Absorption Process and Weak Cosmic Censorship Conjecture In Kerr Black hole
aa r X i v : . [ g r- q c ] S e p Absorption Process and Weak Cosmic Censorship Conjecture In Kerr Black hole
Shun Jiang ∗ Department of Physics, Beijing Normal University,Beijing, 100875, China (Dated: September 17, 2020)In this paper, we investigate the Weak Cosmic Censorship Conjecture in nearly extremal Kerrblack hole by absorbing particles . In previous work, they ignore the process of black hole absorbingparticles. They assume the whole particle can be absorbed and the overspinning occurs. However,it is questionable whether the whole particle can be absorbed. Therefore, we will investigate it inthis paper. We consider the absorption process of a particle with finite size. During this absorptionprocess, the black hole’s parameters will change. This change will prevent the rest part of particleenter black hole. We show the part entering black hole can not overspin a nearly extremal Kerrblack hole. Different from Sorce and Wald, we solve the overspinning problem for Kerr black holewithout radiation effect or self-force effect. Further, we use this process to a general black hole.Under three assumptions, we show if Weak Cosmic Censorship Conjecture is valid for an extremalblack hole by absorbing a particle, it will be valid for the corresponding nearly extremal black hole.In many situation, the extremal black holes satisfy Weak Cosmic Censorship Conjecture and thecorresponding nearly extremal black holes don’t satisfy it. Therefore, if those black holes satisfythree assumptions, these violations may be solved by absorption process.
I. INTRODUCTION
Naked singularity may destroy the predictability of general relativity. To avoid it, Penrose proposed the WeakCosmic Censorship Conjecture(WCCC) [1]. Weak Cosmic Censorship Conjecture asserts that singularity should bealways hidden behind black hole event horizons. A general proof is still lacking. There are several method to testWCCC. One may test WCCC by particles. In 1974, Wald gave a gedanken experiment to test WCCC [2]. By throwinga particle to extremal black hole, he showed the particle with enough angular momentum to destroy horizon can notbe absorbed by black hole. However, by adding a charged particle to nearly extremal charged black hole, Hubenyfound WCCC was invalid [3]. Similarly, T.Jacobson and T.P.Sotiriou found WCCC is invalid for nearly extremal Kerrblack hole [4]. The invalidity was resolved by considering the self-force or back-reaction [5][6]. For other works tocheck WCCC with particles, see [7–26].One may also test WCCC by external field. The scattering of an external field to a black hole have severaldifferent features compared with those observed by particles. One of them is superradiance. When frequency andcharge of external field satisfy some conditions, external field will extract conserverd quantities from black holes. Inconsideration of scalar, Dirac and Maxwell fields, there are various tests for WCCC [27–43]. Recently, a new versionof the gedanken experiments proposed by Sorce and Wald [44]. For other works to check WCCC with new version ofthe gedanken experiments, see[45–53].There is an important difference between particle and field. When we use field to test WCCC, the black hole absorbsfield and the parameters can change continously. However, when we use particles, this is a discontinous process andthe black hole’s parameters can not change continously. In fact, one always ignore the absorption process and assumethe whole particle can enter black hole. However, if we assume particle is a compact object with mass µ , accordingto Schwarzschild black hole radius, the particle size is >µ . One may imagine an absorption process: when part ofparticle enters black hole, the black hole’s parameters change and the rest part may not satisfy the condition to enterthe black hole. Therefore, a finite size particle may be divided into two parts. And the part enters black hole maynot over-charge or over-spin the black hole. We will consider this picture in this paper. We will show particle will bedivided into two parts and the part entering black hole can not over-spin black hole. Furthermore, we also considera general black hole in this absorption process. Under three assumptions, we show when extremal black hole can notbe over-charged or over-spun, the corresponding nearly extremal black hole can not be over-charged or over-spun.This conclusion is useful: most cases of overcharging or overspinning occur for nearly extremal black hole and thecorresponding extremal black hole always satisfy WCCC. Therefore, it may solve these invalid problems. ∗ Electronic address: [email protected]
This paper is organized as follows. In Sec. II, we review the overspinning problems in Kerr black hole. In Sec. III,we use this absorption process to solve the overspinning problems in Kerr black hole. In Sec. IV, we discuss a generalblack hole using this absorption process. In Sec. V, we give a brief conclusion.
