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The Aharonov-Bohm effect is the phase $q\Phi /\hslash$ acquired when a charge $q$ encircles magnetic flux $\Phi$, even when the magnetic field vanishes along the charge's path.

In the magnetic Aharonov-Bohm effect, a coherent electron wave takes two routes around inaccessible magnetic flux. Recombination converts their relative wave phase into displaced interference fringes. In 1959, Yakir Aharonov and David Bohm proposed a shielded solenoid geometry and identified the obstruction to removing the magnetic vector potential throughout the accessible region.^[ref]

Use International System of Units, electric charge $q$, and magnetic minimal coupling $H=(-i\hslash \nabla -q\mathbf{A}{)}^2/(2m)$. Let ${\gamma }_1,{\gamma }_2$ run from the same source $S$ to the same detector point $D$. The oriented closed curve $C$ follows ${\gamma }_1$ and then ${\gamma }_2$ backwards. For a static magnetic vector potential, the magnetic contribution to the relative wave phase isHere ${\Phi }_C={\int }_{\Sigma }\mathbf{B}\cdot d\mathbf{S}$ is the oriented magnetic flux through a spanning surface. Equal or calibrated nonmagnetic propagation phases are assumed. For an electron, $q=-e$, where $e>0$ is the elementary charge; reversing $C$ reverses the sign. The observable is the gauge-invariant holonomy $e^{i\delta }$, rather than a unique local value of the magnetic vector potential.

The accessible region is a multiply connected region: $C$ cannot contract to a point without entering the excluded flux region. Stokes theorem does not set the line integral to zero, because a spanning surface must cross that excluded region, where the magnetic field need not vanish. There is real magnetic flux in the apparatus; only the accessible paths are idealized as free of the magnetic field. The exterior of a finite straight solenoid also permits routes around its ends, so its exact topology and leakage differ from those of an infinite solenoid or an excluded solid torus.

An explicit model makes the obstruction visible. Outside an infinite flux tube of radius $a$, choose cylindrical coordinates $(r,\theta ,z)$ andIts curl vanishes there, but a loop with winding number $n$ has ${\oint }_C\mathbf{A}\cdot d\mathbf{l}=n\Phi$. Locally $\mathbf{A}=\nabla (\Phi \theta /(2\pi ))$; the angular coordinate changes by $2\pi$ on a circuit. For generic $q\Phi /\hslash$, that candidate gauge removal changes the wave function's single-valued boundary condition. On a simply connected region, a smooth static magnetic vector potential with zero curl instead has a globally defined scalar primitive and is a pure gauge potential. Thus local zero magnetic field and globally removable magnetic vector potential are different conditions.

For an original interference example, take mutually coherent scalar amplitudes with constant individual intensities $I_1,I_2$ and an imposed carrier phase $ky$, as in off-axis holography. ThenThe interference fringe spacing is $d=2\pi /k$. An electron enclosing $\Phi =h/(4e)$ has $\delta =-\pi /2$, shifting maxima by $d/4$ for this orientation. Increasing the magnetic flux by $h/e$ changes $\delta$ by $-2\pi$ and restores the same pattern: the electron Aharonov-Bohm effect has magnetic flux quantum $h/e$. This uses $h=2\pi \hslash$, the Planck constant. It does not require the source's magnetic flux to vary continuously during a single exposure.

For $I_1=1$, $I_2=1/4$, the extrema are $I_{\max }=9/4$, $I_{\min }=1/4$, and the interferometric visibility is $4/5$. A fixed Aharonov-Bohm phase translates these fringes without changing their contrast. Averaging unresolved fluctuating phases can reduce the contrast; a darkened pattern alone cannot identify enclosed magnetic flux. These formulas assume coherence, indistinguishable alternatives and stable beam envelopes.

The appeal of the Aharonov-Bohm effect is that a local Lorentz force calculation along either ideal path misses information carried by their global arrangement. It gives a physical example of holonomy and supplies the encircling phase used in Chern--Simons flux attachment. This alone does not establish intrinsic topological order. The static comparison also matters: changing magnetic flux can induce an electric field, so a time-dependent experiment requires the scalar-potential contribution as well as the displayed spatial line integral.

