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Statistical homogeneity makes the two-point correlation depend on the displacement $\mathbf{u}=\mathbf{r}-{\mathbf{r}}^{\prime }$. Substituting the Fourier transforms and changing variables from $(\mathbf{r},{\mathbf{r}}^{\prime })$ to $(\mathbf{u},{\mathbf{r}}^{\prime })$ givesThe last integral is a Dirac delta function, soThe plus sign occurs because neither Fourier transform has undergone complex conjugation. For a real magnetic field, the version with ${\hat{B}}_j({\mathbf{k}}^{\prime }{)}^{\ast }$ instead has $\delta (\mathbf{k}-{\mathbf{k}}^{\prime })$. Homogeneity determines dependence on the displacement vector; dependence only on its norm needs statistical isotropy as well.

Write the transverse displacement as $\mathbf{s}$ and take the common line-of-sight interval to be $[0,L]$. The rotation measure is a line integral of the magnetic field, so its correlation isSubstitute the Fourier inversion of the solenoidal isotropic spectral tensor. Keeping the finite observation length gives the exact windowed expressionThe allowed extension of the longitudinal displacement integral to the whole real line is the long-path projection of a magnetic correlation. It applies when $L$ is much larger than the correlation length: equivalently, the squared window becomes $2\pi L\delta (k_z)$ inside the integral. The Dirac delta function sets $k_z=0$, leavingStatistical isotropy makes this a function of $s=|\mathbf{s}|$. The displayed two-dimensional relation is the intended long-path approximation; the preceding windowed formula accounts for finite-length edge effects.

Put $a=c/(4\pi en)$ and retain terms linear in $\delta \mathbf{B}$. The electron magnetohydrodynamics induction equation becomesHere the curl of $(\nabla \times \delta \mathbf{B})\times B_0\hat{\mathbf{z}}$ reduces to $B_0{\partial }_z\nabla \times \delta \mathbf{B}$, since the background magnetic field is constant and the divergence of a curl vanishes. For a plane wave $\delta \mathbf{B}=\mathbf{b}{e}^{i(\mathbf{k}\cdot \mathbf{r}-\omega t)}$, the linear equation and the solenoidal constraint areOn the transverse plane, the operator $i\mathbf{k}\times$ has eigenvalues $\pm k$ by the helicity decomposition of a transverse Fourier mode. Equivalently, squaring the equation and using the vector triple product gives ${\omega }^2=a^2{B}_0^2{k}_{\parallel }^2{k}^2$. Since $a{B}_0=v_A{d}_i$, where $v_A$ is the Alfvén speed and $d_i$ the ion skin depth,The two branches have opposite circular polarizations. The specified electron-only induction model describes a whistler wave; the paper calls these branches kinetic Alfvén waves. The following cascade calculation uses the specified model and its dispersion relation, without adding the pressure response needed for the usual kinetic Alfvén interpretation.

Multiply the Fokker-Planck equation by $B^n$ and integrate over $B>0$ and all $\sigma$, assuming the required boundary terms vanish. Integration by parts givesThe mixed moment is not determined by $\mathbb{E}[B^n]$: the current strain and the accumulated magnetic field are correlated. Zero mean strain does not imply $\mathbb{E}[\sigma B^n]=0$, so this is an unclosed moment equation.

Keep the strain dependence by defining $P_n(\sigma ,t)={\int }_0^{\infty }{B}^{n}P(B,\sigma ,t)dB$. The magnetic drift term satisfiesThe strain derivatives commute with the $B$ integral. HenceThis weighted density obeys a tilted Ornstein-Uhlenbeck Fokker-Planck equation. Its integral over $\sigma$ is the desired moment, while retaining $\sigma$ makes the evolution closed.

Put $I=R{B}_{\varphi }$ and define ${\Delta }_{\ast }\psi ={\psi }_{RR}-R^{-1}{\psi }_R+{\psi }_{zz}$. Direct cylindrical differentiation givesThe azimuthal component of the Lorentz force is proportional to $I_z{\psi }_R-I_R{\psi }_z$. A force-free magnetic field therefore has parallel meridional gradients of $I$ and $\psi$, implying $I=f(\psi )$ on connected regular flux surfaces. ConsequentlyWith this dependence the poloidal part of $\nabla \times B$ is $f^{\prime }(\psi )B_p$. Its full cross product with $B$ isVanishing force therefore requires ${\Delta }_{\ast }\psi +f{f}^{\prime }=0$ wherever the flux gradient is nonzero, extending by regularity through ordinary isolated nulls. Finally, evaluating the three-dimensional cylindrical divergence gives $R^2\nabla \cdot (R^{-2}\nabla \psi )={\Delta }_{\ast }\psi$, soThis derives the Grad-Shafranov equation for a force-free magnetic field. The function $f$ is set by boundary/current information; it is not an additional fixed constant. Its single-valued global form presumes the usual connected flux-surface labeling. The equation also makes $\nabla \times B=f^{\prime }(\psi )B$, explicitly showing that current is parallel to the 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.

