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Use the Euler-Bernoulli beam equation for a uniform slender elastic beam, small deflection and small slope, neglecting axial tension, shear deformation, rotational inertia of a cross-section and damping. Let $E$ be Young's modulus, $I={\int }_A{z}^{2}dA$ the second moment of area about the neutral axis, and $\rho$ the mass per unit length. If density instead denotes mass per volume, replace this $\rho$ by ${\rho }_{\mathrm{bulk}}A$. The bending stiffness is $EI$; the small-slope curvature is $y_{xx}$. The bending and kinetic energies areTwice integrating the variation of $U$ by parts gives the restoring load $-EI{y}_{xxxx}$ per unit length. Newton's second law therefore givesAt a pivot, displacement is fixed but slope is free; zero end bending moment gives $EI{y}_{xx}=0$. This also follows from the natural boundary term $EI{y}_{xx}\delta y_x$ in the variation.

With separation of variables, $y=XT$ givesChoose the time normalization $T(0)=1$. Initial rest removes the sine time term, so $T(t)=\cos \omega t$. The spatial characteristic roots are $\pm ik,\pm k$, givingThe two conditions at $x=0$ give $B+D=0$ and $-B+D=0$, hence $B=D=0$. At $L$, addition and subtraction of $X(L)=X^{\prime \prime }(L)=0$ give $C\sinh kL=0$ and $A\sin kL=0$. Nontrivial positive-frequency modes therefore have $C=0$ and $k_n=n\pi /L$. Thus the simply supported beam normal modes areThe zero-frequency cubic solution also vanishes under these four endpoint conditions, so $n=0$ is not an additional mode.

For the specified release, small slope requires $|a|L\ll 1$. Expand the initial deflection in the Fourier sine series. Its coefficients areWriting ${\omega }_1={\pi }^2\sqrt{EI/\rho }/L^2$, the resulting motion isOnly odd normal modes occur because the initial profile is symmetric about the midpoint, and their amplitudes decrease as $n^{-3}$. The frequencies grow as $n^2$, so the motion is dispersive and is not a single shape multiplied by one cosine. Nevertheless the ideal undamped solution is exactly periodic with period $2\pi /{\omega }_1$. At $t=\pi /(2{\omega }_1)$ all odd-mode cosines vanish, so the bar is instantaneously straight; at $t=\pi /{\omega }_1$ it has the opposite initial deflection. The first mode dominates displacement, with third-mode amplitude only $1/27$ of the first.

There is initial curvature incompatibility at a moment-free beam end: the prescribed parabola has $y_{xx}(x,0)=2a$, whereas a moment-free pivot requires zero curvature at an endpoint. Therefore the data do not define a classical solution with curvature continuous through the two end-time corners. They do define a finite-energy weak solution after release of the forces or end moments that prepared the initial shape. The sine expansion represents that solution: its displacement reproduces the initial parabola, each mode satisfies the released endpoint conditions, and ${\sum }_n{n}^4|A_n{|}^2<\infty$ gives finite bending energy. One must not differentiate the initial series twice at its endpoints and assign the interior parabola's curvature there. This qualification leaves the modal displacement prediction intact.

Displacement current makes Ampère's law compatible with charge conservation. The macroscopic Maxwell equations areTaking the divergence of the last equation and using Gauss's law gives $\nabla \cdot {\mathbf{J}}_f+{\partial }_t{\rho }_f=0$. Omitting ${\partial }_t\mathbf{D}$ would wrongly require the conduction current to be divergence free during charging. The displacement-current density ${\partial }_t\mathbf{D}$ has current-density units, but need not represent free charges crossing a region.

The surface independence of Ampère circulation during capacitor charging illustrates the need directly. A surface spanning the circuit can cut the wire or pass between the plates. For ideal parallel plates, $Q={ϵ}_{0}AE$ andThe corrected circulation of $\mathbf{H}$ is independent of that surface choice. Free conduction current stops at a plate; displacement current supplies the gap contribution.

