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For a solenoidal vector field $J$ with compact support and suitable integrability, apply the divergence theorem to $x_i{x}_{j}J$. It gives zero boundary term and hence the displayed antisymmetry. Equivalently, for any constant vector $C$, the vector triple product identity yields $\int (C\cdot r)JdV=\frac{1}{2}C\times \int (J\times r)dV$. A piecewise smooth field must have no normal jump at its support boundary; otherwise its weak divergence contains a surface contribution and the argument fails.

Two scalar labels constant along a magnetic field line are Euler potentials when their crossed gradients give the magnetic field. The magnetic vector potential $\mathbf{A}=\alpha \nabla \beta$ proves that this representation is a solenoidal vector field. Conversely, two independent local line labels yield $\mathbf{B}=f(\alpha ,\beta )\nabla \alpha \times \nabla \beta$; replacing $\beta$ by ${\int }^{\beta }f(\alpha ,b)db$ absorbs the factor. This is a local construction; a global regular single-valued pair requires compatible magnetic field line topology. Fixing boundary labels preserves their paired-footpoint connections in an ideal variation.

In a bounded regular domain, fix the global extension or circulation convention for the magnetic vector potential. A bounded inverse of curl on tangent solenoidal vector fields gives $\Vert \mathbf{A}{\Vert }_2\le C_D\Vert \mathbf{B}{\Vert }_2$. The Cauchy-Schwarz inequality then gives $|H_M|\le C_D\Vert \mathbf{B}{\Vert }_2^2=2{\mu }_0{C}_D{E}_B$. One construction extends the tangent field by zero outside the domain and uses the whole-space Biot-Savart law; its kernel of size $|\mathbf{x}-\mathbf{y}{|}^{-2}$ is integrable over a bounded region, giving the required $L^2$ bound. Nonzero conserved magnetic helicity therefore prevents complete decay of magnetic energy.

For a solenoidal vector field carrying current only in a conductor contained in a sphere of radius $R$, match the exterior field to a decaying potential field and include exterior magnetic energy. The smallest free-decay eigenvalue of the enclosing sphere is ${\pi }^2/R^2$, from its dipolar poloidal mode. The corresponding variational estimate gives the displayed bound. Continuity and the usual insulating-interface magnetic conditions are part of the admissible field class. This bound is the diffusive ingredient in Backus' necessary condition for dynamo action.

The magnetic helicity integrates the dot product of a magnetic vector potential and its magnetic field. It measures linkage and twisting of magnetic flux. For a solenoidal vector field $\mathbf{B}$ tangent to the boundary, it is unchanged by the gauge transformation $\mathbf{A}\mapsto \mathbf{A}+\nabla \chi$, since the change is ${\int }_V\nabla \cdot (\chi \mathbf{B})dV={\int }_{\partial V}\chi \mathbf{B}\cdot d\mathbf{S}=0$. If its density satisfies ${\partial }_{t}h=\nabla \cdot (f\mathbf{B}-h\mathbf{u})$ with both vector fields tangent to the boundary of a fixed volume, the divergence theorem gives conservation of its integral. This is an instance of helicity conservation by tangent boundary conditions.

In a steady axisymmetric magnetohydrodynamic wind, axisymmetric pressure and Newtonian gravity exert no azimuthal torque. The azimuthal momentum equation is $\rho {\mathbf{u}}_p\cdot \nabla (s{u}_{\varphi })={\mu }_0^{-1}{\mathbf{B}}_p\cdot \nabla (s{B}_{\varphi })$. Mass conservation and the solenoidal vector field condition imply $\nabla \cdot {\mathbf{F}}_J=0$. With magnetohydrodynamic mass loading $\rho {\mathbf{u}}_p=\kappa {\mathbf{B}}_p$, the transported angular momentum per unit poloidal magnetic flux is $\ell =\kappa s{u}_{\varphi }-s{B}_{\varphi }/{\mu }_0$. It is constant on connected regular poloidal flux surfaces. The usual specific magnetohydrodynamic angular-momentum invariant is $L=\ell /\kappa$, provided $\kappa \ne 0$; confusing these two normalizations loses a factor of mass loading in the wind torque.

