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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.

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

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.

Suppress the common time factor $e^{i\omega t}$, take $\omega >0$, and fix the spatial Fourier transform conventionA plus transform is supported on $x>0$ and analytic above $\Gamma$; a minus transform is supported on $x<0$ and analytic below it. For the outgoing radiation condition with this time convention, initially take $k_0=\omega /c_0-i\epsilon$, $\epsilon >0$, and pass to the limit at the end. Thus $+k_0$ lies below the contour and $-k_0$ above it.

Choose $\gamma =(k^2-k_0^2{)}^{1/2}$ to have positive real part on the real transform line. Its branch cuts run from $+k_0$ into the lower half-plane and from $-k_0$ into the upper half-plane, without crossing $\Gamma$; downward and upward vertical rays are suitable. In the zero-absorption limit, $\gamma >0$ for real $|k|>k_0$ and $\gamma =+i\sqrt{k_0^2-k^2}$ between the branch points. This ensures that $e^{-\gamma |y|}$ represents decay or outgoing radiation, rather than an incoming exterior field.

Evenness in $y$ and the Helmholtz equation give the transformed fieldsAt $y=b$, the one-sided normal derivatives of the scattered field agree for $x<0$, because that part of the interface is open. For $x>0$ they both vanish by rigidity. Their common trace is therefore a minus function, denoted $V^{-}(k)={\partial }_y{\Phi }^{-}{|}_{y=b}$. ConsequentlyThe jump in the total scattered transform isFor $x<0$, continuity of the total mass density requires the scattered jump to cancel the incident jump, so $\varphi (b^{+})-\varphi (b^{-})=e^{i{k}_{0}x}$. Its minus transform is $-i/(k+k_0)$. The unknown plate-side jump is the plus function $J^{+}=[{\Phi }^{+}{]}_{b^{-}}^{b^{+}}$. Thus $J=J^{+}-i/(k+k_0)$, and the Wiener-Hopf equation follows:

Use the Wiener-Hopf factorization $L=L^{+}{L}^{-}$, with factors analytic and nonzero in their designated half-planes. Their analytic continuations allocate outgoing modal zeros to $L^{+}$ below the contour, and the opposite zeros to $L^{-}$ above it. Set $C=L^{+}(-k_0)$. Multiplication by $L^{+}$ and pole subtraction giveThe pole in the upper expression is removable by its numerator. The two expressions analytically continue to the common entire function, which is zero under the stipulated edge/growth assumption. Henceand the transformed fields areA constant reciprocal rescaling of the factors does not alter $C{L}^{-}$ or the physical field.

Inside the guide, $\cosh (\gamma y)$ and $\gamma \sinh (\gamma b)$ are even entire functions of $\gamma$, while $L^{-}$ is analytic in the lower half-plane. Thus continuation across the lower branch cut changes none of the interior transform: that cut is removable. The exterior expression retains the cut, corresponding to radiation into the open exterior.

For $x>0$, the inverse-transform contour closes downwards, clockwise. Away from modal cutoffs, its enclosed singularities are the outgoing simple poleswith positive real part for propagating modes and negative imaginary part for decaying modes; $k_0$ is the $n=0$ root. The opposite roots lie above the contour or cancel against zeros of $L^{-}$. Symmetry excludes odd transverse modes. At $n\ge 1$, $\gamma =in\pi /b$, and differentiation of $D(k)=\gamma \sinh (\gamma b)$ givesEach inverse-transform contribution is $-i$ times its residue. Therefore the outgoing modes of an open rigid acoustic waveguide areThe original time factor multiplies this expression. Thus every cut-on mode propagates in the positive $x$ direction; cut-off modes are evanescent waves decaying into the guide, not additional backward waves. Strictly, a complete mode sum includes these evanescent modes as well as propagating ones. The displayed simple-pole amplitudes apply away from exact cutoffs; cutoff values use the outgoing limiting-absorption continuation before taking the limit. The reflected plane-wave amplitude relative to the unit incident wave is $A_0$ above. It has the expected negative sign in the long-wavelength leading kernel approximation $L\simeq b(k^2-k_0^2)$.

Take a static, nonrotating, nonmagnetic spherical equilibrium with $\nabla P=-\rho \nabla \Phi$ and ${\nabla }^2\Phi =4\pi G\rho$. Use a fluid displacement $\mathbf{\xi }(\mathbf{r})e^{-i\omega t}$. Write $p_1,{\rho }_1,{\varphi }_1$ for Eulerian fluid perturbations. The Lagrangian pressure perturbation and corresponding mass density change obey $\Delta P=p_1+\mathbf{\xi }\cdot \nabla P$, $\Delta \rho ={\rho }_1+\mathbf{\xi }\cdot \nabla \rho$. Linearizing mass conservation, the Euler equations for an inviscid fluid and Poisson equation for Newtonian gravity givesThere is no equilibrium acceleration to multiply a perturbed mass density. The adiabatic process condition, with composition carried by the parcel, closes the system:These are the complete linear adiabatic stellar oscillation equations, including the perturbation of self-gravity. Neglect of heat exchange is appropriate when oscillation periods are short compared with relevant thermal relaxation times; it does not determine nonadiabatic excitation or damping.