In vacuum the two time-dependent curl equations support waves. Taking a curl of Faraday's law in a source-free region yieldsThe corresponding magnetic equation is identical. For a plane wave, $\mathbf{k}\cdot \mathbf{E}=\mathbf{k}\cdot \mathbf{B}=0$ and $\mathbf{B}=\mathbf{k}\times \mathbf{E}/\omega$. The fields carry energy through the Poynting flux. Dropping displacement current everywhere would remove these freely propagating solutions.^[ref]

In a metal, conduction and displacement compete. For a uniform ohmic material with ${\mathbf{J}}_f=\sigma \mathbf{E}$, constant $ϵ,\mu$, and transverse waves,Use $e^{i\mathbf{k}\cdot \mathbf{r}-i\omega t}$. The dispersion relation and conduction-to-displacement current ratio areWhen $\sigma \gg \omega ϵ$, $k\simeq (1+i)/\delta$ with $\delta =\sqrt{2/(\mu \sigma \omega )}$: fields are screened within a skin depth and energy is dissipated by Joule heating. This quasistatic limit neglects displacement current locally compared with conduction; it does not invalidate the full Maxwell law. At sufficiently high frequency, conductivity and permittivity are frequency dependent, so a constant-$\sigma$ approximation must be replaced by a material model.

In an unmagnetized cold plasma, electron inertia gives a cutoff. With immobile ions and negligible collisions, $m_e\dot{\mathbf{v}}=-e\mathbf{E}$ and $\mathbf{J}=-n_{e}e\mathbf{v}$ give $\sigma (\omega )=i{n}_e{e}^2/(m_e\omega )$. For the transverse wave cutoff in a cold unmagnetized plasma, combining this carrier current with vacuum displacement current givesTransverse waves propagate above ${\omega }_p$ and are evanescent below it. The group velocity is $c\sqrt{1-{\omega }_p^2/{\omega }^2}<c$ even though the phase velocity exceeds $c$. The cold model also has longitudinal charge-density oscillations at ${\omega }_p$, distinct from transverse light waves. Collisions add damping, and pressure or magnetic fields modify these simple dispersions. To avoid carrier-current double counting in effective permittivity, once the carrier response is included in ${ϵ}_{\mathrm{eff}}$ it must not be counted a second time as a separate conductivity. Displacement current therefore connects circuit continuity, vacuum radiation, conductor screening and plasma propagation within one consistent field theory.^[ref]

A normal mode is a motion in which all coordinates have a fixed relative pattern and oscillate at one frequency. Near a stable equilibrium, let the quadratic kinetic and potential energies be $T={\dot{\mathbf{q}}}^{\mathsf{T}}\mathbf{M}\dot{\mathbf{q}}/2$ and $V={\mathbf{q}}^{\mathsf{T}}\mathbf{Kq}/2$. With positive-definite mass matrix $\mathbf{M}$, the equations and harmonic ansatz giveThe stiffness and mass matrices are symmetric. For distinct frequencies, their symmetry gives $({\omega }_s^2-{\omega }_r^2){\mathbf{a}}_r^{\mathsf{T}}\mathbf{M}{\mathbf{a}}_s=0$, so the mode vectors are orthogonal in the mass inner product. A basis can also be chosen in a degenerate eigenspace. Expanding $\mathbf{q}={\sum }_s{Q}_s{\mathbf{a}}_s$ decouples the equations into ${\ddot{Q}}_s+{\omega }_s^2{Q}_s=0$; initial displacements and velocities determine the coefficients. Positive stiffness gives real frequencies, a zero eigenvalue can describe a free symmetry motion, and a negative one signals instability rather than an oscillatory mode.^[ref]

For two equal masses between fixed walls with three equal springs, the stiffness matrix is $k\left(\begin{array}{cc}2& -1\\ -1& 2\end{array}\right)$ and the mass matrix is $m1$. The patterns $(1,1)$ and $(1,-1)$ have frequencies $\sqrt{k/m}$ and $\sqrt{3k/m}$. An initial displacement of one mass excites both modes and can produce beats.

For a uniform taut string fixed at both ends, the wave equation ${\partial }_t^{2}y=c_s^2{\partial }_x^{2}y$, with $c_s=\sqrt{\mathcal{T}/{\mu }_{\ell }}$, has patterns and angular frequenciesThe boundary conditions select the spectrum. Fourier expansion of arbitrary admissible initial data provides the mode amplitudes, illustrating the extension from a finite set of coordinates to a field.

For the normal modes of inductively coupled LC circuits, take two identical circuits with self-inductance $L$, capacitance $C$ and mutual inductance $\mathcal{M}$, where $|\mathcal{M}|<L$. The capacitor charges obey $L{\ddot{q}}_1+\mathcal{M}{\ddot{q}}_2+q_1/C=0$ and the exchanged equation. In-phase and opposite-phase charge patterns haveThe positive inductance matrix for two identical coils is the physical counterpart of positive mass. Here inductance plays the role of the mass matrix. Mechanical, wave and electrical examples therefore share the same eigenvalue structure. Weak forcing excites modes according to its spatial overlap with each pattern; damping broadens the resonances, while strong nonlinearity can invalidate a fixed linear-mode decomposition.

