= Solution
In a circular <binary star>, the distances from the centre of mass are $a_1=aM_2/M$ and $a_2=aM_1/M$. Summing the two orbital angular momenta gives
$$
J=M_1a_1^2\Omega+M_2a_2^2\Omega
=\boxed{\frac{M_1M_2}{M}a^2\Omega}.
$$
Equivalently, $J=\mu\sqrt{GMa}$ by <Kepler third law>.
First consider a rapid conservative perturbation. Both $M=M_1+M_2$ and $J$ are fixed, while $dM_1=-dM_2$. With $q=M_2/M_1$,
$$
0=d\log J=(1-q)d\log M_2+\frac12d\log a,
$$
so
$$
\frac{d\log a}{d\log M_2}=2(q-1).
$$
The <Roche lobe> formula then gives its mass-radius exponent
$$
\zeta_L=\frac{d\log R_L}{d\log M_2}
=\frac13+2(q-1)=2q-\frac53.
$$
After mass loss, <dynamical stability of binary mass transfer> requires the donor to shrink relative to its lobe. Since $d\log M_2<0$, this means $\zeta_{\rm ad}>\zeta_L$, or
$$
\boxed{q<\frac{3\zeta_{\rm ad}+5}{6}}.
$$
For stable secular <conservative binary mass transfer>, put $m=\dot M_2/M_2$. The angular-momentum and contact conditions are
$$
-\alpha=(1-q)m+\frac12\frac{\dot a}{a},
\qquad
\beta=\frac13m+\frac{\dot a}{a}.
$$
Elimination of $\dot a/a$ gives
$$
\boxed{-\frac{\dot M_2}{M_2}
=\frac{3(2\alpha+\beta)}{5-6q}}.
$$
Here <magnetic braking of a binary star> removes orbital angular momentum, while the donor's expansion maintains <Roche-lobe overflow>.
In a <cataclysmic variable>, hydrogen-rich material accumulates on a degenerate white dwarf. Degeneracy prevents initial expansion from regulating its temperature, so nuclear ignition produces the <thin-shell instability> and a <classical nova>. Nuclear burning of hydrogen to helium releases about $0.007mc^2$, whereas the binding energy at a white-dwarf surface is only of order $GM_1m/R_1$, typically a few $10^{-4}mc^2$. Even modest coupling can therefore eject all the newly accreted envelope without disrupting the white dwarf.
Finally suppose every transferred mass element is expelled by <isotropic re-emission from a binary star>. Then $\dot M_1=0$, $\dot M=\dot M_2$, and expelled matter carries the white dwarf's specific angular momentum $j_1=(M_2/M)^2a^2\Omega$. Hence
$$
\frac{\dot J}{J}=-\alpha+\frac{q^2}{1+q}m.
$$
On the other hand, logarithmic differentiation of $J=M_1M_2M^{-1/2}G^{1/2}a^{1/2}$ and of the Roche-lobe radius gives
$$
\frac{\dot J}{J}
=\left(1-\frac{q}{2(1+q)}\right)m
+\frac12\frac{\dot a}{a},
\qquad
\beta=\frac{m}{3(1+q)}+\frac{\dot a}{a}.
$$
Eliminating the separation produces
$$
\boxed{-\frac{\dot M_2}{M_2}
=\frac{3(1+q)(2\alpha+\beta)}
{5+3q-6q^2}}.
$$
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