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Test fields cannot destroy extremal black holes José Natário, Leonel Queimada and Rodrigo Vicente 6 1 Center for Mathematical Analysis, Geometry and Dynamical Systems, 0 Mathematics Department,Instituto SuperiorTécnico, 2 Universidade deLisboa, Portugal l u J Abstract 5 1 Weprovethat (possibly charged) test fieldssatisfying thenullenergy conditionattheeventhorizoncannotoverspin/overchargeextremalKerr- ] c Newman or Kerr-Newman-anti de Sitter black holes, that is, the weak q cosmic censorship conjecture cannot be violated in the test field approx- - imation. The argument relies on black hole thermodynamics (without r g assuming cosmic censorship), and does not depend on the precise nature [ of the fields. We also discuss generalizations of this result to other extremal black holes. 2 v 9 1 Introduction 0 8 6 In the wake of the proofs of the singularity theorems in general relativity [1, 2, 0 3], Penrose formulated the weak cosmic censorship conjecture [4, 5], according . 1 to which, generically, the singularities resulting from gravitational collapse are 0 hidden from the observers at infinity by a black hole event horizon. Penrose’s 6 expectation was that, independently of what might happen inside black holes, 1 the evolution of the outside universe would proceed undisturbed. : v To test this conjecture, Wald [6] devised a thought experiment to destroy i extremal Kerr-Newman black holes, already on the verge of becoming naked X singularities, by dropping charged and/or spinning test particles into the event r a horizon. Both him and subsequent authors [7, 8] found that if the parameters of the infalling particle (energy, angular momentum, charge and/or spin) were suited to overspin/overcharge the black hole then the particle would not go in, in agreement with the cosmic censorship conjecture. Similar conclusions were reachedby analyzing scalarand electromagnetic test fields propagatingin extremalKerr-Newmanblack hole backgrounds[9, 10, 11, 12]. In this case, the fluxes of energy, angular momentum and charge across the event horizon were found to be always insufficient to overspin/overchargethe black hole. Some of these results have been extended to higher dimensions [13] and also to the case when there is a negative cosmologicalconstant [14, 15]. More recently, it was noticed that Wald’s thought experiment may produce nakedsingularities when appliedto nearly extremalblack holes [16, 17, 18, 19]. 1 However, in this case the perturbation cannot be assumed to be infinitesimal, andsobackreactioneffectshavetobetakenintoaccount;whenthisisdone,the validity of the cosmic censorship conjecture appears to be restored [20, 21, 22, 23, 24]. It can also be argued that the third law of black hole thermodynamics [25], for which there is some evidence [26, 27, 28], forbids subextremal black holes from ever becoming extremal, and so, presumably, from being destroyed. Nonetheless, this cannot be taken as a definitive argument, since, for instance, extremal Reissner-Nordstro¨m black holes can be formed by collapsing charged thin shells [29]. In this paper, we consider arbitrary (possibly charged) test fields propagating in extremal Kerr-Newman or Kerr-Newman-anti de Sitter (AdS) black hole backgrounds. Apart from ignoring their gravitationaland electromagnetic backreaction, we make no further hypotheses on these fields: they could be any combinationof scalar,vector or tensor fields, chargedfluids, sigma models, elastic media, or other types of matter. This also includes test particles, since they can be seen as singular limits of continuous media [30, 31]. We give a generalproofthatifthe testfieldssatisfythe nullenergyconditionatthe event horizon then