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For constant alpha tensor and positive magnetic diffusivity $\eta$, the steady anisotropic alpha-squared dynamo equation isFor a nonzero Fourier mode with wavevector $\mathbf{K}=(k,l,m)$, this becomes $\eta K^2\mathbf{B}=i\mathbf{K}\times (\mathbf{\alpha }\mathbf{B})$. Rotational symmetry of the alpha tensor in the horizontal plane lets us set $\mathbf{K}=(q,0,m)$, with $q^2=k^2+l^2$. The component equations areSubstitute the first and third into the second. A nonzero steady amplitude requires ${\eta }^2{K}^4={\alpha }_0^2{m}^2+{\alpha }_0{\alpha }_1{q}^2$. Conversely, this relation supplies a nonzero amplitude through the same component equations, and the solenoidal magnetic-field constraint $q{B}_x+m{B}_z=0$ is automatically satisfied. Thus the steady-mode condition isFor the first form, no division by $m^2+\delta q^2$ is needed. The quotient applies when that denominator is nonzero, in particular for $q>0$ and $0<\delta <1$.

Put $Q=q^2>0$ and $r=m^2\ge 0$. The derivative of $F(r)=(Q+r{)}^2/(r+\delta Q)$ isFor $0<\delta <1/2$, the derivative changes from negative to positive at $r=(1-2\delta )Q$. For $1/2\le \delta <1$ it is nonnegative throughout the allowed half-line and the minimum is at $r=0$. Therefore the optimized uniaxial alpha dynamo threshold isBoth expressions agree at $\delta =1/2$. If the horizontal boundary conditions permit $q=0$, a nonzero mode instead has ${\gamma }^2=m^2$ and the infimum is zero as $m\to 0$; it is not attained by a nonzero wavevector. The exactly uniform mode $q=m=0$ has no diffusive or alpha curl term and must be treated separately.

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.

Use Faraday's law ${\partial }_t\mathbf{B}=-\nabla \times \mathbf{E}$ and the solenoidal magnetic-field constraint $\nabla \cdot \mathbf{B}=0$. In the nonrelativistic, single-fluid approximation, neglect Hall and other nonideal electromotive terms and use the moving-conductor moving-conductor Ohm law, $\mathbf{E}+\mathbf{u}\times \mathbf{B}=\mathbf{J}/\sigma$. Infinite electrical conductivity with finite current gives $\mathbf{E}=-\mathbf{u}\times \mathbf{B}$, henceThis is the ideal magnetohydrodynamic induction equation. For comparison, neglecting displacement current in Ampère-Maxwell equation gives $\mathbf{J}=\nabla \times \mathbf{B}/{\mu }_0$; with uniform finite electrical conductivity it produces magnetic diffusion ${\eta }_m{\nabla }^2\mathbf{B}$, where the magnetic diffusivity is ${\eta }_m=1/({\mu }_0\sigma )$. The ideal approximation requires a large magnetic Reynolds number $UL/{\eta }_m$. Dropping displacement current is useful for this finite-conductivity comparison, but Faraday's law and the ideal Ohm relation already suffice for the ideal induction equation.

Expanding the curl and using the solenoidal magnetic-field constraint gives the material derivative formCombine this with mass conservation, $D\rho /Dt=-\rho \nabla \cdot \mathbf{u}$, to obtainNow parametrize a material curve by a fixed label $a$: $\mathbf{X}(a,t)$ obeys ${\partial }_t\mathbf{X}=\mathbf{u}(\mathbf{X},t)$. Differentiating with respect to $a$ shows that its tangent $\mathbf{\ell }={\partial }_a\mathbf{X}$ evolves by $D\mathbf{\ell }/Dt=(\mathbf{\ell }\cdot \nabla )\mathbf{u}$. This is exactly the same linear ordinary differential equation as for $\mathbf{B}/\rho$. Initially parallel tangents remain parallel by uniqueness, with a label-dependent proportionality factor constant along each particle trajectory. Thus magnetic field lines are transported as material curves, wherever the field and fluid flow are smooth and the field is nonzero. This is the field-line form of magnetic flux freezing.

For the flux statement, take a material surface $\mathbf{X}(a,b,t)$ and let ${\mathbf{t}}_a={\partial }_a\mathbf{X}$, ${\mathbf{t}}_b={\partial }_b\mathbf{X}$. Both tangents obey $D{t}_i/Dt=({\partial }_j{u}_i)t_j$. Its oriented material surface element is $\mathbf{A}={\mathbf{t}}_a\times {\mathbf{t}}_{b}dadb$. Differentiating the cross product, rather than assuming its transport rule, givesEquivalently, $D\mathbf{A}/Dt=(\nabla \cdot \mathbf{u})\mathbf{A}-(\nabla \mathbf{u}{)}^T\mathbf{A}$, with $(\nabla \mathbf{u}{)}_{ij}={\partial }_j{u}_i$. Contracting this derived rule with the induction equation gives a pointwise cancellation:Integrating over the fixed material labels therefore proves conservation of flux through an open material surface:The surface need not be closed; its boundary is carried with the fluid. The result follows from material transport, rather than from the zero flux through a closed surface.