The displayed solenoidal vector field has $\nabla \times \mathbf{b}=k\mathbf{b}$, ${\nabla }^2\mathbf{b}=-k^2\mathbf{b}$ and ${\mathbf{e}}_z\times \mathbf{b}=-i\mathbf{b}$. Thus the non-gradient part of its linearized Lorentz force about $\mathbf{B}=Jr{\mathbf{e}}_{\varphi }$ vanishes. Choosing $p_1=-\mathbf{B}\cdot \mathbf{b}/{\mu }_0$ cancels the magnetic-pressure gradient, and $\mathbf{u}=0$, $\omega =0$ is a neutral perturbation. It is regular at the axis and realizes the endpoint $m=2$, $\alpha =0$. A boundary condition can exclude this growing-in-radius field, but without such a condition strict positivity at the endpoint cannot be asserted.

Choose two independent labels for the magnetic field lines, constant along each line. These are Euler potentials: $\mathbf{B}\cdot \nabla \alpha =\mathbf{B}\cdot \nabla \beta =0$. Where the two gradients are independent, the vector perpendicular to both is $\nabla \alpha \times \nabla \beta$, so initially writeThe solenoidal condition gives $0=\nabla \cdot \mathbf{B}=(\nabla \alpha \times \nabla \beta )\cdot \nabla f$. Thus $f$ is constant along a field line and is locally a function of its labels. Redefining one label, for example $\tilde{\beta }(\alpha ,\beta )={\int }^{\beta }f(\alpha ,b)db$, absorbs this factor. Dropping the tilde givesThis representation automatically has zero divergence, since it is also $\nabla \times (\alpha \nabla \beta )$.

For an magnetic arcade, assign the labels on one polarity of the footpoint surface and transport them along each line to its other footpoint. They specify both line identity and magnetic flux: on a transverse surface, flux coordinates can be chosen so that the directed flux element is $d\alpha d\beta$. In particular,The boundary labels at paired footpoints must agree, so prescribed boundary functions encode a compatible footpoint mapping, not independently arbitrary values at the two ends.

The construction assumes a smooth family of nonintersecting lines that can be labelled from a transverse boundary patch, with suitable patches at nulls or coordinate singularities. A local Euler-potential representation does not establish the existence of two globally single-valued regular labels for every possible solenoidal field; closed, recurrent or more complicated field topology can obstruct such a global choice. The line-tied magnetic arcade geometry supplies the relevant simplifying assumption here.

Use the original scanned boundary condition. The scan fixes the normal-field amplitude at the corner $x=z=0$. The supplied TeX drops the condition $x=0$ and would instead prescribe a constant normal field on the entire bottom. Those are different boundary-value problems; the solution below uses the legible original scan.

Choose a positive integer $n$, put $k=n\pi /a$, and seek a separated decaying mode. A suitable flux function isThis is a decaying linear force-free magnetic arcade. The first and third curl components then satisfy the force-free equation automatically; its second component requires $-({\partial }_x^2+{\partial }_z^2)A={\lambda }^{2}A$. ThusDirect differentiation gives $\nabla \cdot \mathbf{B}=0$ and $\nabla \times \mathbf{B}=\lambda \mathbf{B}$. Both horizontal components vanish at $x=0,a$, and $B_z(0,0)=B_0$. The magnetic field on the bottom is $B_0\cos kx$, not a uniform $B_0$.

A strictly decaying real normal mode requires $q>0$, soFor the fundamental, single-arcade choice $n=1$, this becomes $|\lambda |<\pi /a$. Equality gives no vertical exponential decay, while a larger magnitude gives an oscillatory rather than decaying vertical dependence for that mode. The data do not explicitly fix the horizontal mode number. If arbitrarily fine higher-mode arcades are permitted, every finite real $\lambda$ admits some integer $n$ with $n\pi /a>|\lambda |$. The fundamental restriction must therefore not be advertised as a necessary bound for every possible field satisfying only the stated pointwise and wall data.

For a fundamental field line, $A$ is constant along its projection in the $(x,z)$ plane. If $0<s<1$ denotes its normalized flux-function value, its arch isThe maximum height is $q^{-1}\log (1/s)$ at $x=a/2$. In the horizontal projection,Thus the vertical projections are arches and the horizontal projections are straight sheared segments joining their paired footpoints. Positive $\lambda$ gives positive horizontal slope, negative $\lambda$ reverses the shear, and $\lambda =0$ gives an unsheared potential arcade. Higher $n$ repeats such cells across the width.

