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During binary mass-ratio reversal, an initially more massive donor star loses enough material that its companion becomes the more massive member. The stars retain their different prior nuclear evolution, resolving the Algol paradox. Conservative binary mass transfer shrinks the orbit before equal masses and widens it after reversal, because $a\propto (M_d{M}_a{)}^{-2}$ at fixed total mass and orbital angular momentum.

For fixed total solar-unit mass $m$ and conserved orbital angular momentum, conservative binary mass transfer preserves $K=P[m_d(m-m_d){]}^3$. Hence $P/m_d^2=K/[m_d^5(m-m_d{)}^3]$. Its logarithmic derivative is $-5/m_d+3/(m-m_d)$, vanishing at $5m/8$ with positive second derivative. The ratio diverges at both endpoints, so this is its global minimum. Comparing this minimum to a pre-giant bound $P_0$ tests whether any such conservative antecedent can overflow before the giant branch; satisfying it is necessary, not a proof of an entire evolutionary history.

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.

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.

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.

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.

Let $\mathbf{r}={\mathbf{r}}_2-{\mathbf{r}}_1$ and $\mathbf{v}=\dot{\mathbf{r}}$. Relative to the centre of mass, the two positions are ${\mathbf{r}}_1=-(M_2/M)\mathbf{r}$ and ${\mathbf{r}}_2=(M_1/M)\mathbf{r}$. Their velocities have the same mass factors. Adding their kinetic energies and angular momenta gives, with reduced mass $\mu =M_1{M}_2/M$,These are the internal orbital quantities, excluding any uniform centre-of-mass motion. Newton's law of universal gravitation givesFor constant masses, differentiating $E$ makes the gravitational work cancel the derivative of the potential energy, so $\dot{E}=0$. Also $\dot{\mathbf{h}}=\mathbf{r}\times \ddot{\mathbf{r}}=0$, since the force is central; hence conservation of angular momentum gives $\dot{\mathbf{J}}=0$.

The eccentricity vector, the normalized attractive Laplace-Runge-Lenz vector, isUsing $\dot{\hat{\mathbf{r}}}=\mathbf{v}/r-\mathbf{r}(\mathbf{r}\cdot \mathbf{v})/r^3$ andshows that its derivative also vanishes. Adding a relative perturbing acceleration $\mathbf{f}$ changes these cancellations only by the perturbing terms. The perturbed Kepler-orbit conservation laws are therefore

For the distant Newtonian gravitational potential, expand each source term about the binary's centre of mass:The dipole vanishes because ${\sum }_i{M}_i{\mathbf{r}}_i=0$. With $\mathbf{a}=\mathbf{r}$, the quadratic mass moment is ${\sum }_i{M}_i{\mathbf{r}}_i{\mathbf{r}}_i=\mu \mathbf{aa}$, so the quadrupole potential of a circular binary isOn the inner binary's fast timescale, the distant companion's position is approximately fixed. Averaging over the inner circular orbit in the common plane gives $⟨{\cos }^2\theta ⟩=1/2$ and henceThis averaged potential function is independent of both time and azimuth. Consequently the companion's energy and axial angular momentum are conserved in the averaged problem. The exact motion has small fast variations; the statement is a secular approximation for a hierarchical system away from an orbital resonance.

For a circular outer orbit, force balance givesso the circumbinary orbital-period correction isConservative binary mass transfer leaves $M$ unchanged, so it does not change the leading monopole force. It changes $\mu$ and $a$, and therefore the quadrupole correction. If the inner orbital angular momentum is conserved, $J_{\mathrm{in}}=\mu \sqrt{GMa}$ gives $a\propto {\mu }^{-2}$ and $Q=\mu a^2\propto {\mu }^{-3}$. The correction is thus largest as the binary becomes strongly unequal in mass and widens, within the hierarchy $a\ll x$.

One must also allow the outer radius to respond. For slow axisymmetric evolution its specific angular momentum $h_3$ remains constant, while its energy need not: the averaged potential now depends on time. A nearly circular outer orbit has $h_3^2=GMx+3GQ/(4x)$ and $P_3=2\pi x^2/h_3$. At fixed $M,h_3$, differentiation givesThus mass exchange inside the binary can produce a small measurable outer-period change even without total mass loss. Nonconservative mass loss additionally changes the leading monopole term.

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.