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Mean-field electrodynamics applies averaging and a response closure to magnetic induction in a fluctuating conducting fluid. The mean equation contains the correlated mean-field electromotive force in the title. Scale separation permits a local expansion into alpha transport, turbulent diffusion and higher gradients; general anisotropic or finite-memory statistics require tensorial and nonlocal responses. A mean-field dynamo is a self-exciting application of this framework, not a consequence of averaging alone.

With no mean flow and constant isotropic coefficients, the mean-field dynamo equation isIf the symbol $\beta$ already denotes the total diffusivity in the chosen mean-field convention, replace ${\eta }_T$ by $\beta$. Let $\nabla \times {\mathbf{B}}_0=\kappa {\mathbf{B}}_0$. This Beltrami field is solenoidal for $\kappa \ne 0$, and taking its curl again gives ${\nabla }^2{\mathbf{B}}_0=-{\kappa }^2{\mathbf{B}}_0$. ThusFor example, ${\mathbf{B}}_0=(\cos kz,-\sin kz,0)$ has $\kappa =k$, whereas reversing the sine sign gives $\kappa =-k$. Choose the helicity sign to agree with $\alpha$. The homogeneous alpha-squared dynamo growth criterion is thenOn an unbounded or suitably periodic domain with freely selectable wavenumber, this gives exponentially growing modes for every $\alpha \ne 0$. Maximizing their growth rate gives ${\kappa }_{\ast }=\alpha /(2{\eta }_T)$ and $s_{\max }={\alpha }^2/(4{\eta }_T)$. In a specified bounded domain, the permitted modes and boundary conditions impose a threshold; the optimal wavelength must also remain large compared with the turbulent scale for mean-field electrodynamics to apply.

A nonzero alpha effect is necessary in this constant-coefficient model. If $\alpha =0$, the mean equation is pure diffusion and the claimed growing modes do not exist. Positivity of $\beta$ alone does not imply growth.

In mean-field electrodynamics, use Reynolds averaging to write ${\mathbf{U}}_{\mathrm{tot}}=\overline{\mathbf{U}}+\mathbf{u}$ and ${\mathbf{B}}_{\mathrm{tot}}=\overline{\mathbf{B}}+\mathbf{b}$, with zero-mean fluctuations. Assume the averaging operation commutes with the derivatives and that resolved quantities vary little across an averaging region. Averaging the induction equation then givesThis is the mean-field electromotive force, a vector correlation with units of velocity times magnetic field. The closure problem is to express this correlation in terms of the resolved field.

In the kinematic regime the flow statistics are prescribed, so the response is linear in the mean magnetic field, after independently excited fluctuation transients have decayed. If the mean-field scale $L$ is much greater than the turbulent scale $\ell$, and its evolution is slow compared with the response memory, expand that response in the local mean magnetic field and its gradients:Without scale separation one generally has a spatially nonlocal and temporally retarded response rather than constant local coefficients.

For homogeneous turbulence the coefficients are independent of position. Invariance under proper rotations makes the rank-two response proportional to ${\delta }_{ij}$, and the isotropic rank-three gradient response proportional to the alternating tensor. Choosing the diffusivity sign convention givesThe alpha effect can regenerate a mean magnetic field, while $\beta$ supplies turbulent magnetic diffusivity. For constant positive $\beta$, $\nabla \times (-\beta \nabla \times \overline{\mathbf{B}})=\beta {\nabla }^2\overline{\mathbf{B}}$ because the magnetic field is solenoidal. Thus the total diffusivity is $\eta +\beta$.

The electromotive vector is a polar vector, whereas the magnetic field is an axial vector. Therefore $\alpha$ is a pseudoscalar and changes sign under reflection; $\beta$ is an ordinary scalar. A nonzero isotropic alpha effect requires broken reflection symmetry, such as nonzero kinetic helicity. If “isotropic” includes reflection symmetry as well as rotations, $\alpha$ must vanish. The dimensions are $[\alpha ]=L/T$ and $[\beta ]=L^2/T$.