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A separated acoustic wave in a straight rigid channel $0<y<H$ has transverse Neumann boundary conditions, giving factors $\cos (n\pi y/H)$ for integers $n\ge 0$. The velocity potential satisfies the wave equation, and its axial factor obeysHere $c$ is the sound speed. Real $q_n$ permits propagation; positive imaginary $q_n$ produces an evanescent wave. A prescribed wall velocity proportional to one transverse factor excites that mode alone in the linear straight-channel model. The physical field is the real part of the complex expression.

The angular spectrum is the transverse Fourier transform of a monochromatic field on a reference plane. In a uniform medium, each transverse wavevector component propagates with a longitudinal wavenumber satisfying the Helmholtz equation. Components beyond the propagating wavenumber range are evanescent waves. The outgoing angular spectrum chooses the longitudinal branch that propagates or decays away from the plane.

A moving boundary excites an internal gravity wave through its kinematic boundary condition. For wavenumber $k$ and angular frequency $\omega$, the vertical velocity field satisfies $W^{\prime \prime }+k^2(N^2/{\omega }^2-1)W=0$. The radiation condition selects outward group velocity; when $\omega >N$, the bounded response is an evanescent wave.

Take a cold, unmagnetized, collisionless plasma with uniform equilibrium electron density $n_e$, fixed neutralizing ions, and a small electromagnetic wave that is a transverse wave. Cold means that thermal-motion corrections are neglected. For electron mass $m_e$, elementary charge magnitude $e$ and electric field $\mathbf{E}=\mathrm{Re}(\hat{\mathbf{E}}{e}^{-i\omega t})$, the linear Newton's second law equation is $m_e\dot{\mathbf{v}}=-e\mathbf{E}$. Thus $\hat{\mathbf{v}}=-ie\hat{\mathbf{E}}/(m_e\omega )$ and the current density is $\hat{\mathbf{J}}=i{n}_e{e}^2\hat{\mathbf{E}}/(m_e\omega )$. Writing the induced electric polarization as $\hat{\mathbf{J}}=-i\omega \hat{\mathbf{P}}$ gives a negative electric susceptibility. With vacuum permittivity ${ϵ}_0$,The plasma frequency ${\omega }_p$ controls the response. For $\omega >{\omega }_p$, the propagating branch has real refractive index $0<n<1$. For $\omega <{\omega }_p$, this homogeneous model has an evanescent wave, not an ordinary propagating wave with a negative real refractive index. In a slowly varying vertical profile, a normally incident ray approaches a turning region where ${\omega }_p(z)=\omega$; the geometrical-ray approximation fails right at that region. At oblique incidence the conserved horizontal wave vector changes the turning condition. A smooth density gradient can therefore return a skywave without a mirror or a layer of metal.

For a homogeneous propagating section, direct differentiation of the dispersion relation givesA phase velocity greater than $c$ does not send information faster than $c$. Here a narrow wave packet has group velocity $v_g$, while the full causal response must still respect special relativity. Even this stripped-down plasma is strongly dispersive near its plasma frequency. In the actual ionosphere, collisions produce absorption, and the magnetic field of Earth changes the response with direction and wave polarization; a single real scalar $n$ cannot capture all of that behavior.

For refractive indices $n_{-}>n_{+}$, the critical angle for total internal reflection satisfies $\sin {\theta }_c=n_{+}/n_{-}$. Above it, the transmitted normal wavenumber is imaginary and the transmitted field is an evanescent wave.

The critical angle for total internal reflection is the boundary between outgoing refraction and an evanescent wave, rather than an angle at which the external electric field abruptly disappears. Write $n_1=n_{-}$ and $n_2=n_{+}$. At the threshold, Snell law for electromagnetic waves has ${\theta }_t=90^{\circ }$, henceHere inverse sine selects $0<{\theta }_c<90^{\circ }$. For $n_1=1.5$ and $n_2=1$, $\sin {\theta }_c=2/3$ and ${\theta }_c\approx 41.8103149^{\circ }$. The threshold is independent of polarization of an electromagnetic wave under the stated assumptions.