Attosecond streaking uses a synchronized reference electric field to change a photoelectron's outgoing momentum. Scanning relative delay produces an energy-versus-delay distribution. A short emission interval samples only a small part of the reference cycle; a longer interval samples a range of kicks and can broaden or reshape the distribution.

For an explicit classical limit, use the electric dipole approximation and choose the magnetic vector potential $A(t)$ along the detection axis with $E(t)=-dA/dt$ and $A(\infty )=0$. Neglect the residual ion's Coulomb force. An electron released at $t_i$ with mechanical momentum $p_0$ obeysIndeed, integrating $\dot{p}=-eE=e\dot{A}$ from release to the end of the field gives the stated minus sign. The useful clock variable here is $A$, rather than the instantaneous $E$ itself. If $A$ is nearly linear across a narrow emission window with temporal standard deviation ${\sigma }_t$, and initial momentum is sharp, thenThis local mapping fails near a zero of $E$, where first-order timing sensitivity vanishes. A real inversion must include the incident bandwidth, initial photoelectron distribution, collection angles and target–field response.

For an original dimensionless example in atomic units, choose $p_0=2$, $m=e=1$ and $A(t_i)=0.10$. Then $p_f=1.9$ and $K_f=1.805$, compared with initial kinetic energy $2$. Reversing $A$ gives $K_f=2.205$: the shifts are asymmetric because of the $A^2/2$ term. A measurement that keeps only the term linear in $A$ would miss $0.005$ in both cases.

For an axisymmetric magnetic vector potential, the conserved angular canonical momentum is $p_{\theta }=\gamma m{\rho }^2\dot{\theta }+q\rho A_{\theta }$. A particle arriving parallel to a solenoid axis from a field-free region has $p_{\theta }=0$. In a uniform field, $A_{\theta }=B\rho /2$ and $\dot{\theta }=-qB/(2\gamma m)$ wherever $\rho \ne 0$. The mechanical angular momentum changes through the fringe while the canonical quantity remains constant. Static magnetic fields also conserve $\gamma m{c}^2$; the azimuthal kick therefore reduces axial kinetic motion rather than increasing total speed.

The Coulomb gauge imposeson the magnetic vector potential. A gauge transformation $A\mapsto A+\nabla \chi$ reaches this gauge when $\chi$ solves

For an ideal infinitely long solenoid with interior magnetic field $B_0\hat{\mathbf{z}}$ and radius $R$, an axisymmetric Coulomb gauge representative of the magnetic vector potential isThis follows from $2\pi r{A}_{\varphi }=\Phi (r)$ by Stokes theorem. The exterior has zero curl but nonzero circulation around the enclosed magnetic flux. The Coulomb gauge condition alone does not eliminate every residual gauge transformation; symmetry and regularity select this customary representative.

In a smooth interior region with locally constant shear ${\Omega }^{\prime }$, a wound azimuthal Fourier mode has radial wavenumber $k_s\simeq -m{\Omega }^{\prime }t$. Integrating its resistive damping rate $\eta k_s^2$ gives the exponential decay in the title. The characteristic time is ${\tau }_c=[3/(\eta m^2{\Omega }^{\prime 2}){]}^{1/3}$. For $m=1$ and ${\Omega }^{\prime }=-{\Omega }_0/d$, ${\Omega }_0{\tau }_c=(3\mathrm{Rm}{)}^{1/3}$, where $\mathrm{Rm}={\Omega }_0{d}^2/\eta$. This is a high-magnetic Reynolds number interior transient on a time long compared with the rotation time but short compared with $d^2/\eta$. Physical magnetic field components may have additional algebraic factors from taking derivatives of the wound magnetic vector potential. A maintained exterior field, an axial core and a circulation-edge boundary layer prevent interpreting this estimate as an exact, uniform, infinite-time solution.

