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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.

Let $W,\Theta ,C$ be the amplitudes of $w,\theta ,b_z$. The temperature and resistive induction equations immediately giveEliminate the total-pressure perturbation by applying $\Delta$ to the vertical momentum equation and subtracting ${\partial }_z$ times its divergence. The resulting vertical equation isIn particular, $imC=-m^{2}W/(s+\zeta \Lambda )$ shows explicitly that magnetic tension opposes the displacement.

Taking the determinant of these three amplitude equations gives the horizontal-field magnetoconvection dispersion relation:This cubic polynomial includes all roots of the assumed poloidal coupled system; using its determinant form avoids discarding possible roots by dividing by $s+\Lambda$ or $s+\zeta \Lambda$. As a check, $Q{m}^2=0$ factors off magnetic diffusion and leaves the standard stress-free Rayleigh-Bénard convection growth equation.

The solenoidal magnetic-field constraint impliesWith $\delta B/B_0=O(ϵ)$ and ${\partial }_z=O(ϵ{\nabla }_{\perp })$, the right-hand side is $O({ϵ}^2{k}_{\perp }{B}_0)$. Thus the leading, order-$ϵ$ perpendicular field is divergence-free. On a simply connected perpendicular patch, or for the nonzero perpendicular Fourier modes of a periodic domain, it has a stream function. Choose its normalization byConsequently the leading reduced electron magnetohydrodynamics representation isThe error denotes an order-${ϵ}^2$ field relative to $B_0$. The parallel perturbation is an independent scalar and is not removed by leading perpendicular solenoidality.

For clarity, the representation is asymptotic rather than an exact decomposition of an arbitrary three-dimensional solenoidal field. If $\delta B_{\parallel }$ depends on $z$, exact solenoidality can be restored by adding ${\nabla }_{\perp }\chi$ to $\delta {\mathbf{B}}_{\perp }$, where ${\nabla }_{\perp }^2\chi =-{\partial }_z\delta B_{\parallel }$. Its amplitude is $O({ϵ}^2{B}_0)$, so it lies beyond the retained field order. A perpendicular harmonic component can be absorbed into the guide field or fixed separately by boundary conditions.

Use a planar flux function $F=B_{0}z+a$ and writeThis automatically satisfies the solenoidal magnetic-field constraint. There is a sign convention to settle: with the usual definition $\mathbf{B}=\nabla \times \mathbf{A}$, the magnetic vector potential is $\mathbf{A}=-F\hat{\mathbf{y}}$. The positive sign printed for $A_y$ instead gives the opposite field, or corresponds to defining $\mathbf{B}=-\nabla \times \mathbf{A}$. We use the standard curl convention and the flux $F$ oriented along the prescribed positive $B_0$; the flux evolution and all transport identities below are unchanged by reversing the overall potential convention.

For planar incompressible flow, substitute this representation into the resistive induction equation. The two in-plane components implyAn additive time-dependent constant in $F$ leaves the field unchanged and removes $C(t)$. Since ${\nabla }^2(B_{0}z)=0$, this givesThis is the scalar advection-diffusion equation underlying the planar anti-dynamo theorem, with the imposed-field source retained.

Put $M(z)=\bar{a}(z)$, $a=M+a^{\prime }$ and $F_c(z)=\overline{w{a}^{\prime }}$. Stationarity of the horizontal averages, $\bar{w}=0$, and incompressibility giveHere the horizontal derivative of the advective flux has zero average. Integration gives $F_c=\eta M^{\prime }+C$. Averaging once more, with $⟨M^{\prime }⟩=0$, identifies $C=⟨w{a}^{\prime }⟩$. ThusDecay or periodic averaging removes the endpoint contribution from the mean derivative. The periodic assumption used later replaces decay at vertical infinity; a nonzero periodic function cannot also decay there.