Definition and equations. A normal mode is a motion in which every component of a linear system has the same time dependence with fixed amplitude ratios and phases. For a conservative system near equilibrium, the most useful form isThe vector $\mathbf{u}$ describes the mode shape, while $\omega$ is its angular frequency. Linear combinations of modes describe arbitrary sufficiently small motions when the modes form a complete basis.

Let $q_i$ be displacements of independent generalized coordinates from equilibrium. Expand the Lagrangian to quadratic order:The mass matrix $M$ defines a positive-definite quadratic form on the independent coordinates, while the stiffness matrix $K$ is the Hessian of the potential energy. Substituting a mode gives the generalized eigenvalue problem for small oscillationsA practical calculation identifies the equilibrium, chooses independent coordinates, computes the quadratic kinetic and potential energies, solves for the eigenvalues and then obtains their eigenvectors. One can equivalently diagonalize the real symmetric matrix $M^{-1/2}K{M}^{-1/2}$. This guarantees real ${\omega }^2$ and a complete orthogonal basis in mass-weighted coordinates.

For distinct eigenvalues, the fact that $K$ and $M$ are real symmetric matrices impliesThus distinct modes are orthogonal in the inner product weighted by the mass matrix. Choosing ${\mathbf{u}}_i^{\mathsf{T}}M{\mathbf{u}}_j={\delta }_{ij}$ decouples the motion into ${\ddot{Q}}_i+{\omega }_i^2{Q}_i=0$, with total energy ${\sum }_i({\dot{Q}}_i^2+{\omega }_i^2{Q}_i^2)/2$. A positive-definite matrix $K$ gives positive eigenvalues and oscillatory stability. Zero modes can describe rigid translations or rotations without restoring force; a negative eigenvalue gives exponential instability rather than an oscillatory mode.

Symmetry. A symmetry preserving both quadratic energies maps modes into modes of the same frequency. Coordinates adapted to symmetry divide the eigenvalue problem into smaller independent blocks. For two identical subsystems, exchanging them gives symmetric and antisymmetric modes. The degeneracy of normal modes means that independent modes share one frequency. Higher-dimensional group representations of a symmetry can enforce such degeneracies; symmetry does not require every pair of modes to be degenerate, and accidental degeneracies can also occur. Within a degenerate eigenspace, orthogonal linear combinations are equally valid mode shapes.

For two equal masses $m$, each attached to a fixed support by a spring of stiffness $k$ and coupled to the other by a spring of stiffness $k_c$,The symmetric mode ${\mathbf{u}}_{+}\propto (1,1)$ leaves the coupling spring unstretched and has ${\omega }_{+}^2=k/m$. The antisymmetric mode ${\mathbf{u}}_{-}\propto (1,-1)$ stretches it and has ${\omega }_{-}^2=(k+2k_c)/m$. Superposing the two gives beats when their frequencies are close.

Continuous systems. The same idea replaces matrices by differential operators and mode vectors by functions satisfying boundary conditions. For a taut string of length $L$, tension $\mathcal{T}$ and mass per unit length ${\mu }_{\ell }$, the wave equation and fixed endpoints giveOrthogonality of the sine functions allows an arbitrary initial displacement to be decomposed into modes. Changing the endpoints changes the spectrum: boundary conditions are part of the physical mode problem.

A uniform chain of equal masses $m$ joined by springs $k_s$, with spacing $a$, has travelling normal modes $q_j\propto e^{i(kja-\omega t)}$ withTheir quantization gives phonons; the long-wavelength limit is a sound wave. Rectangular electromagnetic cavity modes satisfy Maxwell's equations and conductor boundary conditions, with $\omega =c|\mathbf{k}|$ in vacuum. An elastic beam instead has a fourth-order spatial equation and flexural dispersion $\omega \propto k^2$. These examples show how restoring forces and geometry determine the frequency spectrum.