Current helicity correlates a magnetic field with its curl, which is proportional to electric current density. It differs from magnetic helicity, which uses a magnetic vector potential. For a solenoidal monochromatic field of wavenumber $k$ in a periodic domain, choose ${\mathbf{a}}_b=\nabla \times \mathbf{b}/k^2$. Then $\nabla \times {\mathbf{a}}_b=\mathbf{b}$ and $H_c=k^2⟨{\mathbf{a}}_b\cdot \mathbf{b}⟩$.

In electrostatics, the electric potential $\Phi$ tells us the interaction potential energy per unit test electric charge in the electric field of prescribed sources. Its most directly useful quantity is an electric potential difference: how much work per unit electric charge an external agent must supply to move that test electric charge slowly between two points. This replaces the bookkeeping of a vector force along a path with the difference of two scalar values. The test electric charge must be small enough not to rearrange the sources appreciably.

For motion from $A$ to $B$, write $\Delta \Phi =\Phi (B)-\Phi (A)$. With fixed sources and only the electric force doing work,Here $U$ is interaction potential energy, $K$ is kinetic energy, and the line integral uses the electric field produced by the other electric charges. For slow transport with unchanged kinetic energy, the external work is $W_{\mathrm{ext}}=\Delta U$, so $\Delta \Phi =W_{\mathrm{ext}}/q$ for nonzero $q$. This external work has the opposite sign to the work done by the electric field. The SI unit is the volt,A volt measures energy per unit electric charge, not energy itself. For example, crossing an electric potential difference of $+5\mathrm{V}$ changes the potential energy of $q=+2\mathrm{nC}$ by $+10\mathrm{nJ}$, but changes that of $q=-2\mathrm{nC}$ by $-10\mathrm{nJ}$. Both test electric charges encounter the same electric potential difference; their opposite signs reverse the energy change.^[ref]

The electrostatic potential is a scalar function of position. The electric field is a conservative vector field, and all three components follow from the negative gradient of the electric potential:The electric field points toward decreasing electric potential. A positive test electric charge feels electric force in that direction; a negative test electric charge feels electric force in the opposite direction, since $\mathbf{F}=q\mathbf{E}$. A freely accelerated negative test electric charge therefore moves toward higher electric potential while lowering its potential energy $U=q\Phi$.^[ref]

Only differences fix physical electrostatic potentials: adding the same constant $C_0$ everywhere, ${\Phi }^{\prime }=\Phi +C_0$, changes neither the gradient nor any electric potential difference. Choosing an electrical reference potential is this elementary gauge transformation. Setting zero at infinity is convenient for finitely many localized point charges; it is not available for an ideal uniform electric field extending to infinity. There we can choose a finite reference point. Assigning zero at ground is a convention, whereas physically grounding an electrical conductor also allows exchange of electric charge to impose a boundary value. Zero electric potential at a point never by itself implies zero electric field.^[ref]

For a fixed point charge $Q$ at the origin in vacuum, choose zero at infinity. Coulomb's law giveswhere ${\epsilon }_0$ is the vacuum permittivity. Use the stipulated approximation $k=8.99\times 10^9\mathrm{N}{\mathrm{m}}^2/{\mathrm{C}}^2$ and $Q=+2\mathrm{nC}$. At $r=0.20\mathrm{m}$,A test electric charge $q=-3\mathrm{nC}$ has $U=-2.697\times 10^{-7}\mathrm{J}$ and radial electric force $F_r=-1.3485\times 10^{-6}\mathrm{N}$, directed inward. Moving it inward from $r_A=0.20\mathrm{m}$ to $r_B=0.10\mathrm{m}$ raises the electric potential from $89.9$ to $179.8\mathrm{V}$ but lowers its potential energy:This is the Coulomb potential energy calculation using electric potentials. The source's singular self-field at $r=0$ is excluded; when finding the force on a source point charge, use only the electric field of the other sources.^[ref]

