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Let the upper-state population lifetime be $T_1=1/\Gamma$, and take the lower energy level to be stable. Spontaneous emission gives survival probability $e^{-\Gamma t}$, hence a field or excited-state probability amplitude proportional to $e^{-\Gamma t/2}{e}^{-i{\omega }_{0}t}$ for $t\ge 0$. Its Fourier transform at angular frequency $\omega$ is proportional toThe squared modulus therefore has the Lorentzian line shapeHere $L$ is normalized over the full mathematical frequency axis. The full width at half maximum in ordinary frequency is $1/(2\pi T_1)$. If both energy levels decay, their population decay rates add in the natural linewidth; pure dephasing adds to the amplitude-decay rate instead.

Impact-regime pressure broadening provides another example. Independent phase-randomizing collisions with rate ${\gamma }_c$ give optical correlation proportional to $e^{-{\gamma }_c|t|}$, whose Fourier transform is a Lorentzian line shape of full width at half maximum $2{\gamma }_c$. This assumes short, uncorrelated impacts: a general static distribution of perturbing fields need not give a Lorentzian line shape.

Fix the qubit polarization convention by $\rho =(I+\vec{P}\cdot \vec{\sigma })/2$, where the Pauli matrices are ordered in the basis $(|1⟩,|2⟩)$ and ${\sigma }_z=|1⟩⟨1|-|2⟩⟨2|$. Thus the ground state is the north pole of the Bloch sphere. In a rotating frame and within the rotating-wave approximation, choose the drive phase and detuning sign so thatFor example, with this ground-state convention one may take $\Delta ={\omega }_L-{\omega }_0$ and choose the transverse coupling positive. The Bloch vector precesses around the effective field $(\Omega ,0,\Delta )$ at angular frequency ${\Omega }_R=\sqrt{{\Omega }^2+{\Delta }^2}$, tracing a circle at fixed projection on that axis. Its length is conserved under coherent unitary time evolution. On resonance the axis is transverse and the Rabi oscillation can fully invert an initially ground-state atom. Away from resonance the axis tilts, andThere is no relaxation towards an equilibrium Bloch vector in this ideal model. An inhomogeneous ensemble can nevertheless lose its averaged qubit polarization through dephasing even without spontaneous emission; the circle description assumes a common drive and detuning.

For $|a{|}^2+|b{|}^2=1$, the stated quantum state givesThe diagonal elements are the two state populations. The off-diagonal elements are quantum coherences: they retain the relative quantum phase and determine interference under a subsequent pulse. This is a pure state, even if both populations are nonzero: ${\rho }^2=\rho$, $\mathrm{Tr}{\rho }^2=1$, and $|\vec{P}|=1$. A mixed state averages projectors over preparation probabilities and generally has $\mathrm{Tr}{\rho }^2<1$. Equal populations alone do not distinguish an equal coherent superposition from an unpolarized mixture.

For use in all three cases, multiplication of the gates gives the Hadamard-phase-Hadamard qubit rotationThe global phase cancels in ${\rho }^{\prime }=U\rho U^{†}$. Thus the Bloch vector map is

The final unlabelled Ramsey spectroscopy comparison applies to the three cases as follows. In an atomic clock, an initial coherent pulse prepares a superposition, a free interval $T$ accumulates a relative quantum phase proportional to detuning times $T$, and a final coherent pulse converts that phase to a population. Here $\alpha$ plays the role of that accumulated phase, with sign and fringe offset set by pulse conventions. In (a), the first Hadamard gate creates the needed superposition and the final gate reveals full-contrast Ramsey interferometry. In (b), the first Hadamard gate instead puts the state in $|1⟩$, so the intervening phase gate has no relative quantum phase to imprint: the output is independent of $\alpha$. In (c), the unpolarized density matrix remains unchanged under every gate, so the ensemble has no Ramsey interferometry contrast. A coherent unitary gate alone cannot initialize a maximally mixed ensemble into a pure clock state.

The Ramsey interferometry analogy concerns preparation, phase accumulation, and readout; a Hadamard gate is not literally a transverse $\pi /2$ rotation on every input. In this basis $H=R_y(\pi /2){\sigma }_z$, and its Bloch sphere action is a $\pi$ rotation about $(\hat{x}+\hat{z})/\sqrt{2}$, up to a global phase. The phase conventions of physical Ramsey spectroscopy pulses must be specified to compare their outputs for arbitrary initial states.

