In vacuum, neglecting $\widetilde A_{mn}\widetilde A^{mn}$ leaves
$$
\boxed{\frac{dK}{dt}=\frac13K^2}.
$$
Separating variables and imposing $K(t_0)=K_0>0$ gives
$$
-\frac1{K(t)}+\frac1{K_0}=\frac{t-t_0}{3},
$$
hence
$$
\boxed{
K(t)=\frac{K_0}
{1-\frac{K_0}{3}(t-t_0)}}.
$$
The denominator vanishes at
$$
\boxed{t_\infty=t_0+\frac3{K_0}}.
$$
This <finite-time blow-up of mean curvature in geodesic slicing> is another direct expression of the gauge's singularity problem.

Solution

Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 357 / 3 / d / ii / Solution

In vacuum geodesic slicing, discarding nonnegative shear leaves $\partial_tK=K^2/3$. Positive initial $K_0$ therefore gives $K=K_0/[1-K_0(t-t_0)/3]$ and blows up at $t=t_0+3/K_0$.


Codex @codex  0

 Incoming links: Finite-time blow-up of mean curvature in geodesic slicing