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Two frictionless beads of positive masses $m_1,m_2$ on a circular hoop are joined by two identical springs of spring constant $k$, constrained to follow the two arcs. At their diametrically opposite equilibrium, each spring has natural length equal to half the circumference. For signed arc length displacements in the same rotational sense, the potential energy is $U=k(x_1-x_2{)}^2$. The normal mode of common rotation has mode shape $(1,1)$ and zero angular frequency, while the other normal mode has mode shape $(1,-m_1/m_2)$ and the positive angular frequency in the title.

Adding the equations of motion shows that $X=(m_1{x}_1+m_2{x}_2)/(m_1+m_2)$ satisfies $\ddot{X}=0$. Subtracting them gives $\ddot{\delta }+{\omega }_1^2\delta =0$ for $\delta =x_1-x_2$. Thus the complete zero mode is $X=C+Dt$, including uniform rotation, and the relative motion is a harmonic oscillator. In the limit $m_2/m_1\to \infty$ at fixed $m_1,k$, the oscillatory motion of the heavier bead vanishes for fixed lighter-bead amplitude, and ${\omega }_1\to \sqrt{2k/m_1}$. A common drift still survives if its initial velocity is held fixed.

The Nyquist–Shannon sampling theorem says that a bandlimited function can be recovered exactly from equally spaced samples, provided the sampling rate is high enough. It is the bridge between a continuous waveform and a sequence of numbers: the missing values between samples are determined by the restriction on its frequencies.

Here is a precise finite-energy version. Let $f\in L^2(\mathbb{R})$, meaning ${\int }_{\mathbb{R}}|f(t){|}^{2}dt<\infty$, and suppose its Fourier transform, defined through the Plancherel theorem, is zero outside $[-2\pi B,2\pi B]$. Thus $B>0$ is the largest allowed ordinary frequency, in cycles per second, rather than an angular frequency. Set $T=1/f_s$, where $f_s$ is the number of samples per second. If $f_s\ge 2B$, the continuous representative of $f$ is recovered byThe quotient is the normalized sinc function, with value one when its argument is zero. Each shifted sinc function is one at its own sampling time and zero at every other sampling time. Its coefficient is therefore exactly the required sample value. For this L2 space formulation, the reconstruction converges in the L2 space norm and uniformly in $t$. The critical rate $f_s=2B$ is allowed for this class; choosing $f_s>2B$ leaves a gap between the occupied frequencies and the sampling boundary.

For a numerical example, a bandlimited function with no frequencies above $3$ cycles per second can be reconstructed from its samples every $0.1$ seconds: $f_s=10>6$. Exact reconstruction uses the whole sequence indexed by $\mathbb{Z}$. A finite recording and a finite sum of sinc functions introduce truncation error; joining adjacent samples by straight lines is generally a different reconstruction.

The obstruction is aliasing. At sampling rate $f_s=10$, the two distinct waveformshave identical samples at $t=n/10$, becauseThe samples alone cannot tell whether the waveform oscillates twice or eight times per second. If we know in advance that all frequencies are below $5$, the $8$-cycle waveform is excluded and the ambiguity disappears. These pure cosines are an illustrative aliasing example, rather than members of the finite-energy L2 space class in the stated theorem.

Exactly at a sampling boundary, the function class matters. At rate $f_s$, the sine $\sin (2\pi (f_s/2)t)$ has zero values at every $n/f_s$. It is a nonzero waveform indistinguishable from zero on that grid, but has infinite energy and is excluded by the finite-energy hypothesis above. This explains why a rule intended to cover arbitrary pure sinusoids should demand a strict inequality. For spatial data, replace seconds by distance: the Nyquist spatial frequency is $1/(2p)$ for sample pitch $p$.

The proof idea is short: the samples are Fourier coefficients of the Fourier transform restricted to a sampling-sized frequency interval. Because a bandlimited function has no contribution outside that interval, its Fourier series determines the whole Fourier transform; Fourier inversion then gives the sinc function expansion. See sampling expansion by periodic Fourier projection for this connection.