Use the ordered basis $(|1⟩,|2⟩)$ and ${\sigma }_z=\mathrm{diag}(1,-1)$, withA pure quantum superposition $|+x⟩=(|1⟩+|2⟩)/\sqrt{2}$ has $\mathbf{R}=(1,0,0)$ on the Bloch sphere. In contrast, a mixed state $\rho =p|1⟩⟨1|+(1-p)|2⟩⟨2|$ has $\mathbf{R}=(0,0,2p-1)$, inside the sphere for $0<p<1$. At equal populations its vector is zero. Equal populations alone therefore do not distinguish a coherent superposition from a mixture; their off-diagonal density matrix elements differ.

Choose the drive phase so the rotating-frame Hamiltonian is $H_{\mathrm{rot}}=\hslash (\Omega {\sigma }_x+\delta {\sigma }_z)/2$, where $\delta =\omega -{\omega }_0$ as specified. The Rabi vector is the angular-velocity vectorHere $\Omega$ is the on-resonance population oscillation angular frequency, not the unhalved coefficient of a cosine coupling. From $\mathbf{R}(0)=\hat{z}$, rotation about $\mathbf{W}$ at $W=\sqrt{{\Omega }^2+{\delta }^2}$ givesFor $\Omega \gg |\delta |$, the rotation axis is nearly transverse, the trajectory nearly a great circle, and transfer is almost complete. For $\Omega =\delta >0$, the axis bisects $+x$ and $+z$, the orbit lies in $R_x+R_z=1$, and $p_2=\frac{1}{2}{\sin }^2(\sqrt{2}\Omega t/2)$ never exceeds $1/2$. Each member remains a pure state in this coherent evolution.

For the Ramsey free interval, let atom $j$ have detuning ${\delta }_j=\delta +{\eta }_j$. Instantaneous ideal pulses give the Ramsey ensemble coherence factor $\chi (T)=⟨e^{i{\eta }_{j}T}⟩$. Averaging the free-evolution vector obtained in step (ii) givesDifferent atoms fan out around the equator, shortening the ensemble Bloch vector even though every individual's vector remains unit length. A broad smooth detuning distribution usually makes the average spiral towards the centre and damps the Ramsey fringes towards $1/2$. The dephasing time is of order $1/\Delta {\omega }_0\sim 5/\delta$. inhomogeneous ensemble dephasing is not necessarily irreversible decoherence of individual atoms; an appropriate spin echo can rephase static frequency offsets.

The given spread does not specify a probability distribution or whether it denotes an RMS width. An exact envelope is therefore not determined. The figure illustrates, rather than infers, a Gaussian distribution with RMS ${\sigma }_{\delta }=0.2\delta$, for which $\chi =e^{-{\sigma }_{\delta }^2{T}^2/2}$. A uniform distribution would instead give a sinc envelope and side lobes. Both have the same qualitative initial loss of contrast on the quoted inverse-spread scale.

In the rubidium-87 ground-state hyperfine structure, $J=1/2$ and $I=3/2$, so $F=1,2$. Taking $H_{\mathrm{hf}}=A\mathbf{I}\cdot \mathbf{J}/{\hslash }^2$ givesThe zero-field splitting is ${\Delta }_{\mathrm{hf}}=2A\simeq h\times 6.8\mathrm{GHz}$. Ignoring the small nuclear magnetic term, the weak-field Landé g-factors are $g_{F=1}\simeq -1/2$ and $g_{F=2}\simeq +1/2$. The diagram labels all three $F=1$ and five $F=2$ substates, with linear shifts $g_F{\mu }_B{m}_{F}B$.

The pair $|1⟩=|F=1,m_F=0⟩$ and $|2⟩=|F=2,m_F=0⟩$ is connected by a magnetic dipole transition with $\Delta F=1$, $\Delta m_F=0$, driven with its oscillating magnetic component along the quantization axis. Both first-order shifts vanish, giving the rubidium-87 clock transition first-order insensitivity near zero field. Its insensitivity does not imply absence of quadratic shifts.

