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An isotropic escaping stellar wind has the donor star's mean specific orbital angular momentum. With ${\dot{M}}_d<0$, retained fraction $f$, negligible spin and no additional wind torque, $\dot{J}=(1-f){\dot{M}}_d{a}_d^2\Omega$. The retained matter redistributes angular momentum internally. Unlike isotropic re-emission from a binary star, the escaping material originates at the donor rather than the accretor. With no retention this reduces to Jeans-mode mass loss.

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.

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 Roche-lobe overflow in the stipulated stellar thermal equilibrium, $R_2=R_L$. Using the mass-radius relation and Kepler's third law,The total mass cancels, givingHere $R_{\odot }$ is the solar radius. This is the Roche-lobe-filling period-density relation specialized to $R_2\propto M_2$.

Neglecting spin, the angular momentum of a circular orbit is $J=M_1{M}_2\sqrt{Ga/M}$. Between classical novae the transfer conserves total mass, so $\dot{M}=0$ and ${\dot{M}}_1=-{\dot{M}}_2$. Its logarithmic derivative givesIn particular, for conservation of angular momentum,The source PDF has $\dot{J}=0$ here; the TeX's $J=0$ is a transcription defect. For $q>4/3$, the Roche-lobe radius response exponent exceeds the equilibrium stellar radius response exponent $1$. Since ${\dot{M}}_2<0$, the Roche lobe then shrinks faster than the donor star; the overfill increases and drives more transfer, giving positive feedback. Equality at $q=4/3$ is marginal in this linear response test.

Two qualifications matter. First, $q>4/3$ is a formal extrapolation outside the stated $q<1$ range, and the supplied Roche lobe approximation need not remain accurate there. Second, true dynamical stability of binary mass transfer compares ${\zeta }_L$ with the adiabatic stellar radius response exponent ${\zeta }_{\mathrm{ad}}$, not ${\zeta }_{\mathrm{eq}}=1$. The stellar thermal equilibrium argument supplies the displayed feedback threshold under the imposed radius law, rather than a necessary and sufficient physical dynamical threshold. For example, the polytropic mass-radius relation for a fully convective adiabatic stellar polytrope with index $3/2$ gives ${\zeta }_{\mathrm{ad}}=-1/3$; within this same Roche lobe approximation its dynamical threshold is $q=2/3$. Thus $q<4/3$ by itself does not guarantee dynamical stability.

Gravitational-wave emission from a binary system removes orbital energy and angular momentum. Magnetic braking of a binary star provides another important loss: a magnetized stellar wind carries away donor spin, while tidal locking couples that spin to the orbit. Tides alone redistribute angular momentum and are not an external sink. Matter expelled from the system can also carry orbital angular momentum, although appreciable continuous mass loss would require nonconservative modifications of the preceding equations. Setting ${\dot{R}}_L/R_L={\dot{R}}_2/R_2$ gives the binary mass-transfer contact equationFor $q<4/3$, a negative external $\dot{J}$ therefore sustains negative ${\dot{M}}_2$ along the stipulated stellar thermal equilibrium contact sequence, provided the system is otherwise stable and can remain thermally relaxed.

During the classical nova, the ejecta carry the white dwarf's specific angular momentum, not zero angular momentum. With ${\Omega }^2=GM/a^3$ and the accretor's distance from the centre of mass $a_1=a{M}_2/M$, isotropic escape givesThe slow ejection relative to the orbital period permits the adiabatic circular orbit approximation; the donor star does not transfer appreciable additional mass during it. Take $\delta M_1=-\delta m$, $\delta M_2=0$ and $\delta M=-\delta m$. Differentiating $J=M_1{M}_2\sqrt{Ga/M}$ givesSince $M/M_1=1+q$, this yieldsThese are first-order formulae with errors of order $(\delta m/M{)}^2$. The orbital widening agrees with Jeans-mode mass loss. The donor star's radius is unchanged to this order by the assumed ejection, so its Roche lobe expands away from it, causing nova-induced binary detachment. This idealization neglects changes in the donor from irradiation or interaction with ejecta; those are not supplied in the model.

Let $\Gamma =-\dot{J}>0$ be the fixed external loss rate. During the detached binary phase both component masses are fixed, and ${\dot{R}}_L/R_L=2\dot{J}/J=-2\Gamma /J$. Closing the fractional gap $4\delta m/(3M)$ takesIn the subsequent semidetached binary, the binary mass-transfer contact equation gives $-{\dot{M}}_2=3M_2\Gamma /[J(4-3q)]$. Accumulating a fresh layer of mass $\delta m$ therefore takesThroughout these first-order estimates $J,M,q$ can be evaluated at the start of the cycle: their fractional changes during it are of order $\delta m/M$. The limit $\Gamma =0$ is excluded, since no finite reconnection time follows without an external shrinkage mechanism.

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.

A fast escaping Jeans-mode mass loss wind can drive a giant into Roche-lobe overflow because its radius changes as it loses mass, even though the wind itself is not substantially captured. This differs from slow Wind Roche-lobe overflow. For $R_d\propto M_d^{-n}$ and $R_L\propto a(M_d/M{)}^{1/3}$, wind alone gives $d\log (R_d/R_L)/dt=\dot{M}[3(1-n)q-(1+3n)]/[3M_d(1+q)]$, where $q=M_d/M_a$ and $\dot{M}<0$ is the escaping wind rate. The displayed inequality is exactly the condition that overfill increases.