Aspects of Magnetized Black Holes

Transcription

1 1 2 haryanto.siahaan@wigner.mta.hu MTA Lendület Holographic QFT Group, Wigner Research Centre for Physics, Konkoly-Thege Miklós u , 1121 Budapest, Hungary and Center for Theoretical Physics, Physics Department, Parahyangan Catholic University, Jalan Ciumbuleuit 94, Bandung 40141, Indonesia 1 Phys.Rev. D ) no.2, and Class.Quant.Grav ) no.15, Funded by The Tempus Public Foundation.

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3 Outline 1 Black Holes Black holes and Einstein-Maxwell system Ernst magnetization Ernst magnetization Magnetized Kerr 2 CFT and AdS/CFT CFT in D and 2 dimensions AdS/CFT correspondence 3 Kerr/CFT Extremal Kerr/CFT c for NHEK Frolov-Throne temperature and Entropy Non-extremal Kerr/CFT Magnetized Kerr/CFT Frolov-Throne temperature and Entropy of mag. black holes 4 Destroying black holes Classical test particle in black hole background Destroying asymp. flat. charged rotating black holes Magnetic field: shield or not?

4 Questions: Does Kerr/CFT conjecture still work for strongly) magnetized black hole? Does strong magnetic field around a black hole give contribution to form a naked singularity?

5 Black holes and Einstein-Maxwell system E.o.M. : R µν = 2F µαf α ν 1 2 gµνf 2 where the Maxwell system is source free, i.e. µf µν = 0. Asymptotically flat solution: Kerr-Newman.

6 Black holes and Einstein-Maxwell system E.o.M. : R µν = 2F µαf α ν 1 2 gµνf 2 where the Maxwell system is source free, i.e. µf µν = 0. Asymptotically flat solution: Kerr-Newman. For an axial and stationary spacetime, Ernst formalism is powerful. The line element and vector field: ds 2 = f dφ + ωdt) 2 + e 2µ dρ 2 + dz 2) + f 1 ρ 2 dt 2 and Ernst eqtns. A = A t dt + A φ dφ E + Ē + 2Φ Φ ) 2 E = 2 E + 2 Φ Φ ) E E + Ē + 2Φ Φ ) 2 Φ = 2 E + 2 Φ Φ ) Φ where E = f Φ Φ + iϕ, Φ = A φ + iãt ϕ,ρ = ρ 1 f 2 ω,z + i ΦΦ,ρ ) Φ Φ,ρ, ϕ,z = ρ 1 f 2 ω,ρ + i ΦΦ,z ) Φ Φ,z à t,ρ = ρ 1 f A t,z ωa φ,z ), Ãt,z = ρ 1 f A t,ρ ωa φ,ρ ) The operator is gradient operator in flat) cylindrical coord.

7 Ernst magnetization Melvin universe with ds 2 = 1 + B2 ρ 2 4 ) 2 ) 2 dt 2 + dρ 2 + dz 2) B2 ρ 2 ρ 2 dφ 2 4 F ρφ = 1 + B2 ρ 2 4 ) 1 ρb This solution satisfies the source free and Einstein-Maxwell equations.

8 Ernst magnetization Melvin universe with ds 2 = 1 + B2 ρ 2 4 ) 2 ) 2 dt 2 + dρ 2 + dz 2) B2 ρ 2 ρ 2 dφ 2 4 F ρφ = 1 + B2 ρ 2 4 ) 1 ρb This solution satisfies the source free and Einstein-Maxwell equations. FUN FACT:

9 Ernst magnetization Harrison transformation to a known stationary and axially symmetric spacetime in Einstein-Maxwell theory 3 : E = E Λ, Φ = 1 Φ BE ), Λ 2 where Λ = 1 + BΦ B2 E 4. The transformation leaves the Einstein-Maxwell eqtns. to be invariant, and source free condition still holds. The components of magnetized line element to the old one: Recall Kerr spacetime f = Λ 2 f, ω = Λ 2 ω + ρf 1 Λ Λ Λ Λ ). ds 2 = Σ Ξ dt2 + dr 2 + dθ2 ) + Ξ sin2 θ Σ dφ ωdt) 2, where Ξ = r 2 + a 2) 2 a 2 sin 2 θ, Σ = r 2 + a 2 cos 2 θ, and = r 2 + a 2 2mr, ωξ = 2mra. 3 F. J. Ernst J.Math.Phys.,17,54; F. J. Ernst and W. J. Wild J.Math.Phys.,17,182.