Subtract the averaged equation from the scalar equation and multiply by $a^{\prime }$. The average of the time derivative is zero by stationarity; the advective term becomes a boundary flux and vanishes by incompressibility and the stated boundary averaging. Terms depending only on $z$ multiply $\overline{a^{\prime }}=0$. Integration by parts in the diffusion term therefore givesSubstituting $F_c=\eta M^{\prime }+C$ and using $⟨M^{\prime }⟩=0$ yields the Zeldovich planar flux balanceBoth terms on the right are dissipative. In a nontrivial stationary forced problem this requires transport opposing the imposed flux gradient.

For finite-energy localized fields, enforcing decay at both vertical infinities literally gives $C=0$ from the mean-flux boundary values; the corresponding integral balance then forces both dissipation terms to vanish. The later nonzero-transport, periodic case therefore replaces that localized boundary setting rather than satisfying both simultaneously.

Now assume $B_0\ne 0$ and a positive turbulent magnetic diffusivity ${\eta }_1$. DefineSince $F_c=-{\eta }_1{B}_{0}f$ and $⟨f⟩=1$, its spatial mean is $C=-{\eta }_1{B}_0$. HenceThe Zeldovich planar flux balance becomesBecause $C\ne 0$, the fluctuation cannot vanish, and a nontrivial fluctuation has $G>0$ under the stated boundary or zero-mean periodic conditions. ThereforeIf ${\eta }_1\gg \eta$, then $\Vert f-1{\Vert }_{L^2}^2<\eta /{\eta }_1\ll 1$: the transport profile must be nearly constant in mean square. This does not, without additional regularity, imply a uniform pointwise bound on narrow spikes.

For the final estimate use the supplied Poincaré inequality in the periodic setting and the Cauchy-Schwarz inequality:Insert the expression for $G$, cancel $B_0^2$, and multiply by $V$. Setting $y=r_{t}V\in [0,1)$ givesConsequently the variance bound for planar turbulent magnetic transport isThe factor $1/4$ comes from maximizing $y(1-y)$, so the weaker supplied Poincare constant already suffices. Under a zero vertical mean at each horizontal location, the usual period-$d$ Fourier bound is even stronger, with $4{\pi }^2/d^2$ in place of ${\pi }^2/d^2$.

Use the same length, thermal-time, velocity and temperature scales as in thermal convection. Write the imposed magnetic field as $B_0\hat{\mathbf{z}}$ and measure its perturbation in units of $B_0$. The physical linear magnetohydrodynamic momentum equation contains magnetic tension $(B_0^2/{\mu }_0){\partial }_z\mathbf{b}$; the gradient part of the Lorentz force density is absorbed into magnetohydrodynamic total pressure.

Divide the momentum equation by the viscous force scale ${\rho }_0\nu \kappa /d^3$. Its acceleration coefficient is $\kappa /\nu =1/\sigma$ and its magnetic coefficient isHere $Q$ is the Chandrasekhar number, comparing magnetic and viscous-diffusive effects, and $\zeta$ is the ratio of magnetic diffusivity to thermal diffusivity. The induction scale is $B_0\kappa /d^2$, so linearization givesComplete the linear system with $\dot{\theta }=w+{\nabla }^2\theta$, $\nabla \cdot \mathbf{u}=0$ and the solenoidal magnetic-field constraint $\nabla \cdot \mathbf{b}=0$. The omitted advective and perturbed-field tension terms are quadratic in the disturbances. Magnetic tension resists bending of the vertical field, while diffusion allows that bending to relax.

For constant alpha tensor and positive magnetic diffusivity $\eta$, the steady anisotropic alpha-squared dynamo equation isFor a nonzero Fourier mode with wavevector $\mathbf{K}=(k,l,m)$, this becomes $\eta K^2\mathbf{B}=i\mathbf{K}\times (\mathbf{\alpha }\mathbf{B})$. Rotational symmetry of the alpha tensor in the horizontal plane lets us set $\mathbf{K}=(q,0,m)$, with $q^2=k^2+l^2$. The component equations areSubstitute the first and third into the second. A nonzero steady amplitude requires ${\eta }^2{K}^4={\alpha }_0^2{m}^2+{\alpha }_0{\alpha }_1{q}^2$. Conversely, this relation supplies a nonzero amplitude through the same component equations, and the solenoidal magnetic-field constraint $q{B}_x+m{B}_z=0$ is automatically satisfied. Thus the steady-mode condition isFor the first form, no division by $m^2+\delta q^2$ is needed. The quotient applies when that denominator is nonzero, in particular for $q>0$ and $0<\delta <1$.

