Rotating Attractors - one entropy function to rule them all Kevin Goldstein, TIFR ISM06, Puri,

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1 Rotating Attractors - one entropy function to rule them all Kevin Goldstein, TIFR ISM06, Puri, talk based on: hep-th/ (Astefanesei, K. G., Jena, Sen,Trivedi); hep-th/ (K.G., Iizuka, Jena & Trivedi); hep-th/ (Sen) and work in progress...

2 Plan Study rotating attractors using Sen s entropy function Motivation? mininal requirements for the attractor mechanism? learn more about of the Entropy of non-susy blackholes? technical points: entropy function formalism still apply with less symmetry? only need near horizon geometry

3 What are blackhole attractors? Context = Theory with gravity, gauge fields, neutral scalars generically appear as (part of) string theory low energy limit of scalars (or moduli) dimensions encode geometry of compactified Attractor mechanism = scalars values fixed at Blackhole s horizon independent of values at infinity So horizon area depends only on gauge charges Entropy depends only on charges works for Extremal (T = 0) blackholes

4 Hand waving number of microstates of extremal blackhole determined by quantised charges entropy can not vary continuously but the moduli vary continuously resolution: horizon area independent of moduli moduli take on fixed values at the horizon determined by charges No mention of SUSY

5 Outline Go through examples of application of entropy function without rotation with rotation Time permitting: blackring Study Lagrangians which generically appear as (the bosonic part of) certain low energy limits of string theory

6 Outline Go through examples of application of entropy function without rotation: Only need near horizon geometry Assume extremal (T = 0) AdS 2 S 2 near horizon symmetries Equations of motion Extremising an Entropy function Value of the Entropy function at extremum = Wald Entropy of Blackhole need to solve algebraic equations Argument is independent of SUSY

7 Outline Go through examples of application of entropy function without rotation with rotation: Only need near horizon geometry Assume AdS 2 U(1) near horizon symmetries Equations of motion Extremising an Entropy function need to solve differential equations Entropy function at extremum = Wald Entropy Argument is independent of SUSY

8 Outline Go through examples of application of entropy function without rotation with rotation Time permitting: a blackring in 5-d Only need near horizon geometry Assume AdS 2 S 1 S 2 near horizon symmetries not the most general ansatz Equations of motion Extremising an Entropy function Entropy function at extremum = Wald Entropy need to solve algebraic equations End up with AdS 3 S 2

9 Step 1 First we look at simple 4-dimensional spherically symmetric black holes later we will compare it with more complicated cases.

10 Entropy Function (Sen) Set up: Gravity, p-form gauge fields, massless neutral scalars L gauge and coordinate invariant - in particular there may be higher derivative terms Extremal = AdS 2 S 2 Near horizon geometry Entropy function: First we consider, f, the Lagrangian density evaluated at the horizon: f[e i,p i,r AdS2,R S2,ϕ s ] = q i = f e i Z H p gl

11 Entropy Function (Sen) Set up: Gravity, p-form gauge fields, neutral scalars L gauge and coordinate invariant - in particular there may be higher derivative terms Extremal = AdS 2 S 2 Near horizon geometry Entropy function: First we consider, f, the Lagrangian density evaluated at the horizon. Now take the Legendre transform of f w.r.t the electric charges: Z «p E = 2π q i e i gl H E = E[q i,p i,r AdS2,R S2,ϕ s ]

12 Entropy Function (Sen) Results: E = 2π Z q i e i H p gl «equations of motion Extremising E Wald Entropy = Extremum of E Fixing q i and p i fixes everything else completely

13 Caveats Entropy function, E, might have flat directions The near horizon geometry is not completely determined by extremisation of E There may be a dependence of the near horizon geometry on the moduli But since these are flat directions the entropy is still independent of the moduli Generalised attractor mechanism Also note that we have assumed that a blackhole solution exists which may not always be the case.

