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A regular level surface of the axisymmetric poloidal magnetic flux function contains both the poloidal and toroidal magnetic field directions. Conservation along field lines can be expressed locally as a function of the flux label on connected regular surfaces. Degenerate points with no poloidal magnetic field and disconnected branches with the same label require separate treatment.

An isolated axisymmetric magnetic field cannot be sustained by ordinary resistive dynamo action with scalar positive magnetic diffusivity. Its poloidal magnetic flux function satisfies a homogeneous scalar advection-diffusion equation with no toroidal-to-poloidal source. The maximum principle excludes regeneration of that flux; after it decays the toroidal field has no lasting source either. This constrains the symmetry of the field. An axisymmetric flow may still generate a nonaxisymmetric field, and a turbulent mean-field electromotive force is not covered by the unmodified scalar-induction proof.

The wind integrals require a steady axisymmetric magnetohydrodynamic wind in ideal magnetohydrodynamics, with no viscous azimuthal torque or imposed toroidal loop voltage. Axisymmetry by itself is insufficient. Assume smooth connected poloidal flux surfaces with nonzero poloidal magnetic field; functions of the flux label below are understood on each such branch. In cylindrical coordinates, the given poloidal magnetic flux function obeysFor a steady ideal wind, the electric field is $\mathbf{E}=-\mathbf{u}\times \mathbf{B}$ and $\nabla \times \mathbf{E}=0$. Its toroidal component satisfies ${\partial }_z{E}_{\varphi }=0$ and ${\partial }_s(s{E}_{\varphi })=0$, so $E_{\varphi }=C/s$. Regularity at the axis, or equivalently the absence of an imposed loop voltage in an annular domain, gives $C=0$. Thus $({\mathbf{u}}_p\times {\mathbf{B}}_p{)}_{\varphi }=0$ and ${\mathbf{u}}_p=f{\mathbf{B}}_p$. The continuity equation and $\nabla \cdot {\mathbf{B}}_p=0$ now yieldConsequently $\kappa$ is the magnetohydrodynamic mass loading. Including the toroidal magnetic field, rewrite the full velocity asSince $\mathbf{u}\times \mathbf{B}=\omega \nabla \chi$, steady ideal induction gives $0=\nabla \omega \times \nabla \chi$. Hence $\omega =\omega (\chi )$, establishingThe invariant $\omega$ is the field-line angular velocity of an axisymmetric wind, not generally the fluid's angular velocity $\Omega$. When an ideal magnetic surface is anchored to a rigidly rotating star, $\omega$ equals the stellar angular velocity. With no poloidal outflow this reduces to Ferraro's law of isorotation.

To calculate the magnetic torque, use the azimuthal momentum equation. Axisymmetric pressure and Newtonian gravity have no azimuthal components; the magnetic pressure has none either. The azimuthal acceleration and magnetic tension contain the curvature terms characteristic of cylindrical coordinates:Multiplication by $s$ gives the Maxwell torque conservation in an axisymmetric wind equationUsing $u_{\varphi }=s\Omega$, $\rho {\mathbf{u}}_p=\kappa {\mathbf{B}}_p$ and ${\mathbf{B}}_p\cdot \nabla \kappa =0$ provesThis is the total angular momentum transport per unit poloidal magnetic flux. The first term is the material contribution and the second is the contribution of the Maxwell stress tensor. The specific magnetohydrodynamic angular-momentum invariant is $L=\ell /\kappa$ when $\kappa \ne 0$; its normalization differs from the paper's $\ell$.

Define the poloidal Alfvén Mach number bySubstituting $\Omega =\omega +\kappa B_{\varphi }/(\rho s)$ into $\ell$ givesAt a smooth crossing of the Alfvén surface, $M_A^2=1$ and the toroidal magnetic field is finite. The Alfvén-surface regularity condition for an axisymmetric wind is thereforeOne can also expose the apparent singularity explicitly:Finite passage through the Alfvén surface requires the numerator to vanish with the denominator. Equality of the numerator zeros is necessary; a globally smooth wind must also satisfy its other dynamical critical conditions. Here $s_A$ is the cylindrical distance from the rotation axis, not automatically the spherical Alfvén radius.