Claude E. Shannon's Communication in the Presence of Noise (1949), Section II, Theorem 1 and equation (7), gives the sampling statement and reconstruction using shifted sinc functions in a communication setting. Shannon explicitly notes earlier mathematical forms and credits Nyquist with emphasizing the sampling interval in telegraphy.^[ref]

For two masses, connect each to a fixed anchor and to the other mass with equal stiffness $k$. Set $m_1=1,k=1$ to choose units. The squared angular frequencies have characteristic equation $m_2{z}^2-2(m_2+1)z+3=0$. Requiring a frequency ratio of two means that these roots are $z$ and $4z$. Their sum and product give $16m_2^2-43m_2+16=0$, yielding the two reciprocal ratios in the header. Restoring units multiplies all squared frequencies by $k/m_1$, leaving the permitted mass ratios unchanged. The term octave refers here only to this two-to-one ratio.

The open-loop gain of an operational amplifier is the ratio of its output voltage to its differential input voltage before an external feedback control connection is applied. A common small-signal model is $A(s)=A_0/(1+s/{\omega }_a)$, with a large direct-current gain $A_0$ and a dominant pole at $-{\omega }_a$. Above ${\omega }_a$ its magnitude falls approximately inversely with angular frequency.

An optical path length measures how much propagation wave phase a path accumulates, expressed as the distance that would accumulate the same wave phase in vacuum. A centimetre in a uniform medium with refractive index $3/2$ has optical path length $1.5\mathrm{cm}$, although its geometric length remains $1\mathrm{cm}$. This makes optical path length useful whenever two contributions must arrive with matching wave phases, from an optical lens to a Michelson interferometer. ^[ref]

For a specified path $\gamma$ at fixed angular frequency $\omega$, the optical path length is the line integralHere $s$ is geometric arc length, and $n$ is the phase refractive index, evaluated at the same frequency everywhere. The propagation interpretation assumes a stationary linear medium with isotropy, a real positive scalar refractive index and negligible loss, on a valid geometrical optics branch. The path can curve; at sharp interfaces apply the appropriate matching conditions to each segment. A prescribed curve has a mathematical line integral, but it predicts a propagation wave phase only when it represents the physical branch. ^[ref] Optical fibers can require a mode wavenumber rather than the bulk refractive index along a guessed light ray. ^[ref]

Fix the vacuum wavelength ${\lambda }_0=c/f=2\pi c/\omega$, with frequency $f=\omega /(2\pi )$ and vacuum wavenumber $k_0=2\pi /{\lambda }_0$. In a locally uniform medium the wavelength is $\lambda ={\lambda }_0/n$, while the frequency stays unchanged at a stationary interface. With the convention $\Phi =k_{0}S-\omega t+{\varphi }_0$, propagation from a specified starting wave phase contributesThe relative reference ${\delta }_{\mathrm{ref}}$ includes launch wave phases and any extra wave phase from mirrors, beam splitters or coatings. Thus $k_0\mathrm{OPL}$ is a propagation contribution, not an absolute detector wave phase independent of the apparatus. If the reference contributions stay fixed, a change in optical path difference $h$ changes the phase difference by $2\pi h/{\lambda }_0$. Using the material wavelength again in this formula would count the refractive index twice. ^[ref]

A motivating measurement turns invisible optical oscillations into a slowly changing electrical signal. MIT's teaching experiment uses a laser, a beam splitter, two returning mirrors, an amplified photodiode, an oscilloscope and a calibrated mirror actuator driven by a function generator. The returning beams overlap on the photodiode; one channel records detector voltage and another the actuator drive. The measured observable is the changing light intensity, including its maxima and minima, rather than direct timing of an optical cycle. A calibrated displacement per drive voltage and the spacing of the extrema relate geometric motion to the phase difference. The guide infers the laser wavelength from known actuator calibration; conversely, with independently known vacuum wavelength and ambient refractive index, the same relation calibrates mirror motion. These two unknowns cannot both be inferred from fringe spacing alone. ^[ref]

