Radiation from the non-extremal fuzzball Borun D. Chowdhury The - PowerPoint PPT Presentation

Radiation 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 SUGRA Example Compactify 5 dimensions out of 10 R 1,9 R 1,4 X T 4 X S 1 Take n 1 D1 branes along S 1 Take n 5 D5 branes along T 4 X S 1 Take n p momentum units along S 1 Form their bound state

Extremal and Non-Extremal Black Holes Extremal Black Holes - minimum mass for given charges M D1 ~ R M D5 ~ R V M P ~ 1/R where volume of S 1 ~ R, T 4 ~ V Extra energy excites anti-charges For large R only anti-momentum is excited → D 1 − D 5 − P ¯ D 1 − D 5 − P + Energy − P

AdS/CFT Flat Space Neck Throat (AdS) CFT CFT is unitary Horizon Where are the states in Singularity gravity ?

CFT - Effective String N=n 1 n 5 total winding number Radius=R 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) R

Four bosonic excitations Two complex fermionic excitations ∆ E = 1 R ∆ J = 1 2 1 ∆ E = nR ∆ J = 1 2

Special states have been made E = 0 J = n 1 n 5 2 in the left and right sector

Smooth Solutions (Extremal) Rotation Horizon disappears - single state Black Hole Smooth Geometry Originally done for 2-charge D1-D5 hep-th/0011217 hep-th/0012025 Later on same was done for 3-charge D1-D5-P hep-th/0406103 hep-th/0405017

Non-extremal smooth geometries Rotation Horizon disappears - single state Non-extremal Non-extremal Smooth Geometry Black Hole hep-th/9603100 hep-th/0504181 hep-th/9705192

Features of this geometry No horizon Completely Smooth J ψ = − m n 1 n 5 J φ = 0 Rotation P = 0 m 2 − 1 M = n 1 n 5 2 R Have ergoregion No global timelike Killing vector Negative energy excitations inside ergoregion

CFT dual to non-extremal smooth geometries J = mn 1 n 5 2 E = m 2 − 1 n 1 n 5 4 R fermionic excitations on left and right sectors in left and All component strings of same winding right sectors number

Classical Instability (hep-th/0512277) Klein-Gordon equation in the smooth background 2 Ψ = 0 ansatz Ψ = exp ( − i ω t + i λ y R + im ψ ψ + im φ φ ) χ ( θ ) h ( r ) In the large R limit angular part reduces to laplacian on S 3

The radial equation solution in outer region X ω R = 1 R ( − l − m ψ m − 2) � l +1 ω I = 1 2 π � Q 1 Q 5 ω 2 R ( l !) 2 4 R 2 R solution in inner region

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 R = 1 R ( − l − 2 − m ψ m )

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 ( − i ω t + i λ y R + im ψ ψ + 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: n L left fermions and n R right of two flavors The string has spin half in each direction = n 1 n 5 ( P L + P R ) = n 1 n 5 ( n L ( n L + 1) + n R ( n R + 1)) ER = n 1 n 5 ( P L − P R ) = n 1 n 5 ( n L − n R )( n L + n R + 1) PR n 1 n 5 ( J L , J R ) = (2 n L + 1 , 2 n R + 1) 2

With J ψ = − J L − J R = n L + n R + 1 and m J φ = J L − J R = n n L − n R ∆ M ADM = 1 2 R ( m 2 + 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 n l =4,n R =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 1 P f = R [1 + 2 + 3 + 4] L 10 = R P L = 20 For total left momentum we have R

For each flavor in the right moving sector 1 P f = R [1 + 2] R 3 = R P R = 6 For total left momentum we have R

spin from left moving fermions of one flavor = 1 J f,ferm 2 × 4 = 2 L base spin of the string = 1 J f,base L 2 total spin of left movers J L = 1 2 + 2 + 2 = 9 2

spin from right moving fermions of one flavor = 1 J f,ferm 2 × 2 = 1 R base spin of the string = 1 J f,base R 2 total spin of right movers J R = 1 2 + 1 + 1 = 5 2

P L + P R = 20 + 6 = 26 = E R R P L − P R = 20 − 6 = 14 = P R R − ( J R + J L ) = − 5 + 9 = = − 7 J ψ 2 = − ( J R − J L ) = 2 J φ

grab 3+1 of these X X 2 flavors 3+1 4( P L + P R ) = 104 = E R 4( P L − P R ) = 56 = P R = − 4( J R + J L ) = − 28 J ψ = − 4( J R − J L ) = 8 J φ

and make a twisted string of length (3+1) R X 2 flavors

X 2 flavors 1 momentum quanta have gone down to 4 R For each flavor in the left moving sector L = 1 4 R [1 + 2 + ... + 16] = 34 P f R P L = 68 total left momentum R