Let ${\rho }_0$ be the reference mass density, $\nu$ the kinematic viscosity, $\kappa$ the thermal diffusivity, $\eta$ the magnetic diffusivity, and ${\alpha }_T$ the coefficient of thermal expansion. The conductive state of Rayleigh-Bénard convection is $T_b=T_0+\Delta T(1-z/d)$. For the Boussinesq approximation, density variation is retained only in buoyancy. Linearizing about this state and the imposed magnetic field $B_0\hat{\mathbf{y}}$ givesHere $w=u_z$ and $p_{\ast }=p^{\prime }+B_0{b}_y/{\mu }_0$ is the perturbation of magnetohydrodynamic total pressure. The gradient part of the Lorentz force has been absorbed into $p_{\ast }$; the remaining term is magnetic tension. The sign in the temperature equation follows from $d{T}_b/dz=-\Delta T/d$.

Use the thermal-diffusion scaling of a convection layer: length $d$, time $d^2/\kappa$, velocity $\kappa /d$, temperature perturbation $\Delta T$, field perturbation $B_0$, and total-pressure perturbation ${\rho }_0\nu \kappa /d^2$. Dropping dimensionless-variable decorations giveswithThese are respectively the Rayleigh number, Chandrasekhar number, Prandtl number, and magnetic-to-thermal diffusivity ratio. In particular, the magnetic coefficient is $\zeta Q$, not $Q$ alone, in these thermal-time units. At $z=0,1$, the boundary conditions are $\theta =w=b_z=0$, ${\partial }_z{u}_x={\partial }_z{u}_y=0$, and ${\partial }_z{b}_x={\partial }_z{b}_y=0$.

For disturbances varying only along the imposed magnetic field, set $k=0$ and $x=m^2>0$. The steady horizontal-field magnetoconvection dispersion relation becomesExpand it in powers of $x$:Its derivative and second derivative areBecause $R$ diverges as $x\to 0$ and $x\to \infty$, the unique minimum is determined byFor large Chandrasekhar number, this equation first gives $x\sim {\pi }^3{Q}^{-1/2}$ and hence the requested strong-field steady magnetoconvection varying along the field selection law:

For the threshold through order unity, the leading $Qx+{\pi }^6/x$ contribution at its optimum is $2{\pi }^3\sqrt{Q}$. More precisely, the stationary equation givesThe correction to $Qx+{\pi }^6/x$ has no first-order contribution because that expression is already stationary at its leading minimizer. Also $3{\pi }^{2}x=O(Q^{-1/2})$ and $x^2=O(Q^{-1})$. ThereforeThe constant term $3{\pi }^4$ must be retained. This restricted family bends magnetic field lines, so its large stationary threshold is compatible with the unrestricted field-aligned minimum in the preceding part.

Write $c_T^2=\mathcal{R}{T}_0$ for the squared isothermal sound speed, distinguishing the gas constant $\mathcal{R}$ from a Rayleigh number, and put $H=c_T^2/g$. A uniform magnetic field has no bulk Lorentz force, so hydrostatic equilibrium and the ideal gas equation giveAt the interface, it is magnetohydrodynamic total pressure, not gas pressure alone, that is continuous:Define ${\Delta }_B=B_1^2-B_2^2>0$. The magnetic pressure discontinuity in an isothermal atmosphere therefore producesIn particular, the upper gas is denser at the interface. Both halves have the same scale height $H$, despite their different pressure normalizations.

For the disturbance calculation, write ${\rho }_{-}=\rho$, ${\rho }_{+}=\rho +\delta \rho$, where $\delta \rho ={\Delta }_B/(2{\mu }_0{c}_T^2)$, and ${\Sigma }_{\rho }={\rho }_{-}+{\rho }_{+}$. Work to leading order in the prescribed short-vertical-scale approximation: the bulk motion is locally incompressible potential flow, while the isothermal equation determines the equilibrium density jump. Its scale separation and small compressive correction will be checked at the maximizing mode.

Let the interface displacement be $h{e}^{st+ikx+imy}$ and $K=\sqrt{k^2+m^2}>0$. Using displacement rather than velocity potentials, write ${\mathbf{\xi }}_i=\nabla {\chi }_i$ and ${\mathbf{u}}_i=s{\mathbf{\xi }}_i$. Incompressibility and irrotational flow give $\Delta {\chi }_i=0$. Decay away from the interface and continuity of its normal displacement fixwith the common horizontal and temporal exponential understood. Both satisfy ${\partial }_z{\chi }_i(0)=h$. Thus the disturbance has vertical extent of order $K^{-1}$.