Real systems. Small-amplitude motion is often approximately linear even when the full equations are nonlinear. Large amplitudes produce frequency shifts, mode coupling and transfer of energy between modes. Friction broadens resonances and makes free oscillations decay. A damping matrix proportional to $M$ or $K$ can retain the undamped mode shapes; general damping need not, and the appropriate eigenfrequencies and modes can be complex. External periodic forcing predominantly excites modes near resonance, with amplitude set by coupling to the forcing and dissipation. Imperfections split ideal degeneracies, while radiation or coupling to a surrounding fluid can remove energy from the chosen subsystem. Normal modes remain useful when linearization, the boundary conditions and the treatment of damping accurately describe the motion being measured.

For the gyrocompass oscillation, use the same local basis and define $\theta$ about the stable orientation, with spin angular momentum directed north at $\theta =0$. Because the vertical is perpendicular to the horizontal axle, the extra vertical angular momentum from the slow azimuthal motion is $I_2\dot{\theta }$. To leading order in Earth's small angular velocity,A constant vertical contribution $I_2{\Omega }_E\sin \lambda$ from Earth's rotation has zero terrestrial time derivative; the transverse terrestrial contributions affect the following equation only at order ${\Omega }_E^2$. Taking the vertical component of the torque equation givesThe terrestrial derivative of $J_s\mathbf{u}$ is horizontal, as is ${\mathbf{\Omega }}_E\times (I_2\dot{\theta }\mathbf{z})$, so neither contributes to this component. Thus the small angle satisfies, using the small-angle approximation $\sin \theta \simeq \theta$,This is simple harmonic motion, $\theta =A\cos ({\omega }_{\mathrm{shm}}t+\varphi )$, with positive squared angular frequency away from the poles. The approximation requires rapid spin, small angular displacement and negligible bearing friction. Near the opposite orientation, putting $\theta =\pi +\delta$ instead yields $\ddot{\delta }-{\omega }_{\mathrm{shm}}^2\delta \simeq 0$, which is unstable. Consequently, the paper's oscillation claim applies to the stable equilibrium, not both north--south spin directions. Without damping the ideal gyrocompass continues to oscillate rather than settling.

For a mass change in a viscously damped spring oscillator, use linear viscous damping $m\ddot{x}+b\dot{x}+kx=mg$, holding the spring constant $k$ and damping coefficient $b$ fixed. Measuring $x$ from $mg/k$ removes the constant force. The envelope is $e^{-\gamma t}$, whereDoubling $m$ halves both $k/m$ and $\gamma$. ThereforeNeglecting damping in the frequency gives $0.5/\sqrt{2}=0.35355\mathrm{Hz}$, indistinguishable at the supplied precision. The release condition sets the phase and new initial amplitude, not these envelope ratios. A different mass-dependent drag law would require its damping coefficient to be specified.

For the spring oscillator,so the total mechanical energy is conserved.

For U-tube liquid oscillation, define positive $y$ as the fall of the left surface, equal to the rise of the right surface. The hydrostatic height difference is $2y$. Moving a volume $Ady$ against this height difference gives $dU=2\rho gAydy$, henceEnergy conservation yields $\rho Al\dot{y}(\ddot{y}+2gy/l)=0$. Equivalently the net pressure force is $-2\rho gAy$, so the equation holds also at turning points:With negligible added volume, $y(0)=0$, $\dot{y}(0)=0.015\mathrm{m}{\mathrm{s}}^{-1}$. ThusThe graph starts at zero with positive slope and first reaches its maximum after $T/4$.

For viscously damped U-tube oscillation, the total damping force is $-\beta l\dot{y}$, not $-\beta \dot{y}$. ConsequentlyThe underdamped initial-value solution is $y=(v_0/{\omega }_d)e^{-\Gamma t/2}\sin {\omega }_{d}t$, where ${\omega }_d=\sqrt{{\omega }_0^2-{\Gamma }^2/4}=6.255{\mathrm{s}}^{-1}$. Setting its amplitude envelope equal to $0.1\mathrm{mm}$ givesThis is the decay of the envelope, not the first time the instantaneous displacement passes through $0.1\mathrm{mm}$.

Axial moment of inertia. In the original diagram $R_1$ is the outer radius and $R_2$ the inner radius. For uniform mass density $\rho$, a coaxial annulus of thickness $dr$ has $dm=2\pi \rho Lrdr$. Therefore the axial moment of inertia of a hollow cylinder isSince $M=\pi \rho L(R_1^2-R_2^2)$, factorizing the fourth-power difference givesFor a solid cylinder $R_2\to 0$, giving $I=M{R}_1^2/2$. For a thin cylindrical shell with fixed mass and $R_1,R_2\to R$, $I=M{R}^2$. The latter limit agrees with every mass element being at radius $R$.