Put $Q=q^2>0$ and $r=m^2\ge 0$. The derivative of $F(r)=(Q+r{)}^2/(r+\delta Q)$ isFor $0<\delta <1/2$, the derivative changes from negative to positive at $r=(1-2\delta )Q$. For $1/2\le \delta <1$ it is nonnegative throughout the allowed half-line and the minimum is at $r=0$. Therefore the optimized uniaxial alpha dynamo threshold isBoth expressions agree at $\delta =1/2$. If the horizontal boundary conditions permit $q=0$, a nonzero mode instead has ${\gamma }^2=m^2$ and the infimum is zero as $m\to 0$; it is not attained by a nonzero wavevector. The exactly uniform mode $q=m=0$ has no diffusive or alpha curl term and must be treated separately.

Take the primitive state vector $\mathbf{U}=(\rho ,p,u_x,u_y,u_z,B_x,B_y,B_z{)}^T$, with $\rho >0$. In the absence of gravity, the ideal magnetohydrodynamic equations are first order: expanding the material derivatives, pressure gradient and Lorentz force makes every term linear in a spatial derivative of $\mathbf{U}$, with coefficients depending on $\mathbf{U}$. Dividing momentum balance by $\rho$ therefore gives a quasilinear partial differential equation system ${\partial }_t\mathbf{U}+A_i(\mathbf{U}){\partial }_i\mathbf{U}=0$ with eight-by-eight matrices. The solenoidal magnetic-field constraint is an additional constraint on initial data; the ideal magnetohydrodynamic induction equation preserves it because the divergence of a curl vanishes.

For a one-dimensional simple wave in magnetohydrodynamics, write $\mathbf{U}=\mathbf{U}(q(x,t))$ with a nonzero state-space tangent. Substitution gives ${\mathbf{U}}_q{q}_t+A_x{\mathbf{U}}_q{q}_x=0$. A nonconstant profile requires the tangent to be a right eigenvector of $A_x$:The chain rule then gives the wave-speed equation:This is the Inviscid Burgers equation, including the special case of constant $v$. Its characteristic curves are $x=a+v_0(a)t$, with $v=v_0(a)$. As long as the mapping remains invertible,Thus a region with $v_0^{\prime }<0$ steepens: faster characteristics catch slower ones, and the first gradient catastrophe occurs at $t_{\ast }=-1/{\min }_a{v}_0^{\prime }(a)$ when the minimum is negative. A genuinely nonlinear compressive simple wave therefore forms a shock wave. Rarefactive profiles can spread instead; steepening is not inevitable for every initial profile.

To continue past this time, use weak solutions of the conservative mass, momentum, fluid total-energy equation and ideal magnetohydrodynamic induction equation. The Rankine-Hugoniot conditions fix the jumps and shock velocity, while physical entropy production selects admissible magnetohydrodynamic shocks. Diffusive shock wave layers are replaced by moving discontinuities, so their microscopic structure need not be explicitly resolved. In particular, the smooth adiabatic pressure equation must be replaced by total-energy conservation when imposing the ideal magnetohydrodynamic shock conditions.

