Alpha-squared dynamo (${\alpha }^2$ dynamo)

An alpha-squared dynamo generates a mean magnetic field using an alpha effect without requiring an additional large-scale shear coupling. For $\mathbf{\alpha }={\alpha }_0\mathrm{diag}(1,1,0)$ and a transverse mean field varying as $e^{iKz}$, the growth rates are $-\eta K^2\pm {\alpha }_{0}K$. A nonzero ${\alpha }_0$ therefore allows growth for sufficiently long wavelengths, provided the domain and scale separation permit them.

An alpha-squared dynamo may use a tensorial alpha effect rather than an isotropic coefficient. For constant alpha tensor $\mathbf{\alpha }$ and magnetic diffusivity $\eta >0$, a steady nonzero Fourier mode with wavevector $\mathbf{K}$ satisfies $\eta K^2\mathbf{B}=i\mathbf{K}\times (\mathbf{\alpha }\mathbf{B})$ and $\mathbf{K}\cdot \mathbf{B}=0$. Anisotropy can change both the critical alpha magnitude and the preferred wavevector direction.

For $\mathbf{\alpha }=\mathrm{diag}({\alpha }_0,{\alpha }_0,{\alpha }_1)$, set $\gamma ={\alpha }_0/\eta$, $\delta ={\alpha }_1/{\alpha }_0$, and $q^2=k^2+l^2$. A steady nonzero Fourier mode obeys the displayed condition when the denominator is nonzero. At fixed $q>0$ and $0<\delta <1$, minimizing over $m^2\ge 0$ gives $m_{\ast }^2=(1-2\delta )q^2$ and ${\gamma }_{\min }^2=4(1-\delta )q^2$ for $\delta <1/2$; for $\delta \ge 1/2$ it gives $m_{\ast }^2=0$ and ${\gamma }_{\min }^2=q^2/\delta$. This is a boundary-constrained threshold of an anisotropic alpha-squared dynamo.