For the pressure and buoyancy modes, separate angular dependence using spherical harmonics:The horizontal amplitude ${\xi }_h$ has dimensions of length. Angular differentiation gives $\nabla \cdot \mathbf{\xi }=[r^{-2}(r^2{\xi }_r{)}^{\prime }-\ell (\ell +1){\xi }_h/r]Y_{\ell }^m$, and tangential momentum gives ${\omega }^{2}r{\xi }_h=p_{\ell }/\rho +{\varphi }_{\ell }$. Define the adiabatic sound speed, stellar buoyancy frequency and Lamb frequency byThe adiabatic mass density relation becomes ${\rho }_{\ell }/\rho =p_{\ell }/(\rho c_s^2)+(N^2/g){\xi }_r$. Substitution produces a radial form of the full oscillation equations for nonzero $\omega$:The apparent $N^2/g$ factor is evaluated through its defining gradient at the centre rather than by dividing two zeros.

Regularity at the centre excludes singular solutions. At a free surface the Lagrangian pressure perturbation vanishes, $\Delta P=0$; outside the star the gravitational perturbation decays as $r^{-\ell -1}$. For a model with mass density tending to zero at its surface, continuity of ${\varphi }_{\ell }$ and its radial derivative gives ${\varphi }_{\ell }^{\prime }(R)=-(\ell +1){\varphi }_{\ell }(R)/R$. If the equilibrium mass density jumps to vacuum, include the displaced-surface mass sheet: the outward-minus-inward derivative jump is $4\pi G\rho (R){\xi }_r(R)$, so the interior condition is ${\varphi }_{\ell }^{\prime }(R)=-(\ell +1){\varphi }_{\ell }(R)/R-4\pi G\rho (R){\xi }_r(R)$. An atmospheric boundary condition can replace the ideal free surface.

These conditions make ${\omega }^2$ an eigenvalue, not an arbitrary local sound frequency. With conservative boundary conditions the adiabatic operator is self-adjoint, giving real ${\omega }^2$; negative values describe instability. The frequencies depend on $P(r),\rho (r),{\Gamma }_1(r)$, self-gravity, stratification and boundaries, as well as angular degree $\ell$ and radial order. In a spherical nonrotating star they are degenerate in $m$. For homologous equilibrium structures, their scale isthus the typical oscillation time measures inverse square root of mean mass density, while individual frequencies probe the interior adiabatic sound speed and stellar buoyancy frequency profiles.

For radial modes, write ${\xi }_r=r\eta$. Eliminating the pressure and gravitational perturbations gives the radial stellar pulsation equationThis is a Sturm-Liouville problem. Multiplication by $\eta$ and integration, with vanishing boundary terms, gives its Rayleigh quotientFor constant ${\Gamma }_1>4/3$, $P^{\prime }<0$ makes both numerator contributions nonnegative. At ${\Gamma }_1=4/3$, a homologous displacement is neutral; for constant ${\Gamma }_1<4/3$ the same trial displacement makes the quotient negative. For the uniform-density stellar model, $\eta =$ constant is an exact mode and ${\omega }^2=(3{\Gamma }_1-4)GM/R^3$. With varying ${\Gamma }_1$, the integral criterion, rather than a universal pointwise threshold, controls radial stability.

The Cowling approximation neglects ${\varphi }_{\ell }$ while retaining the equilibrium gravitational field. It is useful for short-wavelength modes, but is not needed for the full derivation above. In a locally slowly varying region, take both remaining amplitudes proportional to $e^{i\int k_{r}dr}$ and retain the leading derivative terms. ThenEliminating either amplitude gives the acoustic-gravity propagation relationPositive $k_r^2$ is oscillatory propagation; negative $k_r^2$ means an evanescent wave. The high-frequency branch, ${\omega }^2>S_{\ell }^2,N^2$, describes stellar acoustic modes, restored chiefly by compressibility and pressure. At frequencies well above $N$, this gives ${\omega }^2\simeq c_s^2[k_r^2+\ell (\ell +1)/r^2]$. The low-frequency propagating branch in stable stratification, ${\omega }^2<N^2,S_{\ell }^2$, instead describes stellar gravity modes, restored by buoyancy. At $\ell =0$ there is no such nonradial gravity-wave cavity.

