In geodesic gauge,

α = 1, β^{m} = 0, D_{m} D_{n} α = 0.

(1)

The BSSN evolution equation reduces to

\partial_{t} K = A_{mn} A^{mn} + \frac{1}{3} K^{2} + 4 π (ρ + S) .

(2)

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

\partial_{t} K \geq 0 .

(3)

The mean curvature can only increase along this geodesic slicing.

ii

 0  0

Solution

 0  0

In vacuum, neglecting

A_{mn} A^{mn}

leaves

\frac{d K}{d t} = \frac{1}{3} K^{2} .

(1)

Separating variables and imposing

K (t_{0}) = K_{0} > 0

gives

- \frac{1}{K ( t )} + \frac{1}{K _{0}} = \frac{t - t _{0}}{3},

(2)

hence

K (t) = \frac{K _{0}}{1 - \frac{K _{0}}{3} ( t - t _{0} )} .

(3)

The denominator vanishes at

t_{\infty} = t_{0} + \frac{3}{K _{0}} .

(4)

This finite-time blow-up of mean curvature in geodesic slicing is another direct expression of the gauge's singularity problem.

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

i

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Solution

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