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Let $g>0$ be the magnitude of the downward gravity and write ${\alpha }_T$ for the thermal expansion coefficient, to distinguish it from the dynamo coefficient of the previous question. The conductive state has $\mathbf{u}=0$, $\mathbf{B}=B_0\hat{\mathbf{z}}$ and $T_0(z)=T_r-\Delta Tz/d$. Its gravitational term is absorbed in a background hydrostatic pressure.

For the dimensional two-dimensional fields, choose the stream function and Cartesian magnetic flux function conventionsThese automatically give zero divergence. Define the Jacobian determinant bracket $J(f,h)=f_x{h}_z-f_z{h}_x$, so $\mathbf{u}\cdot \nabla h=J(\psi ,h)$. The full magnetic flux function, including the imposed field, must be retained in the magnetic force; using the perturbation alone would lose its linear restoring term.

The scalar vorticity is $(\nabla \times \mathbf{u}{)}_y=-{\nabla }^2\psi$. Also $(\nabla \times \mathbf{B}{)}_y=-{\nabla }^{2}A$, and henceThe buoyancy perturbation is $g{\alpha }_T(T-T_0)\hat{\mathbf{z}}$, whose curl in the $y$ direction is $-g{\alpha }_T(T-T_0{)}_x$. Taking the curl of the momentum equation and multiplying by minus one eliminates pressure and therefore givesWe have restored ${\mu }_0$ in this dimensional expression. The printed magnetic force uses units in which ${\mu }_0=1$, or equivalently a correspondingly rescaled field.

Use length scale $d$, time scale $d^2/\kappa$, velocity scale $\kappa /d$, stream-function scale $\kappa$, magnetic flux-function scale $B_{0}d$, and temperature perturbation scale $\Delta T$. Thus the dimensionless total flux function isHere $x,z$ now denote coordinates divided by $d$, and the same $T$ notation in the second identity denotes the physical temperature. The dimensionless parameters for thermal-diffusion scaling of planar magnetoconvection areThese are the Rayleigh number, Chandrasekhar number, Prandtl number and magnetic-to-thermal diffusivity ratio. The nondimensional magnetic-force coefficient is $B_0^2{d}^2/({\mu }_0{\rho }_r{\kappa }^2)=\sigma \zeta Q$, so omitting the factor $\zeta$ would use a different magnetic parameter.

The resistive induction equation becomes $A_t+J(\psi ,A)=\zeta {\nabla }^{2}A$. Since $J(\psi ,x)=-{\psi }_z$, its perturbation equation has source $+{\psi }_z$. Similarly, advection of the conductive temperature gradient gives $J(\psi ,-z)=-{\psi }_x$, producing the source $+{\psi }_x$ in the perturbation heat equation. The resulting nonlinear equations are thereforeThis derives both forcing signs from the background fields, rather than selecting them by a stream-function convention after the fact.

For the specified roll mode put $K=\pi /\lambda$ and $p_L=K^2+{\pi }^2$, the positive eigenvalue of minus the Laplacian on each spatial factor. Writeand take $\psi =(p_L/K)a{f}_s$, $\Theta =b{f}_T$, $\chi =\lambda c{f}_A$, with amplitudes depending on $\tau =p_{L}t$. These factors satisfy the printed boundary conditions: sine factors make the normal velocity and $B_x=-{\chi }_z$ vanish at the required faces; the second derivatives of the velocity factors give the stress-free conditions; $\Theta$ vanishes at the top and bottom and ${\Theta }_x$ vanishes at the sides.

Discard quadratic perturbation terms. Since ${\nabla }^{2}A={\nabla }^2\chi$, the magnetic force linearizes to $J(x,{\nabla }^2\chi )=({\nabla }^2\chi {)}_z$. The needed derivatives areThe heat and induction equations give $b^{\prime }=a-b$ and $c^{\prime }=a-\zeta c$. In the vorticity equation, its left side is $-(p_L^3/K)a^{\prime }{f}_s$ and its three right-side contributions are $\sigma [(p_L^3/K)a-RKb+\zeta Q{p}_L\pi \lambda c]f_s$. Consequently, withthe three-amplitude vertical-field magnetoconvection system isPrimes now mean $d/d\tau$. They are genuine time derivatives, as the original PDF shows.

