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For the charge-generated electric field, use the electric potentialThe vector potential is time independent, so its usual additional contribution $-{\partial }_t\mathbf{A}$ is zero. The continuity equation now givesThus the second term in part (iii) is exactly the displacement current contribution ${\mu }_0{\epsilon }_0{\partial }_t\mathbf{E}$, andThis is the Ampère-Maxwell equation. The constructed field also has zero curl, consistent with the time-independent magnetic field and Faraday's law. A static addition to the electric field does not change this conclusion; the integral construction selects the fields sourced by the given densities rather than an independently added radiation field.

Write $\Omega =e^{2\rho }$, $F={\partial }_x{A}_2-{\partial }_y{A}_1$, and $s=|\Phi {|}^2$, so the physical magnetic field is $B=F/\Omega$. Use finite-energy Abelian Higgs vortex boundary data: the field tends to unit magnitude at the ideal boundary, its phase winds $N$ times, and the boundary limits in the decomposition exist. This vacuum asymptotic behavior is the setting in which the stated degree gives an energy bound. We retain the current boundary term; degree alone must not be silently equated with magnetic flux on an open disk.

Define $j_i=\mathrm{Im}(\bar{\Phi }{D}_i\Phi )$. The gauge-covariant derivative of a charged scalar field satisfies $[D_x,D_y]\Phi =-iF\Phi$. Differentiating the current givesOn the other hand,Combining the two identities and completing the square in the magnetic and potential terms yields the conformal-surface vortex square completionAll integrations on the noncompact metric disk can first be made over $|z|<R<1$ and then passed to the limit. By the Stokes theorem, the two boundary contributions combine to $\frac{1}{2}\oint (A+j)$. At the vacuum boundary write $\Phi =e^{i\chi }$. Then $j=d\chi -A$, so this combined integral is $\frac{1}{2}\oint d\chi =\pi N$, even if the current term does not vanish separately. The usual additional condition of tangential covariant decay makes that current term vanish and yields magnetic flux quantization of an Abelian Higgs vortex, but is not needed if the combined boundary term has this phase limit. Thus for $N\ge 0$, $V\ge \pi N$. Reversing both signs in the square completion gives $V\ge -\pi N$ for $N\le 0$, henceFor positive degree, equality is attained precisely when the Bogomolny vortex equations areDistinguishing the physical field $B$ from the coordinate curvature coefficient $F$ prevents an erroneous conformal factor.

For the proposed field, set $q=x^2+y^2$ and choose the smooth connection one-formIndeed, with ${\partial }_{\bar{z}}=\frac{1}{2}({\partial }_x+i{\partial }_y)$ and $A_{\bar{z}}=\frac{1}{2}(A_1+i{A}_2)$,The second Bogomolny vortex equation follows fromThus both squares vanish, including at the simple zero $z=0$ by smoothness. At $r\to 1$, $\Phi \to e^{i\vartheta }$ and $A=2r^{2}d\vartheta /(1+r^2)\to d\vartheta$, proving degree one. Moreover, $j_{\vartheta }=4r^2(1-r^2)/(1+r^2{)}^3\to 0$, so the current boundary term really vanishes for this solution. The magnetic flux is explicitlyConsequentlyThis is the hyperbolic one-vortex at curvature minus one half; its finite energy follows directly from the vanishing squares and the finite flux.

A Poisson manifold is a smooth manifold with a smooth antisymmetric bivector $\pi =\frac{1}{2}{\pi }^{ij}{\partial }_i\wedge {\partial }_j$ whose function bracket $\{f,g\}={\pi }^{ij}({\partial }_{i}f)({\partial }_{j}g)$ satisfies the Jacobi identity. The bracket is automatically a derivation in each argument. A symplectic manifold has a nondegenerate differential two-form $\omega$ satisfying $d\omega =0$, that is, a closed differential form; nondegeneracy forces its dimension to be even. It determines a nondegenerate Poisson structure. We use matrix convention $\pi =-{\omega }^{-1}$, so that ${\iota }_{X_H}\omega =dH$ corresponds to $X_H(f)=\{f,H\}$.