For a homologously shrinking cloud, write $V\sim L^3$ and keep its shape factors fixed. Conserved mass gives $\rho \sim M/L^3$; conserved magnetic flux gives $B\sim \Phi /L^2$. Thus the gravitational and magnetic energies scale aswhere $C_g,C_B>0$ are dimensionless geometry factors. Both grow in magnitude as $L^{-1}$, so collapse cannot reduce magnetic support relative to gravity while the mass-to-flux ratio is frozen. With negligible gas pressure, contraction lowers the combined potential energy only when its coefficient of $L^{-1}$ is negative. Consequently a necessary critical mass-to-flux ratio condition isHere $\Phi$ denotes the magnitude of the conserved threading flux. The numerical coefficient depends on geometry and boundary conditions; the scaling argument does not determine it or make the condition sufficient in the presence of other support.

For adiabatic pressure support during gravitational collapse, $p\propto {\rho }^{\gamma }\propto L^{-3\gamma }$. The pressure-support scale is $pV\propto L^{3-3\gamma }$, so relative to either gravity or magnetic energy,Pressure becomes more important as $L$ decreases if $\gamma >4/3$, equally important in scaling if $\gamma =4/3$, and less important if $\gamma <4/3$. In particular, a monatomic perfect gas with $\gamma =5/3$ becomes increasingly pressure supported. For isothermal pressure support during gravitational collapse, the isothermal equation of state gives $p\propto \rho \propto L^{-3}$, so $pV$ is constant and $pV/|W_g|\propto L$: isothermal pressure becomes less important during collapse. The same comparisons hold against magnetic support because its energy has the same $L^{-1}$ scaling as gravity.

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.

Differential circulation of a weak magnetic field winds its flux function into progressively smaller radial scales. In a smooth eddy with angular-velocity gradient of order $U/L^2$, its radial wavenumber grows as $k_r\sim Ut/L^2$. Resistive damping becomes substantial when $\eta {\int }_0^t{k}_r^{2}dt\sim 1$, giving $t\sim (L^4/(\eta U^2){)}^{1/3}=(L/U){\mathrm{Rm}}^{1/3}$. This estimate presumes differential rotation and negligible magnetic feedback; a rigidly rotating eddy does not create this phase mixing. Finite magnetic diffusivity then permits weak interior field and boundary concentration, unlike exact ideal flux freezing.

For solenoidal velocity and magnetic field, with constant magnetic diffusivity $\eta$, the resistive induction equation impliesThe product rule gives $D_{t}P=\mathbf{x}\cdot D_t\mathbf{B}+\mathbf{u}\cdot \mathbf{B}$ and $\Delta P=\mathbf{x}\cdot \Delta \mathbf{B}+2\nabla \cdot \mathbf{B}$. Consequently the radial velocity, through $Q$, is the source term for the radial magnetic field.

// Target: fluid-mechanics.bigb

Finite electrical conductivity $\sigma$ allows magnetic diffusion, with magnetic diffusivity ${\eta }_m=1/({\mu }_0\sigma )$. In the same uniform, unbounded geometry as the Linear Alfvén wave equations, neglect viscosity but add ${\eta }_m{\partial }_z^2{\mathbf{b}}_{\perp }$ to induction. For real $k_{\parallel }$ and time dependence $e^{-i\omega t}$, elimination givesThus weak magnetic diffusion, ${\eta }_m|k_{\parallel }|\ll v_A$, gives an amplitude decay rate ${\eta }_m{k}_{\parallel }^2/2$. For ${\eta }_m|k_{\parallel }|>2v_A$, both temporal roots are purely decaying; they no longer describe an oscillating traveling mode. This temporal calculation at fixed real wavenumber is different from a driven vessel at fixed real frequency, where a complex spatial wavenumber describes penetration and attenuation. Viscosity also dissipates kinetic energy, and walls introduce transverse gradients and additional losses.

For an axial length $L$, comparing the magnetic diffusion time $L^2/{\eta }_m$ with the transit time $L/v_A$ gives the ratio $L{v}_A/{\eta }_m$. A large value favors propagation before diffusion, but a narrow vessel can introduce a shorter transverse diffusion scale. In the actual 1949 analysis, finite electrical conductivity and vessel dimensions restrict the useful frequency interval; above its upper limit the response becomes a skin wave.^[ref]

For constant magnetic diffusivity, apply the product rule to $M_{ij}=B_i{B}_j$ and the incompressible resistive induction equation. The identity ${\nabla }^2(B_i{B}_j)=B_i{\nabla }^2{B}_j+B_j{\nabla }^2{B}_i+2{\partial }_n{B}_i{\partial }_n{B}_j$ proves the displayed formula. Omitting the gradient-product term gives the wrong tensor evolution. Its average generally requires information beyond the averaged magnetic second-moment tensor.

