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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.

For constant isotropic $\alpha ,\beta$ and zero mean flow, the mean induction equation isLet $\overline{\mathbf{B}}=\hat{\mathbf{B}}{e}^{\sigma t+i\mathbf{k}\cdot \mathbf{x}}$, with $\mathbf{k}\cdot \hat{\mathbf{B}}=0$. On this two-dimensional transverse space, $i\mathbf{k}\times$ has eigenvalues $\pm k$. For example, if $\mathbf{k}=k{\mathbf{e}}_z$, the vectors ${\mathbf{e}}_x\pm i{\mathbf{e}}_y$ have these curl eigenvalues. The homogeneous alpha-squared dynamo growth criterion follows fromChoose the circular polarization of the mode to match the sign of $\alpha$. This is an alpha-squared dynamo: the alpha effect supplies both parts of the regeneration loop without differential rotation.

If arbitrarily long wavelengths are permitted, every nonzero alpha has a growing band $0<k<|\alpha |/{\eta }_T$. Maximizing the quadratic givesprovided $k_{\ast }\ell \ll 1$ so the mean-field approximation is valid. A uniform field has $k=0$, zero curl and zero growth; a constant alpha cannot amplify that uniform mode. In a periodic domain of size $L$, the lowest nonzero wavenumber is $2\pi /L$, giving $|\alpha |L/{\eta }_T>2\pi$. Other boundaries and geometries give different critical constants, but the general scale criterion is $|\alpha |L/{\eta }_T$ sufficiently large. Nonzero kinetic helicity alone is therefore not a sufficient finite-domain dynamo criterion: its induced alpha must overcome diffusion on an available mean-field scale.