ast })/(2c_s^2)}$, joining a subsonic flow to a supersonic flow, or the reverse. This establishes the plane-parallel isothermal sonic transition at a potential maximum. For a globally defined crossing, the same maximum must also be high enough for the Bernoulli relation to remain real over the intended domain.

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.

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.

For steady spherical polytropic flow, write $u=u_r$ and use the polytropic equation of state $p=K{\rho }^{\gamma }$, with $K>0$. Mass conservation and radial Euler momentum equation areThe mass equation gives ${\rho }^{\prime }/\rho =-2/r-u^{\prime }/u$. Substitute it into momentum balance to obtainAt a sonic point the derivative coefficient vanishes. A smooth finite-slope solution must make the numerator vanish there too:Finally $dp/\rho =d[\gamma K{\rho }^{\gamma -1}/(\gamma -1)]$. Integrating momentum gives the Bernoulli equationThis is kinetic energy plus specific enthalpy plus gravitational potential per unit mass. The polytropic stellar wind selects a particular transonic branch of these equations.

The outward mass loss rate is $\dot{M}=4\pi r^2\rho u>0$. Put $\beta =4/(\gamma +1)$, $\alpha =4(\gamma -1)/(\gamma +1)$ and $y=\rho r^{\beta }$. Both terms in the Bernoulli equation then have the same radial scaling: $\beta (\gamma -1)=4-2\beta =\alpha$. Eliminate $u$ and multiply by $r^{\alpha }$ to obtainFor positive terminal speed, $C>0$. Differentiation gives $F^{\prime }=r^{\alpha -2}[\alpha Cr-(1-\alpha )GM]$. It has an interior minimum precisely when $0<\alpha <1$, or $1<\gamma <5/3$, atAt a sonic point, the Bernoulli equation and $GM/r_c=2c_{s,c}^2$ give $C=c_{s,c}^2(5-3\gamma )/[2(\gamma -1)]$, recovering this radius. The limiting $C=0$, $\gamma =5/3$ case has no isolated interior minimum selected by these formulas.

The function $G$ diverges at both ends of $y>0$ and has a unique minimum whereThis condition is $u_c^2=c_{s,c}^2$. Thus a smooth crossing requires minimum matching at a polytropic sonic point: $F(r_c)=G(y_c)$. If the minimum of $F$ is lower, an interval has no real positive solution; if it is higher, the two $y$ branches stay separate and cannot cross the sonic value. When the minima agree, their positive second derivatives give $G^{\prime \prime }(y_c)(y-y_c{)}^2\sim F^{\prime \prime }(r_c)(r-r_c{)}^2$, allowing a finite-slope branch to pass between them. The wind with finite positive terminal speed takes the lower-$y$ branch outside $r_c$, because $\rho \sim \dot{M}/(4\pi \sqrt{2C}{r}^2)$ and $\beta <2$ make $y\to 0$.

At the matching point,Insert these into $\dot{M}=4\pi r_c^2{\rho }_c{c}_{s,c}$:The positive-$C$, $1<\gamma <5/3$ range is part of this finite-radius transonic wind result, rather than an unrestricted formula for every $\gamma >1$.

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.