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An alpha-Omega mean-field dynamo couples toroidal-field generation by the Omega effect to poloidal-field regeneration by the alpha effect. A simple Fourier model is $\dot{A}={\alpha }_{0}f(t)B-\eta k^{2}A$, $\dot{B}=ik\Omega A-\eta k^{2}B$, where $A$ is a component of the vector potential for the poloidal field and $B$ is a toroidal-field amplitude. For bounded $|f|\le 1$, the bounded-modulation alpha-Omega growth estimate controls the maximum amplification.

For a twice continuously differentiable function, the curl of its gradient vanishes. In components, ${ϵ}_{ijk}{\partial }_j{\partial }_{k}f=0$ because the second derivatives are symmetric in $j,k$ while the Levi-Civita symbol is antisymmetric. Thus adding a gradient to a vector potential preserves the represented curl. No topological assumption is needed for this identity, unlike the converse assertion that every curl-free vector field has a global potential.

A uniformly charged rotating sphere has an exactly dipolar exterior magnetic vector potential. For a concentric equatorial loop of radius $b$ outside the sphere, Stokes theorem gives ${\Phi }_b=2\pi b{A}_{\phi }(b,\pi /2)={\mu }_{0}m/(2b)$. If its effective source electric current is defined by $m=\pi a^2{I}_{\mathrm{eff}}$, its mutual inductance is ${\mu }_0\pi a^2/(2b)$. The finite-source vector potential is regular through the spanning disk; an exterior point-dipole expression should not be integrated through its singular origin.

Write the superconducting order parameters as ${\psi }_j=|{\psi }_j|e^{i{\theta }_j}$, using a field of pair electric charge $q=-2e$. For a specified path across the superconducting weak link, defineHere $\mathbf{A}$ is the magnetic vector potential. To check electromagnetic gauge invariance, use a scalar function $\chi$ and transformThe change of ${\theta }_2-{\theta }_1$ is $q({\chi }_2-{\chi }_1)/\hslash$, exactly cancelled by the change of the line integral. Thus $\varphi$ is unchanged although each local quantum phase changes. This is the same charge/sign bookkeeping discussed in gauge-coupling sign in Ginzburg-Landau superconductivity. Different physical paths can enclose magnetic flux, so fixing the path matters.

For stationary reservoirs with equal internal pair chemical potentials and a time-independent vector potential, the local Schrödinger equation gives ${\dot{\theta }}_j=-q{U}_j/\hslash$ after dropping the common internal term. With electric potentials $U_j$ and $V=U_2-U_1$, subtraction gives $\dot{\varphi }=-qV/\hslash =2eV/\hslash$. For a general circuit, $V$ denotes the corresponding measured electrochemical voltage and the electromagnetic contribution must be treated consistently. A bare difference of phases without the vector potential term is insufficient in an arbitrary gauge.

A spinless charged particle in $B_z=\alpha x$ admits vector potential $A_y=\alpha x^2/2$. Separating a plane wave along y produces a one-dimensional quartic potential energy. For nonzero $q\alpha$, $k/(q\alpha )>0$ gives two minima at $\pm \sqrt{2\hslash k/(q\alpha )}$; the opposite sign gives one minimum at zero. The field gradient creates confinement even though there is no scalar potential. The quartic potential is even, so its nondegenerate eigenstates have definite spatial parity.

For a charged particle minimally coupled to a magnetic vector potential, the mechanical momentum is $\mathbf{\pi }=\mathbf{p}-q\mathbf{A}$, with $\mathbf{p}$ the canonical momentum. It is the mass times velocity in nonrelativistic mechanics.

For a spinless particle, minimal coupling gives $H=(\mathbf{p}-q\mathbf{A}{)}^2/(2m)+q\varphi$. Choose the vector potential $\mathbf{A}=(0,\alpha x^2/2,0)$ and $\varphi =0$. Its curl is $\mathbf{B}=\alpha x\hat{\mathbf{z}}$ and $[p_y,A_y]=0$, soThe translation separation in a magnetic field gradient follows from $[H,p_y]=0$: a simultaneous energy eigenstate can be written $\Psi =L_y^{-1/2}{e}^{iky}{\Phi }_k(x)$. The periodic boundary condition requires $e^{ik{L}_y}=1$, hence $k=2\pi n/L_y$, $n\in \mathbb{Z}$. The remaining Schrödinger equation isThis magnetic-gradient quartic confinement has a positive quartic coefficient for $g\ne 0$. If $k/g<0$, it is a single well at zero, with $V_k(0)={\hslash }^2{k}^2/(2m)$. Its harmonic curvature givesThus the two lowest states are respectively even and odd, with one central peak or two opposite-sign lobes, and approximately $E_j=V_k(0)+(j+1/2)\hslash {\omega }_s$. They are not a nearly degenerate pair.

