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For a pressure-free flowing poloidal flux surface, subtracting its angular velocity times the azimuthal momentum equation from the mechanical energy equation shows that ${ϵ}_J$ is constant. It equals poloidal kinetic energy plus $(u_{\varphi }-R\omega {)}^2/2+\Phi -R^2{\omega }^2/2$. This is the total magnetohydrodynamic Bernoulli invariant minus $\omega$ times the total magnetohydrodynamic angular-momentum invariant, not the total energy alone.

In a steady axisymmetric magnetohydrodynamic wind, axisymmetric pressure and Newtonian gravity exert no azimuthal torque. The azimuthal momentum equation is $\rho {\mathbf{u}}_p\cdot \nabla (s{u}_{\varphi })={\mu }_0^{-1}{\mathbf{B}}_p\cdot \nabla (s{B}_{\varphi })$. Mass conservation and the solenoidal vector field condition imply $\nabla \cdot {\mathbf{F}}_J=0$. With magnetohydrodynamic mass loading $\rho {\mathbf{u}}_p=\kappa {\mathbf{B}}_p$, the transported angular momentum per unit poloidal magnetic flux is $\ell =\kappa s{u}_{\varphi }-s{B}_{\varphi }/{\mu }_0$. It is constant on connected regular poloidal flux surfaces. The usual specific magnetohydrodynamic angular-momentum invariant is $L=\ell /\kappa$, provided $\kappa \ne 0$; confusing these two normalizations loses a factor of mass loading in the wind torque.

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.

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.

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.

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.

The flux-function form solves $\nabla \cdot \mathbf{B}=0$ identically. With $k=k(\psi )$, the poloidal relation $\rho {\mathbf{u}}_p=k{\mathbf{B}}_p$ and axisymmetry solve steady mass conservation. The ideal induction equation is solved by the velocity representation with the field-line angular velocity $\omega =\omega (\psi )$. The azimuthal momentum equation integrates once to the magnetohydrodynamic angular-momentum invariant $\ell =\ell (\psi )$.

What remains is the pressure or entropy equation and the two poloidal components of momentum. Projecting poloidal momentum along a field line produces the Bernoulli invariant in part (c); projecting across magnetic surfaces produces the transfield or Grad-Shafranov equation that determines their shape. An equation of state and boundary conditions complete the problem. The stellar gravitational potential is prescribed, so no self-gravity Poisson equation remains to solve.