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A normal mode has all displacement coordinates proportional to a common time dependence, with fixed relative amplitudes. For a nonzero frequency the time dependence is sinusoidal. A zero-frequency mode is a free displacement or uniform drift rather than an oscillation.

Use the printed mass ordering $M,m,M$. Displacements are measured from the unstretched equilibrium positions, givingEquivalently,This is the generalized eigenvalue problem for normal modes of three masses connected by two equal springs. Substitute $\mathbf{x}=\mathbf{u}{e}^{i\omega t}$ to obtain $\mathsf{K}\mathbf{u}={\omega }^2\mathsf{M}\mathbf{u}$. Reflection symmetry gives an antisymmetric vector $(1,0,-1)$ and symmetric vectors $(1,b,1)$. The former has ${\omega }^2=k/M$. For a symmetric vector, the end and centre equations give $M{\omega }^2=k(1-b)$ and $m{\omega }^{2}b=2k(b-1)$. Their solutions are $b=1$, $\omega =0$, or $b=-2M/m$, ${\omega }^2=k/M+2k/m$. HenceThese are all three eigenmodes. The last two keep the centre of mass fixed and are orthogonal in the mass-weighted inner product. The general zero-mode contribution is $(A+Bt)(1,1,1)$, accounting for both a centre-of-mass shift and a conserved total momentum. There are no fixed end springs in this system.

The original amplitude profiles illustrate $M=2m$; the symmetric internal shape is rescaled for display. The formula above gives every positive mass ratio.

Two further physical examples are standing-wave modes of a string fixed at both ends, with spatial shapes $\sin (n\pi x/L)$ and frequencies $n\pi c/L$, and electrical normal modes of two coupled LC circuits, in which the capacitor charges oscillate at fixed relative amplitudes. Independent normal mode coordinates simplify both systems in the same way as the spring-chain eigenvectors.

A Fourier series separates a periodic physical signal into harmonics. For example, a periodic voltage applied to a linear circuit can be treated one frequency at a time; the harmonic responses then superpose to give the output waveform.

Write $w=\Delta T$ for the pulse width and $\delta =w/T$ for its duty cycle, with $0<w<T$. The waveform is a train of positive rectangular pulses centred on $t=jT$, each with area one. It is even about $t=0$, so every sine coefficient vanishes. Cosine orthogonality givesand, for $n\ge 1$,This is the Fourier series of a periodic rectangular pulse:The actual constant term in the series is $a_0/2=1/T$, not $a_0$. As usual for a jump discontinuity, the series converges to the average of the one-sided values at each pulse edge, namely $1/(2w)$, irrespective of the value assigned at that isolated endpoint. Elsewhere it converges to the rectangular waveform.

Counting the constant as the first term, the first three terms areso their sum isIf “first three” instead means the first three nonconstant harmonics, add $a_3\cos (6\pi t/T)$; the original plot shows that alternative partial sum as well. The three-term approximation rounds the narrow peaks and ripples between them; many harmonics are needed to represent a sharp pulse.

The original sketches use the illustrative duty cycle $\delta =0.15$ and cover $-T\le t\le T$. The coefficient panel also displays the first spectral zero.

For $n\delta \ll 1$, expand the sinc function:Thus the low-index coefficients, including $a_0$, have nearly equal values, although their constant-series contribution still has the factor $1/2$. The first sinc zero occurs at $n\delta =1$, or $n=T/w$. If this is not an integer, the closest integer coefficient is near zero rather than exactly zero: writing $n\delta =1+\epsilon$ gives $a_n\simeq -(2/T)\epsilon$. This pulse-width scale of Fourier coefficients means that a narrower pulse requires significant harmonics extending to proportionally larger indices, the reciprocal time-width/frequency-width relation.

Convolution describes the response of a linear time-invariant system to an arbitrary input. If $h$ is its impulse response and $f$ its input,This is a weighted sum of responses to shifted impulses. For integrable functions the operation is commutative and associative. A causal system has $h(t)=0$ for $t<0$, restricting the integral to past inputs. With the Fourier convention $F(\omega )=\int f(t)e^{-i\omega t}dt$, the convolution theorem gives $G=HF$.

For example, an RC low-pass frequency response has $h(t)={\tau }_0^{-1}{e}^{-t/{\tau }_0}\Theta (t)$ and $H(\omega )=(1+i\omega {\tau }_0{)}^{-1}$, where $\Theta$ is the Heaviside step function. Fast input variations are smoothed. The same mathematics describes optical blurring by a point spread function and many other linear response measurements.

