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Hall magnetohydrodynamics retains bulk ion inertia but freezes magnetic flux into the electron velocity ${\mathbf{u}}_e=\mathbf{u}-\mathbf{j}/(en)$. With constant density, negligible electrical resistivity and electron inertia, Gaussian units give $\mathbf{j}=c\nabla \times \mathbf{B}/(4\pi )$ and ${\partial }_t\mathbf{B}=\nabla \times [(\mathbf{u}-c\nabla \times \mathbf{B}/(4\pi en))\times \mathbf{B}]$. The ion velocity still satisfies the magnetohydrodynamic momentum equation. Neglecting that velocity gives electron magnetohydrodynamics, a further approximation rather than a synonym for the complete Hall system.

The inviscid magnetohydrodynamic momentum equation balances inertia, pressure, gravitational acceleration and Lorentz force density. The magnetic force separates into magnetic tension and magnetic pressure. It is coupled to the continuity equation and the ideal magnetohydrodynamic induction equation.

The magnetic second-moment tensor is $M_{ij}=B_i{B}_j$. This is the magnetic dyadic called the Maxwell tensor in the paper; the isotropic part of the usual Maxwell stress tensor is already contained in the magnetohydrodynamic total pressure. The solenoidal magnetic-field constraint givesConsequently the magnetohydrodynamic momentum equation in part a becomes $D_t{u}_i=-{\partial }_{i}p+{\partial }_j{M}_{ij}$, where $D_t={\partial }_t+u_n{\partial }_n$ is the material derivative. Applying its product rule and the ideal magnetohydrodynamic induction equation givesThus the evolution uses only $\mathbf{u},M,p$, together with ${\partial }_i{u}_i=0$. To determine $p$, take the divergence of the magnetohydrodynamic momentum equation. Since ${\partial }_t{\partial }_i{u}_i=0$ and $u_j{\partial }_j{\partial }_i{u}_i=0$,This is the pressure Poisson equation for incompressible flow, with boundary conditions supplied by the imposed normal acceleration and normal stress, or by periodicity or decay at infinity as appropriate. For example, the normal component of the magnetohydrodynamic momentum equation requires $\mathbf{n}\cdot \nabla p=\mathbf{n}\cdot (\nabla \cdot M-D_t\mathbf{u})$ at a prescribed boundary. The pressure is determined up to a spatially constant function of time.

For equivalence to one resolved magnetic field, the initial $M$ must actually be $\mathbf{B}{\mathbf{B}}^T$ for a divergence-free vector field. In particular it is a symmetric positive semidefinite matrix of matrix rank at most one. The tensor equations also admit more general stresses; the averaged $M$ in part d need not have matrix rank one.

With zero velocity and no time dependence, the magnetic second-moment tensor equation holds automatically, but the magnetohydrodynamic momentum equation still requiresThe magnetic tension must therefore be a pressure gradient. Locally, its necessary integrability condition ison a simply connected region this is also sufficient for a scalar $p_0$, subject to the global boundary conditions. In physical units the same balance is $\nabla P_0={\mathbf{J}}_0\times {\mathbf{B}}_{\mathrm{phys},0}$: the Lorentz force density is balanced by mechanical pressure, rather than necessarily vanishing. A zero mean magnetic field does not make the local magnetic tension zero, so an arbitrary tangle is not automatically a static equilibrium.

Average the equations in part b and make exactly the prescribed scale-separation approximation: the retained perturbations vary on lengths much greater than $\ell$, and perturbations at the tangle scale are omitted. Products of two retained perturbations can then be neglected. Writing $\mathbf{v}=\delta \mathbf{u}$, $m=\delta M$ and $\pi =\delta p$, the linearized equations areNo unmodelled small-scale response has been inserted into this closure.

