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Use two plane waves, $|\mathbf{k}⟩$ and $|\mathbf{k}-\mathbf{G}⟩$, as a basis near their Bragg degeneracy. The potential has Fourier coefficients $V_{\pm \mathbf{G}}$; its zero Fourier coefficient vanishes. All other plane waves are off resonance and are neglected in the leading nearly-free electron model. With ${\epsilon }_{\mathbf{k}}={\hslash }^2{k}^2/(2m_e)$, the Schrödinger equation reduces to the Hermitian matrix problemSetting the characteristic polynomial equal to zero givesSolving this quadratic equation givesFor the real cosine potential its coefficient $V_G$ is real; the absolute-value form also allows an arbitrary choice of origin and corresponding Fourier phase.

At the Brillouin-zone boundary, $\mathbf{k}\cdot \mathbf{G}=G^2/2$, the free energies are equal. The two standing waves have different probability densities relative to the lattice potential and are split byRepeating this avoided crossing at the reciprocal lattice boundaries joins smooth branches into energy bands. In one dimension a nonzero Fourier coefficient opens a gap at the corresponding crossing. In higher dimensions a direct gap at one wavevector need not give a common forbidden energy interval at every wavevector: different bands may still overlap in energy. The present two-dimensional example remains metallic even though the crossings are locally gapped.

For the rectangular lattice, choose real-space primitive vectors $a\hat{\mathbf{x}}$ and $b\hat{\mathbf{y}}$. The reciprocal lattice vectors areDefine the half-widths of the first Brillouin zone by $K=\pi /a$ and $L=\pi /b$. Their values and the zone areA primitive reciprocal cell can be drawn with a corner at the origin; its centred translate is the first Brillouin zone here. Its area is $4{\pi }^2/(ab)=37.9600\mathrm{n}{\mathrm{m}}^{-2}$.

Including both spin states, counting the occupied free-electron states per unit area givesConsequentlyThe occupied disk has precisely the area of the first Brillouin zone, but their shapes differ: $K<k_F<L$. It protrudes through both vertical zone boundaries and does not reach the horizontal ones. Thus two electrons per cell do not imply a filled first band in the absence of a separating band gap.

Fold the protruding caps by $\pm {\mathbf{G}}_x$. The original circle and the translated circle intersect at $k_x=K$, $k_y=\pm \sqrt{k_F^2-K^2}$. The potential mixes these states and replaces each crossing by an avoided crossing. The lower branch reconnects into two open sheets running across successive zones in the $k_x$ direction; the upper branch forms a closed lens-shaped pocket about a vertical zone boundary. In a reduced-zone sketch its two halves sit at opposite edges, but periodic identification makes them one closed sheet. This is rectangular-lattice Fermi-surface reconstruction.

The largest pocket in this weak-potential construction is the lens-shaped intersection of two disks enclosed by the intersecting free circles. Each has radius $k_F$ and their centres are separated by $2K$. If $\theta =\arccos (K/k_F)$, two sectors minus their two triangles give the lens bound on a reconstructed Fermi pocket:Opening a gap raises the upper branch at fixed chemical potential and shrinks its pocket. The reported quantum oscillation area satisfiesIt is therefore consistent with this closed sheet. Onsager quantization gives $F=\hslash A_k/(2\pi e)\simeq 601\mathrm{T}$. An orbit on this closed contour can return to itself under the Lorentz force and occupy quantized Landau levels. The open sheets do not enclose a finite orbit area for a perpendicular field in this two-dimensional model. The observation is consistent with the reconstructed pocket, without uniquely determining a completely arbitrary band structure.

