= Plane-parallel Cowling displacement-pressure equations
{title2=$\xi'=-gk^2\xi/\omega^2-(k^2/\omega^2-c^{-2})\Pi/\rho$}
Take depth $z$ positive downward, vertical <fluid displacement> $\xi$ positive upward, and <Lagrangian pressure perturbation> $\Pi$. In a <plane-parallel atmosphere> with $p'=\rho g$, the <Cowling approximation> and adiabatic relation give
$$
\xi'=-\frac{gk^2}{\omega^2}\xi-\left(\frac{k^2}{\omega^2}-\frac1{c^2}\right)\frac\Pi\rho,\qquad
\Pi'=-\rho\left(\omega^2-\frac{g^2k^2}{\omega^2}\right)\xi+\frac{gk^2}{\omega^2}\Pi.
$$
Indeed the Eulerian perturbations are $p_1=\Pi+\rho g\xi$ and $\rho_1=\Pi/c^2+\rho'\xi$. Horizontal momentum gives $\boldsymbol\xi_h=i\boldsymbol k p_1/(\rho\omega^2)$; combine this with mass conservation and vertical momentum to eliminate the horizontal displacement. Reversing the sign convention for vertical displacement reverses both displacement-pressure coupling signs.
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