Examples of genuinely nonsymplectic Poisson geometries include the zero bracket on ${\mathbb{R}}^3$ and the Lie-Poisson bracket $\{x_i,x_j\}={ϵ}_{ijk}{x}_k$ on ${\mathbb{R}}^3$. Their three-dimensional underlying manifold cannot carry a nondegenerate two-form. In the latter case the Poisson rank is two away from the origin and zero at the origin, with spheres and the origin as symplectic leaves. Even in even dimension a given Poisson structure need not come from a symplectic form: $\pi =x{\partial }_x\wedge {\partial }_y$ on ${\mathbb{R}}^2$ satisfies Jacobi but vanishes along $x=0$. The statement concerns this degenerate structure, not a claim that ${\mathbb{R}}^2$ admits no other symplectic structure.

For any candidate bivector, applying the Jacobi identity to coordinate functions gives the necessary conditionIt is sufficient as well: expanding $\{f,\{g,h\}\}$ and its cyclic companions cancels all terms with second derivatives by antisymmetry of $\pi$. What remains is $J^{ijk}({\partial }_{i}f)({\partial }_{j}g)({\partial }_{k}h)$. This is the coordinate Jacobi condition for a Poisson bivector.

For a symplectic form, differentiate $\omega \pi =-I$ to obtain ${\partial }_{\ell }\pi =\pi ({\partial }_{\ell }\omega )\pi$. Substituting this into the cyclic expression givesbecause $d\omega =0$. Thus the inverse symplectic structure always satisfies the Jacobi identity. Conversely, for a nondegenerate Poisson structure, multiplying by three inverse matrices proves closure of its associated two-form.

In momentum space, choose Hamiltonian $H(\mathbf{p})=E(\mathbf{p})$ and the constant Poisson bivectorIts Poisson bracket is $\{f,g\}=e\mathbf{B}\cdot (\nabla f\times \nabla g)$. The Hamiltonian vector field therefore givesexactly the prescribed motion. The sign here uses the signed charge $e$ as printed; an electron's physical charge may be negative. Since the magnetic field is uniform, every derivative of ${\pi }^{ij}$ vanishes, so every term in $J^{ijk}$ is zero. Jacobi holds independently of the dispersion function $E$. This is the momentum-space Poisson structure for a uniform magnetic field.

For $e\mathbf{B}\ne 0$ this bivector has rank two and $C(\mathbf{p})=\mathbf{B}\cdot \mathbf{p}$ is a Casimir function of a Poisson manifold: $\{C,f\}=0$ for every $f$. Its symplectic leaves are the planes $C=c$. For axes with $\mathbf{B}=B_0{\mathbf{e}}_3$, the leaf form is $\omega =(e{B}_0{)}^{-1}d{p}_1\wedge d{p}_2$ in our inverse-sign convention. Also $\dot{E}=\{E,E\}=0$, so trajectories on a Fermi surface lie on its intersections with these planes, typically curves. The energy surface is invariant for this Hamiltonian, but need not be a Poisson submanifold; the bracket has been defined on the whole momentum space. If $e\mathbf{B}=0$, the Poisson structure and motion vanish.

For an axisymmetric magnetic field, the solenoidal condition becomes ${\partial }_R(R{B}_R)+{\partial }_z(R{B}_z)=0$. Locally introduce a poloidal magnetic flux function $\Psi$ byThe divergence condition is exactly the integrability condition for these two derivatives. The additive constant in $\Psi$ fixes the reference flux. For a field regular on the axis choose $\Psi (0,z,t)=0$. Then the magnetic flux through a horizontal disk of radius $R$ is $2\pi \Psi (R,z,t)$, and the flux between two such surfaces is $2\pi \Delta \Psi$.

Since ${\mathbf{B}}_p\cdot \nabla \Psi =0$ and ${\partial }_{\varphi }\Psi =0$, the level surfaces of $\Psi$ contain the complete magnetic field, including its toroidal magnetic field component. Thus regular levels are poloidal flux surfaces. Their meridional contours trace the projected field lines. The representation is local on regular regions; disconnected flux surfaces with the same label need not share additional physical data.

The azimuthal component of $\mathbf{u}\times \mathbf{B}$ is $u_z{B}_R-u_R{B}_z=-{\mathbf{u}}_p\cdot \nabla \Psi /R$. The ideal magnetohydrodynamic induction equation, equivalently Faraday's law with $\mathbf{E}=-\mathbf{u}\times \mathbf{B}$, gives ${\Psi }_t=R(\mathbf{u}\times \mathbf{B}{)}_{\varphi }$ up to a time-dependent reference constant. Fixing that reference gives the material advection of an axisymmetric magnetic flux function:Axisymmetry removes the azimuthal contribution to the material derivative. This is magnetic flux freezing expressed using one scalar flux label.