they cannot overspin/overcharge the black hole. This is done by firstestablishing,inSection2,atestfieldversionofthesecondlawofblackhole thermodynamics for extremal Kerr-Newman or Kerr-Newman-AdS black holes (which does not assume cosmic censorship). We use this result in Section 3, together with the Smarr formula and the first law, to conclude the proof. This laststeprequirestheblackholetobeextremal,andcannotbeextendedtonear- extremal black holes. In Section 4 we discuss generalizations of our result to otherextremalblackholes,includinghigherdimensionsandalternativetheories of gravity. We follow the conventionsof[32, 33]; in particular,we use a systemof units for which c=G=1. 2 Second law for test fields In this section we prove that a version of the second law of black hole thermodynamics holds inthe case of(possibly charged)test fields propagatingona backgroundKerr-NewmanorKerr-Newman-AdSblackhole(eithersubextremal orextremal). Thiscalculationissimilartotheonein[34],butwedonotassume cosmiccensorship,i.e.wedonotassumethatthe blackholeis notdestroyedby interacting with the test field. WestartbyrecallingtheKerr-Newman-AdSmetric,giveninBoyer-Lindquist coordinates by ∆ asin2θ 2 ρ2 ds2 =− r dt− dϕ + dr2 ρ2 (cid:18) Ξ (cid:19) ∆ r ρ2 ∆ sin2θ r2+a2 2 + dθ2+ θ adt− dϕ , (1) ∆ ρ2 (cid:18) Ξ (cid:19) θ 2 where ρ2 =r2+a2cos2θ; (2) a2 Ξ=1− ; (3) l2 r2 ∆ =(r2+a2) 1+ −2mr+q2; (4) r (cid:18) l2(cid:19) a2 ∆ =1− cos2θ (5) θ l2 (see for instance [35]). Here m, a and q denote the mass, rotation and electric charge parameters, respectively. These parameters are related to the physical mass M, angular momentum J and electric charge Q by m ma q M = , J = , Q= . (6) Ξ2 Ξ2 Ξ The cosmological constant is Λ = −3, and so the Kerr-Newman metric can l2 be obtained by taking the limit l → +∞. To avoid repeating ourselves, we will present all calculations below for the Kerr-Newman-AdS metric only; the corresponding formulae for the Kerr-Newmanmetric can be easily retrievedby making l→+∞. TheKerr-Newman-AdSmetric,togetherwiththeelectromagnetic4-potential qr asin2θ A=− dt− dϕ , (7) ρ2 (cid:18) Ξ (cid:19) isasolutionoftheEinstein-MaxwellequationswithcosmologicalconstantΛ. It admits a two-dimensional group of isometries, generated by the Killing vector fields X = ∂ and Y = ∂ . ∂t ∂ϕ Weconsiderarbitrary(possiblycharged)testfieldspropagatinginthisback- ground. Apart from ignoring their gravitationaland electromagnetic backreaction,wemakenofurtherhypothesesonthefields: theycouldbeanycombination of scalar,vector or tensor fields, chargedfluids, sigma models, elastic media, or othertypesofmatter. Sincethefieldsmaybecharged,theirenergy-momentum tensor T satisfies the generalized Lorentz law1 ∇ Tµν =Fναj , (8) µ α where F = dA is the Faraday tensor of the background electromagnetic field and j is the charge current density 4-vector associated to the test fields. Using the symmetry of T and the Killing equation, ∇ X +∇ X =0, (9) µ ν ν µ we have ∇ (TµνX )=Fναj X . (10) µ ν α ν 1SeetheAppendixforacompleteexplanationoftheoriginandmeaningofthisequation. 3 On the other hand, using the charge conservation equation, ∇ jµ =0, (11) µ we obtain ∇ (jµAνX )=jµ(∇ Aν)X +jµAν∇ X µ ν µ ν µ ν =jµ(F ν +∇νA )X −jµAν∇ X µ µ ν ν µ =Fµνj X +j (Xν∇ Aµ−Aν∇ Xµ). (12) µ ν µ ν ν Since A is invariant under time translations, we have L A=0⇔[X,A]=0⇔Xν∇ Aµ−Aν∇ Xµ =0, (13) X ν ν and so from (10) and (12) we obtain ∇ (TµνX +jµAνX )=0. (14) µ ν ν This conservation law suggests that the total field energy on a given spacelike hypersurface S extending from the black hole event horizon H+ to infinity (Figure 1) should be E′ = (Tµν +jµAν)XνNµdV3, (15) Z S whereN isthefuture-pointingunitnormaltoS. However,intheKerr-Newman- AdS case the non-rotatingobservers at infinity