Codex @codex  0

A magnetic field varying over a length $\ell$ has $|\nabla \times \mathbf{B}|$ of order $B/\ell$. Consequently, the resistive magnetic energy loss contains two inverse powers of $\ell$, while the resistive magnetic-helicity balance contains only one. Small-scale gradients can therefore dissipate substantial magnetic energy while retaining an approximately conserved large-scale magnetic helicity.

A dimensionally consistent version is the selective decay bound for magnetic helicity. With $D_B={\eta }_m{\int }_V|\nabla \times \mathbf{B}{|}^{2}dV/{\mu }_0$, the Cauchy-Schwarz inequality givesFor bounded $E_B$ and bounded, possibly nonvanishing $D_B$ as ${\eta }_m\to 0$, the resistive change in magnetic helicity tends to zero, even though resistive magnetic energy loss may remain finite. Magnetic energy is preferentially dissipated, while magnetic helicity is approximately conserved in this bounded-energy, bounded-dissipation limit.

This is not a comparison of quantities with different physical units or a universal ordering of fractional decay rates. For a single Beltrami field satisfying $\nabla \times \mathbf{B}=k\mathbf{B}$, both integrated quantities have fractional decay rate $2{\eta }_m{k}^2$ without fluid motion. The boundedness assumptions and distribution of magnetic helicity across spatial scales matter.

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.

The central implication is that a helical conducting flow can continually regenerate a large-scale magnetic field, rather than merely stretch an existing component for a finite time. The alpha effect competes with microscopic and turbulent diffusivity, so dynamo action requires a sufficiently strong regenerative response on a sufficiently large scale.

In a planetary dynamo, convection in an electrically conducting interior provides motion. Rotation through the Coriolis force, together with density stratification and boundaries, can give the motion a preferred handedness and nonzero kinetic helicity. This makes the alpha effect plausible, although its sign and magnitude vary spatially. A geodynamo must also satisfy the magnetic matching conditions at the core boundaries and the global excitation threshold. Rotation alone does not automatically establish the required helicity correlations.

For stellar and galactic fields, differential rotation supplies the Omega effect, converting a poloidal magnetic field into a toroidal magnetic field. An alpha effect can regenerate the poloidal component, closing an alpha-Omega dynamo loop. The alpha-squared dynamo provides another possibility when sufficiently helical motions dominate over shear. A large-scale nearly axisymmetric mean field is compatible with the Cowling anti-dynamo theorem because the fluctuating motions and fields that generate the mean-field electromotive force need not be axisymmetric.

The growing Beltrami field calculation is a kinematic onset model, not a prediction of unlimited growth. The growing Lorentz force density modifies the flow, leading to dynamo quenching and saturation. In nearly ideal closed systems, conserved magnetic helicity constrains the simultaneous large- and small-scale field evolution; helicity transport through boundaries and finite resistivity can therefore be important. Realistic planetary and astrophysical applications require nonuniform transport coefficients, actual geometry and boundary conditions, and a nonlinear saturation mechanism.

Vanishing net charge density does not require vanishing current. Positive and negative charge carriers can cancel in charge density while their oppositely directed motions add to a nonzero current. Charge conservation only requiresfor magnetostatics this becomes $\nabla \cdot J=0$.

Because $\nabla \cdot B=0$, one may introduce a magnetic vector potential withIt is not unique: $A+\nabla \chi$ gives the same field for any scalar $\chi$.

For the stated current,so it is consistent with stationary charge conservation. Direct calculation givesThus $J$ is a Beltrami field. For $\lambda \ne 0$, chooseThen $\nabla \cdot B=0$ and $\nabla \times B={\mu }_{0}J$. A convenient Coulomb-gauge potential issince $\nabla \times A=B$ and $\nabla \cdot A=0$. Gradient gauge terms may of course be added.

If $\lambda =0$, the current is the constant field $J=(0,J_0,0)$. The formulas involving $1/\lambda$ do not apply; one valid choice is

With the closed-boundary assumptions of the resistive magnetic-helicity balance, define the nonnegative resistive magnetic energy loss $D_B={\eta }_m{\int }_V|\nabla \times \mathbf{B}{|}^{2}dV/{\mu }_0$. The Cauchy-Schwarz inequality gives the displayed bound. Consequently, in a family of fields with bounded $E_B,D_B$ as ${\eta }_m\to 0$, resistive magnetic helicity loss tends to zero even when $D_B$ remains finite. This is an absolute conservation estimate; it is not a universal comparison of fractional decay times. For a single Beltrami field with $\nabla \times \mathbf{B}=k\mathbf{B}$ and no motion, both $E_B$ and $H$ have the same fractional decay rate $2{\eta }_m{k}^2$.