14 Summary of rotating attractors Scalars constant on horizon (non-dyonic) S BH = 2π p (V eff ) 2 + J 2 Φ V eff = 0 Scalars vary over horizon (Φ = Φ(θ)) (dyonic) E has no flat directions Θ r Θ Π r Θ Π r E has flat directions Θ 0 Θ Π 2 Θ Π r r r 2 2 2

15 Simple example: spherically symmetric case L = R h rs ( Φ)g µν µ Φ s ν Φ r f ij ( Φ)g µρ g νσ F µν (i) F ρσ (j) 1 f 2 ij ( Φ)ɛ µρνσ F µν (i) F ρσ (j)

16 Simple example: spherically symmetric case L = R h rs ( Φ)g µν µ Φ s ν Φ r f ij ( Φ)g µρ g νσ F µν (i) F ρσ (j) 1 f 2 ij ( Φ)ɛ µρνσ F µν (i) F ρσ (j) The effect of the f ij term is basically q i q i 4 f ij p j and for simplicity we will neglect it

17 Simple example: spherically symmetric case L = R h rs ( Φ)g µν µ Φ s ν Φ r f ij ( Φ)g µρ g νσ F µν (i) F ρσ (j) Ansatz: AdS 2 S 2 near horizon geometry ds 2 = v 1 ` r 2 dt 2 + dr 2 /r 2 + v 2 dω 2 2 F i rt = e i F i θφ = pi 4π sinθ Φ r = u r (const.)

18 Entropy function Wish to calculate: E = 2π(q i e i f) = 2π Z q i e i dθdφ p «gl Calculate the action: f[ e, p, u,v 1,v 2 ] = (4π)(v 1 v 2 ) 2 2 2e i e j + f ij (u r ) v 2 v 1 v1 2 2pi p j (4π) 2 v 2 2 ««Calculate the conjugate variables: q i = f «e j e = (16π)(v 1v i 2 )f ij (u r ) v1 2 e j = 1 16π «v1 v 2 «f ik q k

19 Entropy function Finally E[ q, p, u,v 1,v 2 ] = 2π 8π(v 2 v 1 ) + v1 v 2 «V eff «Notation V eff = 1 2π» p i f ij ( u)p j q if ij ( u)q j

20 Entropy function Finally E[ q, p, u,v 1,v 2 ] = 2π 8π(R 2 S R2 AdS ) + R2 AdS R 2 S!V eff! Notation V eff = 1 2π» p i f ij ( u)p j q if ij ( u)q j

21 Equations of Motion Then the equations of motion are equivalent to extremising the entropy function: E = 0 V eff = 0 Φ I Φ I E = 0 8π v 1 2 v V eff(φ I ) = 0 1 E = 0 8π + v 1 v 2 2 v V eff(φ I ) = 0 2 So and v 1 = v 2 = 8πV eff S BH = 2πV eff

22 Rotation What is the generalisation of an AdS 2 S 2 near horizon geometry for rotating blackholes? Take a hint from the near horizon geometry of extremal Kerr Blackholes (Bardeen, Horowitz) SO(2,1) U(1)

23 Recall: SO(2,1) S 2 Ansatz «ds 2 = v 1 r 2 dt 2 + dr2 + v r 2 2 dθ 2 + v 2 sin 2 θ dφ 2 ϕ s = u s 1 2 F µν (i) dx µ dx ν = e i dr dt + pi sinθ dθ dφ 4π

24 SO(2,1) U(1) Ansatz «ds 2 = v 1 (θ) r 2 dt 2 + dr2 +β 2 dθ 2 +v 2 (θ)sin 2 θ (dφ αrdt) 2 r 2 ϕ s = u s (θ) A i = e i rdt + b i (θ)(dφ αr dt) Horizon has spherical topology v 2 (θ) at poles 1 p i = Z dθdφf (i) θφ = 2π(bi (π) b i (0)).

25 SO(2,1) U(1) Ansatz «ds 2 = v 1 (θ) r 2 dt 2 + dr2 +β 2 dθ 2 +v 2 (θ)sin 2 θ (dφ αrdt) 2 r 2 ϕ s = u s (θ) 1 F (i) 2 µν dx µ dx ν = (e i αb i (θ))dr dt + b i (θ)dθ (dφ αr dt) Horizon has spherical topology v 2 (θ) at poles 1 p i = Z dθdφf (i) θφ = 2π(bi (π) b i (0)).

26 SO(2,1) U(1) Ansatz «ds 2 = Ω 2 e 2ψ r 2 dt 2 + dr2 + βdθ 2 + e 2ψ (dφ αrdt) 2 r 2 ϕ s = u s (θ) 1 F (i) 2 µν dx µ dx ν = (e i αb i (θ))dr dt + b i (θ)dθ (dφ αr dt) Horizon has spherical topology e 2ψ at poles sin 2 θ p i = Z dθdφf (i) θφ = 2π(bi (π) b i (0)).