The azimuthal component itself is generally not an advected scalar. Including the cylindrical basis-vector derivatives in the induction equation givesUsing mass conservation, $D\rho /Dt=-\rho \nabla \cdot \mathbf{u}$, and $DR/Dt=u_R$, this becomes the toroidal magnetic-field winding equation:Thus $B_{\varphi }/(\rho R)$ is conserved following the flow if the angular velocity is constant along the poloidal magnetic field, including the case of no rotation. Otherwise differential rotation winds poloidal field into toroidal field.

There is also an explicit flux-conservation form,Integrating this over a meridional region converts its change of toroidal flux to boundary terms. Hence such a flux is conserved when the indicated boundary transport vanishes. More generally magnetic flux freezing conserves flux through every material surface, but a meridional coordinate surface need not remain material when the fluid rotates. Toroidal flux conservation therefore does not imply that $B_{\varphi }$ itself is materially constant.

Write the total velocity as $\mathbf{U}=\Omega y\hat{\mathbf{x}}+\mathbf{u}$ and the magnetic field as $B_0\hat{\mathbf{x}}+\mathbf{b}$. For solenoidal fields the resistive induction equation is $({\partial }_t+\mathbf{U}\cdot \nabla )\mathbf{B}=\mathbf{B}\cdot \nabla \mathbf{U}+\eta {\nabla }^2\mathbf{B}$. The first-order smoothing approximation drops the fluctuating terms $\mathbf{b}\cdot \nabla \mathbf{u}-\mathbf{u}\cdot \nabla \mathbf{b}$, retaining shear exactly. The uniform field is not stretched by the background shear because ${\partial }_x(\Omega y)=0$. The other retained stretching terms are $\mathbf{b}\cdot \nabla (\Omega y\hat{\mathbf{x}})=\Omega b_y\hat{\mathbf{x}}$ and $B_0{\partial }_x\mathbf{u}$. HenceThe same phase cancellation as in part (i), together with ${\nabla }^2{e}^{i\mathbf{k}\cdot \mathbf{x}}=-K^2{e}^{i\mathbf{k}\cdot \mathbf{x}}$, proves that a response can be taken as $\mathrm{Re}(\mathbf{c}(t)e^{i\mathbf{k}(t)\cdot \mathbf{x}})$, whereIf $\mathbf{k}\cdot \mathbf{c}=0$ initially, its derivative is $-\eta K^2\mathbf{k}\cdot \mathbf{c}$, so magnetic solenoidality is preserved.

For $j=y,z$ the response equation is ${\dot{c}}_j+\eta K^2{c}_j=i{k}_x{B}_0{v}_j$. Its integrating-factor solution isThe velocity amplitudes and wavevector vary on the shear timescale. In the fast-diffusion limit $\eta (k_x^2+k_y^2)\gg |\Omega |$, and after the free transient, the integral is concentrated within time $(\eta K^2{)}^{-1}$ of $t$. Replacing the slowly varying factors there by their current values gives the quasistatic magnetic response of a helical shearing wave:Complex conjugation introduces a minus sign in $c_j^{\ast }$. With precisely the EMF normalization specified in the question,A literal spatial average of the product of the two real Fourier waves would be half this quantity; no extra factor of $1/2$ belongs in the question's defined $\mathcal{E}$.

Use $A=k_x^2+k_z^2$ and part (i) to obtain $\mathcal{E}(t)\simeq -B_0{H}_{0}A/[\eta (A+{\Omega }^2{k}_x^2{t}^2{)}^2]$. For the time integral, take the driven response without a free magnetic transient and suppose the fast-diffusion approximation holds through the contributing interval, in particular at $t=0$; $\eta k_x^2\gg |\Omega |$ is a sufficient uniform condition. For $\Omega k_x\ne 0$, substitute $z=|\Omega k_x|t/\sqrt{A}$. Since ${\int }_{-\infty }^{\infty }(1+z^2{)}^{-2}dz=\pi /2$, this givesThe integral identity follows immediately from $z=\tan \chi$, reducing it to ${\int }_{-\pi /2}^{\pi /2}{\cos }^2\chi d\chi$. If $k_x=0$, the forcing and $H$ vanish and the driven EMF is zero. If $\Omega =0$ but $H_0\ne 0$, the leading EMF is constant rather than time-integrable; the finite integral relies on nonzero shear.

