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Put $K=(k^2+m^2{)}^{1/2}$ and $P=p^{\prime }/{\rho }_0$. Acting on the Fourier mode, $D_0=-i(\hat{\omega }-Uk)=-i\omega$. Continuity and horizontal momentum give $u=-mw/k$ and $P=\omega u/k$. The buoyancy equation gives $g{\rho }^{\prime }/{\rho }_0=i{N}^{2}w/\omega$. Substitution into vertical momentum yieldsThe two signs describe the two frequency branches; when the intrinsic frequency is required positive, $\omega =N|k|/K$. For a stationary disturbance $\hat{\omega }=0$, so $\omega =-Uk>0$. With $U>0$ this requires $k<0$ and selects the minus sign, $\hat{\omega }=Uk-Nk/K$. The stationary dispersion condition isDifferentiate the unconstrained dispersion before imposing stationarity. The stationary mountain-wave group velocity is obtained from the ground-frame dispersion; before imposing stationarity it isOn the stationary locus this becomesDifferentiating after substituting $K=N/U$ as though all neighbouring modes were stationary would incorrectly erase the frequency gradients.

For upward energy propagation from the lower boundary, choose $m<0$, since $k<0$ and ${\hat{c}}_{gz}>0$. It requires real nonzero $m$, henceThe downstream ray inclination above horizontal isIt is fixed for a given monochromatic forcing, with the two signs of $m$ giving the upward/downward rays.

Frame qualification for the requested phase angle: the normal intrinsic phase velocity is ${\mathbf{c}}_p=\omega (k,m)/K^2$. Because $\omega =-Uk$, the boxed stationary group velocity is exactly $-{\mathbf{c}}_p$: its angle to intrinsic phase propagation is $\pi$, and it is perpendicular to the stationary phase lines. By contrast the intrinsic group velocity ${\mathbf{c}}_g=\widehat{{\mathbf{c}}_g}-U\hat{\mathbf{x}}$ satisfies ${\mathbf{c}}_g\cdot (k,m)=0$, so its angle to intrinsic phase propagation is $\pi /2$. In the ground frame the phase velocity of a strictly stationary pattern is zero and has no nonzero propagation direction. For example $k=-0.6N/U$, $m=-0.8N/U$ gives $\widehat{{\mathbf{c}}_g}=(0.36U,0.48U)$ and ${\mathbf{c}}_p=(-0.36U,-0.48U)$. These distinctions prevent applying the usual zero-flow perpendicularity rule to two velocities expressed in different frames.

A mountain width is not an exactly specified sinusoidal wavelength. Taking its characteristic horizontal inverse length as $|k|\sim 1/a$ with $a=10\mathrm{km}$ gives $|k|\sim 10^{-4}{\mathrm{m}}^{-1}$. Since $N/U=10^{-3}{\mathrm{m}}^{-1}$, the stationary vertical wavenumber isThis is close to the long-horizontal-scale/hydrostatic estimate $2\pi U/N=6.28\mathrm{km}$. If instead the quoted width is identified with a sinusoidal wavelength ${\lambda }_x=10\mathrm{km}$, then $|k|=2\pi /{\lambda }_x$ and ${\lambda }_z\simeq 8.1\mathrm{km}$. Thus several kilometres, roughly $6$–$8\mathrm{km}$ under these two common width conventions, is the justified estimate; the terrain profile would fix its actual forcing spectrum.

For the inverse-width choice, both $k$ and $m$ are negative for upward energy radiation. The stationary phase lines $kx+mz=\text{constant}$ tilt upstream with increasing height, while energy propagates upward and slightly downstream. Numerically $\widehat{{\mathbf{c}}_g}\simeq (0.10,0.995)\mathrm{m}{\mathrm{s}}^{-1}$, with ray inclination about $84.3^{\circ }$. Intrinsic phase propagation points upstream and downwards; the ground pattern itself remains fixed. The original sketch below uses this explicit convention.

A ground observer may see stationary lenticular or wave clouds where ascent cools sufficiently moist air, with continuing air flow through a cloud whose position remains almost fixed. Descending regions can clear or warm. An aircraft crossing the pattern encounters alternating updraughts and downdraughts; gliders can exploit the lift. Strong amplitudes, shear or rotors may cause turbulence, but a linear gravity wave can also be smooth, and clear-air waves need not produce any visible cloud. Mountain height, humidity and nonlinear stability are not supplied, so neither a vertical-wind amplitude nor a definite atmospheric precipitation forecast follows.

