I just noticed something interesting with the angular momentum of a close stellar binary. This question is somewhat related to another question of mine, but the question here is clearly different :

Conservation of spin angular momentum in a close binary system

To start the question, I copy-paste most part of that question here :

Consider a model of a close stellar binary, of mass $m_1$ and $m_2 < m_1$, moving on circular orbits around the system's barycenter. Both star's rotation is permanently tidal locked. The total angular momentum relative to the barycenter is conserved : \begin{equation}\tag{1} \vec{L} = \vec{L}_1 + \vec{L}_2 + \vec{S}_1 + \vec{S}_2 = \textit{cste}, \end{equation} where $\vec{S}_i$ is the spin of a star (i.e. its angular momentum around its own center of mass). All these vectors are aligned. Using Newton's theory of gravitation, we can prove that \begin{equation}\tag{2} L_{\text{orbital}} = || \vec{L}_1 + \vec{L}_2 || = \frac{m_1 \, m_2}{m_1 + m_2} \, \sqrt{G (m_1 + m_2) a}, \end{equation} where $a = r_1 + r_2$ is the distance between both stars. Lets write $M \equiv m_1 + m_2$ to simplify. Also, since the stars are tidal locked ; $\omega_{\text{rot}} = \omega_{\text{orbital}} \equiv \omega$ : \begin{equation}\tag{3} S_{\text{tot}} = || \vec{S}_1 + \vec{S}_2 || = (I_1 + I_2) \, \omega, \end{equation} where $I_i$ is the inertia moment of a star around its center. If the stars are approximately spherical, then $I_i = \alpha_i \, m_i \, R_i^2$, where $\alpha_i \approx \frac{2}{5}$ (or any number smaller than 1). From Kepler's third law, the orbital angular velocity is \begin{equation}\tag{4} \omega = \sqrt{\frac{GM}{a^3}}. \end{equation} So the total angular momentum of both stars, including their rotation, is this curious beast : \begin{equation}\tag{5} L = G m_1 \, m_2 \sqrt{\frac{a}{G M}} + (\alpha_1 \, m_1 \, R_1^2 + \alpha_2 \, m_2 \, R_2^2) \sqrt{\frac{G M}{a^3}}. \end{equation}

Now here's the curious thing : Expression (5) is a non-trivial function of the distance $a$, and it has a minimum value at \begin{equation}\tag{6} a_{0} = \sqrt{3 M \Big( \frac{\alpha_1 \, m_1 \, R_1^2 + \alpha_2 \, m_2 \, R_2^2}{m_1 \, m_2} \Big)}. \end{equation}

I never heard of this before. So naturally the question is

What is the physical interpretation of $a_0$, and what does the minimal angular momentum $L_0 \equiv L(a_0)$ represents ? Is this value realized in some astrophysical process ?

  • $\begingroup$ To me it seems like a curious artifact of tidal locking. When two stars are far from each other, the angular momentum is dominated by terms with respect to the barycenter, and when they are close together, they have to spin quick enough to ensure tidal locking. So in both limits to zero and infinity we land with high angular momentum, and this seems to be the minimum that has to be there due to continuity. $\endgroup$ – xletmjm Jun 6 '17 at 23:58
  • $\begingroup$ @xletmjm, well yes, I agree, but is there a real astrophysical situation for which the minimal value of angular momentum is favorished ? Is there a tendency to go to that special value ? $\endgroup$ – Cham Jun 7 '17 at 0:21
  • $\begingroup$ I am very likely to be wrong, but I would not be surprised if the thing that you calculated corresponded to total angular momentum of a circular, tidal locked orbit. When it comes to the the question whether this state is in any way favourised - I don't think so. Systems tend to evolve towards smallest total energy, not angular momentum. Consider collisions/interactions: linear momentum is not minimised, it is kept constant. Why would angular momentum need be minimised, when linear momentum is not? $\endgroup$ – xletmjm Jun 7 '17 at 1:37

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