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With the polytropic equation of state and substitutionswe haveThe pressure-support equation therefore becomesChooseChanging from $r$ to $\xi$ then yields the Lane-Emden equationThese are the Lane-Emden variables for a stellar polytrope.

In the narrow polar flux tube, the magnetic field lines are almost radial and the transverse width is much smaller than the radial scale. The Lorentz force density has no component along the magnetic field, so the longitudinal magnetically channelled accretion is hydrodynamic. Let $v(r)>0$ denote inward speed. Mass conservation, the Euler equations for an inviscid fluid, and the polytropic equation of state giveUsing the adiabatic sound speed $c^2=\gamma K{\rho }^{\gamma -1}$ and $A\propto r^3$,The Bernoulli equation iswhere the right-hand side comes from matching to a nearly stationary reservoir with negligible gravitational potential. The fixed magnetic field provides transverse confinement; neglecting magnetic forces in this one-dimensional equation concerns their longitudinal component.

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 ${\rho }_s=e^{5/2}{\rho }_0$.

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 one-dimensional Isentropic Euler equations areFor the polytropic equation of state,The continuity equation therefore implieswhile the momentum equation becomesAdding and subtracting these equations givesThus the Riemann invariantsare constant along the characteristic curves

The Bernoulli function for steady unmagnetized flow without gravity isFor a polytropic equation of state, $h=c_s^2/(\gamma -1)$. Evaluation at the critical point found in part b givesConsequentlySince $C_B$ is conserved along a streamline, that streamline has a unique critical speed.

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

Put $K(r)=e^{s(r)}$. The perfect gas relation in the question becomes the polytropic equation of stateAt fixed $r$, vertical hydrostatic balance isAfter one integration,Defining the midplane adiabatic sound speed by $c_{s0}^2=\gamma P_0/{\rho }_0=\gamma K{\rho }_0^{\gamma -1}$ givesThis is a vertically truncated polytropic atmosphere; the physical perfect-gas case has $\gamma >1$.

For $p=K{\rho }^{1+1/n}$ with $K,n>0$, the specific enthalpy relative to vacuum is $Q=(n+1)K{\rho }^{1/n}=(n+1)p/\rho$. It obeys $\nabla Q=\nabla p/\rho$ and $n{D}_{t}Q=-Q\nabla \cdot \mathbf{u}$. The vacuum reference is fixed when these equations are combined with the polytropic equation of state.

A polytropic flow uses a polytropic equation of state. When $K$ and the adiabatic exponent are constant, a smooth streamline has specific enthalpy $h=\gamma K{\rho }^{\gamma -1}/(\gamma -1)$ and adiabatic sound speed $c_s^2=\gamma K{\rho }^{\gamma -1}$. Spatially varying entropy can make $K$ depend on streamline; the constant-$K$ spherical wind is a special case.

In harmonic stellar gravity, an ideal gas with a polytropic equation of state of index $n>0$ has a quadratic temperature profile and finite semi-thickness $H^2=2(n+1)\mathcal{R}{T}_0/{\Omega }^2$. This differs from the polytropic vertical structure of a self-gravitating disk.

The dipolar flux-tube area has $A(r)=A_{\star }(r/R_{\star }{)}^3$. Applying transonic accretion in a power-law tube with $n=3$ and a polytropic equation of state givesHere $c_0$ and ${\rho }_0$ are the reservoir sound speed and mass density, and $c_s$ is the sound speed at the sonic point. The mass accretion rate isFor two equal polar caps of surface angular radius ${\theta }_{\star }\ll 1$, $A_{\star }\simeq 2\pi R_{\star }^2{\theta }_{\star }^2$. One loaded cap gives half the total rate. The finite-radius transonic branch exists for $1<\gamma <7/5$, must cross outside the star ($r_s>R_{\star }$), and requires ${\theta }_{\star }^2{r}_s/R_{\star }\ll 1$ for the sonic region to remain a narrow flux tube.

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

For positive finite opacity and nonvacuum ideal gas satisfying radiative diffusion, zero flux implies $dT/dz=0$. Write $P=\mathcal{R}\rho T$, where $\mathcal{R}=k_B/({\mu }_m{m}_p)$ for constant mean molecular weight ${\mu }_m$ in proton-mass units. If a polytropic equation of state $P=K{\rho }^{\gamma }$ with constant $K>0$ is also imposed, then $T=(K/\mathcal{R}){\rho }^{\gamma -1}$. For $\gamma \ne 1$, constant temperature would force constant density, contradicting vertical hydrostatic equilibrium $P^{\prime }=-\rho {\Omega }^{2}z$ on an open interval away from the midplane. The familiar finite polytropic slab for $\gamma >1$ is therefore a solution of the hydrostatic/adiabatic closure, not of this finite-opacity zero-radiative-flux system. For $\gamma =1$, the consistent isothermal profile is $\rho ={\rho }_m\exp [-{\Omega }^2{z}^2/(2\mathcal{R}{T}_m)]$, with no finite surface.