The transcription's alternative uniform-bottom condition would be inconsistent with exponential decay and zero side flux for $B_0\ne 0$: solenoidality gives $d{\int }_0^a{B}_{z}dx/dz=-B_x(a,z)+B_x(0,z)=0$, whereas a uniform nonzero bottom would give a conserved flux $a{B}_0$ that cannot decay to zero. This integral check is an independent reason not to silently solve that mistranscribed problem.

Apply the divergence theorem to $A(r)=(C\cdot r)(C^{\prime }\cdot r)J(r)$. Since $C,C^{\prime }$ are constant and $J$ is a solenoidal vector field, the product rule givesIts boundary flux vanishes when $J$ has compact support, or when its normal component vanishes on the boundary of the integration volume. Integrating therefore provesFor a discontinuous field, the divergence-free hypothesis must include its boundary jump, in the sense of a weak derivative. A field that merely has zero divergence at interior points can still have a nonzero surface source; that weaker hypothesis would not justify discarding the flux. The result is the antisymmetric first moment of a localized solenoidal field.

Linearizing the ideal magnetohydrodynamic induction equation about a static magnetic field and using $u^{\prime }={\partial }_t\xi$ gives the fluid Lagrangian displacement-generated magnetic field perturbation $b=\nabla \times (\xi \times B)$. The two solenoidal vector field conditions reduce the curl identity toIts divergence vanishes because it is a curl. Directly, the two products of first derivatives in ${\partial }_i{b}_i$ cancel after exchanging their indices, while the remaining terms differentiate $\nabla \cdot \xi$ or $\nabla \cdot B$, both zero. Similarly the integrated linear continuity equation gives ${\rho }^{\prime }=-\nabla \cdot (\rho \xi )$, so $\boxed{{\rho }^{\prime }=-\xi \cdot \nabla \rho }$.

Let $\Pi =p+B^2/(2{\mu }_0)$ be the magnetohydrodynamic total pressure. Magnetostatic equilibrium and the momentum equation in the linearized ideal magnetohydrodynamic equations areThere is no perturbed gravitational potential here because the field is fixed. Multiply by ${\xi }_i^{\ast }$ and integrate. The pressure term vanishes by the assumed boundary cancellation and $\nabla \cdot {\xi }^{\ast }=0$. Thuswhere $\mathcal{M}=-\int {\xi }_i^{\ast }(B_j{\partial }_j{b}_i+b_j{\partial }_j{B}_i)dV$.

Integrate the first magnetic term along $B$, using $\nabla \cdot B=0$, then insert the expression for $b$. This gives the positive term $\int |(B\cdot \nabla )\xi {|}^{2}dV$ and two mixed derivative terms. For clarity, the troublesome term transforms asThe first term on the right cancels the other mixed term. Combining the remaining derivative with the term containing ${\partial }_j{B}_k{\partial }_k{B}_i$ givesDifferentiating the equilibrium relation now combines the gradient of mass density and magnetostatic equilibrium terms:We obtain the incompressible magnetic displacement energy identityThe two Hessian matrices are real symmetric, making their Hermitian form evaluations real. The last term is a squared norm, and $I>0$ for a nonzero displacement in fluid with positive mass density. Thus $\boxed{{\sigma }^2\text{ is real}}$. It may be negative, in which case one temporal branch has exponential growth; real squared frequency is not itself a proof of stability.

For the atmosphere with a horizontal magnetic field, $\Phi =gz$ has zero Hessian matrix and the equilibrium equation isConsequently the interchange stability of an incompressible magnetized atmosphere is governed by the numeratorFor ${\xi }_z=\sin ky$, the magnetic tension term is zero: these fluid Lagrangian displacements interchange neighboring magnetic field lines without varying along them. A mass density increasing with height gives negative energy and instability; decreasing mass density gives positive energy for these vertical interchanges. If ${\rho }_z\le 0$ everywhere, the displayed form is nonnegative for every admitted incompressible displacement. If the gradient changes sign, suitably localized divergence-free interchange trials detect an unstable inverted layer; For example, take ${\xi }_z=f(z)e^{iky}$ and ${\xi }_y=i{f}^{\prime }(z)e^{iky}/k$, with $f$ supported in that layer. Their divergence is zero, and neither component varies along the magnetic field.