10 Magnetized Kerr ) ds 2 = Σ Λ 2 Ξ dt2 + dr 2 + dθ2 + Ξ sin2 θ 2 Σ Λ 2 Λ 0 2 dφ ω dt), where ω = 16mra+ω Br,θ)B 4, ω 8Ξ B r, θ) = 4a 3 m 3 r 3 + cos 4 θ ) cos +2am 2 r 4 2 θ 3 ) ) a 2 r cos 2 θ 2 cos 4 θ ) a cos 4 θ )) +amr r 2 + a 2) r cos 2 θ cos 4 θ ) a cos 2 θ 3 cos 4 θ )). ) Re Λ = 1 + B2 r 2 + a 2) sin 2 θ + 2a2 mr sin 4 θ, 4 Σ ) Im Λ = B2 cos θ ) 2am 2 + sin 2 θ + 2a3 m sin 4 θ. 4 Σ A = Φ 0 ω ) Φ 3 dt + Φ3 dφ Φ 0 = a { 4a 4 m 2 + 2a 4 m r 24a 2 m 3 r 24a 2 m 2 r 2 4a 2 mr 3 12m 2 r 4 6mr 5 8Ξ 12rm r 2 + a 2) )) )} cos 2 θ + 2mr 3 + a 2 4m 2 6mr cos 4 θ, { 1 Φ 3 = {4ΞB 8Σ Λ 2 sin 2 θ + B 4 Σ r 2 + a 2) 2 sin 4 θ + 4a 2 mr r 2 + a 2) sin 6 θ +4a 2 m r ) ) )}} sin 2 θ cos 2 θ + a cos 2 θ,

11 CFT in D and 2 dimensions CFT in D dimensions CFT is invariant under translations, rotations, dilations, and special conformal transformations. The conformal invariance of CFT rules the two and three point funtions, e.g. the two point function C 12 φ 1 x 1 ) φ 2 x 2 ) = x 1 x 2 h. 1+h 2

12 CFT in D and 2 dimensions CFT in 2 dimensions The fields transform from z z as φ z, z ) φ z, z) = n= ) εnln + ε n ln φ z, z), where l n = z n+1 z and l n = z n+1 z. The generators obey [l n, l m] = n m) l n+m, [ ln, l m ] = n m) ln+m, [ ln, l m ] = ) Globally defined only for n = ±1, 0, and the algebra is SL2, C) [l ±1, l 0 ] = l ±1, [l +, l ] = 2l 0. Central extension of 3.1): [L m, L n] = m n) L m+n + c 12 m 3 m ) δ m+n,0. Entropy formula in CFT 2 S Cardy = π2 3 c LT L + c R T R ).

13 AdS/CFT correspondence Gravity theory in the bulk is dual to a CFT with no gravity living on the boundary. Maldacena, Adv.Theor.Math.Phys ) )

14 AdS/CFT correspondence Gravity theory in the bulk is dual to a CFT with no gravity living on the boundary. Maldacena, Adv.Theor.Math.Phys ) ) Witten s prescription Witten, Adv.Theor.Math.Phys ) ) exp φ 0 O = Z grav φ 0 ). CFT An AdS/CFT test using scalar fields ) O x O x ) δ2 Z grav φ 0 ) ) δφ ) 1 0 x δφ0 x φ0 =0 x x 2D.