For this question use the normalized magnetic scalar potential $\mathcal{U}$ defined by $\mathbf{B}=({\mu }_0/4\pi )\nabla \mathcal{U}$. The previous question instead includes ${\mu }_0/4\pi$ in its scalar potential; its $U$ equals $({\mu }_0/4\pi )\mathcal{U}$. This normalization change in the paper must be kept separate from the kernel calculation.

Take the radial component of the Geselowitz formula. On the spherical boundary, ${\mathbf{n}}^{\prime }$ is parallel to ${\mathbf{r}}^{\prime }$, andThus the surface-current term contributes no radial magnetic field. The primary Biot-Savart law term gives, along a fixed direction $\hat{\mathbf{r}}$,Choose $\mathcal{U}\to 0$ at infinity. Integrating this equation fixes the full potential along every ray, givingTo express it as a source-position derivative, defineWith $\mathbf{W}=\mathbf{Q}\times {\mathbf{r}}_0$, differentiating the integrand givesso $\mathcal{U}=\mathbf{W}\cdot {\nabla }_{{\mathbf{r}}_0}\varphi$ as required.

Let $b=|{\mathbf{r}}_0|$ and $P=|\mathbf{r}-{\mathbf{r}}_0|$. An elementary antiderivative gives the spherical current-dipole logarithmic kernelFor $r>a>b$, both logarithm arguments are positive. Its radial derivative is $-1/(rP)$ and its limit at infinity is zero, which verifies the preceding integral. At $b=0$ use the continuous limit $\varphi =1/r$.

For a direct check, differentiating along $\mathbf{W}$ leaves $b$ unchanged and gives $\mathbf{W}\cdot {\nabla }_{{\mathbf{r}}_0}P=-\mathbf{W}\cdot \mathbf{r}/P$. Sinceone obtainsIts gradient is the Sarvas formula. The sphere radius and electrical conductivity disappear from this exterior magnetic result because spherical symmetry removed the radial contribution of the return currents. The original derivation and spherical-conductor assumptions are discussed in Sarvas's paper.

The vacuum Maxwell equations are $\nabla \cdot E=\nabla \cdot B=0$, $\nabla \times E=-B_t$ and $\nabla \times B=E_t/c^2$. Taking the curl of Faraday's equation and using $\nabla \times (\nabla \times E)=\nabla (\nabla \cdot E)-{\nabla }^{2}E$ gives $-{\nabla }^{2}E=-E_{tt}/c^2$. The same calculation for Ampère's equation gives ${\nabla }^{2}B=B_{tt}/c^2$. Hence each Cartesian component satisfies the electromagnetic wave equation.

Model the walls as a perfect conductor, with no oscillating fields in its interior. The tangential electric field is zero at the surface, otherwise it would drive unbounded conductor currents. The normal magnetic field is zero, since the divergence-free field has a continuous normal component and the interior oscillating field is zero. Thus $n\times E=0$ and $n\cdot B=0$. Normal $E$ and tangential $B$ may be nonzero because surface charge and surface current support their jumps.

The PDF writes the full magnetic field as $B=B_0(x,y)\hat{z}{e}^{i(kz-\omega t)}$. Taken literally, this cannot be a nonzero travelling wave: $\nabla \cdot B=ik{B}_z$, so for $k\ne 0$ it forces $B_z=0$. The subsequent requests require transverse components. The consistent intended reading is $B_z=B_0(x,y)e^{i(kz-\omega t)}$, with transverse fields determined by the Maxwell equations. Use that reading below.

