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Let ${\rho }_0$ be a constant reference density and define total buoyancy acceleration by $\sigma =-g(\rho -{\rho }_0)/{\rho }_0$. Absorb the reference hydrostatic pressure into a kinematic pressure $p$. The ideal nonrotating Boussinesq equations areFor a resting density profile ${\rho }_b(z)$, the background satisfies $p_b^{\prime }(z)={\sigma }_b(z)$ and the squared buoyancy frequency isStable stratification has $N^2>0$. The Boussinesq approximation requires fractional density variations to be small over the domain and parcel excursions, so density can be replaced by ${\rho }_0$ in inertia while its small variation is retained in the gravitational force. For an approximately uniform density gradient over height $H$, this requires $N^{2}H/g\ll 1$. The stipulated incompressibility alone does not justify neglecting density variation in inertia.

Put $b=\sigma -{\sigma }_b$ and $p^{\prime }=p-p_b$. To linear order the two-dimensional internal gravity wave equations areThe exact impermeability condition on $z=\eta (x,t)$ is the kinematic boundary conditionTaylor expansion about $z=0$ permits $w(x,0,t)={\eta }_t$ at leading order if the boundary slope and its displacement relative to the vertical wave scale are small: $|kϵ|\ll 1$ and $|mϵ|\ll 1$. The second condition controls both the omitted evaluation correction $\eta w_z$ and the product $u{\eta }_x$, since continuity gives $u\sim (m/k)w$. Hence for the harmonic forcing,The propagating calculation assumes $k\ne 0$ and $0<|\omega |<N$; a time-independent boundary does no oscillatory work on fluid initially at rest. If $k=0$ but $\omega \ne 0$, incompressibility forces the vertical velocity to be independent of height, so the unbounded column has no finite-energy radiating internal-wave solution. The propagating wave-generation problem therefore needs the stated nonzero horizontal wavenumber.

For a plane mode with phase $\theta =kx+mz-\omega t$, let $U,W,P,B$ be its complex amplitudes. The linear equations giveSubstituting into $-i\omega W=-imP+B$ gives the dispersion relationNo waves enter from above. The appropriate radiation condition therefore selects upward vertical group velocity, not upward phase propagation:The boundary-forced internal gravity wave satisfying the boundary condition isThe displacement is $ϵ\sin \theta$, consistent with $b=-N^2$ times the displacement. For a right-moving boundary choose $k>0$, $\omega >0$ without loss of generality. Then $m<0$, and the constant-phase lines of an internal gravity wave have slope $dz/dx=-k/m>0$: the crests rise toward the right. Their acute angle with the horizontal isThe group velocity is along these crests in the upward-right direction, whereas the wavevector and phase propagation point downward-right.

The physical pressure perturbation is ${\rho }_0{p}^{\prime }$. Pressure work at the lower boundary, to second order in amplitude, is its period-mean pressure times vertical boundary velocity. Static hydrostatic terms average to zero over a cycle. ThereforeAs an independent energy check, the wave's mean kinetic-plus-buoyancy energy density is ${\rho }_0{ϵ}^2{N}^2/2$, and multiplication by $c_{gz}$ gives exactly this upward energy flux.

For the exact horizontal mean-momentum equation, incompressibility writes the horizontal momentum equation asAverage over a horizontal period. Require the pressure and velocity fluxes to be periodic, or more generally their horizontal derivative averages to vanish; in particular there must be no imposed mean pressure gradient. ThenHorizontal averaging also makes ${\partial }_z\overline{w}=0$; the zero mean boundary displacement rate gives $\overline{w}=0$. These conditions exclude an independent mean mass flux or horizontal force. The mean equation applies in the fluid, with flattening of the small moving boundary understood at the order used above.

For slow switch-on, define $A(z,t)=ϵ(t-z/c_{gz})$, using $ϵ(s)=0$ for $s<0$. The outgoing internal-wave envelope radiation condition is $A_t+c_{gz}{A}_z=0$. Replace $ϵ$ in the oscillatory fields by $A$. A modulation time $T$ must satisfy $|\omega |T\gg 1$ and $|m|c_{gz}T\gg 1$, with a frequency bandwidth sufficiently narrow to linearize the dispersion relation. The exact boundary derivative also contains $\dot{ϵ}\sin (kx-\omega t)$; it is smaller by the slow-modulation parameter and is beyond this leading envelope approximation.

To leading slow-modulation order, the wave momentum flux isSince ${\partial }_z{A}^2=-c_{gz}^{-1}{\partial }_t{A}^2$, integrating the mean-momentum equation from rest gives the internal-wave momentum depositionThis is second order in wave amplitude, with smaller relative slow-modulation corrections. Mean-flow feedback on the leading wave is higher order. The mean current is created as the wave envelope arrives; an already uniform wavetrain has no local flux divergence.