Introduce the fluid displacement $\mathbf{\xi }$ by $\mathbf{v}={\partial }_t\mathbf{\xi }$ and choose $\nabla \cdot \mathbf{\xi }=0$. Integrating the magnetic second-moment tensor equation giveswhere $s$ is a time-independent integration tensor set by the initial stress. For a pure oscillatory normal mode, $s=0$. Its restoring force is therefore $v_A^2{\nabla }^2\mathbf{\xi }$: the isotropic magnetic tangle has the same transverse response as an incompressible elastic medium with shear modulus ${\rho }_0{v}_A^2$.

For a plane wave proportional to $e^{i(\mathbf{k}\cdot \mathbf{r}-\omega t)}$, with $\omega \ne 0$, incompressibility gives $\mathbf{k}\cdot \hat{\mathbf{v}}=0$. The tensor equation and magnetohydrodynamic momentum equation becomeDotting the second equation with $\mathbf{k}$ yields $\hat{\pi }=0$ for $k\ne 0$, since $k_i{k}_j{\hat{m}}_{ij}=0$. Substitution then givesThere are two independent transverse polarizations. The phase speed is $v_A$ in every direction, and the two signs of $\omega$ give group velocities $\pm v_A\hat{\mathbf{k}}$. These magnetoelastic waves in an isotropic magnetic tangle exist because a shear deformation distorts the second moment of the tangled magnetic field, producing a restoring magnetic tension even though its mean is zero.

For comparison, a uniform magnetic field with Alfvén velocity ${\mathbf{B}}_0$ gives the incompressible Alfvén wave relation ${\omega }^2=(\mathbf{k}\cdot {\mathbf{B}}_0{)}^2$. Its frequency depends on the direction of $\mathbf{k}$, and energy propagates along ${\mathbf{B}}_0$. Here the averaged restoring response is isotropic: there is no preferred mean-field direction. Both systems resist transverse deformation through magnetic tension, rather than through compressive sound.

The propagating branch is not the entire perturbation space. For example, $\mathbf{v}=0$, $m_{ij}=f(\mathbf{r}){\delta }_{ij}$ and $\pi =f(\mathbf{r})$ solve the linearized equations without propagation, for any small, slowly varying $f$. Such a static stress perturbation in magnetoelastic wave equations balances a pressure gradient. Thus the requested wave conclusion applies to the oscillatory branch; it must not exclude stationary initial stress perturbations.

Start in an inertial frame with the inviscid magnetohydrodynamic momentum equation,Here $\rho$ is mass density, $P$ is the isotropic fluid pressure, $\mathbf{v}$ is inertial-frame velocity, $\mathbf{J}$ is electric current density, $\mathbf{B}$ is the magnetic field, and $\Phi$ is gravitational potential per unit mass. In the low-frequency approximation to Ampère's law, $\mathbf{J}=\nabla \times \mathbf{B}/{\mu }_0$. The vector calculus identityseparates the Lorentz force density into magnetic tension and a magnetic pressure gradient. Transform the material acceleration to a rigid frame rotating with constant angular velocity ${\mathbf{\Omega }}_0$, using the stated frame transformation. The rotating-frame magnetohydrodynamic momentum equation isThe final terms are the Coriolis force and centrifugal force densities; $\mathbf{u}$ is the velocity relative to the rotating frame, and $D_t={\partial }_t+\mathbf{u}\cdot \nabla$ uses that frame.

This requires a smooth conducting-fluid continuum with nonrelativistic motion, a constant magnetic permeability ${\mu }_0$, the solenoidal magnetic-field constraint, a negligible displacement current and negligible net electric-charge force. Viscous forces and anisotropic fluid stresses are omitted. A time-dependent frame would add an Euler force. Incompressible flow is not needed for this force equation. Finite electrical resistivity does not change its displayed instantaneous magnetic force; the later frozen-field calculation additionally uses ideal magnetohydrodynamics. A differentially rotating fluid does not define a single rigid frame with spatially varying ${\mathbf{\Omega }}_0$: choose a fixed local reference rate and retain the fluid's shear in $\mathbf{u}$.

