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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.

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.

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 $2I$.