Deconvolution tries to recover $f$ from measured $g$ and known $h$. In noise-free data one formally sets $F=G/H$ where $H\ne 0$. Zeros of $H$ lose corresponding information; small values amplify measurement noise. If the measured spectrum is $G+N$, the direct estimate has error $N/H$, which can be very large at frequencies strongly suppressed by the instrument. An inverse is therefore not automatically unique or stable.

One regularized spectral deconvolution reconstruction isThis minimizes $|HF-G{|}^2+\lambda |F{|}^2$ at each frequency and limits noise amplification, at the cost of bias and lost resolution. Choosing the regularization requires assumptions about noise and the desired signal; deconvolution cannot restore components that the data do not determine without such additional information.

A normal mode is a motion in which all displacement coordinates oscillate at one angular frequency with fixed amplitude ratios. For small conservative oscillations, the symmetric stiffness and mass matrices give the generalized eigenvalue problem for small oscillations, $\mathsf{K}\mathbf{u}={\omega }^2\mathsf{M}\mathbf{u}$. For positive mass and a stable equilibrium, its mass-orthogonal eigenvectors span the motion. A superposition of these modes, with phases and amplitudes fixed by initial displacements and velocities, gives the general solution; different modes need not share a phase.

Use upward end displacements. The spring compressions are $x_0-x_1$ and $x_0-x_2$, while the beam's centre rises by $q=(x_1+x_2)/2$. The potential energy isStatic force balance gives $2k{x}_0=mg$, cancelling the linear terms and leavingFor a uniform beam, its central moment of inertia in the vertical plane is $m{l}^2/12$. To first order in the tilt, $\theta =(x_2-x_1)/l$. Its translational and rotational kinetic energies areThus $\mathsf{M}=(m/6)\left(\begin{array}{cc}2& 1\\ 1& 2\end{array}\right)$ and $\mathsf{K}=k1$. The normal modes of a rigid beam on two end springs have eigenvectorsThe first translates without tilt; the second rocks about a stationary centre. Ordinary frequencies are ${\omega }_s/(2\pi )$ and ${\omega }_a/(2\pi )$.

Assume the system had settled before the end load leaves, and put that load at end 1. Its extra static spring load is $Mg$ at that end and zero at the other. Therefore $x_1(0)=-Mg/k$, $x_2(0)=0$, and both initial velocities vanish because no impulse is imparted. Decomposing into the two normal modes givesOnly the symmetric mode contributes to the centre position:The datum $z_{\mathrm{eq}}$ is the unloaded beam's equilibrium height; no absolute height is supplied. The rocking mode is present even though it cancels out of this coordinate. These results use small tilt, $Mg/(kl)\ll 1$, ideal light vertical springs and negligible damping.

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.

The shell theorem gives enclosed halo mass $4\pi \rho r^3/3$. Define $\alpha =4\pi \rho /3$. Then $F_r=-GMm/r^2-\alpha Gmr$. Interior to the uniform halo its potential energy may be writtenOuter shells alter the additive constant, not this interior force. An actually infinite uniform mass distribution has no finite potential referenced to infinity; only local potential differences are needed here.

The central force has zero torque, so $\mathbf{J}=\mathbf{r}\times m\dot{\mathbf{r}}$ is conserved and the motion is planar. With $J=m{r}^2\dot{\varphi }$, the effective radial potential and equation areFor circular radius $r_0$, $J=m{r}_0^2\Omega$ givesSet $r=r_0+\xi$ at fixed $J$. The derivative of the right-hand side at equilibrium is $-3{\Omega }^2+2GM/r_0^3-\alpha G$, soBoth frequencies are positive, but radial and azimuthal motion no longer close with the Kepler ratio. At first order in eccentricity $e$, one may write $r=r_0(1-e\cos \omega t)$ and $\varphi =\Omega t+(2\Omega e/\omega )\sin \omega t$. For a weak halo this is a nearly Keplerian ellipse whose apsis rotates at angular rate $\Omega -\omega$.

Put ${\Omega }_K=\sqrt{GM/r_0^3}$ and $\epsilon =\alpha r_0^3/M\ll 1$. Expansion yieldsThe precession is retrograde. In cycles per unit time the precession frequency is $\dot{\varpi }/(2\pi )$; per radial period the apsidal shift is $2\pi (\Omega /\omega -1)\simeq -3\pi \epsilon$. This weak-halo interpretation avoids confusing a focus-centred Kepler ellipse with the centred ellipse of a purely harmonic potential.