All fields have the same $z,t$ factor. Put $\Delta =(\omega /c{)}^2-k^2$. The harmonic Maxwell equations are $\nabla \times E=i\omega B$ and $\nabla \times B=-i\omega E/c^2$. Their $x$ and $y$ components givetogether with ${\partial }_y{E}_z-ik{E}_y=i\omega B_x$ and $ik{E}_x-{\partial }_x{E}_z=i\omega B_y$. Substitution and solving, for $\Delta \ne 0$, givesThe corresponding electric components areFor a transverse electric mode of a rectangular waveguide, $E_z=0$. The electromagnetic wave equation becomesAt $x=0,a$, the tangential component $E_y=-i\omega {\partial }_x{B}_z/\Delta$ vanishes, so ${\partial }_x{B}_0=0$; at $y=0,b$, the corresponding condition $E_x=0$ gives ${\partial }_y{B}_0=0$. These are Neumann boundary conditions, and the normal-$B$ conditions give the same conclusions through the displayed transverse magnetic components. Separation giveswhere $m,n\ge 0$ are integers and $(m,n)\ne (0,0)$. The constant candidate is not a physical oscillating mode: its transverse derivatives vanish, so it has $E=0$ and Faraday's equation forces $B=0$ for $\omega \ne 0$. More generally a hollow single-conductor guide has no transverse-electromagnetic zero-cutoff mode: its transverse electrostatic potential is harmonic with the same constant boundary value, hence constant, leaving zero transverse field.

The dispersion relation is ${\omega }^2=c^2(k^2+{\kappa }_{mn}^2)$. The waveguide cutoff frequency for this mode is ${\omega }_{mn}=c{\kappa }_{mn}$; below it $k$ is imaginary and the field is evanescent. Since $a\ge b$, the lowest cutoff occurs at $(m,n)=(1,0)$:For a mode propagating in the positive $z$ direction ($k>0$), above its own cutoff ${\omega }_{mn}$, the phase velocity and group velocity areThese waveguide phase and group velocities have product $c^2$. The superluminal phase speed describes motion of a constant phase, while a narrow wave packet travels at the subluminal group velocity; it does not transmit information faster than light. At cutoff $k=0$ there is no travelling mode with nonzero longitudinal flux.

For $ab\ne 0$, a magnetic field line stays at fixed $R$ and satisfies $d\varphi /dz=B_{\varphi }/(R{B}_z)=a/b$. It is therefore a helix with axial pitch $2\pi b/a$; the sign of $a/b$ fixes its handedness. The axial magnetic flux is $\pi b{R}_0^2$, and the toroidal magnetic field supplies the twist. If $a=0$ the field is untwisted; if $b=0$ it consists of azimuthal circles rather than helical lines carrying axial flux.

A convenient globally regular magnetic vector potential has $A_R=0$ andIndeed, the cylindrical curl gives $B_z=R^{-1}d(R{A}_{\varphi })/dR$ and $B_{\varphi }=-d{A}_z/dR$, with $B_R=0$. The tangential potential components are continuous at $R_0$, so their derivatives introduce no unwanted delta-function magnetic field at the tube boundary. The $C$ term is a constant axial potential.

The most general regular potential is this one plus $\nabla \Lambda (R,\varphi ,z)$ for an arbitrary single-valued gauge function. A putative extra $C_{\varphi }/R$ term in the inner $A_{\varphi }$ is excluded by regularity at the axis; the exterior $1/R$ circulation is already fixed by the enclosed magnetic flux. Since the potential is globally defined across the axis and exterior, any remaining difference is a curl-free vector field, hence a gradient. The constant $C$ can itself be absorbed into a gauge function proportional to $z$.

For the displayed gauge family, inside the tubeand henceFor example, requiring $A_z=0$ outside sets $C=a{R}_0^2/2$ and gives $\boxed{H_m/L=\pi ab{R}_0^4/2}$. The equally regular choice $C=0$ gives zero. More generally, for a length-$L$ segment the gauge transformation contributesIn particular, $\Lambda =\Delta Cz$ shifts $H_m/L$ by $\pi b{R}_0^2\Delta C$. This gauge dependence of helicity in an open magnetic flux tube is possible because field lines penetrate the end caps when $b\ne 0$. There is no contradiction with part (b), whose no-penetration condition fails at those caps. A gauge or end-cap convention is needed to give a unique value per unit length. For $b=0$ the ordinary helicity is zero and this end-cap ambiguity disappears.

Write $m_0={\mu }_0\rho$ and treat the imposed mean magnetic field as uniform on the forcing scale. In first-order smoothing, retain terms linear in the fluctuating fields in their equations, while keeping their quadratic correlation when calculating the mean-field electromotive force. The inertial fluctuation, nonlinear induction term and quadratic fluctuating Lorentz force density are neglected in the steady response. The linear magnetic force isAbsorb the gradient into the total pressure. The resulting equations arewith $\nabla \cdot \mathbf{u}=\nabla \cdot \mathbf{b}=0$. Taking the divergence of the first equation gives ${\nabla }^{2}P=0$ for solenoidal forcing. In a periodic or homogeneous Fourier calculation its nonconstant part therefore vanishes.