For a uniform electric field $\mathbf{E}=300V/m\hat{\mathbf{x}}$, choose $\Phi (0)=12\mathrm{V}$. ThenMoving from $x_A=0$ to $x_B=0.04\mathrm{m}$ gives $\Delta \Phi =-12\mathrm{V}$. For $q=+2\mathrm{nC}$ the change in potential energy is $-24\mathrm{nJ}$ and the electric field does $+24\mathrm{nJ}$ of work. For $q=-2\mathrm{nC}$, the same prescribed displacement gives $\Delta U=+24\mathrm{nJ}$: the external agent must supply that work for slow transport. This uniform electric field approximates the interior of a parallel-plate capacitor sufficiently far from the edges; the fringing field of a capacitor limits the approximation.^[ref]

An equipotential is a level set $\Phi =\text{constant}$. On a smooth equipotential with nonzero gradient, every tangent displacement has $d\Phi =\nabla \Phi \cdot d\mathbf{l}=0$, so the electric field is normal to the equipotential. Thus moving a test electric charge along an equipotential requires no net electrical work. Around a point charge, the equipotentials are spheres; the circles in the diagram are their sections, not an assumption of two-dimensional Coulomb's law. A uniform electric field has parallel plane equipotentials. For contours separated by the same small electric potential difference, smaller normal separation means a stronger electric field, approximately $|\mathbf{E}|\approx |\Delta \Phi |/\Delta n$. At a point where $\nabla \Phi =0$, the electric field is zero and has no direction to call normal.^[ref]

The superposition principle adds electric potentials as signed scalars. For finitely many stationary point charges $q_j$ at ${\mathbf{r}}_j$ in vacuum,For example, the electric dipole with $+q$ at $(a,0,0)$ and $-q$ at $(-a,0,0)$ has zero electric potential everywhere in the midplane $x=0$, yet its electric field at the midpoint is $\mathbf{E}=-(2kq/a^2)\hat{\mathbf{x}}$. With $q=2\mathrm{nC}$ and $a=0.10\mathrm{m}$, this is $-3596N/C\hat{\mathbf{x}}$. At $(2a,0,0)$, the same model gives $\Phi =2kq/(3a)\approx 119.867\mathrm{V}$ and $E_x=8kq/(9a^2)\approx 1598.222N/C$. Cancellation of electric potentials is quite different from cancellation of vector electric fields.

A motivating measurement is conductive-paper electric potential mapping. A documented teaching experiment uses slightly conductive paper on an insulating board, two circular electrodes in highly conducting ink, a steady supply, and a digital voltmeter. Connect the supply to the two electrodes, keep the voltmeter common lead on the negative electrode, and use its other lead as a movable probe. Adjust the supply using the measured electric potential difference between electrodes; the published procedure uses $10\mathrm{V}$. The observable is the voltmeter reading $\Phi (P)-\Phi (\mathrm{COM})$ at the probe position $P$. Locate several points reading $2\mathrm{V}$, record their positions on a separate map, and join them into an equipotential; repeat at other readings. Lift the probe between contacts so it does not damage the paper. This procedure makes the electrical reference potential concrete: every reading compares two contacts.^[ref]

From neighboring readings infer an electric field component by a finite difference, $E_x\approx -[\Phi (x+h)-\Phi (x)]/h$. Reconstruct two in-plane components with displacements along two independent directions. Alternatively, draw the electric field normal to mapped equipotentials, toward lower electric potential. The second diagram illustrates the same measurement principle with bar electrodes spanning a uniform sheet, insulating top and bottom edges, and a synthetic model $\Phi (x)=(50V/m)x$. The model readings $4.0\mathrm{V}$ at $x=0.08\mathrm{m}$ and $5.0\mathrm{V}$ at $x=0.10\mathrm{m}$ imply $E_x=-50V/m$. These values are calculated illustrations, not reported experimental data. The documented circular-electrode experiment similarly estimates the electric field from readings at successive centimetre marks.^[ref]

Useful controls include checking the voltmeter zero, checking that the electric potential difference along each painted electrode is negligible, and repeating reference readings to detect supply drift. Reversing the supply should reverse the signs relative to the same reference contact. Contact resistance, leakage electric currents, and touching the sheet can distort the map. A voltmeter has finite input electrical resistance and draws a small electric current; voltmeter loading changes the quantity being measured if that electrical resistance is insufficiently large relative to the sheet's effective source electrical resistance.^[ref][ref]