For ${\xi }_z=\sin kx$, magnetic field lines are bent. Averaging over a full horizontal wavelength gives the trial Rayleigh quotientunder the stipulated admissible boundary/integrability conditions. It is a trial quotient, not a claim that this displacement is an exact eigenfunction for every profile. mass density decreasing upward is stable here too. For inverted mass density, magnetic tension can stabilize sufficiently short waves along the magnetic field, but does not remove the transverse interchange instability. Locally the two contributions have the familiar scale $-g{\rho }_z/\rho +k^2{v}_A^2$, with the Alfvén speed $v_A=B/\sqrt{{\mu }_0\rho }$.

Take the equilibrium mass density to be constant and absorb any conservative equilibrium force into the pressure. The linearized ideal magnetohydrodynamic induction equation, using the solenoidal vector field conditions, isFor $\mathbf{B}=Jr{\mathbf{e}}_{\varphi }$, differentiation of the cylindrical basis is important. Acting on a vector whose cylindrical components have azimuthal dependence $e^{im\varphi }$ gives $(\mathbf{B}\cdot \nabla )\mathbf{u}=imJ\mathbf{u}+J{\mathbf{e}}_z\times \mathbf{u}$. Also $(\mathbf{u}\cdot \nabla )\mathbf{B}=J{\mathbf{e}}_z\times \mathbf{u}$. These basis contributions cancel in induction, giving $i\omega \mathbf{b}=imJ\mathbf{u}$.

The perturbation of the Lorentz force isBoth of the first two terms now contribute a basis-rotation term. With $\tilde{p}=p_1/\rho +\mathbf{B}\cdot \mathbf{b}/({\mu }_0\rho )$, the linearized ideal magnetohydrodynamic momentum equation is thereforeThe magnetic part of $\tilde{p}$ is the perturbation of magnetic pressure, not an extra restoring force to retain after projecting out pressure.

Put $D={\omega }^2-m^2{\tilde{J}}^2$, where $\tilde{J}=J/\sqrt{{\mu }_0\rho }$. For $mJ\ne 0$, substitute $\mathbf{u}=\omega \mathbf{b}/(mJ)$ into the momentum equation and multiply by $mJ$. Taking a curl removes the modified pressure. Sincewe obtain the uniform-current toroidal-field curl reductionA second curl, using $\nabla \times \nabla \times \mathbf{b}=-{\nabla }^2\mathbf{b}$, givesThe divided expression assumes $D\ne 0$. If $mkJ\ne 0$, the undivided curl equation shows that $D=0$ would force $\mathbf{b}=0$, so this restriction loses no such magnetic mode. The $k=0$ degenerate limit can instead be taken in the undivided equations; it has no growing branch of the kind considered below. Axisymmetric $m=0$ disturbances are likewise excluded by the velocity elimination, and induction gives no nonzero-frequency magnetic disturbance in this particular equilibrium.

For a vector Laplacian eigenmode with ${\alpha }^2+k^2>0$, elimination givesThus both squared frequencies are real. The lower branch satisfies ${\omega }_{-}^2\ge {\tilde{J}}^2(|m{|}^2-2|m|)$, so it is positive for $|m|>2$. For $|m|=2$ it is positive when $\alpha \ne 0$, but the printed strict-positivity statement needs an endpoint qualification: $|m|=2$, $\alpha =0$ permits a neutral mode unless boundary conditions exclude it.

This is a genuine counterexample to the unqualified endpoint, not merely a loose algebraic bound. The neutral quadrupolar perturbation of a uniform-current toroidal fieldis regular at the axis, is solenoidal, and has $\nabla \times \mathbf{b}=k\mathbf{b}$, hence ${\nabla }^2\mathbf{b}=-k^2\mathbf{b}$. Also ${\mathbf{e}}_z\times \mathbf{b}=-i\mathbf{b}$, so the non-gradient part of the linearized Lorentz force vanishes for $m=2$. The gas-pressure perturbation $p_1=-\mathbf{B}\cdot \mathbf{b}/{\mu }_0$ cancels its magnetic pressure gradient, as required by $\tilde{p}=0$. Taking its real part gives a real disturbance. Decay at radial infinity or an appropriate homogeneous radial boundary condition would exclude it, restoring strict positivity for the allowed eigenmodes; no such condition is printed.