For a quantitative rough coupling estimate, use the supplied fixed-chemical-potential approximation and retain the two waves coupled by ${\mathbf{G}}_x$. WriteTheir upper eigenvalue isSetting it equal to $C{k}_F^2$ gives the pocket boundaryIts intersection with $k_y=0$ occurs at $q=\pm q_m$, whereThe two-wave estimate of a Bragg-gap coupling from a pocket area therefore usesAt $v=0$ this reproduces the lens bound on a reconstructed Fermi pocket; the pocket disappears at $v=C(k_F^2-K^2)=0.23786\mathrm{eV}$. Numerical integration gives $A(0.075\mathrm{eV})=6.22\mathrm{n}{\mathrm{m}}^{-2}$ and $A(0.100\mathrm{eV})=5.54\mathrm{n}{\mathrm{m}}^{-2}$. Matching the measured area gives $v\simeq 0.0931\mathrm{eV}$, so the justified rough result isThis estimate neglects remote plane-wave admixtures and other Fourier coefficients. It also holds the chemical potential fixed as requested; an exact fixed-density calculation would determine it self-consistently after reconstruction. The numerical root is a way to implement the approximate model, not a measurement of $V_G$ to four significant figures.

In the Lorentz oscillator model, each participating bound electron is displaced by an applied electric field against a restoring force. The model replaces complicated interband transitions by a driven damped harmonic oscillator. If $u$ is the electron displacement along the field and the electron charge is $-e$, its equation of motion isHere ${\omega }_0$ describes binding and $\gamma >0$ describes loss of coherent motion. For $E(t)=\mathrm{Re}(E_0{e}^{-i\omega t})$, the complex amplitude isThe induced electric dipole moment is $-eu$, so the electric polarization amplitude for number density $n$ isUsing $D={ϵ}_{0}E+P={ϵ}_0{ϵ}_{r}E$ givesThe paper's $ϵ$ is this dimensionless relative permittivity, rather than the absolute permittivity ${ϵ}_0{ϵ}_r$. Independent classes of bound electrons contribute additive susceptibilities, usually with different resonance frequencies and optical oscillator strengths.

Writing $D_{\omega }=({\omega }_0^2-{\omega }^2{)}^2+{\gamma }^2{\omega }^2$ separates the response intoAt zero frequency the real part is $1+{\omega }_p^2/{\omega }_0^2$ and the imaginary part vanishes. Near resonance the imaginary part has a positive photon absorption peak and the real part changes rapidly, passing through $1$ at ${\omega }_0$. Above resonance the real part can become negative when damping is sufficiently weak; at very high frequency it approaches $1$ from below while the imaginary part tends to zero. Finite damping removes the undamped divergence.

For a nonmagnetic material take the passive square-root branch $N=\eta +i\kappa =\sqrt{{ϵ}_r}$, with $\kappa \ge 0$. A plane wave then has an intensity attenuation coefficient $\alpha =2\omega \kappa /c$. Thus the imaginary permittivity expresses dissipation, whereas the actual photon absorption coefficient is determined by the imaginary refractive index. The normal-incidence power reflectance isA large refractive-index mismatch produces wave reflection even without loss. In the undamped model ${ϵ}_r<0$ between ${\omega }_0$ and $\sqrt{{\omega }_0^2+{\omega }_p^2}$, so the field is evanescent and wave reflection is total in a semi-infinite sample. With damping the response includes both wave reflection and photon absorption; a peak in the imaginary permittivity should not simply be identified with a reflectance peak.

To infer the short-wavelength background from $R=0.35$, assume weak photon absorption and take the physical semiconductor branch $\eta >1$. ThensoThis is an estimate of the valence-electron background in a frequency window above the doped-carrier response and below the effective bound-electron resonance. One reflectance value alone does not determine both parts of a complex dielectric constant. Even for real ${ϵ}_r$, a reciprocal-index branch gives the same $R$; the ordinary background index greater than unity selects the branch used here.

Using the supplied conventional cubic lattice parameter, the participating valence-electron number density isUsing $m=m_e$ for the stipulated free-electron comparison givesAn angular frequency “in eV” means its quantum energy $\hslash \omega$, not $h\omega$. No doped-carrier effective mass should be inserted into this valence-electron count.

