adiation from the non-extremal fuzzball Borun D. Chowdhury The Ohio State University The Great Lakes Strings Conference 2008 work in collaboration with Samir D. Mathur (arxiv:0711.4817)
Plan Describe non-extremal fuzzball and dual CFT state Describe instability of the above solution Give CFT/ microscopic description of the instability
Structure of Black Holes Singularity Flat Space Neck Throat (AdS) Horizon Horizon Singularity
Black Holes in SUGA Example Compactify 5 dimensions out of 10 1,9 1,4 X T 4 X S 1 Take n1 D1 branes along S 1 Take n5 D5 branes along T 4 X S 1 Take np momentum units along S 1 Form their bound state
Extremal and Non-Extremal Black Holes Extremal Black Holes - minimum mass for given charges MD1 ~ MD5 ~ V MP ~ 1/ where volume of S 1 ~, T 4 ~ V Extra energy excites anti-charges For large only anti-momentum is excited D1 D5 P + Energy D1 D5 P P
AdS/CFT Flat Space Neck Throat (AdS) CFT CFT is unitary Horizon Singularity Where are the states in gravity?
CFT - Effective String adius= N=n1n5 total winding number 1+1 (4,4) CFT with target space deformation of orbifold (T 4 ) N /S N Four bosonic excitations carrying spacetime indices Four real fermions in each sector spin of fermions in CFT = angular momentum SO(4) ~ SU(2)L X SU(2)
Four bosonic excitations Two complex fermionic excitations E = 1 J = 1 2 E = 1 n J = 1 2
Special states have been made E = 0 J = n 1n 5 2 in the left and right sector
Smooth Solutions (Extremal) otation Horizon disappears - single state Black Hole Smooth Geometry Originally done for 2-charge D1-D5 Later on same was done for 3-charge D1-D5-P hep-th/0011217 hep-th/0012025 hep-th/0405017 hep-th/0406103
Non-extremal smooth geometries otation Horizon disappears - single state Non-extremal Black Hole hep-th/9603100 Non-extremal Smooth Geometry hep-th/0504181 hep-th/9705192
Features of this geometry No horizon J ψ = m n 1 n 5 J φ = 0 P = 0 M = m2 1 2 n 1n 5 Completely Smooth otation Have ergoregion No global timelike Killing vector Negative energy excitations inside ergoregion
CFT dual to non-extremal smooth geometries J = m n 1n 5 2 E = m2 1 4 n 1n 5 in left and right sectors fermionic excitations on left and right sectors All component strings of same winding number
Classical Instability (hep-th/0512277) Klein-Gordon equation in the smooth background ansatz Ψ = 0 Ψ = exp( iωt + iλ y + im ψψ + im φ φ)χ(θ)h(r) In the large limit angular part reduces to laplacian on S 3
The radial equation solution in outer region X ω = 1 ( l m ψm 2) solution in inner region ω I = 1 2π (l!) 2 ( ω 2 Q 1 Q 5 4 2 ) l+1
Interpretation of the wave solution flat space and AdS decouple the wave splits off into two parts the energy of excitations in AdS is ω = 1 ( l 2 m ψm)
Like tunneling of excitations out of a box However here the box initially had no excitations Simultaneously excitations produced inside and outside the box Schiff-Schnyder-Weinberg effect Energy conservation?
More general case ω l m ψ m + m φ n λ m ψ n + m φ m 2(N + 1) = where J ψ = m n 1 n 5 J φ = n n 1 n 5 Ψ = exp( iωt + iλ y + im ψψ + im φ φ)χ(θ)h(r)
Proposal: the instability vertex operator - twists (l+1) strings to one - annihilates and creates bosons and fermions on the strings - produces graviton in the bulk energy, momentum, angular momentum conserved
Model: nl left fermions and n right of two flavors The string has spin half in each direction E = n 1 n 5 (P L + P ) = n 1 n 5 (n L (n L + 1) + n (n + 1)) P = n 1 n 5 (P L P ) = n 1 n 5 (n L n )(n L + n + 1) (J L, J ) = n 1n 5 2 (2n L + 1, 2n + 1)
With m = n L + n + 1 n = n L n and J ψ = J L J J φ = J L J M ADM = 1 2 (m2 + n 2 1)n 1 n 5 J ψ = mn 1 n 5 J φ = nn 1 n 5 n p = nmn 1 n 5 matches gravity so we understand the CFT state
Instabilities: Explicit example nl=4,n=2, l =3 start with l+1=4 loops 2 flavors, 4 fermions each X 3+1=4 2 flavors, 2 fermions each
For each flavor in the left moving sector P f L = 1 [1 + 2 + 3 + 4] = 10 For total left momentum we have P L = 20
For each flavor in the right moving sector P f = 1 [1 + 2] = 3 For total left momentum we have P = 6
spin from left moving fermions of one flavor J f,ferm L = 1 2 4 = 2 base spin of the string J f,base L = 1 2 total spin of left movers J L = 1 2 + 2 + 2 = 9 2
