Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 311 / 1

Every metric component depends only on $r$, so translations of $t$ and $\varphi$ leave the metric unchanged. Equivalently,Thus $k={\partial }_t$ and $m={\partial }_{\varphi }$ are Killing vector fields, generating stationarity and axial symmetry respectively.

Define Ingoing BTZ coordinates byThus $dt=dv-dr/f$ and $d\varphi =d\chi -\Omega dr/f$, so the potentially singular terms cancel:andThe metric becomesFor $r_{+}>r_{-}$, antiderivatives may be chosen asandwith the extremal case obtained by taking the limit. The transformed metric contains neither $1/f$ nor any other singular coefficient at $r=r_{+}$; since $f$, $\Omega$, and $r^2$ are analytic there, it gives an analytic extension across the outer horizon.

Let a future-directed causal tangent in the ingoing chart beThe chosen time orientation has $\dot{v}\ge 0$; equality occurs only for the ingoing radial null direction, which has $\dot{r}<0$. Causality requiresFor $r_{-}<r<r_{+}$ one has $f<0$. The first and third terms are then nonnegative, so if $\dot{v}>0$ the inequality forces $\dot{r}<0$. If $\dot{v}=0$, it forces $\dot{\chi }=0$, and future direction again gives $\dot{r}<0$. Thus $r$ strictly decreases along every future causal curve in this region. No such curve can cross outward through $r=r_{+}$ into $r>r_{+}$, which proves the one-way causal disconnection characteristic of an event horizon.

The normal to a surface of constant $r$ is $n_a=(dr{)}_a$, and the inverse ingoing metric givesHence $r=r_{+}$ is a null hypersurface. LetThis constant linear combination of Killing vector fields is Killing. Its squared norm iswhich vanishes at $r_{+}$. More strongly, lowering its index in the ingoing metric gives ${\xi }_a=(dr{)}_a$ on the horizon. It is therefore the null normal and generator, so the surface is a Killing horizon.

Using ${\nabla }_a({\xi }^2)=-2\kappa {\xi }_a$ on a Killing horizon, the squared term contributes no first derivative at $r_{+}$ andDifferentiating $f$ at its simple outer root givesso the surface gravity isIt vanishes in the extremal case $r_{+}=r_{-}$.

When $r_{-}=0$, $\Omega =0$ and $f=(r^2-r_{+}^2)/L^2$. A circle $S$ at fixed $v,r$ has circumference $2\pi r$. Choose its future null normals asFor a one-dimensional transverse surface, each null expansion is the logarithmic derivative of its length:Inside the horizon, $0<r<r_{+}$ implies $f<0$, so both expansions are negative. Every such circle is therefore a trapped surface, specifically a Trapped circle inside a nonrotating BTZ black hole.

The Penrose singularity theorem concludes future null-geodesic incompleteness; it does not require a divergent curvature invariant. This spacetime supplies exactly that distinction. For a null geodesic in the static coordinates, the Killing constantsand the null condition giveAs $r\to 0$, $f\to -r_{+}^2/L^2$. A radial geodesic with $J=0$ reaches $r=0$ linearly in finite affine parameter. If $J\ne 0$, then $|\dot{r}|\sim r_{+}|J|/(Lr)$, so it again reaches zero in finite affine parameter. The anti-de Sitter quotient ends these geodesics at $r=0$ even though local curvature remains finite. Thus the trapped circles lead to the BTZ causal singularity, satisfying the theorem through geodesic incompleteness rather than curvature blowup.