vacuum $|0\rangle$ to the state $|1 n;\,\omega lm\rangle =a^{(1 n)\dagger}_{\omega l m}|0\rangle$ by the classical current density $j^\mu$ is given by $${\cal A}^n_{\omega lm} = i\int d^4 x\sqrt{-g}j^\mu 
\langle 1 n ;\, \omega lm |\hat{A}_\mu|0\rangle
\,.  \label{Amp}$$ Let us recast Eq. (\[Amp\]) as $${\cal A}^{n}_{\omega lm}\equiv 
2\pi i\, {\cal T}^{n}_{\omega lm}\delta(\omega - E)\,,
\label{Amp2}$$ where ${\cal T}^{n}_{\omega lm}$ will be calculated later. The corresponding transition rate, i.e. transition probability per asymptotic proper time, is expressed as $${\cal R}^n_{\omega lm} = 2\pi |{\cal T}^{n}_{\omega lm}|^2
\delta(\omega - E)\,.$$ Now we take into account that in the Unruh vacuum there is a thermal flux of photons coming out of the horizon with inverse temperature $\beta = 8\pi M$ (with no particles coming from infinity). In the limit $E\to 0$, only the absorption and stimulated emission of these thermal photons contribute to the response rate. For this reason, we treat only the modes coming out of the horizon ($n = \rightarrow$). The total response rate due to these modes with fixed angular momentum, in which absorption, spontaneous and stimulated emissions are all taken into account, is $$R_{Elm} = \int_0^{+\infty} d\omega {\cal R}^{\rightarrow}_{\omega lm}
\left( \frac{2}{e^{\beta\omega} - 1} + 1\right)
= 2\pi{\rm coth}\frac{\beta E}{2}|{\cal 