A short proof using phase matching at a planar wave interface reveals all three cases at once. Choose a Cartesian coordinate system with boundary $z=0$, incident medium $z<0$, and transmitted medium $z>0$. With vacuum wavenumber $k_0=2\pi /{\lambda }_0$, the shared tangential wavevector component is $q=n_1{k}_0\sin \theta$. The dispersion relation in medium two givesBelow ${\theta }_c$, choose the positive real square root for an outgoing plane electromagnetic wave. At ${\theta }_c$, $k_{2z}=0$: there is a grazing wave, with zero normal average power but no exponential confinement in $z$. Above ${\theta }_c$, choose $k_{2z}=i\kappa$ with $\kappa >0$, so the wave amplitude decays into the second half-space.

If $n_1\le n_2$, no nongrazing angle of incidence produces this threshold. As $n_2/n_1\to 1$ from below, ${\theta }_c\to 90^{\circ }$; as that ratio tends to zero, ${\theta }_c\to 0$. These limits explain why a larger refractive index contrast permits confinement over a wider range of internal angles.

The cutoff frequency of a membrane waveguide is the lowest frequency at which a specified transverse normal mode can propagate longitudinally. For fixed edges separated by $b$, ${\omega }_{c,n}=n\pi v/b$ and $f_{c,n}={\omega }_{c,n}/(2\pi )$. Below cutoff the longitudinal wavenumber is imaginary and a bounded excitation is an evanescent wave. Above it, $v_g=v\sqrt{1-({\omega }_{c,n}/\omega {)}^2}$ and $v_p{v}_g=v^2$.

An electromagnetic waveguide confines an electromagnetic wave across its axis while allowing it to travel along the axis. Conducting walls or changes of refractive index select transverse normal modes: reproducible cross-sectional field patterns with their own longitudinal wavenumbers. A hollow metal pipe and an optical fiber are different realizations of this principle. Their usefulness is to deliver electromagnetic energy along a chosen route instead of letting it spread freely.

For a uniform vacuum-filled conducting guide, write a normal mode's axial and time dependence as $e^{i(\beta z-\omega t)}$. Here $\omega >0$ is angular frequency, $\beta$ is the longitudinal wavenumber, and $k_{\perp }^2$ is the transverse eigenvalue selected by the wall boundary conditions. The electromagnetic wave equation separates intoThe geometry consumes part of the available squared wavenumber: when $\omega /c<k_{\perp }$, this branch is an evanescent wave rather than a propagating normal mode. The waveguide cutoff frequency is therefore a property of a specified branch and geometry. A transverse electromagnetic mode in a suitable two-conductor geometry can have $k_{\perp }=0$ and no positive geometric cutoff. An optical fiber needs its own dielectric boundary conditions, rather than the metal-wall formulas below.

The rectangular electromagnetic waveguide makes these ideas explicit with a single sine profile. The scalar-potential conducting waveguide modes give a broader construction. At an open aperture, a confined field can radiate into the electromagnetic aperture angular spectrum; it need not remain one guided normal mode.

The evanescent field in total internal reflection oscillates and advances in wave phase along the boundary while its wave amplitude decays away from it. With the coordinates above and time dependence $e^{-i\omega t}$, a TE polarization has complex amplitudeThe physical electric field is $\mathrm{Re}[\mathbf{E}(x,z)e^{-i\omega t}]$. The decaying sign excludes a field that grows without bound at $z\to \infty$. Dielectric-interface boundary conditions determine $A$; for nongrazing incidence it is generally nonzero.^[ref]

Zero normal energy flux follows directly from Maxwell equations. The wavevector is $(q,0,i\kappa )$; the relation $\mathbf{H}=(\mathbf{k}\times \mathbf{E})/(\omega {\mu }_0)$ givesConsequently, the cycle-averaged Poynting vector isThe electric field and $H_x$ are a quarter-cycle apart, so their product has zero cycle average. Tangential energy flux survives. This is a local calculation for the single decaying evanescent wave in the infinite second half-space.