For a tube carrying outward mass flux $d\dot{M}=\kappa {\mathbf{B}}_p\cdot d\mathbf{S}$, the outward angular momentum flux isThus the stellar torque is ${\dot{J}}_{\star }=-\int \omega s_A^{2}d\dot{M}$, with the integral over the escaping wind. Magnetic stresses give each mass element an effective angular-momentum lever arm equal to the cylindrical Alfvén crossing radius. In a strongly sub-Alfvénic region the fluid approximately corotates with the magnetic surface; beyond that region it can carry the accumulated angular momentum away. For an outward field and positive $\omega$, a super-Alfvénic expanding tube with $s>s_A$ has $B_{\varphi }<0$, so the magnetic stress transports positive angular momentum outwards. If $s_A$ greatly exceeds the stellar surface lever arm, a modest mass-loss rate can cause substantial wind-driven magnetic braking of a solar-type star. An unmagnetized outflow would instead carry approximately the material angular momentum fixed at its launch radius.

Write the poloidal magnetic flux function and toroidal amplitude explicitly in spherical components:Dotting the magnetostatic equilibrium equation with $\mathbf{B}$ gives $\mathbf{B}\cdot \nabla p=0$. Since all fields are axisymmetric, the poloidal magnetic field is tangent to the contours of $\chi$ in a meridional section. Thus, locally on regular connected flux surfaces, $p=p(\chi )$. This is a local flux-surface statement; disconnected surfaces with the same numerical value of $\chi$ need not have the same pressure without further global assumptions.

Taking the azimuthal component of the Lorentz force givesIts numerator is the meridional Jacobian determinant of $\psi$ and $\chi$. Vanishing therefore implies locally $\psi =\psi (\chi )$ on the same regular surfaces. DefineA direct curl calculation now givesSince ${\mathbf{e}}_{\varphi }\times {\mathbf{B}}_{\mathrm{pol}}=\nabla \chi /(r\sin \theta )$, the remaining force isConsequently the spherical magnetostatic pressure balance equation following from the stated physical force balance isThe PDF's displayed scalar equation has the wrong sign and omits ${\mu }_0$ for its stated pressure p. This is not corrected by setting ${\mu }_0=1$, which leaves the sign discrepancy. For example, $\chi =r^4{\sin }^2\theta$, $\psi =0$, $p=p_0-10\chi /{\mu }_0$ satisfies the stated vector force balance and the corrected scalar equation, because ${\Delta }_{\ast }\chi =10r^2{\sin }^2\theta$; it fails the printed scalar equation. One could obtain the printed form by replacing its pressure variable by $-{\mu }_0{p}_{\mathrm{physical}}$, but that is not the variable in the given force balance.

Now set $\psi =\alpha \chi$, $p=p_0+P\chi$, and $\chi =F(r){\sin }^2\theta$. The angular operator gives $-2F{\sin }^2\theta /r^2$, soFor the trigonometric form in the question, direct substitution yieldsFor $\alpha \ne 0$, this is a solution providedThe same coefficient relation makes the trigonometric pair regular at the origin: $\cos \alpha r-\sin \alpha r/(\alpha r)=-{\alpha }^2{r}^2/3+O(r^4)$. Thus all components of the magnetic field are finite there. With $p_0=0$, this explicitly constructs the requested proportional-pressure family; adding $p_0$ does not change force balance. The coefficient c is the one called C in the final sentence of the PDF.

For a confined linear-pressure spherical magnetic equilibrium, the condition that the entire magnetic field vanish on $r=a$ is precisely $F(a)=F^{\prime }(a)=0$. In components these conditions readThey eliminate both the radial and meridional components; the toroidal component vanishes too because $\psi =\alpha \chi$. For a nontrivial solution with $A\ne 0$, substituting $B=-A/\alpha$ givesThe last equation selects the allowed nonzero values of $\alpha a$. Equivalently, when division is legitimate, $\tan x=3x/(3-x^2)$; the undivided equation also handles vanishing denominators correctly. Since $\chi (a,\theta )=0$, $\boxed{p(a,\theta )=p_0}$ is independent of latitude. In the strictly proportional convention $p=P\chi$, this is zero. The zero-field solution is a trivial possibility; the displayed conditions describe the nontrivial confined family.