Use the real magnetic field $\mathcal{B}$ in the coupled equations; its leading periodic approximation will be $\mathrm{Re}(\mathbf{B}{e}^{i\omega t})$. In the conducting incompressible flow, the magnetohydrodynamic momentum equation, resistive induction equation and solenoidal magnetic-field constraint areHere $p$ is pressure, $\mathbf{j}$ is electric current density, and $\eta$ and $\nu$ are magnetic diffusivity and kinematic viscosity. The moving-conductor Ohm law gives $\mathbf{E}=-\mathbf{u}\times \mathcal{B}+\eta \nabla \times \mathcal{B}$.

In the insulating exterior, neglecting displacement current, the magnetic field is curl-free away from the wire and solenoidal. The imposed electric current density suppliesThe exterior electric field also satisfies Faraday's law. At the rigid wall, impose the no-slip boundary condition $\mathbf{u}=0$. At finite conductivity, the interface conditions for electromagnetic fields give continuous normal magnetic field, continuous tangential magnetic field for equal permeability and no prescribed singular surface current, and continuous tangential electric field. The normal electric current density is zero at an insulating wall; it is automatic for the present two-dimensional fields with current only along $z$. The induced fields decay far from the wire and wall, and initial conditions, or selection of the periodic state after transients, complete the specification. In the limiting perfect conductor approximation a surface current can emerge, permitting a tangential-field jump.

The fully coupled solution need not remain monochromatic: a harmonic magnetic field produces both a mean and a twice-frequency Lorentz force density, and the resulting velocity can generate further harmonics. The following parts consistently use the leading magnetic response with fluid motion neglected.

Let $K$ be the kinetic energy, $M$ the magnetic energy and $E=K+M$:Dot the magnetohydrodynamic momentum equation with $\mathbf{v}$ and integrate. Incompressibility and the zero boundary velocity remove the advective and pressure work. The viscous term integrates to $-\rho \nu {\int }_D|\nabla \mathbf{v}{|}^2$. ThusThe ideal magnetohydrodynamic induction equation gives, with no boundary contribution,The Lorentz force density therefore transfers energy between the field and the fluid without creating total energy. Since $\nabla \cdot \mathbf{v}=0$ and $\mathbf{v}=0$ on the boundary, another integration by parts gives $\int |\nabla \mathbf{v}{|}^2=\int |{\mathbf{\omega }}_v{|}^2$, where ${\mathbf{\omega }}_v=\nabla \times \mathbf{v}$ is vorticity. The viscous magnetic-relaxation energy identity isThe conserved nonzero magnetic helicity prevents all the field energy from disappearing. Fix a bounded inverse-curl convention for the magnetic vector potential, so that $\Vert \mathbf{A}{\Vert }_2\le C_D\Vert \mathbf{B}{\Vert }_2$. The Cauchy-Schwarz inequality gives the magnetic-energy lower bound from helicity,Consequently $E$ decreases to a finite positive limit andUnder the intended regular long-time interpretation of smooth vorticity, the dissipation cannot keep producing peaks of fixed height and arbitrarily short duration. More precisely, if $d(t)=\Vert {\mathbf{\omega }}_v(t){\Vert }_2^2$ is uniformly continuous, uniformly continuous integrable dissipation tends to zero proves $d(t)\to 0$. The Poincaré inequality and the identity above then imply $\Vert \mathbf{v}(t){\Vert }_2\to 0$.

To identify the asymptotic state, suppose the regular relaxation has a limit, or work on a compact invariant limiting set on which the energy is continuous. On such a set the energy is constant, so the displayed identity forces ${\mathbf{\omega }}_v=0$, and the zero boundary velocity forces $\mathbf{v}=0$. The induction equation is then stationary, and the momentum equation becomesThis is magnetostatic equilibrium, with nonzero field allowed and required here by conserved magnetic helicity. It is not necessarily a force-free magnetic field: a pressure gradient can balance the magnetic force. If the field develops a current sheet, equilibrium is understood with the corresponding interface force balance.

There is a mathematical qualification to the smoothness premise. Smoothness at each finite time alone does not prove uniform long-time bounds, uniform continuity of dissipation or convergence to one limiting field. The energy calculation proves finite total dissipation unconditionally for a smooth solution; the settling conclusion is the regular-relaxation argument just given. A rigorous global convergence theorem requires that additional long-time control. In particular, smooth vorticity does not require smooth electric current density in the limiting state.