For the Alfvén waves, impose the one-dimensional solenoidal magnetic-field constraint, so $B_x$ is constant and $\delta B_x=0$. Put ${\mathbf{B}}_{\perp }=(B_y,B_z)$ and $c=v-u_x$. The transverse components of the linearized ideal magnetohydrodynamic equations giveThe longitudinal momentum equation additionally gives $\delta p+{\mathbf{B}}_{\perp }\cdot \delta {\mathbf{B}}_{\perp }/{\mu }_0=0$. The Alfvén eigenvectors have $\delta \rho =\delta p=\delta u_x=0$, hence their transverse polarization is perpendicular to ${\mathbf{B}}_{\perp }$. For $s=\pm 1$, their speed and explicit right Alfvén characteristic eigenvectors areFor ${\mathbf{B}}_{\perp }\ne 0$ one may choose $(h_y,h_z)=(B_z,-B_y)$, up to a nonzero scalar factor. If ${\mathbf{B}}_{\perp }=0$, either transverse polarization is allowed and this characteristic speed is degenerate. These expressions describe the two propagating Alfvén branches for $B_x\ne 0$; if $B_x=0$, they coalesce with advected, nonpropagating transverse disturbances.

Integrating along an Alfvén simple wave leaves $\rho ,p,u_x,B_x$ unchanged and gives $d{\mathbf{u}}_{\perp }=-sd{\mathbf{B}}_{\perp }/\sqrt{{\mu }_0\rho }$ and ${\mathbf{B}}_{\perp }\cdot d{\mathbf{B}}_{\perp }=0$. Consequently the finite-amplitude nonlinear Alfvén wave relations areThe constant-magnitude condition is essential: otherwise a varying magnetic pressure would drive longitudinal compression. An arbitrary smooth phase profile gives an explicit family, $f_3=B_{\perp }\cos \Theta (x-v_{s}t)$ and $f_4=B_{\perp }\sin \Theta (x-v_{s}t)$, with $f_1,f_2$ as above. Direct substitution in the transverse momentum and ideal magnetohydrodynamic induction equation gives $(u_x-v_s){\mathbf{u}}_{\perp }^{\prime }=B_x{\mathbf{B}}_{\perp }^{\prime }/({\mu }_0\rho )$ and $(u_x-v_s){\mathbf{B}}_{\perp }^{\prime }=B_x{\mathbf{u}}_{\perp }^{\prime }$, confirming the solution without relying on the eigenvector argument. Longitudinal momentum holds because $p+B_{\perp }^2/(2{\mu }_0)$ is constant. The speed $v_s$ is independent of amplitude along this family: these linearly degenerate characteristic fields translate without steepening.

Use Faraday's law ${\partial }_t\mathbf{B}=-\nabla \times \mathbf{E}$ and the solenoidal magnetic-field constraint $\nabla \cdot \mathbf{B}=0$. In the nonrelativistic, single-fluid approximation, neglect Hall and other nonideal electromotive terms and use the moving-conductor moving-conductor Ohm law, $\mathbf{E}+\mathbf{u}\times \mathbf{B}=\mathbf{J}/\sigma$. Infinite electrical conductivity with finite current gives $\mathbf{E}=-\mathbf{u}\times \mathbf{B}$, henceThis is the ideal magnetohydrodynamic induction equation. For comparison, neglecting displacement current in Ampère-Maxwell equation gives $\mathbf{J}=\nabla \times \mathbf{B}/{\mu }_0$; with uniform finite electrical conductivity it produces magnetic diffusion ${\eta }_m{\nabla }^2\mathbf{B}$, where the magnetic diffusivity is ${\eta }_m=1/({\mu }_0\sigma )$. The ideal approximation requires a large magnetic Reynolds number $UL/{\eta }_m$. Dropping displacement current is useful for this finite-conductivity comparison, but Faraday's law and the ideal Ohm relation already suffice for the ideal induction equation.

Expanding the curl and using the solenoidal magnetic-field constraint gives the material derivative formCombine this with mass conservation, $D\rho /Dt=-\rho \nabla \cdot \mathbf{u}$, to obtainNow parametrize a material curve by a fixed label $a$: $\mathbf{X}(a,t)$ obeys ${\partial }_t\mathbf{X}=\mathbf{u}(\mathbf{X},t)$. Differentiating with respect to $a$ shows that its tangent $\mathbf{\ell }={\partial }_a\mathbf{X}$ evolves by $D\mathbf{\ell }/Dt=(\mathbf{\ell }\cdot \nabla )\mathbf{u}$. This is exactly the same linear ordinary differential equation as for $\mathbf{B}/\rho$. Initially parallel tangents remain parallel by uniqueness, with a label-dependent proportionality factor constant along each particle trajectory. Thus magnetic field lines are transported as material curves, wherever the field and fluid flow are smooth and the field is nonzero. This is the field-line form of magnetic flux freezing.

