# How do changing binary separation and Roche Lobe size of the donor affect the long-term evolution of a binary under conservative mass transfer?

I computed that for a binary system that undergoes conservative mass transfer from the Roche-lobe filling Start 2 to Star 1 (i.e. $$\dot M_2<0, \dot M_1=-\dot M_2$$ and the change in angular momentum $$\dot J=0$$) it holds that: $$\frac{\dot a}{a}=-2\frac{\dot M_2}{M_2}(1-q)\ ,\tag{*}$$ where $$q=\frac{M_2}{M_1}$$ and $$a$$ is the binary separation.

Additionally, I know that the size of the Roche Lobe of the donor star (i.e. star 2) denoted $$R_{L,2} \propto a$$.

My question is how do these two things affect the long-term evolution of a binary?

These are my thoughts. I would appreciate to tell me if they are correct, complement them and address the doubts.

$$q$$ is also varying because the masses of the stars are varying, more specifically $$q$$ is decreasing because M2 is decreasing and M1 is increasing.

From (*), since $$\dot M_2<0,$$ we have that the sign of $$a$$ is the same as that of $$1-q$$ therefore I think I have two cases:

(i) $$0, which means that the donor has less mass that the accreting object. Is that possible? Shouldn't a more massive donor be a more logical situation?

Anyway, in this case, we have that $$a$$ and $$R_{L,2} \propto a$$ increase. But I think that since $$R_{L,2}$$ is the region around star 2 in the binary system within which orbiting material is gravitationally bound to that star (using the definition in https://en.wikipedia.org/wiki/Roche_lobe), isn't the fact that this region is increasing helping the system remain bounded while the increasing separation $$a$$ is doing the contrary? Which wins then?

(ii) $$q>1$$ which means that the donor has more mass that the accreting object and from (*), $$a$$ and $$R_{L,2} \propto a$$ decrease,... what does this mean? The decrease of $$a$$ means that the system is merging, right? but what does the decrease in $$R_{L,2}$$ mean? Besides, I am not sure if they actually merge since $$q$$ is also decreasing, eventually, $$q$$ will end up between 0 and 1 so the sign of $$1-q$$ will flip to the previous case and they will start to separate again.

Therefore, whatever the initial situation, it means that the system will separate in the end, unless my analysis in case (i) is wrong and for some reason a start to decrease at some point in which case it would look like there is an oscillatory behaviour.