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The smooth ideal-MHD system couples mass conservation, momentum, adiabatic pressure transport and the ideal magnetohydrodynamic induction equation, subject to the solenoidal magnetic-field constraint. Without gravity it is a quasilinear partial differential equation system. Across shocks, conservative total energy replaces smooth adiabatic pressure transport.

At a spherical interface with an insulating exterior, equal magnetic permeability on both sides and no surface current, all components of the magnetic field are continuous. The solenoidal magnetic-field constraint then makes $P=\mathbf{B}\cdot \mathbf{x}$ and ${\partial }_{r}P$ continuous. A current-free exterior has $\Delta P=0$. For an isolated field decaying at infinity, its degree-$l$ spherical harmonic satisfies ${\partial }_r{P}_{lm}=-(l+1)P_{lm}/a$ at a sphere of radius $a$. This boundary condition incorporates the exterior field without setting the radial field to zero.

// Target: fluid-mechanics.bigb

A magnetic skin layer is the thin region penetrated by an oscillating magnetic field in a conductor. For locally tangential boundary amplitude $C(x)$, negligible material motion and variations along the boundary much slower than those across it, the resistive induction equation gives $B_x=C(x)e^{-(1+i)y/\delta }$ with time convention $e^{i\omega t}$. The solenoidal magnetic-field constraint requires the smaller component $B_y=C^{\prime }(x)e^{-(1+i)y/\delta }\delta /(1+i)$ to leading order. The tangential-field approximation therefore does not mean that the normal component is exactly zero at finite magnetic diffusivity.

For component-first velocity gradient $L_{ij}={\partial }_j{u}_i$, ideal incompressible magnetic field stretching gives the displayed evolution of $M_{ij}=B_i{B}_j$. It has the same upper-convected derivative as a polymer conformation tensor with negligible relaxation. Finite polymer relaxation time introduces a local return toward thermal equilibrium, whereas magnetic diffusion acts through spatial gradients and preserves the solenoidal magnetic-field constraint. The analogy does not identify those dissipative mechanisms.

Use the source's Gaussian units and neglect the displacement current, as in nonrelativistic magnetohydrodynamics. Write $\Delta p=p_{\perp }-p_{\parallel }$ for the pressure anisotropy. The pressure tensor is $\mathsf{P}=p_{\perp }\mathsf{I}-\Delta p\mathbf{bb}$, so its force density isAmpère's law and the solenoidal magnetic-field constraint give the Lorentz force decompositionAdding these two force densities and using $\mathbf{BB}=B^2\mathbf{bb}$ yieldsThe first term is the gradient of the modified magnetohydrodynamic total pressure; the second is the anisotropic Maxwell stress tensor contribution together with the directional pressure force. Their effective coefficient is the pressure-anisotropy-modified magnetic tension, $\mathcal{T}=B^2/(4\pi )+\Delta p$. At fixed magnetic field, positive pressure anisotropy increases resistance to bending, while negative pressure anisotropy reduces it. When $\mathcal{T}$ becomes negative, bending experiences an amplifying force rather than a restoring force. Gradients of $\Delta p$ can also drive motion, so the curvature interpretation refers specifically to the bending contribution.

Set $\Delta p_0=p_{\perp 0}-p_{\parallel 0}$ and ${\mathcal{T}}_0=\Delta p_0+B_0^2/(4\pi )$. Distinguish the scalar magnetic field strength perturbation $\delta B$ from the vector perturbation $\delta \mathbf{B}$. To first order,where the last identity is the solenoidal magnetic-field constraint. For a Fourier mode with the stated time convention, putThe linear momentum conservation equation, after division by $i$, isUse the solenoidal magnetic-field constraint in its $z$ component. The coefficient of $\hat{\mathbf{z}}$ reduces to $-k_{\parallel }[\delta p_{\parallel }+\Delta p_0\delta B/B_0]$. ThusBoth scalar pressure perturbations act within the $xz$ plane. The $y$ component therefore decouples, independently of their closure. The ideal magnetohydrodynamic equations giveEliminating the amplitudes gives the Alfvén wave dispersion relationFor this Shear Alfvén wave, the velocity and fluid displacement are along $y$, perpendicular to both $\mathbf{k}$ and the equilibrium magnetic field. Let $\mathbf{\xi }={\xi }_y\hat{\mathbf{y}}$; thenA real snapshot ${\xi }_y={\xi }_0\cos (\mathbf{k}\cdot \mathbf{r}-\omega t)$ bends the magnetic field lines sideways, with $\delta B_y=B_0{\partial }_z{\xi }_y$. There is no first-order compression or change of magnetic field strength: $\nabla \cdot \mathbf{\xi }=0$, $\delta \rho =0$ and $\delta B=0$. Scalar pressure perturbations are unnecessary in this transverse subsystem. For a traveling stable branch, $\delta B_y/B_0=-k_{\parallel }{u}_y/\omega$, so the velocity and transverse magnetic field perturbations are in phase or opposite phase according to the branch. At $k_{\parallel }=0$ the bending force and this angular frequency vanish.