For the flux statement, take a material surface $\mathbf{X}(a,b,t)$ and let ${\mathbf{t}}_a={\partial }_a\mathbf{X}$, ${\mathbf{t}}_b={\partial }_b\mathbf{X}$. Both tangents obey $D{t}_i/Dt=({\partial }_j{u}_i)t_j$. Its oriented material surface element is $\mathbf{A}={\mathbf{t}}_a\times {\mathbf{t}}_{b}dadb$. Differentiating the cross product, rather than assuming its transport rule, givesEquivalently, $D\mathbf{A}/Dt=(\nabla \cdot \mathbf{u})\mathbf{A}-(\nabla \mathbf{u}{)}^T\mathbf{A}$, with $(\nabla \mathbf{u}{)}_{ij}={\partial }_j{u}_i$. Contracting this derived rule with the induction equation gives a pointwise cancellation:Integrating over the fixed material labels therefore proves conservation of flux through an open material surface:The surface need not be closed; its boundary is carried with the fluid. The result follows from material transport, rather than from the zero flux through a closed surface.

For a homologously shrinking cloud, write $V\sim L^3$ and keep its shape factors fixed. Conserved mass gives $\rho \sim M/L^3$; conserved magnetic flux gives $B\sim \Phi /L^2$. Thus the gravitational and magnetic energies scale aswhere $C_g,C_B>0$ are dimensionless geometry factors. Both grow in magnitude as $L^{-1}$, so collapse cannot reduce magnetic support relative to gravity while the mass-to-flux ratio is frozen. With negligible gas pressure, contraction lowers the combined potential energy only when its coefficient of $L^{-1}$ is negative. Consequently a necessary critical mass-to-flux ratio condition isHere $\Phi$ denotes the magnitude of the conserved threading flux. The numerical coefficient depends on geometry and boundary conditions; the scaling argument does not determine it or make the condition sufficient in the presence of other support.

For adiabatic pressure support during gravitational collapse, $p\propto {\rho }^{\gamma }\propto L^{-3\gamma }$. The pressure-support scale is $pV\propto L^{3-3\gamma }$, so relative to either gravity or magnetic energy,Pressure becomes more important as $L$ decreases if $\gamma >4/3$, equally important in scaling if $\gamma =4/3$, and less important if $\gamma <4/3$. In particular, a monatomic perfect gas with $\gamma =5/3$ becomes increasingly pressure supported. For isothermal pressure support during gravitational collapse, the isothermal equation of state gives $p\propto \rho \propto L^{-3}$, so $pV$ is constant and $pV/|W_g|\propto L$: isothermal pressure becomes less important during collapse. The same comparisons hold against magnetic support because its energy has the same $L^{-1}$ scaling as gravity.

For a rigid frame with constant angular velocity $\mathbf{\Omega }$, add the Coriolis force and centrifugal force to the inviscid magnetohydrodynamic momentum equation. A constant magnetic permeability and solenoidal magnetic-field constraint split the Lorentz force density into magnetic pressure and magnetic tension. This nonrelativistic conducting-fluid approximation neglects the displacement current, net electric-charge force and viscosity. Differential rotation belongs to the fluid's velocity field, not to a spatially varying rigid-frame rotation vector. A time-dependent frame requires the additional Euler force.

Near the axis of a solenoid, the magnetic vector potential $A_{\theta }=\rho B(z)/2$ givesThe radial fringe component enforces the solenoidal magnetic-field constraint. A sharp axial step in $B$ must retain this fringe as a distributional field; a purely axial field that jumps to zero violates Gauss's law for magnetism. The hard-edge model is a convenient limit for near-axis trajectories, not a literal monopole sheet.