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In charge-free, current-free vacuum, Maxwell's equations areTaking the curl of the third equation and using the fourth, the curl of the curl identity, and the zero divergence of the electric field givesThe magnetic flux density obeys the same wave equation. For the plane wave factor $e^{i(\omega t-kx)}$, the spatial Laplacian contributes $-k^2$ and the second time derivative contributes $-{\omega }^2$. Thusfor propagation in the positive $x$ direction. The zero-divergence conditions require $E_{0x}=B_{0x}=0$, while the curl equations require ${\mathbf{B}}_0=(k/\omega )\hat{\mathbf{x}}\times {\mathbf{E}}_0$. These conditions show that the proposed plane waves satisfy all four Maxwell's equations, rather than only their derived wave equations. Physical electric fields and magnetic fields are the real parts of the complex expressions.

A pure stationary state of a time-independent Hamiltonian operator is $\Psi (\mathbf{r},t)=\psi (\mathbf{r})e^{-iEt/\hslash }$. Its probability density and all expectations of time-independent observables are constant. This does not require zero probability current.

For real scalar potential energy and no vector potential, substitute $i\hslash \dot{\Psi }=-({\hslash }^2/2m){\nabla }^2\Psi +V\Psi$ and its complex conjugate into $\dot{\rho }={\Psi }^{\ast }\dot{\Psi }+{\dot{\Psi }}^{\ast }\Psi$. The potential terms cancel:Thus the probability current is $\mathbf{j}=(\hslash /m)\mathrm{Im}({\Psi }^{\ast }\nabla \Psi )$. In a stationary state $\nabla \cdot \mathbf{j}=0$: flow may circulate without changing the density.

Here only the azimuthal phase contributes an imaginary logarithmic derivative. Since $\nabla \varphi =\hat{\mathbf{\varphi }}/(r\sin \theta )$,There is no radial or polar flow; the azimuthal current is independent of $\varphi$, so its divergence vanishes. The apparent coordinate singularity at $\sin \theta =0$ cancels in the final expression.

The z-component of orbital angular momentum follows from the moment of the mass current:Indeed $A^2(24a_0^5)(4/3)(2\pi )=1$. This agrees with $L_z=-i\hslash {\partial }_{\varphi }$ acting on the wavefunction.

displacement current completes Ampère's law:In vacuum $\mathbf{D}={ϵ}_0\mathbf{E}$. Taking the divergence and using $\nabla \cdot \mathbf{D}={\rho }_{\mathrm{free}}$ gives $\nabla \cdot {\mathbf{J}}_{\mathrm{free}}+{\partial }_t{\rho }_{\mathrm{free}}=0$, exactly the continuity equation. Without the added term a time-dependent charge distribution would contradict the curl equation.

A charging parallel-plate capacitor illustrates the physics. For plate area $A$, $Q={ϵ}_{0}AE$, so the gap displacement current is $I_d={ϵ}_{0}A\dot{E}=\dot{Q}$, equal to the conduction current in its leads. An Ampère contour therefore gives the same magnetic circulation whether its spanning surface crosses the wire or the gap. No electrons have to cross an ideal vacuum gap. In matter, ${\partial }_t\mathbf{D}$ includes the polarization response as well as the vacuum term. Combined with Faraday induction, the displacement term permits source-free electromagnetic waves.

stress describes internal force per area. A cut with unit normal $\hat{\mathbf{n}}$ has traction $\mathbf{t}=\mathbf{\sigma }\hat{\mathbf{n}}$; its normal component is $\hat{\mathbf{n}}\cdot \mathbf{\sigma }\hat{\mathbf{n}}$ and its remainder is shear traction. force balance constrains the stress divergence, while ordinary local angular-momentum balance makes the Cauchy stress tensor symmetric in a medium without body couples.

strain describes deformation rather than force. For small displacement gradients,The antisymmetric displacement-gradient part is a rigid rotation to first order and contributes no infinitesimal strain. Engineering shear strain is ${\gamma }_{xy}=2{ϵ}_{xy}$, so it must not be confused with the tensor component. stress has units of pressure; strain is dimensionless.