For a solenoidal vector field $\mathbf{B}$ and constant magnetic diffusivity $\eta$, the magnetic induction equation isIt combines Faraday's law with the moving-conductor relation for electrical conductivity, neglecting displacement current. With incompressible flow, its advective form is ${\partial }_t\mathbf{B}+(\mathbf{u}\cdot \nabla )\mathbf{B}=(\mathbf{B}\cdot \nabla )\mathbf{u}+\eta {\nabla }^2\mathbf{B}$.

For a magnetic vector potential $\mathbf{A}$, a solenoidal magnetic field $\mathbf{B}=\nabla \times \mathbf{A}$ and constant magnetic diffusivity ${\eta }_m$, write ${\partial }_t\mathbf{A}=\mathbf{u}\times \mathbf{B}-{\eta }_m\nabla \times \mathbf{B}-\nabla \varphi$. In a fixed closed volume with zero helicity surface flux, integration by parts makes the two differentiated factors in $H={\int }_V\mathbf{A}\cdot \mathbf{B}dV$ equal; the scalar-potential boundary term vanishes when $\mathbf{B}\cdot \mathbf{n}=0$. One sufficient boundary gauge is $\mathbf{n}\times \mathbf{A}=0$; another standard setting is integration over all space with sufficient decay. Orthogonality of $\mathbf{u}\times \mathbf{B}$ to $\mathbf{B}$ gives the displayed identity. Unlike the resistive magnetic energy loss, $\dot{H}$ is not sign definite. The zero normal magnetic field also makes $H$ gauge invariance under single-valued smooth gauge transformations.

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.

For a planar magnetic field written as $\mathbf{B}=\nabla \times (A\hat{\mathbf{z}})$ and a circular velocity $\mathbf{u}=s\Omega (s)\hat{\mathbf{\varphi }}$, the resistive induction equation reduces, after choosing the additive gauge transformation, to ${\partial }_{t}A+\Omega {\partial }_{\varphi }A=\eta {\nabla }^{2}A$. A Fourier mode $A=a(s,t)e^{im[\varphi -\Omega (s)t]}$ therefore obeys the equation in the title. Indeed, each radial derivative acting on the exponential replaces ${\partial }_s$ by ${\partial }_s-imt{\Omega }^{\prime }$, while the azimuthal Laplacian contributes $-m^2/s^2$. At zero magnetic diffusivity, $a$ is constant in time, but the physical magnetic field can still be amplified algebraically by differential rotation.

For fixed real positive angular frequency $\omega$, a transverse Alfvén wave proportional to $e^{i(kz-\omega t)}$ in a uniform fluid has $k^2={\omega }^2/(v_A^2-i{\eta }_m\omega )$ when magnetic diffusivity ${\eta }_m$ is retained and viscosity is neglected. Choose the branch with positive real and imaginary parts for propagation and attenuation along $+z$. For $ϵ\ll 1$, $k=(\omega /v_A)(1+iϵ/2+O({ϵ}^2))$. The amplitude attenuation length is $L_A=1/\mathrm{Im}k$, whereas the energy attenuation length is $L_E=1/(2\mathrm{Im}k)$. With wavelength $\lambda =2\pi /\mathrm{Re}k$, $L_A/\lambda \sim 1/(\pi ϵ)$ and $L_E/\lambda \sim 1/(2\pi ϵ)$. Fixed-real-wavenumber temporal Resistive damping of an Alfvén wave is a different boundary-value prescription.

With thermal diffusivity $\kappa$ and magnetic diffusivity $\eta$, the standard Chandrasekhar number uses $\nu \eta$ in its denominator. A thermal-diffusion alternative uses $\nu \kappa$ and is $\zeta =\eta /\kappa$ times the standard parameter. In thermal-time magnetoconvection equations, the linear magnetic-force coefficient is $\sigma Q_{\kappa }=\sigma \zeta Q_{\mathrm{Ch}}$. Specifying the convention prevents a missing diffusivity factor in a dispersion relation.

For a bounded conducting sphere with uniform magnetic diffusivity and an insulating exterior, an incompressible flow tangent to concentric spheres cannot sustain dynamo action. Neither the velocity nor the magnetic field needs to be axisymmetric. The radial magnetic scalar has a homogeneous advection-diffusion equation, so the poloidal field decays. Once it is absent, the toroidal magnetic potential also has a homogeneous scalar advection-diffusion equation. This is a theorem about spherical flow geometry, not about a general cylindrical flow with an axial component.

Turbulent magnetic diffusivity is an effective transport coefficient produced by correlated small-scale velocity and magnetic field. It supplements microscopic magnetic diffusivity in a mean-field dynamo description. Its identification requires a specified mean-field geometry and averaging procedure; in a planar flux problem, the correlation $\overline{w{a}^{\prime }}=-{\eta }_t{B}_{0}f(z)$ describes transport relative to a prescribed uniform field.