For constant $N^2$, substitution of the vertical cosine and horizontal plane wave givesThe top rigid-lid condition is satisfied by $\cos (mz)$. To select a discrete vertical mode one must also impose the corresponding no-flow, adiabatic condition at the flat bottom, ${\psi }_z^{\prime }(-H)=0$. It requires $mH=n\pi$. The mode $n=0$ is barotropic; the gravest nonzero baroclinic mode has $m=\pi /H$. HenceThe lid condition alone is insufficient: $\cos [\pi z/(2H)]$ also has zero derivative at the lid but violates the bottom condition. Constant depth motivates the additional physical boundary, which must be used to obtain the printed denominator.

Let $a_d=\pi |f_0|/(HN)=L_d^{-1}$ and set $l=0$. The zonal phase velocity and group velocity areThe phase curve is even in $k$, with limiting value $U-\beta /a_d^2$ at zero and approach to $U$ from below at large $|k|$. The group curve is also even: it starts at the same long-wave limit, equals $U$ at $|k|=a_d$, reaches $U+\beta /(8a_d^2)$ at $|k|=\sqrt{3}{a}_d$, and approaches $U$ from above at large $|k|$. The normalized curves appear in the bottom panels of the preceding figure. For $k=0$, the phase speed means the limit $\omega /k$; there is no zonal crest propagation to define directly at exactly zero wavenumber.

Take a Northern-Hemisphere cyclonic vortex, with positive relative vorticity. For a fixed-depth incompressible layer, conservation of absolute vorticity givesThe anticlockwise flow on the east side moves parcels northward, increasing their planetary vorticity and creating negative relative vorticity anomalies. On the west side it moves parcels southward and creates positive relative vorticity anomalies. Initially,so the positive anomaly grows on the western flank while the eastern flank loses positive vorticity. The cyclonic pattern consequently shifts westward. The induced anomalies themselves drive meridional flow and further relative vorticity anomalies, transmitting a Rossby wave pattern through the surrounding fluid. The restoring process is this exchange between planetary vorticity and relative vorticity, rather than a parcel executing a permanently westward translation.

The sketch shows the initial tendency, with the generated signs separated from the initial positive core. An isolated disturbance contains many wavevectors and disperses; westward phase velocity does not imply that all its wave energy travels westward. A Southern-Hemisphere cyclonic vortex reverses the signs of the initial and generated relative vorticities, while the same beta plane mechanism remains.

For $U=0$ and fixed $\omega >0$, completing the square in the dispersion relation givesThese Rossby-wave isofrequency circles lie in the $k\le 0$ half-plane and touch the origin, where the dispersion relation is undefined. Their radii decrease as frequency increases. The zero-frequency line is $k=0,l\ne 0$.

The group velocity is normal to the constant-frequency curves, toward increasing frequency:At $l=0$, the group velocity is eastward; close to $k=0$ it is westward. For the positive-frequency circles its arrows point inward toward the circle centre. Its meridional component is southward in $k<0,l>0$ and northward in $k<0,l<0$.

For $k\ne 0$, the zonal phase velocity, meaning propagation at fixed $y$, isIt is westward everywhere it is defined. The zonal group velocity is eastward for $|k|>|l|$, westward for $|k|<|l|$, and zero on $|k|=|l|$. There is no region of eastward zonal phase velocity. At $k=0$, $\omega /k$ is undefined, though its continuous limiting expression is westward and $c_{gx}=-\beta /l^2$. Using instead the normal-to-wavecrest phase velocity vector gives ${\mathbf{c}}_p=\omega (k,l)/(k^2+l^2)$; its zonal component is also westward for $k\ne 0$, but is not the same scalar as $\omega /k$.

For a nonzero zonal wavenumber $k$, stationarity requiresA real propagating solution therefore requires a westerly $U>0$. For $0<|k|<K_s$ there are two meridional wavenumbers:The source's assertion of two stationary solutions needs these qualifications: at $|k|=K_s$ they coincide, and at $|k|>K_s$ there is no real meridionally propagating solution. For example $k=2\sqrt{\beta /U}$ gives $l^2=-3\beta /U$. Modes with $k=0$ are a separate zonally invariant degeneracy and do not have a defined zonal phase velocity.

