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>.

Solution

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

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/iii/paper-357/3/d/i/solution