A stellar acoustic mode is trapped between an inner turning point near $\omega =S_{\ell }$ and an outer reflecting region. Low-$\ell$ modes penetrate deeply; radial modes reach the centre. Higher-degree modes turn farther out. Standing waves require the WKB quantization conditionwhere the phase $\alpha$ depends on the central or turning-point behaviour and surface reflection. For high radial order and small degree, the leading acoustic travel-time result is the large frequency separationThe $\ell /2$ term is the leading central angular phase shift; smaller frequency separations depend on detailed interior gradients. The near-surface mass density stratification sets the acoustic cutoff frequency. In a plane-parallel isothermal atmosphere with mass density scale height $H_{\rho }$, ${\omega }_{\mathrm{ac}}\simeq c_s/(2H_{\rho })$. Modes below this cutoff can reflect and form a cavity; waves sufficiently above it escape and need an outgoing-wave boundary condition. Nonadiabatic damping, driving and rotation alter real-star mode properties, but the adiabatic frequency problem isolates their dependence on the equilibrium structure.

The WKB approximation treats the vertical wavenumber as locally constant over one vertical wavelength. Write $w=\Re \{A(z)e^{ikx+i{\int }^{z}m(s)ds}\}$, with $|m^{\prime }|\ll m^2$ and $|A^{\prime }|\ll |mA|$. For the stationary internal-wave WKB solution, $A\propto m^{-1/2}$ away from singular or turning levels. A small terrain slope and small perturbations are also needed for Linearized Boussinesq equations.

Let $p$ be the perturbation fluid pressure divided by reference mass density, and $b$ the buoyancy perturbation. The steady Linearized Boussinesq equations and incompressibility condition areFor an $e^{ikx}$ component, elimination gives $u=-w^{\prime }/(ik)$, $b=-N^{2}w/(ikU)$ and $p=(U{w}^{\prime }-U^{\prime }w)/(ik)$. Substitution into vertical momentum gives the exact stationary Taylor–Goldstein equationThe leading WKB approximation therefore identifiesThe kinematic boundary condition fixes the horizontal wavenumber to $k=k_T$ for the sinusoidal forcing.

For explicit heights assume a constant positive buoyancy frequency $N$, $0<k{U}_0<N$, and $|S|/N\ll 1$. The latter is a large gradient Richardson number. The linear shear flow has $U^{\prime \prime }=0$, so the propagation condition is $|U|<N/k$.

For $S<0$, the first obstruction is the critical level of an internal gravity wave atAs $z\uparrow z_c$, $m\sim N/U\to \infty$. The critical-height approach of a stationary wave in linear shear has upward group velocity $c_{gz}\sim k{U}^2/N$, so the travel time to $z_c$ diverges: the height is a supremum, approached in the ideal ray model. Here $|m^{\prime }|/m^2\to |S|/N$, which can remain small; it would be incorrect to say that the short-wavelength WKB approximation necessarily fails solely because $m$ diverges. Nevertheless $w\propto U^{1/2}$ and displacement $w/(ikU)\propto U^{-1/2}$, so a finite disturbance eventually violates linearity. Dissipation and overturning also matter near this singular level.

For $S>0$, the turning height of a stationary wave in linear shear isAt $z_t$, $m=0$ and the WKB approximation fails. An Airy function matching region describes reflection and an upper evanescent wave. Thus $z_t$ is the highest height of vertical propagation, rather than a sharp cut-off of every disturbance.

The problem initially allows $N(z)$ to vary. Without the extra constant-$N$ assumption, replace the explicit turning height by the first root of $N(z)=k|U_0+Sz|$ encountered on a propagating branch. A declining $N(z)$ can cause a turning level before the negative-shear critical level. The two closed-form heights require the stated constant-stratification assumption.

For a finite potential energy discontinuity, the stationary state $\psi$ and its first derivative are continuous at $a$. Integrating the Time-independent Schrodinger equation across a shrinking interval about $a$ gives continuity of ${\psi }^{\prime }$; a jump in $\psi$ would produce a distributional derivative incompatible with a finite potential energy.

For $E>V_0$, put $k=\sqrt{2mE}/\hslash$, $q=\sqrt{2m(E-V_0)}/\hslash$ and use unit incident amplitude:There is no wave incident from the right. The matching equations $1+r=t$, $k(1-r)=qt$ give $r=(k-q)/(k+q)$ and $t=2k/(k+q)$. Comparing reflected and incident probability currents gives the reflection coefficientThe transmitted fraction is $T=(q/k)|t{|}^2=4kq/(k+q{)}^2$, so $R+T=1$.