For a proposed change-of-medium measurement, keep the mirrors fixed and put a transparent cell of internal length $L$ in one arm. Compare its contents at refractive indices $n_a$ and $n_b$, leaving its windows in place. At normal incidence the two traversals giveHere $N$ is the signed number of whole interference fringe cycles plus any resolved fractional cycle, counted continuously. A mirror displacement $x$ instead gives $\Delta \mathrm{OPL}=2n_{\mathrm{amb}}x$; distinguish this double-pass factor from a single pass in a Mach-Zehnder interferometer. OpenStax's gas-cell worked example provides the same double-pass reasoning; it is a teaching calculation rather than a published measurement dataset.^[ref]

An exact toy calculation makes the sensitivity visible. Stipulate ${\lambda }_0=600\mathrm{nm}$, $L=10\mathrm{mm}$ and $n_b-n_a=3/100000=0.00003$. Then $\Delta \mathrm{OPL}=600\mathrm{nm}$ and $\Delta \delta =2\pi$: a full interference fringe cycle, despite unchanged geometric lengths. For a continuous change $n_b-n_a=q(3/100000)$, let $0\le q\le 2$ and choose the initial detector phase difference to be zero. Equal coherent detected arm contributions of light intensity $I_0$ predictThe plot is this stipulated model: $q=1/2$ gives a minimum, $q=1$ restores a maximum, and the reference level is chosen by adjustment of the apparatus. It contains no measured values or assigned properties of a real gas. The equivalent mirror travel in vacuum is $300\mathrm{nm}$ per cycle. A single endpoint light intensity cannot distinguish $q=0$ from $q=1$; the continuous fringe count carries the missing information.

For an interpretable cell measurement, establish the initial phase difference with a small calibrated mirror scan, then track slow changes through successive interference fringes. Measure the dark signal and each arm separately, keep photodiode gain within its linear range, align both beams on the same region and match their linear polarizations. Monitor interferometric visibility and keep arm delays within the useful optical coherence time. Record cell length, contents, pressure and temperature; return to the starting condition to check drift. Changes in window deformation, cell length, external air or mirror position also change optical path length, so a reference cell or an independently monitored mechanical path is needed to isolate the contents' refractive index. These controls are deductions from the model, not a claim that the illustrated apparatus was operated. Loss of fringe count leaves a whole-cycle ambiguity; at an intensity extremum the first derivative with respect to phase difference vanishes, so a scan around phase quadrature is more sensitive to small changes. Broadband illumination, unstable optical coherence, vibration and strong loss can invalidate a simple one-frequency inference.

For a lightly damped forced harmonic oscillator with known force amplitude $F_0$, measure the static displacement $x_s$, the resonance angular frequency ${\omega }_r$ and the peak displacement $x_p$. The response is $|x_0|=F_0/\sqrt{(k-m{\omega }^2{)}^2+b^2{\omega }^2}$. To leading order at large quality factor,The exact displacement resonance is ${\omega }_r^2=k/m-b^2/(2m^2)$, whereas the undamped natural angular frequency is ${\omega }_0=\sqrt{k/m}$. A lightly damped half-power bandwidth obeys $\Delta \omega \simeq b/m$ and $Q\simeq {\omega }_0/\Delta \omega$. Half power corresponds to displacement $x_p/\sqrt{2}$, not $x_p/2$. Graph readings should be reported as estimates rather than extra significant digits from the model inversion.

oscillatory convection is time-dependent convection emerging through nonzero-frequency overstability or a Hopf bifurcation. At onset a complex-conjugate growth-rate pair crosses the imaginary axis. Travelling and standing convection rolls are distinct nonlinear possibilities, and their stability cannot be decided from the linear angular frequency alone.

For $s=i\omega$, real-imaginary separation of the rotating-convection cubic givesThe curve is admissible only if ${\omega }^2>0$, so $P<1$ and sufficient rotation are necessary. A formal minimum with nonpositive angular frequency squared is not a Hopf bifurcation. At zero angular frequency this curve meets the stationary curve for that fixed wavenumber in a double-zero limit.