Use a steady axisymmetric flow governed by ideal magnetohydrodynamics and take the Newtonian gravitational potential to be axisymmetric. Work locally on connected regular magnetic flux surfaces with nonzero poloidal flow. As usual for a wind without an imposed toroidal loop voltage, $E_{\varphi }=0$; regularity on an included rotation axis guarantees this condition. It is needed to exclude steady cross-field flow in an annular domain.

Since $\nabla \varphi ={\mathbf{e}}_{\varphi }/R$, the poloidal magnetic flux function givesThe poloidal magnetic field is divergence-free. The zero toroidal electric field, with $\mathbf{E}=-\mathbf{u}\times \mathbf{B}$, gives ${\mathbf{u}}_p\parallel {\mathbf{B}}_p$. Define $k$ by $\rho {\mathbf{u}}_p=k{\mathbf{B}}_p$. Steady mass conservation then givesThus $k=k(\psi )$ is the magnetohydrodynamic mass loading, the mass flux per unit poloidal magnetic flux. Subtracting $(k/\rho )\mathbf{B}$ from the full velocity leaves only an azimuthal component. The cross product isThe steady ideal magnetohydrodynamic induction equation requires $0=\nabla \times (\omega \nabla \psi )=\nabla \omega \times \nabla \psi$. Consequently the field-line angular velocity is $\omega =\omega (\psi )$ and

Neither gas pressure nor axisymmetric gravity exerts an azimuthal torque. Multiplying the azimuthal momentum equation by $R$ combines the cylindrical curvature terms into angular-momentum transport:Insert $\rho {\mathbf{u}}_p=k{\mathbf{B}}_p$ and use ${\mathbf{B}}_p\cdot \nabla k=0$. It follows thatso the magnetohydrodynamic angular-momentum invariant isThe two terms account respectively for matter and Maxwell stress tensor transport of angular momentum.

Finally, subtracting $\gamma$ times the logarithmic mass density equation from the logarithmic pressure equation gives $D\ln (p/{\rho }^{\gamma })/Dt=0$. The specific entropy of a fixed-composition ideal gas is a function of $p/{\rho }^{\gamma }$. Steadiness and axisymmetry reduce its advection to ${\mathbf{u}}_p\cdot \nabla s=0$. HenceThese flux-label functions are local on connected regular surfaces; disconnected components with the same numerical label need not share constants without an additional matching condition.

For an axisymmetric magnetic field, the solenoidal condition becomes ${\partial }_R(R{B}_R)+{\partial }_z(R{B}_z)=0$. Locally introduce a poloidal magnetic flux function $\Psi$ byThe divergence condition is exactly the integrability condition for these two derivatives. The additive constant in $\Psi$ fixes the reference flux. For a field regular on the axis choose $\Psi (0,z,t)=0$. Then the magnetic flux through a horizontal disk of radius $R$ is $2\pi \Psi (R,z,t)$, and the flux between two such surfaces is $2\pi \Delta \Psi$.

Since ${\mathbf{B}}_p\cdot \nabla \Psi =0$ and ${\partial }_{\varphi }\Psi =0$, the level surfaces of $\Psi$ contain the complete magnetic field, including its toroidal magnetic field component. Thus regular levels are poloidal flux surfaces. Their meridional contours trace the projected field lines. The representation is local on regular regions; disconnected flux surfaces with the same label need not share additional physical data.

The azimuthal component of $\mathbf{u}\times \mathbf{B}$ is $u_z{B}_R-u_R{B}_z=-{\mathbf{u}}_p\cdot \nabla \Psi /R$. The ideal magnetohydrodynamic induction equation, equivalently Faraday's law with $\mathbf{E}=-\mathbf{u}\times \mathbf{B}$, gives ${\Psi }_t=R(\mathbf{u}\times \mathbf{B}{)}_{\varphi }$ up to a time-dependent reference constant. Fixing that reference gives the material advection of an axisymmetric magnetic flux function:Axisymmetry removes the azimuthal contribution to the material derivative. This is magnetic flux freezing expressed using one scalar flux label.

