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A convective envelope is an outer stellar region in which convection carries a substantial fraction of the energy and mixes material. Its nearly adiabatic response to rapid mass loss is important for the dynamical stability of binary mass transfer.

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.

Since the donor fills its Roche lobe, use $R_2=R_L$ and substitute its equilibrium mass-radius relation into the lobe approximation:Kepler's third law then givesso the Roche-lobe period-mass relation for a linear donor radius law isThe total binary mass cancels, rather than being assumed equal to the white-dwarf mass.

The orbital radii about the center of mass are $a_1=a{M}_2/M$ and $a_2=a{M}_1/M$. Summing the two orbital angular momenta givesWith transferred matter retained by the accretor, total mass is fixed. Logarithmic differentiation givesFor $\dot{J}=0$, the separation response is $\dot{a}/a=2(q-1){\dot{M}}_2/M_2$. Since $R_L\propto a{M}_2^{1/3}$ at fixed total mass,The second relation uses the supplied equilibrium donor sequence. As the donor loses mass, ${\dot{M}}_2<0$. If $q>4/3$, its Roche lobe shrinks fractionally faster than this donor radius, so the excess filling increases and drives more transfer: it is positive feedback. If $q<4/3$, infinitesimal loss instead reduces overfill on that assumed sequence, so some continuing driver is needed to retain contact.

There are two limits to the literal dynamical interpretation. The given initial range is $q<1$, so $q>4/3$ is an extrapolated instability test, not a state inside that range. Also, a true dynamical stability of binary mass transfer test needs the donor's adiabatic stellar radius response exponent ${\zeta }_{\mathrm{ad}}$, rather than its thermal-equilibrium exponent. In general,Thus $4/3$ is the result for the assumed ${\zeta }_{\ast }=1$ response. A fully convective idealized donor has ${\zeta }_{\mathrm{ad}}=-1/3$, giving a dynamical threshold $2/3$, already below $4/3$. The equilibrium radius law alone should not be used to assert a universal dynamical threshold.

One continuing driver is magnetic braking of a binary star. A magnetized wind carries away the donor's spin angular momentum; tides that maintain near-synchronous rotation replenish that spin from the orbit, removing orbital angular momentum. With an external $\dot{J}<0$, the lobe response becomesEquating this to the donor response proves the binary mass-transfer contact equation in this model:A negative torque therefore drives mass loss while maintaining contact for $q<4/3$, subject to the chosen donor response and braking assumptions.

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.

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.

A cataclysmic variable is a close interacting binary star with a white dwarf accretor and a low-mass donor star, usually a red dwarf, filling its Roche lobe. Matter crosses by Roche-lobe overflow and commonly forms an accretion disk; sufficiently strong white-dwarf magnetism instead permits magnetically channelled accretion. Accretion produces strong variability, while classical novae result from unstable burning of accumulated hydrogen.

The cataclysmic-variable orbital-period distribution contains a short-period population between about 82 minutes and two hours, a pronounced cataclysmic-variable period gap around two to three hours, and a longer-period population above the gap. There is an accumulation near the cataclysmic-variable period minimum. These describe ordinary hydrogen-rich systems; evolved donors and helium-rich systems can have shorter periods. Selection by luminosity and outburst activity affects the observed counts, so $N(P)$ is not simply the intrinsic distribution.

For the contact red dwarf, $R_2=R_L\propto a(M_2/M{)}^{1/3}$. Using its supplied thermal-equilibrium mass-radius relation, $R_2\propto M_2$, gives $a\propto M_2^{2/3}{M}^{1/3}$. Kepler's third law then givesindependent of $M_1$. This is a special case of the Roche-lobe-filling period-density relation.

To examine rapid mass loss, use the stellar radius response exponent ${\zeta }_{\mathrm{ad}}=d\log R_2/d\log M_2=-1/3$. During conservative binary mass transfer, $M$ and $J=M_1{M}_2\sqrt{Ga/M}$ are fixed, and $d{M}_1=-d{M}_2$. ThereforeThe change in overfill is $d\log (R_2/R_L)=({\zeta }_{\mathrm{ad}}-{\zeta }_L)d\log M_2$. Since $d\log M_2<0$, overfill grows, and mass loss runs away, when ${\zeta }_{\mathrm{ad}}<{\zeta }_L$. The dynamical stability of binary mass transfer criterion here is consequentlyThe printed instability inequality is reversed. With the defined $q=M_2/M_1$, the specified radius laws imply the inequality above; $q<q_{\mathrm{crit}}$ is the stable side. Equality is marginal, and the supplied Roche-radius approximation restricts this calculation to its stated mass-ratio range.