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.

For the circular MHD wave polarization, $B_x^2+B_y^2=a^2{B}_0^2$ is constant. Choose any constant ${\rho }_0>0,p_0>0$, put $v_a=B_0/\sqrt{{\mu }_0{\rho }_0}$ and takeThis gives a circularly polarized nonlinear Alfvén wave of arbitrary amplitude. All fields depend only on $z,t$, and $u_z=0$, so $\mathbf{u}\cdot \nabla \mathbf{u}=0$ and $\nabla \cdot \mathbf{u}=0$. Constant mass density and pressure then satisfy continuity and the adiabatic pressure equation, and $\nabla \cdot \mathbf{B}=0$ because $B_z$ is constant.

The remaining ideal magnetohydrodynamic induction equation and transverse magnetohydrodynamic momentum equation areSince ${\partial }_t=-v_a{\partial }_z$ on the traveling profiles and ${\mathbf{u}}_{\perp }=-(v_a/B_0){\mathbf{B}}_{\perp }$, both identities hold exactly when $v_a^2=B_0^2/({\mu }_0{\rho }_0)$. The longitudinal magnetohydrodynamic momentum equation holds because $p+|{\mathbf{B}}_{\perp }{|}^2/(2{\mu }_0)$ is constant. The circularly polarized wave is therefore an exact nonlinear solution even though the fluid is compressible. Compressibility permits mass density changes; it does not require every possible motion to compress the gas.

Take $B_0=B_z>0$, $v_a=B_0/\sqrt{{\mu }_0{\rho }_0}$ and $\zeta =k(z-v_{a}t)$. The transverse ideal magnetohydrodynamic induction equation and magnetohydrodynamic momentum equation are satisfied by $u_x=-B_x/\sqrt{{\mu }_0{\rho }_0}$, $u_y=u_z=0$, with constant mass density. However, the longitudinal magnetohydrodynamic momentum equation also requiresHere $B_x^2=a^2{B}_0^2{\cos }^2\zeta$ is not constant. Balancing its gradient would requireThe adiabatic pressure equation instead gives ${\partial }_{t}p=0$, since this velocity is divergence-free and has no component along the spatial variation. No choice of $p_{\ast }(t)$ removes the spatially varying time derivative. Thus a nontrivial finite-amplitude linearly polarized sinusoidal Alfvén wave is not an exact pure traveling wave in a compressible ideal fluid. The magnetic-pressure obstruction to a linearly polarized Alfvén wave is quadratic in $a$, so the wave is a valid linear approximation and drives a compressive response at higher order. In an incompressible model pressure is a constraint variable; it is not governed by the same compressible adiabatic evolution law.

Allowing longitudinal flow and varying mass density does not rescue a regular one-dimensional periodic traveling solution with precisely this transverse magnetic field. To see this, let $U=u_z-v_a$ and use the traveling coordinate. Integrated continuity, induction and transverse momentum givewhere $m,A,D$ are constants. Eliminating $u_x$ givesAt any zero of $B_x$, finite mass density and velocity require the right side to vanish. Wherever $B_x\ne 0$, the coefficient then vanishes and fixes $\rho ={\mu }_0{m}^2/B_0^2$, constant; continuity fixes $U$, and adiabatic entropy transport fixes $p$. The longitudinal magnetohydrodynamic momentum equation again contradicts the varying $B_x^2$. The alternative $m=0$ makes transverse momentum force $B_x$ constant. Trivial cases such as $a=0$ or $k=0$ are excluded from the nontrivial wave conclusion.