The azimuthal component itself is generally not an advected scalar. Including the cylindrical basis-vector derivatives in the induction equation givesUsing mass conservation, $D\rho /Dt=-\rho \nabla \cdot \mathbf{u}$, and $DR/Dt=u_R$, this becomes the toroidal magnetic-field winding equation:Thus $B_{\varphi }/(\rho R)$ is conserved following the flow if the angular velocity is constant along the poloidal magnetic field, including the case of no rotation. Otherwise differential rotation winds poloidal field into toroidal field.

There is also an explicit flux-conservation form,Integrating this over a meridional region converts its change of toroidal flux to boundary terms. Hence such a flux is conserved when the indicated boundary transport vanishes. More generally magnetic flux freezing conserves flux through every material surface, but a meridional coordinate surface need not remain material when the fluid rotates. Toroidal flux conservation therefore does not imply that $B_{\varphi }$ itself is materially constant.

Use the aligned poloidal flow from part (i), and define the magnetohydrodynamic mass loading $\eta =\rho \alpha$, so $\rho {\mathbf{u}}_p=\eta {\mathbf{B}}_p$. Axisymmetric mass conservation and $\nabla \cdot {\mathbf{B}}_p=0$ give ${\mathbf{B}}_p\cdot \nabla \eta =0$. Introduce the poloidal magnetic flux function $\Psi$ by ${\mathbf{B}}_p=R^{-1}\nabla \Psi \times {\mathbf{e}}_{\varphi }$, and write $u_{\varphi }=R\Omega$. ThenThe steady induction equation says the curl of this expression is zero. Hence $\nabla {\Omega }_F\times \nabla \Psi =0$, wherethe field-line angular velocity ${\Omega }_F$ is constant on each connected magnetic surface and along its field lines.

To control the difference between matter rotation and field-line rotation, use the azimuthal component of the ideal magnetohydrodynamic momentum equation. The inertial and magnetic terms are respectively $\rho R^{-1}{\mathbf{u}}_p\cdot \nabla (R{u}_{\varphi })$ and $({\mu }_{0}R{)}^{-1}{\mathbf{B}}_p\cdot \nabla (R{B}_{\varphi })$; pressure and gravity have no azimuthal derivative. Multiplying by $R$ and using constant mass loading givesThus the magnetohydrodynamic angular-momentum invariant $L=R^2\Omega -R{B}_{\varphi }/({\mu }_0\eta )$ is constant along the line when $\eta \ne 0$. Let $v_{Ap}=|{\mathbf{B}}_p|/\sqrt{{\mu }_0\rho }$ and define $M_{Ap}^2=|{\mathbf{u}}_p{|}^2/v_{Ap}^2={\mu }_0{\eta }^2/\rho$. Eliminating $B_{\varphi }$ between the two invariants givesFor a highly sub-Alfvénic poloidal flow, $M_{Ap}^2\ll 1$, this is $\Omega \simeq {\Omega }_F$ along a nonsingular field-line segment. This proves sub-Alfvénic isorotation in a steady axisymmetric wind, with the correction explicitly displayed. If the poloidal flow is exactly zero, the induction equation directly gives ${\mathbf{B}}_p\cdot \nabla \Omega =0$, namely Ferraro's law of isorotation. The approximate claim presumes regular invariants: small $M_{Ap}$ alone is not a uniform relative bound if $L/R^2$ is allowed to become arbitrarily large.

An anti-dynamo theorem identifies a restriction that prevents a conducting fluid from regenerating its magnetic field against resistive losses. The hypotheses matter: throughout the following arguments take positive uniform magnetic diffusivity, smooth fields, no imposed magnetic source, and the stated isolated or energy-closed boundary conditions. An anti-dynamo theorem does not rule out transient magnetic amplification, winding of an imposed field, or an externally supplied magnetic flux. The resistive induction equation is the common starting point:

First consider Backus' necessary condition for dynamo action. For a conductor contained in a sphere of radius $a$, include the potential field in the insulating exterior when defining the total magnetic energy:For an incompressible flow, with boundary conditions eliminating the mechanical surface term, such as a no-slip boundary condition for the fluid, integration by parts giveswhere $e_{ij}=({\partial }_i{u}_j+{\partial }_j{u}_i)/2$ is the rate-of-strain tensor. Its antisymmetric part does no stretching, since $B_i{B}_j$ is symmetric. If $S$ bounds the largest eigenvalue of $e$ throughout the flow, the first term is at most $2S{E}_B$.

