If we have a Reissner-Nordström black hole with Q > 0, then what charge (positive or negative) must an infalling charged particle have in order for the particle to be able to obtain negative energies at the horizon?
$\begingroup$
$\endgroup$
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
The energy of a charged particle with charge q, infalling into a Reissner-Nordström black hole with charge Q is given by $\epsilon = (p-qA)\cdot \zeta_{(t)}$. In this expression $p = p_\mu dx^\mu$ is the momentum of the particle, $A = A_\mu dx^\mu = -\frac{Q}{r}dt$ is the electromagnetic gauge field, and $\zeta_{(t)} = \zeta^\mu_{(t)} \partial_\mu$ is the timelike Killing vector, hence we can take $\zeta^\mu_{(t)} = [1,0,0,0]$, so that $\zeta_{(t)} = \partial_t$. Note $dx^\mu(\partial_\nu) \equiv \frac{\partial x^\mu}{\partial x^\nu} = \delta^\mu_\nu$, hence $dt(\partial_t) = 1$ and $dr(\partial_t) = 0$.
We will take the metric to be $ds^2 = f(r)dt^2-f^{-1}(r)dr^2$ and we will ignore $\theta$ and $\phi$ by assuming a purely radial trajectory for the particle as it approaches the horizon at $r = r_h$, where $f(r_h) \equiv 0$. We do not need to specify $f(r)$ for our problem but it is the standard one for the Reissner-Nordström black hole.
Now we evaluate $\epsilon = (p_t - qA_t) = (p_t + \frac{qQ}{r}) = (f(r)p^t + \frac{qQ}{r})$, where we used $p_t = g_{tt}p^t = f(r)p^t$. Given $ds^2 = d\tau^2 = f(r)dt^2-f^{-1}(r)dr^2$ and $p^\mu \equiv \frac{dx^\mu}{d\tau}$, we can use the first to write $\frac{dt}{d\tau} \equiv \dot{t} \equiv p^t = (\frac{1}{f(r)}+\frac{(p^r)^2}{f^2(r)})^{1/2}$. Substitution gives $\epsilon = (f(r) + (p^r)^2)^{1/2} + \frac{qQ}{r}$. Given $\epsilon$ is a constant of the motion we can evaluate it at $r = r_h$, where $f(r_h) \equiv 0$ and $p^r(r_h) = 0$ from $d\tau^2 = f(r)dt^2-f^{-1}(r)dr^2$.
We have $\epsilon(r_h) = \frac{qQ}{r_h}$, from which we see that if we want $\epsilon < 0$, we need $qQ < 0$ and so if $Q > 0$ then $q < 0$.
For some discussion see page 10 of https://arxiv.org/abs/1008.3657.