Here $\alpha$ is the thermal expansion coefficient and $\rho$ is the mass density. Let $K=|\mathbf{k}|$, $h^2=K^2-k_z^2$, $q=\mathbf{B}\cdot \mathbf{k}$, and $a^2=q^2/({\mu }_0\rho )$, the squared Alfvén frequency. We initially consider a nonzero frequency. The temperature and magnetic equations for a Fourier mode giveIn particular, the minus sign in the magnetic relation follows from $-i\omega \hat{\mathbf{b}}=iq\hat{\mathbf{u}}$. Substitution into the magnetohydrodynamic momentum equation, with $P=p/\rho$, givesFor incompressible flow, $\nabla \times (\hat{\mathbf{z}}\times \mathbf{u})=-{\partial }_z\mathbf{u}$. Taking the vertical component of the curl eliminates pressure and gives, with $\hat{\zeta }=(i\mathbf{k}\times \hat{\mathbf{u}}{)}_z$,Taking another curl and its vertical component uses $\nabla \times \nabla \times \mathbf{u}=-{\nabla }^2\mathbf{u}$ and $(\nabla \times \nabla \times (\theta \hat{\mathbf{z}}){)}_z=h^2\hat{\theta }$. Thusor, after eliminating temperature,The two equations for vertical velocity and vertical vorticity have a nontrivial solution only when their determinant vanishes. This proves the stratified magneto-Coriolis wave dispersion relationNo division by $\Delta$ was necessary, so a possible $\Delta =0$ is retained. Division by $\omega$, however, was necessary. Stationary disturbances must be checked directly in the original equations: for example, with $q\ne 0$, $\mathbf{u}=0$ and an appropriate solenoidal magnetic perturbation can balance temperature-induced buoyancy and pressure. Such static balances are separate from the nonzero-frequency wave branches.

Magnetoconvection is the competition between thermally driven convection and the force and transport of a conducting fluid's magnetic field. An imposed magnetic field resists motions that bend its field lines through magnetic tension, while magnetic diffusion permits slippage. The competition changes the onset threshold, preferred spatial scale, temporal behaviour and finite-amplitude organization. The geometry matters: a disturbance nearly invariant along a horizontal imposed magnetic field can avoid much of its tension, whereas an impermeable layer with a vertical imposed magnetic field cannot generally avoid bending it.

A useful quantitative example is a nonrotating Boussinesq layer of depth $d$, heated from below with temperature drop $\Delta T>0$ and threaded by a vertical magnetic field $B_0\hat{\mathbf{z}}$. Let $\nu$, $\kappa$, $\eta$ denote kinematic viscosity, thermal diffusivity and magnetic diffusivity. In thermal-diffusion units define the Rayleigh number, Prandtl number, magnetic-to-thermal diffusivity ratio and Chandrasekhar number byUse stress-free impermeable boundaries, fixed temperature perturbation zero, and magnetic boundary conditions keeping the field vertical. More explicitly, the normal modes below satisfy $w={\partial }_z^{2}w=\theta =b_x=b_y={\partial }_z{b}_z=0$ at $z=0,1$. For a horizontal wavenumber $a>0$, take $w,\theta \propto \sin \pi z$, $b_z\propto \cos \pi z$, and put $s=a^2+{\pi }^2$. The linear temperature and resistive induction equation give $(\lambda +s)\theta =w$ and $(\lambda +\zeta s)b_z={\partial }_{z}w$. Applying the solenoidal projection to the magnetohydrodynamic momentum equation removes pressure; the magnetic coupling is $Q\zeta {\partial }_z\mathbf{b}$, and the vertical projected buoyancy has coefficient $R{a}^2/s$. Substituting the temperature and magnetic responses gives the vertical-field magnetoconvection dispersion relationThese stated boundary conditions make the single vertical mode exact in this model; other boundaries change numerical thresholds.

For steady marginal convection, set $\lambda =0$. The threshold curve isThe imposed magnetic field adds a positive penalty. Differentiating with respect to $x=a^2$ shows that the minimizing wavenumber obeysAt $Q=0$ this gives $a_c=\pi /\sqrt{2}$ and $R_c=27{\pi }^4/4$. With increasing $Q$, the root moves to larger $x$ because the left-hand side is strictly increasing for $x\ge {\pi }^2/2$. The strong-field wavenumber selection in magnetoconvection limit isThus a strong vertical imposed field raises the steady threshold and selects narrower convection cells. At small horizontal wavenumber the incompressibility constraint makes a given vertical motion require large horizontal displacement and magnetic bending. Increasing horizontal wavenumber reduces this magnetic penalty, but increases viscous and thermal diffusion. Their balance selects the narrow cells. This statement is specific to the vertical-field geometry: wavenumber selection in oblique-field magnetoconvection can instead favour nearly field-invariant motions.

