Some bits and pieces on angular momentum:

Angular momentum is that which is conserved in rotationally invariant systems, just like energy is that which is conserved in time translation invariant systems and momentum is that which is conserved in space translation invariant system. This is the essence of Noether's theorem. The analogue in QFTs are Ward-Takahashi-(like) identities, essentially generalizing the Noether conservation laws as operator equations.

The normal approach to such conservation laws in GR is to consider Killing vectors, see also this old question - and so, if you have a Killing vector fields corresponding to "rotations", you have found what angular momentum is in that theory.

Asking what angular momentum is in a setting where it is not conserved by symmetries would be rather pointless, since it would then not tell you anything about the system as such, so I do not think you should be bothered that no one writes downs a "general" form for the angular momentum, because it would not be conserved in all theories, and as such not very useful.

Discussing angular momentum in a most general GR/QFT setting is quite challenging, and there are various expositions on this. One discussing matter fields minimally coupled to GR is here - look around eq. $(57)$ for the generalization of the conservation of total angular momentum - it is the joint conservation of the "ordinary" angular momentum obtained by the antisymmetriation of the energy-momentum tensor and "spin terms".

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.

## Best Answer

Yes indeed to all your questions: mutually orbitting binaries do spin down, the system's orbital angular momentum thus decreases with time and the loss of energy and angular momentum is almost certainly owing to the emission of gravitational waves.

Look up the Hulse-Taylor binary system: its spin-down has been carefully observed and measured since its discovery in 1974 and the observed spin down carefully compared with the spindown foretold by General Relativity (one calculates, by GTR, the gravitational wave power emitted). So far, as you will quickly learn, the agreement between GTR gravitational wave power loss model and the observations has been astounding. This is considered by mainstream astronomy to be very strong evidence for the existence of gravitational waves and was the first experimental evidence for them. Early this year, direct observation of gravitational waves in the early cosmos is thought to have been made by the BICEP2 experiment when frozen ripples in the CBR.

Edit: as Warrick pointed out (thanks Warrick), a worthwhile piece of physics history to add is that Russell Hulse and Joseph Taylor were awarded the 1993 Nobel prize for physics for their analysis of the system, the wording of the prize was