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Dynamo action is growth or sustained maintenance of a magnetic field by conducting-fluid motion despite magnetic diffusion. In the kinematic problem the velocity is prescribed and the resistive induction equation is linear in the magnetic field. Magnetic feedback through the Lorentz force density matters once the field becomes dynamically important.

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.

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.

Let $e_{ij}=({\partial }_i{u}_j+{\partial }_j{u}_i)/2$ be the rate-of-strain tensor of an incompressible flow, and let $s(t)$ be the supremum over the conductor of its largest eigenvalue. State Backus' necessary condition for dynamo action with its magnetic boundary conditions: an isolated bounded conductor of uniform positive magnetic diffusivity $\eta$, surrounded by an electrical insulator with a decaying potential exterior field, and no imposed magnetic field or boundary energy input. For definiteness take a no-slip boundary condition on the fluid, which eliminates the stretching surface term. If the conductor lies within a sphere of radius $R$ and $S={\sup }_{t}s(t)$, a necessary condition for a nondecaying dynamo isThe constant is the free-decay spectral bound for an insulating exterior, not a universal constant for every magnetic boundary condition. The condition is necessary, not sufficient, and involves maximum stretching rather than an rms velocity.

To see both the condition and the growth-rate bound, include exterior magnetic energy:This follows from the resistive induction equation and integration by parts, with the stated boundary assumptions. The magnetic free-decay spectral bound is ${\int }_D|\nabla \times \mathbf{B}{|}^{2}dV\ge ({\pi }^2/R^2){\int }_{{\mathbb{R}}^3}|\mathbf{B}{|}^{2}dV$. For a sphere its lowest mode is the dipolar poloidal free-decay mode; enclosing a smaller conductor gives the same valid lower bound. Since $B_i{e}_{ij}{B}_j\le s(t)|\mathbf{B}{|}^2$, we obtainIntegrating this differential inequality gives decay whenever $S<\eta {\pi }^2/R^2$. More generally the exponential rate of the field norm, rather than of its squared energy, satisfiesSimply discarding the nonnegative resistive dissipation already proves the requested maximum-strain bound. The energy exponent is twice the field-amplitude exponent.

For the alpha-Omega dynamo model, write $a=|{\alpha }_0|$, $w=|\Omega |$, $k\ge 0$ and $d=\eta k^2$. Direct differentiation givesFor $a,kw>0$, use the weighted energy estimate for two coupled modes and form the positive weighted norm $N=kw|A{|}^2+a|B{|}^2$. The inequality $N\ge 2\sqrt{akw}|A||B|$ givesFor each fixed $k$ and model parameters, this norm is equivalent to the amplitude norm; its square-root exponential rate is therefore bounded by $h(k)=\sqrt{awk}-\eta k^2$, uniformly over all admissible $f(t)$. Maximizing over $k$ givesThe exponent is also achievable in order of magnitude. Choose the admissible constant $f=1$. The growing eigenvalue of the two-component system has real part $-\eta k^2+\sqrt{awk/2}$. Its maximum occurs at $k=(aw/(32{\eta }^2){)}^{1/3}$ and equals $32^{-10/3}(a^2{w}^2/\eta {)}^{1/3}$. Thus the bounded-modulation alpha-Omega growth estimate has the scalingThis means the maximum over allowed modulations and wavenumbers, not that every bounded modulation grows; $f=0$ supplies no regenerating alpha coupling.

The Omega effect rapidly makes toroidal field from poloidal field, but exponential dynamo action also requires the slower alpha effect to regenerate the poloidal component. The coupled amplification rate is of order $\sqrt{a\Omega k}$ rather than $\Omega$; shortening the wavelength to increase it also increases magnetic diffusion as $\eta k^2$. Their optimal balance gives $k\sim {\Omega }^{1/3}$ and growth $\sim {\Omega }^{2/3}$. The Backus' necessary condition for dynamo action estimate controls stretching alone and does not incorporate this regeneration bottleneck. A shear without regeneration can give transient amplification but not this sustained exponential feedback.