A normal mode is a pattern in which all coordinates share one frequency and fixed relative amplitudes. For a conservative linear system, it solves a generalized eigenvalue equation with stiffness matrix $K$ and mass matrix $M$. For this free three-mass spring chain,Solving $K\mathbf{u}={\omega }^{2}M\mathbf{u}$, for $\alpha >0$, givesTheir overall normalization is arbitrary; distinct modes are orthogonal in the mass inner product. The zero mode is rigid translation because neither end is fixed. In the antisymmetric mode the middle mass stays still, so changing its mass cannot alter the frequency. The symmetric internal mode keeps the centre of mass fixed through its middle displacement $-2/\alpha$. Its frequency rises without bound as $\alpha \to 0$ and approaches $\sqrt{k/m}$ as $\alpha \to \infty$, becoming degenerate with the antisymmetric frequency in that limit.

For equal masses put ${\omega }_1=\sqrt{k/m}$ and ${\omega }_2=\sqrt{3k/m}$. Positions are equilibrium positions plus displacementsThe translational mode needs the linear-in-time term; a cosine at zero frequency alone would miss arbitrary centre-of-mass velocity.

Taking the springs initially at equilibrium, the impulse gives $\dot{\mathbf{q}}(0)=(v,0,0)=(v/3){\mathbf{u}}_0+(v/2){\mathbf{u}}_1+(v/6){\mathbf{u}}_2$. Their squared Euclidean norms are $3,2,6$, so their initial kinetic energies are $m{v}^2/6,m{v}^2/4,m{v}^2/12$. Relative to total $m{v}^2/2$,Subsequent motion conserves each mode's total energy, exchanging its kinetic and spring contributions. The zero-mode share remains centre-of-mass kinetic energy.

An applied electric field displaces bound charges or reorients permanent dipoles, producing electric polarization $\mathbf{P}$, the dipole moment per unit volume. It gives bound charges ${\rho }_b=-\nabla \cdot \mathbf{P}$ and ${\sigma }_b=\mathbf{P}\cdot \mathbf{n}$. Definefor a linear isotropic dielectric. Then $\nabla \cdot \mathbf{D}={\rho }_f$ separates free charge from the bound-charge response. Electronic, ionic and orientational electric polarization respond on different timescales; real permittivity is frequency dependent, and its dissipative part describes dielectric loss.

A uniform dielectric filling a parallel-plate capacitor raises its capacitance to $ϵA/d$. At fixed charge this reduces field energy $Q^2/(2C)$; at fixed voltage the stored energy $C{V}^2/2$ increases because the voltage source supplies charge and energy. Insulation and energy storage are useful applications, limited by loss and dielectric breakdown. Static interface conditions are continuous tangential $\mathbf{E}$ and a jump in normal $\mathbf{D}$ equal to free surface charge.

Magnetization $\mathbf{M}$ is magnetic dipole moment per unit volume. In SI units,in a linear isotropic response. The internal field matters: for a uniformly magnetized ellipsoid it is ${\mathbf{H}}_{\mathrm{int}}={\mathbf{H}}_{\mathrm{applied}}-{\mathbf{N}}_d\mathbf{M}$. Sample shape therefore affects an apparent magnetic susceptibility.

Diamagnetism opposes the applied field, giving small negative magnetic susceptibility from the induced orbital response. It is often only weakly temperature dependent. Paramagnetism aligns pre-existing moments and gives positive magnetic susceptibility; independent localized moments obey approximately $\chi =C/T$ in weak fields before saturating. Pauli paramagnetism of conduction electrons is a different, weakly temperature-dependent mechanism.

Ferromagnetism involves collective exchange-driven alignment and spontaneous magnetization below the Curie temperature. A specimen can nevertheless have nearly zero total moment because magnetic domains point in different directions. Field-driven domain-wall motion and rotation change the net magnetization. Antiferromagnets have compensating ordered sublattices below a Néel temperature; ferrimagnets have oppositely ordered but unequal sublattice moments and retain a net moment. Ferrites are useful examples of electrically resistive ferrimagnetic core materials.

