# Paradox from mass transfer in a close stellar binary

I'm considering a simple binary system made of two stars, of mass $m_1$ and $m_2 < m_1$, on circular orbits around their center of mass. Using Newton's theory of gravitation, it is easy to prove the following formulae for the total mechanical energy and angular momentum of the whole binary ($a \equiv r_1 + r_2$ is the separation of the two masses) : \begin{align} E \equiv K + U &= -\, \frac{G m_1 \, m_2}{2 a}, \tag{1} \\[12pt] L \equiv L_1 + L_2 &= \frac{m_1 \, m_2}{M} \, \sqrt{G M a}. \tag{2} \end{align} Suppose that one star is transfering matter to the other star, without any loss. Conservation of matter implies $M = m_1 + m_2 = \textit{cste}$, so $\dot{m}_2 = -\, \dot{m}_1$. Now, many papers/lectures I've found state that angular momentum is still conserved : $\dot{L} = 0$. This gives an equation for the rate of change of the distance $a$ (the orbits are slowly evolving, while remaining approximately circular) : $$\tag{3} \frac{\dot{a}}{a} = -\, 2 \Big( \frac{\dot{m}_1}{m_1} + \frac{\dot{m}_2}{m_2} \Big).$$ See for example these references :

Equation (3) implies the following : $$\tag{4} a(t) = a(0) \Big( \frac{m_1(0) m_2(0)}{m_1(t) m_2(t)} \Big)^2.$$

My trouble comes from conservation of energy, if the system is really assumed to be isolated (no loss/gain of wathever). Then the derivative of equation (1) above gives this (from $\dot{E} = 0$) : $$\tag{5} \frac{\dot{a}}{a} = \frac{\dot{m}_1}{m_1} + \frac{\dot{m}_2}{m_2},$$ which is incompatible with equation (3) (unless of course $\dot{m}_1 = -\, \dot{m}_2 = 0$, but then there is no mass transfer).

So what is going on here ? Why angular momentum should be conserved while energy is not ? Why not the reverse, i.e. that energy is conserved but not angular momentum ? The authors of the lectures cited above says nothing about the binary's mechanical energy.

EDIT : About the total angular momentum, if we add the contributions from the size and rotation spin of the stars, we get the true conserved total angular momentum. I'm assuming tidal locked stars, with parallel rotation and revolution axes : $\omega_{\text{rot} \, 1} = \omega_{\text{rot} \, 2} = \omega_{\text{orbital}} \equiv \omega$ : \begin{align} L_{\text{tot}} &= L_{\text{orbital}} + S_1 + S_2 \nonumber \\[12pt] &= \frac{m_1 \, m_2}{M} \, \sqrt{G M a} + (I_1 + I_2) \, \omega, \tag{6} \end{align} where $\omega$ is given by Kepler's third law : $$\tag{7} \omega = \sqrt{\frac{G M}{a^3}}.$$ I think that the spin contributions and their time variations are negligible in front of the orbital angular momentum, since $I_1 \propto m_1 \, R_1^2$, and $R_1 \ll a$ (same for star 2).

Any clue on this ?

• Neither angular momentum nor energy is really conserved. Kinetic and potential energy can turn into heat, and angular momentum can be transferred to the rotations of the two stars. So the question is: why is the angular momentum the quantity that is approximately conserved? – Peter Shor Jun 5 '17 at 15:27
• Yes, I guess you can put it this way. I know that we may need to take into account the size and rotation of both stars, by adding proper terms to energy AND angular momentum (rotation kinetic energies, gravitational binding energies, spin angular momentum, ...). But then why (2) is approximately conserved, while (1) isn't ? – Cham Jun 5 '17 at 15:31
• My guess: the atmosphere of the one star is attracted to the other star, and the molecules fall on it and go "thud" when they hit it (not really ... it's made of gas), thus generating heat. But these molecules are nearly equally likely to fall on it from all directions, and so they don't change its angular momentum much. – Peter Shor Jun 5 '17 at 16:04
• Maybe the solution is actually "trivial" : there is heat created, which (1) doesn't account (even if we add the stars rotation kinetic energy and self-gravitational binding potential energy). So mechanical energy isn't conserved. But the whole system conserve angular momentum since there is no "heat of angular momentum". If there is energy escaping to infinity (as "heat"), it's approximately isotropic, so doesn't bring out some angular momentum. – Cham Jun 5 '17 at 16:13
• The angular momentum could go from the stars orbiting around each other into each star's individual rotation. You have to show that this doesn't have much effect overall. – Peter Shor Jun 5 '17 at 16:29

Your conclusion based on $\dot{E}$ and $(3)$ is wrong. They imply that $\frac{\dot{m_1}}{m_1} + \frac{\dot{m_2}}{m_2}$ is zero which implies $\frac{\dot{a}}{a} = \dot{a} = 0$ or that the sum of the distances to the center of mass stay the same. It also implies $\frac{\dot{m_1}}{\dot{m_2}} = -\frac{m_1}{m_2}$, in other words the ratio of the rate of mass flows is negative the ratio of the masses. Conservation of angular momentum and of energy is enough for a sort of "Conservation of distance".