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A Newtonian binary chirp is a leading-order model in which loss of orbital energy to gravitational radiation makes an almost circular binary system shrink and its gravitational-wave frequency rise. Here is a derivation with explicit SI units. Let the constant component masses $m_1,m_2$ be in kilograms, $M=m_1+m_2$, $\mu =m_1{m}_2/M$, separation $a$ in metres, and orbital angular frequency $\Omega$ in radians per second. Newtonian gravity, circular motion and the quadrupole formula giveThe factor $\pi$ rather than $2\pi$ appears because the dominant mass quadrupole moment repeats twice per orbit. Since $dE/da>0$, radiative energy loss implies $\dot{a}<0$, henceThe quantity $G\mathcal{M}/c^3$ is a time in seconds. Thus $K$ has units ${\mathrm{s}}^{5/3}$ and $K{f}^{11/3}$ has units ${\mathrm{s}}^{-2}$, as required for frequency change per second. This model assumes weak gravity, slow relative motion, negligible eccentricity of a conic, no spin corrections and adiabatic evolution over an orbit. It does not describe merger or ringdown.

Separating variables gives a finite formal coalescence time $t_c$ and accumulated wave phase $\Phi$:Indeed $d\Phi /dt=2\pi f$; the formal divergence at $t_c$ signals extrapolation outside the approximation. Over a finite frequency interval, the number of gravitational wave cycles is $N=3(f_0^{-5/3}-f_1^{-5/3})/(5K)$. This accumulated phase is why a chirp mass can be constrained even when distance and inclination remain entangled.

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.

Set $\mathbf{r}={\mathbf{x}}_1-{\mathbf{x}}_2$, $M=M_1+M_2$ and $\mu =M_1{M}_2/M$, the reduced mass. Newton's law of universal gravitation gives ${\ddot{\mathbf{x}}}_1=-G{M}_2\mathbf{r}/r^3$ and ${\ddot{\mathbf{x}}}_2=G{M}_1\mathbf{r}/r^3$. Subtraction establishesFor fixed masses, the orbital energy and angular momentum areTaking derivatives gives $\dot{E}=\mu \mathbf{v}\cdot (\ddot{\mathbf{r}}+GM\mathbf{r}/r^3)=0$ and $\dot{\mathbf{J}}=\mu \mathbf{r}\times \ddot{\mathbf{r}}=0$. These are conservation of energy and conservation of angular momentum in the unperturbed Kepler orbit. To see directly that the orbital eccentricity is fixed, put $\mathbf{h}=\mathbf{r}\times \mathbf{v}$ and $\mathbf{n}=\mathbf{r}/r$. The eccentricity vector isSince $\dot{\mathbf{h}}=0$, the cross product identity $\mathbf{r}\times (\mathbf{r}\times \mathbf{v})=\mathbf{r}(\mathbf{r}\cdot \mathbf{v})-r^2\mathbf{v}$ gives $\ddot{\mathbf{r}}\times \mathbf{h}=GM\dot{\mathbf{n}}$, so $\dot{\mathbf{e}}=0$. Its magnitude $e$ is the orbital eccentricity. Equivalently,Thus the semi-major axis and orbital eccentricity of a bound noncollision Kepler orbit are constant, including $e=0$.

Take the centre of mass as origin. The two positions are ${\mathbf{x}}_1=(M_2/M)\mathbf{r}$ and ${\mathbf{x}}_2=-(M_1/M)\mathbf{r}$. For $x\gg r$, a Taylor expansion givesThe mass dipole vanishes because ${\sum }_A{M}_A{\mathbf{x}}_A=0$, and the second mass moment tensor is ${\sum }_A{M}_A{x}_{Ai}{x}_{Aj}=\mu r_i{r}_j$. The gravitational quadrupole potential of a point-mass binary is consequentlyBoth printed tensors are trace-free. Contracting them with the Kronecker delta summed over three spatial dimensions givesThis proves the requested expression through second order. In particular the printed convention is $q_{ij}=\frac{3}{2}{Q}_{ij}$, where $Q_{ij}=\mu (r_i{r}_j-r^2{\delta }_{ij}/3)$ is the standard mass quadrupole moment; the radiation coefficient must be adjusted accordingly. No mass dipole term may be retained in centre of mass coordinates.

To evaluate the quadrupole formula without assuming a circular orbit, put $k=GM$, $u=\dot{r}=\mathbf{n}\cdot \mathbf{v}$ and ${\mathbf{v}}_{\perp }=\mathbf{v}-u\mathbf{n}$. Differentiating the gravitational acceleration gives the kinematic jerkFor $S_{ij}=r_i{r}_j$, the product rule then givesInserting these into the printed mass quadrupole moment convention yieldsThe two tensors in brackets are orthogonal in their tensor contraction, because $\mathbf{n}\cdot {\mathbf{v}}_{\perp }=0$. Their squared norms are $6$ and $2v_{\perp }^2$, respectively. HenceThe instantaneous quadrupole luminosity of a Kepler binary is thereforeThe radial derivative $\dot{r}$ is not the vector speed. The bracket equals $v_{\perp }^2+u^2/12$, so the radiated luminosity is nonnegative for every instantaneous velocity.

For a slowly evolving circular orbit, there is no preferred orbital phase at which to excite a persistent eccentricity vector. More precisely, the rotating mass quadrupole moment emits at twice the orbital frequency, with angular harmonic number two; its gravitational-wave energy and angular-momentum balance is $\dot{E}=\Omega \dot{J}$. The circular Kepler orbit sequence has $J=\mu \sqrt{GMa}$ and $dE/dJ=\Omega$, so this loss is tangent to the sequence. At leading adiabatic order it preserves zero secular orbital eccentricity. This is a circular gravitational-wave inspiral, with a small radial drift rather than an exactly fixed-radius Newtonian circle. The energy flux alone would not establish circularity without this symmetry and angular momentum balance.

To leading radiation order use $r=a$, $u=0$ and ${\mathbf{v}}^2=GM/a$ in the loss formula. Combining it with $E=-GM\mu /(2a)$ givesKeeping the fixed masses and leading quadrupole formula, integration givesThis formal coalescence time displays the strong $a_0^4$ dependence: gravitational-wave emission from a binary system matters far more for close binary stars than for wide ones. The orbital period decreases as $a^{3/2}$ and the gravitational-wave frequency increases, producing a chirp. A detached binary can be driven into Roche-lobe overflow; further evolution then depends on mass-transfer stability. A sufficiently close compact pair may merge. The weak-field, slow-motion, point-mass approximation ceases to apply before literal $a=0$; finite stellar radii or strong relativistic effects determine the final interaction.

The time-dependent components of the second mass moment tensor contain $\sin (2\Omega t)$ and $\cos (2\Omega t)$. Taking two time derivatives for the retarded quadrupole field leaves that frequency unchanged. Hence the gravitational-wave frequency isHere ${\omega }_{\mathrm{GW}}$ is angular frequency and $f_{\mathrm{GW}}$ counts cycles per unit time. If $I_1=I_2$, the mass quadrupole moment is time independent, so the leading gravitational wave amplitude vanishes and there is no emitted quadrupole frequency to measure.

With $u=t-r$, direct integration givesTherefore the time-dependent spatial field isThe gravitational-wave frequency is $2\omega$. On the positive $z$-axis this matrix is transverse and traceless. It is a rotating combination of the plus polarization and cross polarization, with amplitude proportional to $1/z$, exactly as for a plane wave propagating in the $z$ direction.