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, $\dddot Q_{ij}=\dddot I_{ij}$. In units $G=c=1$, the <quadrupole formula> gives
$$
\langle P\rangle=\frac15\left\langle\dddot Q_{ij}\dddot Q_{ij}\right\rangle.
$$
Its normalization can also be seen from the <gravitational-wave energy flux>: $h^{\mathrm{TT}}_{ij}=2\Lambda_{ij,kl}\ddot Q_{kl}/r$, where $\Lambda$ is the <transverse-traceless projector>. With $S_{ij}=\dddot Q_{ij}$ and $P_{ij}=\delta_{ij}-n_i n_j$, the projection is $S^{\mathrm{TT}}_{ij}=P_{ik}P_{jl}S_{kl}-P_{ij}P_{kl}S_{kl}/2$. For a symmetric trace-free $S$,
$$
S^{\mathrm{TT}}_{ij}S^{\mathrm{TT}}_{ij}
=S_{ij}S_{ij}-2n_iS_{ij}S_{jk}n_k+\frac12(n_iS_{ij}n_j)^2.
$$
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 $\int |S^{\mathrm{TT}}|^2d\Omega=8\pi S_{ij}S_{ij}/5$. Integrating the flux yields $\langle P\rangle=(8\pi)^{-1}\int\langle |S^{\mathrm{TT}}|^2\rangle d\Omega=\langle S_{ij}S_{ij}\rangle/5$, as above.

Now differentiate the components from part (i). With $\Delta=I_1-I_2$,
$$
\begin{aligned}
\dddot I_{11}&=4\Delta\Omega^3\sin(2\Omega t),&
\dddot I_{22}&=-4\Delta\Omega^3\sin(2\Omega t),\\
\dddot I_{12}=\dddot I_{21}&=-4\Delta\Omega^3\cos(2\Omega t),&
\dddot I_{33}&=0.
\end{aligned}
$$
The off-diagonal component occurs twice in the contraction. Thus
$$
\dddot I_{ij}\dddot I_{ij}
=32\Delta^2\Omega^6[\sin^2(2\Omega t)+\cos^2(2\Omega t)]
=32\Delta^2\Omega^6.
$$
It is already time independent, so averaging gives
$$
\boxed{\langle P\rangle=\frac{32}{5}\Omega^6(I_1-I_2)^2.}
$$
This 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.

Solution

Past exam of the mathematics course of the University of Cambridge / 2018 / iii / Paper 309 / 3 / b / iii / Solution

{title2=$P=\frac{32G}{5c^5}\Omega^6(I_1-I_2)^2$}

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.

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 Incoming links: Gravitational radiation from a rotating triaxial body