Dynamics of Charged and Rotating NUT Black Holes in Rastall Gravity
Hadyan L. Prihadi, Muhammad F. A. R. Sakti, Getbogi Hikmawan, Freddy P. Zen
aa r X i v : . [ g r- q c ] A ug August 28, 2019 0:28 WSPC/INSTRUCTION FILE ws-ijmpd
Prepared for Submission to International Journal of Modern Physics Dc (cid:13)
World Scientific Publishing Company
Dynamics of Charged and Rotating NUT Black Holes in RastallGravity
Hadyan L. Prihadi ,a , Muhammad F. A. R. Sakti ,b , Getbogi Hikmawan , c , and Freddy P.Zen , d Theoretical High Energy Physics, Department of PhysicsInstitut Teknologi Bandung, Jl. Ganeca 10 Bandung, Indonesia. Indonesia Center for Theoretical and Mathematical Physics (ICTMP)Institut Teknologi Bandung, Jl. Ganesha 10 Bandung, 40132, Indonesia. a [email protected], b m.fi[email protected] c getbogihikmawan@fi.itb.ac.id, d fpzen@fi.itb.ac.id Received Day Month YearRevised Day Month YearIn this work, we generalized the Kerr-Newman-NUT black hole solution in Rastall gravityfrom Ref. 1. Here we are more focused on the black hole dynamics such as the eventhorizons, ergosurface, ZAMO, thermodynamic properties, and the equatorial circularorbit around the black hole such as static radius limit, null equatorial circular orbit, andinnermost stable circular orbit. We present how the NUT and Rastall parameter affectsthe dynamic of the black hole.
Keywords : black hole, equatorial circular orbit, event horizon, Rastall gravity, thermo-dynamics.PACS numbers:
1. Introduction
In general relativity, Einstein field equation (EFE) gives many solutions that rep-resent a geometrical structure of space-time. Many EFE solutions give rise to anobject called the black hole, i.e. a region of space-time that possesses a coordinatesingularity called event horizon. Some of the black hole solutions that have beenfound are the static and spherically symmetric Schwarzschild black hole, the rotat-ing Kerr black hole, the charged Reissner-Nordstr¨om black hole, and the rotatingand charged Kerr-Newman black hole. Decades after the formulation of generalrelativity, the solutions to Einstein’s equations have been found to be even morenumerous, e.g. the black hole solution surrounded by Quintessence field, i.e. a modelof a scalar field that explains the expansion of the universe by governing a negativepressure,
3, 13 the Kiselev black hole and its rotating counterparts.
5, 6
Other stud-ies involving scalar field in cosmological scale can also be found in Ref. 14. There ugust 28, 2019 0:28 WSPC/INSTRUCTION FILE ws-ijmpd Hadyan L. Prihadi, Muhammad F. A. R. Sakti, Getbogi Hikmawan, Freddy P. Zen also exists a black hole solution form Einstein-Maxwell-Dilaton theory or even itshigher dimensional extension, and many more. Aside from the mathematical for-mulations, a black hole has also been found as a real astronomical object recently, e.g. the observation of gravitational waves from a merging of two black holes byLIGO, and the first ever real black hole image captured by the event horizon tele-scope. Other interesting black hole solutions is the one that came from a modified the-ory of gravity, such as Rastall gravity which said that theory of general relativitydoes not need to always fulfill the conservation of the energy-momentum tensor ∇ µ T µν = 0, but in general it could be ∇ µ T µν = λa ν for some vector a ν and aproportional constant λ and we recover the usual conservation law when λ → a ν also needs to be related to the space-time curvature,therefore the most suitable result is a ν = ∇ ν R , where R is the Ricci scalar. Variousblack hole solutions in Rastall gravity have been found, e.g. the static black hole and the charged rotating black hole solutions. Similar works involved in Rastallbackground recently can be found in Ref. 15.To make the solution more general, here we add the NUT parameter
16, 17 to acharged rotating black hole solution in Rastall gravity. The NUT parameter can beinterpreted as the twisting parameter of the space-time or the gravito-magnetic monopole. However, the physical interpretation of the NUT parameter is still de-batable until now. Similar work has been done in Ref. 1, but here we give even moregeneral result, by considering the θ dependence of the horizon, and focused more inthe thermodynamic stability and the geodesic of a particle around the black hole.In this work, we begin with obtaining the static and spherically symmetric so-lution in Rastall gravity, and generate the rotation and NUT parameter using theNewman-Janis algorithm in section 2, and we investigate the horizon and er-goregion structure in section 3, as well as the zero angular momentum observer(ZAMO). Thermodynamic properties are investigated in section 4, and its thermo-dynamic stability are also studied. In section 5, we worked on the equatorial circularorbit of a time-like and null-like particle, i.e. to study its static radius limit, nullequatorial circular orbit, and the innermost stable circular orbit (ISCO). Finally, insection 6, we summed up all of the topics with concluding remarks.