are rotating with respect to the Killing vector field X with angular velocity a Ω∞ =− , (16) l2 and so, as shown in [36], the physical energy should be computed with respect to the non-rotating Killing vector field a K =X +Ω∞Y =X− Y, (17) l2 that is, the physical energy is actually E = (Tµν +jµAν)KνNµdV3. (18) Z S This correction was implemented for test particles in [14]. The need for the corresponding correction in the calculation of the physical black hole mass has been stressed in [37, 38]. Note that in the Kerr-Newman case Ω∞ = 0, and no correction is needed. Analogously,but now without ambiguity,the total field angularmomentum on a spacelike hypersurface S extending from the event horizon to infinity is L=− (Tµν +jµAν)YνNµdV3, (19) Z S 4 i+ i+ H+ I+ H+ S 1 S 1 H H i0 I S 0 S 0 H− I− H− i− i− Figure 1: Penrose diagrams for the region of outer communication of the Kerr- Newman (left) and Kerr-Newman-AdS (right) spacetimes. where the minus sign accounts for the timelike unit normal. Consider now two such spacelike hypersurfaces, S0 and S1, with S1 to the future of S0 (Figure 1). We assume reflecting boundary conditions in the Kerr- Newman-AdScase,sothatallfluxesvanishatinfinity. Theenergyabsorbedby the black hole across the subset H of H+ between S0 and S1 is then ∆M = (Tµν +jµAν)KνNµdV3− (Tµν +jµAν)KνNµdV3, (20) Z Z S0 S1 whereas the angular momentum absorbed by the black hole across H is ∆J =− (Tµν +jµAν)YνNµdV3+ (Tµν +jµAν)YνNµdV3. (21) Z Z S0 S1 Recall that the angular velocity of the black hole horizon is aΞ Ω = , (22) H r2 +a2 + where r+ is the largest root of ∆r = 0. This means that the (future-pointing) Killing generator of H+ is Z =X +Ω Y =K+ΩY, (23) H where Ω=ΩH −Ω∞ (24) is precisely the thermodynamic angular velocity, that is, the angular velocity that occurs in the first law for Kerr-Newman-AdS black holes [37]. Therefore, we have ∆M −Ω∆J = (Tµν +jµAν)ZνNµdV3− (Tµν +jµAν)ZνNµdV3. (25) Z Z S0 S1 5 Because Z is also a Killing vector field, ∇ (TµνZ +jµAνZ )=0, (26) µ ν ν andsothe divergencetheorem,appliedtothe regionboundedbyS0, S1 andH, yields ∆M −Ω∆J = (Tµν +jµAν)ZνZµdV3 (27) Z H (we use −Z as the null normal2 on H). Since on H + er AµZ =− + =−Φ, (28) µ r2 +a2 + where Φ is the horizon’s electric potential, we have jµAνZνZµdV3 =−Φ jµZµdV3. (29) Z Z H H Using againthe divergence theorem, this time together with the charge conservation equation (11), we obtain jµAνZνZµdV3 =−Φ jµNµdV3+Φ jµNµdV3. (30) Z Z Z H S0 S1 Now the total charge on a spacelike hypersurface S extending from the event horizon to infinity is − jµNµdV3, (31) Z S wheretheminussignaccountsforthetimelikeunitnormal. Therefore,denoting by ∆Q the electric charge absorbed by the black hole across H, we have jµAνZνZµdV3 =Φ∆Q, (32) Z H and so equation (27) can then be written as ∆M −Ω∆J −Φ∆Q= (TµνZµZν)dV3. (33) Z H Since Z is null on H, we have the following test field version of the second law of black hole thermodynamics. 2RecallthatthedivergencetheoremonaLorentzianmanifoldrequiresthattheunitnormal isoutward-pointing when spacelike and inward-pointing when timelike. When the normal is nullit isnon-unique, and the volume element depends onthe choice of normal; itshouldbe past-pointing inthe future null subset of the boundary, and future-pointing in the past null subsetoftheboundary. 