For its linear stability, seek amplitudes proportional to $e^{s\tau }$. Eliminating $b,c$ or taking the determinant giveswhereAssume the physical range $\sigma >0$, $\zeta >0$, $q\ge 0$. The Routh-Hurwitz stability criterion gives strict decay exactly when $A_2>0$, $A_3>0$ and $A_1{A}_2>A_3$. At $r=0$ these hold. As $r$ increases, the first loss of stability can occur through a zero eigenvalue or a conjugate imaginary pair.

The stationary threshold isAt this value, $P(s)=s(s^2+A_{1}s+C)$ withIf $C>0$, the other two roots have negative real parts and the zero root is simple. Its derivative with respect to $r$ is $ds/dr=\sigma \zeta /C>0$. This is stationary onset; the reflection symmetry of the roll permits a generic pitchfork bifurcation. The linear calculation does not determine whether the nonlinear branch is supercritical or subcritical.

For oscillatory onset set $s=i\omega$, $\omega \ne 0$. Separating real and imaginary parts gives ${\omega }^2=A_2$ and $A_1{A}_2=A_3$. Solving the latter for $r$ yieldsA genuine Hopf bifurcation onset requires ${\omega }_H^2>0$, soAt this threshold $P(s)=(s+A_1)(s^2+{\omega }_H^2)$, giving the third root $-A_1<0$. The pair crosses transversely: differentiating $P(s,r)=0$ at $i{\omega }_H$ gives

The first instability of the conductive state is therefore stationary at $r_s$ if $\zeta \ge 1$, or if $0<\zeta <1$ and $q<q_{\ast }$. It is oscillatory at $r_H<r_s$ if $0<\zeta <1$ and $q>q_{\ast }$. At $q=q_{\ast }$, the two thresholds meet and $P(s)=s^2(s+A_1)$: the frequency is zero and neither a simple pitchfork nor a nonzero-frequency Hopf test applies. The oscillatory marginality in vertical-field magnetoconvection formula must not be used where its frequency squared is negative.

For conversion back to physical parameters, $R_{s,H}={\lambda }^2{p}_L^3{r}_{s,H}/{\pi }^2$, while the dimensional Hopf frequency is $(\kappa p_L/d^2){\omega }_H$. These are onset conditions for the specified spatial mode. Identifying the globally first mode would additionally require comparison over the allowed spatial modes. Nonlinear saturation and the stability of the bifurcating finite-amplitude states are not determined by this linear three-amplitude system alone.

Magnetic flux concentration and flux expulsion describe redistribution of an existing field, not necessarily creation of net flux. For a weak field in a prescribed conducting flow, the resistive induction equation isIts ideal part combines transport, stretching and compression:Thus a converging flow can intensify a component, and stretching can increase magnetic energy even if the signed net flux is zero. At large magnetic Reynolds number $\mathrm{Rm}=UL/\eta$, diffusion is initially slow on the eddy scale, so magnetic flux freezing carries field lines with the flow. These statements do not imply that an ideal closed eddy can simply lose its initially enclosed flux.

The missing step is the production of small scales by differential circulation. In a two-dimensional incompressible eddy, the Cartesian magnetic flux function obeysDifferent streamlines have different circulation rates. An initially smooth field is wound into fine folds; diffusion and reconnection then smooth the variation of $A$ inside the eddy. As $A$ becomes nearly uniform there, its gradient and hence the interior magnetic field become small. The surviving flux lies mainly near the circulation boundary and adjacent separatrices, where stronger gradients remain. This is flux expulsion. With exactly zero diffusivity, field topology and the flux of material loops remain frozen, so a relaxed flux-free interior is not obtained by pure advection alone. Rigid rotation also fails to generate the differential winding responsible for this process.