Adding $U$ to the zonal group velocity and imposing stationarity givesThus a localized midlatitude disturbance radiates stationary Rossby wave activity downstream, along one poleward and one equatorward branch. For $k>0$ in the Northern Hemisphere, $l_{+}$ gives the northeastward branch and $l_{-}$ the southeastward branch. With strictly constant $U,\beta$, these wave rays are straight. Changing latitude cannot bend them within that constant-coefficient model.

To discuss a planet, extend it locally to slowly varying $\beta (\varphi )=2\Omega \cos \varphi /a_p$ and possibly $U(\varphi )$. The conserved zonal wavenumber in a locally zonally uniform medium then hasFor constant $U$ and northward propagation in the Northern Hemisphere, decreasing $\beta$ reduces $l$ until $l=0$: the wave ray becomes zonal at a turning latitude and can refract equatorward. The equatorward branch initially encounters larger allowed $K_s$ and becomes more meridional. This local midlatitude description eventually needs replacement near the equator or wherever the background varies rapidly.

Orographic forcing by a mountain range or persistent spatially localized heating can supply such a disturbance. Its steady forcing selects the ground-stationary part of the response. The corresponding local total wavelength isFor fixed $U$ it increases toward the pole as $\beta$ decreases. Locally the zonal integer scale has upper bound $n_{\max }\simeq a_p\cos \varphi K_s=\sqrt{2\Omega a_p{\cos }^3\varphi /U}$, so fewer planetary-scale zonal waves fit around higher-latitude circles. This is not an unconditional trend in actual weather: changing $U$, its curvature, stratification and the forcing spectrum can change the dominant Rossby wave wavelengths.

Put $\eta =h-h_0$ and $c=\sqrt{g^{\prime }{h}_0}$. With $u=0$ and complex dependence $e^{i({\omega }_{K}t-{\ell }_{K}y)}$, the northward momentum equation and continuity equation giveThus ${\omega }_K^2=c^2{\ell }_K^2$. Write the signed phase velocity as $c_y={\omega }_K/{\ell }_K=\pm c$, so $v=(c_y/h_0)\eta$. The eastward momentum equation is the transverse geostrophic balance $g^{\prime }{\eta }_x=f_{0}v$. HenceThe decaying sign is selected by which side of $x_b$ contains the ocean. For an eastern boundary the ocean occupies $x<x_b$, so decay offshore requires $f_0/c_y>0$. Consequently the Kelvin wave moves northwards in the northern hemisphere and southwards in the southern hemisphere, with the coast respectively on its right and left. The positive offshore decay length isThe sign of $f_0$ reverses the propagation direction and the sign of $v$ relative to $\eta$, but not the speed or offshore decay length. The real physical fields are the real parts of the complex solution. At the equator itself this local f-plane coastal solution loses its finite trapping scale.

Coastal Kelvin-wave radiation of long Rossby waves requires a common phase convention to match the two waves. The supplied Rossby wave formula conventionally multiplies $e^{i(k_{R}x+{\ell }_{R}y-{\omega }_{R}t)}$, while the coastal expression uses $e^{i({\omega }_{K}t-{\ell }_{K}y)}$. Since physical fields are real, matching may use the complex-conjugate representative; in either convention the boundary forcing fixes the absolute frequency and alongshore wavenumber. Thus $|{\omega }_R|=|{\omega }_K|$ and $|{\ell }_R|=|{\ell }_K|$.

For $k_R^2+{\ell }_R^2\ll L_d^{-2}$, the shallow-water Rossby-wave dispersion relation reduces toChoosing $k_R>0$ recovers the printed magnitude relation. Both the zonal phase velocity and zonal group velocity are approximately $-\beta L_d^2$, so the long Rossby waves carry the coastal signal westwards into the basin.

The coastal condition $|{\ell }_K|L_d\ll 1$ alone does not guarantee a long zonal wavelength. Since $|{\omega }_K|=c|{\ell }_K|$, matching requires $|k_R|=c|{\ell }_K|/(\beta L_d^2)$ and hence the stronger consistency conditionFor example, at $10^{\circ }$ a coastal wavenumber with $|{\ell }_K|L_d=0.05$ gives $|k_R|L_d\simeq 0.56$, which is not asymptotically small even though the meridional wavelength is long. Thus the source's proposed excitation is possible at sufficiently low frequency; it is not an automatic consequence of its meridional long-wave assumption.