To check reflected power independently, put $a=n_1{k}_0\cos \theta >0$. The existing transverse-electric Fresnel reflection coefficient becomes $r=(a-i\kappa )/(a+i\kappa )$. Its numerator and denominator are complex conjugates, so $|r{|}^2=1$, while $1+r=2a/(a+i\kappa )\ne 0$ supplies the surface electric field relative to the incident wave amplitude. Thus a unit optical reflectance does not require a zero transmitted field. The same normal power balance holds for TM polarization; its reflected wave phase generally differs from the TE polarization phase. A mixture can therefore change its polarization of an electromagnetic wave on reflection even though all normal incident power returns.

A nonzero probability density inside a barrier does not by itself imply transport. In one dimension, the probability current of a wavefunction is $J=(\hslash /m)\mathrm{Im}({\psi }^{\ast }{\psi }^{\prime })$. A single evanescent wave $C{e}^{-\kappa x}$ has ${\psi }^{\ast }{\psi }^{\prime }=-\kappa |C{|}^2{e}^{-2\kappa x}$, a real quantity, hence $J=0$. This is what happens on the inaccessible side of a potential barrier extending to infinity with $E<V_0$.^[ref]

For the finite-barrier wavefunction $A{e}^{\kappa x}+B{e}^{-\kappa x}$, the cross terms instead giveThe relative quantum phase of the two probability amplitudes permits a nonzero, spatially constant probability current, even though neither exponential function separately carries it. The probability continuity equation gives ${\partial }_{x}J=0$ for this stationary state. Matching the wavefunction and its derivative also makes $J$ continuous at both faces, soHere $R=|r{|}^2$ is the quantum reflection probability, and $T=J_{\mathrm{tr}}/J_{\mathrm{inc}}$ is the transmission probability. The net exterior probability current equals the interior probability current; it is not lost as the wavefunction decreases. A physical incident wave packet supplies finite total probability, while the stationary state describes the energy component used to compute its transmitted flux.

Frustrated total internal reflection occurs when a nearby second high-index dielectric couples to the evanescent field in total internal reflection, allowing power to cross a finite low-index gap. The extra boundary changes the dielectric-interface boundary conditions; the gap field need not be a single decaying evanescent wave. Consequently, the zero normal Poynting vector calculation for two infinite half-spaces does not forbid transmission in this three-medium geometry.

Harvard’s actual demonstration uses two right-angle glass prisms, a helium-neon laser, a small clamp and two perpendicular screens as detectors. The contact faces are cleaned and placed opposite each other, with the clamp protecting the corners. Increasing pressure narrows the intervening air film; the observed beam routing changes gradually from total internal reflection toward onward transmission. The broad-gap reflecting configuration supplies the comparison for the narrow-gap transmitting configuration. This is a qualitative readout: the page explicitly says the gap is not measured.^[ref]

The inference is that the lower-index region can mediate electromagnetic energy transfer when another boundary is nearby. The demonstration does not determine a numerical evanescent penetration depth, resolve gap uniformity, or establish a three-layer transmission coefficient. For the constructed $50^{\circ }$ example, $\delta \approx 178\mathrm{nm}$ identifies the scale over which the isolated-interface field remains appreciable; substituting $e^{-2g/\delta }$ for a gap transmission fraction would omit the changed boundary fields and their interference. A quantitative gap calculation must solve the full Maxwell equations with both boundaries.

Geometrical optics describes propagation by rays and Snell's law, neglecting diffraction on scales much larger than the wavelength.

A light ray is a path in an approximation, not a physical thread of zero width. In a transparent, homogeneous, isotropic medium it is straight and perpendicular to a wavefront; boundaries redirect it by the law of reflection or Snell's law. The approximation is useful when the wavelength is small compared with relevant apertures, surface curvature scales, and distances over which the refractive index changes. For an illustrative wavelength of $500\mathrm{nm}$ and a $10\mathrm{mm}$ aperture, their ratio is $5\times 10^{-5}$. That small ratio supports ray tracing for image location, but does not eliminate diffraction when calculating the fine structure of a focused spot. At a wave caustic, geometrical optics can fail despite large optical elements. ^[ref].