The phase mixing in magnetic flux expulsion estimate makes the enhanced role of small diffusion explicit. If the angular-velocity gradient is of order $U/L^2$, a wound radial wavenumber grows as $k_r\sim Ut/L^2$. The accumulated damping exponent is of orderIt becomes order one atThis is much shorter than the large-scale diffusion time $(L/U)\mathrm{Rm}$ at large $\mathrm{Rm}$, but longer than one turnover time. It is a kinematic differential-shear estimate, not a universal time for every turbulent or rapidly evolving convection cell.

A related concentration mechanism occurs at converging flow boundaries. For a locally vertical bundle, conservation of flux gives $B\mathcal{A}\simeq \mathrm{constant}$, so reducing its cross-section increases $B$. In a simple converging boundary region with strain rate $U/L$, balancing strain and transverse diffusion givesThis is the thickness estimate for that local strain-diffusion model; it need not be the thickness or energy scaling of every closed-eddy flux-expulsion problem. Stronger field eventually exerts Lorentz force feedback and changes the flow. The equipartition magnetic field $B_{\mathrm{eq}}=\sqrt{{\mu }_0\rho }U$ is a useful indication that the kinematic assumption can fail, not a universal upper bound on concentrated fields.

At the Sun's photosphere, hot upflows form solar granules, and their spreading horizontal motion transports weak vertical field toward cooler intergranular lanes and downflows. A thin-layer illustration, neglecting vertical transport terms, is ${\partial }_t{B}_z+{\nabla }_h\cdot ({\mathbf{u}}_h{B}_z)=0$. It gives $D_h{B}_z/Dt=-B_z{\nabla }_h\cdot {\mathbf{u}}_h$: horizontal convergence intensifies the vertical flux density. This surface transport is related to the redistribution by convective eddies, but is not identical to the closed-streamline mechanism in the preceding planar example. Larger-scale supergranulation also gathers magnetic elements toward its boundaries, helping organize the solar magnetic network.

Concentration can be further intensified by compressible thermal dynamics. In the thin-tube picture of convective collapse, an unstable downward flow drains gas, reduces internal pressure and allows contraction at approximately conserved flux. Lateral balance $p_i+B^2/(2{\mu }_0)\simeq p_e$ then permits field stronger than the local kinetic equipartition estimate. This evacuation mechanism is excluded by a strict Boussinesq approximation. It is not a description of every evolving surface concentration: gas pressure, dynamic stresses and magnetic curvature can all matter when thin-tube equilibrium fails. The resulting strong, intermittent magnetic elements are associated with intergranular structure; their radiative appearance also depends on thermal structure and opacity. The organized fields of sunspots require a broader account of flux supply and magnetic feedback, rather than attribution to weak-field expulsion alone.

Thus convection redistributes and concentrates flux, finite diffusion enables classical expulsion, and nonlinear magnetic and thermal feedback determines the strong-field outcome. An expulsion argument alone neither supplies the original solar field nor establishes a self-sustaining solar dynamo.

Choose a reference circular orbit at radius $r_0$, rotate the frame with $\mathbf{\Omega }=\Omega (r_0){\mathbf{e}}_z$, and introduce local Cartesian coordinates $x=r-r_0$, $y=r_0(\varphi -\Omega t)$ and vertical coordinate $z$. On scales small compared with $r_0$, neglect curvature of the coordinate directions but retain the leading differential rotation. The background velocity relative to the rotating frame isThus $A=3\Omega /4$ for a Keplerian disk. The shearing sheet retains Coriolis acceleration and the linear tidal acceleration. The radial shearing-sheet tidal potential is $-2\Omega A{x}^2$, whose force $4\Omega Ax{\mathbf{e}}_x$ balances the background Coriolis acceleration. If vertical tidal gravity is included, its equilibrium contribution is balanced by a background pressure $p_0(z)$ and disappears when that equilibrium is subtracted.