15 Extremal Kerr/CFT NHEK r λr + r 0 near horizon extreme Kerr) 4 Strominger et al, PRD )) ) ds 2 = 2GJΩ r 2 )dτ 2 + dr r 2 + dθ2 + Λ 2 dϕ + rdτ) 2. The Killing vectors 5 of NHEK; J 0 = 2 τ, J 1 = 2r sin τ 1 + r 2 τ 2 cos τ 1 + r 2 r + 2 sin τ 1 + r 2 ϕ, 2r cos τ J 2 = 1 + r 2 τ 2 sin τ 1 + r 2 r 2 cos τ 1 + r 2 ϕ, J U1) = ϕ The commutation between J s : SL2, R) U1) symmetry [ J2, J ] 0 = 2 J 2, [ J1, J ] 0 = 2 J 1, [ J1, J ] 2 = 2 J 0 4 Ω 2 1+cos 2 θ, Λ 2 2 sin θ 1+cos 2 θ. 5 µζν + ν ζ µ = 0

16 c for NHEK Brown and Henneaux 1986 discovery 6 : The algebra of the canonical generators of symmetry of AdS 3 spacetime 7 with a set of boundary conditions adpoted at spatial infinity turns out to be just the Virasoro algebra CFT 2 ) with c = 3l/2. 6 Comm. Math. Phys. 104, No ), S = d 3 x g R 2l 2)

17 c for NHEK Brown and Henneaux 1986 discovery 6 : The algebra of the canonical generators of symmetry of AdS 3 spacetime 7 with a set of boundary conditions adpoted at spatial infinity turns out to be just the Virasoro algebra CFT 2 ) with c = 3l/2. Boundary condition for NHEK: L ζ g µν = h µν O r 2) O 1) O r 1) O r 2) O 1) O 1) O r 1) O r 1) O r 1) O r 1) O r 1) O r 2) O r 2) O r 1) O r 2) O r 3). where accordingly ζ µ µ = rε φ) r + ε φ) φ. Conserved charges and central term: Q ζ = 1 ) d 2 x 16π µν µ ζ ν ν ζ µ ) Q ζ = dq ζ = Q ζ V V { } Qζ, Q ξ = Q[ζ,ξ] k g ) ζ gµν, L ξ g µν V k g g ζ gµν, hµν) = 64π ε µναβk µν ζ dx α dx β k µν ζ = ζ ν µ h ζ ν ρh µρ + h 2 ν ζ µ h νρ ρζ µ + ζ ρ ν h µρ µ ν) The central charge for NHEK: c = 12J. 6 Comm. Math. Phys. 104, No ), S = d 3 x g R 2l 2)

18 Frolov-Throne temperature and Entropy The spacetime outside of the Schwarzschild black hole is populated by quantum fields in the thermal state Hartle-Hawking state) weighted by a Boltzmann factor e ω/t H. For Kerr black holes, the corresponding thermal state Frolov-Throne state) is weighted by the Boltzmann factor e ω mω H )/T H. In the limits T H 0 and ω mω H, e ω mω H )/T H e m/t L) where T L = 1/2π Frolov-Thorne temp.). Cardy formula and Bekenstein-Hawking entropy S Cardy = π2 3 ct = S BH

19 Frolov-Throne temperature and Entropy The spacetime outside of the Schwarzschild black hole is populated by quantum fields in the thermal state Hartle-Hawking state) weighted by a Boltzmann factor e ω/t H. For Kerr black holes, the corresponding thermal state Frolov-Throne state) is weighted by the Boltzmann factor e ω mω H )/T H. In the limits T H 0 and ω mω H, e ω mω H )/T H e m/t L) where T L = 1/2π Frolov-Thorne temp.). Cardy formula and Bekenstein-Hawking entropy S Cardy = π2 3 ct = S BH The conformal symmetry which NHEK exhibits and the matching of entropy hint the duality of extremal Kerr black hole and CFT 2. However, the specific dual CFT 2 is unknown yet OPEN PROBLEM). Suppose this Kerr/CFT true, it only works for rotating black hole. How about Schwarzschild?

