Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 357 / 2

Set $A=1-2M/R=(R-2M)/R$ andSubstitution into the Schwarzschild metric givesThe cross coefficient and radial coefficient simplify toTherefore

With $R=r+M$ and $dR=dr$,Defining the conformal factor byalso gives $R^2={\psi }^4{r}^2$. HenceThe constant-time spatial metric is therefore conformally flat.

Comparison with the 3+1 decomposition of spacetime gives the diagonal spatial metricThe mixed metric coefficient is ${\beta }_r=g_{tr}=M/r$. Raising its index with ${\gamma }^{rr}=r^2/(r+M{)}^2$ yields the shift vectorNowand $g_{tt}=-{\alpha }^2+{\beta }_i{\beta }^i=-(r-M)/(r+M)$. The positive lapse function is consequently

The spatial metric is stationary and only ${\beta }^r=b(r)=Mr/(r+M{)}^2$ is nonzero. The evolution equation therefore saysthe Lie derivative of the spatial metric along the shift vector. For the radial component,soFor the angular components,and spherical symmetry suppliesAll off-diagonal components vanish.

Contracting with the inverse spatial metric gives the mean curvatureThus

Stationarity makes ${\partial }_t\alpha =0$, whileThe left side of the Bona--Masso slicing condition is thereforeIts right side isEquality requiresSince $\alpha =r/(r+M)$, this is the Stationary Schwarzschild Bona--Masso slicing functionConsequently ${\lim }_{\alpha \to 1}f(\alpha )=0$, as expected in the asymptotically flat region $r\to \infty$.