Write the physical magnetic field as ${\mathbf{B}}_{\mathrm{phys}}$ and use the constant mass density ${\rho }_0$ to introduce Alfvén velocity units, $\mathbf{B}={\mathbf{B}}_{\mathrm{phys}}/\sqrt{{\mu }_0{\rho }_0}$. If $P$ is the fluid's mechanical pressure, the magnetohydrodynamic total pressure per unit mass density is $p=P/{\rho }_0+B^2/2$. The ideal magnetohydrodynamic equations then reduce toThe first equation is momentum conservation: the magnetic pressure has been included in $p$, leaving magnetic tension explicitly. The second is the ideal magnetohydrodynamic induction equation, expressing magnetic flux freezing. The last two equations impose incompressibility and the solenoidal magnetic-field constraint.

The model assumes a smooth conducting-fluid continuum with constant mass density and constant magnetic permeability, nonrelativistic velocity, and scales on which a single-fluid description is appropriate. Negligible electrical resistivity removes magnetic diffusion; negligible viscosity removes viscous momentum diffusion. Equivalently, the magnetic Reynolds number and the Reynolds number are large on the scales being retained. The low-frequency approximation neglects the displacement current; Hall and other kinetic corrections to Ohm's law are absent. Incompressibility is an imposed approximation, commonly justified when the Mach number is small. There is no independently evolved thermodynamic pressure: $p$ enforces the incompressibility condition. We omit external body forces; a conservative one could instead be absorbed into $p$.

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.

The common part of the magnetohydrodynamic-viscoelastic analogy is stretching and advection of a line element. With the component-first velocity gradient $L_{ij}={\partial }_j{u}_i$, the ideal tensor equation isIts left side is an upper-convected derivative, the same deformation law that appears for a polymer conformation tensor when its relaxation is neglected.

There are two important physical differences. First, the solenoidal magnetic-field constraint prevents a magnetic field line from having an isolated source or sink inside the fluid. A polymer strand can have molecular endpoints, so there is no corresponding no-endpoint constraint on the strands. Second, a stretched polymer has a preferred relaxed distribution of configurations and a finite polymer relaxation time; it can recoil relative to the solvent through molecular drag and thermal motion. An ideal magnetic field has no analogous local tendency to return to a prescribed coil size. Real polymers also have a finite contour length, a limitation absent from the ideal field-stretching law.

For example, a normalized Hookean dumbbell conformation equation has the dissipative extensionThis is local relaxation toward an isotropic thermal configuration. It remains nonzero for a spatially uniform stretched $C$. By contrast, constant magnetic diffusivity ${\eta }_m$ adds ${\eta }_m{\nabla }^2\mathbf{B}$ to the resistive induction equation. The product rule gives the resistive evolution of the magnetic second-moment tensor:The magnetic correction involves spatial gradients, vanishes for a uniform magnetic field, and does not create field-line endpoints: ${\partial }_t(\nabla \cdot \mathbf{B})={\eta }_m{\nabla }^2(\nabla \cdot \mathbf{B})$. It permits slippage and magnetic reconnection while preserving the no-monopole constraint. After averaging, its gradient-product term generally needs additional information beyond the averaged $M$. One cannot replace it indiscriminately by the local polymer relaxation term, or even by ${\eta }_m{\nabla }^{2}M$ alone.