6 Theorem 2.1. If the energy-momentum tensor T corresponding to any collec- tion of test fields propagating on a Kerr-Newman or Kerr-Newman-AdS black hole background satisfies the null energy condition at the event horizon and appropriate boundary conditions at infinity then the energy ∆M, angular momentum ∆J and electric charge ∆Q absorbed by the black hole satisfy ∆M ≥Ω∆J +Φ∆Q. (34) It should be stressed that (34) is valid for extremal black holes, and it does not assume cosmic censorship, i.e. it does not assume that the Kerr-Newman- AdSmetricwithphysicalmassM+∆M,angularmomentumJ+∆J andelectric chargeis Q+∆Q representsa black hole rather than a nakedsingularity. Note that this scenario where the test fields interact with the geometry and change the values of the black hole charges is not in contradiction with the test field approximation, since the change is supposed to be infinitesimal. 3 Proof of the result We can now prove our main result. Theorem 3.1. Test fields satisfying thenullenergycondition at theevent horizon and appropriate boundary conditions at infinity cannot destroy extremal Kerr-Newman or Kerr-Newman-AdSblack holes. More precisely, if an extremal black hole is characterized by the physical quantities (M,J,Q), and absorbs energy, angular momentum and electric charge (∆M,∆J,∆Q) by interacting with the test fields, then the metric corresponding to the physical quantities (M +∆M,J + ∆J,Q+ ∆Q) represents either a subextremal or an extremal black hole. Proof. The physical mass of a Kerr-Newman or Kerr-Newman-AdS black hole, givenin (6), is completely determined by the blackhole’s eventhorizonareaA, angular momentum J and electric charge Q through a Smarr formula M =M(A,J,Q). (35) From the first law of black hole thermodynamics, we know that this function satisfies κ dM = dA+ΩdJ +ΦdQ, (36) 8π where κ is the surface gravity of the event horizon [25, 35, 37]. The condition for the black hole to be extremal is ∂M κ=0⇔ (A,J,Q)=0, (37) ∂A which can be solvedto yield the area of an extremal black hole as a function of its angular momentum and charge, A=Aext(J,Q). (38) 7 The mass of an extremal black hole with angular momentum J and electric charge Q is then Mext(J,Q)=M(Aext(J,Q),J,Q). (39) AKerr-Newman-AdSmetriccharacterizedbyM,J andQwillrepresentablack hole if M ≥Mext(J,Q), and a naked singularity if M <Mext(J,Q). We have ∂M ∂A ∂M ∂M ∂A ∂M ext ext dMext = + dJ + + dQ (cid:18)∂A ∂J ∂J (cid:19) (cid:18)∂A ∂Q ∂Q(cid:19) κ ∂A κ ∂A ext ext = +Ω dJ + +Φ dQ (cid:18)8π ∂J (cid:19) (cid:18)8π ∂Q (cid:19) =ΩdJ +ΦdQ, (40) where all quantities are evaluated at the extremal black hole. Consider now an extremal black hole with angular momentum J, electric charge Q and mass M = Mext(J,Q). After interacting with the test fields, its angular momentum is J +∆J, its electric charge is Q+∆Q and its mass is, using (34) and (40), M +∆M ≥M +Ω∆J +Φ∆Q =Mext(J,Q)+∆Mext =Mext(J +∆J,Q+∆Q). (41) In other words,the final mass is abovethe mass of an extremalblack hole with the same angularmomentum andelectric charge,meaning thatthe finalmetric does not represent a naked singularity, that is, the black hole has not been destroyed. 4 Discussion In this paper we proved that extremal Kerr-Newman or Kerr-Newman-AdS black holes cannot be destroyed by interacting with (possibly charged) test fields satisfying the null energy condition at the event horizon and appropriate boundary conditions at infinity. This includes as particular cases all previous resultsofthis kind obtainedfor scalarandelectromagnetictestfields [9,10,11, 12]. The corresponding results for test particles [6, 7, 8] can also be considered particular cases, since particles can be seen as singular limits of continuous media [30, 31]. It is interesting to note that if the null energy condition is not satisfiedthenthe weakcosmiccensorshipconjecturemayindeedbe violated,as shown in [39, 40] for Dirac fields. Our proof depends only on certain generic features of the Kerr-Newman or Kerr-Newman-AdS metric, and can therefore be adapted to other black holes. In fact, Theorem 3.1 can be generalized as follows. Theorem 4.1. Consider a family of charged and spinning black holes in some metric theory of gravity, with suitable asymptotic regions, and test fields propagating in these backgrounds, such that: 8 1. There existsan asymptotically timelike Killing vector fieldK,determining the black hole’s physical mass, and angular Killing vector fields Y , yield- i ing the black hole’s angular momenta, such that event horizon’s Killing generator is Z =K+ Ω Y , (42) i i Xi where Ω are the thermodynamic angular velocities (that is, the angular i velocities that occur in the first law). 