The low-frequency limit of the bound-electron contribution obeysConsequentlyThis effective resonance is much larger than the quoted $0.2\mathrm{eV}$ band gap. There is no contradiction: the single oscillator has been assigned the strength of all sixteen participating valence electrons per conventional cell, whereas the band gap is only the lowest interband excitation threshold. A real spectrum distributes optical oscillator strength over many transitions. Equating the oscillator resonance to the lowest threshold would misrepresent the assumed background response. Here “low frequency” means low relative to ${\omega }_0$; the static response of the doped conducting sample also contains its free carriers.

The additional mobile carriers in doped indium antimonide contribute a Drude frequency response, obtained by setting their restoring frequency to zero. At infrared frequencies well below the effective valence resonance, the sum is approximatelyAt the shorter wavelengths the carrier term is small and $R$ approaches the background value. As wavelength increases, the negative free-carrier contribution lowers the real permittivity towards unity. For weak damping, refractive-index matching to vacuum gives the Drude-Lorentz reflectivity minimum atThis minimum is near ${ϵ}_r^{\prime }=1$, not its zero. At the slightly lower screened plasma frequency ${\omega }_{pc}/\sqrt{{ϵ}_b}$, the real permittivity crosses zero; at longer wavelengths it becomes negative and the material becomes strongly reflective. Finite damping prevents a perfectly vanishing minimum and rounds the edge. At still lower frequency the large conducting response also drives $R$ towards unity.

For fixed carrier effective mass and background permittivity, ${\omega }_{pc}\propto \sqrt{n_c}$ and ${\lambda }_{\min }\propto n_c^{-1/2}$. The observed shift of the minimum towards shorter wavelength as doping increases is therefore the expected free-carrier trend. The figure is an original model illustration, not a digitization or quantitative fit of the supplied curves; the carrier effective mass, damping and their density dependence have not been specified.

Use the $e^{-i\omega t}$ convention, so passive photon absorption has positive imaginary permittivity. For a single Lorentz oscillator model resonance, write the photon energy as $E=\hslash \omega$ andHere $S$ has units of energy squared, $\Gamma =\hslash \gamma$ is a damping energy, and ${ϵ}_b$ is an approximately constant background. The corresponding real and imaginary parts areThus the real part starts above the background, rises on the low-energy side of the resonance, drops rapidly through ${ϵ}_r^{\prime }={ϵ}_b$ at $2\mathrm{eV}$, and approaches the background from below at larger energy. The imaginary part starts at zero, has a positive photon absorption peak near $2\mathrm{eV}$, and decays above it. For weak damping its peak is close to the resonance, but not exactly fixed there for arbitrary damping. At asymptotically large energy the corrections scale as ${ϵ}_r^{\prime }-{ϵ}_b\sim -S/E^2$ and ${ϵ}_r^{\prime \prime }\sim S\Gamma /E^3$.

The resonance position is fixed at 2 eV; peak height, width and background are not specified. The figure chooses an explicit finite damping and oscillator strength only to illustrate the shape over 0–5 eV. In particular, a negative real permittivity above resonance is possible for suitable strength and damping, but is not forced by the resonance energy alone. Setting damping exactly to zero would give a singular idealized response instead of a finite-width photon absorption peak.

Choose the $e^{-i\omega t}$ convention. The steady displacement in the Lorentz oscillator model has complex amplitudeThe induced electric dipole moment is $p_{\omega }=q{u}_{\omega }$, so the microscopic electric polarizability is $\alpha (\omega )=q^2/[m({\omega }_T^2-{\omega }^2-i\gamma \omega )]$. For $N$ identical oscillators per unit volume, $P_{\omega }=N{p}_{\omega }={ϵ}_0{\chi }_{\omega }{E}_{\omega }$. ConsequentlyHere the dimensionless quantity called polarizability in the paper is the electric susceptibility; the single-electron polarizability $\alpha$ has different units. The sign of the electron charge disappears because both the driving force and dipole moment contain $q$.

