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Two equal masses $m$ orbiting their centre of mass at radius $R$ have separation $2R$. Newton's law of universal gravitation gives the displayed orbital frequency. Their mass quadrupole moment varies at twice that frequency, producing gravitational radiation with two gravitational wave polarizations. For inclination $\iota$, its leading plus and cross amplitudes are proportional to $1+{\cos }^2\iota$ and $2\cos \iota$, respectively. Face-on radiation is circularly polarized; edge-on radiation is linearly polarized.

For two masses $M$ at opposite points of a circular orbit, each at radius $R$ from the centre of mass, ${\Omega }^2=GM/(4R^3)$. The trace-free mass quadrupole moment oscillates at twice the orbital frequency, and its third-derivative squared contraction is $128M^2{R}^4{\Omega }^6$. The quadrupole formula gives the displayed phase-independent luminosity. Using separation $2R$ rather than individual radius $R$ avoids a factor-of-32 error.

In units $c=1$, a slowly moving isolated source has trace-free mass quadrupole moment $Q^{ij}=\int T^{00}(x^i{x}^j-{\delta }^{ij}|\mathbf{x}{|}^2/3)d^{3}x$. The retarded linearized gravity solution and the stress-energy quadrupole identity give the displayed physical spatial strain, with $\mathbf{n}=\mathbf{x}/r$ and the transverse-traceless projector $\Lambda$. If the spatial metric is written $g_{ij}=-{\delta }_{ij}-2E_{ij}$ for metric signature $(+---)$, its trace-free Lorenz-gauge coefficient before the transverse projection is $E_{ij}=G{\ddot{Q}}_{ij}(t-r)/r$. Thus $H^{\mathrm{TT}}=2E^{\mathrm{TT}}$. Far distance alone does not permit dropping source-dependent retardation: the quadrupole approximation also requires the source size to be small compared with the radiation wavelength.

For a binary star viewed far from its centre of mass, let $\mathbf{r}$ be the separation and $\mu$ the reduced mass, $M=M_1+M_2$ the total mass, $r=|\mathbf{r}|$ and $x=|\mathbf{x}|$. The Taylor expansion of the point-mass Newtonian gravitational potential has zero dipole because ${\sum }_A{M}_A{\mathbf{x}}_A=0$. Its second moment is ${\sum }_A{M}_A{x}_{Ai}{x}_{Aj}=\mu r_i{r}_j$, soThis holds for an instantaneous separation in any orbit, not only a circular orbit. The mass quadrupole moment convention $Q_{ij}=\mu (r_i{r}_j-r^2{\delta }_{ij}/3)$ gives ${\Phi }_2=-3G{Q}_{ij}{x}_i{x}_j/(2x^5)$.

A gravitational wave is a propagating disturbance of spacetime curvature produced by changing mass-energy multipole moments. In general relativity, its two transverse gravitational wave polarizations alternately stretch and squeeze freely falling separations. The leading radiation from a slowly moving source is governed by its mass quadrupole moment, rather than by its total mass alone.

The experimental breakthrough was GW150914: observed on 14 September 2015 and published on 11 February 2016. The LIGO Scientific Collaboration and Virgo Collaboration reported a merging pair of black holes, using the two LIGO detectors at Hanford and Livingston; Virgo was being upgraded and did not observe this event.^[ref]

The 2017 Nobel Prize in Physics recognized Rainer Weiss, Barry C. Barish and Kip S. Thorne for contributions to LIGO and gravitational wave observation. Rainer Weiss developed interferometric detection and its noise analysis; Barry C. Barish led the project into construction and operation; Kip S. Thorne developed its theoretical and astrophysical foundations. The experiment depended on a large collaboration.^[ref]

