 Solution

In a circular binary star, the distances from the centre of mass are $a_1=a{M}_2/M$ and $a_2=a{M}_1/M$. Summing the two orbital angular momenta givesEquivalently, $J=\mu \sqrt{GMa}$ by Kepler third law.

First consider a rapid conservative perturbation. Both $M=M_1+M_2$ and $J$ are fixed, while $d{M}_1=-d{M}_2$. With $q=M_2/M_1$,soThe Roche lobe formula then gives its mass-radius exponentAfter 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 ${\zeta }_{\mathrm{ad}}>{\zeta }_L$, or

For stable secular conservative binary mass transfer, put $m={\dot{M}}_2/M_2$. The angular-momentum and contact conditions areElimination of $\dot{a}/a$ givesHere 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 $0.007m{c}^2$, whereas the binding energy at a white-dwarf surface is only of order $G{M}_{1}m/R_1$, typically a few $10^{-4}m{c}^2$. 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 $j_1=(M_2/M{)}^2{a}^2\Omega$. HenceOn the other hand, logarithmic differentiation of $J=M_1{M}_2{M}^{-1/2}{G}^{1/2}{a}^{1/2}$ and of the Roche-lobe radius givesEliminating the separation produces