On a constant-$t$ asymptotically flat hypersurface, transform the large-$r$ spatial metric to asymptotically Cartesian coordinates. Its leading perturbation is
$$
h_{ij}=\delta_{ij}+q_{ij}+O(r^{-2}),
\qquad
q_{ij}=\frac{2M}{r}n_in_j
=\frac{2Mx_ix_j}{r^3}.
$$
The rotation parameter $a$ first affects the spatial metric at orders that vanish in the <Arnowitt-Deser-Misner energy> surface limit. Direct differentiation gives
$$
\partial_jq_{ij}=\frac{2M}{r^2}n_i,
\qquad
\partial_iq_{jj}=-\frac{2M}{r^2}n_i.
$$
Hence
$$
n^i(\partial_jh_{ij}-\partial_ih_{jj})
=\frac{4M}{r^2}+O(r^{-3}).
$$
Using $\int_{S_r^2}dA=4\pi r^2$ in the ADM surface integral yields
$$
\boxed{E_{ADM}
=\frac1{16\pi}(4\pi r^2)\frac{4M}{r^2}=M}.
$$
Thus the Kerr parameter $M$ is its total mass in the rest frame.

Solved by gpt-5.6-sol high.

Solution

Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 311 / 4 / d / Solution

{c}
{title2=$E_{ADM}$}
{wiki=ADM_formalism}

The Arnowitt-Deser-Misner energy is the asymptotic surface charge generating time translations on an asymptotically flat spatial hypersurface. In asymptotically Cartesian coordinates,
$$
E_{ADM}=\frac1{16\pi}\lim_{r\to\infty}\int_{S_r^2}n^i(\partial_jh_{ij}-\partial_ih_{jj})\,dA.
$$

= ADM energy
{c}
{synonym}


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