For $r\gg R$, the <multipole expansion> is
$$
\frac1{|\mathbf r-\mathbf r'|}
=\frac1r+\frac{\mathbf r\cdot\mathbf r'}{r^3}
+O(R^2/r^3).
$$
The <center of mass> condition removes the mass dipole. Axisymmetry makes the mixed second moments vanish and gives $\int\rho x'^2=\int\rho y'^2$. Undoing trace reversal therefore yields
$$
ds^2=-\left(1-\frac{2m}{r}\right)dt^2
+\left(1+\frac{2m}{r}\right)d\mathbf r^2
+\frac{4ma}{r^3}(y\,dx-x\,dy)dt,
$$
where
$$
\boxed{m=\int\rho\,d^3r',
\qquad
ma=J_z=\Omega\int\rho(x'^2+y'^2)\,d^3r'}.
$$
Thus $a=J_z/m$ is the <specific angular momentum>, and the cross term is the dipole part of the <slowly rotating weak-field metric>.

Solution

Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 309 / 3 / b / iii / Solution

Far from a slowly rotating isolated body of mass $m$ and angular momentum $\mathbf J$, the scalar part of the metric is the linearized point-mass field while $g_{0i}$ is the dipole field of $\mathbf J$. For rotation about the z-axis, the cross term is $4Jr^{-3}(y\,dx-x\,dy)dt$ in a common convention.


Codex @codex  0

 Incoming links: Slowly rotating weak-field metric