Extraction of Rotational Energy from Kerr-MOG Black Hole

We study the energy extraction by the Penrose process in Kerr-MOG black hole~(BH). We derive the gain in energy for Kerr-MOG is $\Delta {\cal E} \leq \frac{1}{2}\left(\sqrt{\frac{2}{1+\sqrt{\frac{1}{1+\alpha}-\left(\frac{a}{{\cal M}}\right)^2}} -\frac{\alpha}{1+\alpha} \frac{1}{\left(1+\sqrt{\frac{1}{1+\alpha}-\left(\frac{a}{{\cal M}}\right)^2} \right)^2}}-1\right) $. Where $a$ is spin parameter, $\alpha$ is MOG parameter and ${\cal M}$ is the Arnowitt-Deser-Misner(ADM) mass parameter. When $\alpha=0$, we find the gain in energy for Kerr BH. For extremal Kerr-MOG BH, we find the maximum gain in energy is $\Delta {\cal E} \leq \frac{1}{2} \left(\sqrt{\frac{\alpha+2}{1+\alpha}}-1 \right)$. We observe that the MOG parameter has a crucial role in the energy extraction process and it is in fact reducing the value of $\Delta {\cal E}$ in contrast with extremal Kerr BH. Finally, we discuss the \emph{Wald inequality and the Bardeen-Press-Teukolsky inequality} for Kerr-MOG BH in contrast with Kerr BH.


Introduction
Perhaps, BH is the most facinating as well as compact objects in the universe. They are responsible for most astrophysical phenomena in the universe. The astrophysical phenomena are solar system formation and evolution, galaxy formation and evolution, stellar dynamics and evolution, the largescale structure of matter in the universe, the evolution of stars and galaxies (which powers the active galactic nuclei), gamma ray bursts, BH mergers and coalescence of binary BHs (which produces the gravitational waves) etc. The existence of this waves have been tested recently by LIGO and Virgo team [1]. It was generally believed that these waves could be produced by merging of two BHs or other compact objects like neutron stars. In near future it could be possible to verify the above test for certain class of BHs in alternative theory of gravity like modified gravity (MOG) [2].
Classically it should not be possible to extract energy from non-spinning BH however it should be possible to extract rotational energy from spinning BH [11,10,12]. The main concerns regarding the extraction of energy due to presence of the ergosphere in the spinning BH whereas the non-spinning BH does not possess such ergosphere. Such facinating process was put forward by Roger Penrose in 1969 [6,7]. This process is now popularly known as Penrose process in Einstein's general relativity. Neverthless, a simple gedanken experiment proves how the ergosphere could be in principle exploited to extract the rotational energy from the BH. To extract both the energy and the charge from a BH one must see the work of Bekenstein [8]. In this work, the author showed how to extract both the charge and the coulomb energy by using Hawking's area theorem [17].
The most striking feature of spinning BH is that the Killing vector ξ µ = ∂ 0 which is time-like at ∞ becomes space-like in the ergosphere (i.e. the toroidal space between the event horizon and the stationarity limit surface on which the components of the axially symmetric metric g 00 = 0). Another striking feature is the existence of particle orbits with negative total energy which could be measured from infinity. This energy is defined as E = −p µ ξ µ , where ξ µ is the four momentum of the test particle. Outside the ergosphere (where ξ µ is time-like) the energy must be positive, however inside the ergosphere (where ξ µ is time-like) the energy has the nature of a spatial component of momentum and have either sign [7,9,11,12].
Penrose first proposed that one should take the advantage of these negative orbits to extract rotational energy from the BH. The process could be understood shortly as follows. In this process, a particle falls into the ergosphere from infinity. Then it decays into two fragments. One fragment escapes to infinity and other fragment plunges through the event horizon into the BH. Both the energy and the momentum conserved in this hypothetical process. Therefore, one may extract the rotational energy from the BH. It should be noted that in the ergosphere, the Killing vector ∂ 0 becomes spacelike as said previously and similarly the conserved component, p 0 , of the four-momentum. Therefore when an observer observes the toroidal space from infinity he/she could be discerned that the energy of the particle becomes negative. Due to this negative energy, one may able to extract both the energy and the angular momentum from the BH. However the area of BH's event horizon never decreases. Either it must be increases or remains constant.
In this Letter, motivated by the work of Penrose process for Kerr BH we would like to extend this work for Kerr-MOG BH. Because it consists of three parameters i.e. namely the spin parameter a, the ADM mass parameter M and the MOG parameter (α). Whereas the Kerr BH consists of only two parameters i.e. the ADM mass parameter and the spin parameter. Due to the presence of the deformation parameter what would be the change in the "gain in energy expression" in extraction process in contrast to the Kerr BH. This is the primary motivation behind this work. We also investigated the Wald inequality which gives us the energy limits on the energy extraction process. Futhermore, we have discussed the Bardeen-Press-Teukolsky inequality. Finally we have considered the reversible extraction of energy and the irreducible mass for Kerr-MOG BH.
The MOG is a kind of dark-matter dominated theory of gravity [2]. It is well known that Einstein's general gravity is the most well-accepted theory of gravity. However it disrupted at short length scales in which one must have considered the quantum effect. The MOG is formulated by scalar fields and massive vector fields that's why the MOG theory is also called the scalar-tensor-vectorgravity (STVG). The MOG theory correctly interpreted the observations of the solar system [2]. It also explains the rotation curves of the cluster of galaxies and the dynamics of the cluster of galaxies. Moreover, the STVG theory correctly describes the power spectrum of matter and the acoustical power spectrum of the cosmic microwave background (CMB) data [2].
The modified action for the STVG theory is equal to the sum of four actions, namely the Einstein-Hilbert action for gravity, the action for massive vector field, the action for scalar fields and the action for pressure less matter. This means that the MOG theory is satisfied the action principle.
We should mention here the several works regarding the MOG theory. First the ADM mass and the spin parameter of Kerr-MOG BH have been computed in [19]. The gravitational lensing properties of Kerr-MOG has been studied in [5]. The BH shadow has been examined in [3]. The Kerr-MOG BH merger and the ringdown radiation have been considered in [18]. The thermodynamics of Kerr-MOG BH has been explicitly studied in [4,23]. The superradiance in Kerr-MOG has been examined in [22] very recently.
The paper is organized as follows. In Sec. 2, we have derived the Penrose process for Kerr-MOG BH. In Sec. 3, we have discussed the energy limits on the Penrose process followed by the work of Wald. The Bardeen-Press-Teukolsky inequality derived in Sec. 4. In Sec. 5, we have introduced the concept of irreducible mass in Kerr-MOG BH. Finally, we have given a brief discussion and outlook in Sec. 6. In Appendix, we have computed the ISCO energy for extremal Kerr-MOG BH.

