Past exam of the mathematics course of the University of Cambridge / 2010 / iii / Paper 54 / 1 / b

Put $r_{+}=\ell \sqrt{M}$ and $f=(r^2-r_{+}^2)/{\ell }^2$. The apparent failure of the nonrotating BTZ black hole coordinates at $r=r_{+}$ is a coordinate singularity. Define the tortoise coordinate and ingoing time byThe Ingoing BTZ coordinates giveAll coefficients are analytic at $r_{+}>0$, and the determinant is $-r^2$, so the metric tensor extends nondegenerately through that radius. The Killing vector field $K={\partial }_v$ agrees with ${\partial }_t$ in the exterior. On $r=r_{+}$, $K^2=-f=0$ and $K_a=(dr{)}_a$. It is therefore both normal and tangent to this null hypersurface, proving that it is a Killing horizon.

One may include both horizon branches and their bifurcation surface explicitly. Set $\kappa =r_{+}/{\ell }^2$, $u=t-r_{\ast }$ and, initially in the exterior, $U=-e^{-\kappa u}$, $V=e^{\kappa v}$. ThenThis analytic extension of a spacetime is regular near $UV=0$, and $K=\kappa (V{\partial }_V-U{\partial }_U)$ extends analytically. The surfaces $U=0$ and $V=0$ form the extended Killing horizon at $r=r_{+}$; $K$ vanishes only at their regular bifurcation circle.

Use the exterior-normalized Killing vector field $K={\partial }_t={\partial }_v$ from the previous part. In Ingoing BTZ coordinates, its only nonzero acceleration component on $r=r_{+}$ isThe defining equation for surface gravity, ${\nabla }_{K}K=\kappa K$, therefore gives the positive future-branch valueThis value uses exactly the prescribed Killing vector field normalization. A constant rescaling of the Killing vector field rescales its surface gravity.

Write $M=-a^2$ with $a>0$. On a constant-time slice, the proper distance from the axis to radius $r$ and the circumference at that radius areHence $C/s\to 2\pi a$. A smooth axis in a Riemannian metric requires the Euclidean limit $C/s\to 2\pi$, so regularity forces $a=1$, or $M=-1$. For $a\ne 1$ the conical singularity of negative-mass BTZ geometry has deficit angle $2\pi (1-a)$; for $a>1$ this is an angular excess. Even integer $a>1$ is singular with the stipulated angular period $2\pi$: a multiple covering does not make the tip a smooth point.

At $M=-1$, set $r=\ell \sinh \chi$. The metric tensor becomeswhich is regular at $\chi =0$ and describes Anti-de Sitter spacetime with nonperiodic time. For every $M<0$, $f=a^2+r^2/{\ell }^2$ stays positive. An outward radial null geodesic obeys $dt/dr=1/f$, and reaches the timelike conformal boundary in finite coordinate time, since ${\int }_0^{\infty }dr/f=\pi \ell /(2a)$. Thus no event horizon hides the conical axis. The geometry is nakedly singular except at $M=-1$. Its local curvature away from the axis is that of Anti-de Sitter spacetime; the argument is a global angular-identification test, not curvature blowup.