At $10^{\circ }$ N or S the beta plane parameter isTaking a representative Pacific basin width $W\simeq 1.5\times 10^7$ m givesA width of $1.4$–$1.7\times 10^7$ m gives roughly $2.0$–$2.4$ years. This is propagation near $10^{\circ }$ latitude, not an equatorial $f=0$ calculation. Slow Rossby wave adjustment supplies a basin-scale delay relevant to El Niño–Southern Oscillation: changing thermocline depth can affect later sea surface temperature and coupled wind forcing. Reflection and coupling may help produce interannual memory, but this travel-time estimate alone does not establish a climate oscillation or its period.

For group velocity from a period-dependent phase velocity, start with $k=\omega /c_n(\omega )$ rather than equating group velocity to phase velocity. Differentiating,For the stated inverse-frequency term, $d{c}_n/d\omega =-a/{\omega }^2$, soThe inequality assumes positive frequency and physical positive phase velocity, with $a>0$. The symbol $a$ here has units of acceleration and is unrelated to the plate thickness in Question 1.

The phase velocity in Figure 1 is measured along the Earth's surface. Spherical geometry makes the local horizontal phase speed at radius $r$ equal to $cr/R_E$. In the ray approximation, squared radial shear wavenumber isThe spherical surface-wave evanescence threshold immediately below the discontinuity is thereforeEquality is the grazing limit. Using $5.08\mathrm{km}{\mathrm{s}}^{-1}$ directly would omit the requested spherical correction. Immediately above the discontinuity the corresponding boundary is $4.87(6371/5961)\simeq 5.20\mathrm{km}{\mathrm{s}}^{-1}$. Hence the interval $5.20<c<5.43\mathrm{km}{\mathrm{s}}^{-1}$ specifically describes propagation just above the jump and evanescence just below it. Slower modes can already be evanescent above $410\mathrm{km}$.

Intersecting the plotted branches with $5.43\mathrm{km}{\mathrm{s}}^{-1}$ gives the following approximate upper periods. Below each upper period, that branch is evanescent immediately beneath the discontinuity:

For these overtones the displayed ranges thus start near $10\mathrm{s}$ and end at their respective crossings. Any still higher curve already above the threshold at the lower plotted period has no evanescent interval within the displayed range. Readings of the closely spaced higher curves are only accurate to about a second, and the plot does not establish their continuation to unshown periods. Nor does the local inequality prove that a mode remains evanescent at every greater depth in an unspecified Earth model.

Seismic surface-wave dispersion. Rayleigh waves combine vertical and longitudinal motion in the vertical plane of propagation; Love waves have horizontal transverse motion and can be guided by a low-shear-speed surface layer. A homogeneous elastic half-space supports a Rayleigh wave with a frequency-independent speed in the ideal elastic limit, but depth-dependent Earth structure makes the surface-wave eigenfunctions and their propagation speeds depend on frequency. This structural dispersion can occur even when each constituent material is itself nondispersive.

Distinguish phase velocity $c=\omega /k$ from group velocity $U=d\omega /dk$, which controls a narrow-band envelope in the appropriate weakly attenuating limit. Longer periods generally have deeper sensitivity, while shorter periods emphasize shallower structure. In a profile whose shear speed increases with depth this often makes long-period waves faster, but low-velocity zones, anisotropy and higher modes can change that trend. Frequency dependence need not be monotonic in every Earth model.

Measure phase delays or wave-packet arrival times across known source–receiver paths for many periods, accounting for source phase, cycle ambiguity, multipathing and mode identification. Compute predicted dispersion from trial elastic profiles and compare with observations. Linearization expresses the sensitivity schematically aswhere $\alpha ,\beta$ are compressional and shear speeds. A regularized inverse problem then estimates depth structure; many crossing paths additionally provide surface-wave tomography. Joint Love wave and Rayleigh wave measurements help distinguish shear-wave polarizations and constrain radial seismic anisotropy. Finite bandwidth, depth averaging and correlated sensitivities mean that one dispersion curve does not uniquely determine a detailed Earth profile.

Seismic ray theory, the eikonal equation and Fermat's principle. In a smoothly varying isotropic medium, insert a high-frequency ansatz $u=A(\mathbf{x})e^{i\omega [T(\mathbf{x})-t]}$ into the scalar wave equation. The leading powers of frequency giveThe travel-time surfaces are wavefronts and rays are their normal trajectories, with slowness vector $\mathbf{p}=\nabla T$. The next-order equation transports amplitude, including geometrical spreading; attenuation is an additional material effect.