For these two $m_F=0$ levels the Breit-Rabi formula gives the quadratic Zeeman shift of a clock transition:where the last expression neglects the nuclear term and sets $g_J\simeq 2$. HenceThis uses ${\mu }_B/h\simeq 1.40MHz/G$ and the supplied $A/h\simeq 3.4\mathrm{GHz}$. With physical moments defined by ${\mu }_i=-\partial E_i/\partial B$, the differential magnetic moment is ${\mu }_2-{\mu }_1\simeq -8.2\times 10^{-3}{\mu }_B$. If effective magnetic moment denotes the positive energy slope instead, that slope difference is positive. The magnitude and physical frequency sensitivity are the same. Retaining the full atomic factors gives the familiar quadratic clock-frequency coefficient about $575Hz/{\mathrm{G}}^2$; the simplified supplied parameters give about $576Hz/{\mathrm{G}}^2$.^[ref]

Use a far-detuned optical dipole trap, such as a red-detuned focused-beam trap, to confine both clock states. Their first-order magnetic moments vanish; their residual quadratic magnetic slopes are opposite, so a simple static magnetic-field minimum does not confine both in the same way. An optical potential acts through the electronic polarizability and can trap both ground hyperfine states. It is not perfectly state independent: differential AC Stark shifts, intensity noise and residual magnetic gradients still need control during Ramsey evolution.

For an ideal weakly driven two-level atom, opposing red-detuned beams give velocity-dependent absorption and a friction force near zero velocity. Random photon recoils also cause momentum diffusion. Their balance gives the minimum Doppler temperatureat detuning $\Delta =-\Gamma /2$ in the low-saturation model, where $\Gamma$ is the spontaneous population-decay rate, equivalently the natural full linewidth expressed as an angular frequency. This limit assumes the atom has no useful ground-state internal structure.

Real atoms possess magnetic sublevels. In a suitable multilevel transition, position-dependent polarization gives different optical couplings and different AC Stark energy shifts for those sublevels. Counterpropagating beams with orthogonal linear polarizations have nearly uniform total intensity but a periodically varying polarization. Thus spatially varying light shifts can exist even without an intensity standing wave.

In Sisyphus cooling, an atom on one ground-state branch climbs a light-shift hill and loses kinetic energy. Near the upper part of the hill, optical pumping tends to transfer population to a lower branch. spontaneous emission makes this transfer dissipative and randomizes the emitted photon direction. The atom can then climb another hill. Finite pumping time is essential: instantaneous local equilibration would remove the delayed internal-state response that produces friction. The original multilevel analysis relates these optical-pumping cycles to the spatially displaced light shifts.^[ref]

For large detuning compared with the natural linewidth, useful scales areup to angular-coupling and Rabi-frequency convention factors. Here $\Omega$ is the optical Rabi angular frequency. The slow-atom capture range is of order $kv\lesssim {\Gamma }_{\mathrm{op}}$. Typical energies in the Sisyphus regime track the light-shift scale rather than only $\hslash \Gamma$, so reducing intensity can lower temperature below the Doppler limit. Capture becomes narrower, however, and recoil heating, imperfect polarization, magnetic fields and scattered light eventually matter. Define recoil energy $E_r=(\hslash k{)}^2/(2m)$; this is an important scale, not a claim that every sub-Doppler method has the same universal final temperature.^[ref]

The experimental test must establish a narrow velocity distribution independently of fluorescence brightness. William D. Phillips's account describes moving a time-of-flight probe above and below the cloud, deliberately heating the atoms to check a previously absent signal, and release-and-recapture comparisons that included gravity and finite beam geometry. These tests separated very cold atoms from an insensitive detector or a faulty thermometer.^[ref] In a modern implementation, field cancellation and polarization checks test the mechanism, while independent expansion thermometry checks the inferred temperature. sub-Doppler cooling exploits internal-state dynamics omitted from the two-level Doppler model; it does not contradict that model within its assumptions.

Taking the Hessian of the potential with respect to mass-weighted displacements givesThe equations of motion are $\ddot{\mathbf{q}}=-D_0\mathbf{q}$, so eigenvalues are squared angular frequencies. Normalized eigenvectors and frequencies areBoth vectors are in the q basis, and $\mu =m_a{m}_b/(m_a+m_b)$. The translation corresponds to equal physical displacements $\delta y_a=\delta y_b$. The stretch has $m_a\delta y_a+m_b\delta y_b=0$ and opposite physical displacements with magnitude ratio $|\delta y_a/\delta y_b|=m_b/m_a$. Ordinary frequencies are $\nu =\omega /(2\pi )$; spectroscopic wavenumbers are $\omega /(2\pi c)$.