The magnetic free-decay spectral bound for the enclosing insulating sphere isThe exterior energy is essential to this variational estimate. Its constant comes from the slowest freely decaying dipolar poloidal mode: the spherical radial matching condition reduces to $j_0(\kappa a)=0$, whose first positive root is $\kappa a=\pi$. Expanding an admissible field in the orthogonal free-decay modes, or using their variational characterization, gives the bound. A smaller conductor can only increase the minimum dissipation quotient for the enclosing-sphere comparison. Combining the two estimates yieldsThusBelow this threshold the Gronwall inequality gives exponential energy decay. Above it the estimate merely permits growth; it does not prove that the geometry regenerates the field. A time-dependent version bounds the field-amplitude exponent by the time average of the maximum strain minus the diffusive rate.

The geometric obstruction in Cowling anti-dynamo theorem is different: an isolated exactly axisymmetric magnetic field cannot be self-maintained. To see the missing regeneration mechanism, write its poloidal magnetic field in terms of the poloidal magnetic flux function $\psi$:The poloidal part of the resistive induction equation becomesThere is no term containing $B_{\varphi }$ that creates new poloidal flux. Fix the gauge by setting $\psi =0$ on the regular symmetry axis, and match to a decaying exterior field with no imposed flux. At a positive interior maximum away from the axis, the advective term vanishes and the diffusive term is nonpositive. The parabolic comparison principle therefore prevents increase of the maximum; it likewise prevents decrease of a negative minimum. In the steady problem the strong maximum principle, regularity at the axis and decay outside exclude a nonzero maximum altogether. Matching to the insulating exterior does not create a maximum at the interface: the exterior flux is harmonic in the corresponding axisymmetric operator, and the two normal-derivative conditions exclude such an interface extremum. The same homogeneous scalar diffusion problem gives decay in time on a bounded conductor with no replenishing boundary data.

Once the poloidal field has decayed, let $b=B_{\varphi }/r$ and let ${\Omega }_f=u_{\varphi }/r$ denote the fluid's angular velocity. Its equation isDifferential rotation can make toroidal field from poloidal field, but it cannot close the reverse loop. When ${\mathbf{B}}_p=0$, this too is a homogeneous advection-diffusion equation, with a regular axis and zero toroidal field at an insulating exterior. Hence it cannot sustain the remaining field. The exactly axisymmetric field, rather than merely an axisymmetric velocity, is excluded. If an exactly axisymmetric field were maintained by a nonaxisymmetric velocity, averaging the linear induction equation in azimuth would replace that velocity by its azimuthal mean and give the same contradiction. An axisymmetric flow can, however, excite a nonaxisymmetric field. Also an axisymmetric averaged field in mean-field dynamo theory has an additional fluctuating electromotive force, so this unmodified scalar proof does not apply to it.

The toroidal-velocity anti-dynamo theorem restricts the flow instead. In a conducting sphere, suppose $\mathbf{u}\cdot \mathbf{x}=0$, so the incompressible flow is tangent to every concentric sphere. No assumption of axisymmetry is made. Define $P=\mathbf{x}\cdot \mathbf{B}$. A direct Cartesian calculation is particularly useful:The last equality uses $\nabla \cdot \mathbf{B}=0$, so ${\nabla }^2(\mathbf{x}\cdot \mathbf{B})=\mathbf{x}\cdot {\nabla }^2\mathbf{B}$. Thus the radial magnetic scalar is a passive diffusing scalar, not a regenerated field. Multiplying by $P$, integrating over the sphere and using tangential incompressible advection givesIn an insulating exterior, a spherical harmonic component of degree $l\ge 1$ has $P\propto r^{-l-1}$, so its boundary contribution is negative: ${\partial }_{n}P=-(l+1)P/a$. The monopole component is absent because the magnetic field is solenoidal and regular. The resulting dissipative scalar problem makes $P$, and hence the poloidal field, decay.

