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If an escaping stellar wind carries specific orbital angular momentum $\lambda J/M$ with constant $\lambda$, then $\dot{J}/J=\lambda \dot{M}/M$ and $J\propto M^{\lambda }$. Combining this with circular-binary orbital angular momentum gives the displayed period invariant. It is a different prescription from donor-wind angular-momentum loss, whose specific angular momentum generally varies with mass ratio.

Under donor-wind angular-momentum loss and the Roche lobe approximation $R_L=0.46a(M_d/M{)}^{1/3}$, the Roche-lobe radius response exponent is affine in retained fraction: ${\zeta }_L=(1-f){\zeta }_0+f{\zeta }_1$, where $q=M_d/M_a$, ${\zeta }_0=1/3-4q/[3(1+q)]$ and ${\zeta }_1=2q-5/3$. Differentiate $\log R_L=\log a+(\log M_d-\log M)/3+\mathrm{constant}$ and the circular-binary orbital angular momentum to obtain this expression.

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.

Let $M=M_1+M_2$. The centre of mass condition gives $a_1=a{M}_2/M$, $a_2=a{M}_1/M$. For circular motion the component angular momenta are $J_i=M_i{a}_i^2\Omega$, soUsing Kepler's third law, $a^3{\Omega }^2=GM$, and $\Omega =2\pi /P_{\mathrm{orb}}$ givesThis is the circular-binary orbital angular momentum. Mass received by the companion redistributes angular momentum within the binary star; only the escaping stellar wind removes it under the prescribed model.

The stellar wind from the donor star has specific angular momentumHere ${\dot{M}}_1<0$, ${\dot{M}}_2=-f{\dot{M}}_1$ and $\dot{M}=(1-f){\dot{M}}_1$. Thus donor-wind angular-momentum loss givesTaking the logarithmic derivative of the period expression for $J$ givesUse $M_2/(M_{1}M)=1/M_1-1/M$ to simplify this toFor constant $f$, integration therefore givesThis is the donor-wind period invariant with fixed retention fraction. If $f$ changes during the evolution, the differential equation remains valid but the integrated invariant is instead $P_{\mathrm{orb}}{M}_2^3{M}^2\exp (3\int fd\log M_1)=\mathrm{constant}$. The fixed-power expression cannot be used with a time-dependent exponent without this modification.

For the Roche lobe, write $q=M_1/M_2$. The equivalent separation expression $J=M_1{M}_2\sqrt{Ga/M}$ givesTaking the logarithmic derivative of $R_L=0.46a(M_1/M{)}^{1/3}$ then adds ${\dot{M}}_1/(3M_1)-\dot{M}/(3M)$, yieldingThis donor-wind Roche-lobe response uses only the instantaneous value of $f$ and thus still holds if it varies.

Define $A(q)=2q+4q/[3(1+q)]-2$ and $B(q)=1/3-4q/[3(1+q)]$. The Roche-lobe radius response exponent is ${\zeta }_L=B+Af$, whereas the donor star's stellar radius response exponent is ${\zeta }_{\ast }=-n$. Keeping $R_1=R_L$ requires the two logarithmic derivatives of radius to agree, soFor $A\ne 0$, this determinesA physical mixture of Roche-lobe overflow and stellar wind from the donor star needs $0<f<1$. Equivalently, $-n$ must lie strictly between the pure-wind response $B$ and the conservative response $B+A=2q-5/3$. The values $f=0$ and $1$ are valid limiting prescriptions, respectively pure stellar wind and conservative mass transfer, rather than mixtures. If $f_{\mathrm{req}}$ lies outside $[0,1]$, no allowed division of the lost mass can maintain contact under this radius law and angular-momentum-loss prescription. This is feasibility of donor-wind binary contact.

The likely direction away from contact follows without assigning a nonphysical fraction. For any actual $f$, let $\Delta =\log (R_1/R_L)$. ThenSince ${\dot{M}}_1<0$, a positive value of $A(f_{\mathrm{req}}-f)$ makes the donor star move inside its lobe and detach; a negative value increases overfill and promotes stronger transfer. Thus if $A>0$, $f_{\mathrm{req}}<0$ gives increasing overfill and $f_{\mathrm{req}}>1$ gives detachment; if $A<0$, these outcomes are reversed. A runaway or a new equilibrium would require the donor star's response or the angular-momentum-loss model to change, possibly with additional driving. The contact calculation alone does not prove a particular nonlinear endpoint.

There is also a genuine exceptional denominator: $A=0$ at $q=(\sqrt{10}-1)/3$. At this ratio both limiting lobe responses coincide. Contact is possible only if $-n=B$, namely $n=(\sqrt{10}-2)/(\sqrt{10}+2)$; then every $f$ gives the same instantaneous radius response. Otherwise no fraction works, and the sign of $(-n-B){\dot{M}}_1/M_1$ determines detachment or increasing overfill.