If $k/g>0$, the two minima are $x=\pm x_0$, $x_0=\sqrt{2\hslash k/g}$, with zero potential there and a central barrier ${\hslash }^2{k}^2/(2m)$. For well-separated minima the lowest parity states of a symmetric double well areHere ${\varphi }_L,{\varphi }_R$ are the localized ground-state Gaussians. Their even combination is the lower state; the odd combination has a node at the origin. Weak quantum tunnelling gives an exponentially small double-well tunneling splitting.

To obtain their common leading energy, put $x=x_0+\delta x$. Since $\hslash k=g{x}_0^2/2$,The double-well oscillator scale in a linear magnetic field therefore givesThe asymptotic condition is $x_0\gg \sqrt{\hslash /(m{\omega }_d)}$, so the barrier separates narrow local oscillator states. It is this scale comparison that makes the source's “large” ratio meaningful.

The signed group velocity of a magnetic-gradient doublet follows by differentiating the band energy with respect to the conserved wavevector:For $q\alpha >0$ the double-well branch has $k>0$ and moves along $+y$; reversing both signs reverses its velocity. To this order the two parity states have the same velocity. Corrections include local anharmonicity and the exponentially small splitting. The spinless Hamiltonian specified here has no extra Zeeman effect term.

In a contractible region, the Maxwell equations $\nabla \cdot \mathbf{B}=0$ and the Poincaré lemma imply the existence of a magnetic vector potential $\mathbf{A}$ such that $\mathbf{B}=\nabla \times \mathbf{A}$. Faraday's law of induction then givesThe expression in parentheses is a gradient there, so a scalar electric potential $\varphi$ can be chosen withThese statements are local, or global when the domain has the required topology and the electromagnetic fields are regular. A nonzero flux through a topologically nontrivial closed surface can obstruct a single globally defined vector potential, even though local expressions remain valid. The electric potential and magnetic vector potential are not unique: $\mathbf{A}\mapsto \mathbf{A}+\nabla \chi$ and $\varphi \mapsto \varphi -{\partial }_t\chi$ leave both electromagnetic fields unchanged.

For the Lorenz gauge condition $\nabla \cdot \mathbf{A}=-c^{-2}{\partial }_t\varphi$, Gauss's law becomesThe Ampere-Maxwell law givesCancel the two gradient terms using the Lorenz gauge condition. HenceBoth potential equations have the same d'Alembert operator, with their respective charge density and current density sources. In a source-free region these are homogeneous wave equations with propagation speed $c$.

Let ${\rho }_q=3q/(4\pi a^3)$ be the uniform charge density and take $\mathbf{\omega }=\omega \hat{\mathbf{z}}$. Rigid rotation produces current density $\mathbf{J}={\rho }_q\mathbf{\omega }\times \mathbf{r}$. The magnetic dipole moment isSpherical symmetry gives $\int r_i{r}_j{d}^{3}r={\delta }_{ij}\int r^2{d}^{3}r/3$, while $\int r^2{d}^{3}r=4\pi a^5/5$. ThereforeIt is parallel to the angular velocity for positive $q$ and antiparallel for negative $q$. The fully signed magnitude statement is $|{\mathbf{m}}_s|=|q||\omega |a^2/5$; the printed magnitude assumes positive $q$ and positive angular speed.

