Alpha effect ($\mathcal{E}=\mathbf{\alpha }\overline{\mathbf{B}}$)

The alpha effect is the part of a mean-field electromotive force linear in the mean magnetic field itself, $\mathcal{E}=\mathbf{\alpha }\overline{\mathbf{B}}+\cdots$. The alpha tensor can be anisotropic; it need not be a scalar multiple of the identity. Its curl can couple transverse mean-field components and produce an alpha-squared dynamo.

In a Parker dynamo wave model, an alpha effect of the form $\alpha ={ϵ}^{-1}{\alpha }_0\sin (\omega t/ϵ)$ has zero time average but can generate a finite averaged coupling. The method of multiple scales gives $\tilde{A}=-({\alpha }_0/\omega )\cos (\omega \tau )\overline{B}$ and $\tilde{B}=-(\Omega {\alpha }_0/{\omega }^2)\sin (\omega \tau ){\partial }_x\overline{B}$, where $\tau =t/ϵ$. Consequently $C=\Omega {\alpha }_0^2/(2{\omega }^2)$. With diffusion operator ${\partial }_x^2-{\ell }^2$, the averaged Fourier mode growth exponents are $-k^2-{\ell }^2\pm |k|\sqrt{\Omega C}$. For freely adjustable real $k$, growth requires $\Omega C>4{\ell }^2$. The result is an asymptotic averaged model for fixed nonzero $\omega$, not an exact replacement at finite $ϵ$.

For isotropic monochromatic forcing, the longitudinal coefficient in a mean-field electromotive force is proportional to ${\int }_0^1{\mu }^2(1+h^2{\mu }^2{)}^{-2}d\mu =[\arctan h-h/(1+h^2)]/(2h^3)$, where $h$ is proportional to mean-field strength. This gives generic $|B{|}^{-3}$ quenching. The natural angular alpha tensor is anisotropic in a nonzero mean field: its transverse coefficient instead contains ${\int }_0^1(1-{\mu }^2)(1+h^2{\mu }^2{)}^{-2}d\mu$ and scales as $|B{|}^{-1}$. Only the longitudinal contraction enters $\mathcal{E}=\mathbf{\alpha }\mathbf{B}$. An effective scalar representation may therefore have every diagonal entry equal to the longitudinal coefficient without being the unique natural response tensor.

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.

For constant isotropic transport coefficients and zero mean flow, a divergence-free Fourier mode has two circular polarizations that diagonalize curl with eigenvalues $\pm k$. The mean induction equation therefore has the displayed growth rates. Growth requires an admissible $k$ with $|\alpha |>(\eta +\beta )k$. Freely variable wavenumber gives $k_{\ast }=|\alpha |/[2(\eta +\beta )]$, provided it respects mean-field scale separation. A uniform field has zero curl and cannot grow by constant alpha alone; boundaries fix the longest available spatial mode.

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.

The alpha tensor is the linear map from a uniform test magnetic field to the corresponding mean-field electromotive force. Its components satisfy ${\mathcal{E}}_i={\alpha }_{ij}{B}_{0j}$. Anisotropic helical flows can produce a real symmetric rank-two response, such as ${\alpha }_0\mathrm{diag}(1,1,0)$.