 Solution

Set $\Delta =0$ and take ${\xi }_r=C{r}^{\ell -1}$. The divergence equation then fixesHydrostatic balance gives $\delta p=-{\xi }_r{p}^{\prime }=\rho g{\xi }_r$. The horizontal momentum equation becomessoThe radial equation gives the same result because ${\omega }^2{\xi }_r=g{\xi }_r^{\prime }+g^{\prime }{\xi }_r$. This is an incompressible stellar surface mode: it changes the shape of the free surface without compressing fluid elements. For $\ell =1$ it is a rigid displacement of the star in the fixed harmonic potential.