The Penrose Process in Kerr-MOG BH
Before describing the Penrose process we should first describe the basic feature of Kerr-MOG BH. It is an axisymmetric class of spinning BH and it is described by the ADM mass parameter (M), spin parameter (a) and a deformation parameter or MOG parameter (α). This parameter α = G−G N G N should be measured deviation of MOG from GR. The basic postulate in MOG theory is that the charge parameter is proportional to the square root of the MOG parameter i.e.
The Kerr-MOG BH metric (in units where c = 1) can be written in Boyer-Lindquist coordinates (t, r, θ, φ) as [3] where where G N is a Newtonian constant. For simplicity, we have to take G N = 1 hereafter and througout the work. The ADM mass and angular momentum are computed in [19] as M = (1 + α)M and J = aM. After substituting these values the ∆ r becomes The BH consists of two horizons namely event horizon (r + ) and Cauchy horizon (r + ). They are denoted as It may be noted that when α = 0, one obtains the horizon radii of Kerr BH. The BH solution exists when M 2 1+α > a 2 . When M 2 1+α = a 2 , one finds the extremal BH. When M 2 1+α < a 2 , one obtains the naked singularity case. The behavior of the outer horizon and inner horizon could be found in the Fig. 1. It follows from the figure that the presence of the MOG parameter could somehow deformed the shape of the horizon radii. The ergosphere is situated at This surface is outer to the event horizon and it coincides with event horizon at the poles θ = 0 and θ = π.
To obtain the radial equation for the geodesic motion of a test particle in Kerr-MOG BH, we have followed the book of S. Chandrashekar [9]. We should also restricted in the equatorial plane. Therefore the Lagrangian density for the geodesic motion of a test particle could be written as The radial equation that governs the geodesic structure of Kerr-MOG BH is given bẏ where ǫ = −1 for time-like geodesics and ǫ = 0 for null geodesics. Also, E corresponds to the energy and ℓ corresponds to the angular momentum of the test particle.  To study the Penrose process one should use the radial geodesic equation i.e. Eq. (7) then Since there is no contribution to E from the kinetic energy part hence one could solve the above equation for both E and ℓ as separately then where where The above equations have been derived using the following important identity Using Eq. (9), one could derive the condition while the value of the energy is negative as discerned by an observer at infinity. With out loss of generality we have taken the value of E = 1 when a particle of unit mass, at rest at infinity. Therefore at the present moment we have considered the positive sign in the right hand side of the Eq. (9). Thus it must be obeyed that the following criterion should be satisfied for E < 0, ℓ < 0 and Using Eq. (11), the above inequality could be written as It immediately suggests that E < 0 if and only if ℓ < 0. Also and Therefore the only possibility in the equatorial plane is that the counter-rotating particles should have negative energy and it happens inside the ergosphere. This ergosphere radius for Kerr-MOG BH has been given in Eq. (5). For extremal Kerr-MOG BH, the ergo-sphere occurs at r e (θ) = M + a sin θ which is exactly same as the ergosphere radius of extreme Kerr BH. What exactly happens in this process is that when a particle at rest at infinity arrives at a point r < a + M in the equatorial plane it has a turning point in such a way thatṙ = 0. At the meeting point r, the particle splits into two photons: one photon crosses the event horizon and is lost when the other one escapes to infinity. We could arrange this process in such a way that the photon which crosses the event horizon has negative energy and the photon which escapes to infinity has more energy than the particle which arrived from infinity. Now let us suppose E (x) = 1, ℓ (x) ; E (y) , ℓ (y) ; and E (z) , ℓ (z) are the energies and the angular momentum of the particle arriving from infinity and of the photons which cross the outer horizon and escape to infinity, respectively.
Since the particles come from infinity and get at r followed by a time-like circular geodesics then it has a turning point at r, its angular momentum, ℓ (x) , could be determined from Eq. (10) by putting ǫ = −1, E = 1. Therefore one gets, Similarly, substituting the value of ǫ = 0 in Eq. (10) one would get the relation between the energy and the angular momenta of the photon which crosses the event horizon and the photon which escapes to infinity as and Now the conservation of energy and angular momentum gives us and After solving the above equations, we find and Putting the values of χ (x) , χ (y) , and χ (z) by using Eqns. (15), (17), we find and In the limit α = 0, one obtains the energy value for Kerr BH. The energy gain ∆E in this process becomes The maximum gain in energy occurs at the event horizon and this value is given by The variation of ∆E with r + could be observed from the Fig. (3). The gain in energy in terms of spin parameter and MOG parameter is This is the key prediction of this work. It is clearly evident that the gain in energy strictly depends upon the MOG parameter. The effect of this parameter could be seen from the energy gain versus spin diagram (Fig. 4). From this diagram, one could say that there is a direct influence of the MOG parameter in the energy extraction process. When α = 0, the energy gain in Penrose process increases while the spin parameter increases. This scenario is quite different when we add the parameter α. In this case the energy gain is very slower than the former case. In-fact, the energy gain is one half of the previous value. When α = 0, one finds the energy value for Kerr BH. For extremal Kerr-MOG BH, the maximum gain in energy is given by This implies that the deformation parameter plays an important role in the energy extraction process, it is in fact decreasing the value of ∆E in comparison with extremal Kerr BH. In Fig. 5, we have plotted 3D diagram of energy gain in Penrose process for various parameter space. From these figures we can easily see that how the deformation parameter affects in the energy extraction process for Kerr-MOG BH.

