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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.

To obtain the specified expansion, use the usual nova-ejection prescription: the escaping material is expelled isotropically from the white dwarf and carries its specific orbital angular momentum, with no additional torque. This is isotropic re-emission from a binary star. Circularity alone does not determine the angular momentum carried by the ejecta; the prescription is an additional physical assumption.

The accretor's orbital radius is $a_1=a{M}_2/M$, so its specific angular momentum satisfiesDuring ejection, take $\delta M_1=-\delta m$, $\delta M_2=0$ and $\delta M=-\delta m$, where $\delta m>0$. ConsequentlyDifferentiate $J=M_1{M}_2\sqrt{Ga/M}$ to first order:Using $M/M_1=1+q$ yieldsThe same lobe law now has changing total mass but unchanged donor mass, soThe donor radius is unchanged on the short eruption timescale, whereas its lobe expands. It therefore falls inside the lobe and Roche-lobe overflow stops in this idealized model. This is nova-induced binary detachment.

The need for the ejection prescription can also be seen algebraically. If the ejecta instead have mean specific angular momentum $j_{\mathrm{ej}}={\eta }_{J}J/M$, the same circular-orbit differentiation givesThe requested coefficient is obtained only when ${\eta }_J=q$. Other prescriptions can produce a different expansion or a contraction; the stated circularity condition does not rule them out. The ejection may be short compared with the transfer time while remaining slow compared with an orbit, consistent with circularity. If truly impulsive, an eccentricity is generally excited unless an additional circularization assumption is made.

Now put $\Gamma =-\dot{J}>0$ for the constant external torque. Keep the masses, mass ratio and torque coefficients fixed to leading order over one small-$\delta m$ cycle; corrections in those coefficients affect the calculated times only at the next order. During detachment there is no transfer, so fixed component masses giveThe fractional lobe gap $4\delta m/(3M)$ closes in timeDuring the semi-detached accumulation phase, use the same external torque and the equilibrium donor contact equation derived in (a):Accumulating the next layer therefore takesDivision cancels the layer mass, torque and angular momentum:The result uses $q<4/3$ for a positive steady accumulation rate, unchanged donor structure during the eruption and the same external torque in the two phases. Here the given $q<1$ ensures a positive denominator. It describes an idealized nova cycle; extra frictional loss, irradiation-driven donor changes or a different ejection angular momentum would change the ratio.

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.

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.