The mechanism of refraction is a change in phase velocity. For transparent isotropic media at the chosen frequency, the refractive index iswhere $c$ is the vacuum speed of light and the angles are measured from the interface normal vector. At a stationary planar interface the frequency and tangential wavevector match across the boundary. Since the wavenumber is $k=n\omega /c$, that matching gives Snell's law. The transmitted wavelength changes while the frequency stays fixed. Here $n$ uses phase velocity, which need not equal group velocity. ^[ref]. ^[ref].

Choose model air with $n_1=1$ and model glass with $n_2=1.5$, without claiming these are exact values for every sample or wavelength. At ${\theta }_1=45^{\circ }$, Snell's law givesThe transmitted light ray bends toward the normal vector; its phase velocity is $c/1.5$, and its wavelength is two thirds of the air value. By reversibility of an optical ray, incidence from glass at $28.1255^{\circ }$ returns the outgoing air angle $45^{\circ }$.

Incidence from glass at $45^{\circ }$ is a different experiment. It would require $\sin {\theta }_2=1.5\sin 45^{\circ }\approx 1.06066$, so there is no real transmitted angle. The critical angle for total internal reflection isAbove this angle, an ideal lossless glass-air boundary produces total internal reflection. The physical optics description still permits an evanescent wave in the air, although no propagating transmitted light ray crosses the boundary. ^[ref]. ^[ref].

The usefulness of geometrical optics is that complicated instruments can be assembled from these local rules. Applying Snell's law at successive curved surfaces changes a spreading bundle into a converging one. The Fermat principle unifies the construction through stationary optical path length; stationarity does not require a global minimum. ^[ref]. Fang's MIT notes explicitly connect this principle to specular reflection, refraction, and thin lens imaging, providing a route from individual paths to complete instruments. ^[ref].

At a step in buoyancy frequency without a jump in background mass density, vertical velocity and fluid pressure are continuous. For a fixed horizontal wavenumber and angular frequency, these are continuity of $W$ and $W^{\prime }$. They determine reflection and transmission between propagating internal gravity waves and evanescent waves.

For an even field between rigid plates at $y=\pm b$, the transverse modes are $\cos (n\pi y/b)$, with longitudinal wave numbers $k_n=\sqrt{k_0^2-(n\pi /b{)}^2}$. With time dependence $e^{i\omega t}$, outgoing modes into $x>0$ use $e^{-i{k}_{n}x}$: choose positive real $k_n$ for cut-on modes and negative imaginary $k_n$ for cut-off modes. A lower-half-plane Fourier inversion extracts these modes by residues. Their decaying components are evanescent waves, not backward-propagating waves. At an exact cutoff one takes the radiation continuation before its limit.

Let $\mu =\pi /H$. The end-wall forcing has transverse factor $\cos (\mu y)$, which already satisfies both fixed-wall Neumann boundary conditions. By linearity and separation of variables, seek its periodic response asSubstitution into the wave equation and the moving-wall condition givesThe acoustic waveguide cutoff is thereforeIt is the lowest nonzero transverse-mode frequency. The uniform transverse mode has no cutoff, but is not excited here because ${\int }_0^H\cos (\pi y/H)dy=0$.

For $0<\omega <{\omega }_c$, put $\alpha =\sqrt{{\mu }^2-{\omega }^2/c^2}>0$. The axial equation has solutions $e^{\pm \alpha x}$. Boundedness removes the growing exponential, and ${\Phi }^{\prime }(0)=ϵ\omega$ fixes the coefficient. ThusThis evanescent wave oscillates in time but decays with penetration length $1/\alpha$. Its axial velocity is proportional to $\cos (\omega t)$, while its excess pressure $p^{\prime }=-{\rho }_0{\varphi }_t$ is proportional to $\sin (\omega t)$. They are in phase quadrature, so their mean product, the axial acoustic energy flux, is zero.