2. Kerr-Newman-NUT Solution in Rastall Gravity
In this section, we introduce the metric solution for a rotating and charged blackhole with NUT parameter in Rastall gravity. Based on Rastall hypothesis whichstated that the energy-momentum tensor T µν in general does not always satisfy theconservation law condition ∇ µ T µν = 0, but rather it should satisfy ∇ µ T µν = λ ∇ ν R (1)where λ is the proportional constant that measures the deviation from Einsteintheory of gravity. We can clearly see that if λ →
0, it recovers the conservation lawugust 28, 2019 0:28 WSPC/INSTRUCTION FILE ws-ijmpd
Dynamics of Charged and Rotating NUT Black Holes in Rastall Gravity of the energy-momentum tensor. Again, when we work in flat space-time, the usualconservation law is recovered. With that condition, the field equation becomes G µν + κλRg µν = κT µν , (2)where G µν = R µν − Rg µν is the usual Einstein tensor, λ is then called the Rastallparameter, and κ = πGc is the proportional constant in Einstein field equation thatconnects Einstein Tensor with energy-momentum tensor. This equation of motionwill be named the Einstein-Rastall field equation hereafter. Static and Spherically Symmetric Solution in Rastall Gravity
To find the solution for a spherically symmetric space-time in Rastall gravity, weconsider the Schwarzschild-like solution of the metric as ds = − f ( r ) dt + dr f ( r ) + r d Ω , (3)where f ( r ) is a function that depends only on coordinate r that will be obtainedby solving the Einstein-Rastall field equation in Eq. (2) with some properly chosenenergy-momentum tensor. Here d Ω = dθ + sin θdφ implies the two dimensionalsphere part of the metric. By plugging the metric from Eq. (3) to the Einstein-Rastall field equation in Eq. (2), we have H tt ≡ G tt + κλR = 1 r ( f ′ r − f ) + κλR, (4) H rr ≡ G rr + κλR = 1 r ( f ′ r − f ) + κλR, (5) H θθ ≡ G θθ + κλR = 1 r (cid:18) rf ′ + 12 r f ′′ (cid:19) + κλR, (6) H φφ ≡ G φφ + κλR = 1 r (cid:18) rf ′ + 12 r f ′′ (cid:19) + κλR, (7)where R is the Ricci scalar for the metric in Eq. (3), given by R = − r ( r f ′′ + 4 rf ′ − f ) . Here we use the prime notation as the derivative with respect to coordinate r , f ′ ≡ df /dr . From Eq. (4) to Eq. (7), we can see that H tt = H rr and H θθ = H φφ .It indicates that we have to look for the energy-momentum tensor that satisfiesboth T tt = T rr and T θθ = T φφ . We will see that the electromagnetic and quintessenceenergy-momentum tensor is a perfect match and the total energy-momentum tensorwill read as T µν = E µν + J µν , (8)where E µν is the electromagnetic energy-momentum tensor, and J µν is for thequintessence part.ugust 28, 2019 0:28 WSPC/INSTRUCTION FILE ws-ijmpd Hadyan L. Prihadi, Muhammad F. A. R. Sakti, Getbogi Hikmawan, Freddy P. Zen
The electromagnetic part of the energy-momentum tensor is given by
1, 12 E µν = 2 κ (cid:18) F µα F αν − F αβ F αβ g µν (cid:19) , (9)where F µν = ∂ µ A ν − ∂ ν A µ is the usual electromagnetic field strength tensor with theelectromagnetic vector potential A µ , that satisfies Maxwell Equation ∇ µ F µν = 0and Bianchi’s identity ∂ [ σ F µν ] = 0. Here we only interested in the electrostaticcharge Q in the radial direction that gives rise to A µ dx µ = − Qr dt → F tr = Qr . (10)From the equation above, the non-zero part of the electromagnetic energy-momentum tensor is given by E tt = E rr = − Q κr and E θθ = E φφ = Q κr . (11)We see that Eq. (11) satisfies the needed condition for the energy-momentum tensor.Now consider the energy-momentum tensor from
1, 12 that also obey the previousEinstein-Rastall symmetry properties as the quintessence part that reads J tt = J rr = − ρ q ( r ) and J θθ = J φφ = 12 (1 + 3 ω ) ρ q ( r ) , (12)where ρ q ( r ) is the perfect fluid density and ω is the parameter from the equation ofstate that relates pressure with density. Then after plugging the known total energy-momentum tensor from Eqs. (11) and (12) to the Einstein-Rastall field equation inEq. (2), we will have two differential equations:1 r ( f ′ r − f ) − κλr ( r f ′′ + 4 rf ′ − f ) = − κρ − Q r (13)and 1 r (cid:18) rf ′ + 12 r f ′′ (cid:19) − κλr ( r f ′′ + 4 rf ′ − f ) = κ ρ (3 ω + 1) + Q r . (14)To solve those equations, we need to consider that we want to recover Reissner-Nordstr¨om solution, the spherically symmetric black hole with electric (and mag-netic) charge, when both quintessence energy-momentum and λ vanishes since thatcondition gives rise to the Einstein-Maxwell field equation. Therefore, we can usethe function f ( r ) = 1 − Mr + Q r + h ( r ) (15)as the proper ansatz.Our next job is to find the unknown function h ( r ), which will be much easier tocompute. We can also take the function h ( r ) in the form of h ( r ) = − α/r ζ for someconstant α and ζ will be a function of λ and ω . After solving Eqs. (13) and (14)ugust 28, 2019 0:28 WSPC/INSTRUCTION FILE ws-ijmpd Dynamics of Charged and Rotating NUT Black Holes in Rastall Gravity with the ansatz given by Eq. (15), we will get the metric for a spherically symmetriccharged black hole with quintessence in Rastall gravity as ds = − (cid:18) − Mr + Q r − N s r ζ (cid:19) dt + dr (cid:18) − Mr + Q r − N s r ζ (cid:19) + r d Ω , (16)where ζ = 1 + 3 ω − κλ (1 + ω )1 − κλ (1 + ω ) . (17)Here we take the constant α = N s and interpret them as the quintessential density. We can also see that metric solution (16) reduces to Kiselev and Reissner-Nordstr¨omblack hole when λ → N s →
