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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.