For $\omega >{\omega }_c$, put $q=\sqrt{{\omega }^2/c^2-{\mu }^2}>0$. The axial solutions are $e^{\pm iqx}$. The outgoing condition selects $e^{iqx}$ for the chosen time convention, andTaking the real part givesIts axial velocity is $ϵ\omega \cos (\pi y/H)\cos (qx-\omega t)$, so the moving-wall condition is satisfied with the correct sign. This is a traveling acoustic waveguide mode carrying energy toward positive $x$. Its dispersion relation isThe larger axial phase velocity does not imply faster energy transport: the group velocity controls that transport.

At $\omega ={\omega }_c$, the axial equation becomes ${\Phi }^{\prime \prime }=0$. The imposed nonzero derivative forces $\Phi =ϵ{\omega }_{c}x+$ a constant, which is unbounded. Thus there is no bounded periodic response at exact cutoff for this infinite, undamped, velocity-driven model. The divergent coefficients as $q\to 0$ or $\alpha \to 0$ also show why the assumed small perturbations fail sufficiently close to cutoff at fixed forcing. Initial conditions can add free transients; the displayed formulas give the selected periodic forced response.

For the proposed plane wave, define the positive squared effective wavenumberThe horizontal Laplacian contributes $-(k^2+l^2){\psi }^{\prime }$ and the vertical stretching operator contributes $-(f_0^2/N^2)m^2{\psi }^{\prime }$. Substitution in the Linear Rossby-wave equation therefore givesThe intrinsic phase velocity is westward for positive $\beta$. A stationary disturbance with $\omega =0$ and $k\ne 0$ must have its westward intrinsic phase velocity exactly cancelled by the eastward background current. Solving for the squared vertical wavenumber yieldsFor Earth, with northward $y$, $\beta =2\Omega \cos \varphi /a>0$ in both hemispheres away from the poles. Thus an oscillatory vertical structure requiresThe background flow must be eastward but not too strong for the chosen horizontal scale. Long horizontal wavelengths have the largest allowed speed. An easterly current, $U<0$, gives $m^2<0$ and an evanescent wave; so does an eastward current exceeding the upper bound. At equality $m=0$ and the vertical group velocity vanishes. A zonally uniform disturbance, $k=0$, is a degenerate stationary pattern with no vertical Rossby wave energy propagation, so it cannot be obtained by cancelling $k$ in the preceding calculation.

A real $m$ alone does not select upward propagation. Differentiate the dispersion relation at fixed $k,l$ to obtain the vertical group velocity:The outgoing radiation condition for a surface source requiresFor the convention $k>0$, take $m>0$. The stationary phase lines then obey $dx/dz=-m/k<0$: they tilt westward with height while wave energy travels upward. The real conjugate component $(-k,-l,-m)$ describes the same physical upward-propagating wave.

The figure suggests a winter extratropical path from the lower terrestrial atmosphere into the stratosphere, where the mean zonal flow is eastward. The summer stratosphere has westward flow and blocks such stationary Rossby waves in this idealization. A zero-wind surface is a stationary-wave critical layer in a shear flow; it also bounds the winter westerlies aloft near the upper part of the diagram. For a slowly varying real wind profile, local propagation needs $m^2(z)>0$ along the path, sufficient surface forcing, and no intervening absorbing critical layer in a shear flow. A layer where the eastward wind becomes too strong is a turning barrier for the given horizontal wavenumbers. The uniform-current result is a local guide, not a complete prediction through a sheared spherical terrestrial atmosphere.

For scale, at $60^{\circ }$ latitude $\beta \simeq 1.14\times 10^{-11}{\mathrm{m}}^{-1}{\mathrm{s}}^{-1}$. A planetary zonal wavenumber $s$ has local $k=s/(a\cos \varphi )$. Taking $l=0$ gives the most permissive limitThe resulting limits are approximately $116$, $29$ and $13\mathrm{m}{\mathrm{s}}^{-1}$ for $s=1,2,3$. Therefore the long planetary waves are favored; the strong core of the winter vortex can exclude shorter ones even though it is westerly. Nonzero meridional wavenumber lowers these limits further. A compressible log-pressure coordinate model would also contain a density-scale-height correction; it is absent from the equation supplied here and should not be silently inserted.