For a slowly evolving circular binary star with separation $a$, reduced mass $\mu$ and orbital angular frequency $\Omega$, the leading gravitational wave losses satisfy $\dot{E}=\Omega \dot{J}$. One direct check uses the standard quadrupole angular-momentum loss lawwhere $Q$ is the mass quadrupole moment and ${ϵ}_{ijk}$ is the Levi-Civita symbol. For $\mathbf{r}=a(\cos \Omega t,\sin \Omega t,0)$, put $A=\mu a^2$; then $Q_{xx}=A(1/6+\cos (2\Omega t)/2)$, $Q_{yy}=A(1/6-\cos (2\Omega t)/2)$, $Q_{xy}=A\sin (2\Omega t)/2$, and $Q_{zz}=-A/3$. Substitution gives ${\dot{J}}_z=-32G{A}^2{\Omega }^5/(5c^5)$ and the quadrupole formula gives $\dot{E}=-32G{A}^2{\Omega }^6/(5c^5)$. Their ratio proves the balance relation. The same relation follows from the rotating wave's angular harmonic number two and frequency $2\Omega$. The additional angular-momentum flux law, not the energy flux alone, is needed to justify preservation of secular circularity.

For two equal point masses at $\pm z$ falling from rest at $\pm z_0$ under Newtonian gravity, in geometrized units one has $\ddot{z}=-m/(4z^2)$ and $\dot{z}=-\sqrt{m(1/z-1/z_0)/2}$. The only nonzero component of the second mass moment tensor is $I_{zz}=2m{z}^2$, and its third derivative is $-m^2\dot{z}/z^2$. The trace-free mass quadrupole moment and quadrupole formula give the displayed power. From infinity, integrating down to $z=z_c$ gives $E=2\sqrt{2}{m}^{9/2}/(105z_c^{7/2})$. Relative to the total initial mass $2m$ the fraction is $\sqrt{2}(m/z_c{)}^{7/2}/105$. The weak-field slow-motion approximation must be distinguished from its extrapolation to a compact endpoint.

Spatial moments of a mass density distribution organize its far gravitational field. Symmetric trace-free parts isolate angular multipoles; the second moment gives the mass quadrupole moment. In slow-motion linearized gravity, changes of the quadrupole produce the leading gravitational radiation; the conserved monopole and centre-of-mass dipole do not radiate independently.

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.

The changing mass quadrupole moment produces an inspiral gravitational wave signal whose frequency and amplitude increase toward merger. The chirp mass controls the leading frequency evolution, while higher-order phasing and the merger carry information about binary mass ratio and spins. Neutron stars also introduce tidal effects. These signals motivate a search with an interferometer such as LIGO; the supplied ranges quantify the paper's proposed detection prospects, without assuming any particular later observing-run performance.

No, for a uniform or otherwise axisymmetric disc in steady rotation. Although individual material elements accelerate, the mass density and mass current at every spatial point remain constant, and the mass quadrupole moment is time independent. There is no gravitational radiation. Axisymmetry of the material distribution is necessary for this conclusion: the source does not explicitly specify the disc's density. A counterexample to an unconditional answer is a disc with two equal dense spots at opposite radii; their rotating mass quadrupole moment produces gravitational waves at twice the rotation angular frequency.

No, for steady rotation of the uniform sphere. Its mass quadrupole moment is zero by spherical symmetry; its rotating mass current is stationary when the rotation axis and rate are fixed. A stationary source does not emit gravitational waves. Rotation by itself is insufficient: a time-dependent nonaxisymmetric distribution, or another changing radiative multipole moment, is required. The conclusion concerns a sphere with the stated constant density and steady motion, rather than a deformed or time-varying rotator.

Yes, for genuine precession of a nonspherical disc. Let its unit normal be $\mathbf{n}=(\sin \alpha \cos \omega t,\sin \alpha \sin \omega t,\cos \alpha )$. Its trace-free mass quadrupole moment has the axisymmetric formThe $xz,yz$ entries vary at $\omega$, and $xx-yy,xy$ at $2\omega$. Their nonzero second and third time derivatives give an oscillating far-field metric perturbation and positive average radiated power. For $\alpha =\pi /2$ the $\omega$ terms vanish but the $2\omega$ terms remain. The precise condition is $\omega \ne 0$ and $\sin \alpha \ne 0$: the literal inequality $\alpha \ne 0$ also allows $\alpha =\pi$, where the normal is fixed and no such gravitational radiation occurs. For a uniform thin disc of mass $M_d$ and radius $b$, $Q_0=-M_d{b}^2/4$, confirming that it is not zero.

