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For a circular Kepler orbit, neglecting stellar spin, the orbital angular momentum of a binary star is $J=\mu a^2\Omega =\mu \sqrt{GMa}$, where $M$ is total mass, $\mu$ is the reduced mass, and $a$ is separation. Kepler's third law also gives $J=G^{2/3}{P}^{1/3}{M}_1{M}_2/[(2\pi {)}^{1/3}{M}^{1/3}]$. Distances from the center of mass are $a{M}_2/M$ and $a{M}_1/M$; adding their individual orbital angular momenta proves the formula.

Writing particle positions relative to the center of mass as ${\rho }_i=r_i-R$ makes ${\sum }_i{m}_i{\rho }_i=0$, so the cross term in the squared velocities vanishes. The total kinetic energy splits into center-of-mass translation and relative motion as displayed. For a rigid body, the latter is $\frac{1}{2}I{\omega }^2$, where $I$ is the moment of inertia about the instantaneous rotation axis through the center of mass.

Let $M=M_1+M_2$, and let $\mu =M_1{M}_2/M$ be the reduced mass. The stellar distances from the center of mass are $a_1=a{M}_2/M$ and $a_2=a{M}_1/M$. For a circular Kepler orbit, ${\Omega }^2=GM/a^3$, so summing the two orbital contributions gives the circular-binary orbital angular momentumThe separation $a$ is inside the square root. This also has the required dimensions of angular momentum.

For the homologous rotating collapse, label a fluid element by $s=r/R$ and its polar angle $\theta$. Its enclosed mass $m(s)$ is constant during stellar homology, as are the dimensionless density profile and the inertia coefficient $\alpha$. Conservation of angular momentum and solid-body rotation giveThe magnitude of the centrifugal acceleration is ${\Omega }^{2}r\sin \theta$, while the Newtonian gravitational field has magnitude $Gm(s)/r^2$. ThereforeUsing only the radial component of centrifugal force replaces $\sin \theta$ by ${\sin }^2\theta$ and leaves the scaling unchanged. The centrifugal force vanishes on the rotation axis; the central point is understood through the limiting field rather than a ratio of two zero forces.

At the equator of the cloud, critical rotation of a spherical cloud means ${\Omega }_c^2{R}_c=GM/R_c^2$, where $R_c=R_{\mathrm{crit}}$. ConsequentlyImmediately after the first fission, each daughter has mass $M/2$ and radius $a/2$ because the spheres touch. Their orbital moment of inertia is $M{a}^2/4$, while their combined spin moment of inertia is $\alpha M{a}^2/4$. With a common spin and orbital frequency ${\Omega }_f=\sqrt{GM/a^3}$, the synchronous fission model for binary formation givesThe common frequency can change during fission; corotation requires equal instantaneous frequencies, not an unchanged frequency from before the rearrangement. Equating the two values of $J$ yieldsFor a nonzero positive inertia coefficient, the necessary and sufficient condition in this algebraic model for $a<R_c$ isFor example, a moment of inertia of a uniform solid sphere has $\alpha =2/5$ and gives $a/R_c=64/49>1$; it fails the required compact-fission condition. A sufficiently centrally concentrated cloud can have smaller $\alpha$.

For the next collapse, neglect exchange of spin with the unchanged outer orbit, so each daughter's spin angular momentum is separately conserved. Immediately after the first fission that spin isFor a daughter of mass $m=M/2$ to reach critical rotation of a spherical cloud at radius $R_{c,d}$, the same relation gives $S_d=\alpha m\sqrt{Gm{R}_{c,d}}$. HenceEquivalently, its centrifugal-to-gravity ratio starts at $1/4$ when its radius is $a/2$ and reaches one after contraction by a factor four. Applying the same fission rule to this daughter produces an inner pair with separationCorotation at this second fission is local to each newly formed inner pair; it is not a single common frequency for the entire four-star hierarchy.

This spin assumption matters. If tidal locking instead kept each shrinking daughter synchronized to the fixed outer orbital period throughout the intervening collapse, its angular velocity would remain fixed. Its centrifugal-to-gravity ratio would then scale as $R_d^3$ and decrease, so it would never reach the proposed second fission. This is a counterexample to extending the corotation assumption through the whole contraction. The printed ratio describes the separately spin-conserving interpretation of the repeated fission hierarchy.

The model can generate a compact hierarchy algebraically, but is not a general physical account of multiple-star formation. In the allowed range $\alpha <1/3$, one has $a^{\prime }/a<1/8$, suggesting well-separated inner and outer scales. However, rapidly rotating gas deforms, touching daughters are tidally distorted, and pressure, gas flows, dissipation and spin-orbit torques cannot generally be ignored. The assumed identical profiles, equal mass splits, instantaneous corotation and later torque-free contractions are restrictive. Star formation can involve gravitational fragmentation and redistribution of angular momentum; the toy budget neither proves that fission occurs nor predicts the population of real multiple systems.

Take $\mathbf{r}={\mathbf{x}}_1-{\mathbf{x}}_2$, $r=|\mathbf{r}|$ and $M=M_1+M_2$. Newton's law of universal gravitation gives ${\ddot{\mathbf{x}}}_1=-G{M}_2\mathbf{r}/r^3$ and ${\ddot{\mathbf{x}}}_2=G{M}_1\mathbf{r}/r^3$. Subtracting yields the relative two-body equationThe reduced mass is $\mu =M_1{M}_2/M$. The orbital energy and angular momentum in the center of mass frame areDifferentiation givesThus both are conserved. The eccentricity vector isFor $\mathbf{h}=\mathbf{J}/\mu$, one has $\ddot{\mathbf{r}}\times \mathbf{h}/(GM)=\dot{\mathbf{r}}/r-\mathbf{r}(\mathbf{r}\cdot \dot{\mathbf{r}})/r^3=d(\mathbf{r}/r)/dt$, so $\dot{\mathbf{e}}=0$. In particular the orbital eccentricity is constant, withFor a bound Kepler orbit, $E=-GM\mu /(2a)$ and $J^2=GM{\mu }^{2}a(1-e^2)$.