Interchanging the Northern and Southern Hemispheres changes the sign of $f_0$, but only $f_0^2$ enters the dispersion relation and vertical group velocity. With the same eastward $x$ and northward $y$, $\beta$ stays positive. Thus the propagation criterion is the same in both hemispheres: it selects the winter westerlies, irrespective of which hemisphere is winter. Actual amplitudes can differ because mountains, land-sea contrasts, forcing spectra and mean winds differ between hemispheres; those source differences do not reverse this propagation criterion.

The contrast is weak: $(n_1-n_2)/n_2=0.01$. Use a scalar weak-guidance Helmholtz equation for a transverse field component, with longitudinal wave number $b$ and vacuum wave number $k_0=2\pi /{\lambda }_0$. Separating variables as $\psi =F_l(r)e^{il\varphi }{e}^{ibz}$ givesA bound mode requires $n_2{k}_0<b<n_1{k}_0$. DefineThe step-index fiber normalized frequency is $V$; $u$ is a dimensionless core transverse wave number. Regularity at $r=0$ selects the Bessel function of the first kind, while decay at infinity selects the Modified Bessel function of the second kind:In this scalar approximation, continuity of the field and its radial derivative givesAt cutoff the evanescent wave extends arbitrarily far into the cladding, so $b\downarrow n_2{k}_0$, $w\downarrow 0$ and $u\to V$. For the first azimuthally varying family $l=1$, the recurrence relations reduce the last equality toAs $w\to 0$, $K_1(w)\sim 1/w$ and $K_0(w)\sim -\ln w$, making the right side tend to zero. Thus the onset is at the first positive zero of $J_0$:The scalar $l=0$ fundamental family has no positive cutoff. The first additional spatial family is the $l=1$ family just found; the next radial $l=0$ and first $l=2$ families require the larger zero $j_{1,1}\simeq 3.832$. This establishes the first multimode boundary rather than merely finding one possible cutoff. The two polarizations of the fundamental family are not counted as two spatial modes. Exact vector interface conditions split some higher-mode degeneracies at finite contrast; the displayed condition is the weak-guidance result intended by the supplied scalar equations.

For a plane wave, ${\partial }_t$ acts as $-i\omega$ and ${\partial }_x$ as $ik$. Substitution givesso, for $\omega \ne {\omega }_d$, the dispersion relation isAssume ${\omega }_n>{\omega }_d>0$ and real angular frequency. The ratio is negative exactly on ${\omega }_d<\omega <{\omega }_n$, giving imaginary wavenumbers and an evanescent wave. Thus the propagating frequency gap is $({\omega }_d,{\omega }_n)$. At ${\omega }_d$ there is no finite-$k$ solution, while ${\omega }_n$ has $k=0$; neither endpoint is an ordinary finite-wavelength propagating mode.

Choose the branch $k=(\omega /c)\sqrt{(\omega -{\omega }_n)/(\omega -{\omega }_d)}$, so the phase velocity is positive in the allowed bands. Direct differentiation gives the phase velocity and group velocityThe opposite spatial branch reverses both signs. On the chosen branch, the requested limits areThe band-edge limits are taken from their propagating sides. More precisely, with $ϵ=\omega -{\omega }_n\downarrow 0$ and $\delta ={\omega }_d-\omega \downarrow 0$,

For the final case, set $K=k_{0}c>0$. Substitution of $T(t)e^{i{k}_{0}x}$ at ${\omega }_d=0$ gives $T^{(3)}+(3iK/2)T^{\prime \prime }+K^2{T}^{\prime }=0$. Its characteristic polynomial factors as $r(r+2iK)(r-iK/2)$, henceThe initial data give $A+B+C=1$, $-2B+C/2=0$ and $-4B-C/4=-5$. Thus $B=1$, $C=4$, $A=-4$, andThe time-independent contribution is essential: static modes lost by cancelling frequency would be missed if one divided the ${\omega }_d=0$ dispersion relation by $\omega$ and retained only its two nonzero-frequency roots.

