Possibility of Charge Addition to the Reissner-Nordstrom Black Hole

     The Reissner-Nordstrom black hole has background field of metric
\[g_{\mu\nu} = \mathrm{diag}\left\{-\left(1-\frac{r_s}{r} + \frac{r_Q^2}{r^2}\right),\,\left(1-\frac{r_s}{r} + \frac{r_Q^2}{r^2}\right)^{-1},\,r^2,\,r^2\sin^2\theta\right\}\]
where
\[r_s = 2GM;\;\; r_Q^2 = \frac{GQ^2}{4\pi}\]
and of electromagnetic field
\[A_t = \frac{Q}{4\pi r}\]
Then, the geodesic is determined by the action
\[S=\int d\tau\left[-m\sqrt{(1-\frac{r_s}{r} + \frac{r_Q^2}{r^2})\dot{t}^2 - \frac{1}{1-\frac{r_s}{r} + \frac{r_Q^2}{r^2}}\dot{r}^2 - r^2\dot{\phi}^2}+\frac{qQ}{4\pi r} \dot{t}\right]\]
assuming the motion is on the plane \(\theta=\frac{\pi}{2}\),
where the normalisation of the proper time is determined by
\[\sqrt{\cdots} \equiv \sqrt{(1-\frac{r_s}{r} + \frac{r_Q^2}{r^2})\dot{t}^2 - \frac{1}{1-\frac{r_s}{r} + \frac{r_Q^2}{r^2}}\dot{r}^2 - r^2\dot{\phi}^2}=1\]

     We may, first, obtain the equation of motion in angle.
\[\frac{\delta S}{\delta \phi} = -\frac{d}{d\tau} \frac{mr^2\dot{\phi}}{\sqrt{\cdots}} = -\frac{d}{d\tau} mr^2\dot{\phi}=0\]
So, we have found a constant of motion, angular momentum, which should be defined as
\[L\equiv m r^2\dot{\phi}\]
Now, we may take variation in direction of time.
\[\frac{\delta S}{\delta t} = -\frac{d}{d\tau} \left[\frac{-m () \dot{t}}{\sqrt{\cdots}}+\frac{qQ}{4\pi r}\right] = -\frac{d}{d\tau} \left[-m () \dot{t}+\frac{qQ}{4\pi r}\right] = 0\]
with the shorten expression
\[() \equiv \left(1-\frac{r_s}{r} + \frac{r_Q^2}{r^2}\right)\]
This equation of motion is also cyclic so we may define the constant of motion
\[E = -m () \dot{t}+\frac{qQ}{4\pi r}\]

     The only remaining equation of equation is related to \(r\), but we may alternate the equation to the normalisation condition, conserving the degree of freedom.
\[1= \cdots = () \dot{t}^2 - ()^{-1} \dot{r}^2 - r^2 \dot{\phi}^2 = \frac{1}{m^2 ()}\left(\frac{qQ}{4\pi r} -E\right)^2-()^{-1} \dot{r}^2-\frac{L^2}{m^2r^2} \]
We may manipulate the equation as similar to the classical mechanics.
\[\frac{1}{2}m \dot{r}^2  + \frac{L^2}{2mr^2}\left(1-\frac{r_s}{r}+\frac{r_Q^2}{r^2}\right) - \frac{1}{2m}\frac{q^2Q^2}{16\pi^2r^2} + \frac{1}{m} \frac{qQE}{4\pi r} = \frac{E^2}{2m}-\frac{m}{2}()\]
By special relativity, you may find \(\frac{1}{2}m \dot{x}^2 = \frac{E^2}{2m}-\frac{m}{2}\) where \(E\) is the total energy of a particle, so \(E\) is the asymtotic energy of the test particle, in this problem.
Thus, this equation is in analogy to the energy conservation law in classical mechanics.

     To simplify the situation and ease to enter the black hole, we may consider the situation \(L=0\).
Then, this seems to be a system with potential \(V(r) = -\frac{q^2Q^2}{32\pi^2 m r^2}+\frac{qQE}{4\pi mr}\), but this term itself contains the total energy \(E\) so we have to consider more carefully. First, from the shape of the potential, we may guess that the test particle may fall to the centre if the particle just overcome the threshold. i.e. \(V(r) = \frac{E^2}{2m}-\frac{m}{2}\) should have no solution. However, the solution is \[ r = \frac{qQ}{4\pi (E\mp m)}\]which always exists, considering the absolute condition \(E\geq m\). Thus, the particle always bounce back to the infinity at the outer critical point \[r = \frac{qQ}{4\pi (E-m)}\]
Not to enter inside of the horizon, the criterion is thus
\[\frac{E-m}{q} \leq \frac{\sqrt{M^2 + \frac{Q^2}{4\pi G}}-M}{Q}\]


For further study: I'll note here http://arxiv.org/abs/1304.6474 which is a part of my project here. I may list tasks to determine charged black hole stability and radiation
*particle generation in strong electric field
*decay rate of particle-anti-particle pair in terms of gas density

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