Axisymmetry permits a purely azimuthal magnetic vector potential, $\mathbf{A}=f(r)\sin \theta \hat{\mathbf{\phi }}$, with $A_r=A_{\theta }=0$. For the source's associated Legendre function convention, $P_1^1(\cos \theta )=+\sin \theta$, since its definition omits the Condon–Shortley phase. The given differential operator reduces toInside the sphere, $J_{\phi }={\rho }_q\omega r\sin \theta$. Since the radial operator maps $r^3$ to $10r$, the particular coefficient is ${\alpha }_0=-{\mu }_0{\rho }_q\omega /10$. Regularity at the origin and decay at infinity giveThere is no surface sheet electric current, so both normal and tangential magnetic field components are continuous at $r=a$. Usingthis is equivalent to continuity of $f$ and $f^{\prime }$. The two equations give $A={\mu }_0{\rho }_q\omega a^2/6$ and $C={\mu }_0{\rho }_q\omega a^5/15$. ThusIn the homogeneous expansion, ${\alpha }_1=A$, ${\beta }_1=C$ in their respective interior and exterior regions; all higher angular orders vanish. Indeed, matching the regular $r^{\ell }$ and decaying $r^{-\ell -1}$ terms with zero source at $\ell >1$ forces both coefficients to zero. The outside magnetic field is exactly the magnetic dipole field of ${\mathbf{m}}_s$, not merely its leading far-field approximation. For a positive moment along $z$, $A_{\phi }={\mu }_0{m}_s\sin \theta /(4\pi r^2)$, which is the printed cross-product magnitude expression.

For the flux coupling of a rotating charged sphere to a circular loop, take a loop of radius $b>a$ in the equatorial plane and use Stokes theorem on the regular full vector potential, evaluated around the rim. Its linked magnetic flux isWith $m_s=\pi a^2{I}_0$, the mutual inductance defined using this effective electric current isIntegrating an ideal magnetic dipole field through the entire disk would incorrectly use its singular expression inside the actual sphere; the rim calculation avoids that error.

For the prescribed time dependence, the quasistatic Faraday's law of induction result and the purely resistive loop approximation givePositive loop electric current is defined to produce a positive $z$ magnetic dipole moment, so the minus sign is Lenz's law. This assumes negligible retardation and $\nu L_b\ll R$, where $L_b$ is the loop's self-inductance. If that is not negligible, the governing equation is instead $L_b{\dot{I}}_b+R{I}_b=-{\dot{\Phi }}_b$; $R$ alone is insufficient for the exact dynamic response.

Choose the outward normal vector on the side of the elliptic cylinder. Parametrize it bySince ${\mathbf{r}}_{\theta }\times {\mathbf{r}}_z=(\frac{1}{2}\cos \theta ,\sin \theta ,0)$ points outward, the parameter rectangle has its bottom edge traversed with increasing $\theta$, its right cut edge with increasing $z$, its top edge with decreasing $\theta$, and its left cut edge with decreasing $z$. After identification, the two cut edges are the same physical curve with opposite directions. Their line integrals cancel for a single-valued vector field. Viewed from above, the bottom rim is counterclockwise and the top rim clockwise.

To apply Stokes theorem to a flux rather than a curl integral, first find a vector potential:This vector potential is smooth everywhere. On either horizontal rim, $d\mathbf{r}$ is horizontal whereas $\mathbf{A}$ is vertical, giving $\mathbf{A}\cdot d\mathbf{r}=0$. For a cut at ${\theta }_0=\pi /4$, $xy=1/4$, so the upward cut contribution is ${\int }_{-1}^1{z}^2/4dz=1/6$ and the downward one is $-1/6$. ConsequentlyA direct check uses the oriented surface element above: $\mathbf{F}\cdot ({\mathbf{r}}_{\theta }\times {\mathbf{r}}_z)=\frac{1}{2}{z}^2\cos 2\theta$, whose integral over a full turn vanishes. Reversing the chosen orientation still gives zero.

First retain the question's units $c=1$. Insert the retarded time Dirac delta:For a point charge, the spatial delta sets $x^{\prime }=x_0(t^{\prime })$. Put $R(t^{\prime })=|x-x_0(t^{\prime })|$, $\hat{R}=(x-x_0)/R$ and $v={\dot{x}}_0(t^{\prime })$. The retarded root $t_{\ast }$ satisfies $t-t_{\ast }=R(t_{\ast })$. The derivative of the delta's argument is $-1+\hat{R}\cdot v$, so for a subluminal trajectory its absolute value is $1-\hat{R}\cdot v$. Performing the time integral givesAll quantities in the denominator are evaluated at the retarded time. The retarded vector potential uses current $j=qv{\delta }^{(3)}(x^{\prime }-x_0)$ and the same Jacobian determinant, givingRestoring physical $c$, the root is $t-t_{\ast }=R_{\ast }/c$, both denominators become $R_{\ast }-v_{\ast }\cdot {\mathbf{R}}_{\ast }/c$, and ${\mu }_0=1/({ϵ}_0{c}^2)$. Thus $\mathbf{A}=v_{\ast }\varphi /c^2$. This is the Liénard–Wiechert potential; the Jacobian determinant, rather than merely substituting retarded position into a static Coulomb formula, accounts for the velocity factor.