Ferromagnetic response is nonlinear and history dependent. A hysteresis loop displays remanent magnetization, magnetic coercivity and saturation. A single constant magnetic permeability cannot describe the entire loop: differential magnetic permeability describes its local slope. The quasistatic energy loss per unit volume and cycle is the loop area $\oint HdB$. Soft magnetic materials have low magnetic coercivity and usually high useful magnetic permeability; hard magnetic materials retain magnetization and suit permanent magnets.

An electromagnet normally uses a soft, high-permeability core to guide flux generated by a coil. For a uniform-section core with length ${\ell }_c$ and air gap $g$, neglecting leakage and fringing,An air gap can dominate the magnetic reluctance even when the core magnetic permeability is large. Core saturation limits the improvement with current, while winding resistance causes heating. For alternating operation, laminations suppress eddy-current loops in metallic cores; high-resistivity ferrites also reduce those losses. Low hysteresis loss is important in repeatedly cycled cores, whereas high magnetic coercivity is valuable in a permanent magnet. Useful design therefore depends on the full magnetic response, electrical losses and operating frequency, rather than the largest quoted magnetic permeability alone.

For torque-free rotation, angular momentum is fixed in inertial space, but angular velocity need not point along it. In principal body axes,and cyclic permutations. A symmetric top with transverse moment $A$ and axial moment $C$ has constant ${\omega }_3$, while ${\omega }_1+i{\omega }_2$ rotates in body coordinates at signed frequency $(C-A){\omega }_3/A$. The symmetry axis precesses around the fixed angular momentum in space at rate $L/A$. These body-frame and space-frame rates describe different motions and need not coincide.

Poinsot construction expresses the same torque-free motion geometrically. The energy ellipsoid $\sum I_i{\omega }_i^2=2T$ is fixed in the body. Its normal at the angular-velocity point is parallel to $\mathbf{L}$, and $\mathbf{L}\cdot \mathbf{\omega }=2T$ places that point in a fixed plane perpendicular to $\mathbf{L}$. The ellipsoid rolls on this invariable plane without slipping: the instantaneous velocity of its contact point is proportional to $\mathbf{\omega }\times \mathbf{\omega }=0$. The contact curve is the polhode on the body ellipsoid and the herpolhode on the plane.^[ref]

For $I_1<I_2<I_3$, linearization about spin $\Omega$ on the third axis givesThe extreme-moment axes are stable; the intermediate one is unstable. The major-axis relaxation theorem adds weak internal dissipation while retaining negligible external torque. At fixed $L$,Energy loss favours rotation about the axis of largest principal moment. Here “major axis” means maximum inertia, not necessarily the body's longest geometric direction. A perfectly rigid torque-free body does not dissipatively settle; its energy is conserved. Distinguishing stability from relaxation also explains why minimum-inertia rotation can be stable in the ideal rigid model.

Forced precession occurs when an external torque changes $\mathbf{L}$, with $\dot{\mathbf{L}}=\mathbf{\tau }$. For steady precession, $\mathbf{\tau }={\mathbf{\Omega }}_p\times \mathbf{L}$. A rapidly spinning, pivoted heavy gyroscope has $L\simeq C{\omega }_3$ along its symmetry axis, givingThe torque is perpendicular to the nearly axial angular momentum, so it mainly changes its direction. For a steady inclined symmetric top, the more complete condition isIt exhibits slow and fast precession branches when real solutions exist. A generic release excites nutation as well: inclination oscillates instead of staying fixed. Axis constraints and bearing forces must be included when specifying which angular momentum and pivot are being used.

A fictitious force accounts for acceleration of the chosen reference frame rather than an additional interaction. For axes rotating at constant $\mathbf{\Omega }$, differentiating a vector uses $(d/dt{)}_{\mathrm{lab}}=(d/dt{)}_{\mathrm{rot}}+\mathbf{\Omega }\times$. Twice applying this givesso moving the last two terms to the force side produces the Coriolis and centrifugal forces. Here $\mathbf{v}$ is measured in rotating axes.

The rotating saddle trap has potential force $(-m{\omega }^{2}x,+m{\omega }^{2}y)$. ThereforeThe trial solution givesBoth modes are bounded for arbitrary sufficiently small initial displacements and velocities precisely when $|\Omega |>\omega$. At equality there is a zero-frequency secular mode; the exactly stationary solution does not establish stability.