Fermat's principle says that the ray makes $\int ds/c$ stationary under path variations with fixed endpoints. Stationarity, rather than always the global minimum, permits multiple arrivals. Its Euler–Lagrange equation, with arclength $s$ and unit tangent $\hat{\mathbf{t}}$, isFor horizontal layering, the horizontal slowness $p_h=\sin i/c$ is conserved, giving Snell's law at interfaces and smooth refraction between them. Here $i$ is measured from vertical. A turning point satisfies $p_{h}c(z)=1$; increasing speed with depth bends a downgoing ray back upwards. Travel-time tomography uses these ray integrals to infer spatial slowness variations.

The approximation requires wavelength to be small compared with structural variation scales. Caustics, turning-point amplitudes, diffraction, scattering and finite-frequency sensitivity require wave-based corrections. Thus seismic ray theory gives an economical geometric description and useful travel times, while surface-wave dispersion retains essential interference and boundary-mode physics.

For $k>0$, phase velocity is the speed of a fixed phase, $c_p=\omega /k$. The group velocity $c_g=d\omega /dk$ transports a narrow-band wave envelope. With $q=kH$,As $H\to 0$ with $k$ fixed, $\tanh (kH)/(kH)\to 1$. ThereforeBoth velocities tend to the shallow-water speed $\sqrt{gH}$ at leading order.

Set $D=({\omega }_0^2-{\omega }^2{)}^2+{\gamma }^2{\omega }^2$. The real part and imaginary part of the complex permittivity areFor positive frequency the loss part is nonnegative. The real part starts at $1+{\omega }_p^2/{\omega }_0^2$, rises below resonance, falls steeply through the resonance region, and approaches one from below at high frequency. The imaginary part is a positive absorption peak near ${\omega }_0$, tending to zero on both sides.

For a nonmagnetic medium write its complex refractive index as $N=n_r+i\kappa$, with $N^2={ϵ}_r$ and optical extinction coefficient $\kappa$. In a transparent region with $\kappa \ll n_r$ and ${ϵ}_r^{\prime }>0$,The real part controls phase velocity and optical dispersion; the loss part controls wave absorption and the cycle-averaged dissipated power density $\omega {ϵ}_0{ϵ}_r^{\prime \prime }|E_0{|}^2/2$. Weak damping gives a narrow line but does not make wave absorption weak at its centre. The transparent-medium approximation applies away from the line.

Assume a cold, unmagnetized, weakly collisional plasma, a small fractional bandwidth and $\omega \gg {\omega }_p$. The dispersion relation is ${\omega }^2=c^2{k}^2+{\omega }_p^2$, with ${\omega }_p^2=n_e{e}^2/({ϵ}_0{m}_e)$. The group velocity and propagation time areThe lower frequencies arrive later. Taking the magnitude of the frequency derivative over $\Delta \omega$ givesFor a nonuniform line of sight, replace $n_{e}L$ by the dispersion measure $\int n_{e}dl$. The light travels at the group velocity, not at the phase velocity.

Landau damping is the collisionless decay of a plasma wave through resonant energy exchange with particles whose velocity is close to the wave's phase velocity $v_{\varphi }=\omega /k$. Particles slightly slower than the wave tend to gain energy from it; particles slightly faster can give energy to it. For the decreasing resonant part of a Maxwell-Boltzmann distribution, the net transfer is from the wave to the electrons, so the wave amplitude decays.

In the linear Vlasov equation description this is collisionless phase mixing, not a requirement for frequent particle collisions. Fine-scale structure develops in the velocity dependence of the perturbation even while the macroscopic electric field decreases. The resonant contribution to the growth rate has the sign of $\partial f/\partial v$ at a positive wave phase velocity: a negative slope produces damping, whereas a sufficiently positive slope can drive an instability. A population comparison must be local to the resonance; merely counting all particles faster and slower than the wave is not the criterion.

The phase velocity is the speed of a constant-phase surface, $v_{\mathrm{ph}}=\omega /k$. The group velocity is the envelope speed of a narrow packet, $v_{\mathrm{gr}}=d\omega /dk$, requiring a local dispersion relation rather than only a single monochromatic wave. In an isotropic transparent medium with $k=n(\omega )\omega /c$,For a vacuum wave, $n\equiv 1$ over the packet bandwidth, so both speeds are $c$ and $\boxed{v_{\mathrm{ph}}{v}_{\mathrm{gr}}=c^2}$. No superluminal signalling is implied by a phase speed in a dispersive material.