Write the total velocity as $\mathbf{u}={\mathbf{u}}_0+\mathbf{v}$. In the rotating-frame magnetohydrodynamic momentum equation, background advection is $({\mathbf{u}}_0\cdot \nabla )\mathbf{v}=-2Ax{\partial }_y\mathbf{v}$ and advection of the background by the perturbation is $(\mathbf{v}\cdot \nabla ){\mathbf{u}}_0=-2A{v}_x{\mathbf{e}}_y$. The uniform-viscosity diffusion of the linear background shear is zero. For a solenoidal magnetic field the Lorentz force per unit mass isAt constant density, absorb the magnetic-pressure gradient into $\psi =(p-p_0)/\rho +B^2/(2{\mu }_0\rho )$. Subtracting the background force balance then gives the stated momentum equation, with the remaining nonlinear advection $\mathbf{v}\cdot \nabla \mathbf{v}$, Coriolis acceleration $2\mathbf{\Omega }\times \mathbf{v}$, magnetic tension and viscous diffusion.

For constant finite electrical conductivity ${\sigma }_e$, put $\eta =({\mu }_0{\sigma }_e{)}^{-1}$. The resistive induction equation with uniform magnetic diffusivity is ${\partial }_t\mathbf{B}+\mathbf{u}\cdot \nabla \mathbf{B}=\mathbf{B}\cdot \nabla \mathbf{u}+\eta {\nabla }^2\mathbf{B}$. Substituting the background shear supplies the term $-2A{B}_x{\mathbf{e}}_y$ and the same background-advection operator as in the momentum equation. The two further conditions areThey express incompressibility and the absence of magnetic monopoles, respectively.

Now take real plane waves with common phase $\vartheta =\mathbf{k}(t)\cdot \mathbf{x}$ and real wavevector. The solenoidal conditions becomeEvery field depends on position only through $\vartheta$. Hence for either transverse vector field $\mathbf{P}$ and either wave field $\mathbf{Q}$, $(\mathbf{P}\cdot \nabla )\mathbf{Q}=(\mathbf{P}\cdot \mathbf{k}){\partial }_{\vartheta }\mathbf{Q}=0$. This proves that all the displayed quadratic advection and tension terms vanish pointwise, including products involving conjugate amplitudes. These waves are therefore exact solutions at finite amplitude, not merely a linearization.

Acting with $D_0={\partial }_t-2Ax{\partial }_y$ on the phase yieldsCanceling its position-dependent part requiresThe remaining amplitude equations areDifferentiating the velocity constraint determines the pressure amplitude:The resistive induction equation preserves its constraint because $\dot{\mathbf{k}}\cdot \tilde{\mathbf{B}}=2A{k}_y{\tilde{B}}_x$ cancels the corresponding shear term in $\mathbf{k}\cdot \dot{\tilde{\mathbf{B}}}$. The specified magnetic field has no spatially uniform part. Thus there is no background magnetic field producing an additional linear magnetic-tension coupling. The scalar $\psi$ is the modified pressure; the physical pressure can include a constant and a second spatial harmonic from $-B^2/(2{\mu }_0)$ while $\psi$ retains the stipulated plane-wave form. This is an exact zero-mean magnetic shearing wave.