0, respectively.Besides the radial function f ( r ), we can also solve for the perfect fluid density ρ q ( r ) from Eqs. (13) and (14) as in
1, 12 ρ q ( r ) = − W s N s κr (2 ζ − ω +1) , (18)where W s = − (1 − κλ )( κλ (1 + ω ) − ω )(1 − κλ (1 + ω )) . (19)By definition, the perfect fluid density ρ q ( r ) needs to be non-negative. Thus, werestrict W s N s to be a negative value, i.e. W s N s ≤
0. In this paper, we consider somevalues of ω such as dust field ( ω = 0), radiation field ( ω = 1 / ω = − / , − / ω = − ω = − /
3) field, we restrict the Rastall parameter to 0 ≤ κλ ≤ / N s . Generating Rotation and NUT Parameter UsingNewman-Janis Algorithm
After obtaining the metric solution for a static, spherically symmetric black hole inRastall gravity, we now generate the rotating and NUT parameter using the previ-ously well-known process called Newman-Janis algorithm. Briefly, the Newman-Janis algorithm contains 4 steps: Eddington-Finkelstein coordinate transformation,complex transformation, complexification, and Boyer-Lindquist coordinate trans-formation.First, we perform the Eddington-Finkelstein coordinate transformation dt = du + f ( r ) − dr from the metric solution (16), such that the metric can be writtenas the form of the advanced null coordinate ds = − f ( r ) du − dudr + r d Ω . (20)Next, we write the inverse metric in terms of complex null tetrad { e µ ( a ) } =( l µ , n µ , m µ , ¯ m µ ) that form the metric tensor component g µν = l µ n ν + l ν n µ − m µ ¯ m ν − ugust 28, 2019 0:28 WSPC/INSTRUCTION FILE ws-ijmpd Hadyan L. Prihadi, Muhammad F. A. R. Sakti, Getbogi Hikmawan, Freddy P. Zen m ν ¯ m µ . The complex null tetrad are l µ = δ µr , (21) n µ = δ µu − f δ µr , (22) m µ = 1 √ r (cid:18) δ µθ + i sin θ δ µφ (cid:19) , (23)¯ m µ = 1 √ r (cid:18) δ µθ − i sin θ δ µφ (cid:19) . (24)Equation (21) to (22) obey the orthonormal condition for complex null tetrads, i.e. all the contractions are zero except for l µ n µ = − m µ ¯ m µ = 1. FollowingNewman-Janis prescription, the complex transformations that generate the rotatingand NUT parameters are ˜ u = u − ia cos θ + i n ln sin θ, (25)˜ r = r + ia cos θ − in, (26)˜ M = M + in. (27)Here M is the black hole mass, a and n are interpreted as the rotating and NUTparameter respectively and it is quite common to consider a as the angular mo-mentum per unit mass of the black hole, i.e. a = J/M . Following this complextransformations, the general vector transformation is x µ → x ′ µ = δ µu + ia cos θ ( δ µr − δ µu ) + in (2 ln sin θδ µu − δ µr ) + δ µr + δ µθ + δ µφ , (28)and using the transformation { e ′ µ ( a ) } = ∂x ′ µ ∂x ν { e µ ( a ) } , the complex null tetrads become˜ l µ = δ µr , (29)˜ n µ = δ µu −
12 ˜ f ( r, θ ) δ µr , (30)˜ m µ = 1 √ r (cid:18) δ νθ + ia sin θ ( δ µu − δ µr ) + i n cos θ sin θ δ µu + i sin θ δ µφ (cid:19) , (31)˜¯ m µ = 1 √ r (cid:18) δ νθ − ia sin θ ( δ µu − δ µr ) − i n cos θ sin θ δ µu − i sin θ δ µφ (cid:19) . (32)We see that the previous radial function f ( r ) becomes a functions r and θ aftercomplexification f ( r ) → ˜ f ( r, θ ) ∈ R by following rules r →
12 (˜ r + ¯˜ r ) = Re˜ r, (33) Mr → (cid:18) ˜ M ˜ r + ¯˜ M ¯˜ r (cid:19) = Re( ˜ M ˜ r ) | ˜ r | , (34) r → | ˜ r | , (35)ugust 28, 2019 0:28 WSPC/INSTRUCTION FILE ws-ijmpd Dynamics of Charged and Rotating NUT Black Holes in Rastall Gravity and the function ˜ f ( r, θ ) becomes˜ f ( r, θ ) = 1 − M r + 2 n ( − a cos θ + n ) − Q Σ( r, θ ) − N s Σ( r, θ ) ζ (36)where Σ( r, θ ) is defined as Σ( r, θ ) ≡ | ˜ r | = r + ( a cos θ − n ) . Using complex nulltetrad from Eq. (29) to Eq. (32), the non-zero component for the metric tensor withupper indices are˜ g uu = a sin θ Σ( r, θ ) + 4 n cos θ Σ( r, θ ) sin θ + 4 an cos θ Σ( r, θ ) , ˜ g uφ = a Σ( r, θ ) + 2 n cos θ Σ( r, θ ) sin θ , ˜ g φφ = 1Σ( r, θ ) sin θ , ˜ g ur = − − a sin θ Σ( r, θ ) − an cos θ Σ( r, θ ) , ˜ g rr = ˜ f + a sin θ Σ( r, θ ) , ˜ g rφ = − a Σ( r, θ ) , ˜ g θθ = 1Σ( r, θ ) . Straightforwardly from the metric tensor component after complexification, the newline element in advanced null coordinate is ds = − ˜ f ( r, θ ) du − (2 a (1 − ˜ f ( r, θ )) sin θ − f ( r, θ ) n cos θ ) dudφ (37)+ 2( a sin θ + 2 n cos θ ) drdφ + Σ( r, θ ) dθ − dudr + sin θ × (cid:20) Σ( r, θ ) + (1 − ˜ f ( r, θ )) × an cos θ + a (2 − ˜ f ( r, θ )) sin θ − ˜ f ( r, θ )4 n cos θ sin θ (cid:21) dφ . Now we are ready to perform the final part of the whole Newman-Janis algorithm:Boyer-Lindquist coordinate transformation. We choose the following transforma-tions to eliminate all the crossing terms and leave dtdφ term to preserve the axialsymmetry. The coordinate transformations are du = dt + ψdr, dφ = dφ + Ξ dr, (38)where ψ and Ξ are undetermined functions. After making all the off-diagonal metricelements vanish except g tφ , we get the expression for ψ and Ξ and it turns out thatthey are both functions of r and θ as ψ ( r, θ ) = − (Σ( r, θ ) + a sin θ + 2 an cos θ )( ˜ f ( r, θ )Σ( r, θ ) + a sin θ ) , (39)Ξ( r, θ ) = − a ( ˜ f ( r, θ )Σ( r, θ ) + a sin θ ) . (40)The θ dependence might rises because we are working with non-vacuum surroundingand modified theory of gravity.