Multiplying by the complex conjugate of the denominator givesFor positive damping and positive frequency, ${ϵ}_r^{\prime \prime }\ge 0$, as required for passive wave absorption in this Fourier convention. At zero frequency, ${ϵ}_r^{\prime }=1+{\omega }_p^2/{\omega }_T^2$ and ${ϵ}_r^{\prime \prime }=0$. Near a weakly damped resonance the imaginary part has a narrow positive peak, while the real part rises on the low-frequency side and then falls steeply through ${ϵ}_r^{\prime }=1$ at ${\omega }_T$. It can become negative above resonance for sufficiently strong oscillator strength and weak damping. At large frequency, ${ϵ}_r^{\prime }=1-{\omega }_p^2/{\omega }^2+O({\omega }^{-4})$ and ${ϵ}_r^{\prime \prime }\sim {\omega }_p^2\gamma /{\omega }^3$: both approach the vacuum values. The absorption maximum is close to, but need not be exactly at, ${\omega }_T$ for finite damping.

Consider a conventional spin-singlet, weak-coupling, three-dimensional superconductor. Set $\hslash =k_B=1$ in the microscopic action and restore electromagnetic units below. An attraction $\lambda >0$ in an energy shell about the Fermi surface gives a Grassmann actionThe attractive sign permits pairing even when the single-particle Fermi gas is otherwise stable. A complex Hubbard–Stratonovich transformation introduces a pair field $\Delta$ proportional at the saddle to $\lambda ⟨{\psi }_{\downarrow }{\psi }_{\uparrow }⟩$. The action is then quadratic in fermions. Integrating them out gives the BCS pairing-field effective actionHere the trace uses one independent Nambu spinor pairing block; a fully doubled particle-hole representation instead carries its corresponding one-half factor. At a uniform saddle the Bogoliubov--de Gennes Hamiltonian can be written asThe negative branch describes the paired vacuum; excitations are Bogoliubov quasiparticles mixing particles and holes. Extremizing the effective action givesThe physical cutoff or a renormalized interaction is part of the microscopic model; an unregulated contact integral must not be used indiscriminately. With constant density of states $N(0)$ per spin and shell half-width ${\Omega }_c$, the zero-temperature weak-coupling solution is ${\Delta }_0\simeq 2{\Omega }_c{e}^{-1/[N(0)\lambda ]}$. The exponentially small scale follows from the logarithmic pairing susceptibility of states near the Fermi surface. For this ideal weak-coupling model, $2{\Delta }_0/(k_B{T}_c)\simeq 3.53$. The microscopic theory relates the gap, pairing and thermodynamic transition.^[ref]

Close to a continuous transition, for fields varying slowly compared with microscopic scales, expand the effective action in powers and gradients of $\Delta$. Its static part is the Ginzburg-Landau theory of superconductivity:The normalization of $\Delta$ fixes the coefficients; here $a(T)$ changes sign at $T_c$. The uniform minimum has $|{\Delta }_0{|}^2=-a/b$ for $a<0$, condensation free-energy density $-a^2/(2b)$, and characteristic coherence-length scale $\xi =\sqrt{K/|a|}$. Above the transition the minimum is at zero. The quartic stabilization and gradient cost are precisely the ingredients of Landau theory for a complex order parameter. This controlled near-transition expansion supplies a bridge between microscopic BCS theory and macroscopic magnetic response; its coefficients and validity depend on the material and fluctuation regime.^[ref]

Write $\Delta =(|{\Delta }_0|+\eta )e^{i\varphi }$. The radial fluctuation has positive local restoring curvature, whereas the phase has no local potential in a neutral theory with global $U(1)$ symmetry. Freezing the amplitude givesThe neutral case supports a gapless collective phase mode. In the charged case the electromagnetic field couples to this phase rigidity. Under $\mathbf{A}\mapsto \mathbf{A}+\nabla \chi$ and $\varphi \mapsto \varphi +q\chi /\hslash$, the displayed combination is invariant. Locally choosing a gauge with constant phase leaves a quadratic term in $\mathbf{A}$. Varying the free energy gives the transverse London current and magnetic screening scale,Combining this relation with static Maxwell equations gives exponential field penetration, the Meissner effect. Single-valuedness of the pair field around a loop with negligible current gives flux in units ${\Phi }_0=2\pi \hslash /|q|=\pi \hslash /e$. These are gauge-invariant physical consequences of the condensate stiffness.

