# Conservation of spin angular momentum in a close binary system

Consider a simple model of a close stellar binary, of mass $m_1$ and $m_2 < m_1$, moving on circular orbits around the system's barycenter (no eccentricity, to simplify things). Both star's rotation is permanently tidal locked. The total angular momentum relative to the barycenter is conserved : $$\tag{1} \vec{L} = \vec{L}_1 + \vec{L}_2 + \vec{S}_1 + \vec{S}_2 = \textrm{constant},$$ where $\vec{S}_i$ is the spin of a star (i.e. its angular momentum around its own center of mass). These vectors are all aligned. Using Newton's theory of gravitation, we can prove that $$\tag{2} L_{\text{orbital}} = || \vec{L}_1 + \vec{L}_2 || = \frac{m_1 \, m_2}{m_1 + m_2} \, \sqrt{G (m_1 + m_2) a},$$ 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 ; $\omega_1 = \omega_2 = \omega_{\text{orbital}} \equiv \omega$ : $$\tag{3} S_{\text{tot}} = || \vec{S}_1 + \vec{S}_2 || = (I_1 + I_2) \, \omega,$$ 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). The orbital angular velocity is $$\tag{4} \omega = \sqrt{\frac{GM}{a^3}}.$$

Now, suppose that the stars are allowed to exchange some matter: $m_1$, $m_2$, $a$, $\omega$, $I_1$ and $I_2$ are now functions of time ($M$ is conserved, though).

If the whole system is isolated, the total angular momentum (1) is conserved.

How can we justify that both $L_{\text{orbital}}$ and $S_{\text{tot}}$ are separately conserved (maybe approximately) ?

• How are you proposing to exchange the matter? There would have to be an accretion disk (with angular momentum). In addition one cannot necessarily assume the radius of gyration is unchanged by accretion. But perhaps you are looking for someone to address all these complications? – Rob Jeffries Jun 10 '17 at 19:40
• @RobJeffries, I'm thinking about a close binary that exchange some mass from one star to the other one. Whatever the mechanism (accretion disk ?), how the spin angular momentum should be related to the orbital angular momentum ? Of course, from (2) and (3) above : $L_{\text{tot}} = L_{\text{orbital}} + S_{\text{tot}}$ is conserved, but this doesn't say if $L_{\text{orbital}}$ and $S_{\text{tot}}$ are conserved separately. I'm expecting they aren't separately conserved, but how should they be related ? – Cham Jun 10 '17 at 22:01

This method is taken from Taylor's Classical Mechanics, in the "Two-Body Central-Force Problems" section. This goes way more in-depth than it needs to, and assumes uniform densities of the stars for ease of calculation (though this is not necessary).

tl:dr: The Lagrangian is independent of each star's angle off the center of mass, and independent of each star's rotational angle, so the total orbital and each star's rotational angular momentum are all independently conserved. This is achieved by the changing mass accounting for the change in rotational speeds. Note that this does not require tidal locking or circular orbits.

Take $\vec{R}$ to be the center of mass position.

$$\vec{R}=\frac{m_1\dot{}\vec{r}_1+m_2\dot{}\vec{r}_1}{m_1+m_2}=\frac{m_1\dot{}\vec{r}_1+m_2\dot{}\vec{r}_2}{M}$$

where $M≡m_1+m_2$.

We can subsequently define $\vec{r}_1=\vec{R}+\frac{m_2}{M}\vec{r}$ and $\vec{r}_2=\vec{R}-\frac{m_1}{M}\vec{r}$, where $\vec{r}=\vec{r}_1-\vec{r}_2$

The kinetic energy is

$$T=\frac{1}{2}(m_1\dot{\vec{r}}^2_1+m_2\dot{\vec{r}}^2_2+\frac{2}{5}m_1 s_1^2\omega_1^2+\frac{2}{5}m_2 s_2^2\omega_2^2)$$

where $s_i$ are the radii of the stars, and $\omega _i$ are the rotational velocities (not orbital). We can reduce this to

$$T=\frac{1}{2}(m_1\dot{\vec{r}}^2_1+m_2\dot{\vec{r}}^2_2+\frac{2}{5}m_1 s_1^2\omega_1^2+\frac{2}{5}m_2 s_2^2\omega_2^2)$$ $$T=\frac{1}{2}[m_1(\dot{\vec{R}}+\frac{m_2}{M}\dot{\vec{r}})^2+m_2(\dot{\vec{R}}-\frac{m_1}{M}\dot{\vec{r}})^2]+\frac{1}{5}(m_1 s_1^2\omega_1^2+m_2 s_2^2\omega_2^2)$$

$$T=\frac{1}{2}[M\dot{\vec{R}}^2+\frac{m_1 m_2}{M}\dot{\vec{r}}^2]+\frac{1}{5}(m_1 s_1^2\omega_1^2+m_2 s_2^2\omega_2^2)$$