5, 12
From this transformations, we finally to the finalugust 28, 2019 0:28 WSPC/INSTRUCTION FILE ws-ijmpd Hadyan L. Prihadi, Muhammad F. A. R. Sakti, Getbogi Hikmawan, Freddy P. Zen result of this section: the Kerr-Newman-NUT black hole in Rastall gravity metricsolution in Boyer-Lindquist coordinate, ds = − ∆ − a sin θ Σ dt + 2Σ (∆ χ − a (Σ + aχ ) sin θ ) dtdφ + 1Σ ((Σ + aχ ) sin θ − χ ∆) dφ + Σ∆ dr + Σ dθ , (41)where ∆ = r + a − n − M r + Q − N s Σ − ζ and χ = a sin θ + 2 n cos θ are bothfunctions of r and θ . This solution is similar to the ordinary Kerr-Newman-NUTblack hole, but with extra terms in the ∆ function that arises from quintessenceand Rastall parameter. Furthermore, metric solution in Eq. (41) will be named theKNN-R solution for short.KNN-R solution also matches previous known black hole solutions, e.g. if theRastall parameter vanishes ( λ → λ, N s →
0, we re-cover the Kerr-Newman-NUT solution. The Kerr-Newman and Kerr solution ariseswhen N s , n → N s , n, Q →
0, respectively. Finally, we arrived to the staticand spherically symmetric Schwarzschild solution when N s , n, Q, a → θ . It indicates that the horizon structure will not be a perfect sphere, aswe will discuss more detail in the next chapter. At first, the dependence of θ seemsquite unphysical. But we can take it because that arises from the external propertiesof the black hole such as the surrounding matter and Rastall parameter, e.g. whenthe Quintessence part vanishes ( N s → θ dependence of thehorizon and return to the Kerr-Newman-NUT solution with a perfectly sphericalhorizon.
3. Horizon and Ergoregion structure
In this section, we study the horizon and ergoregion structure of the KNN-R blackhole that depends on the values of
M, a, Q, n, N s , ω, and λ . Finding a solution forthe horizon and ergoregion is by determining the root(s) of ∆ and g tt , respectively.Occasionally, we are able to solve it analytically but it often needs numerical com-putation since we are dealing with a nonlinear equation. Furthermore, we will alsosee that it is possible to have more than one horizons. Because g rr = ∆ = 0 atthe horizon, we will have g µν σ ( r ) µ σ ( r ) ν = 0, where σ ( r ) µ is a normal of the horizon’shypersurface and it becomes a null vector. Hence, the horizon is a null hypersurface. Black Hole Horizon
Black hole horizon is the null surface that determined by the coordinate singularity∆ = 0. Since ∆ is a function of r and θ , the r solution is θ dependence. Thus, theKNN-R black hole horizons are determined by zeros of r + a − n − M r + Q − N s Σ − ζ = 0 . (42)ugust 28, 2019 0:28 WSPC/INSTRUCTION FILE ws-ijmpd Dynamics of Charged and Rotating NUT Black Holes in Rastall Gravity From here we expect to have the inner ( r − ), outer ( r + ), and even cosmologicalhorizon ( r q ), if any, with r − < r + < r q . Analytical expressions for the horizonsolution are given by Table 1. It shows all possibilities for a quadratic solution.The value of ω is chosen specifically to match the common surrounding matterssuch as: dust field ( ω = 0), radiation field ( ω = 1 / ω = − / , − / Hence the Rastall parameter has been chosen arbitrarily to matchthe solutions.
Table 1. Some possibilities of ω and κλ values that yield tothe quadratic analytical horizon. The ”+” and ”-” sign indicatesouter and inner horizons, respectively. ω , κλ Horizon ( r ± )0, 1 / M (1 − N s ) ± √ M − (1 − N s )( a − n + Q − N s ( a cos θ − n ) )(1 − N s ) /
3, any M ± p M − ( a − n + Q − N s ) − /
3, 0 M (1 − N s ) ± √ M − (1 − N s )( a − n + Q − N s ( a cos θ − n ) )(1 − N s ) Furthermore, we are able to see the horizon solution aside from the analyticalexpression. We want to solve Eq. (42) numerically so we can use the parametersrather freely. Here we presents inner, outer, and cosmological horizon solutions forarbitrary parameter in Table 2, where ω = − and ω = − It shows thatsome suitable parameters are able to wipe out inner horizon. We can see a detailedbehavior of the ∆ function with respect to r in Fig. 1 and Fig. 2. In Fig. 1 (left),we can see that the cosmological horizon r q is increasing, but the outer horizonis decreasing as the rotation parameter a increases. On the different case withoutcosmological horizon (right), we also get the same behavior of the outer horizon.In Fig. 2, we have the solutions for extremal and non-extremal black hole. We alsohave curves that does not intersect with ∆ = 0 line, and that represents the nakedsingularity since they don’t have any horizons. Black holes with a naked singularityis prohibited by the Cosmic Censorship Conjecture, so it is unphysical.ugust 28, 2019 0:28 WSPC/INSTRUCTION FILE ws-ijmpd Hadyan L. Prihadi, Muhammad F. A. R. Sakti, Getbogi Hikmawan, Freddy P. Zen
Table 2. Table showing horizon solutions for M = 1 , N s = 0 . , Q = 0 .
09 and for various valueof ω, κλ, a, and n at the north pole ( θ = 0). The ” − ”sign indicates that there is no possible solution for r − . ω, κλ a n r − r + r q − , , − , The behavior of KNN-R black hole horizon can gives us some important prop-erties. As shown later in section 3 and 4, the thermodynamics and the particlegeodesic of this black hole depends strongly on the horizon’s behavior. Black holeswith no cosmological horizon r q are easier to be investigated, it only depends onthe outer horizon r + . Fig. 1. Plot of ∆ function with respect to coordinate r at the north pole ( θ = 0). Here we set M = 1 , Q = 0 . , N s = 1 , n = 0 .
