Past exam of the mathematics course of the University of Cambridge / 2001 / iii / Paper 36 / 2 / c / Solution

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.