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 $\mathbf{u}$.

To first order in fluid velocity divided by the speed of light, the Lorentz transformation of electromagnetic fields gives ${\mathbf{E}}^{\prime }=\mathbf{E}+\mathbf{u}\times \mathbf{B}$. Applying Ohm's law in the local fluid frame gives $\mathbf{E}+\mathbf{u}\times \mathbf{B}=\eta \mathbf{J}$. Neglecting the displacement current in Ampère's law, $\mathbf{J}=\nabla \times \mathbf{B}/{\mu }_0$. Substitution in Faraday's law givesHere ${\eta }_m$ is the magnetic diffusivity; the paper's $\eta$ is electrical resistivity, rather than magnetic diffusivity. Expanding the curl with the solenoidal magnetic-field constraint gives the resistive induction equation in its compressible form:The left side is the material derivative of the magnetic field. The three right-hand terms respectively describe field-line stretching and reorientation, amplification or dilution by compression or expansion, and magnetic diffusion. In particular, compression has $\nabla \cdot \mathbf{u}<0$ and amplifies a field component perpendicular to the compression.

The assumptions are nonrelativistic bulk motion, the low-frequency approximation permitting omission of the displacement current, a conducting-fluid continuum, isotropic local Ohm's law with spatially constant scalar electrical resistivity, constant magnetic permeability ${\mu }_0$, and $\nabla \cdot \mathbf{B}=0$. Hall, electron-pressure and other non-Ohmic corrections to the generalized Ohm's law are excluded. Incompressible flow is not required: imposing it would remove the compression term.

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 $\delta =\sqrt{2\eta /\omega }$ and $q=(1+i)/\delta$. Within the magnetic skin layer, differentiation across the layer dominates differentiation along the wall by the factor $b/\delta$. The leading resistive induction equation is thereforeBecause $\eta q^2=i\omega$ and $\mathrm{Re}q>0$, its decaying solution isThe solenoidal magnetic-field constraint determines the first smaller component,Corrections to $B_x$ from longitudinal diffusion have relative order $(\delta /b{)}^2$. Thus the zero normal boundary field from the perfect conductor calculation is a leading-order condition, not an additional exact condition at finite $\eta$. A smaller exterior-field correction matches $B_y$.

To leading order the electric current density is $j_z=qC{e}^{-qy}/{\mu }_0$. For two real harmonic fields, the period average of their product is half the real part of the product of one amplitude with the conjugate of the other. Thus the mean Lorentz force in a magnetic skin layer isIts normal component isThe first possible tangential contribution from the smaller $B_y$ vanishes: it is $-C{C}^{\prime }{e}^{-2y/\delta }\mathrm{Re}(q/q^{\ast })/(2{\mu }_0)$, and $q/q^{\ast }=i$. This cancellation matters in computing the tangential streaming flow.

Choose a magnetic vector potential $\mathbf{A}$ with $\nabla \times \mathbf{A}=\mathbf{B}$, and define the total magnetic helicity byThe value is unchanged by a single-valued gauge transformation $\mathbf{A}\mapsto \mathbf{A}+\nabla \chi$, becauseHere both the solenoidal magnetic-field constraint and the tangent boundary field are essential. In a multiply connected domain, choose the vector potential by extending the tangent field by zero to all space and fixing its decay there, or fix the equivalent circulation convention. This avoids silently varying the non-gradient harmonic ambiguity of a vector potential on the restricted domain.

The ideal magnetohydrodynamic induction equation permitsDifferentiating $H_M$ and using integration by parts for curl,The volume product vanishes algebraically, the gauge flux vanishes since $\mathbf{B}\cdot \mathbf{n}=0$, and the last boundary term vanishes since $\mathbf{v}=0$. Thus magnetic helicity is conserved even though viscosity dissipates mechanical energy.

Through magnetic flux freezing, magnetic field lines move with the fluid and retain their linkage during every smooth ideal motion. For two thin closed flux tubes carrying fluxes ${\Phi }_1,{\Phi }_2$, their mutual contribution is $2L{\Phi }_1{\Phi }_2$, where $L$ is their linking number; self-contributions measure twisting and writhing of the tubes. This explains the topological content of magnetic helicity. Conservation of this one scalar is a consequence of the frozen structure, not a complete description of all that structure.