II. WEAK COSMIC CENSORSHIP CONJECTURE IN KERR BLACK HOLE
In this section, we briefly review the overspinning problems in Kerr black hole. Historically, Jacobson and Sotiriou[4] firstly find a nearly extremal Kerr black hole can be over-spun if two conditions are met. One condition is thegeodesic trajectory of the test particle is timelike at the horizon. The other one is J f > M f where J f is the finalblack hole’s angular momentum and M f is the final black hole’s mass. The first condition is lax. It allows deeplybound orbits and one may preferably avoid those orbits. Jacobson and Sotiriou also acknowledge this issue, theygive two examples show particles from afar can overspin Kerr black hole. However, they don’t give the full rangeof overspinning orbits where deeply bound orbits are disallowed. In [54], the authors firstly give the full range ofoverspinning orbits excluding deeply bound orbits. Therefore, we will review the overspinning condition in [54]. Theinitial configuration features a Kerr black hole of mass M and angular momentum J where J = M a < M . Theyassume a pointlike test particle of rest mass µ ≪ M is sent in on a geodesic and the particle’s orbit is in the equatorialplane. Therefore the angular momentum is aligned with the spin of black hole and it seems most favourable foroverspinning. Because there are two killing vector fields, we can define energy E and angular momentum L , whichare constants along the geodesic. For the geodesic approximation to make sense, they assume the particle energy µE and angular momentum µL satisfy µE ≪ M and µL ≪ J . For a nearly extremal black hole, we have aM = 1 − ǫ (1)where ǫ ≪
1. Let µ α denote the particle’s four-velovity. In Boyer-Lindquist coodrdinates ( t, r, θ, φ ), we have µ θ = 0and two constants can be written as ˙ µ t = 0 , ˙ µ φ = 0 (2)where an overdot denotes differentiation with respect to the proper time. They can be written as E = − ξ αt µ α = − µ t (3) L = ξ αφ µ α = µ φ (4)where ξ αt = ∂ αt (5) ξ αφ = ∂ αφ (6)They are killing vector fields associated with time-translation and rotational symmetries of Kerr black hole. Thepair( E , L ) parametrizes the family of geodesic and the normalization µ α µ α = − r = B ( r )[ E − V − ( L, r )][ E − V + ( L, r )] (7)where B ( r ) = 1 + a ( r + 2 M ) /r , and V ± ( L, r ) = 2
M aLBr (1 ± r Br [ L ( r − M ) + r ∆]4 M a L ) (8)with ∆ = r − M r + a . As shown in [54], V − is manifestly negative definite, so the factor B ( r )( E − V − ) in Eqs. (7)is manifestly postive definite. Therefore, V + plays the role of an effective potential for the radial motion. The allowedrange for E is E ≥ V + ( L, r ), with an equality meaning radial turning points.The effective potential V + ( L, r ) is important to overspinning problems. Therefore, we will pay attention to it. Thestationary points of V + ( L, r ) outside the black hole correspond to circular orbits. They should satisfy the conditions E = V + , ∂ r V + = 0 (9)Using Eqs. (8), we can solve E and L in terms of the circular-orbit radius r = R . The E = E c ( R ) and L = L c ( R )can be written as E c ( R ) = 1 − R − + ˜ a ˜ R − p − R − + 2˜ a ˜ R − (10) L c ( R ) = ˜ R (1 − a ˜ R − + ˜ a ˜ R − )1 + 3 ˜ R − + 2˜ a ˜ R − (11)where ˜ R = R/M ,˜ a = a/M , ˜ L = L/M .The number of stationary points of V + and their location depend on L . When L is below a critical value L isco ( a ), thereare none stationary points outside black hole. There are two for L > L isco ( a ): a maximum value of V + represnting anunstable circular orbit, and a minimum represnting a stable one. The critical value L isco ( a ) represents the innermoststable circular orbit(ISCO). It is given by L isco = L c ( R isco ), where the ISCO radius R isco can be found by solvingEqs. (9) and ∂ r V + ( r, L ) = 0. The ISCO is also the most outer boundary of the region of unstable circular orbits.To make them explicit, we show there is a relation for event horizon’s radius R eh , photon’s unstable circular orbits R ph , particle’s unstable circular orbit’s radius R u , ISCO radius R isco and stable circular orbit’s radius R s . It can bewritten as R eh < R ph < R u < R isco < R s (12)As shown in [54], the unstable circular orbits relate to the overspinning problems. Because particles in stable orbitscan’t enter black hole. The unstable circular orbits may be divided into bound( E <
