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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.

For Roche-lobe overflow, $R_2=R_L$. Cubing the Roche lobe approximation and using the donor's mass-radius relation givesKepler's third law therefore yields the Roche-lobe period-mass relation for a linear donor radius law:The component distances from the centre of mass are $a_1=a{M}_2/M$ and $a_2=a{M}_1/M$. Summing their angular momenta gives the circular-binary orbital angular momentumDuring conservative binary mass transfer, $\dot{M}=0$ and ${\dot{M}}_1=-{\dot{M}}_2$. Logarithmic differentiation at constant angular momentum gives $\dot{a}/a=-2(1-q){\dot{M}}_2/M_2$. Thus the Roche-lobe radius response exponent givesTheir difference is $d\log (R_L/R_2)/dt=(2q-8/3){\dot{M}}_2/M_2$. Since the donor loses mass, this is positive for $q<4/3$: the donor shrinks relative to its Roche lobe, so transfer shuts off without an external driver. The Roche lobe itself expands only for $q<5/6$; for $5/6<q<4/3$ it shrinks more slowly than the donor.

Gravitational-wave emission from a binary system provides such a driver: the orbit emits gravitational waves carrying energy and angular momentum, and the shrinking Roche lobe can maintain contact. Allowing $\dot{J}<0$, the binary mass-transfer contact equation becomesIt gives a negative transfer rate for $q<4/3$.

For the classical nova, assume isotropic re-emission from a binary star: ejecta carry the accretor's specific angular momentum $j_1=a_1^2\Omega$. The eruption lasts many orbital periods, so the orbit responds approximately adiabatically and remains nearly circular; an instantaneous asymmetric kick would be a different model. To first order,Differentiating the circular-binary orbital angular momentum formula during the ejection givesThe accompanying Roche lobe change is thereforeThe donor does not change mass in the eruption and moves inside its expanded Roche lobe: this is nova-induced binary detachment.

Put $\Gamma =-\dot{J}>0$ for the constant external loss. During detachment the component masses are fixed, so the Roche lobe shrinks at fractional rate $2\Gamma /J$. Closing the fractional gap $4\delta m/(3M)$ takes $t_d=2J\delta m/(3M\Gamma )$. During the subsequent semidetached binary phase, the binary mass-transfer contact equation gives $|{\dot{M}}_2|=3M_2\Gamma /[J(4-3q)]$, so accumulating the next eruption's mass takes $t_s=J(4-3q)\delta m/(3M_2\Gamma )$. ConsequentlyBoth times are evaluated to first order in $\delta m/M$, with the same equilibrium donor mass-radius relation; changes of $q,J$ and the transfer rate within one cycle affect higher orders.

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.

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.

Treat the stars as point masses for the orbital motion, neglecting their spin angular momenta. Let $M=M_1+M_2$. Their distances from the centre of mass are $a_1=a{M}_2/M$ and $a_2=a{M}_1/M$. The sum of their orbital angular momenta isThus the circular-binary orbital angular momentum is

For conservative binary mass transfer, both $M$ and $J$ are fixed. In terms of the binary mass ratio $q=M_1/M_2$, $M_1=Mq/(1+q)$ and $M_2=M/(1+q)$, henceMass conservation also gives $d{M}_2=-d{M}_1$, soDifferentiating $J\propto M_1{M}_2{a}^{1/2}$ at fixed $M,J$ gives $d\log a/d\log M_1=2(q-1)$. Therefore the stipulated Roche lobe law has the Roche-lobe radius response exponent

Define the overfill by $\Delta =\log (R_1/R_L)$. On a time scale short compared with the slow change of $K$, the donor's response is $d\log R_1=\beta d\log M_1$. ThusDuring mass loss $d\log M_1<0$. If ${\zeta }_L>\beta$, mass loss increases the overfill and therefore drives more mass loss: this is positive feedback, giving more rapid Roche-lobe overflow. If ${\zeta }_L<\beta$, mass loss decreases the overfill and can balance the slow expansion due to stellar evolution. More explicitly, with $k_{\mathrm{ev}}=d\log K/dt>0$, contact requiresA slow negative transfer rate can satisfy this only on the stable side $\beta >{\zeta }_L$. The rapid-transfer condition is thereforeIts time scale depends on the physical donor response represented by $\beta$; it need not be a dynamical time scale if $\beta$ is a thermal-equilibrium mass-radius exponent.

