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The alpha effect is the part of a mean-field electromotive force linear in the mean magnetic field itself, $\mathcal{E}=\mathbf{\alpha }\overline{\mathbf{B}}+\cdots$. The alpha tensor can be anisotropic; it need not be a scalar multiple of the identity. Its curl can couple transverse mean-field components and produce an alpha-squared dynamo.

An alpha-squared dynamo may use a tensorial alpha effect rather than an isotropic coefficient. For constant alpha tensor $\mathbf{\alpha }$ and magnetic diffusivity $\eta >0$, a steady nonzero Fourier mode with wavevector $\mathbf{K}$ satisfies $\eta K^2\mathbf{B}=i\mathbf{K}\times (\mathbf{\alpha }\mathbf{B})$ and $\mathbf{K}\cdot \mathbf{B}=0$. Anisotropy can change both the critical alpha magnitude and the preferred wavevector direction.

For $\mathbf{C}=\mathbf{u}\times (\mathbf{B}\cdot \nabla )\mathbf{u}$, the cross product identity $(\mathbf{u}\times \nabla \times \mathbf{C}{)}_i=u_j{\partial }_i{C}_j-u_j{\partial }_j{C}_i$ and integration by parts give ${\mathcal{E}}_i^{(2)}=-⟨{\partial }_i{u}_j{C}_j⟩$. Its alpha tensor isInterchanging $i,j$ reverses the cross product, showing that the coefficient is an antisymmetric matrix. Such a coefficient represents a turbulent magnetic pumping contribution rather than the symmetric part of the alpha effect.

// Target: fluid-mechanics.bigb

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.

Write $m_0={\mu }_0\rho$ and treat the imposed mean magnetic field as uniform on the forcing scale. In first-order smoothing, retain terms linear in the fluctuating fields in their equations, while keeping their quadratic correlation when calculating the mean-field electromotive force. The inertial fluctuation, nonlinear induction term and quadratic fluctuating Lorentz force density are neglected in the steady response. The linear magnetic force isAbsorb the gradient into the total pressure. The resulting equations arewith $\nabla \cdot \mathbf{u}=\nabla \cdot \mathbf{b}=0$. Taking the divergence of the first equation gives ${\nabla }^{2}P=0$ for solenoidal forcing. In a periodic or homogeneous Fourier calculation its nonconstant part therefore vanishes.

For each Fourier mode with wavevector $\mathbf{q}$, $|\mathbf{q}|=k$, define $\ell =\mathbf{B}\cdot \mathbf{q}$ andSolving the two coupled linear equations gives the monochromatic coupled magnetic and velocity responseThe magnetic coupling suppresses modes with large projection of their wavevector along the imposed field.

To calculate the mean-field electromotive force, group each pair of opposite wavevectors into a real mode and average its spatial phase. Put $\mathbf{n}=\mathbf{q}/k$, and define its forcing correlationsFor any solenoidal real mode $\mathbf{v}=\mathbf{a}\cos (\mathbf{q}\cdot \mathbf{x})+\mathbf{c}\sin (\mathbf{q}\cdot \mathbf{x})$, its two amplitude vectors are perpendicular to $\mathbf{q}$. Direct multiplication givesAlso, integration by parts gives $⟨(L\mathbf{g})\times (L\mathbf{f})⟩=-{\ell }^2⟨\mathbf{f}\times \mathbf{g}⟩$, where $L=\mathbf{B}\cdot \nabla$. Expanding the response products therefore yieldsSumming over independent real modes, or integrating over their angular spectrum, is the required general answer for $⟨\mathbf{u}\times \mathbf{b}⟩$.

At zero field, $\mathcal{E}(0)={\sum }_{\mathbf{q}}{\mathbf{C}}_{\mathbf{q}}/(\nu \eta k^4)$. Thus the stated zero-field condition removes the unweighted sum of the cross correlations. For anisotropic correlated forcing it does not necessarily remove their differently weighted sum at nonzero field. The displayed general formula retains that allowed contribution. It disappears if cross forcing is absent mode by mode, or under the jointly isotropic statistics assumed for the final part.

