The component distances from the <centre of mass> are $a_1=aM_2/M$ and $a_2=aM_1/M$, where $M=M_1+M_2$. Their orbital angular momenta add to
$$
\begin{aligned}
J
&=M_1a_1^2\Omega+M_2a_2^2\Omega\\
&=\boxed{\frac{M_1M_2}{M}a^2\Omega}.
\end{aligned}
$$

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 hence
$$
\frac{\dot J}{J}
=-W\frac{j_1}{J}
=-W\frac{M_2}{M_1M}.
$$
Using <Kepler third law> to write $J=M_1M_2\sqrt{Ga/M}$ and differentiating shows
$$
\frac{\dot a}{a}=\frac{W}{M},
\qquad
\boxed{aM=\text{constant}}.
$$
Since $P^2\propto a^3/M$, it follows that
$$
\boxed{PM^2=\text{constant}}.
$$
This is <Jeans-mode mass loss>.

Now allow transfer to star 2 at rate $\dot M_2>0$ while the wind continues. Then
$$
\dot M_1=-W-\dot M_2,
\qquad
\dot M_{\rm total}=-W.
$$
The same <conservation of angular momentum> calculation gives
$$
\frac{\dot a}{a}
=\frac{W}{M}
+2\dot M_2\left(\frac1{M_1}-\frac1{M_2}\right).
$$
The donor response $R\propto M_1^{-n}$ is
$$
\frac{\dot R}{R}=n\frac{W+\dot M_2}{M_1},
$$
while
$$
\frac{\dot R_L}{R_L}
=\frac{\dot a}{a}
+\frac13\left(\frac{\dot M_1}{M_1}
-\frac{\dot M_{\rm total}}M\right).
$$

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 to
$$
\boxed{q<\frac{1+3n}{3(1-n)}}.
$$
If 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 gives
$$
\boxed{\dot M_2
=\frac{1+3n-3(1-n)q}
{(1+q)(5-3n-6q)}\,W}.
$$
In the paper's signed notation $W=-\dot M$, this is exactly
$$
\boxed{\dot M_2
=-\frac{1+3n-3(1-n)q}
{(1+q)(5-3n-6q)}\,\dot M}.
$$
If
$$
5-3n-6q<0,
$$
the 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.

Solution

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

{c}

In Jeans-mode mass loss, an isotropic wind escapes from one binary component carrying that star's specific orbital angular momentum. If the companion accretes negligibly, a circular binary obeys $aM=\text{constant}$ and $PM^2=\text{constant}$.


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 Incoming links: Jeans-mode mass loss