An isotropic linear elastic solid obeysTwo independent moduli are required. The shear modulus $G$ and bulk modulus $K=\lambda +2G/3$ must be positive for stability. Young's modulus and Poisson ratio satisfyFor a tensile specimen with free side faces, ${\sigma }_{xx}=F/A$ gives ${ϵ}_{xx}={\sigma }_{xx}/E$ and ${ϵ}_{yy}={ϵ}_{zz}=-\nu {\sigma }_{xx}/E$. Pure shear gives $\tau =G\gamma$; hydrostatic loading gives $\Delta V/V=-p/K$. The incompressible limit has $\nu \to 1/2$ at finite shear stiffness, not vanishing shear resistance. Crystals can be anisotropic and need a fourth-rank elastic stiffness tensor instead of the isotropic two-constant law.

elastic deformation is recoverable on unloading; it need not remain linear at large strain. Beyond yield, irreversible rearrangements such as dislocation motion cause plastic strain, often followed by hardening. Ductile samples can neck before fracture; brittle failure has much less plastic deformation. Engineering stress uses initial area, while true stress uses the changing instantaneous area. A decreasing engineering stress during necking therefore need not mean that local true stress decreases. stress concentrations, flaws and loading geometry can determine failure before a uniform-material strength is reached. Viscoelastic materials also show rate dependence, creep and stress relaxation; their losses cannot be described by a static Hooke law alone.

A tensile test should measure force and gauge-length extension independently, correct machine compliance, align the grips, and control strain rate and temperature. Unloading distinguishes recoverable strain from a residual plastic extension. Separate shear or hydrostatic tests determine moduli that one axial measurement does not fix. Compression specimens need friction and buckling checks. Work per volume is $\int {\sigma }_{ij}d{ϵ}_{ij}$; the recoverable part is elastic energy, while a dissipative cycle produces a loss rather than creating energy. The original loading/unloading curve below is a chosen constitutive sketch, not measured material data.

At fixed positive pressure, the extrapolated term $C_{mp}\ln (T/T_0)$ tends to $-\infty$ as $T\to 0^{+}$. That prediction is not physically realistic for an equilibrium sample. The classical ideal-gas description with constant heat capacity cannot extend unchanged to absolute zero. A real substance generally undergoes phase changes, and quantum statistics and thermal freeze-out change its heat capacity and entropy. For a nondegenerate equilibrium ground state, the third law gives entropy approaching zero; residual ground-state degeneracy can give a finite residual value rather than this negative divergence. This is a breakdown of the extrapolation, not a violation of the third law by a real gas.

At fixed positive temperature, $-R\ln (p/p_0)$ tends to $+\infty$ as $p\to 0^{+}$. This is a physically meaningful unbounded-volume limit for a fixed mole number: $V_m=RT/p\to \infty$, so the number of accessible translational configurations grows without bound. For an ideal gas the entropy difference is $R\ln (V_{m,2}/V_{m,1})$ at fixed temperature. No finite vessel reaches an infinite entropy, and $p=0$ at finite temperature is not a finite-volume one-mole state. The divergence is approached by dilution; it does not conflict with the third law, which concerns a different temperature limit.

The microscopic magnetic response becomes a macroscopic field problem through the magnetization $\mathbf{M}$, the magnetic dipole moment per unit volume. Definein magnetostatics. Bound currents are represented by ${\mathbf{J}}_b=\nabla \times \mathbf{M}$ and surface current ${\mathbf{K}}_b=\mathbf{M}\times \hat{\mathbf{n}}$. Normal $B$ is continuous across an interface, while $\hat{\mathbf{n}}\times ({\mathbf{H}}_2-{\mathbf{H}}_1)={\mathbf{K}}_{\mathrm{free}}$. It is generally incorrect to impose continuity of every component of both fields. A linear isotropic material has $\mathbf{M}=\chi \mathbf{H}$ and ${\mu }_r=1+\chi$, but a ferromagnet usually requires a nonlinear, history-dependent constitutive relation.