The Wald inequality
It is very important to investigate what is the energy limits in the Penrose process for Kerr-MOG BH? In this section, we would try to resolve this issue. Wald [21] was first able to derive this limits. He also derived an inequality which explains the origin and the limitations of this process. To do this let us consider a particle, with a four velocity U µ and specific energy E, breaks up into fragments. Let ε be the specific energy and u µ be the four-velocity of one of the fragments. Now we want to derive the limits on ε.
Choose an orthonormal tetrad-frame, e µ b , in which U µ coincides with e µ 0 and the remaining spacelike basis vectors are e µ (ζ) (ζ = 1, 2, 3): In this frame where v (ζ) are the spatial components of the three-velocity of the fragment η = 1 √ 1−|v| 2 and |v| 2 = v (ζ) v (ζ) . Since the spacetime has time-like Killing vector ξ = ∂ 0 then it could be represent in tetradframe as Now the conserved quantity energy E could be represent in terms of Killing vector as and Therefore one obtains Using Eq. (29), one could obtain the specific energy of the fragment as where ϑ is the angle between the three-dimensional vectors v (ζ) and ξ µ . Using Eq. (32) and Eq. (33), one could write the Eq. (34) as ε = ηE + η|v| E 2 + g 00 cos ϑ .
This equation provides the inequality This is called the famous Wald inequality. For Kerr-MOG BH this inequality becomes It has been shown that the maximum energy that a particle describing a stable circular orbit (See Appendix: Eq.(83)) is For ε to be negative, it is thus necessary that Otherwise, the fragments must have relativistic energies which becomes possible before any extraction of energy by the above process.