For $|m|=1$ the lower branch is unstable precisely whenAt equality it is marginal. With the time convention $e^{i\omega t}$, the growing member has $\omega =-is$ andThe other branch remains oscillatory. Actual admissibility of a chosen $\alpha$ depends on the radial domain and its boundary conditions.

Let $\eta >0$ denote the uniform magnetic diffusivity. The resistive induction equation for a solenoidal vector field in an incompressible flow isHere the radial compression and axial stretching have zero total divergence: $r^{-1}{\partial }_r(-\omega r^2)+{\partial }_z(2\omega z)=0$. A purely axial field independent of $z$ is automatically solenoidal, and the stretching term is $2\omega B_z\hat{\mathbf{z}}$. Thus the ansatz closes under the equation. For its angular mode, with $m$ a nonnegative integer,This is magnetic concentration by an incompressible stagnation flow.

Put $q=r/g(t)$ and $D_{m}b=b^{\prime \prime }+q^{-1}{b}^{\prime }-m^2{q}^{-2}b$. Substitution and division by $f$ giveA convenient self-similar magnetic mode in a stagnation flow is obtained by making the two dimensionless separation coefficients constant. ChooseThe constant $\lambda$ labels the radial modes. More generally the first right side can be $\beta \eta$ and the radial drift term $\beta q{b}^{\prime }$; the displayed equations choose the positive-width normalization $\beta =1$.

If $g(0)=g_0>0$, the width isFor $\omega =0$ its continuous limiting expression is $g^2=g_0^2+2\eta t$. Since $\eta /g^2=\omega +g^{\prime }/g$, the amplitude isFor the compressive case $\omega >0$, the width tends to the diffusion-strain balance scale $\sqrt{\eta /\omega }$ rather than collapsing to zero.

Let $I_b={\int }_0^{\infty }b(q{)}^{2}qdq$ be finite, and let $C_m={\int }_0^{2\pi }{\cos }^2(m\varphi )d\varphi$, equal to $2\pi$ for $m=0$ and $\pi$ otherwise. The magnetic energy per unit axial length isso the complete time dependence isThe angular integral is only a constant; zero signed magnetic flux does not force $I_b$ or $E$ to vanish.

For $m=0$, smoothness on the axis requires $b(0)$ finite and $b^{\prime }(0)=0$. Set $s=-q^2/2$ and $b=h(s)$. The radial equation becomes the Kummer differential equationIts regular solution iswhere $M$ is the confluent hypergeometric function of the first kind. The second independent solution has a logarithmic singularity at the axis and is inadmissible for a smooth field. For real $\lambda >0$, the regular solution vanishes at infinity: generically it has a $q^{-\lambda }$ tail; at positive even integer $\lambda$ it instead has a Gaussian times a polynomial. Thus decay at infinity alone does not select a unique separation parameter. Generically finite magnetic energy further requires $\lambda >1$.

The simplest finite-flux solution is $\lambda =2$, for which $M(1,1,s)=e^s$ andDirect differentiation gives $b^{\prime \prime }+q^{-1}{b}^{\prime }=(q^2-2)b$ and $q{b}^{\prime }=-q^{2}b$, verifying the radial equation. Its axial magnetic flux is $2\pi b_{0}f{g}^2=2\pi b_0{f}_0{g}_0^2$, while its magnetic energy grows by $g_0^2/g^2$ if the original width is larger than $\sqrt{\eta /\omega }$. For positive $\eta$ this Gaussian amplification saturates, rather than giving indefinite exponential growth. The other rapidly decaying regular axisymmetric modes have $\lambda =2(n+1)$ and profiles $e^{-q^2/2}{L}_n(q^2/2)$, where $L_n$ solves $y{L}_n^{\prime \prime }+(1-y)L_n^{\prime }+n{L}_n=0$.

