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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.

Use a circular orbit, as in the stated Roche lobe approximation, with orbital angular speed $\Omega$ and negligible stellar spin. The center of mass distances are $a_1=a{M}_2/M$ and $a_2=a{M}_1/M$. Summing the two orbital angular momenta givesKepler's third law, ${\Omega }^2{a}^3=GM$, makes this $J=M_1{M}_2\sqrt{Ga/M}$. In conservative mass transfer, both $M$ and $J$ are fixed, while $d{M}_2=-d{M}_1$. ThusThe Roche-lobe radius response exponent is consequentlyDuring a dynamical mass-loss perturbation, the deep interior and luminosity do not have time to change. The supplied giant structure therefore has stellar radius response exponent ${\zeta }_{\ast }=-0.27$. Its fractional overfill changes by $d\ln (R/R_L)=({\zeta }_{\ast }-{\zeta }_L)d\ln M_1$. The binary mass ratio uses donor mass divided by accretor mass. Since $d\ln M_1<0$, self-limiting Roche-lobe overflow requires ${\zeta }_{\ast }>{\zeta }_L$. Hence the dynamical stability condition isAt equality the linear restoring response vanishes. This conservative mass-transfer critical mass ratio uses the given radius exponent, rather than imposing the different fully convective $-1/3$ approximation.

The initially more massive component normally evolves first. If it first overflowed only after acquiring the given giant response, it would have $q>1$ and the mass loss would be dynamically unstable: its radius grows while its Roche lobe initially contracts. Starting overflow earlier can avoid this outcome. A main sequence or early post-main-sequence donor star can have a radiative envelope with a stabilizing contraction response. Transfer then reverses the binary mass ratio before the donor develops the giant structure. This is the route to an Algol binary through Case A mass transfer during core hydrogen burning, or early case B mass transfer after core hydrogen exhaustion but before a deep giant envelope develops.

For the initial Roche-lobe overflow to occur before the base of the giant branch, its lobe must be smaller than the base-of-branch stellar radius. Combining $R_L=0.462a(M_1/M{)}^{1/3}$ with Kepler's third law cancels the companion dependence:Writing $m_1=M_1/M_{\odot }$, substitute the mass-radius relation to obtain the pre-giant period limit:We use the specified $0.8\mathrm{days}$ below. This is a pre-giant Roche-lobe-filling period limit, not a condition on the current stripped donor's radius.

At fixed total mass, Kepler's third law gives $P\propto a^{3/2}$, while fixed orbital angular momentum gives $a\propto (M_1{M}_2{)}^{-2}$. Therefore the conservative period invariant isTo find the smallest possible pre-giant period ratio, use solar-unit masses $m_1,m_2$ and fixed $m=m_1+m_2$. Let $K=P(m_1{m}_2{)}^3$. Along this period-product invariant for conservative mass transfer,The denominator has logarithmic derivative $5/m_1-3/(m-m_1)$, with a strictly negative derivative of its own. Its unique maximum, hence the unique minimum of $F$, occurs atSince $F$ diverges at either endpoint, this is the global minimum period-to-donor-mass ratio for conservative evolution.

For the quoted Algol binary, $K=3(1\times 3{)}^3=81\mathrm{days}$ and $m=4$. Choosing the minimizing progenitor gives $m_{1,i}=2.5$, $m_{2,i}=1.5$ andIts pre-giant upper limit is $0.8(2.5{)}^2=5\mathrm{days}$, comfortably above this initial period. Subsequent conservative mass transfer of $1.5M_{\odot }$ produces the stated masses and period. Thus the data permit the proposed early-overflow formation route.

For RT Lac, instead $K=5(0.5\times 1.5{)}^3=2.109375\mathrm{days}$ and $m=2$. At its most favorable conservative progenitor, $m_{1,i}=1.25$, $m_{2,i}=0.75$ andNo conservative progenitor meets the pre-giant overflow condition. An initially more massive giant donor also fails the dynamical-stability condition derived above. Within this stable Algol-type formation model, the current RT Lac system therefore requires nonconservative mass transfer.

A plausible history is substantial envelope loss from the system, through a stellar wind or escaping overflow, possibly with a common envelope episode. Escaping matter also carries orbital angular momentum, so the total-mass and period-product invariants no longer constrain its initial orbit. A more massive progenitor can then shed its envelope and reach the present low donor mass without requiring the excluded conservative sequence. The supplied final data do not determine a unique mass-loss or angular-momentum-loss history.