20 Non-extremal Kerr/CFT The near horizon limit of a non-extremal Kerr black hole is Rindler spacetime. The conformal structure is hidden in the low energy limit, Mω 1, of scalar wave Φ = e iωt+imφ R r) S θ) equation in the near region rω 1, [ r r ) + 2Mr+ω am)2 r r +)r + r ) 2Mr ω am) 2 r r )r + r ) ] Rr) = ll +1) Rr). 4.2) The l.h.s. of 4.2) can be rewritten as a squared Casimir of SL2, R) L SL2, R) R group Strominger et al, Phys.Rev. D ) ), ω + r r + = exp2πt R φ + 2n R t), ω r r + = exp2πt L φ + 2n L t) r r r r y = r + r expπt R + T L )φ + n R + n L )t) r r H 2 = H H 1H 1 + H 1 H 1 ), H 2 = H H1 H 1 + H 1 H1 ). H 1 = i +, H 0 = iω y y ), H 1 = iω + ) ω + y y y 2 ) H 1 = i, H 0 = iω y y ), H 1 = iω ) 2 + ω y y y 2 +) With T {L/R} = r+{+/ }r and the assumption the c is smoothly unchanged, the 4πa Cardy formula gives S Cardy = π2 3 c LT L + c R T R ) = 2πMr + = S BH.

21 Magnetized Kerr/CFT In the NH of extremal mag. Kerr NHEMK), the metric is ds 2 = Σ + Λ 2 + ) ) 1 + r 2) dt 2 + dr 2 1+r 2 ) + dθ2 + A+ sin2 θ Λ 0 1 B 2 4 m 4) 2 + dφ + rdt Σ + Λ 2 + Λ Σ + = m cos 2 θ ), A + = 4m 2 Λ 2 + = 1+B 2 m 2) B 2 m 2) 2 cos 2 θ 1+cos 2 θ and the vector field is A µdx µ = PQLθ)rdt + Lθ)dφ., Λ = Λ 2 + θ=0 In Einstein-Maxwell theory, the charges come from the diffeomorphisms L ζ g µν and L ζ A µ. As r, NHEMK behaves just like NHEK, so the same bound. conds. can be used. For A µ, we can impose A µ r ) [ O r), O r 1), O 1), O r 2)]. Central charges: c g = 12m 2 1 B 2 m 2) and c A = 0. Recall: [L m, L n] = m n) L m+n + c 12 m 3 m ) δ m+n,0.

22 Frolov-Throne temperature and Entropy of mag. black holes We can use the same Frolov-Throne state, i.e. weighted by the Boltzmann factor e ω mω H )/T H. In the limits T H 0 and ω mω H, e ω mω H )/T H e m/t L) where T L = 1+B4 M 4 2π1 B 4 M 4 ). T L can be negative for strong B) Cardy formula and Bekenstein-Hawking entropy S Cardy = π2 3 12m2 1 B 2 m 2) 1 + B 4 M 4 2π 1 B 4 M 4) = S BH It seems the strong magnetization breaks the hidden conformal symmetry, due to the non-separability of wave equation for a scalar perturbation, i.e. magnetized Kerr metric does not possess the Killing-Yano tensor.

23 Classical test particle in black hole background We can go back the the asymptotically flat black holes in Einstein-Maxwell theory, i.e. Kerr-Newman black hole. Horizons: r ± = M ± M 2 a 2 Q 2, where a = JM 1. Extremal state: M 2 = a 2 + Q 2. Dynamics of classical test particle around this black hole is dictated by the geodesic equation: du µ ds + Γµ αβ uα u β = q m F µν u ν Effectively, the e.o.m. can be obtained from Two conserved quantities: E = L ṫ ) = m g tt ṫ + g tφ φ qa t L = 1 2 mg αβẋα ẋ β + qa µẋ µ Naked singularity formation condition for Kerr-Newman):, L = L ) φ = m g tφ ṫ + g φφ φ + qa φ. ) J + L 2 i). M + E) 2 < Q + q) 2 + and ii). V eff = ṙ 2 M + E 2 < 0 r > r+

