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

Hydrostatic equilibrium requiresFor the polytrope $p=K{\rho }^{1+1/n}$, the specific enthalpy isThus $h+\Phi$ is constant. Since $\rho =p=0$ at $r=R$, this constant is ${\Omega }^2{R}^2/2$, andfor $0\le r\le R$, with both fields zero outside the model star.

Let $\mathbf{\xi }$ be the Lagrangian displacement. Linearized mass conservation and adiabaticity giveThe fixed potential has no Eulerian perturbation, so the linearized momentum equation isFor the stated spherical harmonic decomposition,Taking radial and horizontal components and using $g={\Omega }^{2}r$ gives

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.

The tidal potential is proportional to the solid spherical harmonic $r^{\ell }{Y}_{\ell }^m$, whose gradient has exactly the spatial form of the mode in part (c). Projecting the forced linear equation onto that eigenfunction gives the oscillator factor ${\omega }_{\ell }^2-{\omega }^2$. Consequentlyand its radial component isTaking the real part gives the stated physical displacement. The amplitude displays tidal resonance of a stellar oscillation: it is enhanced near $|\omega |={\omega }_{\ell }$ and formally diverges in this undamped linear model. Physical damping makes the peak finite and supplies a phase shift through resonance.