For the far-field multipole expansion, set the center of mass to zero. Then ${\mathbf{x}}_1=(M_2/M)\mathbf{r}$ and ${\mathbf{x}}_2=-(M_1/M)\mathbf{r}$. At $x=|\mathbf{x}|\gg r$, expand each point-mass Newtonian gravitational potential:Repeated indices are summed. The dipole vanishes since ${\sum }_A{M}_A{\mathbf{x}}_A=0$, and ${\sum }_A{M}_A{x}_{Ai}{x}_{Aj}=\mu r_i{r}_j$. The second-order gravitational quadrupole potential of a point-mass binary is thereforeBoth tensors specified in the question are trace-free. Direct contraction givesso to the requested orderFor equal masses the cubic multipole also vanishes, but it need not vanish for unequal masses.

To evaluate the quadrupole formula, use $k=GM$, $\mathbf{v}=\dot{\mathbf{r}}$, $\mathbf{n}=\mathbf{r}/r$, $u=\dot{r}=\mathbf{n}\cdot \mathbf{v}$, and $\mathbf{w}=\mathbf{v}-u\mathbf{n}$. The acceleration and its derivative areThe third derivative of $r_i{r}_j$ is ${\dot{b}}_i{r}_j+r_i{\dot{b}}_j+3b_i{v}_j+3v_i{b}_j$. Its trace is $d^3(r^2)/d{t}^3=-2ku/r^2$. Substituting in $q_{ij}=\mu (3r_i{r}_j-r^2{\delta }_{ij})/2$ givesBecause $\mathbf{n}\cdot \mathbf{w}=0$, the cross tensor contraction is zero. The squared norms of the two bracketed pieces are $6u^2$ and $72w^2$, respectively. Thusand the instantaneous energy loss isThe parenthesis is $w^2+u^2/12\ge 0$, as required for positive radiated power. This is the instantaneous quadrupole luminosity of a Kepler binary with loss sign for the orbital energy.

A slowly evolving initially circular binary is expected to follow a quasi-circular inspiral at leading radiation order. One needs angular-momentum loss as well as energy loss for this conclusion: the rotating quadrupole has frequency $2\Omega$ and azimuthal harmonic number two, so its fluxes satisfy the gravitational-wave energy and angular-momentum balance $\dot{E}=\Omega \dot{J}$. Along the circular family,The flux relation is precisely the tangent condition for remaining on that family; equivalently it keeps $e^2=1+2E{J}^2/(G^2{M}^2{\mu }^3)$ equal to zero at leading secular order. The shrinking orbit is not an exact fixed-radius circle: its small radial drift is neglected at this adiabatic order.

For a circular orbit of instantaneous separation $a$, $u=0$ and $v^2=GM/a$. ThereforeDifferentiating $E=-GM\mu /(2a)$ at fixed masses now givesWriting $C=64G^3{M}^2\mu /(5c^5)$, the circular gravitational-wave inspiral obeys $a^4=a_0^4-4Ct$. The formal point-mass coalescence time is $a_0^4/(4C)$, and $\dot{P}/P=(3/2)\dot{a}/a$ gives a decreasing orbital period and increasing gravitational-wave frequency.

Gravitational radiation tightens a detached close binary and becomes more effective at small separation. It can bring a stellar donor into Roche-lobe overflow or drive compact objects toward merger. Finite radii, tides and mass transfer can then alter the evolution; the fixed-mass inspiral law is not the entire evolution law once these matter. Near merger the weak-field, slow-motion quadrupole formula also ceases to be sufficient.

In the center of mass frame, put the two positions at $(0,0,z(t))$ and $(0,0,-z(t))$, with $z>0$ during infall. Their separation is $2z$, so Newtonian gravity givesMultiplying by $\dot{z}$ and integrating givesThe negative square root is essential for the falling branch. All second mass moment tensor components except $I_{zz}=2m{z}^2$ vanish. Differentiating three times,Substitution yields ${\stackrel{...}{I}}_{zz}=-m^2\dot{z}/z^2$, hencewith every other component of ${\stackrel{...}{I}}_{ij}$ zero. The distance entering each acceleration is $2z$, which accounts for the factor one quarter in the equation of motion.

Let the uniform mass density be $\rho =3M/(4\pi a^3)$. In spherical polar coordinates about the rotation axis, the perpendicular distance is $r\sin \theta$ and the volume element is $r^2\sin \theta drd\theta d\varphi$. The moment of inertia is thereforeIn the given rigid body kinetic energy decomposition, $\frac{1}{2}M|\dot{\mathbf{R}}{|}^2$ is the translational kinetic energy of the whole mass moving at its center of mass velocity. The term $\frac{1}{2}I{\omega }^2$ is the rotational kinetic energy about the center of mass. The mixed term vanishes because the mass-weighted relative positions sum to zero.

For rolling without slipping, the point of contact has zero instantaneous velocity relative to the stationary surface. Its rotational velocity is opposite the center of mass velocity and has magnitude $a|\omega |$. Hence $V=a|\omega |$. Substituting this and the moment of inertia of a uniform solid sphere givesThe rotational axis is through the center of mass and parallel to the surface, perpendicular to the direction of rolling.