The Euler-Lagrange equations areUsing time independence of the vector potential gives $m{\ddot{r}}_i=(e/c){\dot{r}}_j({\partial }_i{A}_j-{\partial }_j{A}_i)$, which is the component form ofTaking its scalar product with the velocity shows $\dot{T}=m\dot{\mathbf{r}}\cdot \ddot{\mathbf{r}}=0$: the kinetic energy is constant, since the magnetic force does no work.

For $\mathbf{B}=(0,0,F^{\prime }(x))$, the equations give $m\ddot{y}=-(e/c)F^{\prime }(x)\dot{x}$ and $\ddot{z}=0$. Thus $mc\dot{y}+eF(x)=-C$ and $\dot{z}=v_z$ are constants of motion. Substitution into $2T/m={\dot{x}}^2+{\dot{y}}^2+{\dot{z}}^2$ yieldsThe constant $D$ in this representation is nonpositive. The gauge choice $\mathbf{A}=(0,F(x),0)$ also makes the conserved transverse canonical momenta explicit.

Put $\mathbf{\pi }=\mathbf{p}-e\mathbf{A}/c=m\dot{\mathbf{q}}$. From Hamilton's equations,For the stated time-independent vector potential, differentiating ${\pi }_i$ gives $m{\ddot{q}}_i=(e/c){\dot{q}}_j({\partial }_i{A}_j-{\partial }_j{A}_i)$, orUse the Poisson bracket convention $[F,G]={\sum }_i(F_{q_i}{G}_{p_i}-F_{p_i}{G}_{q_i})$. Then direct differentiation givesFor $\mathbf{A}=(0,0,F(r))$ with $r=\sqrt{x_1^2+x_2^2}$, the Hamiltonian is independent of $x_3$, so ${\dot{p}}_3=0$. The radial force in the horizontal plane is ${\dot{p}}_r=e(p_3-eF/c)F^{\prime }/(mc)$. For circular motion the radial acceleration is $-r{\Omega }^2$, and $p_r$ is the mechanical radial momentum because the horizontal vector potential vanishes. ThusA circular orbit at that radius is possible when the right side is nonnegative; positive values give either sense of horizontal rotation, and the longitudinal velocity is the constant $(p_3-eF(r)/c)/m$ along that orbit.

For a twice differentiable vector potential $\mathbf{B}$, commuting partial derivatives in $\mathbf{J}=\nabla \times \mathbf{B}$ gives the necessary conditionThe potential is not unique: for any sufficiently smooth scalar $\psi$, $\mathbf{B}+\nabla \psi$ has the same curl, since the curl of a gradient is zero. This is the usual gradient freedom of a vector potential. A divergence-free condition alone need not suffice globally on every domain; here only its necessity is requested.

Stokes theorem states that for an oriented surface with compatible boundary orientation,For the integrand in this problem, a convenient globally smooth vector potential isThe boundary of the upper ellipsoid is the ellipse $z=0$. The outward normal on the upper surface induces counterclockwise orientation viewed from above, so parameterize it as $\mathbf{r}(\theta )=(a\cos \theta ,b\sin \theta ,0)$, $0\le \theta \le 2\pi$. Along this curve the first component of $\mathbf{B}$ is zero and $dz=0$, leavingSince the last angular integral is $3\pi /4$,The result is independent of the vertical semiaxis $c$. As a sign check, the field is divergence-free, so closing the cap with the downward-oriented base disc makes its flux the positive integral of $x^2$ over that disc, which has the same value.

Choose the vector potential $A=(-{\int }_0^{y}B(s)ds,0,0)$, whose curl is $(0,0,B(y))$, and a spin-down wavefunction $\Psi =e^{-ikx}\chi (y)(0,1{)}^{\mathsf{T}}$, independent of $z$. DefineThe covariant $x$ derivative acts as $-i(k+a)$, and ${\sigma }_3$ acts as $-1$. The Pauli equation consequently reduces toThe sign of the plane-wave phase was chosen to match the printed $k+a$ convention; choosing $e^{ikx}$ simply reverses the label $k$.

