Introduction
There are several ways to extract the energy from a black hole [4–6]. The Penrose process states that in the ergoregion of a rotating black hole the particles with negative energy might exist due to collision or decay. As a result, the energy of escaping particle is bigger than the energy of the original particle. This process is possible only in the ergoregion where the Killing vector $\tfrac{{\rm{d}}}{{\rm{d}}{t}}$ becomes spacelike. In spherical symmetry, i.e. for Schwarzschild and Reissner-Nordstrom black holes, the Killing vector $\tfrac{{\rm{d}}}{{\rm{d}}{t}}$ is timelike outside the event horizon and the Penrose process is impossible. However, in the Reissner-Nordstrom case, one can consider the analogy of the Penrose effect because there are charged particles with negative energy [7]. These particles can exist only in the generalized ergoregion [8, 9]. The border of this region is called the generalized ergosphere and it depends on the test particle parameters. Thus, even in the spherically-symmetric case, there is the analogue of the Penrose process for charged black holes [8, 10].
In the present work, we consider a dynamical charged black hole of which the exterior geometry is described by a charged Vaidya solution [11, 12]. The Vaidya metric describes a dynamical spacetime instead of a static spacetime as the Schwarzschild or Reissner-Nordstrom metrics do. In the real world, astronomical bodies gain mass when they absorb radiation and they lose mass when they emit radiation, which means that the space-time around them is time-dependent. In dynamical spherically-symmetric spacetimes, in general, there is only one conserved quantity - angular momentum L. However, for a certain choice of the mass M(v) and charge Q(v) functions, there is additional symmetry due to a conformal Killing vector field. We calculate the location of the generalized ergosphere where the charged particles with negative energy can exist. After that, we consider the Penrose effect and estimate the efficiency of this process.
This paper is organized as follows. In section 2 we consider the charged Vaidya spacetime, transform it to conformally static coordinates, and calculate the location of the generalized ergosphere. In section 3 we show that the Penrose process can occur in the generalized ergoregion and in section 4 we estimate the efficiency of the process. Section 5 is the conclusion. The system of units G = c = 1 will be used throughout the paper.
Charged Vaidya spacetime
The charged Vaidya spacetime in advanced Eddington-Finkelstein coordinates has the following form:
$\begin{eqnarray}{{\rm{d}}s}^{2}=-\left(1-\displaystyle \frac{2M(v)}{r}+\displaystyle \frac{{Q}^{2}(v)}{{r}^{2}}\right){{\rm{d}}{v}}^{2}+2{\rm{d}}{v}{\rm{d}}{r}+{r}^{2}{\rm{d}}{{\rm{\Omega }}}^{2}.\end{eqnarray}$
Here M(v) and Q(v) are mass and charge functions that depend upon the time v, dΩ2 is the metric on the unit two sphere. The electric charge has the dimension of the length and is related to the electric charge q of usual dimension by the relation:
$\begin{eqnarray}{Q}^{2}=\displaystyle \frac{{q}^{2}g}{4\pi {\varepsilon }_{0}{c}^{4}}.\end{eqnarray}$
The metric (2.1 ) violates the weak energy condition [13, 14] at the region $r\lt {r}_{C}=\tfrac{Q\dot{Q}}{\dot{M}}$ (The overdot means the partial derivative with respect to time v -$\dot{M}=\tfrac{\partial M}{\partial v}$). However, particles cannot cross this boundary due to the Lorentz force [15]. We restrict the consideration by the assumption that r > rC. If we put $M(v)\equiv M={\rm{const}}.$ and $Q(v)\equiv Q={\rm{const}}.$ then the metric (2.1 ) becomes Reissner-Nordstrom in advanced Eddington-Finkelstein coordinates. If we consider neutral particles then there are no particles with negative energy outside the event horizon $r={r}_{h}=M+\sqrt{{M}^{2}-{Q}^{2}}$. The same statement is valid in metric (2.1 ) [16]. The situation changes when one considers the non-geodesic motion of the charged particles [7, 8].
The metric (2.1 ) is spherically symmetric and we can consider the motion in the equatorial plane $\theta =\tfrac{\pi }{2}$. From the spherical symmetry, we have the constant of the motion - the angular momentum per mass L:
$\begin{eqnarray}L={r}^{2}\displaystyle \frac{{\rm{d}}\varphi }{{\rm{d}}\lambda },\end{eqnarray}$
where λ is the affine parameter.
