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A simplified conservative model with a fully convective donor gives long-lived stable Roche-lobe overflow only for donor-to-accretor mass ratio $q<2/3$. This follows from the Roche-lobe radius response exponent and the donor's adiabatic stellar radius response exponent $-1/3$. The boundary is model-dependent; radiative donors, evolved core structure and mass or angular-momentum loss change it. Van Rensbergen and collaborators' Algol population study includes reported Algol mass ratios above $0.7$ and discusses observational uncertainties.

With fixed total mass and orbital angular momentum and $R_L\propto a(M_d/M{)}^{1/3}$, the Roche-lobe radius response exponent is $2q-5/3$. A donor with stellar radius response exponent ${\zeta }_{\ast }$ is stable to infinitesimal mass loss when ${\zeta }_{\ast }>2q-5/3$, giving $q<({\zeta }_{\ast }+5/3)/2$. The threshold depends on the donor response.

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.

A donor star with stellar radius response exponent ${\zeta }_{\ast }$ can maintain exact contact in the donor-wind Roche-lobe response model for some retained fraction $0\le f\le 1$ precisely when ${\zeta }_{\ast }$ lies between ${\zeta }_0$ and ${\zeta }_1$. This follows because the attainable Roche-lobe radius response exponents form that closed interval. If its endpoints coincide, all fractions work when ${\zeta }_{\ast }$ equals the common value and none work otherwise. With negative donor mass-loss rate, $d\log (R_d/R_L)/dt=({\zeta }_{\ast }-{\zeta }_L){\dot{M}}_d/M_d$ distinguishes detachment from increasing overfill. Dynamical stability of binary mass transfer requires the adiabatic donor response.

For a circular binary with negligible spin, the two orbital radii about the center of mass are $a_1=a{M}_2/M$ and $a_2=a{M}_1/M$. Summing their orbital angular momenta givesUsing Kepler's third law, ${\Omega }^2=GM/a^3$, also gives $J=M_1{M}_2\sqrt{Ga/M}$. Under conservative binary mass transfer, $M$ and $J$ stay fixed, so $a\propto (M_1{M}_2{)}^{-2}$. Since $d\log M_2/d\log M_1=-q$, the Roche-lobe radius response exponent isFor the supplied giant response, luminosity stays fixed during the mass-loss perturbation and ${\zeta }_{\ast }=d\log R/d\log M_1=-0.27$. The change of the logarithmic overfill is $({\zeta }_{\ast }-{\zeta }_L)d\log M_1$. Because the donor loses mass, this reduces overfill and stabilizes transfer precisely when ${\zeta }_{\ast }>{\zeta }_L$. Thus the stated model requiresThis is a response criterion within the given radius law; a different adiabatic donor response would change the dynamical threshold.

The initially more massive star evolves first. If it waits until it has the supplied giant structure while still having $q>1$, conservative overflow is unstable by this criterion. If Roche-lobe overflow instead begins while the donor is on the main sequence or shortly after central hydrogen exhaustion, its largely radiative-envelope response can allow transfer before a deep convective giant envelope develops. Mass loss and accretion can then reverse the mass ratio before it reaches the giant state. This is the binary mass-ratio reversal route to an Algol binary, through Case A mass transfer or early case B mass transfer.

At lobe filling, the lobe approximation and Kepler's law give the pre-giant Roche-lobe-filling period limitIt increases with the donor's radius. Requiring $R_L<R_{\mathrm{BGB}}=1.68R_{\odot }(M_1/M_{\odot }{)}^{5/3}$ therefore yieldsThe companion mass cancels in this lobe approximation.

Since $P\propto a^{3/2}$ at fixed total mass, the separation scaling gives $\boxed{P(M_1{M}_2{)}^3=\text{constant}}$. Use solar-unit masses $m_i=M_i/M_{\odot }$ and $m=m_1+m_2$, and let $K=P(m_1{m}_2{)}^3$. ThenThe derivative vanishes at $m_1=5m/8$, and its derivative is $5/m_1^2+3/(m-m_1{)}^2>0$, while the ratio diverges at both endpoints. Hence this is the global minimum, the conservative pre-giant period-ratio minimum.