Let ${\gamma }_{\alpha \beta }={\delta }_{\alpha \beta }$ be the positive Euclidean spatial background metric. Write a tensor-sector metric perturbation as $h_{00}=h_{0\alpha }=0$, $h_{\alpha \beta }=-2E_{\alpha \beta }$, with ${\gamma }^{\alpha \beta }{E}_{\alpha \beta }=0$ and ${\partial }^{\alpha }{E}_{\alpha \beta }=0$. Spatial indices here are raised with $\gamma$, and $\square ={\partial }_t^2-\Delta$.

Substitute the linearized Levi-Civita connection into the cover's curvature definition. To first order,In the transverse-traceless tensor sector $h=0$ and ${\partial }^c{h}_{ac}=0$, so $\delta C=0$ andThe tensor part of the Linearized Einstein equations therefore givesIn a calculation strictly confined to the tensor sector, the source on this equation means its tensor projection; it is already trace-free there.

For a generic compact source, trace-free is not the same as transverse-traceless. For example, the static compactly supported spatial stress $T^{ij}={\partial }^i{\partial }^j\chi -{\gamma }^{ij}\Delta \chi$, with smooth compactly supported $\chi$, is divergence-free, but its trace-free part has divergence $2{\partial }^j\Delta \chi /3$, generally nonzero. There is a precise way to reproduce the unprojected trace-free formula used in the question while retaining this distinction. In Lorenz gauge in linearized gravity, the trace-reversed metric perturbation ${\bar{h}}_{ab}=h_{ab}-{\eta }_{ab}h/2$ obeysDefine its spatial trace-free half-strain byThen its wave equation has exactly the displayed trace-free source $\delta {\hat{T}}^{\alpha \beta }=\delta T^{\alpha \beta }-{\gamma }^{\alpha \beta }\delta T^{\mu }{}_{\mu ,\mathrm{spatial}}/3$. The physical tensor radiation is obtained afterwards with the transverse-traceless projector. Without that projection, this $E$ is a Lorenz-gauge trace-free coefficient, not necessarily a transverse tensor mode. At leading far-zone order, the same directional projector is applied to both sides of the quadrupole field formula below.

The flat contracted Bianchi identity implies stress-energy conservation, ${\partial }_b\delta T^{ab}=0$. LetCompact support removes boundary terms. Differentiate and integrate by parts once:A second time derivative, another integration by parts and symmetry of the stress-energy tensor giveTake the spatial trace-free part. The mass quadrupole moment satisfies the stress-energy quadrupole identityInserting this into the assumed leading retarded quadrupole field givesThis is the requested spatial half-strain convention; the measured transverse-traceless strain is $H_{\alpha \beta }^{\mathrm{TT}}=2E_{\alpha \beta }^{\mathrm{TT}}$. The formula selects the retarded source field and does not include an independently added homogeneous gravitational wave. Dropping the source-dependent part of retardation also needs the long-wavelength source approximation, besides $R/r\ll 1$.

For a slowly moving weakly self-gravitating source, the virial theorem or the dynamical balance $v^2/R\sim GM/R^2$ gives $v^2\sim ϵ=GM/R$. A varying nonspherical mass quadrupole moment has characteristic size $M{R}^2$ and dynamical time $\tau \sim R/v$, so $\ddot{Q}\sim M{R}^2/{\tau }^2\sim M{v}^2$. Consequently the quadrupole strain scaling of a weak self-gravitating source isThis is a characteristic scale, not a nonzero lower bound: a stationary or exactly spherical source does not radiate. Slow motion gives $R/\tau \sim \sqrt{ϵ}\ll 1$, justifying the long-wavelength approximation; for a radiation-zone strain one also takes $r\gg \tau$.

Let $I^{ij}(t)=\int T^{00}(t,\mathbf{x})x^i{x}^j{d}^{3}x$, using spatial indices $i,j$. Stress-energy conservation, symmetry of the stress-energy tensor and integration by parts giveand a second derivative givesAll boundary terms vanish by compact support. This proves the stress-energy quadrupole identity,The support is understood to remain bounded over the time interval used for differentiating these integrals.

