 Solution

Use the secular, circular-orbit approximation: the wind escapes fast enough to avoid appreciable interaction with the companion, while the stellar masses change slowly enough that orbital elements describe the evolving binary star. Neglect spin angular momentum. Let $\omega =(GM/a^3{)}^{1/2}$ be the orbital angular speed. The donor's distance from the centre of mass is $a_1=a{M}_2/M$, and the circular-binary orbital angular momentum isA spherically symmetric wind in the donor frame has zero mean additional angular momentum about the donor. Averaging its angular momentum about the binary centre of mass therefore gives the donor's orbital specific angular momentum, $j_1=a_1^2\omega$. Since $\dot{M}<0$ denotes mass removed from the system, ${\dot{J}}_{\mathrm{orb}}=j_1\dot{M}$, andThis is donor-wind angular-momentum loss in the Jeans-mode mass loss approximation.

First allow only wind loss, so ${\dot{M}}_1=\dot{M}$ and ${\dot{M}}_2=0$. The logarithmic derivative of $J_{\mathrm{orb}}$ givesSolving yields $\dot{a}/a=-\dot{M}/M$. By Kepler's third law, $P^2=4{\pi }^2{a}^3/(GM)$, so $\dot{P}/P=-2\dot{M}/M$. Integrating givesWind loss thus expands the separation and lengthens the orbital period. These are secular relations; an impulsive loss on an orbital time could instead generate eccentricity and would need a different analysis.

To test contact, put $D=d\log (R_1/R_L)/dt$. The given short-time stellar radius response exponent gives ${\dot{R}}_1/R_1=-n{\dot{M}}_1/M_1$. Differentiating the Roche lobe approximation givesFor wind alone this becomes ${\dot{R}}_L/R_L=\dot{M}/(3M_1)-4\dot{M}/(3M)$. Hence, with the binary mass ratio $q=M_1/M_2$,Because $\dot{M}$ is negative, increasing overfill, $D_{\mathrm{w}}>0$, occurs precisely whenThis is wind-driven Roche-lobe overflow. For the reverse strict inequality, $D_{\mathrm{w}}<0$ and the donor star retreats inside its Roche lobe, producing a detached binary. At equality the wind is neutral to contact at this order.

Now permit an additional conservative binary mass transfer rate $x={\dot{M}}_2$ to the mass-gaining star. The total escaping wind remains $\dot{M}$, while ${\dot{M}}_1=\dot{M}-x$. Transfer retains orbital angular momentum within the binary under the assumed neglect of spins, so the same wind-loss formula for ${\dot{J}}_{\mathrm{orb}}$ applies. Logarithmic differentiation now givesSubstituting this and ${\dot{M}}_1=\dot{M}-x$ into the two radius derivatives givesSteady contact requires $D=0$. If the wind is driving overflow and $6q<5-3n$, the transfer contribution reduces overfill and a positive contact rate exists:The numerator and denominator are positive in this regime, making ${\dot{M}}_2>0$. This explicitly establishes both the contact transfer rate with Jeans-mode wind loss and its stabilizing sign.

If the wind drives overflow but $6q>5-3n$, positive transfer instead increases $R_1/R_L$: the mass loss required to relieve overfill makes the overfill worse. The formal contact formula would require ${\dot{M}}_2<0$, which is not donor-to-companion transfer. There is no stabilizing positive contact rate; overflow runs away. For a red giant this can lead to a common envelope, followed by envelope ejection into a tighter binary or a merger. The criterion concerns the supplied short-time radius response; the actual runaway timescale and outcome require the donor's dynamical and thermal response. At $6q=5-3n$ transfer is neutral to contact at this order, so it cannot balance a nonzero wind driver. If the wind already causes detachment, the transfer instability condition alone does not initiate overflow.