The Euler-Lagrange equations of the quadratic Lagrangian areThe mechanical mass matrix is the identity. A normal mode $\mathbf{\theta }=\mathbf{a}\cos (\omega t+\varphi )$ therefore satisfies the eigenvalue problem $\mathsf{K}\mathbf{a}={\omega }^2\mathbf{a}$. Direct multiplication givesThese are normal modes of two spring-coupled equal pendulums. The corresponding normal mode coordinates $u={\theta }_1+{\theta }_2$ and $v={\theta }_1-{\theta }_2$ obey $\ddot{u}+u=0$ and $\ddot{v}+(1+2k)v=0$.

For the in-phase normal mode, the bobs undergo identical displacements, leaving their separation exactly $r$. The spring stays at its natural length and exerts no force, so this angular frequency is independent of $k$. The value $1$ is the individual simple pendulum angular frequency $\sqrt{g/l}$ in the chosen units.

Now take zero torque and $\mathbf{\omega }(0)=(1,0,1)$. The axial angular velocity is ${\omega }_3=1$ and the complex transverse component obeys $\dot{z}=i\nu z$ with $z(0)=1$ and $\nu =(I_3-I_1)/I_1$. Hence the angular momentum components in the principal body frame areThe transverse projection therefore traces a circle with signed angular frequency $\nu$. Its magnitude is constant, $L=\sqrt{I_1^2+I_3^2}$.

In the inertial frame, zero torque means that $\mathbf{L}$ is fixed. Usingand the derivative of a body-fixed basis vector, the symmetry axis satisfiesThus there is uniform torque-free precession of a symmetric top about the fixed angular momentum, at a constant cone angle $\beta$ and space angular frequencyThe cone angle stays constant, with no nutation. The body-frame and space-frame frequencies of free symmetric-top precession differ: $\nu$ describes rotating components, whereas ${\Omega }_{\mathrm{space}}$ describes the symmetry axis in space. The formulas also cover $I_3=I_1$, where the angular momentum components in the principal body frame are constant.

The supplied sound speed determines the spring-to-mass ratio without needing either parameter separately:ConsequentlyThe curve is even in $k$, rises linearly from the origin with one-sided slopes $\pm 4.0$ km/s, and flattens at both zone edges. Its maximum is ${\omega }_m=4.0\times 10^{13}$ rad/s, or ordinary frequency ${\nu }_m={\omega }_m/(2\pi )=6.366$ THz. Keeping angular frequency distinct from cycles per second avoids a factor-of-$2\pi$ error. Repeating the curve outside the first zone is a representation of equivalent crystal momenta, not the discovery of additional modes.

A phonon is a quantum of a normal mode of Bravais lattice vibration. A harmonic mode of angular frequency $\omega$ has energy $\hslash \omega (n+1/2)$, and adding one phonon increases its energy by $\hslash \omega$. The phonon carries crystal momentum $\hslash \mathbf{q}$, defined modulo a reciprocal lattice vector.

The first Brillouin zone is the Wigner-Seitz cell of the reciprocal lattice about the origin: the set of wavevectors at least as close to the origin as to any other reciprocal lattice point. Its boundaries are reciprocal-space perpendicular bisectors. It gives one representative of each wavevector modulo reciprocal lattice translations.

Quantum statistics matter when thermal energy is comparable with a characteristic vibrational level spacing. The chain's natural angular frequency is ${\omega }_0=\sqrt{\kappa /m}=5.7471\times 10^{12}{\mathrm{s}}^{-1}$, givingFactors of order unity depend on which vibrational scale is used. The largest chain phonon angular frequency is $2{\omega }_0$, corresponding to $87.8\mathrm{K}$. An isolated two-atom bond has reduced mass $m/2$ and angular frequency $\sqrt{2}{\omega }_0$, corresponding to $62.1\mathrm{K}$. Thus quantum corrections become appreciable on the scale of several tens of kelvin, and all modes are classical only when $T$ is comfortably above the largest relevant vibrational scale. Below it, high-frequency phonons are progressively frozen out and equipartition fails.

