Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 317 / 4

In the synchronously rotating reference frame, stationary fluid obeyswhere $\Psi$ is the sum of the two gravitational potentials and the centrifugal potential. The circular binary's Kepler third law gives ${\Omega }^2=G(M_1+M_2)/a^3=GM/a^3$, hence

Uniform composition and the assumed central condensation make star 1 approximately barotropic, so one may define the specific enthalpyHydrostatic balance becomes $\nabla (h+\Psi )=0$, or $h+\Psi =\text{constant}$ throughout the star. Since $h$ is a monotone function of $P$ and $\rho$, constant-$P$ and constant-$\rho$ surfaces are equipotential surfaces of $\Psi$.

The Roche lobe of star 1 is the volume around it bounded by the critical closed equipotential passing through the inner Lagrange point $L_1$. Material on a lower potential surface remains confined to star 1; at the critical surface a path opens toward star 2.

Let $\mathbf{g}=-\nabla \Psi$ and take $d\mathbf{S}$ outward. For a nearly spherical interior surface, the divergence theorem givesThe self-gravity term contributes $-4\pi G{M}_1$. The companion lies outside $S$, so its potential is harmonic inside and contributes zero net flux. For the centrifugal term, ${\nabla }^2(-{\Omega }^2{s}^2/2)=-2{\Omega }^2$, so its acceleration has divergence $2{\Omega }^2$ and contributes $2{\Omega }^{2}V$. Using $V=4\pi r^3/3$ and $S=4\pi r^2$,The sign is the outward-normal component; the dominant self-gravity is inward and therefore negative.

Let $g=|\nabla \Psi |$ on the equipotential $S(\Psi )$. The volume between neighboring equipotentials separated by $d\Psi$ isBecause density is constant on each equipotential,The volume-equivalent radius is defined by $V=4\pi r^{\prime 3}/3$, so $dV=4\pi r^{\prime 2}d{r}^{\prime }$. It follows immediately that

Hydrostatic balance gives $dP/d\Psi =-\rho$, whileConsequentlyDefining the geometric correctionputs this in the one-dimensional stellar-structure formFor a sphere, $S=4\pi r^{\prime 2}$ and $g=Gm/r^{\prime 2}$, so $f_P=1$ as required.

Because $T=T(\Psi )$,The outward normal radiative flux is thereforeIntegrating over the equipotential givesPart (iii) gives $dm/d\Psi =\rho S⟨g^{-1}⟩$, and henceSubstitution yieldsThis is the radiative equation in the equipotential stellar-structure approximation.

At the critical Roche surface, $g=0$ at $L_1$, making $⟨g^{-1}⟩$ singular. Once star 1 overfills its Roche lobe, the relevant equipotentials are no longer nested closed surfaces belonging to that star: they open through the $L_1$ saddle and connect to star 2. Matter then undergoes Roche-lobe overflow with finite velocity, so advective momentum and energy transport replace hydrostatic and purely radiative equilibrium near the nozzle. Density, temperature, and composition need not remain uniform on an equipotential. The one-dimensional volume coordinate $r^{\prime }(\Psi )$ and its correction factors therefore cease to describe the three-dimensional mass-transfer flow.