1) and unbound(
E > E c ( R ) = 1, giving ˜ R ibco =[1 + (1 − ˜ a ) / ] . For a nearly extremal black hole, we find˜ R eh < ˜ R ph < ˜ R ibco < ˜ R isco ˜ R eh = 1 + √ ǫ + O ( ǫ )˜ R ph = 1 + r ǫ + O ( ǫ )˜ R ibco = (1 + ǫ ) ˜ R isco = 1 + (2 ǫ ) / + O ( ǫ / ) (13)With those relations, we can discuss deeply bound orbits. For paticle’s angular momentum L > L isco , there is amaximum value of V + locating at R max < R isco . We want the particle absorbed by black hole can clear such peak ofthe effective potential. We can achieve it by assuming the test particle’s intially position r out satisfies r out > R isco (14)As shown in [54], with this condition, we can exlcude the deeply bound orbits.Now, we consider the overspinning problem. For unstable circular orbits, the energy E and angular momentum L c satisfy ˜ L c ( E ) = 2 E + (6 E − ǫ (15)With condition (14), a necessary and sufficient condition for a falling particle with specific energy E and angularmomentum L to be captured by black hole is ˜ L < ˜ L c ( E ) (16)The overspinning condition becomes ( M + µE ) < aM + µL (17)Using ˜ a = 1 − ǫ and introducing η = µ/M , this condition becomes ǫ + ηW + η E < W = 2 E − ˜ L (19)Giving ( E , η , ǫ ), Eqs. (18) sets a lower bound on ˜ L and Eqs. (16) sets a upper bound on ˜ L . Our aim is to find thecomplete range of ( η , E , ˜ L ) for which the conditions (14) (16) and (18) are simultaneously satisfied.As shown in [54], for sufficiently small ǫ ( ǫ < / L ≤ L isco can not satisfy Eqs. (14) or Eqs. (18). Therefore weneed the orbits with L > L isco . For such an orbit with given ( E , η ), ˜ L is bounded from above via Eqs. (16) and belowvia Eqs. (18): ǫ + 2 ηE + η E < η ˜ L < η ˜ L c ( E ; ǫ ) (20)The permissible interval length is η ∆ L = − ǫ − η [2 E − ˜ L c ( E ; ǫ )] − η E . Using Eqs. (15), we find η ∆ L = − ǫ + ηǫ p E − − η E (21)Overspinning conditions are satisfied if and only if we find ( E , η , ǫ ) for which η ∆ L >
0. Considering η ∆ L in Eqs. (21)as a quadratic function of η , there is a maximum value maxη ∆ L = ǫ ( E − E (22)It is positive only for E >
1. Therefore, all bound orbits( E ≤ η makes η ∆ L > ǫη − ( E ) < η < ǫη + ( E ) (23)where η ± = 1 √ E [ p E − ± p E −
1] (24)To summarize, the overspinning conditions can be written as
E > η ˜ L c ( E, ǫ ) − η ∆ L ( E, η, ǫ ) < η ˜ L < η ˜ L c ( E, ǫ ) (26) ǫη − ( E ) < η < ǫη + ( E ) (27) III. THE ABSORPTION PROCESS IN KERR BLACK HOLE
In Sec. II, we review the overspinning problem in Kerr black hole by absorbing a test particle. There are severalassumptions in this progress. Let us focus on these assumptions.First, one assume the particle is a pointlike particle. However, as shown in Section. II, the particle’s mass µ whichcan overspin is ∼ ǫ . If we consider the particle as a compact object in the real world instead of a black hole, usingSchwarzschild black hole radius, we find the particle’s size should ≫ ǫ .Second, when one consider WCCC by absorbing a particle, one may assume black hole has several parameters( x , y , z )and particle has several parameters(∆ x ,∆ y ,∆ z ). They ignore the absorption process and after absorption the blackhole parameters becomes ( x + ∆ x , y + ∆ y , z + ∆ z ). This is a discontinuous process and it holds when particle is apointlike particle. However, we have shown the particle is finite size( ≫ ǫ ). One may imagine the absorption process:When part of particle enter event horizon, the black hole may have some changes and the condition for particle toenter it will change. Therefore, the rest part of the particle may not satisfy this condition and it can’t enter the blackhole. This process may solve overspinning problem without self-force or radiation effect.As mentioned above, the particle size may not be small enough and it may lead some effect on WCCC. Therefore,let us use this absorption process to investige WCCC. Firstly, giving a nearly extremal Kerr black hole with mass M ,angular momentum J . We have aM = JM = 1 − ǫ (28)where 0 < ǫ ≪