For decay, take the real part of the scalar product of the velocity-amplitude equation with its complex conjugate. Pressure does no work because $\mathbf{k}\cdot \tilde{\mathbf{v}}=0$, and the real part of Coriolis work is zero. ThereforeThe resistive induction equation similarly givesusing $2|u_x{u}_y|\le |\mathbf{u}{|}^2$. Integrating these differential inequalities, defineThenA non-axisymmetric wave has $k_y\ne 0$. For nonzero differential shear and positive $\nu ,\eta$, the negative cubic terms dominate the linear upper bound on shear work. ConsequentlyThis cubic-exponent decay of a nonaxisymmetric shearing wave allows transient energy growth before the winding to short radial wavelengths makes diffusion dominant. The magnetic amplitudes can also be written explicitly aswhich confirms the damping independently. Positivity of the dissipative coefficients is necessary: with $k_z=0$, an inviscid vertical velocity amplitude can remain constant, and an ideal vertical magnetic amplitude can remain constant when $\eta =0$. The asserted decay applies to the viscous, finitely conducting model; taking an ideal limit before the long-time limit changes the conclusion.

Choose the planar magnetic field convention $\mathbf{B}=\nabla \times (A{\mathbf{e}}_z)$, so $B_s=s^{-1}{A}_{\varphi }$ and $B_{\varphi }=-A_s$. The resistive induction equation, with constant magnetic diffusivity and a suitable time-dependent gauge constant removed, reduces to advection and diffusion of this Cartesian magnetic flux function:For the specified differential rotation, its mth azimuthal Fourier mode obeysThe physical field is the real part of the complex mode. Substitute $a=s^{m}g(t)\exp [-im{\Omega }_{0}t-i{s}^{2}f(t)]$. A direct radial differentiation givesOn the left, rigid rotation cancels, leaving $\dot{g}/g-i{s}^2\dot{f}+im\omega s^2$. Equating the constant and quadratic coefficients yieldsThe initial condition fixes $f(0)=0$ and $g(0)=1$, since the constant C was already factored out. Set ${\mu }^2=4i\eta m\omega$. Then the exact quadratic-shear magnetic flux solution follows by differentiation:Indeed $\dot{f}=m\omega {\mathrm{sech}}^2(\mu t)$ and $4i\eta f^2=m\omega {\tanh }^2(\mu t)$. Also $\dot{g}/g=-(m+1)\mu \tanh (\mu t)=-4i\eta (m+1)f$. For an unspecified initial normalization the second formula has a multiplicative constant $C^{\prime }$, but here $C^{\prime }=1$.

For $\eta >0$ and $\omega >0$, choose $\mu =(1+i)\sqrt{2\eta m\omega }$. The asymptotic condition is $\mathrm{Re}(\mu )t\gg 1$: then $f$ tends to the finite constant $m\omega /\mu$ and $g\sim 2^{m+1}\exp [-(m+1)\mu t]$. At any fixed s, differentiating A to obtain the magnetic field multiplies it by factors involving $m/s$ and $sf$, which have no additional exponential time dependence. Therefore its envelope decays asUsing the paper's magnetic Reynolds number definition gives the precise scalingThus the decay time scales as ${\mathrm{Rm}}^{1/2}/{\Omega }_0$ for positive ${\Omega }_0$ and fixed m. The mechanism is magnetic phase mixing under differential rotation: winding produces progressively fine radial structure on which magnetic diffusion acts efficiently. For negative $\omega$, choose the square root with positive real part, replacing $\omega$ by $|\omega |$ in the decay time. If $\omega =0$, the initial harmonic mode simply rotates, with $f=0$, $g=1$; there is no shear-enhanced decay. The printed positive-root and Reynolds-number expressions implicitly use $\omega >0$.

Interpret the two-dimensional dependence as independence of $z$, allowing the velocity to have a vertical component $U=u_z(x,y)$. Write $u_h=(u_x,u_y)$ and $B_h=(A_y,-A_x)$. The resistive induction equation, with a gauge fixing the potential's zero mean, givesThe first is a homogeneous advection-diffusion equation. Multiply it by $A$, average over the periodic box and use integration by parts. Advection contributes zero because $\nabla \cdot u_h=0$; there are no surface terms. ThusMultiplying by $e^{2\eta k^{2}t}$ makes its derivative nonpositive. With the initial norm convention $⟨A^2(0)⟩=A_0^2$, this provesThe Poincaré inequality uses the zero-mean assumption; a spatially constant flux potential would otherwise have no gradient. Zero mean remains preserved by this incompressible periodic equation.