Long-term contact is maintained by angular momentum transport out of the orbit. Gravitational-wave emission from a binary system provides one loss mechanism. Magnetic braking of a binary star uses the donor's stellar wind: its magnetic field enforces approximate corotation out to the Alfvén radius, so the wind carries much more specific angular momentum than a nonmagnetic surface outflow. At that radius, roughlyThe geometry contributes order-one factors to the torque. Even when ${\dot{M}}_w\ll |{\dot{M}}_2|$, the large lever arm can remove substantial spin angular momentum. Tidal locking couples donor spin to orbital motion, so the wind torque ultimately brakes the orbit. In practice classical novae can also eject matter; the conservative calculations here isolate the stated idealization.

For the prescribed $\dot{J}/J=-\alpha$, now use the supplied equilibrium law $R_2\propto M_2$ for the slowly evolving donor. Contact implies $\dot{a}/a=(2/3){\dot{M}}_2/M_2$, assuming negligible net wind mass loss. Taking the logarithmic derivative of $J$ givesSince $P\propto M_2$ along this equilibrium sequence,For the dynamically stable mass ratios, this is negative. The result uses thermal equilibrium as well as slow hydrostatic evolution; evolution slow compared only with the dynamical time does not by itself guarantee the radius law $R_2\propto M_2$.

In the disrupted magnetic braking model, the donor above the gap loses mass fast enough to be inflated relative to thermal equilibrium. When the donor becomes a fully convective star, the model assumes a substantial reduction in magnetic braking. The red dwarf then contracts toward equilibrium on its Kelvin-Helmholtz cooling time, becoming smaller than its Roche lobe; accretion largely stops. Gravitational-wave emission from a binary system continues to reduce the separation, carrying the detached binary through the gap until contact is restored near its lower edge.

This explanation requires strong braking immediately above the gap: the donor must already be out of equilibrium and inflated, so there is room to contract and detach when braking drops. With nearly fixed component masses during detachment, $R_L\propto a\propto P^{2/3}$, giving the illustrative inflation factorA weak braking torque that kept the donor in thermal equilibrium just above the gap would not produce this detachment. The model requires a decrease in the torque, not disappearance of the donor's magnetic field.

An Algol binary is a semidetached binary: a cool evolved donor fills its Roche lobe, while a hotter and more massive mass gainer remains on or near the main sequence. Roche-lobe overflow feeds the companion through a stream, sometimes forming an accretion disk. A suitably inclined system is an eclipsing binary. The prototype's periodic dimming led John Goodricke to propose an occulting companion in 1783; his original observations record an early eclipse interpretation.

The Algol paradox arises if the present masses are assumed to have been constant: the lower-mass star is the more evolved one, even though coeval isolated stars of higher mass normally exhaust core fuel sooner. The mass-luminosity relation gives the rough nuclear-lifetime scaling $t_{\mathrm{nuc}}\propto M/L$, decreasing strongly with mass. The resolution is binary mass-ratio reversal. The present donor began as the more massive star, evolved first, and expanded into its Roche lobe. Transferring much of its envelope made it less massive and made its initially less massive companion the present mass gainer. Their current masses therefore do not reveal their original evolutionary ordering.

On a Hertzsprung-Russell diagram, both components start on the zero-age main sequence. The initially more massive donor leaves the main sequence first, moving toward lower effective temperature and higher luminosity as a subgiant or red giant. During envelope stripping, it remains oversized and overluminous for its decreasing mass because its evolved core continues to supply energy. The accretor gains mass, moves to higher effective temperature and luminosity, and can undergo stellar rejuvenation if fresh hydrogen mixes into its core. After substantial envelope removal, the donor contracts to a hot stripped star; a sufficiently low-mass helium core ultimately becomes a helium white dwarf. The schematic below separates the two identities through the transfer episode rather than relabelling them when their masses cross.