The energy per mass expression for the particle with an electric charge per mass Q* reads:
$\begin{eqnarray}E(v)=\left(1-\displaystyle \frac{2M}{r}+\displaystyle \frac{{Q}^{2}}{{r}^{2}}\right)\displaystyle \frac{{\rm{d}}{v}}{{\rm{d}}\lambda }-\displaystyle \frac{{\rm{d}}{r}}{{\rm{d}}\lambda }+\displaystyle \frac{{{QQ}}^{* }}{r}.\end{eqnarray}$
And using the fact that gikuiuk = − 1 we formally obtain:
$\begin{eqnarray}\begin{array}{l}{\left(\displaystyle \frac{{\rm{d}}{r}}{{\rm{d}}\lambda }\right)}^{2}={\left(E(v)-\displaystyle \frac{{{QQ}}^{* }}{r}\right)}^{2}-\left(1-\displaystyle \frac{2M}{r}+\displaystyle \frac{{{QQ}}^{* }}{{r}^{2}}\right)\\ \,\times \left(\displaystyle \frac{{L}^{2}}{{r}^{2}}+1\right).\end{array}\end{eqnarray}$
Now, one should express the energy to understand the possible minimum of energy.
$\begin{eqnarray}{E}_{\max }=\displaystyle \frac{{{QQ}}^{* }}{r}+\sqrt{\left(1-\displaystyle \frac{2M}{r}+\displaystyle \frac{{{QQ}}^{* }}{{r}^{2}}\right)\left(\displaystyle \frac{{L}^{2}}{{r}^{2}}+1\right)+{\left(\displaystyle \frac{{\rm{d}}{r}}{{\rm{d}}\lambda }\right)}^{2}}.\end{eqnarray}$
We chose the 'plus' sign due to the fact that the energy E must be positive for particles at infinity in the case Q* = 0. One can have the negative energy E only in the case Q* < 0 i.e. a test particle has the opposite electric charge than a black hole. The energy E has its minimum if we put $\tfrac{{\rm{d}}{r}}{{\rm{d}}\lambda }=0$ and L = 0, substituting these conditions into the equation above, one can obtain the generalized ergosphere: $\begin{eqnarray}-{{QQ}}^{* }\geqslant {r}^{2}\sqrt{1-\displaystyle \frac{2M}{r}+\displaystyle \frac{{{QQ}}^{* }}{{r}^{2}}}.\end{eqnarray}$
One should understand that we cannot take the charge of a black hole Q as large as one wants due to the fact that we have the condition of black hole formation M ≥ Q. Otherwise, one has the naked singularity which we do not consider here.However, metric (2.1 ) is time-depended and as a result, the energy is not a constant. So, to investigate the Penrose process, equation (2.5 ) does not suit us.
In the general case, the charged Vaidya spacetime does not possess any additional symmetry. However, for the particular choice of the mass and charge function, the metric (2.1 ) admits the conformal Killing vector [17]. In this case, M and Q must have the following form [18]:
$\begin{eqnarray}\begin{array}{rcl}M(v) & = & \nu v,\nu \gt 0,\\ Q(v) & = & \mu v,\mu \ne 0.\end{array}\end{eqnarray}$
To avoid the naked singularity, one should demand ν ≥ μ. Thus, by doing the following coordinate transformation [19]:
$\begin{eqnarray}\begin{array}{rcl}v & = & {r}_{0}{{\rm{e}}}^{\tfrac{t}{{r}_{0}}},\\ r & = & {{R}{\rm{e}}}^{\tfrac{t}{{r}_{0}}}.\end{array}\end{eqnarray}$