For the dynamics, the neutral phase and density mode is reorganized by the Coulomb interaction and the dynamical electromagnetic field. In a bulk three-dimensional system the long-wave longitudinal mode has a plasma gap, rather than remaining the neutral sound mode. This is the Anderson-Higgs mechanism in a superconductor: a phase coordinate becomes part of the electromagnetic collective response instead of surviving as an independent massless physical particle. Local gauge transformations are redundancies of the description, so a literal breaking of a local redundancy is not an observable. Phase coherence, current response and magnetic screening provide the physical statements.^[ref]

A static Landau functional alone cannot determine that full dynamical spectrum. A two-dimensional film, for example, need not have the same long-wave plasma dispersion as a bulk charged system. Nor does a positive radial curvature automatically guarantee a sharp Higgs amplitude resonance: coupling to quasiparticles and other modes can damp it. The chosen mean-field phase is also a thermodynamic symmetry-breaking description; an isolated finite system can be formulated with definite particle number. The microscopic pairing saddle, the near-transition Landau expansion and the electromagnetic Anderson-Higgs response describe successive scales of the same superconducting state, with distinct assumptions at each step.

For $\varphi =-\Omega$, $p_1=m({\dot{x}}_1-R\Omega \sin \Omega t)$ and $p_2=m({\dot{y}}_1+R\Omega \cos \Omega t)$. The equations are ${\ddot{x}}_1+{\omega }^2{x}_1=R{\Omega }^2\cos \Omega t$, ${\ddot{y}}_1+{\omega }^2{y}_1=R{\Omega }^2\sin \Omega t$. For $\Omega \ne \omega$,At resonance the forced terms are $R\omega t\sin \omega t/2$ and $-R\omega t\cos \omega t/2$. Constants depend on the marble's initial position and velocity. A general cup orientation is obtained by multiplying the vector solution by $\mathcal{R}(-\alpha t)$; the physical inertial solution is unchanged by relabeling the cup axes. Resonant growth eventually invalidates the small-horizontal-motion/neglected-vertical-speed approximation.

Rigid walls require zero normal velocity, so ${\varphi }_x=0$ at $x=0,L$ and ${\varphi }_y=0$ at $y=0,b$. Separation of variables gives spatial modes $\cos (m\pi x/L)\cos (n\pi y/b)$, for nonnegative integers $m,n$. The wave equation therefore givesThe $(0,0)$ spatially constant mode has zero frequency and is not an ordinary positive-frequency acoustic oscillation.

A spatially uniform boundary forcing excites the $n=0$ transverse mode. Put $\varphi =X(x)\cos \omega t$. Then $X^{\prime \prime }+(\omega /c_0{)}^{2}X=0$ and $X^{\prime }(L)=0$, hence $X=C\cos [\omega (L-x)/c_0]$. The condition $X^{\prime }(0)=a$ fixes $C$ and givesArbitrary free oscillations can be superposed. The denominator vanishes at longitudinal resonances. Near them the lossless steady harmonic response ceases to be a uniformly small-amplitude approximation; damping from viscosity, heat transfer or wall losses regularizes the response, and large amplitudes can require nonlinear equations of motion or a more accurate moving-boundary model. At exact resonance a switched-on ideal forcing produces secular growth rather than a bounded harmonic particular solution.