Oscillatory marginality in vertical-field magnetoconvection occurs because magnetic and temperature perturbations need not relax at the same rate. To exhibit the phase balance, substitute $\lambda =i\omega$ with $\omega \ne 0$ and write the neutral condition asThe imaginary part must vanish because $R$ is real. Dividing that imaginary part by $\omega s$ givessoA nonzero oscillation consequently requires $0<\zeta <1$ and a sufficiently strong imposed magnetic field. In this regime thermal diffusion can erase the temperature anomaly faster than magnetic diffusion relaxes bent field lines. The magnetic restoring force can outlast the buoyant anomaly, reverse the motion, and maintain the phase lag needed for oscillatory instability. The real part gives the oscillatory neutral curveOnly its portions with ${\omega }^2>0$ are physical. The first onset is the lower of the minimized steady and admissible oscillatory thresholds; the existence of an oscillatory neutral solution at a given wavenumber alone does not establish that it is first. For example, at fixed $P>0$ and $0<\zeta <1$, both minimizing wavenumbers grow like $Q^{1/6}$ for large $Q$, while $R_s\sim {\pi }^{2}Q$ and $R_o\sim D{\pi }^{2}Q$ with $D=\zeta (P+\zeta )/(1+P)<1$. At those oscillatory minima, the positive order-$Q$ contribution to ${\omega }^2$ dominates its order-$Q^{2/3}$ negative contribution, so the oscillatory branch is genuinely admissible and becomes the lower threshold. An overstability therefore precedes steady onset in this limit. More generally nonlinear convection can oscillate even when linear onset is steady; the linear calculation is not a classification of every time-dependent state.

The nonlinear regime introduces a different feedback. The magnetic Reynolds number $\mathrm{Rm}=Ud/\eta$ measures induction relative to diffusion for a finite convective velocity $U$. At large but finite $\mathrm{Rm}$, persistent convection rolls can redistribute magnetic flux. In a two-dimensional incompressible roll, a flux function $\mathcal{A}$ obeysCirculation repeatedly stretches the flux pattern, building gradients on which even small magnetic diffusivity acts efficiently. Over the flux-rearrangement time this permits flux expulsion: the interior of a vigorous roll becomes weakly magnetized, while magnetic flux is concentrated near its edges. The weak-field interior then suffers less magnetic inhibition and can support broader, stronger convection than the linear narrow-cell prediction for the original uniform magnetic field. Exact zero resistivity would preserve material flux and cannot by itself justify the relaxed flux-expelled state.

The weak interior field also follows from a steady leading-order balance for a coherent two-dimensional roll. Let $\psi$ be its streamfunction and suppose its interior streamlines are nested closed curves with $\nabla \psi \ne 0$. In the small-diffusivity interior, $\mathbf{u}\cdot \nabla {\mathcal{A}}_0=0$ implies ${\mathcal{A}}_0=F(\psi )$. Integrate the full steady flux equation over the region enclosed by one streamline. Because $\mathbf{u}\cdot \mathbf{n}=0$ there, the advection integral vanishes; the divergence theorem then gives $\oint {\partial }_n\mathcal{A}ds=0$. In a regular interior expansion this requires $F^{\prime }(\psi )\oint {\partial }_n\psi ds=0$. The normal derivative of $\psi$ has one nonzero sign around a regular nested streamline, so $F^{\prime }(\psi )=0$. Hence the leading interior flux function is constant, and its spatial gradient, which gives the two-dimensional magnetic field, vanishes. The mismatch with external flux is accommodated by magnetic boundary layers. This argument assumes the coherent steady roll and a regular interior; it does not prove that every turbulent flow expels flux.

