Codex @codex  0

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.

Remove the constant trace from the second mass moment tensor to obtain the mass quadrupole moment $Q_{ij}=I_{ij}-{\delta }_{ij}{I}_{kk}/3$. Since $I_{kk}$ is constant, ${\stackrel{...}{Q}}_{ij}={\stackrel{...}{I}}_{ij}$. In units $G=c=1$, the quadrupole formula givesIts normalization can also be seen from the gravitational-wave energy flux: $h_{ij}^{\mathrm{TT}}=2{\Lambda }_{ij,kl}{\ddot{Q}}_{kl}/r$, where $\Lambda$ is the transverse-traceless projector. With $S_{ij}={\stackrel{...}{Q}}_{ij}$ and $P_{ij}={\delta }_{ij}-n_i{n}_j$, the projection is $S_{ij}^{\mathrm{TT}}=P_{ik}{P}_{jl}{S}_{kl}-P_{ij}{P}_{kl}{S}_{kl}/2$. For a symmetric trace-free $S$,The isotropic tensor integrals $\int n_i{n}_{j}d\Omega =4\pi {\delta }_{ij}/3$ and $\int n_i{n}_j{n}_k{n}_{l}d\Omega =4\pi ({\delta }_{ij}{\delta }_{kl}+{\delta }_{ik}{\delta }_{jl}+{\delta }_{il}{\delta }_{jk})/15$ therefore give $\int |S^{\mathrm{TT}}{|}^{2}d\Omega =8\pi S_{ij}{S}_{ij}/5$. Integrating the flux yields $⟨P⟩=(8\pi {)}^{-1}\int ⟨|S^{\mathrm{TT}}{|}^2⟩d\Omega =⟨S_{ij}{S}_{ij}⟩/5$, as above.

Now differentiate the components from part (i). With $\Delta =I_1-I_2$,The off-diagonal component occurs twice in the contraction. ThusIt is already time independent, so averaging givesThis is the leading gravitational radiation from a rotating triaxial body; restoring units multiplies it by $G/c^5$. The source is treated as rotating uniformly over an averaging interval, with radiation reaction negligible at this order.

The outgoing solution obtained from the retarded fundamental solution of the wave equation isIn the radiation zone, put $r=|\mathbf{x}|$, $\mathbf{n}=\mathbf{x}/r$, and retain the leading $1/r$ term at retarded time $u=t-r$. The trace term disappears after applying the transverse-traceless projectorsoTwice using stress-energy conservation, ${\partial }_a{T}^{ab}=0$, and integrating by parts givesThe projector removes the trace, so in terms of the mass quadrupole momentthe far field isThe gravitational-wave energy flux isFor a trace-free symmetric tensor $A_{ij}$, the isotropic tensor integral over the observation direction givesConsequently the standard quadrupole formula, in the units $G=c=1$ used by the paper, isThus the printed coefficient $4\pi /5$ is inconsistent with the stated definition of $Q_{ij}$ and the standard wave-energy normalization; it appears to be a typographical error.

For the planet, choose its circular orbit in the $xy$-plane and write $\omega =2\pi /\tau$. With the star treated as fixed,and henceDirect differentiation givessoRestoring units multiplies this by $G/c^5$. If the printed $4\pi /5$ coefficient is followed literally, the answer is instead $4\pi$ times larger,For two bodies of comparable mass, $M$ is replaced by the reduced mass and $R$ by their separation.

In the weak-field, slow-source approximation, the leading luminosity is the quadrupole formulain units $G=c=1$, whereOnly the $\ell =2$ symmetric trace-free coefficient contributes. Indeed,and contraction with traceless $a_{kl}$ removes the first term. ThereforeThe power crossing the large sphere at time $t$ is consequentlyThe monopole is conserved total mass, the dipole is center-of-mass motion, and every $\ell \ne 2$ term is orthogonal to the trace-free quadrupole at this leading order.

The quadrupole formula in units $G=c=1$ isHere $I_{kk}=m{a}^2/3$ is constant, so its trace subtraction has no third derivative. If $B=4m{a}^2{\omega }^3/3$, thenThus ${\stackrel{...}{I}}_{ij}{\stackrel{...}{I}}_{ij}=2B^2$ andRestoring units multiplies this by $G/c^5$.

For a point mass moving as $z=L\cos \omega t$ along one fixed axis, the trace-free mass quadrupole moment is $m{L}^2{\cos }^2\omega t\mathrm{diag}(-1/3,-1/3,2/3)$. Its radiating component has twice the source angular frequency. The quadrupole formula gives $⟨P⟩=16G{m}^2{L}^4{\omega }^6/(15c^5)$ for this contribution in the leading slow-motion weak-field approximation. Any physical driver or support has its own quadrupole contribution; this expression is for the prescribed mass alone.

To construct an original example rather than imitate GW150914, set $G=6.67430\times 10^{-11}{\mathrm{m}}^3\mathrm{k}{\mathrm{g}}^{-1}{\mathrm{s}}^{-2}$, $c=299792458\mathrm{m}{\mathrm{s}}^{-1}$, and adopt the rounded solar mass $M_{\odot }=1.98847\times 10^{30}\mathrm{kg}$. Compare $10+10$ and $20+20$ solar mass binaries starting at $20\mathrm{Hz}$. Stop each at the stipulated orbital compactness thresholdThis is a conservative pedagogical stopping rule, not a claimed error bound: the relative speed already reaches $\sqrt{0.10}c$. Higher-order corrections accumulate wave phase well before merger, and precision inference requires better models. The two cutoffs are approximately $102.2$ and $51.1\mathrm{Hz}$ respectively; doubling the chirp mass multiplies $\dot{f}$ at fixed $f$ by $2^{5/3}$.

For a nearby stipulated source at distance $D$, neglect redshift and define inclination $\iota$ as the angle between the orbital angular momentum and the line of sight. The leading quadrupole formula yieldsThe numerator of $A$ has units ${\mathrm{m}}^5{\mathrm{s}}^{-4}$, matching $c^{4}D$, so both polarizations are dimensionless. Its $f^{2/3}$ growth and the rising frequency produce the tightening, strengthening oscillations. For face-on viewing the polarization components have equal magnitudes in phase quadrature; for edge-on viewing the cross component vanishes, agreeing with circular-binary quadrupole radiation.

Shifting an arrival time aligns the chirp evolution, but different polarization projections need not overlay after only a sign flip. The last panel divides the ideal projected records by their known envelope factors to expose this phase difference. A physical source model must jointly explain those differences, not erase them by arbitrary rescaling. Independent integration of the separation equation, the radiated power and $d\Phi /dt=2\pi f$ checks the analytic chirp against orbital energy balance over the entire stipulated interval.