For the semi-infinite duct expand $F(y)={\sum }_{n\ge 0}{F}_n\cos (n\pi y/b)$. Each axial mode hasFor a propagating mode a particular solution is ${\varphi }_n=(F_n/k_n)\cos (n\pi y/b)\cos (k_{n}x-\omega t)$; differentiating at zero gives the prescribed $F_n\cos (n\pi y/b)\sin \omega t$. A nonzero higher transverse coefficient propagates when $|\omega |>n\pi c_0/b$.

The outgoing radiation condition selects $c_0^2{k}_n/\omega >0$, its positive axial group velocity. Thus $\boxed{\omega k_n>0}$. This sign is selected by the absence of waves incident from infinity, not by the differential equation alone. If $F$ has a nonzero mean, its $n=0$ component can propagate even below every higher-mode cutoff.

For $F=\cos (\pi y/b)$ and $0<|\omega |<\pi c_0/b$, put $\kappa =\sqrt{(\pi /b{)}^2-{\omega }^2/c_0^2}$. The decaying solution isIt is evanescent and carries no mean energy to infinity. These mode and sign conditions are summarized by the radiation condition for a waveguide mode.

On periodic ring states the angular-momentum index $m$ is an integer. At the stated resonance,so $U_T$ is the identity on every orbital state. After $N$ steps from the normalized ground state,Its phase varies in space but its position probability density remains uniform. Periodicity permits integration by parts, givingConsequentlyEquivalently, the angular-momentum probabilities are $P_m=J_m(N\eta {)}^2$, with ${\sum }_m{m}^2{J}_m(z{)}^2=z^2/2$. The mean angular momentum remains zero while its variance grows as $N^2$. This quantum resonance of the kicked rotor gives coherent ballistic energy growth. The Bessel second-moment identity connects the momentum spreading to that quadratic energy law; an incoherent random walk would instead give linear energy growth.

First assume ${\omega }_0\ne 0$ and regard $\theta$ modulo $2\pi$. The section $\theta =0$ has return time $T=2\pi /|{\omega }_0|$. Solving the resonantly forced oscillator from this section givesAt $t=T$, the sine vanishes and the cosine is one, while differentiating the secular term adds $\gamma T/2$ to $y$. Hence the Poincaré map isFor $\gamma >0$, every iterate has $y_n=y_0+n\gamma T/2\to \infty$. Thus every oscillator trajectory is unbounded, regardless of its initial amplitude. For an arbitrary section phase ${\theta }_0$, the return map translates by $(\gamma T\sin {\theta }_0/(2{\omega }_0),\gamma T\cos {\theta }_0/2)$, which is still nonzero. At $\gamma =0$ the map is the identity, and all $(x,y)$ motions are bounded harmonic oscillations, including the zero-amplitude equilibrium.

The zero-frequency case allowed by the printed parameter range is different. If ${\omega }_0=0$, then $\theta ={\theta }_0$ is constant and $x=x_0+y_{0}t+\frac{1}{2}\gamma \cos {\theta }_0{t}^2$. There is no phase-crossing Poincaré section. For $\gamma >0$, choosing $\cos {\theta }_0=0$ and $y_0=0$ gives bounded equilibria, contradicting an unqualified “all trajectories” statement. For $\gamma =0$ at zero frequency, free-particle motion is unbounded unless $y_0=0$. The resonance conclusion therefore requires the natural nonzero-frequency assumption.

For the $n$th normal mode, the denominator just obtained isIts forcing coefficient is nonzero for $c\ne 0$ and $\omega >0$. Therefore the periodic response grows like $({\omega }_n^2-{\omega }^2{)}^{-1}$ as $\omega$ approaches ${\omega }_n$: this is resonance. At exact resonance there is no bounded periodic particular solution in the linear inviscid model; the resonant amplitude instead contains a secular term proportional to $t\sin ({\omega }_{n}t)$ or $t\cos ({\omega }_{n}t)$. Physical damping and nonlinear effects eventually matter. Away from resonance, undamped free oscillations can persist alongside the periodic particular solution.