88 and vary the rotation parameter a . We have the solution withcosmological horizon r q (left) and the one that does not have cosmological horizon r q (right). Ergoregion Structure
An axisymmetric black hole, such as this one, will have a Killing vector in theform of K µ = ξ µ ( t ) + Ω H ξ µ ( φ ) , where Ω H is the angular velocity of the zero angularugust 28, 2019 0:28 WSPC/INSTRUCTION FILE ws-ijmpd Dynamics of Charged and Rotating NUT Black Holes in Rastall Gravity r at θ = 0. In thisfigure we represent the extremal solution, i.e. the solution with only one horizon. The extremalblack hole solution is given by the green-dashed line on both images. momentum observer (ZAMO) that will be investigated further, ξ µ ( t ) and ξ µ ( φ ) arevectors associated with time and rotational translation invariance, respectively. Anull time-like hypersurface is then given by ξ µ ( t ) ξ µ ( t ) = 0, which leads to g tt = − (cid:18) ∆ − a sin θ Σ (cid:19) = 0 , (43)as the static limit surface and forms the ergosurface of the black hole, i.e. theboundary between ergoregion and an asymptotically flat time-like region. Therefore,the r solution for r + a cos θ − n − M r + Q − N s Σ − ζ = 0 , (44)gives the ergosurface. Even though it also depends on coordinate θ , this surfacediffers from the horizon and it coincides at the north and south pole. Equation (44)can also gives us more than one solutions.Since previously we have been working only on equatorial case, now we representhow the θ dependence act on the horizon and ergoregion structure, given by Fig. 3.We can see from the figure that the horizon size of a KNN-R black hole surroundedby dust field is raising along with the increasing of NUT parameter. The innerhorizon of the black hole is barely noticeable. We also have the cosmological horizonat κλ = 9 /
40 (here we take 9 /
40 instead of 1 / r q is getting smaller when the NUT parameter is increasing.It is possible to have an ergosphere radius smaller than the cosmological horizon. Zero Angular Momentum Observer
Inside the ergosphere, we can have an observer that seems to be rotating around theblack hole seen by a distant observer but with vanishing angular momentum, calledthe Zero Angular Momentum Observer or ZAMO for short. The angular velocity ofugust 28, 2019 0:28 WSPC/INSTRUCTION FILE ws-ijmpd Hadyan L. Prihadi, Muhammad F. A. R. Sakti, Getbogi Hikmawan, Freddy P. Zen
Fig. 3. Plot showing the structure of the horizon and ergosurface with surrounding dust field ω = 0, M = 1, Q = 0 . a = 0 .
9, and N s = 0 . n and Rastall parameters.The red-dashed and blue-solid line indicated the ergosurface and horizon, respectively. As theNUT parameter n increases, r + is slightly bigger, but r − gets smaller. However, we discover theexistence of r q when κλ = 1 / ugust 28, 2019 0:28 WSPC/INSTRUCTION FILE ws-ijmpd Dynamics of Charged and Rotating NUT Black Holes in Rastall Gravity ZAMO is given by Ω = − g tφ g φφ . (45)Using the KNN-R line element (41), we obtain thatΩ = a (Σ + aχ ) sin θ − χ ∆(Σ + aχ ) sin θ − χ ∆ , (46)and the angular velocity of ZAMO at the horizon isΩ H = ar h + n + a , (47)where r h can be an inner, outer, or cosmological horizon. In many cases, the outerhorizon is used to determine the angular velocity of a black hole. Unlike Newtonianmechanics, here we can have non-vanishing angular velocity even though the angularmomentum is zero. This can happen because in General Relativity, the angularmomentum corresponds to the space-time geometry. This phenomena also oftencalled the dragging frame effect. We can also write the KNN-R black hole line element (41) in terms of ZAMOas ds = − (cid:20) ∆Σ sin θ (Σ + aχ ) sin θ − χ ∆ (cid:21) dt + Σ∆ dr + Σ dθ + 1Σ ((Σ + aχ ) sin θ − χ ∆)( dφ − Ω dt ) . (48)
4. Thermodynamic Properties
One of the most important discussions about black hole dynamics is the aspect ofthermodynamics. Since the area law, i.e. the law of non-decreasing area of a blackholes, has been discovered, physicist formulated the classical thermodynamics of ablack hole that connects the black hole parameters such as Mass, Spin, and Chargewith the thermodynamical quantities. The entropy of a black hole, formulated byBekenstein and Hawking, is represented by a quarter of the black hole horizonarea. In this section, we only discuss a slowly rotating KNN-R black hole, and assumethat a << n to get rid with the θ dependence of the horizon since in this limit,the ∆ function is hardly depends on θ . We want to avoid the θ dependence of theHawking Temperature, i.e. the temperature of the horizon, as well. This assumptionis valid since in Fig. 3, the θ dependence of the horizon is also hardly apparent evenfor a not-really-slowly-rotating KNN-R black hole.ugust 28, 2019 0:28 WSPC/INSTRUCTION FILE ws-ijmpd Hadyan L. Prihadi, Muhammad F. A. R. Sakti, Getbogi Hikmawan, Freddy P. Zen
Entropy and Surface Gravity
