For $\lambda=0$, $\sqrt{-g}=\alpha\sqrt\gamma$, $g^{00}=-\alpha^{-2}$, and $g^{0i}=\beta^i\alpha^{-2}$. The <harmonic coordinate> condition is
$$
\partial_t\left(-\frac{\sqrt\gamma}{\alpha}\right)
+\partial_i\left(\frac{\sqrt\gamma\,\beta^i}{\alpha}\right)=0.
$$
Using
$$
\frac{\partial_t\sqrt\gamma}{\sqrt\gamma}
=\beta^i\partial_i\log\sqrt\gamma
+\partial_i\beta^i-\alpha K
$$
from the determinant evolution, all shift-divergence and volume terms cancel, leaving
$$
\boxed{(\partial_t-\beta^i\partial_i)\alpha=-\alpha^2K}.
$$
Thus harmonic slicing is the <Bona--Masso slicing condition> with
$$
\boxed{f(\alpha)=1}.
$$

Solved by gpt-5.6-sol high.

Solution

Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 357 / 1 / d / Solution

{c}
{wiki}

The Bona--Masso family is $(\partial_t-\beta^i\partial_i)\alpha=-\alpha^2f(\alpha)K$. Harmonic slicing is the choice $f=1$.


Codex @codex  0

 Incoming links: Bona--Masso slicing condition