For rapidly rotating saddle confinement with $\Omega \gg \omega >0$, real mode ellipses can be writtenThey are nearly circular clockwise motions at rates ${ϵ}_{\pm }\simeq \Omega \pm {\omega }^2/(2\Omega )$. Rotating these axes back to the laboratory cancels the large angular rate: the leading laboratory motion is a slow approximately harmonic ellipse with frequency ${\omega }^2/(2\Omega )$, plus small rapid micromotion. It results from the changing saddle force, not a static potential minimum. At next order the ellipse precesses prograde at ${\omega }^4/(8{\Omega }^3)$. The rotating-potential model and this guiding-centre distinction are analysed in the primary work cited here.^[ref]

A normal mode is a small oscillation with fixed relative displacement amplitudes, all coordinates oscillating at one frequency. linearization gives $M\ddot{\mathbf{y}}=-K\mathbf{y}$, so mode shapes solve $K\mathbf{u}={\omega }_n^{2}M\mathbf{u}$.

Write $C=q^2/(4\pi {ϵ}_0)$. force balance on an outer ion gives $m{\omega }^2{x}_0=C/x_0^2+C/(2x_0{)}^2$, henceA pair separated by $d$ has quadratic energy $C(y_j-y_i{)}^2/d^3$, equivalent to spring constant $2C/d^3$. The adjacent-pair constants are $8m{\omega }^2/5$, and the outer-pair constant is $m{\omega }^2/5$. Adding the trap stiffness givesIn the three-ion axial normal modes, the translation mode leaves every separation unchanged. The antisymmetric breathing mode leaves the central ion fixed; the remaining vector is orthogonal to both. Direct multiplication yieldsFor breathing excitation after a trap-frequency quench, put ${\omega }^{\prime }=\omega +\delta \omega$ and $x_0^{\prime }=x_0(\omega /{\omega }^{\prime }{)}^{2/3}$. The displacement from the new equilibrium is solely along the breathing vector and the initial velocity is zero. To first order in the fractional quench,The frequency shown is the linearized frequency about the new equilibrium. Nonlinear amplitude corrections are beyond this small-quench result.

Molecules. In a dilute classical gas, collisions establish a Maxwell-Boltzmann distribution. The normalized speed distribution and its first two moments areMomentum transfer to a wall gives $p=nm\overline{v^2}/3=n{k}_{B}T$. Translational internal energy is $U=3N{k}_{B}T/2$; rotational and vibrational degrees of freedom add energy when thermally populated, without changing the dilute ideal-gas pressure law. Classical equipartition gives $k_{B}T/2$ per quadratic contribution, but quantum level spacing can freeze out modes.

Only the molecules moving towards a surface contribute to its one-sided number flux:For an effusive molecular energy flux, faster molecules are preferentially sampled, giving mean translational energy $2k_{B}T$ per crossing molecule and $F_E=2k_{B}TJ$. This differs from multiplying $J$ by the bulk mean $3k_{B}T/2$. Collisions and finite molecular interactions produce transport and departures from the ideal-gas equation of state.

Photons. All photons travel at $c$ in vacuum, so their thermal distribution is over frequency, not over a range of speeds. The two polarizations and Bose-Einstein occupation with zero chemical potential giveHere the spectral quantity is $u_{\nu }$ and the total energy density is $u$. Momentum flux gives $p=u/3$, while isotropic angular averaging gives the one-sided energy flux $F_E=cu/4={\sigma }_{\mathrm{SB}}{T}^4$ through a small aperture of a black-body cavity. The photon number flux is similarly $c{n}_{\gamma }/4$.

Both pressures arise from momentum transfer, but their equations of state differ: a classical molecular gas has $pV=N{k}_{B}T$ with fixed molecule number, whereas a photon gas has $pV=U/3$, $U=a_{R}V{T}^4$ and an equilibrium photon number that changes with temperature. Photon creation and absorption by the walls are essential to the zero-chemical-potential equilibrium description.

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.

In Lagrangian mechanics, choose independent generalized coordinates $q_i$ compatible with the mechanical constraints, write $L=T-U$, and apply the Euler-Lagrange equationEquivalently, the physical path makes the action stationary under variations with fixed endpoints. For ideal holonomic constraints, their unknown reaction forces need not appear in the reduced equations. Direct use of Newton's laws of motion instead resolves vector forces and reactions and enforces the constraints separately. Both approaches describe the same mechanics; the scalar Lagrangian is often shorter in curvilinear coordinates. Nonconservative generalized forces must be added separately. Here a drive maintains the prescribed azimuthal angular velocity $\Omega$.