For $0<E<V_0$, write $\kappa =\sqrt{2m(V_0-E)}/\hslash$ and retain only the decaying right-hand solution $t{e}^{-\kappa (x-a)}$. Matching gives $r=(k-i\kappa )/(k+i\kappa )$, which has modulus one. The evanescent wave carries no transmitted probability current. At $E=V_0$, the bounded right-hand solution is constant and matching again gives $r=1$. ThusClassically, a particle is wholly transmitted for $E>V_0$ and wholly reflected for $E<V_0$; there is no classical partial reflection above the step. At the threshold it has zero right-hand speed and no transmitted flux, a marginal case requiring a convention about motion exactly on the discontinuity. The quantum result approaches $R=0$ as $E/V_0\to \infty$.

Let $p$ denote the fluid pressure perturbation divided by the reference mass density, and let $b=-g{\rho }^{\prime }/{\rho }_0$ be the buoyancy perturbation. The buoyancy frequency satisfies $N^2=-g{\hat{\rho }}_z/{\rho }_0$. Linearization requires small boundary slope $k|{\eta }_0|\ll 1$, small displacements compared with the vertical scales of the disturbance and background, and advection small compared with the oscillatory acceleration. For a propagating internal gravity wave this includes $|m{\eta }_0|\ll 1$; boundary slope alone is insufficient when $|m|\gg k$. The Boussinesq approximation also requires small relative mass density differences over the region of interest. These conditions must hold for the resulting disturbance, including any amplification by resonance.

The inviscid Linearized Boussinesq equations and linear kinematic boundary condition areWrite the velocity field as $(u,w)=\mathrm{Re}\{(U(z),W(z))e^{i(kx-\omega t)}\}$. Incompressibility and horizontal momentum balance giveThe vertical momentum balance therefore yieldsFor a real vertical wavenumber $m$, this gives the dispersion relationWhen $0<\omega <N$, define $\ell =k\sqrt{N^2/{\omega }^2-1}$. The radiation condition selects $m=-\ell$, because energy must leave the boundary upward. The boundary-forced internal gravity wave isIts wavevector is $\mathbf{K}=(k,m)$, its phase velocity is ${\mathbf{c}}_p=\omega \mathbf{K}/(k^2+m^2)$, and its group velocity isThus the phase velocity points down and right, while the group velocity points up and right. They are perpendicular. Internal-wave polarization makes the particle motion an oscillation along the group velocity direction, with zero first-order mean transport. The constant-phase lines of an internal gravity wave are also parallel to the group velocity. Their inclination $\vartheta$ above the horizontal satisfies $\sin \vartheta =\omega /N$.

When $\omega >N$, set $\kappa =k\sqrt{1-N^2/{\omega }^2}$. Boundedness at infinity selects the evanescent waveThe horizontal and vertical velocity components are in quadrature: fluid particles describe small ellipses, and the response decays over ${\kappa }^{-1}$. There is no upward time-averaged energy flux, because $P$ and $W$ are in quadrature. A real vertical group velocity is not defined for this evanescent wave. The pattern travels horizontally with phase velocity $\omega /k$.

For $N=0$ the response is the evanescent wave with $\kappa =k$; the ratios $\omega /N$ should not be used. At the cutoff $\omega =N>0$, the bounded harmonic solution has $W=-i\omega {\eta }_0$, $U=0$: it neither decays nor has nonzero upward group velocity. It is the limiting cutoff response, rather than a localized radiating disturbance. A uniform nonzero buoyancy frequency in an infinitely deep Boussinesq approximation is itself a local idealization of the background mass density.

The arrows distinguish the phase velocity, group velocity, and oscillatory particle motion of the internal gravity wave. The right panel shows the decay envelope and particle ellipses of the evanescent wave.

The central stratified layer supports upward and downward internal gravity waves, whose counterpropagating components form a standing wave. The unstratified regions support evanescent waves, and the upper disturbance decays as $e^{-k(z-2H)}$. Consequently the central layer is a stratified internal-wave guide, rather than a source of propagating energy at infinity.

The largest response occurs at its trapped normal mode frequencies. They can be specified without solving the forced problem. Set the boundary forcing to zero to find a free normal mode. The lower solution is proportional to $\sinh (kz)$, so continuity of fluid pressure and vertical velocity givesIn the central layer writeThe upper Robin boundary condition then gives the exact trapped-mode conditionThe left side increases strictly from zero to infinity, so there is one positive root for each $n$; the frequencies accumulate at zero. This is constructive phase matching after reflection at both ends. The phase shifts from the evanescent waves matter: simply imposing integer half-wavelengths across the stratified layer is generally incorrect.

In the ideal inviscid model, exact resonant forcing has no bounded steady harmonic solution. The undamped normal mode grows secularly under sustained forcing. Weak viscosity or other losses would produce large finite peaks near the displayed frequencies; the linear approximation eventually fails if the disturbance becomes too large.

The illustrated free normal mode has an oscillatory central region and evanescent wave tails. Its lower tail reaches the fixed zero-displacement boundary; its upper tail decays to infinity.