For a resonator, the quality factor is $2\pi$ times the stored energy divided by the energy dissipated per cycle, using a specified steady-state stored-energy convention. A series RLC circuit at resonance has $Q=L{\omega }_0/R$, because its maximum magnetic energy is $L{I}_M^2/2$ and one cycle of Joule heating dissipates $\pi R{I}_M^2/{\omega }_0$.

As an original stipulated low-field model, take the axial magnetic field intensity $H$ in $A/m$ and prescribeAll strain components are dimensionless, and their trace vanishes. This is an even, quasistatic, unsaturated toy law with a fixed measuring axis. It is distinct from the fixed-magnitude directional model in saturated isotropic magnetostrictive strain. In particular, one must not replace local unit magnetization by normalized net magnetization in a many-domain specimen and assume this law follows.

For $H(t)=H_0\sin (2\pi ft)$, $H_0=20kA/m$, and $f=50\mathrm{Hz}$, a trigonometric identity givesThe axial strain ranges from zero to $20\times 10^{-6}$, with mean $10\times 10^{-6}$ and a $100\mathrm{Hz}$ component of amplitude $10\times 10^{-6}$. The cosine coefficient is negative with the chosen time origin. For $L=100\mathrm{mm}$ the length ranges from zero to $2\mu \mathrm{m}$ relative to $H=0$. There is no $50\mathrm{Hz}$ term in this unbiased model.

A static bias $H_b$ changes the conclusion. Expanding $\kappa [H_b+H_0\sin (2\pi ft){]}^2$ produces a fundamental-frequency term $2\kappa H_b{H}_0\sin (2\pi ft)$ as well as the doubled-frequency term. This is an incremental linear response about a state with a chosen magnetic axis; it does not contradict isotropic linear magnetostriction vanishes at an unbiased isotropic reference. Real hysteresis, thermal expansion from Joule heating, and mechanical resonance can add further components or change their phases. Frequency alone cannot identify the cause of motion: Joule heating also contains an even-in-electric current contribution.

For $u^{\prime \prime }+u=ϵu^2$, a leading oscillation $U_0=A(T)e^{it}+\overline{A}(T)e^{-it}$ has slow time $T={ϵ}^{2}t$. The first-order forcing contains only zero and double frequencies. At second order the fundamental forcing is $(10/3)|A{|}^{2}A-2i{A}_T$; eliminating resonance gives the displayed equation. Its modulus is constant and its phase drifts. A leading real amplitude $R=2|A|$ therefore has frequency $1-5R^2{ϵ}^2/12+O({ϵ}^3)$. This replaces a secular phase term by a uniformly modulated oscillation.

When the free angular-momentum phases all recur at the driving period, successive identical kicks add coherently. For the full revival $\hslash T/(2I)=2\pi$ and an initial zero-angular-momentum state, the kinetic energy after $N$ kicks is ${\hslash }^2{\eta }^2{N}^2/(4I)$. This ballistic law is specific to the stated resonance and boundary conditions.

At normal incidence, alternating layers with indices $n_H,n_L$ and optical thicknesses $n_H{d}_H=n_L{d}_L={\lambda }_0/4$ give reinforcing reflected waves near design wavelength ${\lambda }_0$. Their periodic transfer matrix has a stop band when its Bloch phase is complex. A finite stack has nonzero transmission and its exterior impedances matter. Introducing a suitably phased central defect can support a transmission resonance within the stop band; a one-dimensional stop band is not a complete three-dimensional photonic gap.

The second-order, angle-averaged evolution of a galactic distribution function caused by weak fluctuating gravitational perturbations. In action-angle variables, the first-order response to a Fourier series harmonic feeds a second-order flux proportional to $\mathbf{k}(\mathbf{k}\cdot {\partial }_{\mathbf{J}}{f}_0)$ and to the perturbation's power. Orbital resonances select where transport is strongest; a finite pulse produces Gaussian resonance broadening when its time envelope is a Gaussian function. The general flux retains time memory until a further short-memory or pulse-integrated approximation is made.

