Mean-field dynamo

A mean-field dynamo evolves a slowly varying spatially averaged magnetic field. Correlations of fluctuating velocity and magnetic field supply a mean-field electromotive force. Its local linear response includes the alpha effect; gradients of the mean field can contribute additional transport terms. The distinction between a uniform test field used to measure a response coefficient and a spatially varying mean field used to obtain growth is essential.

For $k(t)=k_0\sqrt{1+(t/T{)}^2}$, factor resistive damping out of the shearing-mode equations by $R=e^{-D}G$, $S=e^{-D}H$, $D=\eta \int k^{2}dt$. They imply $G^{\prime \prime }={\alpha }^2{k}^{2}G$, so a growing WKB approximation has $G\propto k^{-1/2}{e}^{|\alpha |\int kdt}$ and $H\simeq i\mathrm{sgn}(\alpha )kG$. Its magnetic energy is proportional to $k\exp [2|\alpha |\int kdt-2\eta \int k^{2}dt]$. At long times the positive quadratic exponent is overtaken by negative cubic resistive damping. A well-separated growing interval exists for sufficiently small positive magnetic diffusivity, with $\eta k_0/|\alpha |\ll 1$ at fixed nonzero $|\alpha |k_{0}T$, and the displayed leading peak time. This is transient amplification, not an asymptotically self-sustaining dynamo.

In a reflection-symmetric mean-field dynamo model, dipole parity can be represented by an even poloidal-potential amplitude and an odd toroidal-field amplitude; quadrupole parity reverses these assignments. The names describe the reflection classes appropriate to the model, not arbitrary electric multipoles. A complex amplitude $P=A+iB$ allows a phase rotation by $i$ to interchange the two classes when the unperturbed equations have that phase symmetry.

In a two-hemisphere mean-field dynamo, dipole and quadrupole subspaces can reduce to $\dot{A}=aB-A$, $\dot{B}=bA-B$, with $(a,b)=({\alpha }_0-{ϵ}_1,i\Omega -{ϵ}_2)$ and $({\alpha }_0+{ϵ}_1,i\Omega +{ϵ}_2)$ respectively. Their eigenvalues satisfy $(\lambda +1{)}^2=ab$. Strict decay is equivalent to $\mathrm{Re}(ab)+[\mathrm{Im}(ab){]}^2/4<1$. Thus dipole decay requires $F_d<1$, while quadrupole decay requires $F_q={\Omega }^2({\alpha }_0+{ϵ}_1{)}^2/4+{ϵ}_2({\alpha }_0+{ϵ}_1)<1$. Their difference is $F_q-F_d={\alpha }_0({\Omega }^2{ϵ}_1+2{ϵ}_2)$, which selects the first unstable parity when starting from a state stable in both subspaces.

An antisymmetric perturbation can split the thresholds of two degenerate mean-field dynamo parity classes. At first order, project the perturbation equation onto the adjoint null mode using the Fredholm alternative. If multiplying the unperturbed complex mode by $i$ leaves the parameter-derivative matrix element unchanged but reverses the perturbation matrix element, the two first-order threshold shifts are equal and opposite.

An alpha-Omega mean-field dynamo couples toroidal-field generation by the Omega effect to poloidal-field regeneration by the alpha effect. A simple Fourier model is $\dot{A}={\alpha }_{0}f(t)B-\eta k^{2}A$, $\dot{B}=ik\Omega A-\eta k^{2}B$, where $A$ is a component of the vector potential for the poloidal field and $B$ is a toroidal-field amplitude. For bounded $|f|\le 1$, the bounded-modulation alpha-Omega growth estimate controls the maximum amplification.

For a mean-field dynamo with uniform alpha, shear $\pi V/d$, diffusivity $\eta >0$ and fixed vertical wavenumber $\pi /d$, a horizontal Fourier mode obeys $(s+\eta (k^2+{\pi }^2/d^2){)}^2=i\alpha \pi Vk/d$. With $D=\alpha V{d}^2/({\pi }^2{\eta }^2)$ and $h=|k|d/\pi$, the growing branch has positive real part exactly when $|D|>2(1+h^2{)}^2/h$. Differentiation gives $3h^4+2h^2-1=0$, hence $h=1/\sqrt{3}$. Equality is a neutral oscillation, not exponential growth. The sign of $D$ reverses wave propagation without changing the growth threshold.