Write horizontal components with a subscript $h$. The steady ideal magnetohydrodynamic induction equation gives $\nabla \times (\mathbf{u}\times \mathbf{B})=0$. Since ${\partial }_z=0$, its horizontal components implyThus $(\mathbf{u}\times \mathbf{B}{)}_z=C$ is a spatial constant. A vanishing axial electric field, $E_z=-C=0$, is needed to infer horizontal alignment. It is satisfied, for example, if the constant electric field vanishes at a boundary or at infinity. With this additional condition and ${\mathbf{B}}_h\ne 0$, define the magnetohydrodynamic mass loading $k$ by $\rho {\mathbf{u}}_h=k{\mathbf{B}}_h$, and put $U=u_z-k{B}_z/\rho$. ThenThe steady continuity equation and the solenoidal magnetic-field constraint giveAlso $\mathbf{u}\times \mathbf{B}=U{\mathbf{e}}_z\times \mathbf{B}$. Using the vector identity for the curl of a cross product, together with ${\partial }_z\mathbf{B}=0$ and $\nabla \cdot \mathbf{B}=0$, its curl is ${\mathbf{e}}_z\mathbf{B}\cdot \nabla U$. HenceThese are steady planar ideal-MHD field-line invariants: both quantities are constant along a magnetic field line.

The zero-electric-field condition does not follow from the printed stationarity assumptions alone. Constant $\rho ,p,s$, $\Phi =0$, $\mathbf{u}=u_0{\mathbf{e}}_x$ and $\mathbf{B}=B_0{\mathbf{e}}_y$, with $u_0{B}_0\ne 0$, satisfy every displayed ideal magnetohydrodynamics equation but have perpendicular horizontal velocity and magnetic field. Their $C=u_0{B}_0$ is nonzero, so no $k,U$ give the stated decomposition. The field-line decomposition is valid on the $E_z=0$ branch; as an unrestricted implication, the first claim has this counterexample. The subsequent calculations use that branch.

Consider the complex field before taking its real part. Put $p=\hat{z}\times k=(-k_y,k_x,0)$ and $b=Rp+S\hat{z}$. Both components are perpendicular to $k$, so the solenoidal magnetic-field constraint holds. The time-dependent phase obeysbecause ${\dot{k}}_y=-k_0/T$ and $u\cdot k=k_{0}y/T$. Moreover $\dot{p}=(k_0/T,0,0)$, and magnetic stretching is $b\cdot \nabla u=(R{k}_0/T,0,0)=R\dot{p}$. Thus the changing basis exactly supplies the shear-stretching term. Finally,and ${\nabla }^2$ acts as $-k^2$. Equating coefficients verifies

Take $k_0,T>0$ and first $\alpha \ne 0$. Remove magnetic diffusion by writing $R=e^{-D}G$, $S=e^{-D}H$, whereThen $\dot{G}=-i\alpha H$, $\dot{H}=i\alpha k^{2}G$, and $\ddot{G}={\alpha }^2{k}^{2}G$. Equivalently, rescaling time by $|\alpha |$ and replacing $H$ by $\mathrm{sgn}(\alpha )H$ gives the pair used in the question. Its growing WKB approximation isThe complex relative phase follows directly from $H=i\dot{G}/\alpha$; the printed $H\sim kG$ is sufficient for magnitudes but omits that phase.

The phase-averaged magnetic energy is $(k^2|R{|}^2+|S{|}^2)/(4{\mu }_0)$, hence for a nonzero growing-mode componentAt $t\gg T$, $k\sim k_{0}t/T$, so its leading exponential isThe positive quadratic term permits strong amplification before the negative cubic diffusion term dominates. The leading maximum occurs near $t_{\mathrm{peak}}=|\alpha |T/(\eta k_0)$. Put $\epsilon =\eta k_0/|\alpha |$ and $b_0=|\alpha |k_{0}T$. A well-separated late-time growing interval requiresThe first ensures $t_{\mathrm{peak}}\gg T$; the second ensures the WKB condition at the peak, since $|\dot{k}|/(|\alpha |k^2)\sim {\epsilon }^2/b_0$. For fixed nonzero $b_0$ both follow by taking sufficiently small positive $\eta$. The peak's leading exponent is $b_0/(3{\epsilon }^2)$, whereas the eventual exponent tends to minus infinity. This is mean-field shearing-wave transient amplification, not sustained dynamo action.

