= Solution
Use the nonrotating linear <Boussinesq approximation> with reference density $\rho_0$ and $N^2=-(g/\rho_0)d\bar\rho/dz>0$:
$$
\partial_tu'=-\frac{\partial_xp'}{\rho_0},\quad
\partial_tv'=-\frac{\partial_yp'}{\rho_0},\quad
\partial_tw'=-\frac{\partial_zp'}{\rho_0}-\frac{g\rho'}{\rho_0},\quad
\partial_t\rho'+w'\bar\rho_z=0,\quad
\nabla\cdot\mathbf u'=0.
$$
Let $K_h^2=k^2+l^2>0$, $K^2=K_h^2+m^2$ and $E=e^{i(kx+ly+mz-\omega t)}$. The horizontal momentum equations give $u'=kp'/(\rho_0\omega)$ and $v'=lp'/(\rho_0\omega)$. Continuity requires $ku'+lv'+mw'=0$. Thus the <internal-wave polarization> is
$$
\boxed{u'=-\frac{km}{K_h^2}w_0E,\qquad
v'=-\frac{lm}{K_h^2}w_0E,\qquad
p'=-\frac{\rho_0\omega m}{K_h^2}w_0E,\qquad
\rho'=i\frac{\rho_0N^2}{g\omega}w_0E.}
$$
Physical perturbations are the real parts. The density is in temporal quadrature with the vertical velocity, because buoyancy responds to the vertical <fluid displacement>. Substitution into the vertical momentum equation determines the <internal gravity wave> <dispersion relation>,
$$
\boxed{\omega^2=N^2\frac{K_h^2}{K^2}.}
$$
A nontrivial propagating wave requires a nonzero horizontal <wave vector> and a nonzero frequency. The formulas are not to be divided by $K_h^2=0$: incompressibility excludes a nonzero oscillatory vertical velocity for a purely vertical <wave vector>.
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