Axisymmetric waves of a homogeneous incompressible disk

In a thin constant-density disk with free surfaces $z=\pm H$, neglect slow radial gradients and let perturbations vary as $e^{ikr-i\omega t}$. With $h=p^{\prime }/\rho$, the local inviscid equations are $-i\omega u-2\Omega v=-ikh$, $-i\omega v+{\kappa }^{2}u/(2\Omega )=0$, $-i\omega w=-h_z$ and $iku+w_z=0$. Eliminating velocity gives $h_{zz}=k^2{\omega }^{2}h/({\omega }^2-{\kappa }^2)$. This local WKB approximation assumes $H/r\ll 1$ and $|kr|\gg 1$ and has both surface and inertial wave branches.

// Target: astrophysics.bigb

For radial and vertical Lagrangian fluid displacements of a local axisymmetric mode, integrate the momentum equations and use incompressible flow plus the free-surface pressure condition of an incompressible disk. The result isFor nonnegative ${\kappa }^2$ and positive ${\Omega }_z^2$, every nontrivial oscillatory mode has real nonnegative ${\omega }^2$. The surface term is the gravitational restoring energy of the displaced vacuum interfaces.

For $0<{\omega }^2<{\kappa }^2$, set $S=kH\omega /\sqrt{{\kappa }^2-{\omega }^2}$. Even-pressure modes satisfy ${\omega }^2=-{\Omega }_z^{2}S\tan S$, while odd-pressure modes satisfy ${\omega }^2={\Omega }_z^{2}S\cot S$. In the case ${\Omega }_z>\kappa >0$, the branches approach $\kappa$ as $kH\to 0$. For each fixed vertical order $n\ge 1$ and large $kH$, $S\to n\pi /2$ and $\omega \sim \kappa n\pi /(2kH)$. Even orders have even pressure and odd orders odd pressure. The limiting boundary condition is zero normal derivative of pressure at both interfaces.

// Target: astrophysics.bigb

For ${\omega }^2>{\kappa }^2$, set $Q=kH\omega /\sqrt{{\omega }^2-{\kappa }^2}$. Even-pressure and odd-pressure modes have $p^{\prime }\propto \cosh (Qz/H)$ and $\sinh (Qz/H)$ and obey ${\omega }^2={\Omega }_z^{2}Q\tanh Q$ and ${\omega }^2={\Omega }_z^{2}Q\coth Q$, respectively. At large $kH$, both become free-surface gravity waves withTheir opposite-parity splitting is exponentially small as the two interfaces decouple.

// Target: astrophysics.bigb

Vertical hydrostatic equilibrium gives $p_0=\rho {\Omega }_z^2(H^2-z^2)/2$. At a vacuum interface, the Lagrangian pressure perturbation vanishes: $p^{\prime }+{\xi }_z{\partial }_z{p}_0=0$. The vertical equation gives ${\xi }_z=p_z^{\prime }/(\rho {\omega }^2)$, henceThe moving interface makes this different from setting the Eulerian perturbation of a fluid variable $p^{\prime }$ to zero on the original surface.

// Target: astrophysics.bigb