For the radiation calculation set $c=1$, write $g_{ab}={\eta }_{ab}+h_{ab}$ and use the trace-reversed metric perturbation ${\bar{h}}_{ab}=h_{ab}-{\eta }_{ab}h/2$ in Lorenz gauge in linearized gravity. With the curvature and field-equation signs in this paper, the linear equation isThe retarded solution with no incoming radiation is consequentlyLet $r=|\mathbf{x}|$ be the distance from the source's centre of mass, and let $d$ be its size. In the radiation zone, $r\gg d$. The leading multipole expansion additionally assumes slow internal motion, or $d$ small compared with the radiation wavelength. These assumptions allow $|\mathbf{x}-{\mathbf{x}}^{\prime }|\simeq r$ in the denominator and the common retarded time $u=t-r$ at leading quadrupole order. ThusDefine the trace-free mass quadrupole moment byTo make the requested normalization explicit, write the trace-free spatial metric coefficient as $h_{ij}^{\mathrm{TF}}=-2E_{ij}$. Trace reversal only adds a spatial trace, so the preceding equations give the far-zone quadrupole strain coefficientThe supplied scan does not contain the referenced perturbation handout. The definition above fixes the normalization of $E$ directly. If $E$ denotes the physical transverse-traceless tensor in that handout, the right-hand side is also to be transversely projected for the observation direction. Trace-free alone does not imply transverse. With $\mathbf{n}=\mathbf{x}/r$, putThe transverse-traceless projector gives $E_{ij}^{\mathrm{TT}}={\Lambda }_{ij,kl}{E}_{kl}$ and the physical spatial strain is $H_{ij}^{\mathrm{TT}}=2E_{ij}^{\mathrm{TT}}$. This also states the result without relying on omitted handout notation. The retarded construction and direction-dependent transverse projection are described in Carroll's general relativity lectures, with the appropriate changes of metric-sign convention.

For the equal-mass binary star, its centre of mass is at the origin and the separation is $2R$. The gravitational force on either star is $G{m}^2/(2R{)}^2$, while its centripetal force is $mR{\omega }^2$. HenceAt leading Newtonian order, $T^{00}$ is the sum of the two point-mass densities. Its moments areso the trace-free mass quadrupole moment isThe Newtonian bound system is treated in the usual leading weak-field quadrupole approximation; an exactly conserved effective source also includes the interaction stresses. Accelerated point-mass dust alone would not satisfy the conservation hypothesis.

Differentiating twice and evaluating at $u=t-r$ givesFor an observer on the $z$ axis this is already transverse and traceless. Its alternating plus polarization and cross polarization have a phase difference of $\pi /2$, giving circularly polarized gravitational radiation at twice the orbital frequency. The amplitude decays as $1/r$, as expected for outgoing radiation in three spatial dimensions.

More generally take the line of sight in the $xz$ plane at inclination $\iota$ to the orbital angular-momentum axis. In the transverse basis ${\mathbf{e}}_{\theta }=(\cos \iota ,0,-\sin \iota )$, ${\mathbf{e}}_{\varphi }=(0,1,0)$, define $A=4Gm{R}^2{\omega }^2/r$. Projection gives the physical gravitational wave polarizationsThus a generic view is elliptically polarized, the face-on view is circularly polarized and the edge-on view is linearly polarized. This is circular-binary quadrupole radiation; an overall sign or transverse-axis rotation only changes the polarization convention.

The separation of the stars is $2R$. Newton's law of universal gravitation and circular acceleration giveTo leading nonrelativistic order, the mass density is the sum of the two translated point-mass Dirac delta distributions. Thuswith all components involving $z$ zero. The trace is $2M{R}^2$, a constant. Define the trace-free mass quadrupole moment $Q_{ij}=I_{ij}-{\delta }_{ij}{I}_{kk}/3$; its third derivatives equal those of $I_{ij}$. If $K=8M{R}^2{\Omega }^3$, thenBoth off-diagonal entries count in the contraction. Therefore ${\stackrel{...}{Q}}_{ij}{\stackrel{...}{Q}}_{ij}=2K^2=128M^2{R}^4{\Omega }^6$, independent of phase. The quadrupole formula gives the equal-mass circular-binary quadrupole luminosityRestoring units, ${\Omega }^2=GM/(4R^3)$ andHere $R$ is each star's radius about the centre of mass, not the separation. The rest-frame prescription supplies the leading mass density; a moving star does not still have zero momentum density and spatial stress. The calculation uses the assumed quadrupole formula and the Newtonian orbit rather than imposing those rest-frame zeros on the moving binary.