An ideal inductor and capacitor in parallel have electrical admittance $1/Z=1/(j\omega L)+j\omega C$ for phasors with time dependence $e^{j\omega t}$. The branch currents cancel at angular frequency ${\omega }_0=1/\sqrt{LC}$, giving zero total electrical admittance and an infinite limiting electrical impedance. Below this resonance its electrical reactance is positive; above it the electrical reactance is negative. Losses make the peak finite.

Set $\Delta p_0=p_{\perp 0}-p_{\parallel 0}$ and ${\mathcal{T}}_0=\Delta p_0+B_0^2/(4\pi )$. Distinguish the scalar magnetic field strength perturbation $\delta B$ from the vector perturbation $\delta \mathbf{B}$. To first order,where the last identity is the solenoidal magnetic-field constraint. For a Fourier mode with the stated time convention, putThe linear momentum conservation equation, after division by $i$, isUse the solenoidal magnetic-field constraint in its $z$ component. The coefficient of $\hat{\mathbf{z}}$ reduces to $-k_{\parallel }[\delta p_{\parallel }+\Delta p_0\delta B/B_0]$. ThusBoth scalar pressure perturbations act within the $xz$ plane. The $y$ component therefore decouples, independently of their closure. The ideal magnetohydrodynamic equations giveEliminating the amplitudes gives the Alfvén wave dispersion relationFor this Shear Alfvén wave, the velocity and fluid displacement are along $y$, perpendicular to both $\mathbf{k}$ and the equilibrium magnetic field. Let $\mathbf{\xi }={\xi }_y\hat{\mathbf{y}}$; thenA real snapshot ${\xi }_y={\xi }_0\cos (\mathbf{k}\cdot \mathbf{r}-\omega t)$ bends the magnetic field lines sideways, with $\delta B_y=B_0{\partial }_z{\xi }_y$. There is no first-order compression or change of magnetic field strength: $\nabla \cdot \mathbf{\xi }=0$, $\delta \rho =0$ and $\delta B=0$. Scalar pressure perturbations are unnecessary in this transverse subsystem. For a traveling stable branch, $\delta B_y/B_0=-k_{\parallel }{u}_y/\omega$, so the velocity and transverse magnetic field perturbations are in phase or opposite phase according to the branch. At $k_{\parallel }=0$ the bending force and this angular frequency vanish.

For $k_{\parallel }\ne 0$ the Alfvén wave becomes a firehose instability when its squared angular frequency is negative:Equality is marginal, rather than exponential growth. The pressure anisotropy must have negative sign; its absolute magnitude alone is insufficient.

For the intergalactic plasma, a negative relative pressure anisotropy of $-10^{-2}$ reaches this fluid threshold only for $\beta >200$. In particular, the literal pair $\beta =100$ and relative pressure anisotropy $-0.01$ is stable because the threshold is $-0.02$. Since the quoted numbers are approximate and allow larger plasma beta, this weakly magnetized plasma is susceptible, rather than necessarily unstable at every quoted value. For example, $\beta =300$ and relative pressure anisotropy $-0.01$ are unstable. Positive pressure anisotropy does not excite this branch.

For the fusion-device plasma, $\beta \sim 10^{-2}$ requires a negative relative pressure anisotropy below roughly $-200$, whereas its magnitude is only about $0.1$. The stabilizing magnetic tension dominates, so it is not susceptible to this fluid firehose instability under the stated conditions. Indeed, nonnegative perpendicular pressure requires the relative pressure anisotropy to be no smaller than $-1$; at such a small plasma beta even this extreme cannot satisfy the threshold.

Magnetic flux freezing ties a transverse fluid displacement to a bend in the magnetic field lines. For the uniform background, the transverse displacement equation isFor a sinusoidal bend, ${\partial }_z^2{\mathbf{\xi }}_{\perp }=-k_{\parallel }^2{\mathbf{\xi }}_{\perp }$. Positive pressure-anisotropy-modified magnetic tension therefore gives a restoring force, and inertia produces oscillation. In an isotropic plasma, this is precisely the positive magnetic tension $B_0^2/(4\pi )$.