For constant coefficients and a transverse wavenumber $\ell \ne 0$, a planar mean-field dynamo retaining both alpha feedback and shear has dispersion $(\sigma +\eta p{)}^2={\alpha }^{2}p+i\alpha \Omega k$, $p=k^2+{\ell }^2$. Marginal stability gives ${\alpha }_m^2=4{\eta }^4{p}^4/(4{\eta }^2{p}^3+{\Omega }^2{k}^2)$. Differentiate with respect to $k^2$: an interior minimum solves $4{\eta }^2{p}^3+{\Omega }^2(3k^2-{\ell }^2)=0$. Its derivative with respect to $k^2$ is positive, so a unique positive minimizing $k^2$ exists only above the displayed shear threshold. Below it the minimum is at $k=0$. The sign of the marginal oscillation frequency is fixed by $2\eta p\omega =\alpha \Omega k$.

In the two-component alpha-Omega dynamo with $|f(t)|\le 1$, apply the weighted energy estimate for two coupled modes with $a=|{\alpha }_0|$, $c=|k\Omega |$, and $d=\eta k^2$. The field-amplitude growth rate is at most $\sqrt{|{\alpha }_0\Omega k|}-\eta k^2$. Optimizing over $k$ gives the upper constant $3/2^{8/3}$ multiplying $({\alpha }_0^2{\Omega }^2/\eta {)}^{1/3}$. The admissible choice $f=1$ gives a lower constant $3/2^{10/3}$, so the maximum possible growth has the displayed scaling. An arbitrary chosen modulation need not grow. The statement concerns the model with freely adjustable wavenumber; physical scale-separation restrictions can constrain that optimization.

Differential rotation or shear stretches a poloidal magnetic field into a toroidal magnetic field. This is the Omega effect. By itself it need not regenerate the poloidal component; the alpha effect provides one possible feedback mechanism in an alpha-Omega dynamo. Without feedback, shear amplification can be transient or algebraic rather than sustained exponential dynamo action.

First-order smoothing neglects the fluctuating nonlinear velocity-magnetic-field product in the fluctuation resistive induction equation, but retains $⟨\mathbf{u}\times \mathbf{b}⟩$ in the mean-field electromotive force. For a constant test field it gives ${\partial }_t\mathbf{b}-\eta {\nabla }^2\mathbf{b}=({\mathbf{B}}_0\cdot \nabla )\mathbf{u}$. Its validity requires the omitted terms to be small compared with the retained forcing and diffusion terms.

Three equal-amplitude transverse waves along orthogonal axes, with the same circular polarization, each satisfy $\nabla \times {\mathbf{u}}_j=\sigma k{\mathbf{u}}_j$, $\sigma =\pm 1$. Distinct spatial phases have zero cross averages. The periodic first-order smoothing response to a uniform test field satisfies $({\partial }_t-\eta \Delta ){\mathbf{b}}_j=(\overline{\mathbf{B}}\cdot \nabla ){\mathbf{u}}_j$. Writing $a=\eta k^2$, the response is ${\mathbf{b}}_j=k{\overline{B}}_j[a{\partial }_{\varphi }{\mathbf{u}}_j-\Omega {\mathbf{u}}_j]/(a^2+{\Omega }^2)$. Since ${\mathbf{u}}_j\times {\partial }_{\varphi }{\mathbf{u}}_j=-\sigma U^2{\mathbf{e}}_j$, summing yields the displayed isotropic alpha effect. It includes the resistive and temporal weighting absent from an unqualified instantaneous helicity rule.

For a locally uniform mean magnetic field, first-order smoothing solves ${\partial }_t\mathbf{b}-\eta \Delta \mathbf{b}=(\overline{\mathbf{B}}\cdot \nabla )\mathbf{u}$. Inserting its causal diffusion propagator into the mean-field electromotive force and taking the trace of the isotropic response tensor gives the displayed relation to kinetic helicity. When diffusion during a correlation time is negligible, it reduces to $\alpha =-{\tau }_H⟨\mathbf{u}\cdot \nabla \times \mathbf{u}⟩/3$. At small magnetic Reynolds number, a quasistatic monochromatic velocity instead gives $\alpha =-⟨\mathbf{u}\cdot \nabla \times \mathbf{u}⟩/(3\eta k_0^2)$. The effective correlation time and diffusion weighting should not be silently discarded.