Internal field differs from externally applied field. For uniform magnetization along a principal axis of an ellipsoid, the demagnetizing field is $-N_{d}M$ and $H_{\mathrm{int}}=H_{\mathrm{applied}}-N_{d}M$. A linear specimen therefore has apparent susceptibility $\chi /(1+N_d\chi )$ relative to the applied field. Shape is especially important for large susceptibility. A long closed magnetic circuit reduces stray fields; a bar magnet still has demagnetizing fields despite the absence of magnetic monopoles. Equivalent magnetic-pole descriptions are bookkeeping for magnetization divergence, not free magnetic charge.^[ref]

In paramagnetism, permanent localized moments are preferentially aligned by the field, against thermal disorder. For independent spin-one-half moments with projections $\pm \mu$, the two Zeeman energies givein the dilute weak-field limit with $B\simeq {\mu }_{0}H$. This is a Curie law; the response saturates at $n\mu$ when alignment dominates. More generally, spin $J$ gives $\chi ={\mu }_{0}n(g{\mu }_B{)}^{2}J(J+1)/(3k_{B}T)$. Interactions can produce a Curie–Weiss response $C/(T-\Theta )$ over a suitable range. Conduction-electron Pauli paramagnetism need not show the localized-moment $1/T$ law.

In diamagnetism, the field induces a magnetic moment opposing it, so the linear susceptibility is negative. The microscopic cause is the change in orbital electronic motion, and a closed-shell insulating material has no permanent atomic moment to orient. This weak response is often only mildly temperature dependent away from other transitions. It also exists as a background in materials with a stronger positive response. Perfect flux expulsion in a superconductor is a distinct collective phenomenon and should not be identified with ordinary weak atomic diamagnetism.

In ferromagnetism, exchange interactions stabilize spontaneous magnetic order below the Curie temperature. A specimen can have zero net moment while its magnetic domains remain locally magnetized in different directions. Domain-wall motion and moment rotation produce a large response; anisotropy and pinning lead to remanence, coercivity and hysteresis. Increasing field eventually approaches saturation. Heating above the ordering transition removes spontaneous order, although an applied field still induces a response. Soft magnetic cores favor easy reversal and small cyclic loss; hard permanent-magnet materials favor retained magnetization and resistance to reversal. These properties concern collective order and barriers, not merely independent moments all following a weak field.^[ref]

A practical distinction requires a controlled field sweep and a separate magnetic readout. A toroidal core with drive winding $N_1$, path length $l$ and current $I$ has approximately $H=N_{1}I/l$. A pickup winding measures ${\mathcal{E}}_2=-N_{2}AdB/dt$; integrating after calibration gives $B$, and $M=B/{\mu }_0-H$. Reversing the sweep reveals a loop whose positive loss area per volume is $\oint HdB$. An air-core/empty-holder reference, known geometry and turn count, offset correction, temperature control and slower sweeps test background, heating and rate-dependent eddy-current losses. A single force or susceptibility reading cannot establish spontaneous order or hysteresis. The original curves below are illustrative responses, not measured material data.

Choose the guide axis as $z$, with $0<x<a$ and $0<y<b$. The electric field is along $y$, parallel to the walls' $b$ dimension. Take a homogeneous lossless nonmagnetic dielectric with relative permittivity $ϵ$, ideal conducting walls, and time dependence $e^{-i\omega t}$. The lowest transverse electric mode of a rectangular waveguide isAt $x=0,a$, this tangential electric field vanishes. At $y=0,b$ the electric field is normal to the wall; a nonzero value is supported by surface electric charge and violates no conducting boundary condition. The field has zero divergence. The associated normal magnetic field also vanishes at each wall, as verified below. The fundamental sine across the wider dimension minimizes the nonzero transverse wave number; ${\mathrm{TE}}_{00}$ is not a physical propagating mode of a hollow simply connected guide.

