Y_{lm} e^{-i\omega t}\,.$$ This gives us $$\begin{aligned}
A_t^{(1'n;\,\omega lm)}
& = & \frac{i\omega r^{2-p}}{l(l+p-1)}\sqrt{\frac{f}{h}}\frac{d}{dr}
\left( \frac{r^p}{\sqrt{fh}}R^{(1n)}_{\omega l}\right)Y_{lm}
e^{-i\omega t}\,, \label{AtCoul} \\
A_r^{(1'n;\,\omega lm)}
& = & \frac{\omega^2 r^{2}}{l(l+p-1)}\frac{1}{f}R^{(1n)}_{\omega l}Y_{lm}
e^{-i\omega t}\,. \label{suppr}\end{aligned}$$

Response rate of a static charge outside a four-dimensional Schwarzschild black hole {#sec:Rate}
====================================================================================

Here we briefly review, in the context of the previous two sections, the calculation of response rate in the Unruh vacuum [@Unruh] of a static electric charge in Schwarzschild spacetime performed in Ref. [@CHM].

The line element of the four-dimensional Schwarzschild spacetime is $$d\tau^2 = f(r)dt^2 - f(r)^{-1}dr^2 - r^2 ds_2^2$$ with $f(r) = 1-2M/r$, where $ds_2^2 = d\theta^2 + \sin^2\theta\,d\phi^2$ is the line element of the unit two-sphere. (The horizon radius is $r_H= 2M$.) We compute the response rate of the static electric charge to the Hawking radiation present in the Unruh vacuum. The charge is placed at $(r,\theta,\phi) = (r_0,\theta_0,\phi_0)$ as described by the current density $$j^\mu = \delta^\mu_{ t}\frac{q}{r^2\sin\theta}
\delta(r-r_0)\delta(\theta - \theta_0)
\delta(\phi-\phi_0) \,. \label{Jt}$$ Since this current density is static, it couples to photons with zero energy. It turns out that the rate of response of this current to a single photon vanishes. However, the Bose-Einstein distribution of the thermal 