spin from right moving fermions of one flavor J f,ferm = 1 2 2 = 1 base spin of the string J f,base = 1 2 total spin of right movers J = 1 2 + 1 + 1 = 5 2
E = P L + P = 20 + 6 = 26 P = P L P = 20 6 = 14 J ψ = (J + J L ) = 5 + 9 = 7 2 J φ = (J J L ) = 2
grab 3+1 of these X 3+1 X 2 flavors E = 4(P L + P ) = 104 P = 4(P L P ) = 56 J ψ = 4(J + J L ) = 28 J φ = 4(J J L ) = 8
and make a twisted string of length (3+1) X 2 flavors
X 2 flavors momentum quanta have gone down to For each flavor in the left moving sector P f L = 1 34 [1 + 2 +... + 16] = 4 total left momentum P L = 68 1 4
X 2 flavors momentum quanta have gone down to For each flavor in the left moving sector P f = 1 4 [1 + 2 +... + 8] = 9 total right momentum P = 18 1 4
The spin from left moving fermions of one flavor J f,ferm L = 1 2 16 = 8 X 2 flavors base spin of the string J f,base = 1 2 total spin of left movers J L = 1 2 + 2 8 = 33 2
The spin from left moving fermions of one flavor J f,ferm = 1 2 8 = 4 base spin of the string X 2 flavors J f,base = 1 2 total spin of left movers J = 1 2 + 2 4 = 17 2
after the twisting we have X 2 flavors E = P L + P = P = P L P = 68 + 18 68 18 J ψ = (J + J L ) = J φ = (J J L ) = 8 = 86 = 50 17 + 33 2 = 25
Account E=2/ for bosons E = 104 P = 56 J ψ = 28 J φ = 8 E = 86 + 2 = 88 P = 50 J ψ = 25 J φ = 8 ω = 16 λ = 6 m ψ = 3 m φ = 0
ω = 16 λ = 6 m ψ = 3 m φ = 0 m = n L + n + 1 = 7 n = n L n = 2 ω l m ψ m + m φ n λ m ψ n + m φ m 2(N + 1) = 16 = 3 ( 3)7 + (0)2 6 ( 3)2 + (0)7 2 Our model gives agreement with grav. energy
Width of instability: growth rate - Earlier result: reproduced Hawking radiation - similar model: fermions and bosons on the string thermally distributed - Take interaction vertex from those calculations - Find decay rate for our non-thermal fermions Γ l = 4π (l!) 2 ( Q1 Q 5 4 2 ) l+1 ω 2(l+1)
But this is the spontaneous part of decay dn dt = Γ Stimulated emission would give dn dt Is it LASE? = Γ(1 + N) No X No mirror
ecall CFT was symmetrized 1+1 (4,4) CFT with target space (T 4 ) N /S N transition between two BECs H int n, k = α N n n + 1 k + 1 n + 1, k + 1 + α N n + 1 n k 1 n 1, k 1 (6.160 N-n excited k scalars n de-excited Scalars escaping: n can only go to n+1 X dn dt = α 2 (N n)(n + 1) α 2 N(n + 1)
- decay of excited state the quanta in the wave grows as dn = γ(n + 1) dt - identify spontaneous emission part to the black hole decay rate - for large N the spontaneous part is negligible and we get N = N 0 e Γ lt - wave grows as Im(ω) = 1 2 Γ l = N 2π (l!) 2 ( Q1 Q 5 4 2 ) l+1 ω 2(l+1) Model agrees with grav. decay width
Conclusion Emission from non-extremal fuzzball found On the CFT side process same as Hawking adiation Hawking pair interpretation Suggests a non-rotating fuzzball could have ergoregion like regions while having net angular momentum zero For generic fuzzball population of each kind of component string is small so the above process will manifest itself as Hawking radiation
ds 2 f = (dt 2 dy 2 ) + H1 H5 M H1 H5 (s p dy c p dt) 2 ( r + H 2 dr 2 ) 1 H5 (r 2 + a 2 1 )(r2 + a 2 + dθ2 2 ) Mr2 ( + H1 H5 (a 2 2 a2 1 )( H 1 + H ) 5 f) cos 2 θ cos 2 θdψ 2 H1 H5 + ( H1 H5 + (a 2 2 a2 1 )( H 1 + H 5 f) sin 2 θ H1 H5 ) sin 2 θdφ 2 + M H1 H5 (a 1 cos 2 θdψ + a 2 sin 2 θdφ) 2 + 2M cos2 θ H1 H5 [(a 1 c 1 c 5 c p a 2 s 1 s 5 s p )dt + (a 2 s 1 s 5 c p a 1 c 1 c 5 s p )dy]dψ + 2M sin2 θ H1 H5 [(a 2 c 1 c 5 c p a 1 s 1 s 5 s p )dt + (a 1 s 1 s 5 c p a 2 c 1 c 5 s p )dy]dφ + H1 H 5 4 i=1 dz 2 i H i = f + M sinh 2 δ i, f = r 2 + a 2 1 sin 2 θ + a 2 2 cos 2 θ,
The D1 and D5 charges of the solution produce a 2-form gauge field given by [6] C 2 = M cos2 θ [(a 2 c 1 s 5 c p a 1 s 1 c 5 s p )dt + (a 1 s 1 c 5 c p a 2 c 1 s 5 s p )dy] dψ H 1 + M sin2 θ [(a 1 c 1 s 5 c p a 2 s 1 c 5 s p )dt + (a 2 s 1 c 5 c p a 1 c 1 s 5 s p )dy] dφ H 1 The angular momenta are given by Ms 1c 1 dt dy Ms 5c 5 (r H 1 H 2 + a 2 2 + Ms2 1 ) cos2 θdψ dφ. 1 and the mass is given by J ψ = πm 4G (5) (a 1c 1 c 5 c p a 2 s 1 s 5 s p ) J φ = πm 4G (5) (a 2c 1 c 5 c p a 1 s 1 s 5 s p ) M ADM = πm 4G (5) (s2 1 + s 2 5 + s 2 p + 3 2 ) Q 1 = gα 3 V n 1 Q 5 = gα n 5 Q p = g2 α 4 V 2 n p