In the rapidly rotating magnetohydrodynamics of a conducting core, the magnetostrophic balance neglects inertia and, initially, viscosity:Take a spherical or suitable axisymmetric impenetrable container, with gravity/buoyancy having no azimuthal component and no applied mechanical torque. Small Rossby number is the slow-motion assumption. It does not imply that an arbitrary magnetic field is compatible with the balance.

Use cylindrical coordinates $(s,\varphi ,z)$ and integrate the azimuthal momentum equation over the fluid portion $C_s$ of a cylinder. The pressure term integrates to zero because it is a periodic $\varphi$ derivative. The Coriolis contribution is proportional to the radial flux ${\int }_{C_s}{u}_{s}sd\varphi dz$. By incompressibility and impermeability it is zero: closing the cylinder with the container's caps and applying the divergence theorem produces no net flux. There is no azimuthal buoyancy term. ConsequentlyThis is Taylor's condition. Equivalently define the torque per radial thickness by $\mathcal{T}(s)=s^2{\int }_{C_s}(J\times B{)}_{\varphi }d\varphi dz$; it must vanish. In a shell, cylinders split by the inner boundary require the corresponding connected-region conditions. These compatibility conditions concern rotating-fluid torque, distinct from force-free magnetic relaxation in a plasma. Their primary origin is Taylor's rotating-fluid analysis.

A magnetostrophic Taylor state allows construction of the flow's ageostrophic part. Taking the curl givesIntegrate along rotation-axis lines and impose impermeability, then use incompressibility and periodicity to reconstruct a particular solution $u_a$. In a sphere the boundary geometry supplies the independent endpoint conditions; a cylinder with parallel planar end normals has additional degeneracies. The homogeneous momentum equation admits any differential cylindrical rotation $u_g=s\omega (s)\hat{\varphi }$. Its Coriolis force is a radial pressure gradient, so momentum balance alone cannot determine $\omega$.

The additional equation is dynamic preservation of the constraint. LetThis is the first variation of the cylindrical Lorentz torque. The resistive induction equation supplies $\dot{B}=\nabla \times (u\times B)+\eta {\nabla }^{2}B$. Define $C_a=\nabla \times (u_a\times B)+\eta {\nabla }^{2}B$ and $L_B[\omega ]=\nabla \times (s\omega \hat{\varphi }\times B)$. Then the geostrophic flow preserving the Taylor constraint solvesFor a given field this is linear in $\omega$; it prescribes the cylindrical rotation needed to keep the field on the Taylor-state manifold. The magnetic boundary conditions must also be imposed on $\dot{B}$. Boundary terms cannot generally be discarded for insulating exteriors; their role in a correct construction is described in Hardy and collaborators' primary analysis.

For example, when magnetic surface torque vanishes, as for compatible $B\cdot n=0$ conditions, the Maxwell stress tensor givesCylindrical rotation induces component changes ${\dot{B}}_s=-\omega {\partial }_{\varphi }{B}_s$ and ${\dot{B}}_{\varphi }=s{B}_s{\omega }^{\prime }-\omega {\partial }_{\varphi }{B}_{\varphi }$. Their product variation integrates to $s{\omega }^{\prime }\int B_s^{2}d\varphi dz$, since the azimuthal derivative integrates to zero. Thus the geostrophic part of the preservation equation reduces toRegularity and compatible boundary conditions fix the nontrivial integration freedom; total angular momentum fixes the solid-body rotation null mode. If $B_s$ vanishes on an entire cylinder, the magnetic coupling coefficient degenerates and generic uniqueness must not be asserted. The simplified expression is conditional on the surface-torque assumption, rather than a general replacement for the boundary-compatible operator equation.