When that radial field is absent, the remaining tangential solenoidal field can be written $\mathbf{B}=\nabla \times (T\mathbf{x})$. Its induction equation reduces to a second passive scalar equation. Indeedand the curl of the last gradient term is zero. Therefore $T_t+\mathbf{u}\cdot \nabla T=\eta {\nabla }^{2}T$, up to an irrelevant radial gauge function, which can be removed by taking the angular mean of $T$ to be zero. Its angular variations vanish at the insulating spherical boundary, so this scalar also decays by its quadratic energy identity. Equivalently, a putative growing magnetic normal mode must first have $P=0$, after which the scalar $T$ equation rules it out. Purely toroidal spherical flow cannot close a dynamo regeneration loop, even if it is nonaxisymmetric. This does not assert that purely poloidal flow is always incapable of a dynamo, or that any azimuthally directed cylindrical flow is governed by the same spherical theorem.

Other restrictions illuminate the same distinction between stretching and regeneration. The planar anti-dynamo theorem has a similar scalar proof in the usual two-dimensional geometry. For planar velocity and magnetic field independent of the third coordinate, write the in-plane field as $\nabla \times (A\hat{\mathbf{z}})$. Its vector potential obeys $A_t+\mathbf{u}\cdot \nabla A=\eta \Delta A$ after fixing the gauge, so $\frac{1}{2}\frac{d}{dt}\int A^2=-\eta \int |\nabla A{|}^2$ under periodic or homogeneous energy-closed boundary conditions. A third field component independent of that coordinate is also a passive scalar. A spatially uniform imposed field may survive in a periodic box but is not regenerated dynamo field. Three-component motions depending on only two coordinates are not covered by this strictly planar formulation.

There are also necessary bounds in terms of velocity amplitude rather than strain. The magnetic work term is bounded by $\Vert \mathbf{u}{\Vert }_{\infty }\Vert \mathbf{B}{\Vert }_2\Vert \nabla \times \mathbf{B}{\Vert }_2$; combining this with the same free-decay inequality gives a velocity-based lower threshold on the magnetic Reynolds number. These bounds, like Backus' necessary condition for dynamo action, are necessary rather than sufficient. Pure motion that only rearranges a scalar magnetic potential cannot evade diffusion merely by becoming faster. Conversely the stagnation-flow energy amplification in Question 2 does not contradict the bounded-body theorems: it uses an unbounded three-dimensional strain field and, for its growing finite-energy modes, a slowly decaying field at infinity. Similarly, the torsional Alfvén waves in Question 3 transfer energy between velocity and a toroidal perturbation about a prescribed background; they are not a construction of that background by dynamo action. The anti-dynamo results identify failed regeneration mechanisms and the assumptions under which those failures are rigorous.

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.

Use the poloidal magnetic flux function convention ${\mathbf{B}}_p=R^{-1}\nabla \Psi \times {\mathbf{e}}_{\varphi }$, so $2\pi \Psi$ is magnetic flux up to a reference constant. Its poloidal magnetic field components areThe prescribed surface flux therefore givesThe constant offset ${\Psi }_1$ does not affect either field component. If instead $\Psi$ denotes the full physical flux rather than flux divided by $2\pi$, both component formulas acquire a common $1/(2\pi )$ factor; the subsequent logarithmic derivative relation is unchanged.

The Gauss's law for magnetism constraint is $R^{-1}{\partial }_R(R{B}_R)+{\partial }_z{B}_z=0$. Since $\alpha$ is constant with radius on the surface, it gives the power-law poloidal field near a disk surface relationFor a smooth one-sided field above the surface, $B_z(R,z)=B_z(R,0)[1-(\delta -1)\tan \alpha z/R]+O(z^2)$. Thus the field initially decreases with height if $\delta >1$, increases if $0<\delta <1$, and has zero first height derivative if $\delta =1$.