2. There exists a Smarr formula relating the black hole’s physical mass M, its entropy S, its angular momenta J and its electric charge Q, i M =M(S,J ,Q), (43) i yielding the first law of black hole thermodynamics, dM =TdS+ Ω dJ +ΦdQ, (44) i i Xi where T is the black hole temperature and Φ is the event horizon’s electric potential. 3. Extremal black holes (that is, black holes with T =0) are characterized by M =Mext(Ji,Q), and subextremal black holes by M >Mext(Ji,Q). 4. The test fields satisfy the null energy condition at the event horizon and appropriate boundary conditions at infinity. Then the test fields cannot destroy extremal black holes. More precisely, if an extremal black hole is characterized by the physical quantities (M,J ,Q), and i absorbs energy, angular momenta and electric charge (∆M,∆J ,∆Q) by inter- i acting with the test fields, then the metric corresponding to the physical quantities (M+∆M,J +∆J ,Q+∆Q)represents either a subextremalor an extremal i i black hole. Itiseasytocheckthatthisresultappliestoblackholesinhigherdimensions [41], including the case ofa negative3 cosmologicalconstant [37]. It can also be usedfor other blackholes, like acceleratedblackholes with conicalsingularities [42] or black holes in alternative theories of gravity [43]. There is, however, no a priori reason why it should apply to arbitrary parametrized deformations of the Kerr metric [44]. Acknowledgments JNwaspartiallyfundedbyFCT/PortugalthroughprojectPEst-OE/EEI/LA0009/2013. LQ gratefully acknowledges a scholarship from the Calouste Gulbenkian Foun- dationprogramNovos Talentos em Matemática. RVwassupportedbyagradu- ateresearchfellowshipfromtheFCT/PortugalprojectEXCL/MAT-GEO/0222/2012. 3Theorem 4.1does not applyto the caseof apositivecosmological constant, because the firsthypothesis isnotsatisfied. 9 Appendix Toobtainequation(8),weobservethatthechargedtestfieldsgenerateanextra electromagnetic field f satisfying the Maxwell equations df =0 and ∇ fµν =−jν. (45) µ The total electromagnetic energy-momentum tensor is then 1 TEM =(F +f )(F α+f α)− g (F +f )(Fαβ +fαβ). (46) µν µα µα ν ν 4 µν αβ αβ Besides the stationary part, due solely to F, one has to consider, in the test field approximation,the cross terms 1 t =f F α+F f α− g F fαβ. (47) µν µα ν µα ν 2 µν αβ We have 1 1 ∇µt =−j F α+f ∇µF α+F ∇µf α− (∇ F )fαβ − F ∇ fαβ µν α ν µα ν µα ν 2 ν αβ 2 αβ ν =−F jα, (48) να where we used (45), the Maxwell equation ∇µF =0, the fact that µα 1 1 f ∇µFνα− (∇νF )fαβ = f ∇αFνβ +∇βFαν −∇νFαβ =0 (49) µα αβ αβ 2 2 (cid:0) (cid:1) (because of the Maxwell equation dF =0), and the fact that 1 1 F ∇µfνα− F ∇νfαβ = F ∇αfνβ +∇βfαν −∇νfαβ =0 (50) µα αβ αβ 2 2 (cid:0) (cid:1) (because of the Maxwell equation df =0). Therefore, in the test field approximation, we have ∇µ T +TEM =0⇔∇µ(T +t )=0⇔∇µT =F jα, (51) µν µν µν µν µν να (cid:0) (cid:1) which is equation (8). One may wonder why not use the conserved current ∇ (TµνK +tµνK )=0 (52) µ ν ν to define the energy of the test field as E′′ = (Tµν +tµν)KνNµdV3. (53) Z S The reason is that this expression accounts for the energy of the interaction betweenthechargedfieldandthebackgroundelectromagneticfieldthroughthe electromagnetic cross terms (47), whereas (18) localizes it on the charges. As 10