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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.