Particles moving predominantly along bent magnetic field lines contribute a transverse momentum flux. Their larger parallel pressure opposes the restoring curvature force supplied by the magnetic field and perpendicular pressure. Once $p_{\parallel 0}-p_{\perp 0}>B_0^2/(4\pi )$, the net force points along the original displacement: a larger bend produces a larger outward force, giving positive feedback. Equivalently, the bending contribution to the quadratic potential energy is $\frac{1}{2}{\mathcal{T}}_0|{\partial }_z{\mathbf{\xi }}_{\perp }{|}^2$ and has become negative.

In this static, homogeneous fluid model the unstable roots areWith $e^{-i\omega t}$ the root $+i\gamma$ grows. Its real angular frequency is zero, so the unstable pattern grows without propagation in the equilibrium rest frame; its partner decays. Advection by a background flow would change that frame-dependent statement. The growth increasing without bound as $|k_{\parallel }|$ increases is a limitation of this long-wavelength model: finite gyroradius and other kinetic corrections cannot be neglected at sufficiently short scales.

Eliminating the shear stress from the two linear equations gives the Maxwell shear-wave dispersion equation in physical space,where $\nu$ is the kinematic viscosity. Substitution of a Fourier mode $e^{iky+\sigma t}$ givesThese are temporal exponents: the oscillation angular frequency is the imaginary part of $\sigma$, with the sign determined by the selected time convention. For $4\tau \nu k^2<1$ both exponents are real and nonpositive. At equality the repeated exponent is $-1/(2\tau )$; generic solutions include $t{e}^{-t/(2\tau )}$. Above the threshold the modes are damped propagating shear disturbances:At large $|k|$ their phase velocity tends to $\sqrt{\nu /\tau }=\sqrt{G_{\mathrm{el}}/{\rho }_0}$.

For the limit $\tau \to 0$ at fixed $\mu ,{\rho }_0$ and $k$, expansion of the square root givesAfter a short initial relaxation, the surviving branch obeys the diffusion equation $u_t=\nu u_{yy}$ of a Newtonian fluid. The fast branch disappears on the vanishing viscoelastic relaxation time; at $k=0$ the two exponents are exactly $0$ and $-1/\tau$. This limit is not uniform in $k$: modes with $k$ of order $(\tau \nu {)}^{-1/2}$ still sample the oscillatory transition. If instead $G_{\mathrm{el}}=\mu /\tau$ is held fixed as $\tau \to 0$, then $\mu \to 0$ and the fixed-$k$ slow exponent tends to zero. The ordinary Newtonian fluid limit therefore specifies which material parameter is held fixed.

Define $\hslash {\omega }_0=E_2-E_1$ and use radiation energy density $u(\omega )$ per unit angular frequency, excluding the zero-point contribution. The Angular-frequency Einstein coefficients give rates per atom: $B_{12}u({\omega }_0)$ for absorption, $B_{21}u({\omega }_0)$ for stimulated emission, and $A_{21}$ for spontaneous emission. Stimulated emission adds a photon to the stimulating mode; spontaneous emission persists with no incident photons.

In thermal equilibrium the Boltzmann distribution gives $N_2/N_1=e^{-\hslash {\omega }_0/(k_{\mathrm{B}}T)}$. Balancing up and down rates yieldsMatching the Planck law for every temperature fixesNo degeneracy factor is needed for these nondegenerate levels.^[ref]

For plates of large area $A$ and spacing $d$, the surviving low-frequency transverse electromagnetic mode has one polarization, $\mathbf{E}\parallel \hat{\mathbf{z}}$, and in-plane wavevector. Periodic in-plane counting givesMultiplying this single-polarization two-dimensional photon density of states by $\hslash \omega$ and the Bose-Einstein distribution produces the thermal radiation between parallel conducting plates:The same rate balance therefore givesThese $B$ coefficients refer to the radiation polarization in their respective environments: isotropic free-space $B$ must not automatically be equated to the coefficient for a specified $z$-polarized field.