Undamped rolling oscillations. Let $x$ be axle displacement from the spring equilibrium. Rolling without slipping imposes $|{\omega }_{\mathrm{roll}}|=|\dot{x}|/R$. The sum of translational and rotational kinetic energy isThe potential energy is $U=k{x}^2/2$. Differentiating $T+U$ in the absence of damping, or applying the force and torque equations below, givesFor the thin shell $I=m{R}^2$, the effective inertia is $2m$, and the angular frequency of the simple harmonic motion isStatic contact friction supplies the required torque but does no work when the contact point is instantaneously stationary.

Damped force balance. Take rightward force as positive and counterclockwise rotation as positive. Write $f$ for the signed static friction at the bottom contact. ThenEliminate $f=-I\ddot{x}/R^2$ and use $I=m{R}^2$:Here $b$ has units $\mathrm{kg}{\mathrm{s}}^{-1}$, so $B$ has units ${\mathrm{s}}^{-1}$ and $C$ units ${\mathrm{s}}^{-2}$. The stated drag force is interpreted as acting through the axle, with no additional viscous torque. An oil bath could also exert such a torque physically, but its magnitude is not specified by this force-only model.

Light damping. Substitution of $z=A{e}^{-pt}$ into the differential equation gives $p^2-Bp+C=0$. In the underdamped regime $B^2<4C$,Taking real linear combinations gives the general damped harmonic oscillator displacementThe constants are fixed by the initial displacement and velocity. For example, release from rest at $x_0$ gives $A_1=x_0$ and $A_2=b{x}_0/(4m{\omega }_d)$. Light damping means $b^2\ll 8mk$, in which case ${\omega }_d\simeq {\omega }_0$ and the envelope decays slowly compared with an oscillation.

Equilibrium and undamped motion. Resolve forces along the incline, taking positive displacement down the slope. At equilibrium, Hooke's law balances the downhill component of weight:If $x$ is measured from that equilibrium position, Newton's second law becomesThus the motion is a harmonic oscillator withGravity shifts the equilibrium extension but drops out of this oscillation frequency because the downhill force component is constant.

Viscous damping. For a damped harmonic oscillator, put $\beta =b/(2m)$. Substitution of $z=A{e}^{-pt}$ in the supplied linear ordinary differential equation givesIn the underdamped regime $\beta <{\omega }_0$, define ${\omega }_d=\sqrt{{\omega }_0^2-{\beta }^2}$. The two characteristic roots are $p=\beta \pm i{\omega }_d$, and the general real displacement isHere $C,D$, or equivalently $A_{\mathrm{env}}$ and $\varphi$, are fixed by the initial conditions. For release from rest at $x_0$, the condition $\dot{x}(0)=0$ requires $D=\beta x_0/{\omega }_d$, not $D=0$:The curve oscillates about zero between exponentially shrinking envelopes $\pm A_{\mathrm{env}}{e}^{-\beta t}$, where $A_{\mathrm{env}}=|x_0|\sqrt{1+(\beta /{\omega }_d{)}^2}$ for this release. The upper panel illustrates that shape; its damping strength is chosen to make the envelope visible.

Frequency change. Viscous damping reduces the angular frequency and lengthens the oscillation period:For light damping, ${\omega }_d\simeq {\omega }_0[1-b^2/(8mk)]$, so the frequency shift is second order in $b$, even though the envelope decay rate is first order in $b$.

Will the block reach the stop? The given parameters giveFor $x_0=-0.260\mathrm{m}$, differentiating the release-from-rest solution givesIt is positive throughout the first half-cycle. The first positive turning point therefore occurs at $t_{+}=\pi /{\omega }_d=0.993460\mathrm{s}$. The turning points of an underdamped release from rest give its unobstructed height:Since $x_{+}>0.250\mathrm{m}$ and the displacement increases continuously up to that point,Solving $x(t)=0.250\mathrm{m}$ gives first contact at approximately $0.91117\mathrm{s}$. The plotted dashed continuation is the hypothetical free motion after that time; the post-impact motion is not determined by the supplied damping model.