For each Fourier mode with wavevector $\mathbf{q}$, $|\mathbf{q}|=k$, define $\ell =\mathbf{B}\cdot \mathbf{q}$ andSolving the two coupled linear equations gives the monochromatic coupled magnetic and velocity responseThe magnetic coupling suppresses modes with large projection of their wavevector along the imposed field.

To calculate the mean-field electromotive force, group each pair of opposite wavevectors into a real mode and average its spatial phase. Put $\mathbf{n}=\mathbf{q}/k$, and define its forcing correlationsFor any solenoidal real mode $\mathbf{v}=\mathbf{a}\cos (\mathbf{q}\cdot \mathbf{x})+\mathbf{c}\sin (\mathbf{q}\cdot \mathbf{x})$, its two amplitude vectors are perpendicular to $\mathbf{q}$. Direct multiplication givesAlso, integration by parts gives $⟨(L\mathbf{g})\times (L\mathbf{f})⟩=-{\ell }^2⟨\mathbf{f}\times \mathbf{g}⟩$, where $L=\mathbf{B}\cdot \nabla$. Expanding the response products therefore yieldsSumming over independent real modes, or integrating over their angular spectrum, is the required general answer for $⟨\mathbf{u}\times \mathbf{b}⟩$.

At zero field, $\mathcal{E}(0)={\sum }_{\mathbf{q}}{\mathbf{C}}_{\mathbf{q}}/(\nu \eta k^4)$. Thus the stated zero-field condition removes the unweighted sum of the cross correlations. For anisotropic correlated forcing it does not necessarily remove their differently weighted sum at nonzero field. The displayed general formula retains that allowed contribution. It disappears if cross forcing is absent mode by mode, or under the jointly isotropic statistics assumed for the final part.

The helicity content can also be expressed directly using the actual fluctuating fields. Define their kinetic helicity $H_u(\mathbf{q})=⟨{\mathbf{u}}_{\mathbf{q}}\cdot \nabla \times {\mathbf{u}}_{\mathbf{q}}⟩$ and current helicity $H_b(\mathbf{q})=⟨{\mathbf{b}}_{\mathbf{q}}\cdot \nabla \times {\mathbf{b}}_{\mathbf{q}}⟩$. Substitute $\mathbf{f}=\nu k^2\mathbf{u}-L\mathbf{b}/m_0$ and $\mathbf{g}=\eta k^2\mathbf{b}-L\mathbf{u}$ into their cross correlation. This gives the exact identityIt shows the competing kinetic and magnetic contributions without introducing a spurious singularity when the factor on the left vanishes. For a monochromatic field, ${\mathbf{a}}_b=\nabla \times \mathbf{b}/k^2$ is a magnetic vector potential, so $H_b=k^2⟨{\mathbf{a}}_b\cdot \mathbf{b}⟩$: current helicity is $k^2$ times magnetic helicity. In the isotropic weak-field limit, with zero-field fluctuations ${\mathbf{u}}_0,{\mathbf{b}}_0$, the scalar alpha effect isThis identifies both signs and their different viscous and magnetic relaxation times.

For jointly isotropic forcing let $h_f,h_g$ now denote the total forcing helicities, and set $S_h=\nu h_g/m_0-\eta h_f$. The cross-forcing contribution is even in $\mathbf{B}$, whereas isotropy requires any vector depending only on $\mathbf{B}$ to be parallel to it and odd under reversal. Hence that contribution vanishes. A natural symmetric alpha tensor isOnly its longitudinal coefficient contributes to $\mathcal{E}={\alpha }_{\parallel }\mathbf{B}$. Introduce $h=|B|/[k\sqrt{m_0\nu \eta }]$ and $A_0=\nu \eta k^4$. Angular integration givesTherefore the strong-field quenching of an isotropic electromotive force isprovided $S_h\ne 0$. If the two helicity contributions cancel exactly, this electromotive force vanishes.