The paper carries a steady electric current, so this is a planar analogue rather than a direct measurement of isolated point charges in three-dimensional vacuum. For constant electrical conductivity, Ohm's law and steady charge conservation give $\mathbf{j}=-\sigma \nabla \Phi$ and $\nabla \cdot \mathbf{j}=0$, hence the two-dimensional Laplace equation away from electrodes. Insulating sheet edges impose zero normal electric current, while finite electrodes fix boundary values. This derivation explains both the analogy and its limits: inhomogeneous paper, finite boundaries, and finite electrodes can change the contours. One should not fit the three-dimensional point charge formula $kQ/r$ to a planar map without testing that model. Small probe spacing resolves spatial variation better, but amplifies uncertainty in the inferred finite difference. The published laboratory instructions explicitly ask students to assess whether their proposed field fit is appropriate.^[ref][ref]

The scalar description has a crucial limit. Faraday's law of induction for a stationary closed path $\Gamma$ sayswhere ${\Phi }_B$ is magnetic flux. Nonzero changing magnetic flux gives nonzero circulation, incompatible with a single-valued electric potential alone. More generally, a scalar electric potential and a magnetic vector potential together giveConsequently, in the presence of induction, two voltmeter lead routes between the same contacts can produce different readings if the enclosed magnetic flux changes. This follows by subtracting the two path line integrals; it is a deduction from Faraday's law of induction, not a claimed outcome of the paper experiment. The steady setup and negligible induction justify interpreting its readings as electric potential differences. Merely introducing a scalar electric potential that depends on time cannot supply a nonzero curl of the electric field.^[ref]

A short mathematical justification connects the statements above. Along a smooth path, the chain rule gives $d\Phi =\nabla \Phi \cdot d\mathbf{l}$. Integrating and using $\mathbf{E}=-\nabla \Phi$ yields the endpoint formula by the fundamental theorem for line integrals, proving path independence. Conversely, vanishing circulation around every closed path lets us define $\Phi (P)=-{\int }_{P_0}^P\mathbf{E}\cdot d\mathbf{l}$ independently of the route, with $\Phi (P_0)=0$. Zero curl is a local test; for a continuously differentiable curl-free vector field, a simply connected open region suffices for a global potential. The explicit finite-source formula already gives a global electric potential away from its source points.^[ref]

The four-potential combines the electric potential and magnetic vector potential as $A^{\mu }=(\varphi /c,\mathbf{A})$. The electromagnetic four-potential $A_{\mu }$ transforms underwhile the antisymmetric field tensor $F_{\mu \nu }={\partial }_{\mu }{A}_{\nu }-{\partial }_{\nu }{A}_{\mu }$ is unchanged.

Two scalar labels constant along a magnetic field line are Euler potentials when their crossed gradients give the magnetic field. The magnetic vector potential $\mathbf{A}=\alpha \nabla \beta$ proves that this representation is a solenoidal vector field. Conversely, two independent local line labels yield $\mathbf{B}=f(\alpha ,\beta )\nabla \alpha \times \nabla \beta$; replacing $\beta$ by ${\int }^{\beta }f(\alpha ,b)db$ absorbs the factor. This is a local construction; a global regular single-valued pair requires compatible magnetic field line topology. Fixing boundary labels preserves their paired-footpoint connections in an ideal variation.

Far outside a localized steady source of size $a$, at distance $r\gg a$, its multipole expansion starts with the magnetic dipole moment if that moment is nonzero. In vacuum, using SI units,where ${\mu }_0$ is the vacuum permeability, $\hat{\mathbf{r}}=\mathbf{r}/r$, and the magnetic field is the curl of the magnetic vector potential. The magnetic dipole field falls as $r^{-3}$, and the magnetic vector potential as $r^{-2}$. The localized steady-current moment identities remove the putative $r^{-1}$ term in the magnetic vector potential. No magnetic monopole is needed. If different current loops cancel the total magnetic dipole moment, a higher term of the multipole expansion supplies the leading distant magnetic field. ^[ref].