Below the waveguide cutoff frequency $f_c=c/(2a\sqrt{ϵ})$, put $\beta =i\kappa$ with $\kappa >0$ to select decay for $z>0$. The evanescent transverse electric mode in a rectangular waveguide then hasFor the fixed frequency in the question,The decay length increases monotonically and diverges as $ϵ\to 4^{-}$; its formal limit at $ϵ\to 0^{+}$ is $a/\pi$. At $ϵ=4$ the mode is at cutoff, with zero axial wave number. For $ϵ>4$, $\beta =(\pi /a)\sqrt{ϵ/4-1}$ is real and the mode propagates without axial attenuation in this ideal lossless model. There is no real finite evanescent decay length to continue beyond cutoff. The paper's graph notation $l$ denotes the same $\ell$ as its preceding formula.

From Faraday's law of induction, $\nabla \times \tilde{\mathbf{E}}=i\omega \tilde{\mathbf{B}}$. Writing $q=\pi /a$ givesThus $B_x$ vanishes on the $x$ walls and $B_y$ vanishes on the $y$ walls, meeting the normal magnetic boundary conditions. The fields also satisfy $\nabla \cdot \mathbf{B}=0$. For real $E_0$ the physical fields are $E_y=E_0\sin (qx)e^{-\kappa z}\cos \omega t$, $B_x=-(\kappa /\omega )E_0\sin (qx)e^{-\kappa z}\sin \omega t$, and $B_z=-(q/\omega )E_0\cos (qx)e^{-\kappa z}\sin \omega t$.

The time-averaged Poynting vector isAll products are imaginary because the electric field and magnetic field are in quadrature. A single decaying evanescent wave stores reactive energy without transmitting average power into a semi-infinite lossless guide. A finite below-cutoff section can transmit power when both independent evanescent solutions are present; that is a different boundary-value problem.

The skin effect also gives exponential field decay, but in a good metal $k$ has both real and imaginary parts and there is a real inward mean Poynting vector. That flux supplies Joule heating $\sigma |\tilde{E}{|}^2/2$. Here attenuation comes from geometric cutoff in a lossless region, with no conductivity and no dissipative power sink; exponential decay alone does not establish absorption.

Maxwell's displacement current is the electric-field contribution needed to make Ampère's law consistent with charge conservation. In macroscopic form,The displacement current density ${\mathbf{J}}_d={\partial }_t\mathbf{D}$ has units $\mathrm{A}{\mathrm{m}}^{-2}$. Since $\mathbf{D}={ϵ}_0\mathbf{E}+\mathbf{P}$, it contains the vacuum field term and, in matter, the polarization-current contribution. It need not represent free charge moving through a material.

Taking the divergence gives $0=\nabla \cdot {\mathbf{J}}_f+{\partial }_t(\nabla \cdot \mathbf{D})$. Gauss's law, $\nabla \cdot \mathbf{D}={\rho }_f$, then yields the continuity equation ${\partial }_t{\rho }_f+\nabla \cdot {\mathbf{J}}_f=0$. Without the displacement term, Ampère's original law would incorrectly require every current distribution to be divergence-free even while charge accumulates.

A charging parallel-plate capacitor makes this concrete. Ignoring fringing, $D=Q/A$, so the displacement current across its dielectric gap is $I_d=A{\partial }_{t}D=dQ/dt$, equal to the conduction electric current in its wire. The integral lawtherefore gives the same answer whether the spanning surface cuts the wire or passes through the gap. Together with Faraday's law, this field term also permits self-consistent electromagnetic waves in vacuum, with speed $c=1/\sqrt{{\mu }_0{ϵ}_0}$.

At every $T>0$, $V>b$,There is no constant-pressure coexistence interval, unstable loop or finite-temperature critical inflection. The isothermal compressibility is positive:Thus this equation of state predicts no liquid-gas phase transition in its physical domain. Its divergence at the excluded-volume boundary is a repulsive constraint, not a liquid-gas coexistence mechanism. This conclusion concerns the stated model and cannot rule out an ordering transition requiring additional physics or an additional thermodynamic variable. Extrapolation to absolute zero is also outside the classical gas model's reliable regime.

