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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.

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.

Interpret perfect conductivity here as the perfectly conducting thermal boundary condition $\theta =0$; impose the stated magnetic verticality separately. Together with the stress-free boundary condition, admissible vertical profiles areeach multiplied by $e^{i\mathbf{k}\cdot \mathbf{x}+\lambda t}$. Solenoidality requires $i\mathbf{k}\cdot {\mathbf{U}}_h=-\pi W$ and $i\mathbf{k}\cdot {\mathbf{B}}_h=\pi Z$. The resistive induction equation preserves both profile choices. In particular ${\mathbf{b}}_h=0$ at each boundary and ${\partial }_z{b}_z=0$ there.

Set $s=k^2+{\pi }^2$ and $p={\pi }^2$. The pressure-projected vertical momentum, thermal and vertical induction equations reduce toTheir determinant gives the vertical-field magnetoconvection dispersion relationIt is best kept as a polynomial rather than canceling a factor that might vanish at a special eigenvalue. For $\lambda =0$,For an oscillatory marginal mode $\lambda =i\omega$ with $\omega \ne 0$, write the polynomial as ${\lambda }^3+a_1{\lambda }^2+a_2\lambda +a_3$. Its coefficients areSeparating real and imaginary parts gives $a_3=a_1{a}_2$ and ${\omega }^2=a_2>0$. Solving yieldsTherefore oscillatory marginality in vertical-field magnetoconvection requiresThe equality has zero frequency and is not an oscillatory marginal mode. For some positive horizontal wavenumber to satisfy this condition, it is necessary and sufficient that $Q>\zeta (1+\sigma ){\pi }^2/[\sigma (1-\zeta )]$, although the associated threshold must still be minimized over admissible modes. When ${\omega }^2>0$, $R^{(o)}<R^{(e)}$ at the same wavenumber; this also follows from $R^{(e)}-R^{(o)}=(\sigma +1+\zeta )s{\omega }^2/(\sigma \zeta k^2)$.