The outer horizon area of KNN-R black hole, when the slowly-rotating assumptionwas applied, is given by A H = Z π Z π √ g θθ g φφ dθdφ (cid:12)(cid:12)(cid:12)(cid:12) r = r + = 4 π ( r + a + n ) ≈ π ( r + n ) . (49)The area forms a spherical shape with a radius, say, r eff = r + a + n ≈ r + n .Here, r + is just the outer horizon of the KNN-R black hole, without having to knowthe explicit formula analytically. Following the Bekenstein-Hawking formula for theentropy of a black hole, we may write (in Planck units) S BH = A H π ( r + a + n ) ≈ π ( r + n ) . (50)The dependence of the other KNN-R black hole parameters, such as Q, N s , ω, κλ ,on the Bekenstein-Hawking entropy is contained explicitly on the outer horizon r + . This expression reduces to the Bekenstein-Hawking entropy for a Schwarzschildblack hole when a, n → κ is defined by κ = − ∇ µ K ν ∇ µ K ν , (51)where K µ = ξ µt + Ω H ξ µφ is a Killing vector for KNN-R space-time. A straightcalculation gives the explicit form of the surface gravity for a slowly-rotating KNN-R black hole, as κ = ∆ ′ ( r + )2( r + a + n ) ≈ ∆ ′ ( r + )2( r + n ) , (52)where prime notation denotes a derivative with respect to r , ∆ ′ ( r ) ≡ d ∆( r ) /dr .Equation (52) vanishes in the extremal case since we will have ∆( r + ) = ∆ ′ ( r + ) = 0. Temperature and Heat Capacity
Aside from the entropy of a black hole, Bekenstein-Hawking formula also givesthe temperature of a horizon, related to its surface gravity, given by (in Planckunits)
12, 28 T = κ π = ∆ ′ ( r + )4 π ( r + a + n ) ≈ ∆ ′ ( r + )4 π ( r + n ) . (53)For a slowly-rotating KNN-R black hole, the temperature will have the form T ≈ π ( r + n ) (cid:20) r + − M ) + N s ( ζ − r + ( r + n ) ζ (cid:21) , (54)ugust 28, 2019 0:28 WSPC/INSTRUCTION FILE ws-ijmpd Dynamics of Charged and Rotating NUT Black Holes in Rastall Gravity where M is the mass of a slowly-rotating KNN-R black hole, M ≈ r + [ r − n + Q − N s ( r + n ) − ζ ] . (55)For the case of vanishing surrounding matter, i.e. N s →
0, we recover theBekenstein-Hawking temperature for a Kerr-Newman-NUT black hole. We also getzero temperature for the extremal case. We can see how the temperature behaveswhen r + varies with different NUT n and Rastall parameter in Fig. 4. The presenceof NUT n parameter makes the temperature slightly increases. As the black holeevaporates, the horizon radius started to decrease and the temperature rises untilit reaches a certain maximum point and continuously falls to absolute zero when itreaches an extremity, i.e. when there is only one horizon. When KNN-R black holeis surrounded by dust field, the temperature reaches a certain asymptotic value if r + is quite large (except when κλ = 9 /
1, 12
Fig. 4. Plot showing the temperature of a KNN-R black hole horizon surrounded by dust field(top) and quintessence field (bottom) when the outer horizon r + varies. Here we use Q = 0 . a ≈
0, and N s = 0 . C = dMdT (cid:12)(cid:12)(cid:12)(cid:12) r = r + = dM/dr + dT /dr + . (56)ugust 28, 2019 0:28 WSPC/INSTRUCTION FILE ws-ijmpd Hadyan L. Prihadi, Muhammad F. A. R. Sakti, Getbogi Hikmawan, Freddy P. Zen
Hence, from Eq. (56) and (55), we have the explicit form of the heat capacity, C ≈ π ( r + n ) ( r + N s α ( r + )) r + ( Q − n )(1 + r ) + N s β ( r + ) , (57)where α ( r + ) and β ( r + ) are functions defined by α ( r + ) = ( r ( ζ −
1) + n )( r + n ) − ζ + n − Q , (58) β ( r + ) = [( r ζ ( ζ − − ( r + n ) / r + n ) − +( r (1 − ζ ) − n )]( r + n ) − ζ . (59)The behavior of heat capacity can be useful to investigate the thermodynamic sta-bility of a black hole. A black hole is said to be thermodynamically stable when theheat capacity is positive-valued, i.e. when C >
C <
0. Fig. 5 shows regions where the black hole is thermodynamicallystable. The value varies for different values of Rastall parameter, and the existenceof NUT parameter also affects the stable region.
Fig. 5. Plot showing the heat capacity of a KNN-R black hole horizon surrounded by dust field(top) and quintessence field (bottom) when the outer horizon r + varies. Here we use Q = 0 . a ≈
0, and N s = 0 .
5. Equatorial Circular Orbits Analysis
In this section, we would like to investigate the motion of a particle (both massiveand massless) orbiting the KNN-R black hole with quintessential background. Inugust 28, 2019 0:28 WSPC/INSTRUCTION FILE ws-ijmpd
Dynamics of Charged and Rotating NUT Black Holes in Rastall Gravity order to do so, first, we need to consider the Lagrangian for following particle thatgiven by L = 12 g µν ˙ x µ ˙ x ν . (60)Here we use the dot notation as the derivative with respect to an Affine parameter τ , ˙ x µ ≡ dx µ /dτ . Metric tensor g µν used in Eq. (60) is given by metric elements inEq. (41), i.e. line element for the KNN-R black hole. Since we are only interestedin circular equatorial geodesic, we will use θ = π/ r = ˙ θ = 0 afterward.We have mentioned before that KNN-R black hole, as an axisymmetric blackhole, will have symmetries under time and axial translation. Hence, we will havetwo conserved quantities, energy E and angular momentum L z that can be derivedstraight from the Lagrangian as − E = ∂ L ∂ ˙ t = g tt ˙ t + g tφ ˙ φ, L z = ∂ L ∂ ˙ φ = g tφ ˙ t + g φφ ˙ φ, (61)and the other momentum components are p r = g rr ˙ r , p θ = 0. The Hamiltonian2 H = − E ˙ t + g rr ˙ r + L z ˙ φ (62)is equal to − , ,
27, 29
Following Eq. (61), and assume that the particle follows a time-likegeodesics, we will have g rr ˙ r = g φφ E + 2 g tφ EL z + g tt L z g tφ − g tt g φφ − , (63)and from the equation above, we can define the effective potential as V eff = g φφ E + 2 g tφ EL z + g tt L z g tφ − g tt g φφ − , (64)and therefore having an explicit expression given by V eff ( r ) = (( r + n + a ) E − aL z ) − ∆( r )(( aE − L z ) + ( r + n ))∆( r )( r + n ) . (65)Here, ∆( r ) = r + a − n − M r + Q − N s ( r + n ) − ζ is a function of r only.Equatorial circular orbits occurs at the zeros and turning points of the effectivepotential, hence we have ˙ r = ˙ θ = 0 which indicates V eff = 0, and ¨ r = ¨ θ = 0which requires ∂ r V eff = ∂ θ V eff = 0.