Integral conservation. Electric charge cannot be created by ordinary transport. For a fixed volume $V$, outward current reduces the enclosed charge:Here ${\rho }_e$ has units $\mathrm{C}{\mathrm{m}}^{-3}$ and $\mathbf{J}$ has units $\mathrm{A}{\mathrm{m}}^{-2}$. Applying the divergence theorem to an arbitrary volume gives the local continuity equationIt is local charge conservation: a net outflow must be accompanied by depletion. In a steady state, $\nabla \cdot \mathbf{J}=0$. Integrating over a small circuit junction then gives Kirchhoff's current law. During a transient the junction can accumulate charge and the incoming and outgoing conduction currents need not yet match.

Consistency with Maxwell's equations. In vacuum, take the divergence of the Ampère-Maxwell equation:Gauss's law, $\nabla \cdot \mathbf{E}={\rho }_e/{ϵ}_0$, reproduces the continuity equation. The displacement current term makes these equations consistent for time-dependent charge distributions, such as a charging capacitor. It is not a stream of free charges across the insulating gap. In matter, $\nabla \cdot \mathbf{D}={\rho }_f$ and $\nabla \times \mathbf{H}={\mathbf{J}}_f+{\partial }_t\mathbf{D}$ give the analogous conservation law for free charge; total charge also includes bound charge.

Origin of diffusion. In a material near local thermal equilibrium, microscopic carriers transfer energy from hotter regions to colder regions. On distances long compared with their mean free path and times long compared with relaxation times, the local constitutive law is Fourier's law,Combining this temperature-gradient-driven flux with local conservation of energy produces a diffusion equation. This description assumes diffusive thermal conduction; it is not generally valid for ballistic transport, and flow would add advective transport.

Energy balance with heating. Choose any fixed volume $V$, let $c$ be its uniform volumetric heat capacity and $H$ its heat generation rate per unit volume. The rate of increase of internal energy equals heat generation minus outward heat flux:Since $V$ is arbitrary, for constant thermal conductivityThe ratio $\kappa /c$ is the thermal diffusivity in ${\mathrm{m}}^2{\mathrm{s}}^{-1}$. Here $c$ is in $\mathrm{J}{\mathrm{m}}^{-3}{\mathrm{K}}^{-1}$, $\kappa$ in $\mathrm{W}{\mathrm{m}}^{-1}{\mathrm{K}}^{-1}$ and $H$ in $\mathrm{W}{\mathrm{m}}^{-3}$, so both right-hand terms have units $\mathrm{K}{\mathrm{s}}^{-1}$. If thermal conductivity varies with position or temperature, the divergence $\nabla \cdot (\kappa \nabla T)$ must be retained rather than replacing it by $\kappa {\nabla }^{2}T$.

In a homogeneous linear medium, write $\mathbf{D}=ϵ\mathbf{E}$ and $\mathbf{B}=\mu \mathbf{H}$. Without free charge or current, Maxwell's equations becomeTaking the curl of Faraday's law, using $\nabla \times (\nabla \times \mathbf{E})=\nabla (\nabla \cdot \mathbf{E})-{\nabla }^2\mathbf{E}$, givesTheir plane-wave solutions have ${\omega }^2=k^2/(\mu ϵ)$ and speed $v=1/\sqrt{\mu ϵ}$; in vacuum this is $c=1/\sqrt{{\mu }_0{ϵ}_0}$. For a wave $\mathrm{Re}[{\mathbf{E}}_0{e}^{i(\mathbf{k}\cdot \mathbf{r}-\omega t)}]$, the divergence equations give $\mathbf{k}\cdot {\mathbf{E}}_0=\mathbf{k}\cdot {\mathbf{B}}_0=0$, while Faraday's law gives ${\mathbf{B}}_0=\mathbf{k}\times {\mathbf{E}}_0/\omega$. Thus the fields and propagation direction form an orthogonal triad, with $B_0=E_0/v$.

