Let $\boldsymbol\xi$ be the <Lagrangian displacement (fluid mechanics)>. Linearized mass conservation and adiabaticity give
$$
\delta\rho=-\boldsymbol\xi\mathbin\cdot\nabla\rho-\rho\nabla\mathbin\cdot\boldsymbol\xi,
\qquad
\delta p=-\boldsymbol\xi\mathbin\cdot\nabla p-\gamma p\nabla\mathbin\cdot\boldsymbol\xi.
$$
The fixed potential has no Eulerian perturbation, so the linearized momentum equation is
$$
\rho\,\partial_t^2\boldsymbol\xi
=-\nabla\delta p-\delta\rho\,\nabla\Phi.
$$
For the stated <spherical harmonic> decomposition,
$$
\Delta\equiv\nabla\mathbin\cdot\boldsymbol\xi
=\frac1{r^2}\frac{d(r^2\xi_r)}{dr}
-\frac{\ell(\ell+1)}{r^2}\xi_h.
$$
Taking radial and horizontal components and using $g=\Omega^2r$ gives
$$
\boxed{-\rho\omega^2\xi_r=-g\delta\rho-\frac{d\delta p}{dr}},
\qquad
\boxed{-\rho\omega^2\xi_h=-\delta p},
$$
$$
\boxed{\delta\rho=-\xi_r\frac{d\rho}{dr}-\rho\Delta},
\qquad
\boxed{\delta p=-\xi_r\frac{dp}{dr}-\gamma p\Delta}.
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 314 / 2 / b / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/iii/paper-314/2/b/solution