Use the usual minor and major symmetries of the elastic stiffness tensor, positive density, and strong ellipticity in elasticity. For the plane elastic wave $u=a{e}^{ik(n\cdot x-vt)}$, substitution into momentum balance gives the acoustic tensor problemThe major symmetry makes $\Gamma$ a real symmetric matrix. Moreover $a^{\mathsf{T}}\Gamma a=c_{ijpq}{a}_i{n}_j{a}_p{n}_q>0$ for nonzero $a$ under strong ellipticity in elasticity. Hence its three eigenvalues are positive and the three positive wave speeds are real. Symmetry alone proves only reality of $v^2$: an unstable isotropic tensor with negative shear modulus has negative transverse $v^2$. The requested reality therefore uses stability as part of the physical elastic-solid assumption.

The elastic slowness surface consists of three sheets, with $s=n/v_{\alpha }(n)$, satisfyingIt is centrosymmetric. Sheets can meet at degenerate polarizations; they need not be spherical or convex in an anisotropic solid. The group velocity is normal to the corresponding smooth sheet: for a sheet tangent $ds$, differentiation of $a^{\mathsf{T}}[c_{ijpq}{s}_j{s}_q-\rho I]a=0$ gives $a_i{c}_{ijpq}{a}_p{s}_{q}d{s}_j=0$. The coefficient vector is proportional to the group velocity, so its scalar product with every tangent vanishes.

Fix horizontal slowness $p=(n_1/c,n_2/c)$ and put $q=s_3$. DefineThe symbol $p$ on a tensor index is an index, whereas $p_{\alpha }$ is a horizontal slowness component. The vertical slowness equation is the vertical slowness sextic of an anisotropic solidThe leading coefficient is $\det T>0$, so it is a degree-six polynomial with real coefficients. Nonreal roots occur in conjugate pairs, leaving between zero and six real roots counted with multiplicity. Generically the counts are $0,2,4,6$; a grazing repeated root can give an odd count of distinct real roots. Geometrically these are the intersections of a vertical line of fixed horizontal slowness with the three sheets.

All four generic counts are actually possible in a stable material. For example, take $\rho =1$ and orthotropic Voigt stiffnessesThe stiffness matrix is positive definite. For horizontal slowness $(p,0)$, the three diagonal acoustic equations areThe choices $p=0.1,0.3,0.8,1.2$ give respectively six, four, two and zero real vertical roots.

For the general depth-dependent amplitude, substitution of the horizontal harmonic factor yieldsFor six simple roots $q_{\ell }$, choose nonzero null vectors $a_{\ell }$ of the corresponding matrix. The complete solution isComplex vertical roots give evanescent waves. At defective repeated roots the usual polynomial-in-depth factors multiplying exponentials must be included, or the solution can be defined by a limiting procedure.

To classify the modes, take $x_3$ positive downwards and $\omega >0$. The reduced traction for a mode isMechanical energy crosses a plane at rate $-{\sigma }_{i3}{u}_{i,t}$. Averaging the real harmonic fields uses $⟨\mathrm{Re}(f{e}^{-i\omega t})\mathrm{Re}(g{e}^{-i\omega t})⟩=\mathrm{Re}(f{g}^{\ast })/2$. With velocity amplitude $-i\omega a_{\ell }$ and traction amplitude $i\omega b_{\ell }$, the time-averaged vertical acoustic energy flux isFor a real propagating mode, write its physical wavevector as $k=\omega (p_1,p_2,q)$. Differentiate $a^{\mathsf{T}}\Gamma (k)a=\rho \omega (k{)}^2{a}^{\mathsf{T}}a$ in $k_3$. The derivatives of $a$ cancel by the eigenvalue equation, while $a^{\mathsf{T}}({\partial }_{k_3}\Gamma )a=2\omega a^{\mathsf{T}}(Tq+R)a$. Thus its vertical group velocity and flux areThus the elastic mode classification by vertical energy flux is upgoing if $J_3<0$, downgoing if $J_3>0$. In anisotropy, the sign of vertical phase slowness alone need not classify energy propagation. For evanescent modes, select $\mathrm{Im}q>0$ for decay into a downward half-space and $\mathrm{Im}q<0$ for decay into an upward half-space. Away from grazing and degeneracy, continuation from zero horizontal slowness, with a small radiation damping if needed, separates three outgoing modes on each side. Superpose only the appropriate three to construct the most general upgoing or downgoing field; limiting modes handle exceptional cases.