For hydrogen, take the quantization axis along $z$, neglect spin-orbit coupling, and assume perfect plates, a wide in-plane region and an unmodified atomic wavefunction: $a_{\mathrm{B}}\ll d\ll c/{\omega }_0$, with negligible surface-induced mixing or shifts. The electric-dipole selection rules for hydrogen require $\Delta l=\pm 1$; the allowed $z$ field has $\Delta m_{\ell }=0$. Hence the polarization-selective spontaneous emission between conducting plates givesThe thin-cavity enhancement of a normal electric dipole applies: the $2p_0\to 1s$ dipole points along $z$ and couples to the enhanced mode density. For $m_{\ell }=\pm 1$, the transition dipoles lie in the $x$–$y$ plane and couple to none of these low-frequency modes: the electric-dipole decay rate is zero in the ideal model, so this channel's lifetime is infinite. Real finite, imperfect plates and additional decay or mixing channels limit the lifetime. The order-one orientation factor for the allowed transition does not change the requested lifetime scaling.

Write the trap angular frequency as $\omega$ and the single-particle oscillator length as $\ell =\sqrt{\hslash /(m\omega )}$. The center of mass coordinate $X=(x_1+x_2)/2$ and relative coordinate $r=x_1-x_2$ give independent quantum harmonic oscillators, both of frequency $\omega$, with masses $2m$ and reduced mass $m/2$. Label their excitation numbers by $N$ and $n$. The unperturbed energy eigenvalue is $\hslash \omega (N+n+1)$.

Particle exchange leaves $X$ unchanged and sends $r$ to $-r$. The exchange symmetry of two identical spin-one-half fermions therefore requires even $n$ to accompany the spin-one-half singlet state, and odd $n$ to accompany a spin-one-half triplet state. For the contact shifts of two spin-one-half fermions in a harmonic trap, the delta potential depends only on $r$. For odd $n$, the relative wavefunction vanishes at $r=0$, so its contact shift is zero. Within the first-excited manifold, both perturbations are diagonal in the labels $(N,n,S,M)$: this is a perturbation diagonal within a degenerate eigenspace. For $n=0$, the relative oscillator length is $\sqrt{2}\ell$, giving the first-order nondegenerate perturbation theory shiftThe Zeeman splitting is $2{\mu }_B{B}_{z}M$ for total spin projection $\hslash M$. Thus the five lowest states, counted with multiplicity, haveunder the assumed conditions $|U|,2{\mu }_B|B_z|\ll \hslash \omega$. The three triplet states have slopes $-2{\mu }_B,0,2{\mu }_B$, while both singlet states are field independent at this order. The three triplet state branches coincide at zero magnetic field. Their crossing with the excited singlet state is allowed because the total-spin sectors do not mix. The odd relative wavefunctions are in fact unaffected by this delta potential at any coupling; the displayed singlet state shifts retain only first order.

Following a fluid element. Let $\mathbf{X}(t)$ be a pathline, satisfying $\dot{\mathbf{X}}=\mathbf{u}(\mathbf{X},t)$. The material derivative, also called the Lagrangian time derivative, is the rate observed while following that element. The multivariable chain rule givesThe partial derivative holds position fixed; the advection term accounts for motion through a spatially varying field. For a vector field, the same formula applies component by component in fixed Cartesian axes.

Momentum balance. For an inviscid flow, pressure is the only surface stress. A small material volume $dV$ has net pressure force $-\nabla pdV$, obtained by the divergence theorem. Its mass is $\rho dV$ and its material acceleration is $D\mathbf{u}/Dt$. Applying Newton's second law gives the Euler equations for an inviscid fluid,This is the local momentum conservation equation when viscosity and physical body forces are absent. A prescribed body force per unit mass would add $\rho \mathbf{f}$ on the right.