The mass has polar speed $a\dot{\theta }$ and azimuthal speed $a\Omega \sin \theta$, so its kinetic energy is $m{a}^2({\dot{\theta }}^2+{\Omega }^2{\sin }^2\theta )/2$. The elastic potential energy is $\mu {\theta }^2/2$, and the height is $a\cos \theta$, giving gravitational potential energy $mga\cos \theta$. Thereforeand the Euler-Lagrange equation givesFor $\mu =2mga$, define $q={\Omega }^{2}a/g$ and $F(\theta )=q\sin \theta \cos \theta -2\theta +\sin \theta$, so $\ddot{\theta }=(g/a)F(\theta )$. Near zero,When $q<1$ the upright equilibrium is linearly stable and the small oscillations have ${\omega }_1^2=g/a-{\Omega }^2$.

For $q>1$, the graph of $F$ initially rises, whereas $F(\pi /2)=1-\pi <0$. There must therefore be a zero inside the interval. More strongly, for $0<\theta <\pi /2$,is strictly decreasing: $\cos \theta$ decreases and $\theta /\sin \theta$ increases because $\sin \theta -\theta \cos \theta >0$. The ratio goes from $q-1>0$ to $1-\pi <0$, giving exactly one zero ${\theta }_0$ in that interval. At the zero, $F^{\prime }({\theta }_0)<0$, so it is a stable equilibrium. The rotating torsion-stabilized inverted pendulum also has a symmetric stable angle $-{\theta }_0$ if both signs of tilt are permitted.

At ${\theta }_0=\pi /4$, setting $F({\theta }_0)=0$ gives $q/2-\pi /2+1/\sqrt{2}=0$, soFor $\theta ={\theta }_0+\xi$, the linearized equation is $\ddot{\xi }=(g/a)F^{\prime }({\theta }_0)\xi$. Since $F^{\prime }=q\cos 2\theta -2+\cos \theta$,At the threshold $q=1$ the linear restoring term vanishes, but the cubic term remains:The upright state remains nonlinearly stable, with no nonzero infinitesimal harmonic frequency. Small finite disturbances produce quartic oscillator motion whose frequency is proportional to its small angular amplitude; its period tends to infinity as that amplitude tends to zero. Zero linear stiffness is not the same as an absent restoring torque.

Use a cold, unmagnetized, collisionless plasma with a sharp interface, and take the incident medium to be vacuum. Its relative permittivity is ${ϵ}_r=1-{\omega }_p^2/{\omega }^2$, so at the stated frequencies $n=\sqrt{1-(10/12{)}^2}=\sqrt{11}/6$. Conservation of tangential wavenumber givesFor oblique total reflection at an unmagnetized plasma, require $k_{z,\mathrm{plasma}}^2<0$, or $\cos {\theta }_i<5/6$. Thus total reflection occurs for ${\theta }_i>33.56^{\circ }$ measured from the normal, up to grazing incidence. At the ideal critical angle ${\theta }_c={\cos }^{-1}(5/6)$ the transmitted normal energy flux also vanishes, so the lossless reflectance reaches unity there. Collisions or an imposed magnetic field change this simple criterion. The constitutive response used here follows the electron-motion derivation in the cited university notes.^[ref]

Coupled oscillators have more than one collective pattern of motion. For two equal masses $m$, each attached to a fixed support by a spring of stiffness $k$ and linked to each other by stiffness $\kappa$, small displacements obeyHere equal linear damping $b$ is included to make finite steady-state responses at resonance. The normal mode coordinates $q_{+}=(x_1+x_2)/\sqrt{2}$ and $q_{-}=(x_1-x_2)/\sqrt{2}$ diagonalize these equations. Their natural angular frequencies areThe symmetric normal mode leaves the coupling spring unextended; the antisymmetric normal mode stretches it and therefore has the higher frequency. With $F_{\pm }=(F_1\pm F_2)/\sqrt{2}$, their complex steady-state responses areThus the drive's spatial pattern matters as well as its frequency. Equal in-phase forces drive only the symmetric normal mode; opposite forces drive only the antisymmetric one; driving a single mass generally excites both. Finite damping broadens the resonance peaks and produces a phase lag. In the ideal undamped system, a resonant drive with nonzero projection onto a normal mode produces secular growth rather than a bounded resonant steady state.