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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.

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.

The second mass moment tensor records the quadratic spatial distribution of mass. Its trace-free part is the mass quadrupole moment $Q_{ij}=I_{ij}-{\delta }_{ij}{I}_{kk}/3$. The mechanical moment-of-inertia tensor instead equals $J_{ij}={\delta }_{ij}{I}_{kk}-I_{ij}$. A rigid rotation $L$ transforms the second mass moment tensor as $I(t)=L(t)I(0)L(t{)}^T$.