With $u=t-r$, direct integration gives
$$
I_{xx}=\frac{ma^2}{3}\cos^2\omega u,
\quad
I_{yy}=\frac{ma^2}{3}\sin^2\omega u,
\quad
I_{xy}=\frac{ma^2}{3}\sin\omega u\cos\omega u,
\quad
I_{zz}=0.
$$
Therefore the time-dependent spatial field is
$$
\boxed{
\bar h_{ij}^{\rm rad}=\frac{4ma^2\omega^2}{3r}
\begin{pmatrix}
-\cos2\omega u&-\sin2\omega u&0\\
-\sin2\omega u&\cos2\omega u&0\\
0&0&0
\end{pmatrix}}.
$$
The <gravitational-wave frequency> is $2\omega$. On the positive $z$-axis this matrix is transverse and traceless. It is a rotating combination of the <plus polarization> and <cross polarization>, with amplitude proportional to $1/z$, exactly as for a plane wave propagating in the $z$ direction.

Solution

Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 309 / 3 / b / i / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2022/iii/paper-309/3/b/i/solution