one obtains the charged Vaidya spacetime in conformally static coordinates: $\begin{eqnarray}\begin{array}{rcl}{{\rm{d}}{s}}^{2} & = & {{\rm{e}}}^{\tfrac{2t}{{r}_{0}}}\left[-\left(1-\displaystyle \frac{2\nu {r}_{0}}{R}+\displaystyle \frac{{\mu }^{2}{r}_{0}^{2}}{{R}^{2}}-2\displaystyle \frac{R}{{r}_{0}}\right){{\rm{d}}{t}}^{2}\right.\\ & & \left.+2{\rm{d}}{t}{\rm{d}}{R}+{R}^{2}{\rm{d}}{{\rm{\Omega }}}^{2}\right].\end{array}\end{eqnarray}$
The conformal Killing vector becomes:
$\begin{eqnarray}v\displaystyle \frac{{\rm{d}}}{{\rm{d}}{v}}+r\displaystyle \frac{{\rm{d}}}{{\rm{d}}{r}}={r}_{0}\displaystyle \frac{{\rm{d}}}{{\rm{d}}{t}}.\end{eqnarray}$
And it is timelike when: $\begin{eqnarray}1-\displaystyle \frac{2\nu {r}_{0}}{R}+\displaystyle \frac{{\mu }^{2}{r}_{0}^{2}}{{R}^{2}}-2\displaystyle \frac{R}{{r}_{0}}\gt 0.\end{eqnarray}$
Now, we have two constants of motion - energy E and angular momentum L :
$\begin{eqnarray}\begin{array}{rcl}E & = & {{\rm{e}}}^{\tfrac{2t}{{r}_{0}}}\left[\left(1-\displaystyle \frac{2\nu {r}_{0}}{R}+\displaystyle \frac{{\mu }^{2}{r}_{0}^{2}}{{R}^{2}}-2\displaystyle \frac{R}{{r}_{0}}\right)\displaystyle \frac{{\rm{d}}{t}}{{\rm{d}}\lambda }-\displaystyle \frac{{\rm{d}}{R}}{{\rm{d}}\lambda }\right]\\ & & +\displaystyle \frac{\mu {Q}^{* }{r}_{0}}{R},\\ L & = & {{\rm{e}}}^{\tfrac{2t}{{r}_{0}}}{R}^{2}\displaystyle \frac{{\rm{d}}\varphi }{{\rm{d}}\lambda }.\end{array}\end{eqnarray}$
Substituting (2.13 ) into the normalization condition gikuiuk = − 1 we obtain:
$\begin{eqnarray}\begin{array}{l}{{\rm{e}}}^{\tfrac{4t}{{r}_{0}}}{\left(\displaystyle \frac{{\rm{d}}{R}}{{\rm{d}}\lambda }\right)}^{2}={\left(E-\displaystyle \frac{\mu {Q}^{* }{r}_{0}}{R}\right)}^{2}\\ \quad -{{\rm{e}}}^{\tfrac{2t}{{r}_{0}}}\left(1-\displaystyle \frac{2\nu {r}_{0}}{R}+\displaystyle \frac{{\mu }^{2}{r}_{0}^{2}}{{R}^{2}}-2\displaystyle \frac{R}{{r}_{0}}\right)\\ \quad \times \left(\displaystyle \frac{{L}^{2}}{{{R}^{2}}^{\tfrac{2t}{{r}_{0}}}}+1\right).\end{array}\end{eqnarray}$
Or, expressing the energy E, we have:2.15 ). By doing this we obtain the region where particles with negative energy can exist:2.16 ) gives the border of the generalized ergosphere. The first thing that one can notice from (2.16 ) is that the generalized ergosphere is temporal and its border is shrinking during the evolution of the system. It means that the charged particles with negative energy can exist only for the same period of time. Hence, all such particles must inevitably fall on the apparent horizon and closed orbits cannot exist for such particles because as soon as the generalized ergosphere disappears only particles with positive energy are allowed.