The approximation is not valid at the center of a finite current loop. For a circular current loop of radius $R$ and electric current $I$, the exact on-axis magnetic field of a circular current loop and the distant approximation areTheir ratio is $B_z/B_{z,\mathrm{dip}}=(1+R^2/z^2{)}^{-3/2}$: at $|z|=10R$, the approximate magnetic field exceeds the exact magnetic field by about $1.50\%$ relative to the exact value. Near the source, retain its geometry rather than replacing it with a point source specified only by its magnetic dipole moment.

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.

For a free particle of mass $m>0$ in nonrelativistic quantum mechanics, choose the potential energy zero and writeApplying the momentum operator $-i\hslash \nabla$ returns $\mathbf{p}\psi$; this fixes the sign convention. The angular frequency is measured relative to this chosen energy zero. Adding a constant energy changes only a time-dependent global phase, leaving the spatial de Broglie wavelength unchanged. No rest-energy term is included in this nonrelativistic quantum mechanics convention.

The plane wave has $|\psi {|}^2=|A{|}^2$ and probability current $\mathbf{j}=(\mathbf{p}/m)|A{|}^2$. Its real part can be drawn as a sine or cosine, but that graph is a component of a probability amplitude, not a transverse displacement. On infinite space this is a generalized momentum eigenstate, not a state satisfying wavefunction normalization. A narrow wave packet instead moves with group velocity ${\nabla }_{\mathbf{k}}\omega =\mathbf{p}/m$; the phase velocity $\omega /|\mathbf{k}|=|\mathbf{p}|/(2m)$ is not its transport speed.

The equality of canonical momentum and kinetic momentum here assumes a field-free region with the magnetic vector potential chosen zero. For a charged particle in a magnetic field, the kinetic momentum is $\mathbf{\Pi }=\mathbf{p}-q\mathbf{A}$. An arbitrary canonical phase gradient must not then be substituted into $m\mathbf{v}$; the free particle formula needs its stated assumptions.

An open flux tube carries magnetic flux through its end caps. Its ordinary magnetic helicity can therefore change under a single-valued gauge transformation. For $\mathbf{B}=aR{\mathbf{e}}_{\varphi }+b{\mathbf{e}}_z$ inside $R<R_0$ and zero outside, a regular family of magnetic vector potentials is $A_{\varphi }=bR/2$, $A_z=C-a{R}^2/2$ inside and $A_{\varphi }=b{R}_0^2/(2R)$, $A_z=C-a{R}_0^2/2$ outside. Their tangential components agree at $R_0$. Inside, $\mathbf{A}\cdot \mathbf{B}=bC$, so the helicity per unit length is $\pi bC{R}_0^2$. Changing $C$ is the gauge transformation $\Lambda =\Delta Cz$, and changes this result whenever $b\ne 0$. The convention $A_z=0$ outside gives $C=a{R}_0^2/2$ and helicity per unit length $\pi ab{R}_0^4/2$, while $C=0$ gives zero. Without an end-cap or gauge convention the ordinary helicity is not unique.

For a localized quantum harmonic oscillator along $z$, let the transverse magnetic vector potential be $A_z(t)=2A_0\cos (\theta -\omega t)$ for $t>0$, with zero potential before the switch. The Coulomb gauge form of minimal coupling gives $H_1=e{A}_z{p}_z/m$. The creation operator matrix element $⟨1|p_z|0⟩=i\sqrt{m\hslash {\omega }_0/2}$ means that only $|1⟩$ is reached at first order from the ground state. Its interaction picture amplitude isIn particular $F_0=t$, so resonant finite-time expressions are nonsingular. The induced electric dipole moment is $-2e\sqrt{\hslash /(2m{\omega }_0)}\mathrm{Re}(e^{-i{\omega }_{0}t}{b}_1)$. It contains a driven component with the dynamic electric polarizability and free-frequency transients required by the initial conditions. The discontinuity in $A_z$ gives an impulsive electric field and an initial mechanical velocity $e{A}_z(0^{+})/m$ even though the initial canonical momentum has zero mean.