Growth from the initial instant is possible explicitly: choose $S(0)=i\mathrm{sgn}(\alpha )k_{0}R(0)$ with $R(0)\ne 0$. The initial logarithmic energy derivative is $2(|\alpha |k_0-\eta k_0^2)>0$ when $\epsilon <1$. Growth is not asserted for every initial polarization, for example an identically zero field. If $\alpha =0$, the amplitudes simply diffuse and the energy is proportional to $[k^2|R_0{|}^2+|S_0{|}^2]e^{-2D}$; shear can still give algebraic transient amplification of an in-plane field for small diffusivity, but a purely vertical field only decays. Positive diffusivity is essential to the final decay.

Use Gaussian units, uniform number density $n$, and ion mass density $\rho =n{m}_i$. Here $e>0$ is the magnitude of the elementary charge, as fixed by the printed current relation. Neglect electron inertia and displacement current. ThenFreezing the magnetic flux into this electron velocity, while retaining the ion momentum equation, gives the closed Hall magnetohydrodynamics systemThe pressure enforces the incompressible flow constraint; taking the divergence of momentum determines it up to the usual boundary-dependent constant. The induction equation preserves the solenoidal magnetic-field constraint. One can absorb $B^2/(8\pi )$ into magnetohydrodynamic total pressure, replacing the Lorentz term by magnetic tension. Keeping ion inertia here is important: electron magnetohydrodynamics with fixed ions will be a later limit, not the complete two-field system.

Statistical isotropy and statistical reflection symmetry allow the spectral tensor to contain only ${\delta }_{ij}$ and $k_i{k}_j$:The solenoidal magnetic-field constraint is essential here. Taking its Fourier transform gives $k_i{\hat{B}}_i=0$, hence $k_i{\hat{C}}_{ij}=0$ and $A+D{k}^2=0$. Thus the spectral tensor is a scalar multiple of the transverse projector of a vector field. Its trace is $2A$, so the prescribed normalization givesThis is the solenoidal isotropic spectral tensor. Without the solenoidal condition, isotropy and reflection symmetry would leave two scalar functions. Without reflection symmetry, an antisymmetric term proportional to $i{ϵ}_{ijm}{k}_m$ could additionally encode magnetic helicity.

There is a minor operator misprint to resolve first. Interpreting $(\mathbf{x}\cdot \nabla {)}^2$ as composition, the Cartesian formula for the angular Laplacian isThe last term is missing from the printed Cartesian expression. For example, its truncated expression maps $x_1$ to $x_1$, whereas a degree-one spherical harmonic must have angular eigenvalue two. In what follows, use the stated angular spectrum $l(l+1)$ and the supplied poloidal-energy identity; these are consistent with the corrected operator.

The divergence theorem and the solenoidal magnetic-field constraint imply zero magnetic flux through every sphere. Consequently $P=r{B}_r$ has zero angular mean and no $l=0$ component. Expand in orthonormal spherical harmonics:By angular orthogonality, withthe gradient norm and the supplied poloidal-toroidal decomposition identity becomeSince $l(l+1)\ge 2$, this proves the poloidal magnetic energy bound by the radial scalar gradient:Equality holds when the radial scalar contains only degree-one angular components.

Define the poloidal and total magnetic energies by $E_P=M_P/(2{\mu }_0)$ and $E_{\mathrm{total}}=M/(2{\mu }_0)$. Combining the radial-flow dynamo energy bound with $D\ge 2M_P$ gives the necessary conditionIt has the same temporal interpretation as the bound in the preceding part; statistically steady versions use the corresponding time-averaged energies. It quantifies why dynamo action with a substantial poloidal fraction requires radial motion, while a purely tangential flow has $Q=0$.

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 ${\partial }_r{P}_{lm}(a)=-(l+1)P_{lm}(a)/a$.