Set $W=k+a$ and $M={\partial }_y+W$. On the square-integrable operator domain with vanishing boundary terms, $M^{†}=-{\partial }_y+W$, andTaking the inner product with $\chi$ gives $ϵ\Vert \chi {\Vert }_2^2=\Vert M\chi {\Vert }_2^2\ge 0$. Thus $\boxed{E\ge 0}$.

A zero mode solves $M\chi =0$, soSince $a(0)=0$ and $a^{\prime }\ge b_0=e{B}_0/\hslash$, integration on either side of zero gives ${\int }_0^{y}a(s)ds\ge b_0{y}^2/2$. Therefore $|{\chi }_k(y)|\le |C_k|e^{-ky-b_0{y}^2/2}$, proving square integrability for every real $k$. Each can be normalized and supplies a distinct momentum label at zero energy. Hence zero energy is infinitely degenerate. In an infinite $x$ direction the individual plane waves are generalized momentum states; square-integrable superpositions of their normalized transverse zero modes give ordinary transverse states by the Fourier isometry. The stipulated zero $z$ momentum is understood per unit longitudinal length, or with periodic longitudinal normalization.

An anti-dynamo theorem identifies a restriction that prevents a conducting fluid from regenerating its magnetic field against resistive losses. The hypotheses matter: throughout the following arguments take positive uniform magnetic diffusivity, smooth fields, no imposed magnetic source, and the stated isolated or energy-closed boundary conditions. An anti-dynamo theorem does not rule out transient magnetic amplification, winding of an imposed field, or an externally supplied magnetic flux. The resistive induction equation is the common starting point:

First consider Backus' necessary condition for dynamo action. For a conductor contained in a sphere of radius $a$, include the potential field in the insulating exterior when defining the total magnetic energy:For an incompressible flow, with boundary conditions eliminating the mechanical surface term, such as a no-slip boundary condition for the fluid, integration by parts giveswhere $e_{ij}=({\partial }_i{u}_j+{\partial }_j{u}_i)/2$ is the rate-of-strain tensor. Its antisymmetric part does no stretching, since $B_i{B}_j$ is symmetric. If $S$ bounds the largest eigenvalue of $e$ throughout the flow, the first term is at most $2S{E}_B$.

The magnetic free-decay spectral bound for the enclosing insulating sphere isThe exterior energy is essential to this variational estimate. Its constant comes from the slowest freely decaying dipolar poloidal mode: the spherical radial matching condition reduces to $j_0(\kappa a)=0$, whose first positive root is $\kappa a=\pi$. Expanding an admissible field in the orthogonal free-decay modes, or using their variational characterization, gives the bound. A smaller conductor can only increase the minimum dissipation quotient for the enclosing-sphere comparison. Combining the two estimates yieldsThusBelow this threshold the Gronwall inequality gives exponential energy decay. Above it the estimate merely permits growth; it does not prove that the geometry regenerates the field. A time-dependent version bounds the field-amplitude exponent by the time average of the maximum strain minus the diffusive rate.

The geometric obstruction in Cowling anti-dynamo theorem is different: an isolated exactly axisymmetric magnetic field cannot be self-maintained. To see the missing regeneration mechanism, write its poloidal magnetic field in terms of the poloidal magnetic flux function $\psi$:The poloidal part of the resistive induction equation becomesThere is no term containing $B_{\varphi }$ that creates new poloidal flux. Fix the gauge by setting $\psi =0$ on the regular symmetry axis, and match to a decaying exterior field with no imposed flux. At a positive interior maximum away from the axis, the advective term vanishes and the diffusive term is nonpositive. The parabolic comparison principle therefore prevents increase of the maximum; it likewise prevents decrease of a negative minimum. In the steady problem the strong maximum principle, regularity at the axis and decay outside exclude a nonzero maximum altogether. Matching to the insulating exterior does not create a maximum at the interface: the exterior flux is harmonic in the corresponding axisymmetric operator, and the two normal-derivative conditions exclude such an interface extremum. The same homogeneous scalar diffusion problem gives decay in time on a bounded conductor with no replenishing boundary data.