27 Symmetries One way to see that the ansatz has SO(2, 1) U(1) symmetries is to check that it is invariant under the Killing vectors, φ and L 1 = t, L 0 = t t r r, L 1 = 1 2 «1 r + 2 t2 t (tr) r + α r φ. can also be seen by thinking of φ as a compact dimension and find that the resulting geometry has a manifest SO(2, 1) symmetry with the conventional generators.

28 Entropy function: We define f[α,β, e,ω(θ),ψ(θ), u(θ), b(θ)] := Z dθdφ p gl The equations of motion are: f α = J f β = 0 f e i = q i δf δb i (θ) = 0 δf δω(θ) = 0 δf δψ(θ) = 0 δf δu s (θ) = 0

29 Entropy function: Equivalently we let E[J, q, b(θ),β,v 1 (θ),v 2 (θ), u(θ)] = 2π (Jα + q e f) The equations of motion: E α = 0 E β = 0 E e i = 0 δe δb i (θ) = 0 δe δv 1 (θ) = 0 δe δv 2 (θ) = 0 δe δu s (θ) = 0

30 Examples Kerr, Kerr-Newman, constant scalars (non-dyonic) Dyonic Kaluza Klein blackhole (5-d 4-d). -(Rasheed) Blackholes in toroidally compactified heterotic string theory -(Cvetic,Youm;Jatkar,Mukherji,Panda)

31 Two derivative Lagrangians L = R h rs ( Φ)g µν µ Φ s ν Φ r f ij ( Φ)g µρ g νσ F µν (i) F ρσ (j) Z E 2π(Jα + q e dθdφ p detg L) Z " = 2πJα + 2π q e 4π 2 dθ 2Ω 1 β 1 Ω 2 2Ωβ 2Ωβ 1 ψ α2 Ω 1 βe 4ψ β 1 Ωh rs ( u)u ru s +2f ij ( u) n βω 1 e 2ψ (e i αb i )(e j αb j ) β 1 Ωe 2ψ b i b j o# +8π 2 ˆΩ 2 e 2ψ sinθ(ψ + 2Ω /Ω) θ=π θ=0.

32 Equations of motion Notation: χ i = e i αb i Ω equation: 4β 1 Ω /Ω+2β 1 (Ω /Ω) 2 2β 2β 1 (ψ ) α2 Ω 2 βe 4ψ β 1 h rs u ru s+2f ij n βω 2 e 2ψ χ i χ j α 2 β 1 e 2ψ χ i χ j o = 0, ψ equation: 4β 1 (Ωψ ) 2α 2 Ω 1 βe 4ψ + 2f ij n 2βΩ 1 e 2ψ χ i χ j 2α 2 β 1 Ωe 2ψ χ i χ j o = 0,

33 u s equation: 2`β 1 Ωh rs u +2 r s f ij nβω 1 e 2ψ χ i χ j α 2 β 1 Ωe 2ψ χ i χ j o β 1 Ω( r h ts )u tu s = 0, b equation: αβf ij Ω 1 e 2ψ χ j α 1 β 1 f ij Ωe 2ψ χ j = 0 β equation: where Z dθi(θ) = 0 I(θ) = 2Ω 1 β 2 (Ω ) 2 2Ω + 2Ωβ 2 (ψ ) α2 Ω 1 e 4ψ +β 2 Ωh rs u ru s+2f ij n Ω 1 e 2ψ χ i χ j + α 2 β 2 Ω(θ)e 2ψ(θ) χ i χ j o

34 Charges: q i = 8π Z dθ ˆf ij βω 1 e 2ψ χ j, J = 2π R π 0 dθ αω 1 βe 4ψ 4βf ij Ω 1 e 2ψ χ i b j

35 Solutions Equations can be solved for some simple cases Kerr, Kerr-Newmann, constant scalars Check known solutions fitted into the frame work: KK blackholes, Toroidal compactification of Heterotic string theory

36 Kaluza-Klein Blackholes Charges = Q, P, J L = R 2( ϕ) 2 e 2 3ϕ F 2 2 types of extremal blackholes Both have SO(2,1) U(1) near horizon geometry non-susy 1. Ergo branch J > P Q Ergo-sphere S = 2π p J 2 P 2 Q 2 E has flat directions 2. Ergo-free branch J < P Q no Ergo-sphere S = 2π p P 2 Q 2 J 2 E has no flat directions

