# Why does mass (gas) transfer between binary stars cause them to move apart?

Some binaries, like Algol binaries, move monotonically apart from each other as one steals gas from the other.

Why?

On a naive level, shouldn't friction and such cause the two stars to move (rotate?, revolve?, etc.) more slowly, and lose velocity, and move closer to each other? To a potential merger?

• Commented Jan 12, 2020 at 22:40

The angular momentum of a star in orbit is $$m r^2 \omega$$. Thus the total angular momentum of a binary system is constant (if no mass is lost from the system) and given by $$J = (m_1 a_1^2 + m_2 a_2^2) \omega,$$ where $$a = a_1 + a_2$$ is the separation, $$m_1 a_1 = m_2 a_2$$, the total mass $$m = m_1+m_2$$ and (from Kepler's 3rd law) $$a^3 \omega^2$$ is constant (call it $$\alpha$$), if the mass lost by one star is accreted by the other.
Thus $$J = m_1 a_1 a \omega = m_1 a_1 a^{-1/2} \sqrt{\alpha}$$ and $$ma_1 = m_1 a_1 + m_2 a_1 = m_2 a$$ thus $$J m\alpha^{-1/2} = m_1 m_2 \sqrt{a}$$
The LHS of the above expression is a constant, so the ratio of final to initial separation after any conservative mass loss is $$\frac{a_f}{a_i} =\left(\frac{m_{1,i} m_{2,i}}{m_{1,f} m_{2,f}}\right)^2$$
The product $$m_1 m_2$$ is maximised (if the total mass is constant), when $$m_1 = m_2$$. Thus if mass is lost by the less massive star, then the final $$m_1 m_2$$ becomes smaller and the binary widens. This is what happens in an Algol-type system -- a less massive red giant loses mass to a more massive main sequence companion and the orbit widens.