At the equilibrium thickness $H_0$,
$$
\Omega_z^2H_0=CH_0^{-\gamma}.
$$
Set $H=H_0+h$ and retain terms linear in $h$. The <linearization> is
$$
\ddot h+(\gamma+1)\Omega_z^2h=0.
$$
Thus the <vertical breathing mode of an astrophysical disk> has angular frequency
$$
\boxed{\omega_H=\sqrt{\gamma+1}\,\Omega_z}.
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 321 / 3 / f / Solution

{title2=$\omega_H$}

The vertical breathing mode is a coherent oscillation of a disk's thickness. For an adiabatic gas with pressure proportional to $H^{1-\gamma}$ in a harmonic vertical potential, $\omega_H=\sqrt{\gamma+1}\,\Omega_z$.


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