photons coming out of the horizon diverges at zero energy. This makes the rates of absorption and stimulated emission indefinite. This ambiguity can be resolved by replacing the current density (\[Jt\]) by $j^\mu = (j^t, j^r, 0, 0)$, where $$\begin{aligned}
j^t & = & \frac{\sqrt{2}\,q \cos Et}{r^2 \sin\theta}
\delta(r-r_0)\delta(\theta-\theta_0)\delta(\phi - \phi_0)\,, \\
j^r & = & \frac{\sqrt{2}q E\sin Et}{r^2 \sin\theta}
\Theta(r-r_0)\delta(\theta-\theta_0)\delta(\phi - \phi_0)\,. \end{aligned}$$ Here, $\Theta(x) = 1$ if $x>0$ and $\Theta(x) = 0$ if $x \leq 0$. (The factor of $\sqrt{2}$ is necessary to make the time average of the squared charge equal to $q^2$.) We take the limit $E\to 0$ in the end, assuming that the rate is continuous at $E=0$. This continuity has been verified in Minkowski spacetime [@HMS1].

Since the current (\[Jt\]) is conserved, it does not interact with pure-gauge particles corresponding to the creation operators $a^{(3 n)\dagger}_{\omega lm}$. It does interact with the states created by $a^{(0 n)\dagger}_{\omega lm}$ but these have zero norm and do not contribute to physical probabilities. Hence, as usual, we need to consider only the physical modes. Note also that only physical modes I are relevant because $A_t=A_r =0$ for physical modes II.

Now, the transition amplitude from the Boulware 