= Solution
Write $M=M_1+M_2$, $a=a_1+a_2$, and $\mu=M_1M_2/M$, the <reduced mass>. In the <center of mass> frame, $a_1=aM_2/M$ and $a_2=aM_1/M$. Both components of the <circular orbit> have the same <angular velocity> $\Omega$. Thus their total orbital <angular momentum> is
$$
J=(M_1a_1^2+M_2a_2^2)\Omega=\mu a^2\Omega.
$$
Using <Kepler's third law>, $\Omega^2=GM/a^3$, and the <orbital period> $P=2\pi/\Omega$, gives the <circular-binary orbital angular momentum>
$$
\boxed{J=\mu\sqrt{GMa}=\frac{G^{2/3}P^{1/3}M_1M_2}{(2\pi)^{1/3}M^{1/3}}.}
$$
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 gives
$$
\boxed{\dot J=(1-f)\dot M_1a_1^2\Omega<0\quad(f<1),}\qquad
\frac{\dot J}{J}=(1-f)\dot M_1\frac{M_2}{M_1M}.
$$
Isotropy 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>:
$$
\frac{\dot J}{J}=\frac13\frac{\dot P}{P}+\frac{\dot M_1}{M_1}+\frac{\dot M_2}{M_2}-\frac13\frac{\dot M}{M}.
$$
Substitution of the <donor-wind angular-momentum loss> and mass rates gives
$$
\frac{\dot P}{P}=3\dot M_1\left[\frac{(1-f)M_2}{M_1M}-\frac1{M_1}+\frac f{M_2}\right]+(1-f)\frac{\dot M_1}{M}
=\left[-\frac{3f}{M_1}+\frac{3f}{M_2}-\frac{2(1-f)}M\right]\dot M_1.
$$
For \b[constant $f$], the last expression is $-3f\,d\log M_1/dt-3\,d\log M_2/dt-2\,d\log M/dt$. Hence the <period invariant for constant-fraction donor-wind mass loss> is
$$
\boxed{PM_1^{3f}M_2^3M^2=\text{constant},\qquad P\propto M_1^{-3f}M_2^{-3}M^{-2}.}
$$
For 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 $PM^2$ is constant, as in <Jeans-mode mass loss>.
Set the <binary mass ratio> $q=M_1/M_2$. Since $a^3=GMP^2/(4\pi^2)$, the <Kepler third law> and the preceding <orbital period> derivative imply
$$
\frac{\dot a}{a}=\frac{\dot M_1}{M_1}\left[2f(q-1)-\frac{(1-f)q}{1+q}\right].
$$
The specified approximation to the <Roche lobe> gives $\log R_L=\log(0.46)+\log a+\frac13(\log M_1-\log M)$. Consequently the <donor-wind Roche-lobe response> is
$$
\frac{\dot R_L}{R_L}=\frac{\dot M_1}{M_1}\left[2f(q-1)+\frac13-\frac{4(1-f)q}{3(1+q)}\right]
=\frac{\dot M_1}{M_1}\left\{f\left[2q+\frac{4q}{3(1+q)}-2\right]+\frac13-\frac{4q}{3(1+q)}\right\}.
$$
This 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 $\zeta_*=-n$. Maintaining <Roche-lobe overflow> in exact contact requires $\dot R_1/R_1=\dot R_L/R_L$, yielding
$$
\boxed{f\left[2q+\frac{4q}{3(1+q)}-2\right]=\frac{4q}{3(1+q)}-n-\frac13.}
$$
For a precise <feasibility of donor-wind binary contact> test, put $\zeta_0=\frac13-\frac{4q}{3(1+q)}$ and $\zeta_1=2q-\frac53$. The <Roche-lobe radius response exponent> is $\zeta_L=(1-f)\zeta_0+f\zeta_1$. Thus \b[a permitted contact fraction exists exactly when]
$$
\boxed{\min(\zeta_0,\zeta_1)\le -n\le\max(\zeta_0,\zeta_1).}
$$
Unless $A=\zeta_1-\zeta_0=2q+4q/[3(1+q)]-2$ 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:
$$
\frac{d}{dt}\log\frac{R_1}{R_L}=(\zeta_*-\zeta_L)\frac{\dot M_1}{M_1}.
$$
If $\zeta_*>\max(\zeta_0,\zeta_1)$, mass loss makes the <donor star> underfill its <Roche lobe>: \b[the system detaches and contact-driven transfer stops]. If $\zeta_*<\min(\zeta_0,\zeta_1)$, mass loss increases the overfilling: \b[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.
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