Once the poloidal field has decayed, let $b=B_{\varphi }/r$ and let ${\Omega }_f=u_{\varphi }/r$ denote the fluid's angular velocity. Its equation isDifferential rotation can make toroidal field from poloidal field, but it cannot close the reverse loop. When ${\mathbf{B}}_p=0$, this too is a homogeneous advection-diffusion equation, with a regular axis and zero toroidal field at an insulating exterior. Hence it cannot sustain the remaining field. The exactly axisymmetric field, rather than merely an axisymmetric velocity, is excluded. If an exactly axisymmetric field were maintained by a nonaxisymmetric velocity, averaging the linear induction equation in azimuth would replace that velocity by its azimuthal mean and give the same contradiction. An axisymmetric flow can, however, excite a nonaxisymmetric field. Also an axisymmetric averaged field in mean-field dynamo theory has an additional fluctuating electromotive force, so this unmodified scalar proof does not apply to it.

The toroidal-velocity anti-dynamo theorem restricts the flow instead. In a conducting sphere, suppose $\mathbf{u}\cdot \mathbf{x}=0$, so the incompressible flow is tangent to every concentric sphere. No assumption of axisymmetry is made. Define $P=\mathbf{x}\cdot \mathbf{B}$. A direct Cartesian calculation is particularly useful:The last equality uses $\nabla \cdot \mathbf{B}=0$, so ${\nabla }^2(\mathbf{x}\cdot \mathbf{B})=\mathbf{x}\cdot {\nabla }^2\mathbf{B}$. Thus the radial magnetic scalar is a passive diffusing scalar, not a regenerated field. Multiplying by $P$, integrating over the sphere and using tangential incompressible advection givesIn an insulating exterior, a spherical harmonic component of degree $l\ge 1$ has $P\propto r^{-l-1}$, so its boundary contribution is negative: ${\partial }_{n}P=-(l+1)P/a$. The monopole component is absent because the magnetic field is solenoidal and regular. The resulting dissipative scalar problem makes $P$, and hence the poloidal field, decay.

When that radial field is absent, the remaining tangential solenoidal field can be written $\mathbf{B}=\nabla \times (T\mathbf{x})$. Its induction equation reduces to a second passive scalar equation. Indeedand the curl of the last gradient term is zero. Therefore $T_t+\mathbf{u}\cdot \nabla T=\eta {\nabla }^{2}T$, up to an irrelevant radial gauge function, which can be removed by taking the angular mean of $T$ to be zero. Its angular variations vanish at the insulating spherical boundary, so this scalar also decays by its quadratic energy identity. Equivalently, a putative growing magnetic normal mode must first have $P=0$, after which the scalar $T$ equation rules it out. Purely toroidal spherical flow cannot close a dynamo regeneration loop, even if it is nonaxisymmetric. This does not assert that purely poloidal flow is always incapable of a dynamo, or that any azimuthally directed cylindrical flow is governed by the same spherical theorem.

Other restrictions illuminate the same distinction between stretching and regeneration. The planar anti-dynamo theorem has a similar scalar proof in the usual two-dimensional geometry. For planar velocity and magnetic field independent of the third coordinate, write the in-plane field as $\nabla \times (A\hat{\mathbf{z}})$. Its vector potential obeys $A_t+\mathbf{u}\cdot \nabla A=\eta \Delta A$ after fixing the gauge, so $\frac{1}{2}\frac{d}{dt}\int A^2=-\eta \int |\nabla A{|}^2$ under periodic or homogeneous energy-closed boundary conditions. A third field component independent of that coordinate is also a passive scalar. A spatially uniform imposed field may survive in a periodic box but is not regenerated dynamo field. Three-component motions depending on only two coordinates are not covered by this strictly planar formulation.

There are also necessary bounds in terms of velocity amplitude rather than strain. The magnetic work term is bounded by $\Vert \mathbf{u}{\Vert }_{\infty }\Vert \mathbf{B}{\Vert }_2\Vert \nabla \times \mathbf{B}{\Vert }_2$; combining this with the same free-decay inequality gives a velocity-based lower threshold on the magnetic Reynolds number. These bounds, like Backus' necessary condition for dynamo action, are necessary rather than sufficient. Pure motion that only rearranges a scalar magnetic potential cannot evade diffusion merely by becoming faster. Conversely the stagnation-flow energy amplification in Question 2 does not contradict the bounded-body theorems: it uses an unbounded three-dimensional strain field and, for its growing finite-energy modes, a slowly decaying field at infinity. Similarly, the torsional Alfvén waves in Question 3 transfer energy between velocity and a toroidal perturbation about a prescribed background; they are not a construction of that background by dynamo action. The anti-dynamo results identify failed regeneration mechanisms and the assumptions under which those failures are rigorous.

