Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 357 (Gravitational Waves and Numerical Relativity)

The inverse metric condition isKeeping only terms linear in the metric perturbation givesThus the linearized inverse metric iswhere indices on $h_{\alpha \beta }$ are raised with the Minkowski metric.

In the Levi-Civita connection, the background metric is constant and replacing $g^{\alpha \rho }$ by its $O(ϵ)$ correction would multiply a derivative of $h$ and produce an $O({ϵ}^2)$ term. Hence the linearized Levi-Civita connection isThe products of two connection coefficients in the Riemann curvature tensor are also quadratic and may be discarded. Lowering its first index with ${\eta }_{\mu \nu }$ gives

Because the coordinate change is ${\tilde{x}}^{\alpha }=x^{\alpha }-{\xi }^{\alpha }$, the perturbation changes at linear order bySubstitution into $R_{\mu \rho \alpha \beta }^{(1)}$ produces terms containing three partial derivatives of $\xi$. Since partial derivatives commute, every term cancels another with the opposite sign. ThereforeThis is the gauge invariance of the linearized Riemann tensor.

All time components of the perturbation vanish. For $\mu =t$ this immediately givesFor a spatial index $\mu =i$,by the stated property. Thus the perturbation is transverse.

Its Minkowski trace is purely spatial:Adding the displayed components and collecting the coefficients of the independent functions $A$, $B$, and $C$ makes each coefficient vanish separately, soTogether with $h_{t\mu }=0$ and ${\partial }^{\nu }{h}_{\mu \nu }=0$, this proves that the wave is in transverse-traceless gauge.

On the positive $z$ axis, $x=y=0$ and $r=z$. Taking this limit in the spatial components giveswith $A,B,C$ evaluated at retarded time $t-r$. A wave propagating toward the observer along the $z$ direction has polarization matrixConsequently the observed gravitational wave polarization amplitudes areThe $B$ mode and the orthogonal combination of $A,C$ do not contribute on this symmetry axis. The observer therefore sees a purely plus-polarized wave; a rotation of the transverse $x,y$ axes would represent the same physical polarization as the corresponding spin-two mixture of plus and cross.

LetThe identity decomposes asApplying this decomposition to both indices of $T_{\mu \nu }$ givesThe four terms are, from the definitions, $S_{\alpha \beta }$, $j_{\alpha }{n}_{\beta }$, $n_{\alpha }{j}_{\beta }$, and $\rho n_{\alpha }{n}_{\beta }$. Hence the 3+1 decomposition of the stress-energy tensor isIt also makes explicit that

Define the acceleration of the normal congruence byBecause $S^{\sigma \gamma }$ is spatial, differentiating
$n_{\sigma }{S}^{\sigma \gamma }=0$ givesInsertinto the contracted derivative and project the free index:Rearranging proves

The Lie derivative of the spatial covector $j_{\alpha }$ along $n^{\mu }$ isTo show that it is spatial, contract with $n^{\alpha }$. Differentiating $n^{\alpha }{j}_{\alpha }=0$ along $n^{\mu }$ giveswhereasThe terms cancel, soA spatial projector therefore acts trivially:

Spatially project stress-energy conservation,and substitute the decomposition from part a. The $\rho n^{\mu }{n}_{\nu }$ term contributes $\rho a_{\alpha }$. The two momentum terms combine intobecause the two contractions with the full extrinsic curvature cancel andFinally, part b givesThusThe momentum equation in a 3+1 decomposition isTherefore

Sinceand $j_{\mu }{n}^{\mu }=0$, the scaling rule for a covector Lie derivative has no extra term:The acceleration is the spatial lapse gradient,Multiplying the momentum equation by $\alpha$ therefore givesFor spatial components, the shift term can be writtenso explicitly

Write the Kruskal–Szekeres plane with $\hat{r}$ horizontal and $\hat{t}$ vertical. The relationshows that constant-$r$ curves are hyperbolae.

For $r>2M$, their right and left exterior branches satisfyFor $0<r<2M$, their future and past interior branches satisfyThe event horizons $r=2M$ are the null diagonalsand the curvature singularities $r=0$ are the spacelike hyperbolaein the normalization stated in the question.

Constant Schwarzschild-$t$ curves are straight rays through the origin. In the exteriors their slopes obeywhile in the interiorsThus the qualitative diagram is the usual four-region Kruskal diagram: two exterior wedges separated from black-hole and white-hole interiors by the two null horizons, with spacelike singularities bounding the interior wedges.

In geodesic slicing, $\alpha =1$ and ${\beta }^i=0$. The normal acceleration therefore vanishes: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 givesthroughout their common domain.

The point $\hat{t}=\hat{r}=0$ is the bifurcation sphere $r=2M$. The observer starts there with $\dot{r}=0$, so its radial-geodesic equationgives $E=0$. During infall,Put $x=r/(2M)$. The proper time to the singularity isThusThe normals of the geodesic foliation reach the physical singularity after finite coordinate time because unit lapse identifies coordinate and normal proper time. Nearby normals can also focus and form coordinate caustics. Consequently geodesic gauge is unsuitable for long-term black-hole evolution: the numerical slice encounters singular behavior in finite time rather than avoiding it.

In geodesic gauge,The BSSN evolution equation reduces toThe 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. ThereforeThe mean curvature can only increase along this geodesic slicing.

In vacuum, neglecting ${\tilde{A}}_{mn}{\tilde{A}}^{mn}$ leavesSeparating variables and imposing $K(t_0)=K_0>0$ giveshenceThe denominator vanishes atThis finite-time blow-up of mean curvature in geodesic slicing is another direct expression of the gauge's singularity problem.