The Bardeen-Press-Teukolsky Inequality
In this section, we shall review what is the lower bound on the magnitude of three velocity between two particles of different specific energies followed by two orbits and collide at some point [14]. Let the two particles have specific energies as E 1 and E 2 . Also let the magnitude of three velocity between two particles be ̟. Suppose we have an orthonormal tetrad-frame as defined previously, in which the two orbits cross with equal and opposite three velocities, +v (ζ) and -v (ζ) so that The four velocities, u µ 1 and u µ 2 of two particles in the said tetrad-frame at the time of collision are . (42) where η = 1 √ 1−|v| 2 . As proceeding previously the space-time allows a time-like Killing vector ξ = ∂ 0 then its representation in tetrad-frame be Now, by definition, so that The specific energies at the time of collision are given by and where ϑ is the angle between the 3-vectors v (ζ) and ξ µ . From the preceeding equations we can write Therefore, = |v| 2 4η 2 ξ 2 (0) + 4η 2 g 00 cos 2 ϑ = |v| 2 (E 1 + E 2 ) 2 + 4η 2 g 00 cos 2 ϑ It indicates that Substituting the value of η, one obtains or re-arranging this equation It follows that and the required lower bound on ̟ according to Eq. (41); and consequently the inequality is called well-known Bardeen-Press-Teukolsky inequality [14].
In case of Kerr-MOG BH, let the particle with the energy E 1 followed by a stable circular geodesics in the equatorial plane then its maximum energy is given in Eq. (38). Since the value of g 00 = 1 − α 1+α and choosing the value of E 2 = 0, the inequality (58) becomes and subsequently the inequality for ̟ is which is in agreement with the result (39) performed from Wald's inequality. In the limit α = 0, one gets the result for Kerr BH. The key conclusion from the two ineqalities are that to achieve effective energy extraction from Penrose process, one should first accelerate the particle pieces to more than √ 1+α 2 times the speed of light by hydrodynamical forces.

The Irreducible Mass & Reversible Extraction of Energy
In a landmark paper "Reversible Transformations of a Charged Black Hole" [16], Christodoulou and Ruffini have derived an important relation between energy of a charged rotating BH and the irreducible mass [15] of the BH. Using similar analogy, in this section we would like to provide the relation between the energy and the irreducible mass for Kerr-MOG BH. It is now well established by fact that the BH area never decreases.
To prove the area of the BH always increases, we could define the "irreducible mass" [13] as For Kerr-MOG BH, it is given by One could derive more general inequality by using Eq. (9) if and only if The inequality should be equality if the process considered occurs at the outer horizon i.e.
(r 2 + + a 2 )r 2 Let a particle with negative energy, −E and an angular momentum, −ℓ approaching towards the outer horizon then the gain in energy δM(= E) and the gain in the angular momentum δJ(= ℓ) under the condition Let us consider the process should take place adiabatically then Therefore the inequality (66) becomes More precisely, this can be written as By the definition of irreducible mass it has been shown that for Kerr BH Using same analogy, one could say that for Kerr-MOG BH It implies that by no continuous process it is impossible to decrease the irreducible mass of a BH. We can also say that by no continuous process it is impossible to decrease the surface area of a BH. Where the surface area of a BH can be defined as We determine the rotational energy as For higher dimensional BH and black ring this has been studied by Nozawa et al. [20]. For extremal Kerr-MOG BH, one gets the ratio as when α = 0, ε Rot ≃ 29 percentage. When α = 0, ε Rot varies as in the Fig. 6. Using Eq. (72), one could say that by "no continuous process can the surface area of a BH be decreased" [9]. This is the outcome of Hawking's area theorem. It should be emphasized that the irreducible mass of a BH never be unchanged and the processes in which it should remain constant are said to be reversible one. We also should noted that by virtue of definition (62), the Christodoulou-Ruffini mass formula for Kerr-MOG BH becomes Now let us pause! What is the physical meaning of this equation. It indicates that if M irr is irreducible one then the second term J 2 4(1+α) 2 M 2 irr gives us towards the contribution of the rotational kinetic energy to the square of the inertial mass of the BH. This means that it is the rotational energy which is being extracted by the Penrose process.

Discussion and Outlook
We examined the Penrose process in Kerr-MOG BH. We derived the gain in energy for Kerr-MOG which is described by the Eq. (26). If α = 0, one obtained the gain in energy for Kerr BH. For extremal Kerr-MOG BH, we derived the maximum gain in energy is ∆E ≤ 1 2 α+2 1+α − 1 . We showed that the MOG parameter has an important role in the energy extraction process and it is in fact reduced the value of ∆E in contrast with extremal Kerr BH. Finally, we described the Wald inequality and the Bardeen-Press-Teukolsky inequality for Kerr-MOG BH in comparison with Kerr BH. It would be an interesting project if one could study the Blandford-Znajek process [24] for this BH where one may extract the rotational energy by electromagnetically from spinning BH.
The first one gives the direct ISCO for extremal Kerr-MOG BH which occurs at r isco = M when J M 2 ≥ 1 √ 2 . After substituting the value of r isco = M in Eq. (77), one can easily obtain the value of ISCO energy for direct orbit (in the extremal limit) In the limit α = 0, one gets the ISCO energy for extremal Kerr BH [14].