In ideal magnetohydrodynamics, magnetic flux freezing gives ${\mathbf{b}}_i=({\mathbf{B}}_i\cdot \nabla ){\mathbf{\xi }}_i$ for these locally incompressible displacements. If ${\pi }_i=p_i^{\prime }+{\mathbf{B}}_i\cdot {\mathbf{b}}_i/{\mu }_0$ is the total-pressure perturbation, linear momentum isSince $({\mathbf{B}}_i\cdot \nabla {)}^2=-({\mathbf{k}}_h\cdot {\mathbf{B}}_i{)}^2$ on the mode, integration in each half-space yieldsAt the displaced interface, continuity of Lagrangian magnetohydrodynamic total pressure requires ${\pi }_1-{\rho }_{-}gh={\pi }_2-{\rho }_{+}gh$. Inserting the two potentials gives the orthogonal-field magnetic Rayleigh-Taylor dispersion relation:The first term is the destabilizing Rayleigh-Taylor instability of the heavier upper gas; the second is stabilizing magnetic tension. A negative right-hand side gives an oscillatory mode rather than growth.

For fixed $K$, the magnetic penalty isIt is minimized by $m=0$: the wavevector is along the weaker upper magnetic field and perpendicular to the stronger lower field. If $B_2\ne 0$, define $A=g\delta \rho /{\Sigma }_{\rho }$ and $C=B_2^2/({\mu }_0{\Sigma }_{\rho })$. The growth law for this orientation is $s^2=AK-C{K}^2$, a concave quadratic with maximum at $K_{\ast }=A/(2C)$. HenceFor this orientation the unstable interval is $0<K<2K_{\ast }$; the maximum lies halfway along it.

The localization condition for the fastest orthogonal-field interface mode follows directly from the vertical decay length:This is the condition for ${\ell }_{\ast }\ll H$ on both sides. It also controls the locally incompressible approximation, sinceAcoustic adjustment is then fast compared with growth, so bulk density perturbations are higher-order corrections. If this localization condition fails, the constant-background approximation cannot be used to identify the physical fastest mode.

Finally, the printed assumptions allow $B_2=0$. In that case $m=0$ eliminates magnetic tension on both sides and gives $s^2=g\delta \rho |k|/{\Sigma }_{\rho }$. There is no finite maximum growth rate or maximizing wavenumber in the ideal sharp-interface model: growth is unbounded as $|k|\to \infty$. The finite maximum and localization formula above therefore require $0<|B_2|<|B_1|$.

In the growth curves, $\theta$ is the angle of the horizontal wavevector from the $x$-axis. The illustrated ratio $B_1^2/B_2^2=101$ gives $K_{\ast }H=25$, and shows how even a small rotation toward the stronger magnetic field lowers the maximum growth.

Use rationalized electromagnetic units with $c=1$ and electrostatic constant $1/(4\pi )$. The four-potential $A^a=(\phi ,A_x,A_y,A_z)$ transforms by the same matrix as the coordinates:Write $r=\sqrt{y^2+z^2}$ and ${\mathbf{e}}_r=(0,y,z)/r$. Gauss's law for the line at rest gives ${\mathbf{E}}^{\prime }=\sigma {\mathbf{e}}_r/(2\pi r)$, ${\mathbf{B}}^{\prime }=0$. A suitable potential is ${\phi }^{\prime }=-\sigma \log (r/r_{\ast })/(2\pi )$, ${\mathbf{A}}^{\prime }=0$. The inverse Lorentz transformation gives $\phi =\gamma {\phi }^{\prime }$, $A_x=\gamma v{\phi }^{\prime }$. The fields from $\mathbf{E}=-\nabla \phi -{\partial }_t\mathbf{A}$ and $\mathbf{B}=\nabla \times \mathbf{A}$ are thereforeThe sign agrees with a line charge moving in the positive $x$ direction: its density is $\gamma \sigma$ and its current $\gamma v\sigma$. In SI units insert $1/{ϵ}_0$ in the electric expressions and replace $\mathbf{B}=\mathbf{v}\times \mathbf{E}$ by $\mathbf{B}=\mathbf{v}\times \mathbf{E}/c^2$.

The two invariants check explicitly:For $|v|\ll 1$, the electric field differs from the static field only at order $v^2$, while the magnetic field is of order $v$ and is the field of the moving line current. These fields are evaluated off the charged axis.

A steady localized electric current density satisfies $\nabla \cdot \mathbf{j}=0$, from charge conservation or the divergence of Ampère's equation. Choose an enclosing region whose boundary lies outside its support; extending the integration region does not change the source integrals. The divergence theorem applied componentwise givesTheir boundary fluxes vanish, proving the localized steady-current moment identitiesFor discontinuous but localized currents the same argument uses the distributional steady conservation law and an outer enclosing surface.

