The azimuthal metric component is
$$
g_{\phi\phi}=r^2e^{-2\gamma}\sin^2\theta.
$$
Since $\gamma=c/r+O(r^{-2})$,
$$
\partial_ug_{\phi\phi}
=-2r^2e^{-2\gamma}(\partial_u\gamma)\sin^2\theta
=-2r(\partial_uc)\sin^2\theta+O(1).
$$
Thus the <Bondi news function> satisfies
$$
\boxed{\partial_uc
=-\frac12\lim_{r\to\infty}
r^{-1}(\sin\theta)^{-2}\partial_ug_{\phi\phi}}.
$$

Solved by gpt-5.6-sol high.

Solution

Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 357 / 2 / b / Solution

{c}
{wiki}

The Bondi news is the retarded-time derivative of the leading asymptotic shear and measures outgoing gravitational radiation at null infinity.


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 Incoming links: Bondi news function