The helicity content can also be expressed directly using the actual fluctuating fields. Define their kinetic helicity $H_u(\mathbf{q})=⟨{\mathbf{u}}_{\mathbf{q}}\cdot \nabla \times {\mathbf{u}}_{\mathbf{q}}⟩$ and current helicity $H_b(\mathbf{q})=⟨{\mathbf{b}}_{\mathbf{q}}\cdot \nabla \times {\mathbf{b}}_{\mathbf{q}}⟩$. Substitute $\mathbf{f}=\nu k^2\mathbf{u}-L\mathbf{b}/m_0$ and $\mathbf{g}=\eta k^2\mathbf{b}-L\mathbf{u}$ into their cross correlation. This gives the exact identityIt shows the competing kinetic and magnetic contributions without introducing a spurious singularity when the factor on the left vanishes. For a monochromatic field, ${\mathbf{a}}_b=\nabla \times \mathbf{b}/k^2$ is a magnetic vector potential, so $H_b=k^2⟨{\mathbf{a}}_b\cdot \mathbf{b}⟩$: current helicity is $k^2$ times magnetic helicity. In the isotropic weak-field limit, with zero-field fluctuations ${\mathbf{u}}_0,{\mathbf{b}}_0$, the scalar alpha effect isThis identifies both signs and their different viscous and magnetic relaxation times.

For jointly isotropic forcing let $h_f,h_g$ now denote the total forcing helicities, and set $S_h=\nu h_g/m_0-\eta h_f$. The cross-forcing contribution is even in $\mathbf{B}$, whereas isotropy requires any vector depending only on $\mathbf{B}$ to be parallel to it and odd under reversal. Hence that contribution vanishes. A natural symmetric alpha tensor isOnly its longitudinal coefficient contributes to $\mathcal{E}={\alpha }_{\parallel }\mathbf{B}$. Introduce $h=|B|/[k\sqrt{m_0\nu \eta }]$ and $A_0=\nu \eta k^4$. Angular integration givesTherefore the strong-field quenching of an isotropic electromotive force isprovided $S_h\ne 0$. If the two helicity contributions cancel exactly, this electromotive force vanishes.

There is a tensor qualification to the printed scaling claim. The natural transverse coefficient instead contains $\frac{1}{2}{\int }_0^1(1-{\mu }^2)(1+h^2{\mu }^2{)}^{-2}d\mu \sim \pi /(8h)$ and scales as $|B{|}^{-1}$. The relation ${\mathcal{E}}_i={\alpha }_{ij}{B}_j$ by itself leaves transverse components undetermined. One may choose the effective representative ${\alpha }_{ij}={\alpha }_{\parallel }{\delta }_{ij}$, for which the requested $|B{|}^{-3}$ scaling holds. It is the observable longitudinal response that is fixed, rather than every component of a unique field-dependent tensor.

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

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

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.

For isotropic monochromatic forcing, the longitudinal coefficient in a mean-field electromotive force is proportional to ${\int }_0^1{\mu }^2(1+h^2{\mu }^2{)}^{-2}d\mu =[\arctan h-h/(1+h^2)]/(2h^3)$, where $h$ is proportional to mean-field strength. This gives generic $|B{|}^{-3}$ quenching. The natural angular alpha tensor is anisotropic in a nonzero mean field: its transverse coefficient instead contains ${\int }_0^1(1-{\mu }^2)(1+h^2{\mu }^2{)}^{-2}d\mu$ and scales as $|B{|}^{-1}$. Only the longitudinal contraction enters $\mathcal{E}=\mathbf{\alpha }\mathbf{B}$. An effective scalar representation may therefore have every diagonal entry equal to the longitudinal coefficient without being the unique natural response tensor.

The first coefficient of the alpha tensor is ${\alpha }_{ij}^{(1)}={ϵ}_{ikl}⟨u_k{\partial }_j{u}_l⟩$. Its contraction with the Levi-Civita symbol isby incompressibility and vanishing averages of total derivatives. Every antisymmetric second-rank tensor in three dimensions is detected by this contraction, so the coefficient is a symmetric matrix. The proof needs no Fourier analysis.

// Target: fluid-mechanics.bigb