At fixed masses, the equal-mass circular-binary quadrupole luminosity grows as $R^{-5}$. The binary's Newtonian binding energy is $-G{M}^2/(4R)$, so reducing the separation increases both the binding and the radiated power. Equivalently, its luminosity scaling iswhich makes the importance of orbital compactness explicit.

Ordinary extended stars cannot remain separate at very small orbital radii: contact, mass transfer and tidal disruption intervene. A neutron star or black hole can remain a compact orbiting object down to separations of order a few gravitational radii, allowing high orbital speeds, rapidly changing mass quadrupole moments and strong gravitational waves. Thus compact, tightly bound binaries are especially efficient emitters. The Newtonian quadrupole formula explains the scaling; precision predictions near merger require relativistic dynamics, where that approximation itself ceases to be reliable.

Here $I_{ij}$ denotes the unreduced second mass moment tensor, distinct from its trace-free mass quadrupole moment. In coordinates referred to the chosen origin,This expression uses $c=1$. In SI units the leading nonrelativistic mass density is $T^{00}/c^2$. For slowly moving point masses, $T^{00}\simeq {\sum }_A{m}_{(A)}{\delta }^{(3)}(\mathbf{y}-{\mathbf{y}}_{(A)}(t))$, soKinetic corrections to the energy density are higher order in the velocity. The second mass moment tensor is not the mechanical moment of inertia tensor, which instead has components ${\delta }_{ij}{I}_{kk}-I_{ij}$.

The trace-free mass quadrupole moment is diagonal:The same constant factors multiply its third derivatives, giving ${\stackrel{...}{Q}}_{ij}{\stackrel{...}{Q}}_{ij}=(2/3)({\stackrel{...}{I}}_{zz}{)}^2$. The quadrupole formula and the preceding infall result therefore giveFor infall from infinity, $P=m^5/(15z^5)$ and $|\dot{z}|=\sqrt{m/(2z)}$. To find the emitted energy, integrate power over time, using $dt=|dz|/|\dot{z}|$:The total initial mass is $2m$, so the radiated fraction is $1/840\simeq 1.19\times 10^{-3}$. The Sun would emit that fraction of its rest energy inunder the stipulated constant luminosity. This is the head-on quadrupole radiation from equal masses prediction with the prescribed stopping rule. At the endpoint $|\dot{z}|=1/2$ and $m/z=1/2$, so extending the slow-motion weak-field formula that far is an extrapolation, not a controlled strong-field prediction.

Take the oscillation to be $z(t)=L\cos \omega t$; a different initial phase does not affect a period average. In the mass-density normalization of the quadrupole formula, the second mass moment tensor of the point mass has only $I_{33}=m{z}^2$ nonzero. Its trace-free mass quadrupole moment is thereforeThe constant part does not radiate. Since $d^3({\cos }^2\omega t)/d{t}^3=4{\omega }^3\sin 2\omega t$, the tensor contraction isUsing $⟨{\sin }^2(2\omega t)⟩=1/2$ in the quadrupole formula gives the averaged positive radiated powerThe mass quadrupole moment oscillates at twice the source's angular frequency, and the source energy loss has the opposite sign. The expression uses the leading slow-motion weak-field approximation, $\omega L\ll c$. With SI stress-energy tensor components, the mass-density integrand is $T_{00}/c^2$; the displayed $c^{-5}$ quadrupole formula uses that mass normalization. This is the point-mass contribution requested by the model: an apparatus maintaining an accelerating mass would also contribute its own mass quadrupole moment.