With small positive viscosity the momentum equation adds $\rho \nu {\nabla }^{2}u$. Its cylinder integral becomes the viscous regularization of the Taylor constraintMagnetic torque need no longer vanish exactly; it can be transmitted by viscous stress. Bulk viscosity is order $E=\nu /(2\Omega L^2)$ at fixed velocity. For ordinary no-slip rotating boundaries, thin Ekman layers have thickness of order $L\sqrt{E}$ and shear stress of order $\rho \Omega LU\sqrt{E}$, larger than the bulk order-$E$ correction. Boundary pumping and this friction help select the otherwise undetermined geostrophic flow. In the inviscid interior only impermeability is imposed; no-slip matching is supplied by the thin layer, so simply imposing all viscous boundary conditions on the zero-viscosity equations is incorrect.

For bounded velocities, the viscous torque tends to zero as $E\to 0$, recovering the leading Taylor condition. A prescribed order-one uncompensated magnetic torque would instead require a velocity growing like $E^{-1/2}$ under Ekman friction, or $E^{-1}$ under bulk friction alone. Such growth eventually invalidates the assumed small Rossby number: the limit is singular, not a regular guarantee that every imposed magnetic field has an inviscid solution. With stress-free boundaries the leading no-slip friction is absent, and bulk stresses, magnetic boundary conditions and angular-momentum normalization require their own treatment. Retaining inertia permits torque-driven torsional motion rather than an exact magnetostrophic constraint. Thus zero viscosity gives a magnetic compatibility condition and an induction-based selection of cylindrical rotation, while small viscosity replaces it by a torque balance whose bounded solutions approach compatible Taylor states.

Use the resistive induction equation with constant magnetic diffusivity,Seek ${\mathbf{B}}_{\mathrm{tot}}=B_0{\mathbf{e}}_z+B(s,z,t){\mathbf{e}}_{\varphi }$. The axisymmetric velocity $\mathbf{u}=s\omega (s,z){\mathbf{e}}_{\varphi }$ is incompressible, and the proposed field is solenoidal. The toroidal magnetic field is parallel to the velocity, so it does not contribute to their cross product. The remaining term isThe uniform axial field has zero vector Laplacian. For the azimuthal part, differentiation of the cylindrical basis gives ${\partial }_{\varphi }^2{\mathbf{e}}_{\varphi }=-{\mathbf{e}}_{\varphi }$. ConsequentlyThere is no source for a radial or axial perturbation, so zero initial values of those components remain zero. Equating the toroidal components provesThe last Laplacian is the scalar axisymmetric Laplacian. The additional $-B/s^2$ term is essential for a vector component in a rotating basis. The source is the Omega effect: rotation that varies along an axial field line winds it into a toroidal magnetic field.

Write $D_t={\partial }_t+\mathbf{u}\cdot \nabla$ for the material derivative. The resistive induction equation, together with the solenoidal conditions on both fields, isSince $D_t\mathbf{x}=\mathbf{u}$, the product rule givesAlso $\Delta P=\mathbf{x}\cdot \Delta \mathbf{B}+2\nabla \cdot \mathbf{B}=\mathbf{x}\cdot \Delta \mathbf{B}$. Substitution proves the radial magnetic induction scalar equation:

In the insulating exterior the quasistatic magnetic field is current-free, so $\nabla \times \mathbf{B}=0$ and $\nabla \cdot \mathbf{B}=0$. Thus $\Delta \mathbf{B}=0$ and $\Delta P=0$ for $r>a$, with decay at infinity for an isolated field. This is an instantaneous exterior Laplace equation, rather than a diffusion equation with the interior value of $\eta$.

Use the usual nonmagnetic-interface assumptions: equal magnetic permeability on both sides and no surface current. The electromagnetic boundary conditions then make $B_r$ and the tangential magnetic field continuous. Hence $P=r{B}_r$ is continuous. In spherical coordinates, the solenoidal magnetic-field constraint readswhere the last term is the angular divergence on the unit sphere. Its right-hand side is also continuous, giving the insulating boundary condition for the radial magnetic scalar:Equivalently, the decaying exterior degree-$l$ spherical harmonic is proportional to $r^{-(l+1)}$, so its boundary amplitude satisfies ${\partial }_r{P}_{lm}(a)=-(l+1)P_{lm}(a)/a$.