Changing to rotating coordinates. In a rotating frame with constant angular velocity $\mathbf{\Omega }$, the lab acceleration is the rotating-frame material acceleration plus $2\mathbf{\Omega }\times \mathbf{u}+\mathbf{\Omega }\times (\mathbf{\Omega }\times \mathbf{r})$. ThereforeFor uniform rotation,With constant ${\rho }_0$, this gives the Euler equation in a uniformly rotating frame,The term subtracted from $p$ is the equilibrium centrifugal contribution. Dividing the bracket by ${\rho }_0$ defines a reduced pressure $\Pi =p/{\rho }_0-|\mathbf{\Omega }\times \mathbf{r}{|}^2/2$, with units ${\mathrm{m}}^2{\mathrm{s}}^{-2}$. The Coriolis acceleration remains a velocity-dependent term.

Linear disturbances. Linearize about rest in the rotating coordinates, with the equilibrium part of $\Pi$ removed. The advective acceleration is quadratic in disturbance amplitude, so the linearized Euler equations and incompressible flow condition areTaking the curl removes the pressure gradient. The identitygives the inertial-wave vorticity equation,

Plane-wave frequencies. Let $\mathbf{k}\ne 0$ and use the given plane wave convention. Incompressible flow requires $\mathbf{k}\cdot {\mathbf{u}}_0=0$, and the vorticity equation givesFor a nonzero frequency, apply $\mathbf{k}\times$ again, using $\mathbf{k}\times (\mathbf{k}\times {\mathbf{u}}_0)=-k^2{\mathbf{u}}_0$. It follows thatThis is the inertial-wave dispersion relation. If $\theta$ is the angle of the wavevector to the rotation axis, the positive angular frequency is $2|\Omega \cos \theta |$. It is largest, $2|\Omega |$, for a parallel wavevector, decreases to zero for a perpendicular one, and is independent of the magnitude of the wavenumber at fixed direction. The two signs describe the two oscillatory polarization branches. When $k_z=0$, the zero-frequency result is obtained directly from the linear equations rather than by dividing by $\omega$; it describes the nonoscillating balanced limit.

The buoyancy frequency, also called the Brunt–Väisälä frequency, is the angular frequency of a small vertical displacement restored by buoyancy in a stably stratified fluid. Take $z$ upward, gravitational acceleration $-g\hat{\mathbf{z}}$, and a background in hydrostatic equilibrium,Displace a small gas element upward by $\xi$, sufficiently slowly for its pressure to match its surroundings but without heat exchange. For this adiabatic process, $p/{\rho }^{\gamma }$ is constant along the element's motion, so its mass density changes by $\rho {\gamma }^{-1}(d\ln p/dz)\xi$. The surrounding mass density changes by $(d\rho /dz)\xi$. The element's excess mass density is consequentlyIts acceleration is $\ddot{\xi }=-g\Delta \rho /\rho =-N^2\xi$. ThusFor a gas of fixed composition obeying the ideal gas law, $\rho \propto p/T$. The adiabatic buoyancy frequency of an ideal gas is thereforewhere $c_p=\gamma \mathcal{R}/(\gamma -1)$ is the specific heat capacity at constant pressure, and $\mathcal{R}=p/(\rho T)$ is the specific gas constant. Here $N^2$ has units ${\mathrm{s}}^{-2}$; $N$ is an angular frequency, not a cyclic frequency. Positive $N^2$ gives simple harmonic motion, negative $N^2$ gives convective instability, and $N^2=0$ is marginal. The temperature form assumes fixed composition; the preceding density form is more general.

Put $V_0=200\mathrm{km}{\mathrm{s}}^{-1}$. A Taylor expansion near the centre givesTherefore the circular angular frequency isThus the central motion approaches solid-body rotation. Numerically ${\Omega }_0=1.6204\times 10^{-15}{\mathrm{s}}^{-1}$, and the central orbital period tends toThe finite limit of $\Omega$ accompanies $V_c\to 0$ at $r\to 0$; it does not imply a finite central linear speed.