The two independent transverse electric components allow linear, circular and elliptical polarization of an electromagnetic wave. Their relative amplitudes and phase determine the polarization state. In a lossless nondispersive medium the energy density and energy flux areA travelling plane wave has equal electric and magnetic energy contributions and time-averaged intensity $⟨S⟩=E_0^2/(2Z)$, where $Z=\sqrt{\mu /ϵ}$ is its wave impedance and $E_0$ denotes peak electric-field amplitude. Reflection and refraction follow by imposing the electromagnetic boundary conditions. Frequency-dependent constitutive response introduces dispersion; conductivity introduces attenuation, so the simple real constant-speed result has a specified domain of validity.

Use the divergence identity for a cross product, followed by Faraday's law of induction and the Ampère-Maxwell equation:With the electromagnetic energy density of part (b), these giveIntegrate over a fixed volume $V$ and apply the divergence theorem, choosing $d\mathbf{S}$ outward:This is Poynting's theorem. The outward electromagnetic energy flux equals minus the power transferred to matter and minus the rate at which electromagnetic energy accumulates inside. For the dissipative electrical conductor in this question, the matter term is Ohmic heating.

In SI units, with microscopic charge density ${\rho }_e$ and current density $\mathbf{J}$, Maxwell's equations can be writtenThe first relates electric flux to charge. The second states that the net magnetic flux through a closed surface is zero. Faraday's law relates electric circulation to changing magnetic flux. The Ampère-Maxwell equation includes both conduction current and displacement current; the latter is essential to time-dependent fields and local charge conservation.

For fixed contours and spanning surfaces, the integral curl equations are $\oint \mathbf{E}\cdot d\mathbf{l}=-d{\Phi }_B/dt$ and $\oint \mathbf{B}\cdot d\mathbf{l}={\mu }_0{I}_{\mathrm{enc}}+{\mu }_0{\epsilon }_{0}d{\Phi }_E/dt$. For a moving material contour the first must include the motional force term; its total emf is $\oint (\mathbf{E}+{\mathbf{v}}_{\mathrm{loop}}\times \mathbf{B})\cdot d\mathbf{l}$. This prevents confusing a changing circuit area with a partial time derivative of a static magnetic field.

Taking the divergence of the Ampère-Maxwell equation and using electric Gauss's law gives ${\partial }_t{\rho }_e+\nabla \cdot \mathbf{J}=0$. In a source-free vacuum, combining the two curl equations gives ${\nabla }^2\mathbf{E}-{\mu }_0{\epsilon }_0{\partial }_t^2\mathbf{E}=0$, and similarly for $\mathbf{B}$, predicting electromagnetic waves of speed $c=1/\sqrt{{\mu }_0{\epsilon }_0}$. The Lorentz force law then specifies how these fields act on a charge. Macroscopic matter descriptions may instead use $\mathbf{D}$ and $\mathbf{H}$ with free sources and constitutive relations; those should not be silently mixed with the microscopic-source equations above.

In the northern hemisphere the coast is on the wave's right when looking in its direction of propagation. In the ocean-$x<0$ convention above it travels northward, with land east of the flow; for the opposite ocean side it travels southward. This follows from combining $g{\eta }_x=f_{0}v$ with a sea-level bulge that decreases offshore, rather than choosing propagation independently of trapping.

For a positive northward bulge, $v=(c/H)\eta$ is northward and strongest at its crest, and there is no cross-shore velocity in the ideal mode. Ahead of the crest the current decreases northwards, so $v_y<0$: horizontal flow convergence increases sea level, ${\eta }_t=-H{v}_y>0$. Behind the crest $v_y>0$: divergence lowers sea level. At the crest itself the alongshore derivative and instantaneous horizontal flow convergence vanish. These signs reverse appropriately for a trough or the opposite coast, while the trapping rule remains the same.

The first panel below sketches a positive bulge, alongshore velocity and the leading convergence/trailing divergence regions. Its cross-shore decay is the pressure-gradient/Coriolis balance, not a current directed into the coast.

