In <geodesic slicing>, $\alpha=1$ and $\beta^i=0$. The normal acceleration therefore vanishes:
$$
n^\rho\nabla_\rho n_\mu
=a_\mu=D_\mu\log\alpha=0.
$$
Hence each integral curve of $n^\mu$ is an affinely parametrized timelike geodesic.

On the initial surface $\hat t=0$, the diagonal Kruskal metric makes the unit normal point purely in the $\hat t$ direction. The observer starts at $\hat r=0$ with $\dot{\hat r}=0$, so its initial unit four-velocity equals that normal. The observer's geodesic and the normal integral curve solve the same geodesic initial-value problem. Uniqueness therefore gives
$$
\boxed{u^\mu=n^\mu}
$$
throughout their common domain.

Solution

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

In geodesic gauge,
$$
\alpha=1,\qquad\beta^m=0,\qquad D_mD_n\alpha=0.
$$
The BSSN evolution equation reduces to
$$
\partial_tK
=\widetilde A_{mn}\widetilde A^{mn}
+\frac13K^2+4\pi(\rho+S).
$$
The first term is a squared norm with respect to the positive-definite spatial metric, the second is nonnegative, and the stated energy condition makes the final term nonnegative. Therefore
$$
\boxed{\partial_tK\geq0}.
$$
The mean curvature can only increase along this <geodesic slicing>.

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

{wiki}

Geodesic slicing chooses unit lapse and zero shift, so the hypersurface normals are affinely parametrized timelike geodesics. It is simple but commonly develops coordinate singularities in finite time.


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