For a transverse monochromatic electromagnetic wave with angular frequency $\omega \ne 0$, ${\partial }_t^2\mathbf{E}=-{\omega }^2\mathbf{E}$. The displayed electric potential and magnetic vector potential therefore give $-\nabla \Phi -{\partial }_t\mathbf{A}=\mathbf{E}$. Transversality gives $\nabla \cdot \mathbf{A}=0$, so the Lorenz gauge holds. For $\mathbf{E}=E_0(\cos \theta ,\sin \theta ,0)$ and $\theta =\omega t-Kz$, taking a curl yields $\mathbf{B}=(K/\omega )\hat{\mathbf{z}}\times \mathbf{E}$. The vacuum Maxwell equations require $K^2={\omega }^2/c^2$. The field has circular polarization, with $|\mathbf{B}|=|\mathbf{E}|/c$ and perpendicular electric and magnetic directions.

In a bounded regular domain, fix the global extension or circulation convention for the magnetic vector potential. A bounded inverse of curl on tangent solenoidal vector fields gives $\Vert \mathbf{A}{\Vert }_2\le C_D\Vert \mathbf{B}{\Vert }_2$. The Cauchy-Schwarz inequality then gives $|H_M|\le C_D\Vert \mathbf{B}{\Vert }_2^2=2{\mu }_0{C}_D{E}_B$. One construction extends the tangent field by zero outside the domain and uses the whole-space Biot-Savart law; its kernel of size $|\mathbf{x}-\mathbf{y}{|}^{-2}$ is integrable over a bounded region, giving the required $L^2$ bound. Nonzero conserved magnetic helicity therefore prevents complete decay of magnetic energy.

The magnetic helicity integrates the dot product of a magnetic vector potential and its magnetic field. It measures linkage and twisting of magnetic flux. For a solenoidal vector field $\mathbf{B}$ tangent to the boundary, it is unchanged by the gauge transformation $\mathbf{A}\mapsto \mathbf{A}+\nabla \chi$, since the change is ${\int }_V\nabla \cdot (\chi \mathbf{B})dV={\int }_{\partial V}\chi \mathbf{B}\cdot d\mathbf{S}=0$. If its density satisfies ${\partial }_{t}h=\nabla \cdot (f\mathbf{B}-h\mathbf{u})$ with both vector fields tangent to the boundary of a fixed volume, the divergence theorem gives conservation of its integral. This is an instance of helicity conservation by tangent boundary conditions.

Choose a magnetic vector potential with $\mathbf{B}=\nabla \times \mathbf{A}$. The resistive induction equation permits the evolution equationA gauge transformation changes $H={\int }_V\mathbf{A}\cdot \mathbf{B}dV$ by a surface integral of the normal magnetic field, which is zero here. Thus the integrated magnetic helicity has gauge invariance under the stated boundary conditions.

Differentiate both factors in $H$. Use the stated vanishing-boundary assumption to remove the helicity surface flux; this can be implemented in the usual closed-boundary gauge $\mathbf{n}\times \mathbf{A}=0$, or by extending the localized field and integrating over all space with sufficient decay. The curl integration by parts identity then givesThe motional term is zero pointwise because $(\mathbf{u}\times \mathbf{B})\cdot \mathbf{B}=0$. The scalar-potential term integrates to zero by $\nabla \cdot \mathbf{B}=0$ and the boundary condition. The resistive magnetic-helicity balance is thereforeUnlike the resistive magnetic energy contribution, this signed integral need not have a fixed sign.

The components of the momentum operator do not generally commute with a position-dependent magnetic vector potential. Expanding minimal coupling with the ordering intact givesActing on a wavefunction shows that $\mathbf{p}\cdot \mathbf{A}-\mathbf{A}\cdot \mathbf{p}=-i\hslash \nabla \cdot \mathbf{A}$. In the Coulomb gauge the divergence vanishes, so the two cross terms agree. Discarding the quadratic term in the weak radiation amplitude givesThe equality of the two operator orderings follows from the gauge condition, not from treating $\mathbf{p}$ and $\mathbf{A}$ as commuting numbers component by component.