For the stated Algol parameters, $m=4$ and $K=3(1\cdot 3{)}^3=81$ days. At the minimum $m_1=2.5,m_2=1.5$, the period is $81/(2.5\cdot 1.5{)}^3=1.536$ days and $P/m_1^2=0.24576$ days, well below $P_0=0.8$ days. This is a possible initially more-massive donor configuration with overflow before the giant branch. The current $q=1/3$ is also below the giant stability threshold. Thus Algol is compatible with the proposed conservative early-overflow route, although this necessary geometric test alone does not prove its full evolutionary history.

For RT Lac, $m=2.4$ and $K=5.1(0.9\cdot 1.5{)}^3$. Its minimum occurs at $m_1=1.5,m_2=0.9$, where the mass product equals its current value, so $P=5.1$ days andNo conservative antecedent on this orbit can begin overflow before the giant branch. Beginning with an initially more massive giant instead encounters the instability just derived. RT Lac cannot follow the stated coeval, conservative Algol-formation model. A plausible alternative is rapid initial transfer with substantial mass and angular-momentum loss, via an outflow, strong wind or envelope-ejection episode. Such nonconservative binary mass transfer changes both total mass and $J$ and removes the period-product constraint. Its present mass ratio $0.9/1.5=0.6$ permits stable current giant transfer. The given data do not determine a unique mass-loss history or establish a specific ejection mechanism.

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.

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 $f=0.46$. At Roche-lobe overflow contact, the supplied equilibrium radius relation givesKepler's third law then impliesThus the Roche-lobe period-mass relation for a linear donor radius law isHere $R_{\odot }$ and $M_{\odot }$ denote the solar radius and solar mass. Cancellation of the total mass is the special Roche-lobe-filling period-density relation combined with $R_2\propto M_2$.

Between eruptions the conservative binary mass transfer has $M$ constant and ${\dot{M}}_1=-{\dot{M}}_2$. Neglecting spin, writeIts logarithmic derivative givesDifferentiating the Roche lobe relation consequently givesFor $\dot{J}=0$, the Roche-lobe radius response exponent is ${\zeta }_L=2q-5/3$. If the donor follows its specified equilibrium radius sequence, ${\dot{R}}_2/R_2={\dot{M}}_2/M_2$. Its logarithmic overflow therefore changes bySince ${\dot{M}}_2<0$, $q<4/3$ gives negative feedback in this equilibrium-response model, whereas $q>4/3$ gives increasing overflow. For $q>4/3$, loss of donor mass shrinks its Roche lobe faster than its modeled stellar radius, increasing transfer and hence amplifying the original disturbance. The $q>4/3$ comparison is a formal extension; the initial system is specified to have $q<1$.

The word dynamically needs a qualification. Thermal equilibrium specifies ${\zeta }_{\mathrm{eq}}=1$, but actual dynamical stability of binary mass transfer depends on the adiabatic stellar radius response exponent ${\zeta }_{\mathrm{ad}}$, which need not equal one. Stability after a rapid small mass loss requires ${\zeta }_{\mathrm{ad}}>{\zeta }_L$, because $d\log M_2<0$. The equilibrium relation alone does not supply this adiabatic response. As a formal counterexample to deriving a universal dynamical threshold from it, a radius model with a dimensionless entropy label $S$, $R(M_2,S)=C{M}_2^2{e}^S$, can have equilibrium $S_{\mathrm{eq}}=S_0-\log (M_2/M_{\ast })$, giving $R_{\mathrm{eq}}\propto M_2$ but ${\zeta }_{\mathrm{ad}}=2$ at fixed $S$. At $q=3/2$, ${\zeta }_L=4/3<2$, so that model is dynamically stable despite the extrapolated $q>4/3$ condition. This shows the missing response hypothesis, not a claim that this toy radius law describes the given low-mass star.