24 Destroying asymp. flat. charged rotating black holes For the limits of static or neutral black holes, i.e. RN or Kerr black holes, these black holes can be overspun or overcharged from its extremal state. V. E. Hubeny, PhysRevD ); T. Jacobson and T. P. Sotiriou, PhysRevLett ) However, when these black holes are near extremal, a classical test particle may destroy their horizon. The last observation holds for a rotating and charged black hole. Even for an extremal rotating and charged black hole, a test particle can destroy its horizon. Gao and Zhang, PhysRevD ); HMS, PhysRevD ). However, there are pros and cons related to this results. Some authors show that if self-force effect is considered, such destruction can never happen. Ref. eg. P. Zimmerman, I. Vega, E. Poisson and R. Haas, PhysRevD ).

25 Destroying asymp. flat. charged rotating black holes Example: rotating charged black hole in string theory with horizons r ± = M b ± M b) 2 a 2 where b = Q 2 /2M. Extremal condition: 2 M + E) 2 = 2 J + L + Q + q) 2. Near-extreme parameter: δ = M b a. After capturing a test particle: δm + 2ME + E 2 < Qq + L + q2. Constraint for the maximum energy. 2 The condition for minimum energy is obtained from the conserved quantity E expression evaluated at r = r +.

26 Magnetic field: shield or not? What is mass, angular momentum, and charged in magnetized spacetime? M. Astorino, G. Compère, R. Oliveri and N. Vandevoorde, PhysRevD ) M = [ M 2 + 2JQB + 2J M2 Q 2 Q 4 ) B 2 + JQ + J 2 M J2 Q ) ] 1 16 M2 Q 4 B 4 2. J = J Q 3 B 3 2 JQ2 B 2 Q 4 8J 2 + Q 4) B 3 J 16 Q = Q + 2JB Q3 B 2 4 2M 2 32 Q2 ) B 3 16J 2 + 3Q 4) B 4 Interesting relation: Christodoulou-Ruffini mass for Kerr-Newman black hole D. Christodoulou and R. Ruffini, Phys. Rev. D4 1971) 3552 Non-negative Entropy: 0 M 4 Q 2 M2 J 2 = M 2 ) M 2 = S 4π + Q 2 π Q4 + 4 J S ) M 2 J2 M 2 Q2 Λ RN,0 2 + Λ ) K, JQB 3

27 Magnetic field: shield or not? We consider the extremal magnetized Kerr only : M 2 = J. Make sure that test particle s geodesic exists on the equatorial plane perpendicular to the rotational and magnetic field axis). Cosmic censor violation constraint for E max ) M + E ) 2 < Q + q ) 2 + Q + q ) J + L ) 2 1 B Minimum energy: E min = L 4 M 4) +4qM 4 B 3 1+2B 4 M 4). 2M1+B 4 M 4 ) 2

28 Magnetic field: shield or not? However, such particle cannot reach the horizon. Illustration

29 Conclusions and future works The extremal Kerr/CFT conjecture holds for a Kerr black hole immersed in a strong magnetic field. The hidden conformal symmetry for magnetized Kerr black hole does not exist. An extremal Kerr black hole immersed by strong magnetic field cannot be destroyed by a classical test particle.

30 Conclusions and future works The extremal Kerr/CFT conjecture holds for a Kerr black hole immersed in a strong magnetic field. The hidden conformal symmetry for magnetized Kerr black hole does not exist. An extremal Kerr black hole immersed by strong magnetic field cannot be destroyed by a classical test particle. Can test particle destroy a near extremal magnetized Kerr black hole? Can one obtain a magnetized black hole system in 4d Einstein-Maxwell-Λ theory? Non-unitary CFT 2 dual non-unitary physics of extremal magnetized Kerr black hole?

31 Acknowledgements My host: Dr. Zoltan Bajnok. The Tempus Public Foundation.