 Solution

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.