Transfer of fluctuation energy to particles selected by a wave-particle resonance. For a weakly damped Langmuir wave, particles near its phase velocity absorb net energy from a stable distribution by Landau damping. Under weak random forcing, a smaller damping rate produces a larger stationary wave intensity; a very long waiting time may be required before that stationary balance holds.

For $x^{\prime \prime }+{\omega }^{2}x=\gamma \cos \theta$, $\dot{\theta }=\omega \ne 0$, use the section $\theta =0$ modulo $2\pi$ and $T=2\pi /|\omega |$. Exact resonance produces the secular term $\gamma t\sin (\omega t)/(2\omega )$, giving the displayed return translation. Nonzero forcing makes every section orbit unbounded. At zero forcing the map is the identity. The assumption $\omega \ne 0$ is essential: at zero frequency no transverse phase section exists and the motion is constant-acceleration or free-particle motion.

Drive a series RLC circuit with electric potential difference $V(t)=V_0\cos (\omega t)$, where $V_0>0$ and $\omega >0$. After the free transient decays, the electric current has the same angular frequency as the source:For fixed positive $R,L,C,V_0$, the electric current amplitude is largest exactly at $\omega ={\omega }_0=1/\sqrt{LC}$. The two electrical reactance contributions cancel there, leaving only the drop from electrical resistance. In particular,Every solution approaches this periodic response because its difference from the displayed solution is a decaying free response. The peak remains at ${\omega }_0$ even for an overdamped RLC response; free ringing is not required for a maximum of driven electric current. ^[ref].

For example, stipulate $L=10\mathrm{mH}$, $C=1\mu \mathrm{F}$, $R=20\Omega$, and $V_0=1\mathrm{V}$. Then ${\omega }_0=10^4{\mathrm{s}}^{-1}$, $f_0={\omega }_0/(2\pi )\approx 1591.55\mathrm{Hz}$, and the peak electric current amplitude is $50\mathrm{mA}$. Holding the source amplitude fixed is essential when comparing responses at different frequencies.

Three frequencies must be distinguished. The free underdamped RLC response rings at ${\omega }_d=\sqrt{{\omega }_0^2-{\alpha }^2}$. Driven electric current peaks at ${\omega }_0$. Driven electric charge has amplitudewhose peak occurs at $\sqrt{{\omega }_0^2-2{\alpha }^2}$ only if $\alpha <{\omega }_0/\sqrt{2}$. Otherwise it decreases from its zero-frequency limit $C{V}_0$, with no positive-frequency peak. The different peak locations follow because electric current is the time derivative of electric charge, multiplying a sinusoidal amplitude by $\omega$. The limiting responses are $I_{\mathrm{amp}}\sim V_{0}C\omega$ as $\omega \to 0^{+}$ and $I_{\mathrm{amp}}\sim V_0/(L\omega )$ as $\omega \to \infty$.

The Q factor measures selectivity:The average Joule heating power is $R{I}_{\mathrm{amp}}^2/2$. Its half-maximum angular frequencies and their separation are exactlyThese follow by setting $|\omega L-1/(\omega C)|=R$. In the numerical example, $Q=5$ and $\Delta \omega =2000{\mathrm{s}}^{-1}$. Reducing $R$ makes a taller, narrower electric current peak; varying $C$ tunes the selected frequency. Measuring the drop $v_R=Ri$ gives a band-pass filter that suppresses very low and very high frequencies while passing a range near resonance. This is useful for selecting a radio signal, though the range must be broad enough to pass its modulation. ^[ref].

For an isotropic elastic amplitude normalized by an incident unit plane wave and outgoing $f{e}^{ikr}/r$, total cross section is $4\pi |f{|}^2$. The optical theorem gives the displayed identity. At a resonance where the real part of $1/f$ vanishes, $f=i/k$ and the cross section is $4\pi /k^2$.