For background shear $\Omega y\hat{\mathbf{x}}$ and uniform field $B_0\hat{\mathbf{x}}$, first-order smoothing gives the magnetic shearing-wave amplitude equationIf magnetic relaxation is fast compared with shear, $c_y,c_z\simeq i{k}_x{B}_0(v_y,v_z)/(\eta K^2)$ after free transients decay. Thus the doubled Fourier-mode electromotive quantity $\mathcal{E}=\mathrm{Re}(v_y{c}_z^{\ast }-v_z{c}_y^{\ast })$ is $-B_{0}H/(\eta K^2)$, where $H=i{k}_x(v_y{v}_z^{\ast }-v_z{v}_y^{\ast })$. The spatially averaged mean-field electromotive force of real waves has an extra factor $1/2$. Using the helicity invariant of an inviscid shearing wave, and assuming the approximation holds uniformly in time with $\Omega k_x\ne 0$, givesIndeed $K^2=A+{\Omega }^2{k}_x^2{t}^2$, $A=k_x^2+k_z^2$, and substitution $z=|\Omega k_x|t/\sqrt{A}$ uses ${\int }_{-\infty }^{\infty }(1+z^2{)}^{-2}dz=\pi /2$.

For a steady forced Fourier mode with $|\mathbf{q}|=k$, coupled viscous and magnetic response gives $\hat{\mathbf{u}}=[\eta k^2\hat{\mathbf{f}}+i(\mathbf{B}\cdot \mathbf{q})\hat{\mathbf{g}}/({\mu }_0\rho )]/D$ and $\hat{\mathbf{b}}=[\nu k^2\hat{\mathbf{g}}+i(\mathbf{B}\cdot \mathbf{q})\hat{\mathbf{f}}]/D$. Their mean-field electromotive force contains both forcing helicities and a possible weighted cross-forcing correlation. Jointly isotropic forcing removes the latter; its vanishing at zero mean field alone is insufficient for anisotropic forcing.

For a velocity Fourier mode with wavenumber $k$, define $a=\eta k^2$, $d=a^2+{\Omega }^2$ and $\mathbf{C}=i({\mathbf{B}}_0\cdot \mathbf{k})\hat{\mathbf{u}}$. A cosine-forced mode has periodic response coefficients $(\mathbf{p},\mathbf{q})=(a\mathbf{C},\Omega \mathbf{C})/d$, while a sine-forced mode has $(-\Omega \mathbf{C},a\mathbf{C})/d$. Homogeneous transients decay as $e^{-at}$. The periodic response is a particular solution, not arbitrary initial data.

A Parker dynamo wave is a traveling or oscillatory mode of a mean-field dynamo coupling poloidal and toroidal magnetic-field components through an alpha effect and shear. A reduced complex-amplitude model can represent fluctuating alpha by a time-dependent coupling. Its periodically switched growth must be calculated from the monodromy matrix of a periodic linear system, rather than by averaging noncommuting generators.

Take a right-handed local Cartesian frame with $x$ towards the equator, $y$ azimuthal and $z$ radially outwards in the northern hemisphere. A local radial shear $U_y=Sz$ and a poloidal magnetic field represented by $A\hat{\mathbf{y}}$ give the alpha-Omega dynamo equations $A_t=\alpha B+{\eta }_T{A}_{xx}$ and $B_t=S{A}_x+{\eta }_T{B}_{xx}$. Substitution of a Fourier mode proportional to $e^{pt+ikx}$ gives the displayed dispersion relation. The growing branch has $\mathrm{Re}p=-{\eta }_T{k}^2+\sqrt{|\alpha Sk|/2}$ and $\mathrm{Im}p=\mathrm{sgn}(\alpha Sk)\sqrt{|\alpha Sk|/2}$. Its phase velocity is $-\mathrm{Im}p/k$: it propagates equatorwards when $\alpha S<0$, polewards when $\alpha S>0$. The result assumes local constant coefficients and no meridional circulation in a star; advection and spatially separated field regeneration can alter migration.

For $\dot{A}=D^{2}f(t)B$ and $\dot{B}=iA$, with $f$ alternating between $+1$ and $-1$ for intervals of length $T$, the monodromy matrix of a periodic linear system has determinant one and trace $\cosh (\sqrt{2}DT)+\cos (\sqrt{2}DT)>2$. Its reciprocal Floquet multipliers give positive dominant Floquet growth rate for $D,T>0$, despite the zero mean of $f$. An exceptional initial state in the stable eigenspace decays.

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 ${\alpha }_0\mathrm{diag}(1,1,0)$.