For the isolated bond, the zero-temperature root-mean-square relative displacement isIt is small enough for a local bond expansion. The classical weakly anharmonic estimate in part d therefore has a useful parametric windowwhere the lower condition means comfortably above the vibrational scale, rather than a sharp threshold. The model's upper bound concerns local anharmonicity and binding, not a prediction of the melting temperature.

Subtracting the three acoustic phonon branches from the six displacement normal modes leavesAt small wavevector, these involve relative motion of the two atomic basis sites rather than a rigid translation of the primitive unit cell. Their angular frequencies remain nonzero at the zone centre. The count uses the primitive unit cell; using all eight chemical atoms in the conventional unit cell would spuriously multiply the number of distinct branches in the first Brillouin zone.

In the diamond cubic structure of silicon, the longitudinal acoustic phonon branch and longitudinal optical phonon branch meet at $X=(2\pi /a,0,0)$. The two atomic basis sites are identical, and the associated crystal symmetry protects their boundary degeneracy of normal modes. In aluminium phosphide, the two sites contain different chemical atoms. The corresponding zinc blende structure lacks the site-exchange symmetry, and these two longitudinal branches generally split at $X$.

A simple way to see the splitting is a diatomic-chain dispersion relation with equal springs, representing successive $[100]$ atomic planes by alternating masses $M_1,M_2$ at spacing $d=a/4$. With effective spring constant $K$, its angular frequencies obeyAt $X$, $kd=\pi /2$, so the inter-sublattice coupling cancels and the two squared angular frequencies are $2K/M_1$ and $2K/M_2$. Equal masses give a contact; unequal masses giveIn aluminium phosphide, the heavier phosphorus plane moves in the lower-frequency edge mode, and the lighter aluminium plane in the higher-frequency mode in this model. Their similar atomic masses explain why the splitting is relatively small. Different electronic environments can also change the force constant matrix. The chain calculation illustrates the mechanism; a quantitative three-dimensional phonon spectrum requires its full dynamical matrix. It does not assert that all transverse acoustic and optical branches meet in silicon.

In the harmonic approximation, each in-plane displacement coordinate contributes one quadratic kinetic energy term and one quadratic restoring potential energy term. The equipartition theorem assigns $k_{B}T/2$ to each. There are two such coordinates per chemical atom, henceHere the fixed-size heat capacity is at constant area in this two-dimensional model. On cooling, phonon excitation freezes out and the heat capacity decreases. For each nonzero angular frequency, the quantum harmonic oscillator contributeswhich is below $k_B$ at finite $x$ and approaches zero as $T\to 0$. Two linearly dispersing in-plane acoustic phonon branches give the familiar two-dimensional Debye model scaling $C_v\propto T^2$ at sufficiently low temperature. Out-of-plane modes are excluded by the specified model.

A phonon is one quantum of a normal vibrational mode of a crystal. In the harmonic approximation, each mode with wavevector $\mathbf{q}$, branch index $s$ and angular frequency ${\omega }_s(\mathbf{q})$ is an oscillator with energies $\hslash {\omega }_s(n+1/2)$. A phonon changes the occupation $n$ by one; it is not a particular atom moving through the solid.

For a three-dimensional primitive cell containing $r$ atoms there are $3r$ branches. Three acoustic branches approach zero frequency as $\mathbf{q}\to 0$: at zero wavevector they represent rigid translations, with all basis atoms displaced together. Their small-wavevector frequencies usually grow linearly with wavevector. The remaining $3r-3$ optical branches have internal relative motion of basis atoms and generally nonzero zone-centre frequencies. The names reflect their coupling to radiation in suitable polar crystals, not a rule that every optical mode must be infrared active. Acoustic and optical branches can each have longitudinal or transverse polarizations.

Write $\mu ={\mu }_0{\mu }_r$ and $ϵ={ϵ}_0{ϵ}_r$ for homogeneous scalar material constants, and use $\mathbf{J}=\sigma \mathbf{E}$. Taking the curl of Faraday's law of induction and substituting the Ampere-Maxwell law givesThe vector calculus identity for a double curl therefore yieldsThe second term represents electric current response, rather than displacement current. A dispersive $\sigma (\omega )$ is used in the frequency domain; its general time-domain counterpart is a temporal response convolution.