27, 29
Equation (65) indicates that the effectivepotential blows up ( V eff → ∞ ) when r approaches the horizon (∆ → E and L z , and is given by E ± = − g tt + g tφ Ω circ. ± q − g tt − g tφ Ω circ. ± − g φφ (Ω circ. ± ) , (66) L z ± = g tφ + g φφ Ω circ. ± q − g tt − g tφ Ω circ. ± − g φφ (Ω circ. ± ) , (67)ugust 28, 2019 0:28 WSPC/INSTRUCTION FILE ws-ijmpd Hadyan L. Prihadi, Muhammad F. A. R. Sakti, Getbogi Hikmawan, Freddy P. Zen where Ω circ. ± = ˙ φ/ ˙ t is the particle’s angular velocity restricted by circular equatorialcondition, and ± denotes that the particle could have a co-rotating (Ω circ. + ) andcounter-rotating (Ω circ. − ) angular velocity with respect to ZAMO. An expression forthe angular velocity Ω circ. ± is given by Ω circ. ± = − ∂ r g tφ ± p ( ∂ r g tφ ) − ∂ r g φφ ∂ r g tt ∂ r g φφ . (68)A straightforward computation using KNN-R metric elements leads us to an explicitform for Ω circ. ± , given byΩ circ. ± = a [2 r (∆( r ) − a ) − ∆ ′ ( r )( r + n )] a [2 r (∆( r ) − a ) − ∆ ′ ( r )( r + n )] + 2 r ( r + n ) ± ( r + n ) p r ∆ ′ ( r )( r + n ) − r (∆( r ) − a ) a [2 r (∆( r ) − a ) − ∆ ′ ( r )( r + n )] + 2 r ( r + n ) , (69)From here, it is possible to compute an explicit formula for E and L z by pluggingΩ circ. ± in Eq. (69) to Eq. (66) and (67), hence the results are quite tedious. We aregoing to show how E and L z varies with respect to r by making a plot, given byFig. 6. It is shown that the NUT parameter gives us an important information. Asthe NUT parameter n increases, a time-like co-rotating particle will have a greaterequatorial circular orbit radius for a given energy. That contradicts the rotationparameter a as shown in Ref. 27, because as the NUT parameter n increases, wealso have greater horizon radius, as given by Fig. 3. Surprisingly, a time-like counter-rotating particle gives the exact same behavior from its counterparts, for a varyingNUT parameter, as shown in Fig. 7. It is also shown that the energy of a co- andcounter-rotating time-like particle approaches infinity at a certain point. This givesus information about the existing of a null circular orbit, as we will discuss further. Static Radius Limit for a Time-like Particle
We already showed that energy E and angular momentum L z of a massive particleon a circular equatorial orbit depends on Ω circ. ± . Hence, both energy E and angularmomentum L z will be real valued if∆ ′ ( r )( r + n ) − r (∆( r ) − a ) ≥ i.e. n →
0, we recover the static radiuslimit condition found in Ref. 29 for a rotating black hole. Equation (70) gives anexistence of a static radius defined by r s , where r ≤ r s are always satisfied for atest time-like particle. At r = r s , an object is located at an unstable-equilibriumposition with L z = 0.
27, 29
In the case of the KNN-R black hole metric given by (41), we can have an explicitformula for Eq. (70), given by2 r ( M r + 2 n − Q ) − M n + N s ζr ( r + n ) − ζ ≥ . (71)ugust 28, 2019 0:28 WSPC/INSTRUCTION FILE ws-ijmpd Dynamics of Charged and Rotating NUT Black Holes in Rastall Gravity E + and angular momentum L z + with respect to r for aKNN-R black hole surrounded by dust field with M = 1 , a = 0 . , Q = 0 . , κλ = 0, and N s = − . E − and angular momentum L z − with respect to rfor a KNN-R black hole surrounded by dust field with M = 1 , a = 0 . , Q = 0 . , κλ = 0, and N s = − .
01. This gives a similar result with the previous co-rotating plot.
Interestingly, Eq. (71) does not depends on the rotation parameter a , as in the caseof the rotating black hole with the quintessential energy. We would like to see thevalues of r s for different values of n and κλ , as shown in Table 3.ugust 28, 2019 0:28 WSPC/INSTRUCTION FILE ws-ijmpd Hadyan L. Prihadi, Muhammad F. A. R. Sakti, Getbogi Hikmawan, Freddy P. Zen
Table 3. Table showing the static radius r s fordifferent values of n and κλ . Here we take M = 1 , Q = 1 , ω = − / N s = 0 . κλ = 0 κλ = 1 / κλ = 1 / κλ = 1 / r s r s r s r s When the KNN-R black hole is surrounded by the quintessence field, an in-creasing n gives an increasing r s as well, for various value of κλ , e.g. we have r s = 13 . n = 0 . r s = 13 . n = 0 .