$\begin{eqnarray}\begin{array}{l}E=\displaystyle \frac{\mu {Q}^{* }{r}_{0}}{R}+{{\rm{e}}}^{\displaystyle \frac{t}{{r}_{0}}}\\ \quad \times \sqrt{\left(1-\displaystyle \frac{2\nu {r}_{0}}{R}+\displaystyle \frac{{\mu }^{2}{r}_{0}^{2}}{{R}^{2}}-2\displaystyle \frac{R}{{r}_{0}}\right)\left(\displaystyle \frac{{L}^{2}}{{R}^{2}{{\rm{e}}}^{\tfrac{2t}{{r}_{0}}}}+1\right)+{{\rm{e}}}^{\tfrac{2t}{{r}_{0}}}{\left(\displaystyle \frac{{\rm{d}}{R}}{{\rm{d}}\lambda }\right)}^{2}}.\end{array}\end{eqnarray}$
One should pick up the plus sign as we did above because the observer, which moves along t line, must measure the positive energy for neutral particles at R = r0. Again, we have the negative energy only if Q* < 0 i.e. the particle and a black hole have opposite electric charges. To obtain the generalized ergosphere one should put $\tfrac{{\rm{d}}{R}}{{\rm{d}}\lambda }=0$ and L = 0 in ( $\begin{eqnarray}-\mu {Q}^{* }{r}_{0}\geqslant {{\rm{e}}}^{\tfrac{t}{{r}_{0}}}R\sqrt{1-\displaystyle \frac{2\nu {r}_{0}}{R}+\displaystyle \frac{{\mu }^{2}{r}_{0}^{2}}{{R}^{2}}-2\displaystyle \frac{R}{{r}_{0}}}.\end{eqnarray}$
The extremal case of inequality (Note that in the Kerr metric, the ergosphere is the surface on which the Killing vector $\tfrac{{\rm{d}}}{{\rm{d}}{t}}$ becomes null and, as a result, it is defined purely geometrically. The generalized ergosphere (2.16 ), on the other hand, depends upon the particle parameters, especially upon the particle charge Q*. As a result, for each particle the border of the generalized ergosphere and the time of its existence are different. This quantity shows only the region where the particle with certain parameters might possess negative energy.
Penrose process
In this section we investigate the energy extraction process which was proposed by Penrose [4]. The mechanism we use is as follows: a negatively charged particle gets into the generalized ergosphere and breaks up into two fragments, one of which escapes to the observer at R = r0 (r0 must be in the region where the conformal Killing vector $\tfrac{{\rm{d}}}{{\rm{d}}{t}}$ is timelike) with more energy than the original particle. The remaining particle with negative energy falls into a black hole which we showed in the previous section. Let us denote the initial particle as subscript 0, the outgoing particle as 2, and the falling particle as 1. We consider the equatorial motion, so the break up point is located in {R*, φ*} point which must be inside of the generalized ergosphere.
The quantities that characterize each particle are related by conservation equations. Charge conservation, for instance, yields:
$\begin{eqnarray}{Q}_{0}^{* }{m}_{0}={Q}_{1}^{* }{m}_{1}+{Q}_{2}^{* }{m}_{2}.\end{eqnarray}$
The conservation of the four-momentum reads:3.18 ) are three different conservation equations with straightforward physical interpretation. The zero component is the conservation of energy:
$\begin{eqnarray}{p}_{0}^{\mu }={p}_{1}^{\mu }+{p}_{2}^{\mu }.\end{eqnarray}$
The equations ( $\begin{eqnarray}{E}_{0}{m}_{0}={E}_{1}{m}_{1}+{E}_{2}{m}_{2},\end{eqnarray}$
the R component corresponds to the conservations of linear momenta: $\begin{eqnarray}{m}_{0}{\left(\displaystyle \frac{{\rm{d}}{R}}{{\rm{d}}\lambda }\right)}_{0}={m}_{1}{\left(\displaystyle \frac{{\rm{d}}{R}}{{\rm{d}}\lambda }\right)}_{1}+{m}_{2}{\left(\displaystyle \frac{{\rm{d}}{R}}{{\rm{d}}\lambda }\right)}_{2},\end{eqnarray}$
and the φ component means the angular momentum conservation: $\begin{eqnarray}{m}_{0}{L}_{0}={m}_{1}{L}_{1}+{m}_{2}{L}_{2}.\end{eqnarray}$
Note that all derivatives in (3.20 ) are evaluated at the break-up point. Also, if we square (3.18 ) by keeping in mind the fact that $| {p}_{i}^{\mu }{| }^{2}={m}_{i}^{2}$ and pμ is timelike and future-directed, we come to the inequality:3.19 ),:3.23 ) shows that E2 > E0 and we have the extraction of the energy from the black hole. As we know that particle 1 with negative energy must fall into a black hole, hence, it will directly decrease the energy associated with the black hole, as in Penrose's original proposal.