For the convention $e^{i(\mathbf{k}\cdot \mathbf{r}-\omega t)}$, the plane wave equation isDecompose the electric field into components parallel and perpendicular to $\mathbf{k}$. With $S=\mu ϵ{\omega }^2+i\mu \sigma \omega$, the longitudinal eigenvalue is $S$ and each transverse eigenvalue is $S-k^2$; thus the determinant is $S(S-k^2{)}^2$. For a transverse electromagnetic wave, $\mathbf{k}\cdot {\mathbf{E}}_0=0$ andwhen ${\mu }_r=1$ and the background ${ϵ}_r\simeq 1$. A real dissipative electrical conductivity generally makes $k$ complex, with the decaying branch chosen for a passive medium.

For a cold, collisionless, unmagnetized plasma, treat ions as stationary and write $e>0$ for the magnitude of the elementary charge. The electron equation of motion is $-i\omega m_e\mathbf{v}=-e\mathbf{E}$, soThis reactive response includes no collision-induced Joule heating. Substitution into the transverse dispersion relation givesThe quantity ${\omega }_p$ is the electron plasma frequency. The refractive index is real above this cutoff; below it, the wave is evanescent. The electron response is represented here by $\sigma (\omega )$, so it must not also be counted a second time in the background permittivity.

For a nonzero longitudinal component, taking the dot product of the matrix equation with $\mathbf{k}$ instead gives $S(\mathbf{k}\cdot {\mathbf{E}}_0)=0$. The same cold plasma response makes $S=({\omega }^2-{\omega }_p^2)/c^2$, henceAt nonzero $k$, the remaining equation forces ${\mathbf{E}}_0$ to be parallel to $\mathbf{k}$, and Faraday's law of induction gives ${\mathbf{B}}_0=0$. These are electrostatic Langmuir waves, with no $k$ dependence of angular frequency in the zero-temperature model. Equilibrium neutrality does not forbid their perturbed charge density; Gauss's law requires such a density when $\mathbf{k}\cdot {\mathbf{E}}_0\ne 0$.

Represent a time-harmonic wave by $A\cos (kx-\omega t+\delta )$, with $k,\omega >0$. The angular frequency $\omega$ is the rate at which the wave phase changes at a fixed position, measured in radians per unit time. The wavenumber $k$ is its spatial rate of change, measured in radians per unit length. One full spatial cycle defines the wavelength $\lambda$, and one temporal cycle defines the wave period $T$, soThe ordinary frequency is $f=1/T=\omega /(2\pi )$; the paper's symbol $\omega$ denotes angular frequency.

Seek the periodic response of a surface gravity wave forced by a traveling bottom, with the same wavenumber $k$ and angular frequency $kc$ as the bottom. The linear equations have constant coefficients in $x,t$, so they preserve each such harmonic. There is no dissipative phase lag in this model; away from resonance, the surface response is in phase or in opposition with the bottom. Thus write $\eta =b\sin \theta$, $\theta =k(x-ct)$, with a signed coefficient $b$. Its physical amplitude is $|b|$. Arbitrary additional free surface gravity waves depend on initial data; the paper's expression selects the forced traveling component.

For $k>0$, $c\ne 0$, use $\varphi =F(y)\cos \theta$. The Laplace equation gives $F^{\prime \prime }-k^{2}F=0$. At $y=0$, the two linearized free-surface boundary conditions implyThese determine both constants in the vertical solution:Its vertical derivative at the bottom isMatching it to $-akc\cos \theta$ and multiplying by $c$ givesand thereforeAn equivalent expression for the velocity potential, which makes the bottom condition immediate, isLet ${\omega }_f=kc$ be the forcing angular frequency, and let ${\omega }_0^2=gk\tanh (kd)$ be the natural dispersion relation from Airy wave theory. ThenThe denominator vanishes at $c^2=(g/k)\tanh (kd)$, when the forcing travels at the free-wave phase velocity. For nonzero forcing there is no bounded periodic forced solution at resonance; the undamped linear response develops secular growth. Below that speed $b<0$, giving a phase reversal, while above it $b>0$. At $c=0$ the bottom is stationary: the rest solution has $\varphi =0$ and $\eta =0$, consistent with the zero limit of $b$, though the formula containing $1/c$ must be interpreted through that limit.