2, with κλ = 0. An increas-ing Rastall parameter makes the static radius r s decreases, e.g. when κλ = 0 and κλ = 1 /
10, we have r s = 14 . r s = 8 . n = 1. Null Equatorial Circular Orbit
For null particles, such as photon, the corresponding velocity vector u µ = dx µ /dτ satisfies u µ u µ = 0. This also indicates that the Lagrangian, given by Eq. (60), isequal to zero. Therefore, from the equatorial circular orbit condition, we have anequation that gives the equatorial null circular orbit, as g tt + 2 g tφ Ω circ. ± + g φφ (Ω circ. ± ) = 0 . (72)This condition happens when the energy E of a time-like particle approaches infinity.Hence, for most cases, r > r γ is always satisfied for a time-like particle, where r γ isdefined as the equatorial null circular orbit, i.e. the root(s) of equation (72). Herewe are only considering the co-rotating null equatorial circular orbit, as the zeros of0 = 2 a [∆( r ) − ( r + n + a )][2 r (∆( r ) − a ) − ∆ ′ ( r )( r + n )]( a [2 r (∆( r ) (73) − a ) − ∆ ′ ( r )( r + n )] + 2 r ( r + n ) ) + (( r + a + n ) − a ∆( r ))( a × [2 r (∆( r ) − a ) − ∆ ′ ( r )( r + n )] + ( r + n ) [2 r ∆ ′ ( r )( r + n ) − r × (∆( r ) − a )]) + { a (∆( r ) − ( r + a + n ))( r + n )[ a [2 r (∆( r ) − a ) − ∆ ′ ( r )( r + n )] + 2 r ( r + n ) ] + 2 a ( r + n )(( r + a + n ) − a ∆( r )) × [2 r (∆( r ) − a ) − ∆ ′ ( r )( r + n )] } p r ∆ ′ ( r )( r + n ) − r (∆( r ) − a ) , where the dependence of other parameters such as Q, N s , ω, κλ, and M containedimplicitly on the ∆( r ) function. From Eq. (73), we can see that the condition (70)ugust 28, 2019 0:28 WSPC/INSTRUCTION FILE ws-ijmpd Dynamics of Charged and Rotating NUT Black Holes in Rastall Gravity also needs to be satisfied for the null equatorial circular orbit to be real-valued, i.e. r γ < r s , thus again confirms that r > r γ . The results for various NUT androtation parameter are given by Table 4 for a KNN-R black hole surrounded bydust field. As we can see, the NUT parameter makes r γ increases, while the rotationparameter makes r γ decreases for the few first, then starting to increase at somepoint. This also corresponds with the increasing and decreasing horizon with respectto those parameters. For example, when a = 0 we have r γ = 2 . r γ = 2 . n = 0 and n = 0 .
22, respectively.
Table 4. The values of r γ for ω = 0 (dust field), κλ = 1 / M = 1 , Q = 0 . , N s = 0 . n = 0 n = 0 . n = 0 . n = 0 . n = 0 . r γ r γ r γ r γ r γ Innermost Stable Circular Orbit (ISCO)
Another important aspect of a particle geodesic around black hole is the InnermostStable Circular Orbit (ISCO). When ∂ V eff ∂r = 0 , (74)the equatorial circular orbit becomes r ISCO and every orbit that smaller than r ISCO will be unstable. The inner radius of an accretion disk around a black hole isassumed as the r ISCO according to the Novikov-Thorne model. From Eq. (65)and using V eff = 0, r ISCO is then the zeros of E (12 r + a + n ) − aL z E − ∆ ′′ ( r )[( aE − L z ) +( r + n )] − ′ ( r ) r − r ) = 0 , (75)where the double prime notation denotes the second derivative with respect to r .As we can see from Fig. 8, r ISCO is increased when the NUT parameter decreases,for both dust and quintessence field case.
6. Conclusions and Discussions
We obtained the Kerr-Newman-NUT black hole solution in Rastall theory of gravity(KNN-R black hole) using Newman-Janis algorithm as the generalized version ofthe solution found in Ref. 1. In this work, we have found that the horizon of theKNN-R black hole is θ dependence, i.e. the shape of the horizon is not perfectlyspherical. We studied the behavior of the horizon and ergosurface of KNN-R blackhole in section 3. The horizon can have more than one solution, e.g. inner horizonugust 28, 2019 0:28 WSPC/INSTRUCTION FILE ws-ijmpd Hadyan L. Prihadi, Muhammad F. A. R. Sakti, Getbogi Hikmawan, Freddy P. Zen
Fig. 8. Plot showing the behavior of r ISCO when the NUT parameter n varies for different value ofRastall parameter for a KNN-R black hole surrounded by dust (left) and quintessence (right) field.Here blue-dot , black-dot-dashed , orange-dashed , and red-solid line represents κλ = 1 / , / , / , r − , outer horizon r + , cosmological horizon r q , or even more horizons. Analyticalsolution of the horizon is shown in Table 1 and the behavior of the ∆ functionis shown in Fig. 1 and Fig. 2. The new NUT parameter affects the shape and ofthe horizon and ergosurface, as shown in Table 2 and Fig. 3, that when the NUTparameter increases, the radius of the horizon also became slightly bigger.In section 4, we studied the thermodynamic properties of the KNN-R black hole.We investigate the black hole entropy using the Bekenstein-Hawking formula. Herewe assume that the black hole is slowly-rotating to get rid with the θ dependence.Area and entropy of KNN-R black hole have been found in Eq. (49) and (50), andcoincides with the entropy of a slowly-rotating Kerr-Newman-NUT black hole with the Rastall parameter affect the horizon radius r + implicitly. We also seehow the black hole temperature and heat capacity varies with respect to the outerhorizon in Fig. 4 and 5. The thermodynamic stability of the black hole also hasbeen studied in this work from the behavior of the heat capacity. Existence of theNUT parameter affects the thermodynamically stable region.In this work, aside from Ref. 1, we also studied some geodesic properties of thisblack hole, i.e. the equatorial circular orbit. From the Hamiltonian formulation, wederived the effective potential, as shown in Eq. (65), and the angular velocity ofa time-like particle with respect to ZAMO, as shown in Eq. (69). From that, weobtained the energy and angular momentum of a time-like particle and it turns outthat when the NUT parameter increased, the particle’s equatorial circular orbit isalso increased. We investigated the static radius limit in Table 3, which showedthat the increasing NUT parameter causes an increasing static radius limit. Thenull equatorial circular orbits are also obtained in Table 4 for various values ofugust 28, 2019 0:28 WSPC/INSTRUCTION FILE ws-ijmpd Dynamics of Charged and Rotating NUT Black Holes in Rastall Gravity rotation and NUT parameter. Finally, we obtained the innermost stable circularorbit (ISCO) and plotted the results in Fig. 8. When the NUT parameter decreases,we found that r ISCO increases.
Acknowledgments
F.P.Z. and G.H. would like to thank Kemenristek DIKTI Indonesia for financialsupports. H.L.P. and M.F.A.R.S. would like to thank the members of TheoreticalPhysics Groups of Institut Teknologi Bandung for the hospitality.
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