X 2 flavors 1 momentum quanta have gone down to 4 R For each flavor in the left moving sector R = 1 4 R [1 + 2 + ... + 8] = 9 P f R P R = 18 total right momentum R

The spin from left moving fermions of one flavor = 1 X 2 flavors J f,ferm 2 × 16 = 8 L base spin of the string = 1 J f,base R 2 total spin of left movers J L = 1 2 + 2 × 8 = 33 2

The spin from left moving fermions of one flavor X 2 flavors = 1 J f,ferm 2 × 8 = 4 R base spin of the string = 1 J f,base R 2 total spin of left movers J R = 1 2 + 2 × 4 = 17 2

after the twisting we have X 2 flavors P L + P R = 68 + 18 = 86 = E R R P L − P R = 68 − 18 = 50 = P R R − ( J R + J L ) = − 17 + 33 = = − 25 J ψ 2 = − ( J R − J L ) = 8 J φ

Account E=2/R for bosons 16 104 86 R + 2 R = 88 ω = E = E = R R R 6 50 56 P = λ = P = R R R J ψ = − 25 J ψ = − 28 m ψ = − 3 J φ = 8 J φ = 8 m φ = 0

16 ω = R 6 = n L + n R + 1 = 7 λ = m R = n L − n R = 2 n m ψ = − 3 m φ = 0 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 � l +1 � Q 1 Q 5 Γ l = 4 π ω 2( l +1) ( l !) 2 4 R 2

But this is the spontaneous part of decay dN dt = Γ Stimulated emission would give dN dt = Γ (1 + N ) X Is it LASER ? No No mirror

Recall CFT was symmetrized 1+1 (4,4) CFT with target space (T 4 ) N /S N transition between two BECs √ √ √ k + 1 | n + 1 , k + 1 � + α ∗ √ √ N − n + 1 √ n H int | n, k � = α N − n n + 1 k − 1 | n − 1 , k − 1 � (6.160 N-n excited k scalars n de-excited X Scalars escaping: n can only go to n+1 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 dt = γ ( N + 1) - 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 l t Model - wave grows as √ N agrees with � l +1 � Q 1 Q 5 Im ( ω ) = 1 2 Γ l = 2 π grav. decay ω 2( l +1) ( l !) 2 4 R 2 width

Conclusion Emission from non-extremal fuzzball found On the CFT side process same as Hawking Radiation 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

f M d s 2 (d t 2 − d y 2 ) + ( s p d y − c p d t ) 2 � ˜ � ˜ = − H 1 ˜ H 1 ˜ H 5 H 5 r 2 d r 2 � � � H 1 ˜ ˜ 2 ) − Mr 2 + d θ 2 + H 5 ( r 2 + a 2 1 )( r 2 + a 2 �� H 5 − f ) cos 2 θ � 1 )( ˜ H 1 + ˜ cos 2 θ d ψ 2 H 1 ˜ ˜ H 5 − ( a 2 2 − a 2 + � ˜ H 1 ˜ H 5 �� H 5 − f ) sin 2 θ � 1 )( ˜ H 1 + ˜ sin 2 θ d φ 2 H 1 ˜ ˜ H 5 + ( a 2 2 − a 2 � ˜ + H 1 ˜ H 5 M ( a 1 cos 2 θ d ψ + a 2 sin 2 θ d φ ) 2 � ˜ + H 1 ˜ H 5 +2 M cos 2 θ � ˜ [( a 1 c 1 c 5 c p − a 2 s 1 s 5 s p )d t + ( a 2 s 1 s 5 c p − a 1 c 1 c 5 s p )d y ]d ψ H 1 ˜ H 5 +2 M sin 2 θ � ˜ [( a 2 c 1 c 5 c p − a 1 s 1 s 5 s p )d t + ( a 1 s 1 s 5 c p − a 2 c 1 c 5 s p )d y ]d φ H 1 ˜ H 5 � 4 ˜ H 1 � d z 2 + ˜ i H 5 i =1 f = r 2 + a 2 H i = f + M sinh 2 δ i , 1 sin 2 θ + a 2 2 cos 2 θ , ˜