Let $⟨\cdot ⟩$ denote the spatial average. Assume periodicity, or an equivalent homogeneous averaging limit in which total derivatives average to zero and integration by parts is valid. Fix the induced field to have zero mean. Expand the solution of the resistive induction equation in small magnetic Reynolds number:The uniform test magnetic field commutes with differentiation. Since $\Delta \mathbf{u}=-\mathbf{u}$, the first equation has solution ${\mathbf{b}}_1=(\mathbf{B}\cdot \nabla )\mathbf{u}$.

The quadratic second-order forcing need not be monochromatic, so one must not simply replace its inverse Laplacian by multiplication by $-1$. Instead use the self-adjointness of the Laplacian under the average:This proves the monochromatic small-Reynolds-number mean electromotive force expansion

For the symmetry of the first-order monochromatic alpha tensor, writeIts antisymmetric part is determined by contraction with the Levi-Civita symbol. The contraction identity givesThe second term vanishes by incompressibility; the first is $⟨{\partial }_j(u_j{u}_p)⟩=0$. Since every three-dimensional antisymmetric second-rank tensor is equivalent to its contracted axial vector, ${\alpha }^{(1)}$ is symmetric. No Fourier series or Fourier transform is required.

To derive the next identity, set $\mathbf{C}=\mathbf{u}\times (\mathbf{B}\cdot \nabla )\mathbf{u}$. Expanding the curl and cross product in components,The average of the second term is zero: integration by parts makes it $-⟨C_i{\partial }_j{u}_j⟩$. Integrating the first term by parts givesExtracting the coefficient of $B_j$ now yieldsInterchanging $i$ and $j$ reverses the cross product, proving the antisymmetry of the second-order monochromatic alpha tensor:The second coefficient of the alpha tensor can therefore contribute turbulent magnetic pumping, while its symmetric part vanishes at this order.

Set $s=0$ in the horizontal-field magnetoconvection dispersion relation. With positive $\sigma ,\zeta$ and nonzero $a_h$, cancellation gives the stationary neutral curveAt fixed $a_h^2$, the Chandrasekhar number contributes a nonnegative penalty proportional to $m^2$. It can be eliminated by choosing $m=0$. It remains to minimize $f(x)=(x+{\pi }^2{)}^3/x$ for $x=k^2>0$:The function diverges at both endpoints and has its unique minimum at $x={\pi }^2/2$. Thus the magnetic-field-aligned rolls evade steady magnetic inhibition:independently of $Q$. If higher vertical sine modes are included, replacing $\pi$ by $n\pi$ gives minimum $27n^4{\pi }^4/4$, so the displayed first vertical mode also gives the smallest steady threshold.

Physically, these convection rolls have axes along the imposed magnetic field. Their perturbations are independent of $y$, so ${\partial }_y\mathbf{u}=0$ in the resistive induction equation and no growing magnetic perturbation is generated. The flow does not bend the field; consequently magnetic tension cannot inhibit these rolls, however large the field strength. This conclusion concerns the requested steady disturbances.

Put $a_h^2=k^2+m^2>0$ and $\Lambda =a_h^2+{\pi }^2$. The plane-layer poloidal-toroidal decomposition supplies the assumed poloidal fields asSince $w=-{\Delta }_h\varphi$ and $b_z=-{\Delta }_h\chi$, potentials proportional to $e^{st+ikx+imy}\sin \pi z$ have the required vertical components. Their horizontal components are proportional to $\cos \pi z$, satisfying the stress-free boundary conditions and the horizontal-field boundary conditions. The double-curl representation automatically enforces incompressibility and the solenoidal magnetic-field constraint.