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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.

Align the polar axis with the magnetic dipole moment. The magnetic dipole field hasThe magnetic-field-line equation gives $dr/(rd\theta )=2\cot \theta$, hence the dipole magnetic-field line is $r=C{\sin }^2\theta$. Normalize the loaded bundle by its small surface angular radius ${\theta }_{\star }$ at $r=R_{\star }$. Its boundary obeysThe dipolar flux-tube area for one polar cap is consequentlyEquivalently, $BA$ is constant by conservation of magnetic flux and $B\propto r^{-3}$ near the axis. Two equal loaded caps double the total area. The $r^3$ scaling is independent of the normalization; if the loading angle is specified at another radius $r_L$, use $A(r)=\pi {\theta }_L^2{r}^3/r_L$ instead.

The magnetic-field-line equation for the dipole isIntegration gives the dipole magnetic-field lineThe line crosses the midplane at $r=R_e$, so $C=R_e$. At the stellar surface,Consequentlyfor a small polar-cap opening angle.

For the homologous spherical flow $\mathbf{u}=q\mathbf{r}$, the ideal magnetohydrodynamic induction equation and $\nabla \cdot \mathbf{B}=0$ give the material derivativeLet $x=r/R(t)$ and use the self-similar ansatzPart (b) gives $q=3\dot{R}/(4R)$, and thereforeThe radial induction equation becomes $x{f}^{\prime }=6f$. The boundary value $f(1)=1$ gives $f=x^6$. The solenoidal vector field condition requireswhich also matches the tangential shock value in part (c). Hence the interior field is

Outside the shock, the uniform-field lines obey $r\sin \theta =\text{constant}$. Inside, the magnetic-field-line equation givessoA sketch therefore shows straight exterior lines refracting at the spherical shock into north-south symmetric curves that bow toward the equatorial interior before leaving through the opposite hemisphere. The field strength falls as $r^6$ toward the centre.