For a plane wave with amplitude $\alpha$, wavenumber $k>0$ and angular frequency $\omega >0$, the dispersion relation $\omega =\omega (k)$ specifies which frequencies the governing equations permit. A point of fixed wave phase obeys $kx-\omega t=\text{constant}$, so the phase speed isIt is the speed of individual crests. For a narrow wave packet, two nearby plane waves with differences $\Delta k$ and $\Delta \omega$ have a slowly varying wave envelope depending on $\Delta kx-\Delta \omega t$. Taking the limit gives the group velocityThis is the envelope's propagation speed and, in the present linear lossless wave model, the energy transport speed for a narrow packet.

A dispersive wave has phase speed dependent on wavenumber, so different components can change their relative phases and alter the packet's shape. A nondispersive wave on the chosen propagation branch has constant phase speed over the relevant band:Indeed $d{c}_p/dk=(c_g-c_p)/k$, so equality throughout a band is equivalent to constant phase speed there. Equality at an isolated wavenumber does not make the entire wave family nondispersive.

Take the wind to blow toward positive $x$, so $U_0\ge 0$. In subsonic flow, both stationary observers record the source angular frequency $\omega$ and, for the imposed density boundary data, the same acoustic density perturbation amplitude ${\rho }_0|A|$. Their spatial records distinguish the wavelengths and directions:The downstream pattern moves at $c_0+U_0$ and the upstream pattern at signed velocity $U_0-c_0$. Thus the downstream observer sees longer wavelengths while the upstream observer sees shorter ones. The interval containing multiple wavelengths lets each measure the spatial spacing and its direction of travel. The arrival delays at $\pm L$ are $L/(c_0+U_0)$ and $L/(c_0-U_0)$ respectively; after arrival their temporal traces differ by these propagation phases, not by temporal frequency.

In supersonic flow, both characteristic curves have positive slopes $dx/dt=U_0\pm c_0$. With an initially undisturbed upstream region, the observer at $-L$ sees no source-generated disturbance. The observer at $+L$ can see two downstream travelling waves, withThe faster signal arrives first, at $L/(U_0+c_0)$; the slower one arrives at $L/(U_0-c_0)$. Once both are present, their different wavenumbers can produce spatial modulation, while the time dependence at a fixed position remains harmonic with angular frequency $\omega$.

There is a qualification to the source data: the single prescribed value of ${\rho }_1(0^{+},t)$ fixes only $A_{+}+A_{-}=A$. It does not determine the two complex amplitudes separately. A second source condition, such as its downstream acoustic particle velocity with amplitude $c_0(A_{+}-A_{-})$, or a specified mass/force source model, is needed. Consequently two downstream branches are allowed; the stated density data alone do not require both to have nonzero amplitude. Extending a formally upstream-going subsonic formula into $x<0$ when $U_0>c_0$ would impose a signal arriving from upstream infinity and violate the causal source interpretation.

As $U_0\to c_0$, one laboratory phase velocity tends to zero. Its wavelength tends to zero, its wavenumber diverges, and its arrival time diverges; a finite-time experiment never attains the assumed established upstream field at fixed $L$. Finite source size, viscosity, thermal losses and eventually molecular scales invalidate this singular inviscid flow limit. With effective acoustic diffusivity ${\nu }_a$, a loss estimate is ${\nu }_a{k}^2/\omega \sim {\nu }_a\omega /(U_0-c_0{)}^2$, which diverges. Moreover the finite-amplitude change in the characteristic speed is of order $c_0|A|$, so uniform linearization requires $|U_0-c_0|\gg c_0|A|$. At exactly sonic background speed, the zero-speed branch cannot support a nonzero-$\omega$ harmonic solution. Nonlinear acoustics and losses must resolve the near-sonic regime; an unlimited spatial compression of the ideal wave is not physical.

No, for a uniform or otherwise axisymmetric disc in steady rotation. Although individual material elements accelerate, the mass density and mass current at every spatial point remain constant, and the mass quadrupole moment is time independent. There is no gravitational radiation. Axisymmetry of the material distribution is necessary for this conclusion: the source does not explicitly specify the disc's density. A counterexample to an unconditional answer is a disc with two equal dense spots at opposite radii; their rotating mass quadrupole moment produces gravitational waves at twice the rotation angular frequency.