The mean imposed vertical magnetic flux is conserved for horizontally periodic boundaries. In a schematic two-region picture, let a fraction $f$ have weak field $B_c$ in vigorous convection and the remainder have field $B_q$. ThenThe intensified field outside the active cells inhibits surrounding convection. This gives magnetic flux separation: vigorous weak-field patches coexist with strong-field regions whose motion is small or occurs on a finer scale. The redistribution is a consequence of nonlinear induction and back reaction, rather than disappearance of the net imposed flux.

This feedback can permit subcritical magnetoconvection. Infinitesimal perturbations decay below the conductive state's linear threshold because they encounter the full imposed field. A finite-amplitude roll can nevertheless persist by first expelling enough flux to reduce its own magnetic restraint. A sufficient finite disturbance can then switch between the stable conductive state and a convective state, producing hysteresis. A schematic amplitude equation makes the distinction between local linear stability and finite-amplitude persistence explicit:Here $\mu$ measures distance from linear onset; this is an illustrative normal form, not a claim that these coefficients have been derived for every magnetic layer. Writing $r=|A{|}^2$, the nonzero equilibria satisfyThey appear at a saddle-node bifurcation at $\mu =-c^2/(4d)$. For $-c^2/(4d)<\mu <0$, the zero state is stable, the lower radial-amplitude branch is unstable, and the upper branch is stable because its radial derivative has sign $c-2d{r}_{+}<0$. Thus finite-amplitude convection can coexist with a linearly stable conductive state below the onset threshold. Large $\mathrm{Rm}$ facilitates the flux-expulsion mechanism but is not a universal sufficient condition: geometry, the imposed magnetic field, the capacity to form coherent recirculating cells, and magnetic back reaction all matter. The broad nonlinear cells in weak-field regions can coexist with much narrower cells in magnetically dominated regions, linking the nonlinear flux distribution back to scale selection.

Use the standard uniform-density incompressible flow version of ideal magnetohydrodynamics, in which $\nabla \cdot \mathbf{u}=0$ and pressure enforces that constraint. The finite-$\gamma$ compressible adiabatic pressure equation is replaced by this pressure constraint. The consequences of retaining that cover-sheet equation literally as an additional condition are given below.

Write $F=f(R,z+v(R)t)$, $G=g(R,z-v(R)t)$, $U=F+G$ and $b=F-G$, where $b$ is the toroidal Alfvén velocity. The azimuthal velocity is $U$, while the Alfvén velocity vector is $b{\mathbf{e}}_{\varphi }+v{\mathbf{e}}_z$. Both solenoidal constraints hold automatically. Since there is no meridional flow, the azimuthal magnetohydrodynamic momentum equation and ideal magnetohydrodynamic induction equation reduce toDifferentiating the traveling arguments verifies both equations for arbitrary smooth $f$, $g$ and $v$. In particular, no condition $v^{\prime }=0$ is needed.

The remaining force balance is essential. Define the total pressure per unit mass, including the gravitational potential, byThe radial acceleration is $-U^2/R$, and magnetic curvature tension contributes $-b^2/R$. The radial and axial momentum equations are consequentlyThey are compatible if and only ifThis is the pressure compatibility of nonlinear torsional Alfvén profiles. It is also sufficient locally: when $FG$ is independent of $z$, chooseA stationary axisymmetric $\Phi$ satisfying the Poisson equation can be absorbed in this pressure. Gravity need not vanish. If the axis is included, smoothness requires $U,b=O(R)$ and a regular axial field; $f,g=O(R)$ and smooth even $v$ give the usual regular extension. The pressure constant may be chosen to make $p$ positive on a bounded region where the other terms are bounded.

