Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 322 / 2 / Solution

The component distances from the centre of mass are $a_1=a{M}_2/M$ and $a_2=a{M}_1/M$, where $M=M_1+M_2$. Their orbital angular momenta add to

Let $W>0$ be the isotropic wind-loss magnitude, so ${\dot{M}}_1=-W$ and ${\dot{M}}_2=0$. The wind carries star 1's specific orbital angular momentum $j_1=a_1^2\Omega$, and henceUsing Kepler third law to write $J=M_1{M}_2\sqrt{Ga/M}$ and differentiating showsSince $P^2\propto a^3/M$, it follows thatThis is Jeans-mode mass loss.

Now allow transfer to star 2 at rate ${\dot{M}}_2>0$ while the wind continues. ThenThe same conservation of angular momentum calculation givesThe donor response $R\propto M_1^{-n}$ iswhile

Before contact, put ${\dot{M}}_2=0$. The wind drives the star farther into its Roche lobe when $\dot{\log }(R/R_L)>0$. With $q=M_1/M_2$, this condition reduces toIf the inequality is reversed, the Roche lobe expands relative to the donor, so wind loss detaches the star and no wind-driven Roche-lobe transfer is sustained; later nuclear expansion may restore contact.

During stable contact, impose $\dot{R}/R={\dot{R}}_L/R_L$ and solve for the transfer rate. Straightforward algebra givesIn the paper's signed notation $W=-\dot{M}$, this is exactlyIfthe stationary response has the wrong sign: transfer enlarges the overfill rather than removing it. Dynamical stability of binary mass transfer is lost, leading to runaway transfer and usually a common envelope or merger.