The mean-field electromotive force is the vector correlation $\mathcal{E}=⟨\mathbf{u}\times \mathbf{b}⟩$. Its curl enters the mean resistive induction equation. It is a local vector with units of velocity times magnetic field, not the scalar circuit voltage called electromotive force. A space-time average of periodic modes is often used to measure its transport coefficients.

Suppose a steady solenoidal velocity satisfies $\Delta \mathbf{u}=-\mathbf{u}$, and spatial averaging permits integration by parts without boundary terms. For a uniform test magnetic field $\mathbf{B}$, the small-magnetic Reynolds number expansion gives ${\mathbf{b}}_1=(\mathbf{B}\cdot \nabla )\mathbf{u}$ and $\Delta {\mathbf{b}}_2=-\nabla \times (\mathbf{u}\times {\mathbf{b}}_1)$. Using the self-adjointness of the Laplacian inside the average,The second equality concerns the averaged mean-field electromotive force; it does not assert that the quadratic forcing itself is monochromatic.

// Target: fluid-mechanics.bigb

For $\mathbf{C}=\mathbf{u}\times (\mathbf{B}\cdot \nabla )\mathbf{u}$, the cross product identity $(\mathbf{u}\times \nabla \times \mathbf{C}{)}_i=u_j{\partial }_i{C}_j-u_j{\partial }_j{C}_i$ and integration by parts give ${\mathcal{E}}_i^{(2)}=-⟨{\partial }_i{u}_j{C}_j⟩$. Its alpha tensor isInterchanging $i,j$ reverses the cross product, showing that the coefficient is an antisymmetric matrix. Such a coefficient represents a turbulent magnetic pumping contribution rather than the symmetric part of the alpha effect.

// Target: fluid-mechanics.bigb

The first coefficient of the alpha tensor is ${\alpha }_{ij}^{(1)}={ϵ}_{ikl}⟨u_k{\partial }_j{u}_l⟩$. Its contraction with the Levi-Civita symbol isby incompressibility and vanishing averages of total derivatives. Every antisymmetric second-rank tensor in three dimensions is detected by this contraction, so the coefficient is a symmetric matrix. The proof needs no Fourier analysis.

// Target: fluid-mechanics.bigb

Turbulent magnetic pumping is an effective transport of the mean magnetic field produced by correlations in inhomogeneous or anisotropic turbulence. Its contribution to the mean-field electromotive force is ${\mathcal{E}}_{\mathrm{pump}}=\mathbf{\gamma }\times \overline{\mathbf{B}}$. In the mean induction equation this enters as $\nabla \times (\mathbf{\gamma }\times \overline{\mathbf{B}})$, the same algebraic form as advection by a velocity $\mathbf{\gamma }$. Thus it supplements the mean material velocity without requiring an equal mean mass flow. Downward pumping can help retain or couple fields in a solar dynamo.

For the ideal magnetic response to a cylindrical cyclonic event, the plane-integrated $x$-component of $\mathbf{u}\times \mathbf{B}$ at duration $t$ isThis follows from $(\mathbf{u}\times \mathbf{B}{)}_x=(rf{B}_z+g{A}_r)\cos \varphi -(g/r)A_{\varphi }\sin \varphi$ and azimuthal integration. If boundary terms vanish, integration by parts gives the equivalent form $\pi B_0{\int }_0^{\infty }{r}^2{g}^{\prime }[\sin (ft)+tf\cos (ft)]dr$. For $f=g=a^2-r^2$ on $r<a$, zero outside, the response isThe value at zero is its continuous limit zero. It is initially $-\pi B_0{a}^{6}t/3$, but changes sign repeatedly at the nonzero roots of $\tan x=x$, with oscillatory envelope of order $t^{-1}$. The flow's kinetic helicity density is $2f^2\ge 0$, showing that the sign of a finite-duration ideal response is not fixed by the instantaneous helicity sign.

A space-time average over periodic modes integrates over the spatial periodic cell and one temporal period. For real spatial Fourier modes sharing a wavevector, $⟨\mathrm{Re}(\mathbf{v}{e}^{i\mathbf{k}\cdot \mathbf{x}})\times \mathrm{Re}(\mathbf{w}{e}^{i\mathbf{k}\cdot \mathbf{x}}){⟩}_x=\mathrm{Re}({\mathbf{v}}^{\ast }\times \mathbf{w})/2$. A matching sine or cosine temporal factor supplies another $1/2$. Opposite wavevectors can have nonzero spatial cross averages, so their temporal coefficients must be checked rather than discarded merely because the wavevectors are distinct.