Strictly, the value $n({\omega }_0)=1$ at one frequency alone does not prove the product identity. For example $n(\omega )=1+q(\omega -{\omega }_0)/{\omega }_0$ gives $v_{\mathrm{ph}}({\omega }_0)=c$ but $v_{\mathrm{gr}}({\omega }_0)=c/(1+q)$. The vacuum interpretation, or an equivalent nondispersive assumption, is necessary.

Write $\omega$ for the nonzero forcing frequency, to distinguish it from the vertical velocity $w$. Put $k_z=\pi /h$ and $Z(z)=\cos (\pi (2z+h)/(2h))=-\sin (k_{z}z)$. Then $Z(0)=Z(h)=0$ and $Z^{\prime \prime }=-k_z^{2}Z$, so it meets the rigid-wall conditions. Substituting $w=\mathrm{Re}(f(x)e^{-i\omega t}Z(z))$ into the scalar wave equation givesorThe radiation sign for an inertial wave must exclude any disturbance arriving from $x=+\infty$. This means an radiation condition for propagating waves, or bounded decay for an evanescent wave.

For $0<|\omega |<2|\Omega |$, put $q=k_z\sqrt{4{\Omega }^2/{\omega }^2-1}$. The solutions are oscillatory. The inertial-wave dispersion relation is ${\omega }^2=4{\Omega }^2{k}_z^2/(k_x^2+k_z^2)$; its horizontal group velocity is $c_{g,x}=-\omega k_x/(k_x^2+k_z^2)$. Positive energy transport into the channel therefore requires $k_x=-\mathrm{sgn}(\omega )q$. The solution isFor a positive frequency, the phase travels towards negative $x$ while the group velocity travels towards positive $x$. Choosing the outgoing sign from the phase velocity alone would give the wrong result.

For $|\omega |>2|\Omega |$, set $\kappa =k_z\sqrt{1-4{\Omega }^2/{\omega }^2}$. Excluding the exponentially growing solution givesThis disturbance decays away from the wavemaker and does not propagate horizontally. At the cutoff the ODE degenerates to $f^{\prime \prime }=0$; zero frequency is not a nonzero forced harmonic mode with this vertical shape.

There is a second printed inconsistency: the stated velocity $u\mathbf{i}+w\mathbf{k}$, if taken literally with no $y$ dependence and no transverse velocity, cannot support a nontrivial rotating inertial wave. Its transverse momentum equation would give $2\Omega u=0$, then continuity and the rigid walls would force $w=0$. The scalar solutions above are the intended inertial-wave solutions when two-dimensional means dependence on $x,z$ while retaining the transverse component $v$. For a mode, $-i\omega v+2\Omega u=0$ explicitly supplies the generally nonzero transverse velocity. Thus the scalar equation can be solved as requested, but the full vector field needs this correction to the printed two-component description.

For this density the Euler-Lagrange field equation with second derivatives givesInsert a plane wave $\varphi =A{e}^{i(kx-\omega t)}$. The second and fourth spatial derivatives give $-k^2$ and $k^4$, respectively, soFor propagating modes, $k^2>\alpha /\beta$ and the phase velocity has magnitude $|\omega /k|=\sqrt{(\beta k^2-\alpha )/\rho }$, which depends on the wave number. Thus the waves are dispersive. For $0<k^2<\alpha /\beta$, the squared frequency is negative: the corresponding solutions grow or decay exponentially rather than propagate with a real frequency. The uniform $k=0$ displacement is a zero-frequency shift mode.

Choose the rightgoing signed dispersion relation $\omega (k)=Nk/\sqrt{k^2+b_n^2}$. The horizontal phase velocity and group velocity areThe leftgoing branch has their negatives. For a convention using positive frequency at both signs of $k$, multiply each displayed speed by $\mathrm{sgn}k$. For every nonzero $k$, $0<c_g<c_p$ on the rightgoing branch; the horizontal group-speed bound for a confined internal wave is $c_g\le N/b_n$, approached as $k\to 0$.

Pavel Cherenkov investigated the visible radiation produced by fast charged particles in transparent materials. The emission's directional and polarization properties distinguished it from ordinary luminescence. Its interpretation as Cherenkov radiation relates the optical signal to a particle moving faster than the local phase velocity of light. His authored lecture connects the discovery with practical detectors, in which the presence and geometry of a light cone provide information about particle speed.^[ref]

A physicist.^[ref]