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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.

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.

For a slowly evolving circular orbit with constant retained fraction $f$ and donor-wind angular-momentum loss, logarithmic differentiation of the circular-binary orbital angular momentum gives $d\log P=-3fd\log M_d-3d\log M_a-2d\log (M_d+M_a)$. Integration gives the displayed invariant. At $f=1$ it becomes the period-product invariant for conservative mass transfer, while at $f=0$ it becomes $P{M}^2=\mathrm{constant}$. A varying retained fraction does not give this simple power-law invariant.