37 Blackholes in Heterotic String Theory on T 6 Charges = Q 1, Q 2, Q 3, Q 4,P 1, P 2, P 3, P 4, J, (Actually 56 P s and Q s) Duality invariant quartic D = (Q 1 Q 3 + Q 2 Q 4 )(P 1 P 3 + P 2 P 4 ) 1 4 (Q 1P 1 + Q 2 P 2 + Q 3 P 3 + Q 4 P 4 ) 2 1. Ergo branch Ergo-sphere S = 2π J 2 + D E has flat directions 2. Ergo-free branch no Ergo-sphere S = 2π J 2 D E has no flat directions

38 Ergo-free branch Θ r Φ π/2 π θ 0.4

39 Ergo-free branch Θ Π r Φ π/2 π θ 0.4

40 Ergo-free branch Θ Π r Φ π/2 π θ 0.4

41 Ergo-free branch Θ 3Π r Φ π/2 π θ 0.4

42 Ergo-free branch Θ Π r Φ π/2 π θ 0.4

43 Scalar Field at Horizon Π 2 Π 3Π 2 2Π Θ 0.3

44 Ergo-branch Θ r

45 Ergo-branch Θ Π r

46 Ergo-branch Θ Π r

47 Ergo-branch Θ 3Π r

48 Ergo-branch Θ Π r

49 Scalar Field at Horizon Π 2 Π 3Π 2 2Π Θ

50 Black ring attractors (5-d) Consider 5-d Lagrangian with massless uncharged scalars coupled to U(1) gauge fields with Chern-Simons terms: L = R h ab ( Φ) µ Φ a µ Φ b f ij ( Φ)F i µνf j µν c ijk ɛ µναβγ F i µνf j αβ Ak γ Action is not gauge invariant Entropy function formalism does not apply similar to BTZ black hole with gravitational Chern- Simons and/or gauge Chern-Simons term compactify ψ (Sen, Sahoo) can apply formalism to dimensionally reduced action Related work: (Kraus, Larsen), (Dabholkar, Iizuka, Iqubal, Sen, Shigemori)

51 Black ring attractors (5-d) Ansatz: AdS 2 S 1 S 2 «ds 2 = v 1 r 2 dt 2 + dr2 + w 2 (dψ + erdt) 2 + v r 2 2 `dθ 2 + sin 2 θdφ 2, Φ s = u s (θ) A i = e i rdt + A i ψ (dψ + erdt) pi cosθdφ, dimensionally reduce calculate entropy function for our ansatz

52 Black ring attractors (5-d) Ansatz: AdS 2 S 1 S 2 «ds 2 = v 1 r 2 dt 2 + dr2 + w 2 (dψ + erdt) 2 + v r 2 2 `dθ 2 + sin 2 θdφ 2, Φ s = u s (θ) 1 F (i) 2 µν dx µ dx ν = (e i + ea i ψ )dr dt + pi sinθ dθ dφ dimensionally reduce calculate entropy function for our ansatz

53 Calculation First we consider, f, the (dimensionally reduced) action evaluated at the horizon f[v 1,v 2,w,e, A ψ, p, e, u] = Z H p gl The equations of motion are f v 1 = 0 f v 2 = 0 f w = 0 f A i ψ = 0 f u a = 0 f e i = q i f e = q

54 Entropy function To calculate the entropy function, E = 2π(qe + q e f), it is convenient to eliminate A i ψ, e and e using their equations of motion Except in special cases, we find, from the A i ψ equation of motion, F i tr = 0

55 Entropy function After some algebra, we obtain the entropy function where E = 32π 3 w v 2 v 1 + v 1 v 2 V eff + V eff = f ij ( u)p i p j ˆq `q 1 4 dij q i q j 3 4 ˆq2 w 4 and d ij is proportional to inverse of c ijk p k.!!,

56 Solution Extremising the entropy function we find s E = 128π 3 ˆq(V eff ) = 32π 3 wv 2 = V =0 A 4G N with and v 1 = v 2 = 4 V 3 eff V =0 w 4 = 9 4 ˆq2 V eff V =0 e 2 w 2 = v 1 AdS 3 S 2

57 Puzzles What is the relationship between the ergosphere and the uniqueness of the Entropy function? Under what conditions do these solutions actually exist? Why are these (bosonic) symmetries sufficient for the attractor mechanism? AdS 2 /CF T, AdS 3 /CF T What does this really tell us about the entropy of non- SUSY blackholes?

58 Thank you