Write $\gamma =b/(2m)$ and $\nu =\sqrt{{\omega }^2-{\gamma }^2}$, where $\omega =\sqrt{k/m}$ is the undamped angular frequency. The linear drag gives the vector damped harmonic oscillator equationFor the original circular initial conditions, ${\mathbf{r}}_0=(c,c)$ and ${\mathbf{v}}_0=\omega (-c,c)$, the exact underdamped solution isUsing $|{\mathbf{r}}_0{|}^2=2c^2$, $|{\mathbf{v}}_0{|}^2=2{\omega }^2{c}^2$ and ${\mathbf{r}}_0\cdot {\mathbf{v}}_0=0$ givesThis exact potential energy is useful because it satisfies the prescribed initial conditions, including $\dot{U}(0)=k{\mathbf{r}}_0\cdot {\mathbf{v}}_0=0$. In the stated weak-damping limit $\gamma \ll \omega$, $\nu \simeq \omega$ and the square bracket differs from one by a small oscillatory term. The leading approximation is thereforeThe radius is approximately $\sqrt{2}c{e}^{-bt/(2m)}$, so the trajectory is approximately a shrinking logarithmic spiral. The exactly exponential result would require slightly adjusted initial velocity; with the specified undamped circular velocity, the small modulation above is real. This is the initially circular damped planar oscillator.

For weak linear damping, write the amplitude envelope as $A(t)=A_0{e}^{-{\omega }_{0}t/(2Q)}$. The amplitude e-folding time is $2Q/{\omega }_0$, and one period is approximately $2\pi /{\omega }_0$. ThusEnergy, being proportional to amplitude squared, decays twice as rapidly in its exponent. Using its e-folding time instead would introduce a factor-of-two error.

Let $a$ be the cube's side and $z$ a small upward displacement. Its displaced water volume decreases by $a^{2}z$, so the change in buoyancy is $-{\rho }_{w}g{a}^{2}z$. With cube mass $m=0.9{\rho }_w{a}^3$,The displacement must be small enough that the upper face remains above water and the bottom remains submerged. This buoyancy estimate neglects fluid added mass, damping and melting.

Different measurements reveal different aspects of electronic band structure; a convincing determination combines their constraints.

Angle-resolved photoemission. In angle-resolved photoemission spectroscopy, photons of known energy eject occupied electrons. Measuring their kinetic energy and emission angle reconstructs binding energy and the component of crystal momentum parallel to the surface:Here $\Phi$ is the sample work function when kinetic energy is referenced to the nearby vacuum; an analyzer energy scale requires its own work-function calibration. Momentum is conserved modulo a surface reciprocal lattice vector, neglecting photon momentum. Tracking the peaks maps occupied dispersion relations and the Fermi surface. The measured intensity also contains the matrix element, the Fermi-Dirac distribution and the spectral function, so an absent bright feature does not establish an absent band. For a three-dimensional solid, the perpendicular momentum requires a final-state model. Inverse photoemission spectroscopy probes unoccupied states by injecting electrons and detecting emitted photons.

Optical spectroscopy. Photon absorption and ellipsometry measure interband transitions, optical selection rules and the complex dielectric response. Because optical photon momentum is small on the scale of the Brillouin zone, direct transitions approximately preserve crystal momentum. Their onset constrains a direct band gap; a threshold involving a phonon can reveal an indirect band gap. Peak positions constrain energy differences between bands, but the response also depends on the joint density of states and optical matrix elements, so it does not uniquely determine every $E_n(\mathbf{k})$.

Quantum oscillations and cyclotron resonance. In clean samples at low temperature, the de Haas-van Alphen effect measures oscillations in magnetization and the Shubnikov-de Haas effect measures oscillations in electrical resistivity as the inverse magnetic field is varied. Their period in inverse magnetic field is $1/F$, where Onsager quantization relates $F$ to an extremal Fermi-surface area perpendicular to the field:Rotating the magnetic field maps the geometry of extremal orbits. The thermal damping determines the cyclotron effective mass,with the appropriate magnitude for a hole orbit. Cyclotron resonance supplies the corresponding dynamical mass through ${\omega }_c=eB/m_c$. These are detailed probes near the Fermi level, rather than measurements of all occupied bands. Diffraction supplies the reciprocal lattice needed to label the measured momenta, but diffraction by itself is not a measurement of electronic dispersion.

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.

A magnetic field quantizes a charged carrier's cyclotron motion into Landau levels. For an isotropic parabolic band, neglecting spin splitting,The degeneracy of each orbital level per unit perpendicular area is $eB/h$ per spin. Increasing the field changes both the level spacing and the number of states in each level. Successive levels cross the Fermi energy, producing an oscillatory density of states and hence quantum oscillations in thermodynamic and transport quantities.