There is a tensor qualification to the printed scaling claim. The natural transverse coefficient instead contains $\frac{1}{2}{\int }_0^1(1-{\mu }^2)(1+h^2{\mu }^2{)}^{-2}d\mu \sim \pi /(8h)$ and scales as $|B{|}^{-1}$. The relation ${\mathcal{E}}_i={\alpha }_{ij}{B}_j$ by itself leaves transverse components undetermined. One may choose the effective representative ${\alpha }_{ij}={\alpha }_{\parallel }{\delta }_{ij}$, for which the requested $|B{|}^{-3}$ scaling holds. It is the observable longitudinal response that is fixed, rather than every component of a unique field-dependent tensor.

A planetary magnetic field is evidence about electrical conductivity, fluid motion and thermal evolution, but its presence alone does not identify a present planetary dynamo. One must separate internal global fields, crustal remanent magnetization and currents driven by the solar wind. The following discussion uses the observational picture available by 2009, including the early MESSENGER flybys and the Voyager, Galileo and Cassini results.

Outside a planet's internal electric currents, the internal component is approximately a potential magnetic field. Its spherical harmonic of degree $\ell$ decreases as $r^{-(\ell +2)}$. Measurements at different positions and altitudes can therefore distinguish a deep, low-degree source from localized crustal anomalies and external magnetospheric currents. Repeated observations, radio emissions, aurorae and particle measurements supply complementary evidence for a magnetosphere, while slow secular variation and records of past reversals test the internal field's evolution.

For Earth, ground observatories and satellite measurements reveal a predominantly dipolar field of several tens of microtesla at the surface, tilted relative to the rotation axis and changing with time. Magnetic minerals in rocks record a much older field and repeated geomagnetic reversals. These observations, together with seismic evidence for a liquid outer core, support an operating planetary dynamo in conducting iron alloy. Core temperatures exceed the relevant Curie temperatures, so ordinary permanent remanent magnetization cannot explain the main field. Thermal convection, cooling and compositional convection associated with inner-core growth supply buoyancy; the Coriolis force organizes the motion. The observed nonaxisymmetric components and time dependence are consistent with a dynamic source, rather than a fixed bar magnet.

Mariner 10 and the first two MESSENGER flybys established a weak but coherent intrinsic field at Mercury, with roughly dipolar structure closely aligned with its rotation axis. It is much weaker than Earth's surface field, but its global character favors a deep source over an explanation solely by localized crustal patches. A planetary dynamo in a fluid part of Mercury's large iron-rich core is the natural interpretation, supported by evidence for a molten core component. Light elements can depress the alloy's freezing temperature, and thermal or compositional buoyancy can maintain motion. Slow rotation is not enough to rule out dynamo action. The limited pre-orbital measurements do not uniquely determine the detailed core structure or the mechanism responsible for the field's weakness.

For Venus, spacecraft including Pioneer Venus found no appreciable internally generated global field. Nevertheless its ionosphere interacts with the solar wind, producing magnetic draping, a bow shock and an induced magnetosphere. Those external signatures are not evidence for an internal planetary dynamo. Venus is comparable to Earth in size and may possess a substantial conducting core; the absence of a detectable dynamo therefore need not mean that all of its core is solid. Insufficient core heat loss or stable stratification can prevent the required convection. Its slow rotation can affect flow organization, but rotation rate by itself is not a sufficient explanation of the contrast with Earth or Mercury.

For Mars, Mars Global Surveyor found strong, spatially localized crustal anomalies, especially in ancient southern terrain, rather than an appreciable present global dipole. This is remanent magnetization, indicating that an ancient field magnetized the crust while it formed and cooled. The distribution of anomalies and weakly magnetized major impact basins constrain when that ancient planetary dynamo operated; crustal magnetization is a record of past activity, not a present global dynamo. Modern interaction with the solar wind also produces an induced magnetosphere. Cooling and changes in core heat extraction can explain the cessation of dynamo-driving convection; the observations alone do not prove complete solidification of the core.

For Jupiter, radio emissions, its intense radiation belts and measurements by Pioneer, Voyager and Galileo demonstrate a strong global intrinsic field and an enormous magnetosphere. The field is broadly dipolar but tilted and contains higher multipoles. It is generated in a deep conducting fluid rather than by a permanent solid magnet: the principal conducting region is metallic hydrogen in the high-pressure interior. Cooling and gravitational energy release maintain convection, while rapid rotation organizes the flow. Plasma supplied by Io and currents in the magnetosphere strongly modify the observed external field, but cannot account for the internally sourced planetary component. Jupiter's large magnetic moment also reflects its large size; a moment ratio should not be confused with the ratio of local surface field strengths.