Integrating the resistive induction equation once introduces a constant $C$:Because $b(L)=b(-L)=0$, integrating this relation again over the channel gives $C=B_{0}Q/(2L)$, where $Q$ is the volumetric flow rate per unit span. Put $m^2=B_0^2/({\mu }_0\rho \nu \eta )$ and $H=mL$, taking $B_0\ge 0$ without loss of generality. Reversing the imposed field reverses $b$ but leaves $u$ and $Q$ unchanged. Eliminating $b^{\prime }$ from the magnetohydrodynamic momentum equation givesIts two zero wall values remove the odd homogeneous solution. Write $u=U[1-\cosh (mz)/\cosh H]$. Integrating the first relation with the two magnetic wall values yieldsTherefore $U=(G{L}^2/\nu H^2)H\coth H$. Substitution and integration produce the velocity and induced field:Here $H=B_{0}L/\sqrt{{\mu }_0\rho \nu \eta }$ is the Hartmann number. The hyperbolic cosine makes $u$ even, while the hyperbolic sine makes $b$ odd. Direct differentiation verifies both coupled equations and all four wall values.

Integrating the velocity gives the fluxThe apparent singularity at $H=0$ is removable. The small-field limit is plane Poiseuille flow:The cubic coefficient of $b$ is its first-order response in $B_0$; $b$ itself vanishes at zero imposed field. For a fixed positive $G$, the flux decreases monotonically as $H>0$ increases. One exact way to see the sketch's monotonicity is the partial-fraction expansion of the hyperbolic cotangent:whose derivative is strictly negative for $H>0$. The flux has a horizontal tangent at $H=0$ and approaches zero asThe flux is even if signed $H$ is used.

For $H\gg 1$, put $\zeta =z/L$. Away from the walls,The velocity has a nearly uniform core, with thin Hartmann layers of thickness $\delta =L/H$ enforcing the no-slip boundary condition. Near the upper wall, with $s=H(1-\zeta )$ fixed, $u/U_c\sim 1-e^{-s}$ and $b/B_s\sim -1+s/H+e^{-s}$; the lower wall follows by even/odd symmetry. Thus $b$ is negative for $0<z<L$, positive for $-L<z<0$, and returns rapidly to zero at both walls. The required sketches show a decreasing flux, a flat velocity core, and an odd induced field with magnetic wall layers.

Use the fully developed flow branch of Hartmann flow. Translation invariance along the walls makes the velocity and induced magnetic field functions of $z$ alone. Incompressible flow and zero normal velocity at the walls give $u_z^{\prime }=0$ and hence $u_z=0$. The zero divergence of the magnetic field makes $B_z$ constant, equal to the imposed $B_0$. There is no forcing in $y$; the homogeneous $y$-components obey the same coupled viscous-resistive equations as the $x$-components, with zero boundary data. Multiplying them by $u_y$ and $b_y/({\mu }_0\rho )$, integrating by parts and adding givesThus $u_y=b_y=0$, establishing the asserted forms on this fully developed flow branch. This is a symmetry reduction of the steady channel model, rather than a claim that every possible flow in a channel is translation invariant.

The current density and Lorentz force density areThe advective acceleration vanishes because $\mathbf{u}\cdot \nabla =u(z){\partial }_x$ acts on fields independent of $x$. The $x$-component of the magnetohydrodynamic momentum equation therefore givesThe $z$-component must also balance: it requires ${\partial }_z(p+b^2/(2{\mu }_0))=0$. Hence a compatible pressure is $p(x,z)=p_{\ast }-\rho Gx-b(z{)}^2/(2{\mu }_0)$; the specified streamwise pressure gradient does not require the ordinary pressure to be uniform in $z$. This accounts for the magnetic pressure of the induced field.

The resistive induction equation with constant magnetic diffusivity gives ${\partial }_t\mathbf{B}=\nabla \times (\mathbf{u}\times \mathbf{B})+\eta {\nabla }^2\mathbf{B}$. Since $\mathbf{u}\times \mathbf{B}=(0,-B_{0}u,0)$, its curl has $x$-component $B_0{u}^{\prime }$. Steadiness consequently givesFinally the no-slip boundary condition gives $u(\pm L)=0$, and the normal magnetic field boundary condition gives $b(\pm L)=0$. The coupled equations and all wall conditions follow directly from momentum balance and magnetic induction.

