Black hole's (gauge) charge and its associated field If you have a black hole characterized by its mass, angular momentum, and charge (say electric charge) then how can this black hole have an associated field (electromagnetic)?
I.e. if the field is mediated by a massless gauge boson (e.g. photon) then how can this charge communicate given that the gauge boson cannot escape the event horizon?  
It seems that there should be no field (?).  However if you place a test charge near the BH then it should feel some force due to the charge of the BH.  But again, this force has to be mediated by a boson which cant escape the BH unless the BH charge exists slightly outside of the horizon
Im very much confused by this.  Please elucidate
 A: The Schwarzschild metric 
$$
ds^2~=~(1~-~2m/r)c^2dt^2~-~(1~-~2m/r)^{-1}dr^2~-~r^2d\Omega^2,
$$
allows us to look at the motion of a massive charge approaching the black hole and the motion of a photon leaving that particle according to a distant observer. We have for the massive particle that 
$$
\left(\frac{ds}{dt}\right)~=~\sqrt{(1~-~2m/r)c^2~-~(1~-~2m/r)^{-1}\frac{dr}{ds}^2}.
$$
This gives the general Lorentz factor
$$
\Gamma~=~\frac{1}{\sqrt{(1~-~2m/r)}}\frac{1}{\sqrt{1~-~(1~-~2m/r)^{-2}\frac{dr}{ds}^2}}.
$$
This then gives the time dilation of a clock on any particle or system approaching the event horizon. It will take a time $T~\rightarrow~\infty$ for the particle to reach the event horizon.
For a null geodesic has $ds^2~=~0$. We now compute the delay time for the radial motion of a photon from a source near the black hole
$$
c\int dt~=~\int_{R>2m}^r(1~-~2m/r)^{-1}dr'~=~\left[r'~+~ln\left(r'~-~2m\right)\right]|_R^r.
$$
This defines the delay coordinate. For $R,~r$ large this is just the time it takes a photon to travel a distance between two points. This is continuous all the way to the horizon. There is then a null geodesic from a particle just above the horizon to the distant region outside. This delay coordinate indicates that the distant observer will witness any particle approach the horizon more slowly and creep endlessly towards the horizon as illustrated by the long delay taken by any photon leaving that particle.
We can now consider the momentum of a null particle from the black hole. From the metric with $ds^2~=~0$ the momentum of a radially directed photon is then
$$
\frac{dr}{dt}~=~(1~-~2m/r).
$$
This approaches zero as the photon is emitted as a distance approaching the horizon. The term $1~-~2m/r$ is a redshift factor for photons. As indicated above it will take an infinite time to observe the charge emitting these photons to reach the event horizon. This holds for a virtual photon as well. We then consider the motion of a particle that emits a photon with the covariant operator $\vec P~=~\vec p~+~\vec A$. The gauge vector potential is then $\vec A~=~(1~-~2m/r)\vec A_0$ for $\vec A_0$ the vector potential in flat spacetime. The redshift factor $1~-~2m/r$ enters into the physics of virtual photons.
As a result the exterior observer detects the electric field of a charge approaching a black hole. In effect the black hole acquires this charge. The distant observer never witnesses the charge actually fall through the event horizon. 
