Past exam of the mathematics course of the University of Cambridge / 2017 / iii / Paper 314 / 4

Let the free surface be at $r=R_1$ and take the Newtonian gravitational potential to vanish at infinity. Mass conservation fixes $R_1=(3M_1/(4\pi {\rho }_1){)}^{1/3}$. The enclosed mass is $m(r)=4\pi {\rho }_1{r}^3/3$, and hydrostatic equilibrium requires $d{p}_0/dr=-{\rho }_{1}Gm(r)/r^2$. With vacuum outside, $p_0(R_1)=0$, givingThe pressure is zero outside. A prescribed constant external pressure would simply add that constant to $p_0$.

The bulk Eulerian and Lagrangian fluid perturbations obey $\delta \rho =0$ because the mass density is uniform and the Lagrangian displacement has zero divergence. The linearized Euler momentum equation in the interior isTaking the curl gives ${\omega }^2\nabla \times \mathbf{\xi }=0$. Thus every nonzero-frequency normal mode has an irrotational flow displacement; the simply connected space formed by the interior admits $\mathbf{\xi }=\nabla U$ and incompressibility gives the Laplace equation ${\nabla }^{2}U=0$. For a regular center, one spherical harmonic component is $U=A_l{r}^l{Y}_l^m$, with $|m|\le l$.

It is essential to retain the surface gravity perturbation of a uniform-density sphere. Although the bulk mass density perturbation is zero, the displaced density discontinuity givesThe Poisson equation implies that $\delta \Phi$ is continuous and its outward radial derivative has the jumpUsing regularity at the center and decay at infinity, writeThe derivative jump is $-(2l+1)C_l{R}_1^{l-1}{Y}_l^m$, so $C_l=-4\pi G{\rho }_{1}l{A}_l/(2l+1)$.

Integrating the bulk Euler momentum equation gives $\delta p/{\rho }_1={\omega }^{2}U-\delta \Phi$ for $l\ge 1$. At the free surface the Lagrangian pressure perturbation vanishes:Combining these relations yields the Kelvin stellar mode frequencyThe $2l+1$ choices of $m$ are degenerate because of spherical symmetry. For $l=1$, $r{Y}_1^m$ is a linear combination of coordinates in a Cartesian coordinate system, so $\nabla U$ is a constant displacement: the zero frequency is rigid translation of the whole isolated body. Such a translation has no restoring force. The constant $l=0$ potential generates no displacement; a radial breathing motion is excluded by incompressibility and a regular center.

The printed irrotational assertion needs the nonzero-frequency qualification, or restriction to the potential incompressible stellar surface modes. It is false for all neutral displacements: $\mathbf{\xi }=\mathbf{\Omega }\times \mathbf{r}$ has zero divergence, zero normal displacement at the surface, and $\delta p=\delta \Phi =0$, so it is a zero-frequency displacement, but $\nabla \times \mathbf{\xi }=2\mathbf{\Omega }\ne 0$. This does not change the Kelvin stellar mode spectrum just derived.

The separation of the two bodies is $a$, as is fixed by the supplied Kepler third law ${\omega }_o^2=G(M_1+M_2)/a^3$. The phrase “orbit of radius $a$” cannot denote the first body's distance from the center of mass in that formula.

Expand the companion's Newtonian potential of a point mass about the first body's center. The constant term produces no force; the linear term produces the uniform acceleration of the center and disappears in the translating frame. The leading differential acceleration therefore comes from the quadrupolar tidal forcingwhere $\mathbf{n}(t)$ points toward the companion. It is a degree-two solid spherical harmonic. For a circular nonrotating-frame orbit, its time-dependent components have $m=\pm 2$ and frequency $2{\omega }_o$, because the squared direction cosines repeat twice per orbit. There is also a static $m=0$ component. The real part of one complex harmonic represents the pair of time-dependent components; its normalization and orientation are absorbed into the order-one coefficient $\psi$. This gives $\Psi =\psi (G{M}_2/a^3)r^2{Y}_2^m$ and $\omega =2{\omega }_o$. The expansion is the leading term for $R_1/a$ small; spin is being neglected.

For a forced potential flow with the same temporal frequency, the Euler momentum equation becomesUse $U=A{r}^2{Y}_2^m$. The surface gravity perturbation of a uniform-density sphere is unchanged, as is the vanishing Lagrangian pressure perturbation boundary condition. Integrating the equation and applying that condition givesSince ${\xi }_r(R_1)=2A{R}_1{Y}_2^m$, the real surface displacement isIndeed $ϵ{\omega }_2^2=2G{M}_2/a^3$, so the normalization is consistent. This is a particular forced response away from the resonance; arbitrary free normal modes can be added.

Put $\mu =M_2/M_1$ and $q=(R_1/a{)}^3$. The tidal resonance of a stellar oscillation condition is $4{\omega }_o^2={\omega }_2^2$, henceThe resonance gives $0\le ϵ<1/2$ for finite nonnegative $\mu$, with actual forcing requiring $M_2>0$. At exact resonance the undamped linearization has no bounded response at the forcing frequency: the time-domain particular solution grows proportionally to $t$ times a sinusoid. Thus the frequency-response formula has a pole, rather than describing a finite resonant displacement. Away from that pole, linearization requires the displayed surface displacement to be small compared with $R_1$; the leading quadrupolar tidal forcing approximation separately requires $R_1/a$ small.