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There is a minor operator misprint to resolve first. Interpreting $(\mathbf{x}\cdot \nabla {)}^2$ as composition, the Cartesian formula for the angular Laplacian isThe last term is missing from the printed Cartesian expression. For example, its truncated expression maps $x_1$ to $x_1$, whereas a degree-one spherical harmonic must have angular eigenvalue two. In what follows, use the stated angular spectrum $l(l+1)$ and the supplied poloidal-energy identity; these are consistent with the corrected operator.

The divergence theorem and the solenoidal magnetic-field constraint imply zero magnetic flux through every sphere. Consequently $P=r{B}_r$ has zero angular mean and no $l=0$ component. Expand in orthonormal spherical harmonics:By angular orthogonality, withthe gradient norm and the supplied poloidal-toroidal decomposition identity becomeSince $l(l+1)\ge 2$, this proves the poloidal magnetic energy bound by the radial scalar gradient:Equality holds when the radial scalar contains only degree-one angular components.

Define the poloidal and total magnetic energies by $E_P=M_P/(2{\mu }_0)$ and $E_{\mathrm{total}}=M/(2{\mu }_0)$. Combining the radial-flow dynamo energy bound with $D\ge 2M_P$ gives the necessary conditionIt has the same temporal interpretation as the bound in the preceding part; statistically steady versions use the corresponding time-averaged energies. It quantifies why dynamo action with a substantial poloidal fraction requires radial motion, while a purely tangential flow has $Q=0$.

Put $a_h^2=k^2+m^2>0$ and $\Lambda =a_h^2+{\pi }^2$. The plane-layer poloidal-toroidal decomposition supplies the assumed poloidal fields asSince $w=-{\Delta }_h\varphi$ and $b_z=-{\Delta }_h\chi$, potentials proportional to $e^{st+ikx+imy}\sin \pi z$ have the required vertical components. Their horizontal components are proportional to $\cos \pi z$, satisfying the stress-free boundary conditions and the horizontal-field boundary conditions. The double-curl representation automatically enforces incompressibility and the solenoidal magnetic-field constraint.

Let $W,\Theta ,C$ be the amplitudes of $w,\theta ,b_z$. The temperature and resistive induction equations immediately giveEliminate the total-pressure perturbation by applying $\Delta$ to the vertical momentum equation and subtracting ${\partial }_z$ times its divergence. The resulting vertical equation isIn particular, $imC=-m^{2}W/(s+\zeta \Lambda )$ shows explicitly that magnetic tension opposes the displacement.

Taking the determinant of these three amplitude equations gives the horizontal-field magnetoconvection dispersion relation:This cubic polynomial includes all roots of the assumed poloidal coupled system; using its determinant form avoids discarding possible roots by dividing by $s+\Lambda$ or $s+\zeta \Lambda$. As a check, $Q{m}^2=0$ factors off magnetic diffusion and leaves the standard stress-free Rayleigh-Bénard convection growth equation.

For an isolated solenoidal magnetic field, zero flux through every sphere removes the degree-zero spherical harmonic of $P=\mathbf{B}\cdot \mathbf{x}$. In a poloidal-toroidal decomposition, let ${\mathcal{L}}^2=-{\Delta }_{S^2}$. The spherical poloidal component obeysIf $P={\sum }_{l\ge 1,m}{p}_{lm}(r)Y_{lm}$, put $D_{lm}={\int }_0^{\infty }[r^2|p_{lm}^{\prime }{|}^2+l(l+1)|p_{lm}{|}^2]dr$. The two sides of the bound are respectively $\sum D_{lm}/[l(l+1)]$ and $\frac{1}{2}\sum D_{lm}$. Since $l(l+1)\ge 2$, the bound follows, with equality for degree-one radial scalars. Together with the radial-flow dynamo energy bound it gives $(q/\eta {)}^2\ge 2E_P/E_{\mathrm{total}}$ for a marginal or growing dynamo field.

// Target: fluid-mechanics.bigb

A spherical poloidal-toroidal decomposition represents a solenoidal vector field with zero flux through each sphere by scalar potentials $S,T$. The second term is tangent to concentric spheres; the first carries the radial component. Requiring zero angular means for the potentials fixes their otherwise irrelevant radial additions. In a plane layer, the analogous fluctuating components are $\nabla \times \nabla \times (\varphi \hat{\mathbf{z}})$ and $\nabla \times (\psi \hat{\mathbf{z}})$, with any horizontally uniform components handled separately. These constructions enforce the divergence-free condition because the divergence of a curl vanishes.

// Target: fluid-mechanics.bigb