The divergence-free poloidal field can be represented by the poloidal magnetic flux functionIn a steady axisymmetric ideal magnetohydrodynamics flow, the azimuthal component of $\nabla \times (\mathbf{u}\times \mathbf{B})=0$ makes ${\mathbf{u}}_p$ parallel to ${\mathbf{B}}_p$. WriteMass conservation and $\nabla \cdot \mathbf{B}=0$ then imply $\mathbf{B}\cdot \nabla k=0$, so the magnetohydrodynamic mass loading $k=k(\psi )$ is constant along each magnetic line.

The poloidal part of $\mathbf{u}\times \mathbf{B}$ isIts curl vanishes only if its coefficient is a flux function, giving the field-line angular velocityThe azimuthal component of momentum conservation is a divergence of matter and magnetic angular-momentum flux. Dividing its field-line constant by the mass loading yields the magnetohydrodynamic angular-momentum invariantThe conservative total-energy equation similarly gives the magnetohydrodynamic Bernoulli invariantFinally, the entropy advection equation and ${\mathbf{u}}_p\parallel {\mathbf{B}}_p$ imply $s=s(\psi )$. Thus $k,\omega ,\ell ,ϵ$, and $s$ are constant along each magnetic field line.

Axisymmetry makes the azimuthal direction geometrically distinct, so the field splits into a meridional poloidal part and an azimuthal toroidal part:The divergence-free condition on the poloidal field permits the poloidal magnetic flux functionor in cylindrical coordinatesThe magnetic flux through a circle of radius $r$ at fixed $z$ isThus, after choosing $\psi =0$ on the axis, $2\pi \psi$ is the enclosed poloidal flux and surfaces of constant $\psi$ are magnetic surfaces.

The representationgivesHence $\mathbf{B}\cdot \nabla \psi =0$: the level sets of the poloidal magnetic flux function are magnetic surfaces. Moreover, the magnetic flux through a circular disc isThus $2\pi \psi$ is the enclosed poloidal magnetic flux up to an additive axial reference.

A poloidal magnetic field lies in meridional planes through the symmetry axis, in contrast to a toroidal magnetic field. For an axisymmetric vector field, it is represented by a poloidal magnetic flux function as ${\mathbf{B}}_p=R^{-1}\nabla \Psi \times {\mathbf{e}}_{\varphi }$.

If $\Psi (R,0)={\Psi }_1+{\Psi }_0(R/R_0{)}^{\delta }$ and the surface inclination to the vertical is a radius-independent $\alpha$, then $B_z=\delta {\Psi }_0{R}^{\delta -2}/R_0^{\delta }$ and $B_R=B_z\tan \alpha$. The solenoidal vector field condition givesThe constant offset in the poloidal magnetic flux function has no effect on the magnetic field.

For steady axisymmetric ideal flow with only poloidal components, induction implies $\mathbf{u}\times \mathbf{B}=C{\mathbf{e}}_{\varphi }/R$. Axis regularity, or zero toroidal electromotive force, sets $C=0$. Continuity then gives $\rho \mathbf{u}=k(\psi )\mathbf{B}$, locally on connected magnetic surfaces of the poloidal magnetic flux function. Without the zero-circulation condition, annular domains can support steady cross-field flow.

For an axisymmetric magnetic field $\mathbf{B}=\psi {\mathbf{e}}_{\varphi }/(r\sin \theta )+\nabla \times [\chi {\mathbf{e}}_{\varphi }/(r\sin \theta )]$, the poloidal magnetic flux function gives $B_r={\chi }_{\theta }/(r^2\sin \theta )$ and $B_{\theta }=-{\chi }_r/(r\sin \theta )$. Force balance implies locally $p=p(\chi )$ and $\psi =\psi (\chi )$ on regular connected flux surfaces. Direct calculation gives $(\nabla \times \mathbf{B})\times \mathbf{B}=-({\Delta }_{\ast }\chi +\psi {\psi }^{\prime })\nabla \chi /(r^2{\sin }^2\theta )$, with ${\Delta }_{\ast }\chi ={\chi }_{rr}+\sin \theta {\partial }_{\theta }({\chi }_{\theta }/\sin \theta )/r^2$. Equating this to ${\mu }_0\nabla p$ proves the displayed magnetostatic equilibrium equation, including its sign and dimensions.