1. For an overspinning particle with mass µ ( η = µ/M ), angular momentum µL and energy µE , itsparameters should satisfy Eqs. (25) (26) and (27). When we consider a absorption process, the black hole mass M will change. Therefore, it is not convenient to use η and ˜ L . We rewrite overspinning conditon Eqs. (25) (26) and (27)using E , L and µ . They can be written as E > µL c − η ∆ L M < µL < µL c (30) ǫη − ( E ) M < µ < ǫη + ( E ) M (31)There are three parameters and if we fix E and µ , the allowed range of angular momentum µL can be fixed. Wewant the length of angular momentum’s interval takes maximum. Because in this case, it will include most kinds ofoverspinning particles. From Eqs. (30), we know this is equal to take maximum value of η ∆ L ( E, η, ǫ ). Using Eqs. (22),we find the η should take the value: η = √ E − E ǫ (32)Using η = µ/M , we find µ = √ E − E ǫM (33)Combining Eqs. (22) and (33), the range of overspinning specific angular momentum L can be written as L = L c − η ∆ L M bµ = L c − ( E − √ E − M bǫ (34)where 0 < b <
1. For convenience, we assume the particle is composed of N same slices where N → ∞ and each slicehas equal energy and angular momentum. We assume the black hole can absorb n slices and the rest part can’t enterit. The black hole mass and angular momentum become M ( x ) = M (0) + xµE (35) J ( x ) = J (0) + xµL (36)where x = n/N , M ( x ) and J ( x ) represent the black hole mass and energy after absorbing n slices. Using Eqs. (33)and (34), one find M ( x ) = M (1 + √ E − E xǫ ) (37) J ( x ) = (1 − ǫ ) M +[2 M E + (6 E − M ǫ ] √ E − E M xǫ − [ E − √ E − M ǫb ] √ E − E M xǫ (38)Using J ( x ) and M ( x ) , ǫ ( x ) can be written as J ( x ) M ( x )2 = 1 − ǫ ( x )2 (39)Using Eqs. (37), we find 1 M ( x )2 = 1 M (1 + √ E − E xǫ ) = (1 − √ E − E xǫ ) M (1 − E − E x ǫ ) = (1 − √ E − E xǫ ) (1 + E − E x ǫ ) M (1 − ( E − E x ǫ ) ) (40)Combing it with Eqs. (38), after some calculation, we find ǫ ( x ) = ǫ r E − E bx − E − E (2 x − x ) (41)where we omit O ( ǫ ) terms.During this absorption process, ǫ ( x ) can represent the state of the nearly extremal black hole: when ǫ ( x ) decreases,it means the nearly extremal black hole develops towards extremal black. When ǫ ( x ) increases, the black hole willdevelop in the opposite direction. Therefore, we want to investigate the developing direction of the black hole. Let f ( x ) = ǫ ( x )2 = (1 + E − E bx − E − E (2 x − x )) ǫ (42)differentiating f ( x ) with x , one find df ( x ) dx = ( E − E b − E − E (2 − x )) ǫ (43)Solving df /dx = 0, we have x min = 1 − E − E − b (44)Becasue f ( x ) is a quadratic function of x and overspinning leads f (1) <
0, we find f ( x min ) <
0. There are twosolutions ( x , x ) for f ( x ) = 0. They satisfy 0 < x < x min < < x . The point x menas black hole absorb x slicesof particle and become an extremal black hole. It easy to see ǫ ( x ) decreases monotonously in the interval [0 , x ].In above disscusion, we show a black hole with mass M ( x ) angular momentum J ( x ) can become more extremalby absorbing slice of particle with mass µ/N energy µE/N and angular momentum µL/N . For convenience, weintroduce an angular momentum L d ( ǫ ( x ) ). We need ǫ ( x ) doesn’t change when a black hole with mass M ( x ) angularmomentum J ( x ) absorbs a slice of particle with mass µ/N energy µE/N and angular momentum µL d /N . It is easyto see L > L d ( ǫ ( x ) ) during this absorption process. This coincide with the result ǫ ( x ) decreases monotonously in theinterval [0 , x ]. This result means black hole will become more extremal by absorbing slice of this kind particle.During this absorption process, the black hole parameters change continuously and condition for a particle to becaptured by the black hole will change. Using Eqs. (15) and (16), We find the capture condition become L < M ( x ) E + (6 E − M ( x ) ǫ ( x ) (45)We can label it by L < L u ( ǫ ( x ) ), where L u ( ǫ ( x ) ) = 2 M ( x ) E + (6 E − M ( x ) ǫ ( x ) (46)It is easy to see for the initial state L d < L < L u . Therefore, at least, black hole can absorb fisrt slice of particle.When black hole absorbs more slices, x will increase and L d ( ǫ ( x ) ), L u ( ǫ ( x ) ) will change. The key point is black holecan absorb how many slices. It is easy to see when L = L u , the slice of particle can not enter black hole. We can solvethis equation and find there is a solution x e where 0 < x