The phase is
$$
p_\rho x^\rho=-\omega t+\omega z=-\omega u.
$$
Because the <transverse-traceless gauge> perturbation has no component along
$$
k=\frac1{\sqrt2}(\partial_t-\partial_z),
$$
only the third term of the <linearized Riemann curvature operator> contributes to the required contraction:
$$
\Psi_4
=-\frac12(k^\rho\partial_\rho)^2
\left(h_{\mu\nu}\bar m^\mu\bar m^\nu\right).
$$
The derivatives and transverse polarization contraction are
$$
k^\rho\partial_\rho e^{-i\omega u}
=-i\sqrt2\,\omega e^{-i\omega u},
$$
$$
H_{\mu\nu}\bar m^\mu\bar m^\nu
=H_+-iH_\times,
\qquad
\bar m=\frac1{\sqrt2}(\partial_x-i\partial_y).
$$
Therefore the <Newman--Penrose scalar Psi4> is
$$
\boxed{
\Psi_4
=\omega^2(H_+-iH_\times)e^{-i\omega u}.
}
$$
Its real and imaginary parts encode the plus and cross <gravitational wave polarizations>, up to the stated <Riemann curvature tensor> and <complex null tetrad> conventions.

Solution

Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 357 / 3 / b / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2024/iii/paper-357/3/b/solution