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

Use signature $(-,+,+,+)$ and the Riemann curvature tensor convention $[{\nabla }_a,{\nabla }_b]X^c=R^c{}_{dab}{X}^d$. Lower the index of the Killing vector field and set $T_{abc}={\nabla }_a{\nabla }_b{V}_c$. Differentiating the Killing equation gives $T_{abc}=-T_{acb}$, while the Ricci identity givesDefine $S_{abc}=R_{cbad}{V}^d$. The first-pair antisymmetry of the Riemann curvature tensor gives $S_{abc}=-S_{acb}$. The pair symmetry and the first Bianchi identity giveThus $D=T-S$ is symmetric in its first two indices and antisymmetric in its last two. These two symmetries force it to vanish:Consequently $T_{abc}=R_{cbad}{V}^d$. Raising $c$ using metric compatibility proves the required second covariant derivative of a Killing vector:

The Levi-Civita connection has symmetric Christoffel symbols, so antisymmetrization removes its connection terms. Hence the Papapetrou electromagnetic field iswhere the last equality uses the Killing equation. As a differential two-form, $F=d{V}^{♭}$, so the identity $d^2=0$ for the exterior derivative proves ${\nabla }_{[a}{F}_{bc]}=0$. Contracting the preceding second covariant derivative of a Killing vector givesA vacuum spacetime here has $R_{ab}=0$, with no cosmological constant. Therefore both source-free Maxwell equations hold:This is a test electromagnetic field on the given vacuum spacetime; its own stress-energy tensor is not included in that background's Einstein field equations.

In Cartesian coordinates $x=r\sin \theta \cos \varphi$, $y=r\sin \theta \sin \varphi$, $z=r\cos \theta$, the rotational Killing vector field and its dual differential one-form areTherefore the Papapetrou electromagnetic field isChoose the electromagnetic field convention $F_{it}=E_i$ and $F_{ij}={ϵ}_{ijk}{B}^k$. There are no time components, and $F_{xy}=2$, so this is a uniform magnetic field along the rotation axis, with no electric field:Multiplying the Killing vector field by $B_0/2$ produces any desired uniform magnetic field strength $B_0$.

Write $f=1-2M/r$ in the Schwarzschild metric. For the stationary Killing vector field $V={\partial }_t$, its dual is $V^{♭}=-fdt$ and the Papapetrou electromagnetic field isOutside the Schwarzschild event horizon, use the static orthonormal coframe $e^{\hat{0}}=\sqrt{f}dt$, $e^{\hat{r}}=dr/\sqrt{f}$, $e^{\hat{\theta }}=rd\theta$, $e^{\hat{\varphi }}=r\sin \theta d\varphi$. Then $F=(2M/r^2)e^{\hat{0}}\wedge e^{\hat{r}}$. With the electric field convention $E_{\hat{i}}=F_{\hat{i}\hat{0}}$ from the preceding part,Thus the stationary Killing vector field gives a Coulomb electric field, directed inward in this convention. If the charge is normalized by $E_{\hat{r}}=Q/r^2$, its charge is $Q=-2M$. A convention with $F\mapsto -F$ reverses that charge; the invariant content is a radial monopole electric field of magnitude $2M/r^2$.

For the axial Killing vector field $V={\partial }_{\varphi }$, $V^{♭}=r^2{\sin }^2\theta d\varphi$, givingIn the same static orthonormal coframe, this is a purely magnetic field withAt large $r$ these are the spherical components of $2{\mathbf{e}}_z$. This is therefore the regular magnetic test field asymptotic to a uniform axial field, distorted by the Schwarzschild black hole. Its net magnetic monopole flux is zero because the integral of $\cos \theta$ over a sphere vanishes. Both electromagnetic fields are regular at the future event horizon: in ingoing Eddington-Finkelstein coordinates, the electric field is $(2M/r^2)dv\wedge dr$, and the magnetic two-form already contains only regular spatial coordinates. Static observers themselves cease to exist on the event horizon.

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.