For Roche-lobe overflow feedback with a constant wind torque factor, retain mass conservation to the stated approximation but allow the prescribed angular-momentum loss. Since $d\log J=fd\log M_1$, differentiation givesThe new Roche-lobe radius response exponent isThe same overfill calculation givesA loss of angular momentum has $f>0$ because ${\dot{M}}_1<0$, and it lowers the critical ratio by making the orbit contract more strongly.

For the specified solar-mass donor exponent $\beta =1/2$, the answer isWithout the wind torque this becomes $q>22/23\simeq 0.957$, so donors slightly less massive than their companions can already lie on the rapid-transfer side of this model. For $0<f<11/10$ the lower threshold is $(22-20f)/23$, and for $f\ge 11/10$ every positive $q$ satisfies the strict inequality. There is no finite upper bound on $q$ from this criterion. Since the paper does not specify $f$, the wind case has a conditional range rather than a unique numerical cutoff.

Put $M=M_1+M_2$. In a circular orbit, the distances from the centre of mass are $r_1=a{M}_2/M$ and $r_2=a{M}_1/M$. Summing the components' angular momentum givesUse Kepler's third law ${\Omega }^2{a}^3=GM$ and $\Omega =2\pi /P_{\mathrm{orb}}$ to obtain the circular-binary orbital angular momentumThe numerator contains orbital period, not the erroneous $M_{\mathrm{orb}}$ token in the converted TeX.

The total mass changes as $\dot{M}=(1-f){\dot{M}}_1$. With the prescribed wind carrying specific angular momentum $\lambda J/M$, the angular-momentum loss law is $\dot{J}/J=\lambda \dot{M}/M$. For constant $\lambda$, integration gives $J\propto M^{\lambda }$. Cubing the orbital formula then proves the binary period invariant for a constant wind angular-momentum factor:This integration does not require $f$ to be constant, although its instantaneous value enters the following response coefficients. In the conservative case $f=1$, both $M$ and $J$ are constant and the usual period-product invariant for conservative mass transfer is recovered.

Kepler's law and the given Roche lobe approximation imply $R_L\propto M_1^{1/3}{P}_{\mathrm{orb}}^{2/3}$, since the total-mass factor cancels. Using the period law therefore givesBecause ${\dot{M}}_2=-f{\dot{M}}_1$, the instantaneous Roche-lobe radius response exponent isFor $x=\log (R/R_L)$, subtracting this response from the prescribed stellar radius evolution givesThe overflow-rate model requires $f>0$ and is used for small positive overfill. It gives ${\dot{M}}_1/M_1=-x/(f{t}_{\mathrm{dyn}})$. Hence the nuclear-driven Roche-overfill relaxation equation is

Let the bracket be $A(q)$. On the short response time, freeze the slowly changing masses and coefficients. If $A>0$, the local solution isThe equilibrium donor mass-loss rate is ${\dot{M}}_1/M_1=-1/(A{t}_{\mathrm{nuc}})$. Thus the required stabilizing mass-ratio condition isWhen $A$ is of order one and $t_{\mathrm{dyn}}\ll t_{\mathrm{nuc}}$, the overfill is small and relaxation is fast, while mass transfer proceeds on a nuclear timescale. A positive $A$ arbitrarily close to zero does not by itself ensure this timescale separation or small overfill.

For an explicit threshold put $B=n+5/3$ and $C_w=2\delta (1-f)/3$. The inequality isIf $B>0$ and $f>0$, its positive root givesFor $f=1$ this reduces to $q<(n+5/3)/2$, the conservative mass-transfer critical mass ratio. For other parameter signs the unsimplified inequality is the general condition.

If $A<0$, a larger overfill causes further growth rather than restoration, on the short scale $f{t}_{\mathrm{dyn}}/|A|$. Transfer runs away towards a dynamical episode, potentially a common envelope or merger; the small-overfill model then fails. At $A=0$ the overfill grows linearly under nuclear expansion and there is no steady regulated contact. These conclusions use the stipulated donor-radius response; an actual dynamical episode requires its adiabatic response and hydrodynamics.

