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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.

Orbital angular momentum carried by the wind. Use a circular orbit of separation $a$, and write $M=M_1+M_2$ and ${\Omega }_o^2=GM/a^3$. The donor's distance from the center of mass is $a_1=a{M}_2/M$. With spins neglected, its specific orbital angular momentum is $j_1=a_1^2{\Omega }_o$, while the circular-binary orbital angular momentum isA fast spherically symmetric stellar wind has no net extra angular momentum in the donor's rest frame and escapes with mean specific orbital angular momentum $j_1$. Thus Jeans-mode mass loss givesInitially take negligible wind capture by the companion, so ${\dot{M}}_1=\dot{M}$, ${\dot{M}}_2=0$ and the total mass derivative is $\dot{M}$. Logarithmic differentiation of $J_{\mathrm{orb}}$ givesCombining the two expressions yields $\dot{a}/a=-\dot{M}/M$, so $aM$ is constant. Kepler's third law then gives $\dot{P}/P=(3/2)\dot{a}/a-(1/2)\dot{M}/M=-2\dot{M}/M$. HenceThese are secular relations for slow changes of the orbit through nearly circular states. A wind being fast compared with orbital speed does not mean its mass-loss timescale is shorter than an orbital period; impulsive removal would require a different orbital calculation.

Whether wind drives contact. Ignore nuclear expansion on the stated short evolutionary timescale. With $R_1\propto M_1^{-n}$ and $R_L\propto a(M_1/M{)}^{1/3}$, wind alone givesBecause $\dot{M}<0$, the ratio increases precisely whenThis is wind-driven Roche-lobe overflow: the escaping wind makes the donor expand relative to its lobe and drives a separate transfer stream. It is not the slow-wind capture mechanism called Wind Roche-lobe overflow. If the inequality is reversed, the donor recedes inside its expanding lobe and detaches; wind loss alone does not maintain a Roche-lobe transfer stream. At equality it gives no leading-order change of overfill, so additional evolution or higher-order changes decide the outcome.

Contact rate and its stability. Now let the transfer stream be accreted conservatively by star 2 while the separate wind still escapes at rate $\dot{M}<0$. Denote the stream rate by ${\dot{M}}_2>0$. ThenOnly the wind removes orbital angular momentum; internal transfer redistributes it. Using the same wind-loss law and differentiating $J_{\mathrm{orb}}$ givesSubstitute this into the donor and lobe radius derivatives. The overfill evolves asMaintaining contact sets the left side to zero and gives the contact transfer rate with Jeans-mode wind loss,When the wind-drive condition holds, its numerator is positive. If $6q<5-3n$, the denominator is also positive and the stream rate is positive; increasing transfer reduces overfill and can balance the wind driver.

If instead $6q>5-3n$, positive transfer increases overfill, which drives still more transfer. The formal contact formula would demand a negative stream rate and is not a physical steady solution. The system develops unstable, runaway mass transfer on the timescale appropriate to the assumed donor response. For a giant with an adiabatically responding envelope, this can lead to a common envelope and rapid orbital evolution, rather than the stable contact sequence; the eventual outcome requires physics beyond these radius laws. At $6q=5-3n$, transfer gives no linear restoring effect, and a nonzero wind driver cannot be balanced by a finite rate within this approximation.