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

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.