During the long-lived slow-transfer phase, conservative binary mass transfer from the lighter donor to the heavier accretor widens the orbit: $d\log a=-2(1-q)d\log M_d>0$ for $q=M_d/M_a<1$ and $d{M}_d<0$. Transfer eventually stops when the donor's shrinking envelope can no longer maintain contact. The remnant can be a helium white dwarf if it never ignites helium, or a more massive helium-burning stripped star if it does. Later the mass gainer also leaves the main sequence. Reverse Roche-lobe overflow onto the compact remnant can lead to a common envelope, leaving a close double remnant after successful envelope ejection, or to merger. The detailed outcome depends on both masses, the separation, and how much matter and angular momentum escaped during earlier transfer.

The approximate mass-ratio boundary in the question has a stability interpretation. For an ideal fully convective donor star with adiabatic stellar radius response exponent ${\zeta }_{\mathrm{ad}}=-1/3$, the conservative Roche-radius approximation gives ${\zeta }_L=2q-5/3$. Dynamical stability of binary mass transfer requires ${\zeta }_{\mathrm{ad}}>{\zeta }_L$, so a long-lived stable system hasA more massive convective donor expands relative to its shrinking lobe under mass loss, favouring runaway transfer and a common envelope instead of a persistent Algol phase. This explains the approximate Algol mass-ratio stability limit in that model. It is not a universal observational boundary: a donor with a radiative envelope or a substantial evolved core has a different adiabatic response, and nonconservative loss changes the Roche-lobe response. Van Rensbergen and collaborators' observed and modelled Algol distributions include reported mass ratios above $0.7$ and discuss uncertainties in their determination. The literal claim that all Algols obey the same cutoff is therefore too strong.

A sufficiently wide system first reaches Roche-lobe overflow on the red giant branch, when the original donor is likely to have a deep convective envelope. Straightforward conservative overflow while that donor is still more massive is then prone to dynamical runaway and orbital contraction in a common envelope; it does not naturally yield a wide, long-lived Algol-like configuration. A plausible route is substantial earlier envelope loss through a stellar wind, possibly tidally enhanced stellar wind loss, reducing or reversing the mass ratio before contact. Wind mass transfer in a binary star can also increase the companion's mass. The lower donor mass, reduced envelope and larger core fraction make later transfer easier to stabilize. Alternatively, a detached pair with the same reversed evolutionary appearance may be interacting only through a wind and need never have undergone overflow. Its current width alone does not uniquely determine the initial orbit, but it indicates that prior mass loss or a more general nonconservative history must be considered, rather than applying the simple conservative convective-donor picture unchanged.

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.

In a circular binary star, the distances from the centre of mass are $a_1=a{M}_2/M$ and $a_2=a{M}_1/M$. Summing the two orbital angular momenta givesEquivalently, $J=\mu \sqrt{GMa}$ by Kepler third law.

First consider a rapid conservative perturbation. Both $M=M_1+M_2$ and $J$ are fixed, while $d{M}_1=-d{M}_2$. With $q=M_2/M_1$,soThe Roche lobe formula then gives its mass-radius exponentAfter mass loss, dynamical stability of binary mass transfer requires the donor to shrink relative to its lobe. Since $d\log M_2<0$, this means ${\zeta }_{\mathrm{ad}}>{\zeta }_L$, or

For stable secular conservative binary mass transfer, put $m={\dot{M}}_2/M_2$. The angular-momentum and contact conditions areElimination of $\dot{a}/a$ givesHere magnetic braking of a binary star removes orbital angular momentum, while the donor's expansion maintains Roche-lobe overflow.

In a cataclysmic variable, hydrogen-rich material accumulates on a degenerate white dwarf. Degeneracy prevents initial expansion from regulating its temperature, so nuclear ignition produces the thin-shell instability and a classical nova. Nuclear burning of hydrogen to helium releases about $0.007m{c}^2$, whereas the binding energy at a white-dwarf surface is only of order $G{M}_{1}m/R_1$, typically a few $10^{-4}m{c}^2$. Even modest coupling can therefore eject all the newly accreted envelope without disrupting the white dwarf.

Finally suppose every transferred mass element is expelled by isotropic re-emission from a binary star. Then ${\dot{M}}_1=0$, $\dot{M}={\dot{M}}_2$, and expelled matter carries the white dwarf's specific angular momentum $j_1=(M_2/M{)}^2{a}^2\Omega$. HenceOn the other hand, logarithmic differentiation of $J=M_1{M}_2{M}^{-1/2}{G}^{1/2}{a}^{1/2}$ and of the Roche-lobe radius givesEliminating the separation produces