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For a rigid body rotating around its third principal axis, the anisotropic part of its second mass moment tensor oscillates at angular frequency $2\Omega$. The quadrupole formula gives the displayed luminosity. Equal transverse principal moments remove this rotating quadrupole and its leading gravitational wave emission.

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.

Use metric signature $-+++$ and transverse-traceless gauge, $g_{ij}={\delta }_{ij}+h_{ij}^{\mathrm{TT}}$. To resolve the source's origin ambiguity, put the detector centre at $z=0$ and the source centre at $z=-r_0$, so the gravitational wave travels in the positive $z$ direction. Reversing that direction replaces $z$ by $-z$. Work in the weak-field approximation, with $\omega l\ll c$, $r_0\gg l,c/\omega$, and detector size $a\ll c/\omega ,r_0$. The radiation reaction is neglected, so the rod maintains the prescribed steady motion.

At the retarded source time $u=t-(r_0+z)/c$, take the source positions as ${\mathbf{x}}_{\pm }=\pm l(\cos (\omega u+\beta ),\sin (\omega u+\beta ),0)$. The arbitrary constant $\beta$ specifies the initial source phase. The second mass moment tensor $I_{ij}={\sum }_{A}M{x}_{Ai}{x}_{Aj}$ obeysThe quadrupole formula $h_{ij}^{\mathrm{TT}}=2G{\ddot{I}}_{ij}^{\mathrm{TT}}/(c^4{r}_0)$ projects away the transverse trace, which is already constant here. Differentiating twice therefore givesAll time and longitudinal metric perturbation components vanish in transverse-traceless gauge. The gravitational-wave frequency is $2\omega$ in angular units; $M$ is each source mass, not their sum, and the detector mass $m$ does not enter the strain. The plane wave approximation holds the amplitude fixed at $r_0$; retaining spherical spreading replaces its denominator by $r_0+z$ near the axis. The retarded quadrupole formula and the distinction between source and wave frequency are derived in ^[ref].

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 ${\delta }_{ij}{I}_{kk}-I_{ij}$.

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.

Apply the continuity equation and integrate by parts, with the stated vanishing boundary terms and finite moments. The second mass moment tensor satisfiesand differentiation again givesInsert the stress-divergence equation from part (a). The two force integrals becomeThe first term is $4K_{ij}$. Hence the magnetized-fluid tensor virial theorem isHere $\mathcal{T}$ is the volume-integrated stress, whose sign differs from some gravitational potential-energy tensor conventions. The derivation also requires the advective mass-moment surface terms to vanish; this is automatic for an isolated sufficiently decaying configuration.

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.

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.

Use geometrized units $G=c=1$ and signature $(-,+,+,+)$. The original PDF starts with the d'Alembert operator $\square {\bar{h}}_{\mu \nu }=-16\pi T_{\mu \nu }$; the local TeX incorrectly transcribes its derivative indices. The harmonic condition is the Lorenz gauge in linearized gravity.

Choose the retarded solution, excluding an incoming homogeneous wave. For a spatially localized source, the Linearized Einstein equations giveLet the source size be $D$ and the characteristic angular frequency be $\omega$. Assume the weak-field approximation, nonrelativistic source velocities, and $D\ll {\omega }^{-1}\ll r$ for a radiative far-zone measurement. Then $|\mathbf{x}-\mathbf{y}|=r-\mathbf{n}\cdot \mathbf{y}+O(D^2/r)$. At leading order in $D/r$ and $\omega D$, one may replace the denominator by $r$ and the retarded time throughout the source by $t-r$, obtaining

To identify this integral, use stress-energy conservation ${\partial }_{\mu }{T}^{\mu \nu }=0$ for a symmetric source tensor. Compact support or sufficient decay permits integration by parts with no boundary flux. Because $T^{00}=T_{00}$ in this signature, the second mass moment tensor satisfiesSpatial indices are raised with ${\delta }_{ij}$, so $T^{ij}=T_{ij}$. Substitution gives the retarded quadrupole field:The assumptions are an isolated conserved source, retarded boundary conditions, weak gravity, a source small compared with the wavelength, and observation far from it. The physical radiative field follows by the transverse-traceless projector applied to this trace-reversed metric perturbation.

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 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.

Write $I_a={\bar{I}}_{aa}$ and use the right-handed rotationIn the integral defining the second mass moment tensor, change variables to $\mathbf{y}=L^{-1}\mathbf{x}$. The determinant is one and $x_i=L_{ik}{y}_k$, givingSet $C=(I_1+I_2)/2$ and $\Delta =I_1-I_2$. Multiplication givesIn particular, $I_{kk}=I_1+I_2+I_3$ is constant. These expressions also show how the rotating anisotropy enters the mass quadrupole moment.

A point mass idealizes a localized body's mass density as $\rho (\mathbf{x},t)=m{\delta }^{(3)}(\mathbf{x}-\mathbf{z}(t))$, using the Dirac delta function. Its second mass moment tensor is $I_{ij}=m{z}_i{z}_j$. The approximation ignores its internal size and structure.

The scalar second mass moment is the trace of the second mass moment tensor. The moment of inertia about a specified axis instead integrates squared distance from that axis. For an isotropic distribution each axial moment of inertia is $2I/3$; the sum over three orthogonal axes is $2I$.