$\begin{eqnarray}{m}_{0}^{2}\geqslant {m}_{1}^{2}+{m}_{2}^{2}.\end{eqnarray}$
The energy which is carried away by particle 2 is, according to ( $\begin{eqnarray}{m}_{2}{E}_{2}={m}_{0}{E}_{0}-{E}_{1}{m}_{1}.\end{eqnarray}$
As we found out in the previous section, a negatively charged particle in the generalized ergosphere that falls into a black hole might possess negative energy. If this is the case, then (The maximum of the energy extraction
The efficiency η of the Penrose process can be defined as:2.15 ) the energy of particle 1 is most negative if the particle is initially at rest i.e.:
$\begin{eqnarray}\eta =\displaystyle \frac{{m}_{2}{E}_{2}}{{m}_{0}{E}_{0}}-1=-\displaystyle \frac{{m}_{1}{E}_{1}}{{m}_{0}{E}_{0}}.\end{eqnarray}$
To maximize we should make E0 as small as possible and ∣E1∣ as large as possible. In other words, we want to extract as much energy as possible starting with as little energy as possible. Let us put E0 = 1 for simplicity. Also, we assume that particle 0 does not possess the angular momentum L0 = 0. From ( $\begin{eqnarray}\begin{array}{rcl}{\left(\displaystyle \frac{{\rm{d}}{R}}{{\rm{d}}\lambda }\right)}_{1} & = & {\left(\displaystyle \frac{{\rm{d}}\varphi }{{\rm{d}}\lambda }\right)}_{1}=0,\\ E & = & \displaystyle \frac{\mu {Q}^{* }{r}_{0}}{R}+{{\rm{e}}}^{\displaystyle \frac{t}{{r}_{0}}}\\ & & \times \sqrt{\left(1-\displaystyle \frac{2\nu {r}_{0}}{R}+\displaystyle \frac{{\mu }^{2}{r}_{0}^{2}}{{R}^{2}}-2\displaystyle \frac{R}{{r}_{0}}\right)}.\end{array}\end{eqnarray}$
From the condition (4.25 ) one can easily see that L0 = L1 = L2 = 0. Now we should evaluate the mass m1, for this purpose, let us express m2 from (3.20 ), keeping in mind that ${\left(\tfrac{{\rm{d}}{R}}{{\rm{d}}\lambda }\right)}_{1}\,=\,0$:
$\begin{eqnarray}{m}_{2}^{2}={\left[\displaystyle \frac{{m}_{0}{\left(\tfrac{{\rm{d}}{R}}{{\rm{d}}\lambda }\right)}_{0}}{{\left(\tfrac{{\rm{d}}{R}}{{\rm{d}}\lambda }\right)}_{2}}\right]}^{2}.\end{eqnarray}$
Now, by using (3.22 ) and (4.26 ), we obtain the following restriction on η:4.27 ) one can see that the maximum energy, which particle 2 can carry out, cannot exceed the absolute value of the energy of the falling particle 1. As we have in the case of the boundary of the generalized ergosphere, the efficiency η depends on the particle charge ${Q}_{1}^{* }$. Note that if the radial velocities of falling and escaping particles are equal, then we do not have any energy extraction at all. To have energy extraction the following condition must be held:
$\begin{eqnarray}\begin{array}{l}\eta \lt -\left[\displaystyle \frac{\mu {Q}^{* }{r}_{0}}{R}+{{\rm{e}}}^{\displaystyle \frac{t}{{r}_{0}}}\sqrt{\left(1-\displaystyle \frac{2\nu {r}_{0}}{R}+\displaystyle \frac{{\mu }^{2}{r}_{0}^{2}}{{R}^{2}}-2\displaystyle \frac{R}{{r}_{0}}\right)}\right]\\ \qquad \times \sqrt{1-\displaystyle \frac{{\left(\tfrac{{\rm{d}}{R}}{{\rm{d}}\lambda }\right)}_{0}^{2}}{{\left(\tfrac{{\rm{d}}{R}}{{\rm{d}}\lambda }\right)}_{2}^{2}}}.\end{array}\end{eqnarray}$
From ( $\begin{eqnarray}\left|{\left(\displaystyle \frac{{\rm{d}}{R}}{{\rm{d}}\lambda }\right)}_{2}\right|\gt \left|{\left(\displaystyle \frac{{\rm{d}}{R}}{{\rm{d}}\lambda }\right)}_{0}\right|.\end{eqnarray}$
The efficiency η has the upper bound if the collision takes place at the vicinity of $R={R}_{\max }$, where Rmax is the solution of the following equation:
$\begin{eqnarray}1-\displaystyle \frac{2\nu {r}_{0}}{{R}_{\max }}+\displaystyle \frac{{\mu }^{2}{r}_{0}^{2}}{{R}_{\max }^{2}}-2\displaystyle \frac{{R}_{\max }}{{r}_{0}}=0.\end{eqnarray}$
It means that the collision takes place at the vicinity of the conformal Killing horizon. If we take into account the condition (4.29 ) then one obtains:
$\begin{eqnarray}\eta \lt -\displaystyle \frac{\mu {Q}_{1}^{* }{r}_{0}}{{R}_{\max }}\sqrt{1-\displaystyle \frac{{\left(\tfrac{{\rm{d}}{R}}{{\rm{d}}\lambda }\right)}_{0}^{2}}{{\left(\tfrac{{\rm{d}}{R}}{{\rm{d}}\lambda }\right)}_{2}^{2}}}.\end{eqnarray}$
So a particle 2 can carry away only a small amount of the energy if the velocities of particles 1 and 2 are of the same order. To evaluate the upper bound of η one needs to know r0 (which is an arbitrary constant. The only restriction, we impose on r0, is that r0 must be in the region where the conformal Killing vector is timelike) and rmax from (4.30 ). However, we can improve this result by noting that ∣ − μr0∣ ≤ νr0 ≤ M where M is a black hole total mass. Then one has:4.30 ). Taking into consideration all these facts, one obtains:
$\begin{eqnarray}\eta \lt \displaystyle \frac{M| {Q}_{1}^{* }| }{{r}_{\max }}\sqrt{1-\displaystyle \frac{{\left(\tfrac{{\rm{d}}{R}}{{\rm{d}}\lambda }\right)}_{0}^{2}}{{\left(\tfrac{{\rm{d}}{R}}{{\rm{d}}\lambda }\right)}_{2}^{2}}}.\end{eqnarray}$
Now, we assume that particle 2 is an ultra-relativistic one and its radial velocity is much large than the velocity of an initial particle. Further, we consider the extremal black hole which means that ν = ∣μ∣. Also, we know that in the case of the usual Vaidya spacetime with a linear mass function, one has a restriction $\nu \leqslant \tfrac{1}{16}$ [17]. For the value $\nu =\tfrac{1}{16}$, one has ${r}_{\max }=2M$ as one of the roots of the ( $\begin{eqnarray}\eta \lt \displaystyle \frac{| {Q}_{1}^{* }| }{2}.\end{eqnarray}$
We obtain the result that the upper bound for η cannot be larger than half of an absolute value of the electric charge per mass of particle 1.We found out that the charged particle with negative energy will fall into a black hole. This particle must be the opposite charge of the black hole. From this fact, we can conclude that The extraction energy decreases the energy associated with the black hole, charge.
Conclusion
In this paper, the analogue of the Penrose process has been considered in the case of charged Vaidya spacetime. Despite the fact that this spacetime is time-depended, it admits an additional symmetry due to a conformal Killing vector field for the special choice of the mass and charge functions. We have found out that the charged particles with negative energy can exist in the generalized ergoregion only if the electric charge of the particle is opposite to the charge of a black hole. We have calculated the border of this region - generalized ergosphere and showed that its location depends upon the particle's charge. This border is shrinking. It means that the generalized ergosphere is temporary and it will disappear. Hence, closed orbits for charged particles with negative energy are impossible.
However, when the generalized ergosphere exists there is a possibility of energy extraction from a black hole via the Penrose process. Since the charged particle must possess the opposite electric charge to the black hole, energy extraction is possible due to reducing the black hole charge. We have estimated the efficiency of this process and found out that energy extraction is possible if the speed of the escaping particle is bigger than the speed of the original one. The upper bound of efficiency can be achieved only if the process occurs in the vicinity of the conformal Killing horizon. The energy which is carried away by the escaping particle is bounded and the efficiency cannot exceed the absolute value of the energy of the falling particle with negative energy.