In Lagrangian mechanics, the energy balance for an autonomous Lagrangian uses $E={\sum }_i{\dot{q}}_i\partial L/\partial {\dot{q}}_i-L$. For the stationary vector potential the velocity-linear magnetic term cancels in this expression, leaving kinetic energy. In cylindrical coordinates,Time independence and the cyclic angular coordinate give two constants:The three Euler-Lagrange equations areThe supplied angular velocity is $\dot{\varphi }=-2eg(z_0)/m$. Set $r=r_0$, $z=z_0$ and use this constant angular velocity. The radial right-hand side is $m{r}_0({\dot{\varphi }}^2+2eg\dot{\varphi }/m)=0$, the angular momentum is constant, and the vertical equation vanishes when $g^{\prime }(z_0)=0$. These equations and the initial data therefore giveFor nonzero radius and charge this is the requested circular orbit.

The specified vector potential gives $\mathbf{B}=(0,0,B)$. Put $\omega =qB/m$, retaining its sign. The Lorentz force law yields $\ddot{x}=\omega \dot{y}$, $\ddot{y}=-\omega \dot{x}$ and $\ddot{z}=0$. For initial velocities $(u,v,w)$ and $\omega \ne 0$, integration givesThe transverse motion is a circle centred at $(x_0+v/\omega ,y_0-u/\omega )$ with radius $\sqrt{u^2+v^2}/|\omega |$. Thus the full orbit is a helix around a line parallel to the magnetic field, with period $2\pi /|\omega |$ and signed axial advance $2\pi w/|\omega |$ per turn. The charge sign determines the sense of rotation. If $w=0$ it is a circle, if $u=v=0$ it is a straight line parallel to the field, and if $qB=0$ all motion is uniform rectilinear motion.

The identity divergence of a curl is zero follows by the symmetry of second partial derivatives and antisymmetry of the Levi-Civita symbol. ThereforeIf $\mathbf{B}=\nabla \times \mathbf{A}$ and $\mathbf{A}$ is in Coulomb gauge, the curl of the curl identity gives $\mathbf{J}=-{\nabla }^2\mathbf{A}$. Apply the supplied scalar Poisson equation representation to each component of the decaying vector potential. In the units of this problem the Coulomb-gauge vector potential of a localized steady current isHere one needs $\mathbf{A}=O(|\mathbf{x}{|}^{-1})$, $\nabla \mathbf{A}=O(|\mathbf{x}{|}^{-2})$ and enough localization of $\mathbf{J}$ for the stated representation and differentiations. A smooth compactly supported current density is a sufficient convenient case. The local differential equations and gauge alone do not impose these conditions: $\mathbf{A}=(-y,x,0)$, $\mathbf{B}=(0,0,2)$, $\mathbf{J}=0$ satisfy them but do not satisfy the displayed integral formula. The formula selects the decaying solution, excluding nondecaying homogeneous harmonic functions.

With $\mathbf{r}=\mathbf{x}-\mathbf{y}$, the gradient of the kernel is ${\nabla }_x(1/|\mathbf{r}|)=-\mathbf{r}/|\mathbf{r}{|}^3$. Since $\mathbf{J}(\mathbf{y})$ is independent of $\mathbf{x}$,Thus taking the curl gives the Biot-Savart law in the problem's normalization:Finally, change variables $\mathbf{z}=\mathbf{x}-\mathbf{y}$ in the integral for $\mathbf{A}$. This keeps the singular kernel independent of the differentiation variable:For smooth localized $\mathbf{J}$, differentiation under the integral sign is legitimate: $1/|\mathbf{z}|$ is locally integrable in three dimensions, and on a bounded neighbourhood of $\mathbf{x}$ the translated sources and their derivatives have common compact support. The integral indeed satisfies Coulomb gauge.

For a twice continuously differentiable vector potential $A$, write the curl with the Levi-Civita symbol:The second derivatives are symmetric in $i,j$, whereas ${ϵ}_{ijk}$ is antisymmetric, so the contraction is zero. Equivalently, writing out the three components shows that each mixed derivative cancels its reversed-order partner. ThusThis is divergence of a curl is zero. The regularity hypothesis is what licenses commutation of the mixed partial derivatives; it must not be omitted for a singular potential without specifying a weaker derivative interpretation.