= Solution
Let $p$ be <kinematic pressure>, with the resting hydrostatic part subtracted, and let $\sigma$ be the <buoyancy perturbation> about background buoyancy $N^2z$. The nonrotating inviscid <Boussinesq equations>, with no buoyancy diffusion, are
$$
\frac{Du}{Dt}=-p_x,\qquad
\frac{Dw}{Dt}=-p_z+\sigma,\qquad
u_x+w_z=0,\qquad
\frac{D\sigma}{Dt}+N^2w=0,
\qquad \frac D{Dt}=\partial_t+u\partial_x+w\partial_z.
$$
Here the third equation is <mass conservation> for <incompressible flow>, and the last expresses <buoyancy> conservation following a parcel. Linearization about rest removes the products of disturbances. For a <plane internal gravity wave> proportional to $e^{i(kx+mz-\omega t)}$, the resulting <Linearized Boussinesq equations> become
$$
-i\omega\widehat u=-ik\widehat p,\qquad
-i\omega\widehat w=-im\widehat p+\widehat\sigma,\qquad
k\widehat u+m\widehat w=0,\qquad
-i\omega\widehat\sigma+N^2\widehat w=0.
$$
With $K=(k^2+m^2)^{1/2}$ and $k\ne0$, the first and third equations give $\widehat p=-m\omega\widehat w/k^2$. Substitution in vertical momentum and buoyancy gives the <dispersion relation>
$$
\boxed{\omega^2=\frac{N^2k^2}{k^2+m^2}.}
$$
Since the <velocity> is the time <derivative> of the <displacement field>, $\widehat{\boldsymbol u}=-i\omega\widehat{\boldsymbol\xi}$. Thus the complete <buoyancy polarization of a plane internal gravity wave> is
$$
\boxed{
\widehat w=\frac{i\omega}{N^2}\widehat\sigma,\quad
\widehat u=-\frac{im\omega}{kN^2}\widehat\sigma,\quad
\widehat\zeta=-\frac{\widehat\sigma}{N^2},\quad
\widehat\xi=\frac{m\widehat\sigma}{kN^2},\quad
\widehat p=-\frac{im}{K^2}\widehat\sigma.}
$$
Both <velocity> and <displacement> are perpendicular to the <wavevector>; <pressure> and <velocity> are in <quadrature> with <buoyancy> and <displacement>.
For upward/rightward <phase velocity>, choose $k,m>0$ and $\omega=Nk/K$. Differentiating this <dispersion relation> gives the <group velocity>, while the vector <phase velocity> points along the <wavevector>:
$$
\boxed{\mathbf c_g=\left(\frac{Nm^2}{K^3},-\frac{Nkm}{K^3}\right),\qquad
\mathbf c_p=\frac{Nk}{K^3}(k,m).}
$$
Their <dot product> is zero and $\mathbf c_p+\mathbf c_g=(N/K,0)$. Draw $\mathbf c_g$ from the tip of $\mathbf c_p$: the <right-triangle geometry of internal-wave velocities> has horizontal hypotenuse $N/K$. This hypotenuse is distinct from the <horizontal phase velocity> $\omega/k$.
At an instant with real <complex amplitude> $\widehat\sigma=A$, write $\phi=kx+mz-\omega t$. The spatial fields are
$$
\sigma=A\cos\phi,\quad
(\xi,\zeta)=\frac A{N^2}(m/k,-1)\cos\phi,\quad
(u,w)=\frac{\omega A}{N^2}(m/k,-1)\sin\phi,\quad
p=\frac{mA}{K^2}\sin\phi.
$$
The <constant-phase lines of an internal gravity wave> slope down to the right, with $dz/dx=-k/m$. <Buoyancy> maxima lie a quarter <wavelength> from <pressure> maxima; the oscillating <displacement> and <velocity> run along the phase lines. The <group velocity> points down and right along them, despite upward/rightward motion of the <wave phase>.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2002/iii/paper-49-internal-wave.png]
{title=Internal-wave buoyancy, displacement, pressure and velocity fields, with downward energy propagation and the phase/group-velocity right triangle}
Let $P=\overline{\xi_xu+\zeta_xw}$, using horizontal averaging over a <wavelength>. This is the <displacement pseudomomentum of an internal gravity wave>, whose sign convention is opposite to $k\overline E/\omega$. For complex coefficients, $\overline{ab}=\tfrac12\operatorname{Re}(\widehat a\widehat b^*)$. The above <wave polarization> consequently gives
$$
P=-\frac{\omega K^2|\widehat\sigma|^2}{2kN^4},\qquad
\overline{\zeta_xp}=\frac{km|\widehat\sigma|^2}{2N^2K^2}.
$$
Since $c_{gz}=-\omega m/K^2$, these obey $\overline{\zeta_xp}=c_{gz}P$. With weak <kinematic viscosity>, each leading monochromatic <velocity> component satisfies $\nabla^2u=-K^2u$, $\nabla^2w=-K^2w$. The dissipative term is therefore $-\nu K^2P$, so the supplied averaged balance reduces to
$$
P_t+\partial_z(c_{gz}P)=-\nu K^2P.
$$
For constant $k,m,N$ and a steady wave envelope, $c_{gz}P_z=-\nu K^2P$. As $P$ is proportional to squared <wave amplitude>, <viscous attenuation of an internal gravity wave> gives
$$
\boxed{|\widehat\sigma(z)|=|\widehat\sigma(z_s)|
\exp\left[-\frac{\nu K^2(z-z_s)}{2c_{gz}}\right].}
$$
Here $c_{gz}<0$, so amplitude decreases below a source at $z_s$: \b[the source is above the receiving level]. The same conclusion follows directly because downward <group velocity> carries <energy> from the source downward. Upward <phase velocity> does not locate the source.
For a slowly varying mean <shear flow> $U(z)$, the local <intrinsic frequency> is $\widehat\omega=\omega-kU(z)$. A local <WKB approximation> uses $\widehat\omega^2=N^2k^2/(k^2+m(z)^2)$, with varying vertical <wavenumber> and <group velocity>. The appropriate transported quantity is <wave action>, or horizontal <wave pseudomomentum> $kE/\widehat\omega$; its vertical flux has viscous loss, and the changing polarization factors must be retained. In particular, $P/|\widehat\sigma|^2$ is no longer constant.
Near a <critical level of an internal gravity wave>, $\widehat\omega\to0$, so $|m|\sim N|k|/|\widehat\omega|$ and $|c_{gz}|\sim\widehat\omega^2/(N|k|)$. The damping per vertical distance, of order $\nu K^2/|c_{gz}|$, grows rapidly. The waves are strongly attenuated over a narrow altitude range and transfer momentum to the mean flow there or before reaching the critical level. The inviscid <WKB approximation> eventually fails; <viscosity> and possibly <internal-wave breaking> regulate the small scales.
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