= Free-surface pressure condition of an incompressible disk
Vertical <hydrostatic equilibrium> gives $p_0=\rho\Omega_z^2(H^2-z^2)/2$. At a vacuum interface, the <Lagrangian pressure perturbation> vanishes: $p'+\xi_z\partial_zp_0=0$. The vertical equation gives $\xi_z=p'_z/(\rho\omega^2)$, hence
$$
p'-\frac{\Omega_z^2}{\omega^2}zp'_z=0\quad\text{at }z=\pm H.
$$
The moving interface makes this different from setting the <Eulerian perturbation of a fluid variable> $p'$ to zero on the original surface.
// Target: astrophysics.bigb
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