A more relevant illustration is an ideal fully convective adiabatic stellar polytrope of index $3/2$. At fixed entropy its polytropic mass-radius relation is $R\propto M_2^{-1/3}$, giving ${\zeta }_{\mathrm{ad}}=-1/3$ and a dynamical critical ratio $q=2/3$ under the same Roche lobe approximation. For example, $q=0.9<4/3$ can be unstable dynamically even though the equilibrium comparison has negative feedback. The requested $4/3$ threshold is the intended result with instantaneous response $R_2\propto M_2$; it is not a universal dynamical criterion. The subsequent cycle calculation uses the supplied equilibrium sequence and presumes a physically stable, sufficiently slow transfer regime.

Two important angular-momentum-loss mechanisms are gravitational-wave emission from a binary system and magnetic braking of a binary star. Gravitational waves directly remove orbital energy and angular momentum. A magnetized stellar wind removes donor spin angular momentum with a large magnetic lever arm; tidal synchronization replenishes that spin from the orbit, so the net effect is orbital angular-momentum loss. Treating the wind mass loss as small compared with the Roche-transfer flow is consistent with the conservative approximation used here; otherwise the mass and angular-momentum equations must both be modified. Neglecting the stored spin does not eliminate the torque transmitted through it.

Such losses shrink the Roche lobe and offset the donor's tendency to move inside it. Maintaining contact on the equilibrium sequence requires ${\dot{R}}_L/R_L={\dot{M}}_2/M_2$, hence the binary mass-transfer contact equation becomesFor $q<4/3$ and $\dot{J}<0$, this drives ${\dot{M}}_2<0$ and maintains the modeled semi-detached binary.

Now take $\delta m>0$ for the mass expelled from the accretor in one classical nova. The white dwarf loses $\delta M_1=-\delta m$, while the donor has $\delta M_2=0$ and the total mass change is $\delta M=-\delta m$. Isotropic ejection in the accretor rest frame supplies no mean kick and carries the accretor's specific orbital angular momentum. Its orbital radius is $a_1=a{M}_2/M$, soThis is the angular-momentum budget for isotropic re-emission from a binary star, without an additional spin or frictional torque on the ejecta. The general logarithmic change of $J=M_1{M}_2\sqrt{Ga/M}$ isUsing $M/M_1=1+q$ gives the first-order nova changesThe donor mass and its assumed equilibrium radius are unchanged during the eruption. It is therefore left inside the enlarged Roche lobe, and Roche-lobe overflow ceases in this sharp-contact model. An extended atmosphere, irradiation-induced donor expansion or additional ejecta torques would require a different treatment.

The stipulated circular orbit is an important idealization. Isotropic loss that is slow compared with the orbital period can preserve circularity. If it is truly impulsive compared with that period, an initially circular point-mass orbit instead acquires eccentricity. With unchanged velocity and position, its new specific orbital energy is ${\epsilon }^{\prime }=GM/(2a)-G(M-\delta m)/a=-G(M-2\delta m)/(2a)$ and its squared specific angular momentum stays $h^2=GMa$. The Kepler orbit energy and eccentricity relations then give $a^{\prime }=a(M-\delta m)/(M-2\delta m)$ and $e^{\prime }=\delta m/(M-\delta m)$. The latter still has the same first-order separation increase. The question assumes circularity; its first-order circular budget is consistent with adiabatic orbital mass loss or with subsequent circularization that changes the separation only at second order. Work symbolically with $\delta m$, since the printed numerical layer mass has no explicit unit.

Finally put ${\mathcal{L}}_J=-\dot{J}>0$, constant over both intervals as specified, and let $J,M,q$ denote their leading-order values over one small cycle. While detached, masses and donor radius are fixed, so ${\dot{R}}_L/R_L=2\dot{J}/J=-2{\mathcal{L}}_J/J$. The fractional gap $4\delta m/(3M)$ closes afterOnce contact resumes, the contact equation gives $-{\dot{M}}_2=3M_2{\mathcal{L}}_J/[J(4-3q)]$. Accumulation of the next layer takesDividing yields the detached-to-semidetached time ratioBoth times are first-order in $\delta m/M$. Corrections to either time are second-order in that small mass fraction, so the displayed ratio has relative corrections of order $\delta m/M$. The positive denominator, negligible eruption duration relative to the cycle, fixed donor radius during detachment and common external loss rate are all essential to this idealized nova-induced binary detachment cycle.

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.