For a free surface at $x_3=0$ with solid occupying $x_3>0$, let $(q_0,a_0)$ be a unit-amplitude incident upgoing mode. The reflected field consists of the three downgoing propagating or depth-decaying modes:Frequency and horizontal slowness are common to all modes, by translation invariance of the surface. Zero surface traction imposes the three equationsHence the free-surface anisotropic reflection amplitudes arewhenever $B$ is invertible. This specifies mode conversion as well as reflection. If $B$ is singular, a nonzero outgoing homogeneous surface mode exists; these are surface-wave poles, including Rayleigh waves in the isotropic case. At such parameters one must use the appropriate limiting radiation solution and check solvability instead of assuming a unique finite inverse.

Assume $U>0$, $k=\alpha >0$, constant $N$, negligible rotation, dissipation and reflection from infinity, small $k{\eta }_0$ and $m{\eta }_0$, and no wave overturning. The terrain is stationary, so laboratory frequency is zero and intrinsic frequency is $\hat{\omega }=-kU$. Propagation requires $kU<N$. PutThe kinematic boundary condition at the linearized bed is $w(x,0)=U{\eta }_x=-kU{\eta }_0\sin (kx)$. The upward radiation condition selects the positive $m$ branch because $c_{gz}=-\hat{\omega }m/(k^2+m^2)>0$. ThusThe internal-wave polarization and every other perturbation follow from this vertical velocity and the linear equations.

In the frame moving with the fluid, $x^{\prime }=x-Ut$, the phase is $k{x}^{\prime }+mz+kUt$, andThe intrinsic energy ray slopes upward and upstream, $dz/d{x}^{\prime }=-k/m$, along a constant-phase line of an internal gravity wave. At fixed intrinsic frequency the spatial wave equation is hyperbolic, with the two characteristic curve line slopes $dz/d{x}^{\prime }=\pm k/m$; the radiation condition selects the upward ray of the appropriate horizontal propagation branch.

In the terrain frame, phase lines remain fixed and the pattern phase velocity is zero. A transported packet hasIts energy path slopes upward and downstream, $dz/dx=m/k$, and need not coincide with a stationary phase line. One should not add $(U,0)$ to the chosen normal intrinsic phase velocity and call that the normal ground-frame phase velocity: tangential motion along a phase line is arbitrary; zero laboratory frequency makes the entire crest pattern stationary. The sketch distinguishes stationary crests, intrinsic characteristics and transported energy.

If $kU>N$, $m$ is imaginary and the bounded field is an evanescent wave; at $kU=N$ the vertical group velocity vanishes and an ordinary upward radiation interpretation degenerates.

For $\omega >N$, the dispersion relation gives $k^2=-m^2{\omega }^2/({\omega }^2-N^2)$. Put $\kappa =m\omega /\sqrt{{\omega }^2-N^2}>0$. Boundedness as $x\to +\infty$ selects $k=i\kappa$, rather than the exponentially growing root. The internal-wave polarization relations now yieldAgain take real parts and set $v=0$. This is an evanescent wave, decaying over distance ${\kappa }^{-1}$; there is no propagating far-field internal gravity wave at this frequency. The pressure amplitude $P=-i\omega u/\kappa$ is in quadrature with the horizontal velocity, so the time-averaged horizontal pressure-work flux vanishes. The equality $\omega =N$ is a singular limiting case and is outside the two frequency ranges considered here.