Under the first-order smoothing approximation, neglect the fluctuating product in the resistive induction equation while retaining its contribution to the mean electromotive force. Since the test field ${\mathbf{B}}_0$ is constant and the velocity is a solenoidal vector field, the fluctuation equation isLet $a=\eta k^2>0$, $d=a^2+{\Omega }^2$, and ${\mathbf{C}}_j=i({\mathbf{B}}_0\cdot {\mathbf{k}}_j){\hat{\mathbf{u}}}_j$. For each Fourier mode, the diffusion equation becomes a scalar linear relaxation equation applied to each vector component. Take its long-time periodic response, with $\Omega \ne 0$. For the first mode,while for the second,Thus the complex response coefficients areThey giveAn arbitrary initial fluctuation also contains a homogeneous diffusive transient. In the forced modes this is $\mathrm{Re}{\sum }_j{\mathbf{r}}_j{e}^{i{\mathbf{k}}_j\cdot \mathbf{x}-at}$; more general initial modes also diffuse. The displayed harmonic expression is the periodic particular solution after those transients, not the most general solution at finite time. The constants $a,d$ and the factors of $i$ are unchanged by taking the final real part.

To justify the helicity relation, write ${\hat{\mathbf{u}}}_j={\mathbf{a}}_j+i{\mathbf{c}}_j$ with real vectors. Solenoidality gives ${\mathbf{k}}_j\cdot {\mathbf{a}}_j={\mathbf{k}}_j\cdot {\mathbf{c}}_j=0$. Their cross product is parallel to ${\mathbf{k}}_j$, andThis is the helicity vector of a solenoidal Fourier mode. In these conventions the spatially averaged kinetic helicity density of $\mathrm{Re}({\hat{\mathbf{u}}}_j{e}^{i{\mathbf{k}}_j\cdot \mathbf{x}})$ is $-H_j{k}^2$; the sign is set by the specified cross-product order.

For equal wavevectors, spatial averaging of two real harmonic fields givesTime averaging gives another factor $1/2$, because $⟨{\cos }^2\Omega t{⟩}_t=⟨{\sin }^2\Omega t{⟩}_t=1/2$ and the mixed temporal average is zero. Consequently each diagonal mode contributes to the mean-field electromotive forceDistinct modes with ${\mathbf{k}}_1\ne \pm {\mathbf{k}}_2$ have no spatially averaged cross term. The printed condition also permits the opposite-wavevector case. The possible cross terms still cancel after time averaging: the out-of-phase response coefficients are ${\mathbf{p}}_2=-\Omega {\mathbf{C}}_2/d$ and ${\mathbf{q}}_1=\Omega {\mathbf{C}}_1/d$, while integration by parts gives $⟨{\mathbf{u}}_1\times ({\mathbf{B}}_0\cdot \nabla ){\mathbf{u}}_2{⟩}_x=⟨{\mathbf{u}}_2\times ({\mathbf{B}}_0\cdot \nabla ){\mathbf{u}}_1{⟩}_x$ for the real spatial fields. This handles all the distinct wavevectors allowed by the question, assuming the usual periodic-cell or whole-space spatial average.

Thus the alpha tensor isBoth averaging factors matter. The alpha tensor is real and symmetric, and this contribution vanishes when the mode kinetic helicity densities vanish.

For the two perpendicular wavevectors and common $H_1=H_2=H$, put ${\alpha }_0=\eta H{k}^4/[2({\Omega }^2+{\eta }^2{k}^4)]$. ThenMean-field dynamo action is possible when ${\alpha }_0\ne 0$ and sufficiently long mean-field wavelengths are allowed. The missing $zz$ entry does not prevent an alpha-squared dynamo. Indeed a slowly varying mean magnetic field transverse to $z$, with dependence $e^{iKz+\lambda t}$, obeysThe two growth rates are $-\eta K^2\pm {\alpha }_{0}K$, so the larger is positive if $0<|K|<|{\alpha }_0|/\eta$. This supplies a brief constructive reason; scale separation also requires $|K|\ll k$. A spatially uniform test field itself has zero curl of its mean-field electromotive force and does not grow. If $H=0$, or if boundary conditions exclude all unstable long wavelengths, this particular alpha effect does not yield growth. The conclusion concerns the stated first-order smoothing approximation; no uncomputed turbulent-diffusion correction is assumed.

Resistive magnetohydrodynamics retains finite magnetic diffusivity in the coupled evolution of a conducting fluid and its magnetic field. The resistive induction equation allows field diffusion relative to the fluid, while the magnetohydrodynamic momentum equation includes the Lorentz force density. It contrasts with the zero-diffusivity limit of ideal magnetohydrodynamics.