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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.

Let $M=M_1+M_2$. The centre of mass condition gives $a_1=a{M}_2/M$, $a_2=a{M}_1/M$. For circular motion the component angular momenta are $J_i=M_i{a}_i^2\Omega$, soUsing Kepler's third law, $a^3{\Omega }^2=GM$, and $\Omega =2\pi /P_{\mathrm{orb}}$ givesThis is the circular-binary orbital angular momentum. Mass received by the companion redistributes angular momentum within the binary star; only the escaping stellar wind removes it under the prescribed model.

The stellar wind from the donor star has specific angular momentumHere ${\dot{M}}_1<0$, ${\dot{M}}_2=-f{\dot{M}}_1$ and $\dot{M}=(1-f){\dot{M}}_1$. Thus donor-wind angular-momentum loss givesTaking the logarithmic derivative of the period expression for $J$ givesUse $M_2/(M_{1}M)=1/M_1-1/M$ to simplify this toFor constant $f$, integration therefore givesThis is the donor-wind period invariant with fixed retention fraction. If $f$ changes during the evolution, the differential equation remains valid but the integrated invariant is instead $P_{\mathrm{orb}}{M}_2^3{M}^2\exp (3\int fd\log M_1)=\mathrm{constant}$. The fixed-power expression cannot be used with a time-dependent exponent without this modification.

For the Roche lobe, write $q=M_1/M_2$. The equivalent separation expression $J=M_1{M}_2\sqrt{Ga/M}$ givesTaking the logarithmic derivative of $R_L=0.46a(M_1/M{)}^{1/3}$ then adds ${\dot{M}}_1/(3M_1)-\dot{M}/(3M)$, yieldingThis donor-wind Roche-lobe response uses only the instantaneous value of $f$ and thus still holds if it varies.

Define $A(q)=2q+4q/[3(1+q)]-2$ and $B(q)=1/3-4q/[3(1+q)]$. The Roche-lobe radius response exponent is ${\zeta }_L=B+Af$, whereas the donor star's stellar radius response exponent is ${\zeta }_{\ast }=-n$. Keeping $R_1=R_L$ requires the two logarithmic derivatives of radius to agree, soFor $A\ne 0$, this determinesA physical mixture of Roche-lobe overflow and stellar wind from the donor star needs $0<f<1$. Equivalently, $-n$ must lie strictly between the pure-wind response $B$ and the conservative response $B+A=2q-5/3$. The values $f=0$ and $1$ are valid limiting prescriptions, respectively pure stellar wind and conservative mass transfer, rather than mixtures. If $f_{\mathrm{req}}$ lies outside $[0,1]$, no allowed division of the lost mass can maintain contact under this radius law and angular-momentum-loss prescription. This is feasibility of donor-wind binary contact.

The likely direction away from contact follows without assigning a nonphysical fraction. For any actual $f$, let $\Delta =\log (R_1/R_L)$. ThenSince ${\dot{M}}_1<0$, a positive value of $A(f_{\mathrm{req}}-f)$ makes the donor star move inside its lobe and detach; a negative value increases overfill and promotes stronger transfer. Thus if $A>0$, $f_{\mathrm{req}}<0$ gives increasing overfill and $f_{\mathrm{req}}>1$ gives detachment; if $A<0$, these outcomes are reversed. A runaway or a new equilibrium would require the donor star's response or the angular-momentum-loss model to change, possibly with additional driving. The contact calculation alone does not prove a particular nonlinear endpoint.

There is also a genuine exceptional denominator: $A=0$ at $q=(\sqrt{10}-1)/3$. At this ratio both limiting lobe responses coincide. Contact is possible only if $-n=B$, namely $n=(\sqrt{10}-2)/(\sqrt{10}+2)$; then every $f$ gives the same instantaneous radius response. Otherwise no fraction works, and the sign of $(-n-B){\dot{M}}_1/M_1$ determines detachment or increasing overfill.

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.