For Saturn, Pioneer, Voyager and Cassini measurements likewise show an intrinsic field and a magnetosphere, with a dipole strikingly close to the rotation axis. A planetary dynamo in conducting metallic hydrogen is supported by the same broad interior argument as for Jupiter. Interior cooling and differentiation can provide power, including energy associated with helium separation. The near alignment does not invalidate the Cowling anti-dynamo theorem: that theorem excludes an exactly axisymmetric isolated dynamo field, not a predominantly axisymmetric exterior with undetected nonaxisymmetric structure in its source region. A stably stratified conducting layer above the active region is one possible way to attenuate the observable nonaxisymmetric components; it is a model, not a uniquely established consequence of the observations.

For Uranus, Voyager 2 in 1986 observed an intrinsic field with a dipole tilted by about $59^{\circ }$ to the rotation axis and significantly displaced from the centre. Higher multipoles are substantial, so a centred dipole is an inadequate description. The field and associated magnetosphere indicate an active planetary dynamo, despite the planet's small intrinsic heat flux. That low luminosity constrains its thermal evolution and may indicate inhibited deep mixing; it does not demonstrate an absence of all conducting-fluid motion. A conducting, volatile-rich fluid shell is a plausible source region. The term ice giant describes bulk composition, not a claim that the dynamo region is ordinary frozen ice.

For Neptune, Voyager 2 in 1989 also found a highly non-dipolar, displaced intrinsic field, with a dipole tilt near $47^{\circ }$. Its field, radio emissions and magnetosphere imply another active planetary dynamo. Its larger intrinsic heat flux than Uranus supplies evidence for substantial continuing interior energy transport. Electrically conducting water-ammonia-rich mixtures at high pressure, with ionic or otherwise mobile charges, can supply the conducting region without requiring the metallic hydrogen layer characteristic of gas giants. Numerical dynamos in convecting shells surrounding more stably stratified interiors can produce strongly nonaxisymmetric, multipolar fields resembling those of the two ice giants. Such models show that their unusual field geometry is compatible with dynamo action, but field measurements do not uniquely identify the shell's composition or thickness.

The maintenance mechanism is governed by the resistive induction equation,A conducting fluid needs a sufficiently large magnetic Reynolds number and a flow geometry that regenerates magnetic flux rather than merely amplifying one component temporarily. Three-dimensional stretching and twisting can couple poloidal and toroidal fields; strictly planar regeneration and a strictly axisymmetric magnetic field are restricted by the relevant anti-dynamo theorems. Rotation influences this regeneration through the Coriolis force, but does not provide an inexhaustible power source merely by being present.

Without regeneration, the characteristic magnetic diffusion time is ${\tau }_{\eta }\sim L^2/\eta$. For a terrestrial core scale of order $10^6$ metres and diffusivity of order one square metre per second, this is of order $10^4$–$10^5$ years, far shorter than geological ages. Long-lived global fields therefore require continual energy input. A schematic magnetic energy balance iswhen boundary energy fluxes vanish. Mechanical work against the Lorentz force density supplies the energy lost as Ohmic heating. Thermal cooling, latent heat release, compositional convection and differentiation supply that mechanical power. Magnetic feedback eventually changes the flow and saturates the field. Strong conductivity without an adequate energy source is not sufficient for a dynamo.

The 2009 observations support active global dynamos in Mercury, Earth, Jupiter, Saturn, Uranus and Neptune; Venus has no detected appreciable intrinsic dynamo field, and Mars preserves evidence of an extinct earlier dynamo in its crust. Induced fields and surviving remanence explain why planets without a current global dynamo can still display magnetic phenomena. The diversity of field geometries and strengths constrains, rather than uniquely solves, the diversity of planetary conducting regions and their convective histories.

Put $\pi =p-eA/c=m\dot{q}$. The Hamilton equations areDifferentiate the mechanical momentum and substitute:With $B=\nabla \times A$ and $E=-c^{-1}{\partial }_{t}A$ in this zero-scalar-potential gauge, this is $\boxed{m\ddot{q}=eE+(e/c)\dot{q}\times B}$, the Lorentz force in the units of the Hamiltonian.