Donor-wind angular-momentum loss ($\dot{J}=(1-f){\dot{M}}_d{a}_d^2\Omega$)

An isotropic escaping stellar wind has the donor star's mean specific orbital angular momentum. With ${\dot{M}}_d<0$, retained fraction $f$, negligible spin and no additional wind torque, $\dot{J}=(1-f){\dot{M}}_d{a}_d^2\Omega$. The retained matter redistributes angular momentum internally. Unlike isotropic re-emission from a binary star, the escaping material originates at the donor rather than the accretor. With no retention this reduces to Jeans-mode mass loss.

In a circular binary star let the donor star lose mass ${\dot{M}}_1<0$, with constant retained fraction $f$ and ${\dot{M}}_2=-f{\dot{M}}_1$. A donor star stellar wind escaping with the donor star's orbital specific angular momentum has $\dot{J}/J=(1-f){\dot{M}}_1{M}_2/[M_1(M_1+M_2)]$. Differentiating $J=G^{2/3}(P_{\mathrm{orb}}/2\pi {)}^{1/3}{M}_1{M}_2(M_1+M_2{)}^{-1/3}$ and using this loss law givesFor constant $f$ this integrates to $P_{\mathrm{orb}}{M}_1^{3f}{M}_2^3(M_1+M_2{)}^2=\mathrm{constant}$. If $f$ varies, the differential relation remains valid but the first factor must be replaced by $\exp (3\int fd\log M_1)$; treating a time-dependent exponent as a fixed power introduces an extra $\dot{f}\log M_1$ term.

Under donor-wind angular-momentum loss and the Roche lobe approximation $R_L=0.46a(M_d/M{)}^{1/3}$, the Roche-lobe radius response exponent is affine in retained fraction: ${\zeta }_L=(1-f){\zeta }_0+f{\zeta }_1$, where $q=M_d/M_a$, ${\zeta }_0=1/3-4q/[3(1+q)]$ and ${\zeta }_1=2q-5/3$. Differentiate $\log R_L=\log a+(\log M_d-\log M)/3+\mathrm{constant}$ and the circular-binary orbital angular momentum to obtain this expression.

A fast escaping Jeans-mode mass loss wind can drive a giant into Roche-lobe overflow because its radius changes as it loses mass, even though the wind itself is not substantially captured. This differs from slow Wind Roche-lobe overflow. For $R_d\propto M_d^{-n}$ and $R_L\propto a(M_d/M{)}^{1/3}$, wind alone gives $d\log (R_d/R_L)/dt=\dot{M}[3(1-n)q-(1+3n)]/[3M_d(1+q)]$, where $q=M_d/M_a$ and $\dot{M}<0$ is the escaping wind rate. The displayed inequality is exactly the condition that overfill increases.

Let Roche-lobe overflow transfer mass conservatively within a circular binary while a separate fast wind escapes with the donor's specific orbital angular momentum. Then ${\dot{M}}_d=\dot{M}-{\dot{M}}_a$ and $\dot{a}/a=-\dot{M}/M+2{\dot{M}}_a(1/M_d-1/M_a)$. Differentiating the donor and Roche lobe radii and enforcing contact gives the displayed rate. It is positive and stabilizing when the wind drives overfill and $6q<5-3n$. When the latter inequality is reversed, transfer increases the overfill, producing runaway rather than a steady negative-transfer solution.

A donor star with stellar radius response exponent ${\zeta }_{\ast }$ can maintain exact contact in the donor-wind Roche-lobe response model for some retained fraction $0\le f\le 1$ precisely when ${\zeta }_{\ast }$ lies between ${\zeta }_0$ and ${\zeta }_1$. This follows because the attainable Roche-lobe radius response exponents form that closed interval. If its endpoints coincide, all fractions work when ${\zeta }_{\ast }$ equals the common value and none work otherwise. With negative donor mass-loss rate, $d\log (R_d/R_L)/dt=({\zeta }_{\ast }-{\zeta }_L){\dot{M}}_d/M_d$ distinguishes detachment from increasing overfill. Dynamical stability of binary mass transfer requires the adiabatic donor response.

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.