In the weak-field, slow-source approximation, the leading luminosity is the <quadrupole formula>
$$
P(t)=\frac15\dddot Q_{ij}(t-R)\dddot Q_{ij}(t-R)
$$
in units $G=c=1$, where
$$
Q_{ij}=\int d^3x\,T_{00}
\left(x_ix_j-\frac13\delta_{ij}r^2\right).
$$
Only the $\ell=2$ symmetric trace-free coefficient contributes. Indeed,
$$
\int d\Omega\,\widehat x_i\widehat x_j
\widehat x_k\widehat x_l
=\frac{4\pi}{15}
(\delta_{ij}\delta_{kl}+\delta_{ik}\delta_{jl}+\delta_{il}\delta_{jk}),
$$
and contraction with traceless $a_{kl}$ removes the first term. Therefore
$$
\boxed{Q_{ij}(t)=\frac{8\pi}{15}
\int_0^\infty r^4a_{ij}(t,r)\,dr}.
$$
The power crossing the large sphere at time $t$ is consequently
$$
\boxed{
P(t)=\frac{64\pi^2}{1125}
\sum_{i,j}\left[
\int_0^\infty r^4
\frac{\partial^3a_{ij}}{\partial t^3}(t-R,r)\,dr
\right]^2}.
$$
The monopole is conserved total mass, the dipole is center-of-mass motion, and every $\ell\neq2$ term is orthogonal to the trace-free quadrupole at this leading order.

Solution

Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 309 / 2 / b / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2021/iii/paper-309/2/b/solution