The magnetic vector potential is a vector field with $\mathbf{B}=\nabla \times \mathbf{A}$, consistent with $\nabla \cdot \mathbf{B}=0$. It is defined up to $\mathbf{A}\mapsto \mathbf{A}+\nabla \chi$. The steady Maxwell equation $\nabla \times \mathbf{B}={\mu }_0\mathbf{j}$ givesIn the Coulomb gauge $\nabla \cdot \mathbf{A}=0$, the equation is ${\nabla }^2\mathbf{A}=-{\mu }_0\mathbf{j}$, whose decaying Green-function solution is the one supplied in the source.

Let $r=|\mathbf{x}|$ and expand its kernel uniformly over the bounded source region:The first term integrates to zero. Put $T_{ki}=\int x_k^{\prime }{j}_i({\mathbf{x}}^{\prime })d{V}^{\prime }$, an antisymmetric tensor by the earlier identity. Define the magnetic dipole moment $m_{\ell }=\frac{1}{2}{ϵ}_{\ell ki}{T}_{ki}$, so $T_{ki}={ϵ}_{ki\ell }{m}_{\ell }$. ConsequentlyorThe displayed dipole term has size $r^{-2}$. If $\mathbf{m}=0$, it vanishes and a higher multipole determines the leading magnetic field instead.

For $r>0$, $\nabla \cdot (\mathbf{x}/r^3)=0$ andTherefore the far magnetic field is

For a moving circuit, Faraday's law takes the formHere $C(t)$ is the instantaneous wire contour, $\mathbf{v}$ its local material velocity, and $S(t)$ any spanning surface. The contour orientation and surface normal vector obey the right-hand convention. The integral on the right is the magnetic flux, differentiated with both the field and circuit position allowed to vary. The electric field term is the electric force per unit electric charge around the wire, and $\mathbf{v}\times \mathbf{B}$ supplies the motional electromotive force. Their sum is the electromotive force driving charge around the moving circuit. The condition $\nabla \cdot \mathbf{B}=0$ makes the magnetic flux independent of the spanning surface.

Choose the initial loop normal vector to be $+{\mathbf{e}}_y$, so rotation about $+{\mathbf{e}}_z$ gives $\mathbf{n}(t)=(-\sin \Omega t,\cos \Omega t,0)$. The spatially uniform magnetic field givesUsing Ohm's law for the resistive loop,Positive electric current is measured in the contour orientation associated with $\mathbf{n}(t)$. Reversing that convention reverses the signed answer. When $\Omega =\omega$, the magnetic flux is constant and the induced electric current vanishes. This uses the usual thin resistive-circuit approximation with self-inductance neglected.

Let the equilibrium ideal gas be at rest, with uniform ${\rho }_0,p_0$ and ${\mathbf{B}}_0=B_0{\mathbf{e}}_z$. This local wave calculation neglects gravity, viscosity and magnetic diffusion. Define the adiabatic sound speed $c\equiv v_s=(\gamma p_0/{\rho }_0{)}^{1/2}$, the Alfvén speed $v_a=B_0/({\mu }_0{\rho }_0{)}^{1/2}$, and a wavevector $\mathbf{k}=k(\sin \theta ,0,\cos \theta )$, where $k=|\mathbf{k}|$ and $\theta$ is its angle to the equilibrium magnetic field. Perturbations have dependence $e^{i\mathbf{k}\cdot \mathbf{x}-i\omega t}$ and phase speed $v=\omega /k$.

The linearized ideal magnetohydrodynamic equations for mass conservation, pressure, induction and momentum giveThe relation between pressure and mass density follows from eliminating $\mathbf{k}\cdot {\mathbf{u}}^{\prime }$ between the two material evolution equations; it applies to the nonzero-frequency wave sector. The induction result automatically has $\mathbf{k}\cdot \mathbf{b}=0$.

The $y$ MHD wave polarization is perpendicular to both the field and wavevector. It has $\delta p=\delta \rho =0$ and obeysThus the transverse Alfvén wave hasIn the $x$-$z$ plane the two remaining velocity components instead obey the symmetric matrix eigenproblemTaking its determinant gives $v^4-(c^2+v_a^2)v^2+c^2{v}_a^2{\cos }^2\theta =0$. Solving this quadratic equation in $v^2$ derives both magnetoacoustic wave phase speeds,The plus sign is the fast magnetosonic wave, the minus sign the slow magnetosonic wave. The radicand is nonnegative, since it equals $[(c^2-v_a^2{)}^2+4c^2{v}_a^2{\sin }^2\theta ]/4$. Both squared speeds are nonnegative. Each nonzero mode has forward and backward propagation; an advected entropy disturbance is a separate zero-frequency mode in this static equilibrium.

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.