The compatible profiles have the following interpretation. At a cylinder with $v\ne 0$, the variables $a=z+vt$ and $c=z-vt$ are independent as $z,t$ vary. Compatibility reads $f_a(R,a)g(R,c)+f(R,a)g_c(R,c)=0$. If both profiles are nonzero, division gives $f_a/f=-g_c/g=\kappa (R)$, independent of both traveling arguments. Thereforelocally on such a region. Alternatively, one profile vanishes identically and the other is unrestricted. Choosing a point where one profile is nonzero shows that a nontrivial opposite profile obeys the same linear ordinary differential equation everywhere on a connected axial interval, so isolated zeros do not create additional overlapping branches.

If $v>0$, a pure $f$ profile travels towards negative $z$, and a pure $g$ profile towards positive $z$, each at the local Alfvén speed $v(R)$. Their velocity and toroidal-field perturbations have equal magnitudes, with respectively the same or opposite signs. They are finite-amplitude torsional Alfvén waves. For a single profile, centrifugal force and magnetic curvature tension cancel, leaving spatially constant $\Pi$; gas pressure compensates the varying magnetic pressure. The waveform is transported without axial steepening. Different cylinder speeds shear its phase radially, magnetohydrodynamic phase mixing.

At linear order both wave families may be superposed, since their interaction product is second order in amplitude. Two arbitrary opposing profiles cannot simply be superposed at finite amplitude: their product would generate an axial variation in the radial force imbalance, incompatible with purely azimuthal velocity. The exponential pair is a formal exception, but is unbounded in an axial direction when $\kappa \ne 0$. If both profiles must be bounded on the whole axial line, $\kappa =0$, so $f=A(R)$ and $g=C(R)$ describe a steady rotating, twisted-field equilibrium. Their pressure gradient balances the remaining centrifugal and magnetic curvature forces. At $v=0$ there is no propagation; the condition is just $f(R,z)g(R,z)=h(R)$, giving additional steady profiles. These cases exhaust the local possibilities under the stated ansatz and domain qualifications.

There is a closure distinction in the wording. Incompressible flow is the formal infinite-sound-speed limit, not the simultaneous imposition of finite-$\gamma$ adiabatic compression and an independently fixed density; see Ogilvie's discussion of simplified fluid models, section 2.8. If the cover-sheet pressure equation is nevertheless imposed in addition, then axisymmetry and purely azimuthal motion give $p_t=0$. In addition to $(FG{)}_z=0$, the necessary and sufficient local pressure-gradient conditions, with stationary $\Phi$, areThey follow by requiring the gradients $p_z/\rho =-{\Phi }_z-\frac{1}{2}(b^2{)}_z$ and $p_R/\rho =4FG/R-{\Phi }_R-\frac{1}{2}(b^2+v^2{)}_R$ to be time independent; the remaining spatially uniform pressure variation is removed by $C(t)$.

For a single traveling profile with $v\ne 0$, these extra conditions require its square to be affine in its traveling argument, $f^2=a(R)(z+vt)+b_0(R)$ or $g^2=a(R)(z-vt)+b_0(R)$, with $v(R)a(R)$ constant across a connected radial region. Smooth bounded profiles on the entire axial line then have $a=0$ and are steady. For two nonzero exponential profiles the extra conditions force $\kappa =0$. Thus arbitrary finite-amplitude traveling pulses are valid for the intended incompressible pressure model, but not generally for the overconstrained finite-$\gamma$ closure. The force compatibility derived above makes the distinction explicit.

Let $L$ be a core-scale length, $B_{\ast }$ a representative strength of the prescribed poloidal magnetic field, and $v_A=B_{\ast }/\sqrt{{\mu }_0\rho }$ the Alfvén speed. A consistent weakly damped, small-amplitude ordering isThe first ratio is the Lehnert number. It makes the wave slow relative to rotation and allows almost geostrophic flow, with the azimuthal velocity approximately constant along a rotation-axis cylinder. The small wave amplitude gives $|B_{\varphi }|\sim \sqrt{{\mu }_0\rho }|U|\ll B_{\ast }$. Poloidal advection then changes neither the background field nor the wave substantially over one wave period. If there are no-slip boundary layers, also require their damping rate $\sqrt{\nu \Omega }/L\ll \sigma$, or otherwise retain their torque; a small bulk viscosity alone does not remove that boundary torque.