e < x . This means slice can not enter it when black hole isstill a near extremal black. This means absorption process will save WCCC. The expression of x e is very complicatedand we will not give exact expression here. However, we will give another method to prove it and this method can beextended to general situations. Therefore, let us focus on it. We first consider the extremal Kerr black hole. For anextremal Kerr black hole with mass M e and angular momentum J e , L u the max value of angular momentum whichparticle can enter it becomes L u = 2 M e E (47)We have defined an angular momentum L d ( ǫ ( x ) ). We need ǫ ( x ) doesn’t change when the black hole with mass M ( x ) angular momentum J ( x ) absorbs a slice of particle with mass µ/N energy µE/N and angular momentum µL d /N . wecalculate L d for extremal black hole J e + µL d N = ( M e + µEN ) (48)To first order of 1 /N , we find L d = 2 M e E (49)We find L u = L d . This is reasonable. Because this means a particle with L > L d which can make black hole moreextremal can not enter the extremal black hole. This is nothing but an extremal Kerr black hole can not be over-spun.Let us analyze this result: we start with a nearly extremal black hole where ǫ = ǫ (0) > L d ( ǫ ) < L < L u ( ǫ ).By absorbing slices, ǫ ( x ) decreases and L d , L u change continuously. However, for extremal black hole ( ǫ ( x ) = 0), wefind L d (0) = L u (0). During this process we have L > L d and use particle’s angular momentum L is a constant, thislead to L > L u (0). Combing the initial state L < L u ( ǫ ) and extremal state L > L u (0), we find there is 0 < x e < x satisfies 0 < ǫ ( x e ) < ǫ (0) and L = L u ( ǫ ( x e ) ). At this point, the near extremal black hole stops absorbing slices. Becasueof ǫ ( x e ) >
0, this means when black hole is a nearly extremal black hole, the slices can not enter it. Therefore, thisparticle can not overspin a black hole. This also means when taking absorption process into account, the black holeeven can not become extremal by absorbing particle. At this time, the rest part of paricle can not clear the peak ofpotential and can not enter black hole. However, there are some slices between horzion and peak of potential. Onemay wonder these slices may have some effects. This will not bother us. Because we consider a particle in the realworld. In this situation, its size is ≫ ǫ and the distance bewteen the peak of potential and horzion is ∼ ǫ . Therefore,those silices between them are too small compared to the whole particle and its effect can be ignored.By considering the absorption process, we show the whole particle can not enter it. In fact, the particle will bedivided into two parts: one can enter black hole and the other can’t. The black hole can not be over-spun by the partentering black hole. We solve the overspinning problem without second order effect. It is easy to see this method maybe useful for a general black hole. Therefore, in next section, we will consider the general case. IV. WEAK COSMIC CENSORSHIP CONJECTURE IN GENERAL BLACK HOLE
In this section, we consider WCCC absorption process for a general black hole. For a general black, we assume ithas energy M and several parameters ( A , B , C ) which can lead to naked singularity. For example, let A representsangular momentum and we go back to the Kerr black hole. We consider a particle with mass µ , energy µE andone general charge of ( A , B , C ). Without loss of generality, we assume particle has charge µA . We use parameter ǫ represents the state of black hole. A nearly extremal black hole means ǫ > ǫ = 0.When a near extremal black hole develop to an extremal balck hole, we need ǫ decreases continously.For a over-charged particle( µ , µE , µA ) to a nearly extremal black hole, when we fix ( µ , E ), there will be an upperboundary A u and lower boundary A d for A . The upper boundary means if A is too large, the particle can not enterthe black hole. We need A d ( ǫ ) satisfies that ǫ doesn’t change when add a slice of particle with ( µ/N , µE/N , µA d /N )into black hole, where N → ∞ .We use the absorption process we develop in Sec. III to consider WCCC. The upper boundary condition and lowerboundary condition can be written as functions of ǫ , we