Let $M=M_1+M_2$, and let $\mu =M_1{M}_2/M$ be the reduced mass. The stellar distances from the center of mass are $a_1=a{M}_2/M$ and $a_2=a{M}_1/M$. For a circular Kepler orbit, ${\Omega }^2=GM/a^3$, so summing the two orbital contributions gives the circular-binary orbital angular momentumThe separation $a$ is inside the square root. This also has the required dimensions of angular momentum.

For the homologous rotating collapse, label a fluid element by $s=r/R$ and its polar angle $\theta$. Its enclosed mass $m(s)$ is constant during stellar homology, as are the dimensionless density profile and the inertia coefficient $\alpha$. Conservation of angular momentum and solid-body rotation giveThe magnitude of the centrifugal acceleration is ${\Omega }^{2}r\sin \theta$, while the Newtonian gravitational field has magnitude $Gm(s)/r^2$. ThereforeUsing only the radial component of centrifugal force replaces $\sin \theta$ by ${\sin }^2\theta$ and leaves the scaling unchanged. The centrifugal force vanishes on the rotation axis; the central point is understood through the limiting field rather than a ratio of two zero forces.

At the equator of the cloud, critical rotation of a spherical cloud means ${\Omega }_c^2{R}_c=GM/R_c^2$, where $R_c=R_{\mathrm{crit}}$. ConsequentlyImmediately after the first fission, each daughter has mass $M/2$ and radius $a/2$ because the spheres touch. Their orbital moment of inertia is $M{a}^2/4$, while their combined spin moment of inertia is $\alpha M{a}^2/4$. With a common spin and orbital frequency ${\Omega }_f=\sqrt{GM/a^3}$, the synchronous fission model for binary formation givesThe common frequency can change during fission; corotation requires equal instantaneous frequencies, not an unchanged frequency from before the rearrangement. Equating the two values of $J$ yieldsFor a nonzero positive inertia coefficient, the necessary and sufficient condition in this algebraic model for $a<R_c$ isFor example, a moment of inertia of a uniform solid sphere has $\alpha =2/5$ and gives $a/R_c=64/49>1$; it fails the required compact-fission condition. A sufficiently centrally concentrated cloud can have smaller $\alpha$.

For the next collapse, neglect exchange of spin with the unchanged outer orbit, so each daughter's spin angular momentum is separately conserved. Immediately after the first fission that spin isFor a daughter of mass $m=M/2$ to reach critical rotation of a spherical cloud at radius $R_{c,d}$, the same relation gives $S_d=\alpha m\sqrt{Gm{R}_{c,d}}$. HenceEquivalently, its centrifugal-to-gravity ratio starts at $1/4$ when its radius is $a/2$ and reaches one after contraction by a factor four. Applying the same fission rule to this daughter produces an inner pair with separationCorotation at this second fission is local to each newly formed inner pair; it is not a single common frequency for the entire four-star hierarchy.

This spin assumption matters. If tidal locking instead kept each shrinking daughter synchronized to the fixed outer orbital period throughout the intervening collapse, its angular velocity would remain fixed. Its centrifugal-to-gravity ratio would then scale as $R_d^3$ and decrease, so it would never reach the proposed second fission. This is a counterexample to extending the corotation assumption through the whole contraction. The printed ratio describes the separately spin-conserving interpretation of the repeated fission hierarchy.

The model can generate a compact hierarchy algebraically, but is not a general physical account of multiple-star formation. In the allowed range $\alpha <1/3$, one has $a^{\prime }/a<1/8$, suggesting well-separated inner and outer scales. However, rapidly rotating gas deforms, touching daughters are tidally distorted, and pressure, gas flows, dissipation and spin-orbit torques cannot generally be ignored. The assumed identical profiles, equal mass splits, instantaneous corotation and later torque-free contractions are restrictive. Star formation can involve gravitational fragmentation and redistribution of angular momentum; the toy budget neither proves that fission occurs nor predicts the population of real multiple systems.

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.