Finally move horizontally with speed $c_p=\omega /k$. The initial uniform relative flow is $-c_p$, with infinite total energy in the unbounded column. Its finite mean-energy change is insteadAt finite time the envelope is absent sufficiently far above, so integrating the momentum equation givesThis is the boundary work and mean-flow energy of an internal wave relation. In the moving frame the boundary pattern is stationary to leading modulation order. It extracts energy from the incident mean flow and converts it to outgoing internal-wave energy; in the laboratory frame the same energy is supplied as boundary work.

In the Boussinesq approximation, replace density by a constant reference density ${\rho }_0$ in inertia and incompressible flow, retaining the small density departure only in the gravitational force. Write $\rho =\bar{\rho }(z)+{\rho }^{\prime }$ and subtract the resting hydrostatic pressure. The upward buoyancy perturbation acceleration and squared buoyancy frequency areThus stable density stratification has $N^2>0$. Let $p$ henceforth denote the pressure anomaly divided by ${\rho }_0$. The ideal, nonrotating Boussinesq equations areThe buoyancy equation follows by materially conserving the total buoyancy $b_0(z)+\sigma$, where $b_0^{\prime }=N^2$; it does not assume constant $N$.

For an $x$-periodic flow of period $P$, define $\bar{a}=P^{-1}{\int }_0^{P}adx$. An infinite-domain average with vanishing end contributions works as well. The necessary pressure condition is no mean zonal pressure gradient, $\overline{p_x}=0$: periodic pressure satisfies it, whereas a pressure containing a term proportional to $x$ does not. Use incompressible flow to write the zonal momentum equation in conservative form,Averaging its zonal derivative terms yields the exact Reynolds stress relation

For two-dimensional small disturbances about rest, the Linearized Boussinesq equations are $u_t=-p_x$, $w_t=-p_z+\sigma$, ${\sigma }_t=-N^{2}w$, and $u_x+w_z=0$. Differentiate the two momentum equations to eliminate pressure: $(u_z-w_x{)}_t=-{\sigma }_x$. Differentiating in $x$ and using $u_{xz}=-w_{zz}$ gives ${\nabla }^2{w}_t={\sigma }_{xx}$. One further time derivative givesThe equation remains valid for $N=N(z)$. No vertical derivative of $N$ was taken: ${\partial }_x^2(N^{2}w)=N^2{w}_{xx}$.

For constant $N$ and $k\ne 0$, a plane internal gravity wave hasBoth signs give the prescribed lower-boundary velocity. The radiation condition excludes energy arriving from infinity. At positive frequency, the vertical group velocity isFor a source below the fluid, energy must propagate upward, so the outgoing solution has $m<0$, even though its vertical phase velocity $\omega /m$ is downward.

To obtain this selection causally, let $T=\mu t$ and $Z=\mu z$, and replace the constant amplitude by $a(Z,T)$. The method of multiple scales treats $(z,t,Z,T)$ as independent during differentiation: the carrier derivatives hold $Z,T$ fixed, while $a_T$ holds $Z$ fixed and $a_Z$ holds $T$ fixed. Only after differentiating do we restrict to $T=\mu t$, $Z=\mu z$. Acting on the complex wave, the operators becomeWith $K^2=k^2+m^2$, the wave equation has residualThe zeroth-order term is the dispersion relation; the first-order term vanishes precisely whenIf $a_0(T)$ is the switched-on boundary amplitude, with $a_0(T)=0$ for $T<0$, the outgoing envelope is $a(Z,T)=a_0(T-Z/b)$ for $b>0$. Its characteristics carry the boundary signal into initially undisturbed fluid. If $b<0$, characteristics instead arrive from the upper half-space; zero initial disturbances and no incoming signal cannot establish the proposed boundary-forced wavetrain. This is the internal-wave envelope radiation condition.

The leading incompressible flow polarization gives $u^{\prime }=-(m/k)w^{\prime }$. Hence, to second order in amplitude and leading order in the slow modulation,There is no meridional transport here. Envelope transport gives ${\partial }_t|a{|}^2=-b{\partial }_z|a{|}^2$, so an initially vanishing mean develops asThese are the leading expressions for internal-wave momentum deposition; higher slow-modulation corrections to the stress are beyond the displayed approximation. In particular, a spatially uniform fully established wavetrain has zero stress divergence, but the front has already produced its mean flow.

The mean wave energy density per unit mass follows from ${\sigma }^{\prime }=-i{N}^2{w}^{\prime }/\omega$:Thus $\bar{u}=(k/\omega )\mathcal{E}$. In the original frame the mean kinetic energy is ${\bar{u}}^2/2=O(a^4)$. In a frame translating at $c=\omega /k$, the undisturbed velocity is $-c$, and the change in mean kinetic energy is $-c\bar{u}+O(a^4)=O(a^2)$. Indeed $\mathcal{E}-c\bar{u}=0$ at this order behind the front. Mean-flow energy is therefore essential at the same order as wave energy in the translating frame. This follows also from the frame transformation of energy, $E\mapsto E-cP$ up to the fixed background constant, and is why conservation of momentum cannot be omitted from the moving-frame conservation of energy budget.