Write $M=M_1+M_2$, $a=a_1+a_2$, and $\mu =M_1{M}_2/M$, the reduced mass. In the center of mass frame, $a_1=a{M}_2/M$ and $a_2=a{M}_1/M$. Both components of the circular orbit have the same angular velocity $\Omega$. Thus their total orbital angular momentum isUsing Kepler's third law, ${\Omega }^2=GM/a^3$, and the orbital period $P=2\pi /\Omega$, gives the circular-binary orbital angular momentum

Take ${\dot{M}}_1<0$, so ${\dot{M}}_2=-f{\dot{M}}_1$ and $\dot{M}=(1-f){\dot{M}}_1$. A parcel in an isotropic stellar wind has, on average, the donor star's orbital velocity. Its wind velocity relative to the donor star averages to zero, so its mean specific orbital angular momentum about the center of mass is $j_1=a_1^2\Omega$. The escaping mass per unit time is $-(1-f){\dot{M}}_1$. With negligible stellar spin, no additional wind torque, and internal redistribution of the angular momentum of retained matter, donor-wind angular-momentum loss therefore givesIsotropy is in the donor star's frame: it does not make the escaping orbital angular momentum vanish. This is also different from isotropic re-emission from a binary star, where matter escapes from the accretor.

Logarithmically differentiate the expression for $J$ along a slowly evolving sequence of circular orbits:Substitution of the donor-wind angular-momentum loss and mass rates givesFor constant $f$, the last expression is $-3fd\log M_1/dt-3d\log M_2/dt-2d\log M/dt$. Hence the period invariant for constant-fraction donor-wind mass loss isFor a time-dependent $f$, the differential equation still holds, but this integrated power law does not. At $f=1$ it reduces to the period-product invariant for conservative mass transfer; at $f=0$, $M_2$ is constant and $P{M}^2$ is constant, as in Jeans-mode mass loss.

Set the binary mass ratio $q=M_1/M_2$. Since $a^3=GM{P}^2/(4{\pi }^2)$, the Kepler third law and the preceding orbital period derivative implyThe specified approximation to the Roche lobe gives $\log R_L=\log (0.46)+\log a+\frac{1}{3}(\log M_1-\log M)$. Consequently the donor-wind Roche-lobe response isThis is a local Roche-lobe radius response exponent, so it remains valid instantaneously even if $f$ varies.

The stellar radius response exponent of $R_1\propto M_1^{-n}$ is ${\zeta }_{\ast }=-n$. Maintaining Roche-lobe overflow in exact contact requires ${\dot{R}}_1/R_1={\dot{R}}_L/R_L$, yieldingFor a precise feasibility of donor-wind binary contact test, put ${\zeta }_0=\frac{1}{3}-\frac{4q}{3(1+q)}$ and ${\zeta }_1=2q-\frac{5}{3}$. The Roche-lobe radius response exponent is ${\zeta }_L=(1-f){\zeta }_0+f{\zeta }_1$. Thus a permitted contact fraction exists exactly whenUnless $A={\zeta }_1-{\zeta }_0=2q+4q/[3(1+q)]-2$ vanishes, the required fraction is $f=(-n-{\zeta }_0)/A$. At $q=(\sqrt{10}-1)/3$, both limiting Roche-lobe radius response exponents coincide: contact is possible for any $f$ only if $n=5/3-2q$, and for no $f$ otherwise.

The sign of the overfilling change resolves the failure of contact:If ${\zeta }_{\ast }>\max ({\zeta }_0,{\zeta }_1)$, mass loss makes the donor star underfill its Roche lobe: the system detaches and contact-driven transfer stops. If ${\zeta }_{\ast }<\min ({\zeta }_0,{\zeta }_1)$, mass loss increases the overfilling: transfer is destabilized, and rapid transfer or a common envelope may result. Calling this a failure of dynamical stability of binary mass transfer specifically requires $-n$ to be the adiabatic stellar radius response exponent; a thermal or equilibrium response concerns a different timescale. Additional angular momentum losses or intrinsic stellar expansion can change these outcomes by changing the contact equation.

For a slowly evolving circular orbit with constant retained fraction $f$ and donor-wind angular-momentum loss, logarithmic differentiation of the circular-binary orbital angular momentum gives $d\log P=-3fd\log M_d-3d\log M_a-2d\log (M_d+M_a)$. Integration gives the displayed invariant. At $f=1$ it becomes the period-product invariant for conservative mass transfer, while at $f=0$ it becomes $P{M}^2=\mathrm{constant}$. A varying retained fraction does not give this simple power-law invariant.