The energy mechanism is transparent from a separate integral calculation. For any sufficiently decaying field of this axial form, integration by parts in the transverse plane givesThe positive term represents work by the axial stretching, with radial compression accounted for in the area integral. In the ideal limit, $g=g_0{e}^{-\omega t}$ and $f=f_0{e}^{2\omega t}$ for any smooth profile, so $E=E_0{e}^{2\omega t}$ even when $m\ne 0$ and the angular mean is zero. For sufficiently weak initial diffusion, this also gives genuine finite-time energy amplification at positive $\eta$.

Even sustained energy growth is possible in this unbounded model if slowly decaying profiles are admitted. For example, with $\omega >0$, take the constant width $g=\sqrt{\eta /\omega }$, $m=1$ and $1<\lambda <2$. The regular radial solution $b=qM((1+\lambda )/2,2,-q^2/2)$ has a $q^{-\lambda }$ tail and finite $I_b$. It gives $E\propto e^{2(2-\lambda )\omega t}$ and zero signed flux through every centered disk. Its absolute flux integral over the whole plane diverges, so the zero-flux statement here means the limit of these disk integrals. This example is not an isolated bounded-body dynamo: the velocity is unbounded at infinity and the field has an algebraic tail. The dynamo property concerns conversion of fluid energy into magnetic energy, not production of a nonzero signed magnetic flux; growth and its boundary assumptions must be specified.

Use suffix notation, with repeated indices summed. Contracting the two Levi-Civita symbols givesApplying the product rule to $h=A_i{B}_i$ and substituting the two evolution equations yieldsIn the second line we renamed dummy indices in $A_i{B}_j{\partial }_j{u}_i$ and then combined product derivatives. HenceNo zero-divergence assumption was used in this local identity.

When both fields are solenoidal vector fields, the product rule for divergence rewrites this as a conservation law:Because the volume is fixed, differentiation under its integral and the divergence theorem giveunder the two tangency conditions. Thus the volume integral is conserved. This is helicity conservation by tangent boundary conditions; when $\mathbf{B}=\nabla \times \mathbf{A}$, the density is the magnetic helicity density.

For the given axisymmetric field, the curl in spherical coordinates givesThere are no $\varphi$ derivatives and $A_r=0$. ConsequentlyThe volume form in spherical polar coordinates is $r^2\sin \theta drd\theta d\varphi$, soHere $B_r=0$ at $r=a$, as required for the magnetic-field tangency condition in the conservation argument; tangency of $\mathbf{u}$ is a separate hypothesis.

Under the first-order smoothing approximation, neglect the fluctuating product in the resistive induction equation while retaining its contribution to the mean electromotive force. Since the test field ${\mathbf{B}}_0$ is constant and the velocity is a solenoidal vector field, the fluctuation equation isLet $a=\eta k^2>0$, $d=a^2+{\Omega }^2$, and ${\mathbf{C}}_j=i({\mathbf{B}}_0\cdot {\mathbf{k}}_j){\hat{\mathbf{u}}}_j$. For each Fourier mode, the diffusion equation becomes a scalar linear relaxation equation applied to each vector component. Take its long-time periodic response, with $\Omega \ne 0$. For the first mode,while for the second,Thus the complex response coefficients areThey giveAn arbitrary initial fluctuation also contains a homogeneous diffusive transient. In the forced modes this is $\mathrm{Re}{\sum }_j{\mathbf{r}}_j{e}^{i{\mathbf{k}}_j\cdot \mathbf{x}-at}$; more general initial modes also diffuse. The displayed harmonic expression is the periodic particular solution after those transients, not the most general solution at finite time. The constants $a,d$ and the factors of $i$ are unchanged by taking the final real part.

To justify the helicity relation, write ${\hat{\mathbf{u}}}_j={\mathbf{a}}_j+i{\mathbf{c}}_j$ with real vectors. Solenoidality gives ${\mathbf{k}}_j\cdot {\mathbf{a}}_j={\mathbf{k}}_j\cdot {\mathbf{c}}_j=0$. Their cross product is parallel to ${\mathbf{k}}_j$, andThis is the helicity vector of a solenoidal Fourier mode. In these conventions the spatially averaged kinetic helicity density of $\mathrm{Re}({\hat{\mathbf{u}}}_j{e}^{i{\mathbf{k}}_j\cdot \mathbf{x}})$ is $-H_j{k}^2$; the sign is set by the specified cross-product order.