label them by A u ( ǫ ) and A d ( ǫ ). If we assume A > A d inthis absorption process, then we can show a WCCC relation between extremal black hole and nearly extremal blackhole. A > A d is our first assumption. Let us first consider what this condition means. For A > A d , when eachslice of over-charge particle enters black hole, the nearly extremal black hole become more extremal. This seemsreasonable for an over-charged particle but we don’t prove it and make it as an assumption. With this assumptionwe can prove that if an extremal black hole can not be over-charged, an near extremal black hole can also not beover-charged in this absorption process. The proof is same as Kerr black hole. If the initial near extremal black holewith ǫ (initial) > µ , µE , µA ), A should satisfy A < A u ( ǫ (intial))and we assume A > A d during this process. It is easy to see when ǫ = 0, the condition for an extremal black hole cannot be over-charged becomes A d (0) ≥ A u (0). Using A > A d , A d (0) ≥ A u (0) and A d ( ǫ (intial)) < A < A u ( ǫ (intial)), wefind A u (0) < A < A u ( ǫ (intial)). Using A is a constant, ǫ decrease in this process and A u ( ǫ ) is a continous funtion of ǫ , we find there is ǫ ( c ) > A = A u ( ǫ ( c )), where 0 < ǫ ( c ) < ǫ (intial). This means rest part of particle can notenter black hole. At this time, the black hole is still nearly extremal. Therefore, a nearly extremal black hole can notbe over-charged when the extremal black hole can not be over-charged. In fact, it can even not become extremal.A similar confusion should be cleared up. We assume the particle’s size is much greater than the distance betweenthe peak of potential and horizon. Therefore, the part between them will not have influence on the result. This isreasonable. For spherical black hole, this distance of many balck holes are 0. For Kerr black hole, we see this distanceis ∼ ǫ and the overspinning particle mass is ∼ ǫ . We consider the particle as a compact object in the real world insteadof a black hole. Using Schwarzschild black hole radius, we find the particle’s size should ≫ ǫ . Therefore, it also willnot affect the result. We assume similar situations occur for other rotating black holes. This is the second assumption.In this proof process, we only consider a particle with one general charge. This is the third assumption. If theparticle has two kind charge, we can’t use this general proof. But it doesn’t mean the absorption process can’t solvethis kind problems. We can also use the method developed in Sec. III to investigate WCCC.If three assumptions holds, this conclusion is useful. First, if extremal black hole can not be over-charged byabsorbing particle, then we don’t need to check nearly extremal black hole. Because it also can not be over-charged.Second, we do not have any limit on interval between upper boundary A u and lower boundary A d mentioned in thesecond paragraph in this section. This interval is important for radiation effect and self-force. Because those aresecond order effect of ǫ in Kerr black hole. However, this absorption process does not depend on it. V. CONCLUSION
In this paper, we consider the absorption process. We use it to check WCCC and solving overspinning problemsin Kerr black hole. By simulating the process of black hole absorbing a particle, we show the whole particle can notenter it. The particle will be divided into two parts: one can enter black hole and the other can’t. The black hole cannot be over-spun by absorbing the part entering black hole. According to it, we find an interesting fact that an nearlyextremal black hole can not become extremal in this absorption process. We also use this method to a general blackhole. Under three assumptions, we find a relation bewteen a general extremal black hole and nearly extremal blackhole: if WCCC is valid for extremal black hole by absorbing particle, it will also be valid for near extremal black holeby absorbing particle. This gives a general relation between extremal black hole and nearly extremal black hole inWeak Cosmic Censorship Conjecture. We also show our method is different from the method considering radiationeffect and self-force effect.
Acknowledgments
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