Use a shallow, homogeneous active upper layer of mean thickness $h_0$ and mass density ${\rho }_1$, over a much deeper, effectively stationary layer with mass density ${\rho }_2>{\rho }_1$. The Boussinesq approximation keeps the small density contrast only in buoyancy, giving reduced gravity $g^{\prime }=g({\rho }_2-{\rho }_1)/{\rho }_0$. The hydrostatic approximation and shallow-water approximation make horizontal velocity nearly depth independent and turn the interface slope into the horizontal force $-g^{\prime }\nabla h$.

Linearize about rest and uniform $h_0$, with $|h-h_0|\ll h_0$, neglecting products of perturbations and convective acceleration. The f-plane approximation sets the Coriolis parameter to constant $f_0$. Neglect wind forcing, viscosity, entrainment and bottom drag. The two horizontal momentum equations then balance local acceleration, the Coriolis force and the pressure gradient. The signs $-f_{0}v$ and $+f_{0}u$ follow from eastward and northward coordinates.

The thickness equation is the linearized conservation of mass law: horizontal divergence decreases $h$, while horizontal convergence increases it. Here $h$ is active-layer thickness, not total ocean depth, and $g^{\prime }$ rather than $g$ gives the restoring force because the motion principally displaces the thermocline.

Take the curl of the Euler equations for an inviscid fluid. The gradient on the right has zero curl. Use the stated vector identity with $\mathbf{F}=\mathbf{\omega }$, $\mathbf{G}=\mathbf{u}$:incompressible flow gives $\nabla \cdot \mathbf{u}=0$, while a curl always has zero divergence, so $\nabla \cdot \mathbf{\omega }=0$. The remaining terms giveFor a two-dimensional velocity independent of $z$, $\mathbf{\omega }=(0,0,v_x-u_y)$ and $(\mathbf{\omega }\cdot \nabla )\mathbf{u}={\omega }_z{\partial }_z\mathbf{u}=0$. Thus the scalar vorticity is conserved following each fluid particle. It can still vary from particle to particle; this is advection, not a claim of spatially uniform vorticity.

Use the two-dimensional streamfunction convention $u={\psi }_y$, $v=-{\psi }_x$. HereThe divergence is $u_x+v_y=-\cos x\sin y+\cos x\sin y=0$, so the flow is incompressible.

In vacuum, the source-free Maxwell equations arePut $\psi =\omega (t-z/c)$ and denote the constant electric amplitude by $E_0$. A corresponding traveling-wave magnetic field isBoth divergences vanish because both the electric field and the magnetic field depend only on $z,t$ and have zero $z$ component. Direct differentiation gives $\nabla \times \mathbf{E}=(\omega E_0/c)(\cos \psi ,\sin \psi ,0)=-{\partial }_t\mathbf{B}$ and $\nabla \times \mathbf{B}=(\omega E_0/c^2)(-\sin \psi ,\cos \psi ,0)=c^{-2}{\partial }_t\mathbf{E}$. A time-independent background magnetic field satisfying the homogeneous equations could be added; the formula is the wave component.

Introduce the magnetic vector potential and electric potential by $\mathbf{B}=\nabla \times \mathbf{A}$ and $\mathbf{E}=-\nabla \varphi -{\partial }_t\mathbf{A}$. These identities automatically give the divergence equation for $\mathbf{B}$ and Faraday's law. For this wave choose $\varphi =0$ and $\mathbf{A}=(c/\omega )\mathbf{B}$; the derivatives just calculated verify both potential equations.

The electromagnetic force is the Lorentz force, $Q(\mathbf{E}+\mathbf{v}\times \mathbf{B})$. On the smooth constraint plane $z=z_0$, the constraint force is normal to the plane. Both $\mathbf{v}$ and $\mathbf{B}$ are tangential, so $\mathbf{v}\times \mathbf{B}$ is normal and is balanced by that constraint force; it does not affect the tangential motion. ThusA particular motion is ${\mathbf{r}}_{\parallel }={\mathbf{r}}_0-[Q{E}_0/(m{\omega }^2)](\cos \psi ,\sin \psi )$, a circle of radiusThe general planar motion also includes a constant drift velocity; the particle moves on a circle when that drift is zero. The constraint force can do no work because it is perpendicular to the velocity.