This lets us define a new quantity, the reduced mass: $\mu≡\frac{m_1 m_2}{M}$. We finally get

$$T=\frac{1}{2}M\dot{\vec{R}}^2+\frac{1}{2}\mu\dot{\vec{r}}^2+\frac{1}{5}(m_1 s_1^2\omega_1^2+m_2 s_2^2\omega_2^2)$$

For the total Lagrangian, taking a potential energy $U=U(r)$, then, we obtain

$$L=T-U=\frac{1}{2}M\dot{\vec{R}}^2+\frac{1}{2}\mu\dot{\vec{r}}^2+\frac{1}{5}(m_1 s_1^2\omega_1^2+m_2 s_2^2\omega_2^2)-U(r).$$ We can see that since the Lagrangian is independent of $\vec{R}$, that $M\ddot{\vec{R}}=0$ or $\dot{\vec{R}}=const.$. This tells us total momentum is conserved, our first conservation law. This is because, in the closed system, $\dot{m_1}=-\dot{m_2}$, so $\dot{M}=0$.

Since $\dot{\vec{R}}=const.$, we can move into the CM rest frame, so $\dot{\vec{R}}=0$.

$$L=\frac{1}{2}\mu\dot{\vec{r}}^2+\frac{1}{5}(m_1 s_1^2\omega_1^2+m_2 s_2^2\omega_2^2)-U(r).$$

Let $\dot{\vec{r}}^2=\dot{r}^2+r^2\dot{\phi}^2$, $\omega_i ^2=\dot{\theta}_i^2$.

$$L=\frac{1}{2}\mu(\dot{r}^2+r^2\dot{\phi}^2)+\frac{1}{5}(m_1 s_1^2\dot{\theta}_1^2+m_2 s_2^2\dot{\theta}_2^2)-U(r).$$

The Lagrangian is independent of $\phi$, so we again obtain a conservation equation.

$$\frac{d}{dt}\frac{\partial L}{\partial \dot{\phi}}=0$$ $$\frac{\partial L}{\partial \dot{\phi}}=\mu r^2\dot{\phi}=const=l_{orbit}$$

This tells us that orbital angular momentum is conserved. Apparently $$\mu r^2 \ddot{\phi} + 2 \mu r \dot{r} \dot{\phi} + \dot{\mu} r^2 \dot{\phi} = 0.$$

If you were curious, $\dot{\mu}=\frac{\dot{m}_2 (m_1-m_2)}{M}=\frac{\dot{m}_1 (m_2-m_1)}{M}$.

Again, since the Lagrangian is independent of both $\theta _1$ and $\theta _2$, each star's individual rotational angular momentum is conserved.

We can find that $$\frac{2}{5}m_i s_i ^2 \dot{\theta}_i = const = l_{rot,i}$$ and apparently $$m_i s_i ^2 \ddot{\theta}_i + 2 m_i s_i \dot{s_i} \dot{\theta}_i + \dot{m}_i s_i ^2 \dot{\theta}_i = 0,$$ giving us our last constraint equation. Solving the Lagrangian for $r$ tells us the equations of motion.

• I think this assumes from the start that there is no mass transfer between both stars. Is that correct ? – Cham Jun 13 '17 at 18:10
• No, I believe this does not assume that. The definition of center of mass and the reduced mass do not assume that $\frac{d}{dt} m=0$, and we do not take $\frac{d}{dt} m=0$ when examining the Lagrange equations. Why do you think it assumes that? I can look more closely at the assumptions. – Pawr Jun 13 '17 at 19:10
• Apparently, there's no mechanism of mass transfer, no kinetic term for it or something else in your lagrangian. I feel this is suspicious. – Cham Jun 16 '17 at 10:01