For equal wavevectors, spatial averaging of two real harmonic fields givesTime averaging gives another factor $1/2$, because $⟨{\cos }^2\Omega t{⟩}_t=⟨{\sin }^2\Omega t{⟩}_t=1/2$ and the mixed temporal average is zero. Consequently each diagonal mode contributes to the mean-field electromotive forceDistinct modes with ${\mathbf{k}}_1\ne \pm {\mathbf{k}}_2$ have no spatially averaged cross term. The printed condition also permits the opposite-wavevector case. The possible cross terms still cancel after time averaging: the out-of-phase response coefficients are ${\mathbf{p}}_2=-\Omega {\mathbf{C}}_2/d$ and ${\mathbf{q}}_1=\Omega {\mathbf{C}}_1/d$, while integration by parts gives $⟨{\mathbf{u}}_1\times ({\mathbf{B}}_0\cdot \nabla ){\mathbf{u}}_2{⟩}_x=⟨{\mathbf{u}}_2\times ({\mathbf{B}}_0\cdot \nabla ){\mathbf{u}}_1{⟩}_x$ for the real spatial fields. This handles all the distinct wavevectors allowed by the question, assuming the usual periodic-cell or whole-space spatial average.

Thus the alpha tensor isBoth averaging factors matter. The alpha tensor is real and symmetric, and this contribution vanishes when the mode kinetic helicity densities vanish.

For the two perpendicular wavevectors and common $H_1=H_2=H$, put ${\alpha }_0=\eta H{k}^4/[2({\Omega }^2+{\eta }^2{k}^4)]$. ThenMean-field dynamo action is possible when ${\alpha }_0\ne 0$ and sufficiently long mean-field wavelengths are allowed. The missing $zz$ entry does not prevent an alpha-squared dynamo. Indeed a slowly varying mean magnetic field transverse to $z$, with dependence $e^{iKz+\lambda t}$, obeysThe two growth rates are $-\eta K^2\pm {\alpha }_{0}K$, so the larger is positive if $0<|K|<|{\alpha }_0|/\eta$. This supplies a brief constructive reason; scale separation also requires $|K|\ll k$. A spatially uniform test field itself has zero curl of its mean-field electromotive force and does not grow. If $H=0$, or if boundary conditions exclude all unstable long wavelengths, this particular alpha effect does not yield growth. The conclusion concerns the stated first-order smoothing approximation; no uncomputed turbulent-diffusion correction is assumed.

Use the usual setting of a continuously differentiable solenoidal vector field defined on all radial segments from the origin, including the origin. More generally an open star-shaped set containing zero suffices. Differentiation under the integral sign givesApply divergence and curl of a cross product with the second field $x$, using $\nabla \cdot x=3$ and $(D\cdot \nabla )x=D$:The scaling identity in part (a) now converts the last expression to a one-variable derivative:Continuity at the origin ensures the lower endpoint is zero. Hence the radial vector potential of a solenoidal vector field isThe order of the cross product matters: $x\times D$ would produce $-B$. This construction is not valid on every punctured or multiply connected domain. For example $B=x/|x{|}^3$ is solenoidal away from zero, but its radial integral diverges at zero and it has nonzero flux through a surrounding sphere. The stated formula therefore implicitly needs the regular radial-domain hypothesis, rather than just local zero divergence.

Let $M$ be the central stellar mass and take positive Keplerian rotation, $\Omega (R)=\sqrt{GM/R^3}$. The initial poloidal magnetic field stays fixed by part (a), while $R{\Omega }^{\prime }(R)=-(3/2)\Omega (R)$. Integrating the toroidal induction equation with $B_{\varphi }(R,0)=0$ givesThe initial field is a solenoidal vector field on the disk region $R>0$, since ${\partial }_R(R{B}_R)=0$. Assuming $B_0\ne 0$, the amplitude ratio isIt reaches one atThus the equality time is the same fraction $1/(3\pi )\approx 0.106$ of the orbital period at every radius. If “orbital time” instead means ${\Omega }^{-1}$, the corresponding fraction is $2/3$. These are kinematic solutions with prescribed rotation and the neglected magnetic back-reaction.