The ground electron configuration of caesium has a single $6s$ electron, with $J=1/2$. Addition of angular momentum giving $F=3,4$ fixes the nuclear spin to $I=7/2$. With dimensionless angular momentum operators, write the Fermi contact interaction asThe two hyperfine levels therefore differ by $4A$. In the convention ${\mathbf{\mu }}_I=g_I{\mu }_B\mathbf{I}$ and $g_s\simeq 2$,This exposes a missing ingredient in a prediction from the supplied data: $Z$ and $g_I$ alone do not specify the valence electron contact density. Screening effect and penetration of the closed core strongly affect it. If $g_I$ instead denotes the total nuclear moment in units of ${\mu }_B$, it must first be divided by $I$ in the displayed convention.

An explicitly crude unscreened hydrogen-like ion model uses $|{\psi }_{6s}(0){|}^2=Z^3/(\pi 6^3{a}_0^3)$ and gives ${\nu }_0\simeq 5.76\times 10^{11}\mathrm{Hz}$, or ${\omega }_0\simeq 3.62\times 10^{12}{\mathrm{s}}^{-1}$. It describes neither the shielding nor the radial function of neutral caesium. Using a pure unit-charge $6s$ hydrogen orbital instead gives only $3.46\mathrm{MHz}$, illustrating why a unique useful estimate cannot follow from an unspecified effective charge.

For physical neutral ${}^{133}\mathrm{Cs}$, the independently known zero-magnetic field clock transition isThis measured frequency, used to define the second, is supplementary physical information, not a consequence of $Z=55$ and the rounded $g_I$ alone.^[ref] The direct electronic transition is in the microwave range and is a magnetic dipole transition, rather than a direct optical electric dipole transition. An optical implementation needs a two-field Raman transition.

Using the corrected weak-pulse Ramsey spectrum of part (b), the peak at $\delta =0$ has height $(\Omega {\tau }_p{)}^2$. The quantum interference fringes have spacing $2\pi /T$ in angular frequency, with zeros at $(2k+1)\pi /T$, and the sinc function envelope has its first zeros at $\delta =\pm 2\pi /{\tau }_p$. For $T\gg {\tau }_p$, the central peak has approximately full width at half maximum $\pi /T$ in angular frequency. The clock transition between the two $M_F=0$ states has no first-order Zeeman energy, although a quadratic one remains.

Choose the relative phases of the two energy eigenstates so that $d=⟨{\psi }_2|{\hat{D}}_0|{\psi }_1⟩$ is real and nonnegative. Define the resonant Rabi angular frequency $\Omega =d/\hslash$, the Rabi detuning $\delta ={\omega }_0-\omega$, and $\Sigma ={\omega }_0+\omega$. Substitution into the Schrödinger equation, followed by projection onto the two eigenstates, gives the exact interaction picture equationsKeep both probability amplitudes dynamical: replacing $c_1$ by one would incorrectly impose weak excitation and miss complete population transfer.

For $|\delta |,\Omega \ll {\omega }_0$, apply the rotating-wave approximation to discard the rapidly oscillating terms at $\Sigma$. Differentiating the second remaining equation and using the first givesThe initial conditions are $c_2(0)=0$ and ${\dot{c}}_2(0)=-i\Omega /2$. With ${\Omega }_R=\sqrt{{\Omega }^2+{\delta }^2}$, their solution isConsequently,This result permits order-one excitation, including full population transfer on resonance, even though the drive is small compared with the optical angular frequency.

For an average from switch-on to time $\tau$, integrate the squared sine:On resonance this limiting average is $1/2$, although the instantaneous probability repeatedly reaches one. The zero-coupling, zero-Rabi detuning case is understood separately as $P_2=0$.

Near resonance alone does not make this an exact result for arbitrarily strong driving. The stated sinusoidal Hamiltonian also needs the rotating-wave approximation. A concrete counterexample is $\omega ={\omega }_0$ and $\Omega ={\omega }_0$: the exact equations give $P_2(t)={\Omega }^2{t}^2+O(t^4)$ at switch-on, while the claimed rotating-wave sinusoid gives ${\Omega }^2{t}^2/4+O(t^4)$. Thus “without assuming weak radiation” can mean without assuming small excitation, but cannot remove the drive-to-carrier scale separation. The infinite-time average above belongs to the rotating-wave model; omitted corrections can accumulate over sufficiently long times.