How Fuzzballs solve the black hole information paradox Work done - - PowerPoint PPT Presentation

How Fuzzballs solve the black hole information paradox

Work done with

• B. Chowdhury, S. Giusto, O. Lunin, A. Saxena,
• Y. Srivastava

The entropy problem

Sbek = A 4G

Can we show that there are states of the hole ?

4 eSbek

(Bekenstein, 72)

Black holes behave as if they have an entropy given by their surface area

But statistical mechanics then says that there should be

4 eSbek

states of the hole for the same mass and charge

(Classical relativity finds that black holes have no hair, so there is only one state)

Hawking radiation The information problem

Large distance (much bigger than planck length)

How can the Hawking radiation carry the information of the initial matter ?

If the radiation does not carry the information, then the final state cannot be determined from the initial state, and there is no Schrodinger type evolution equation for the whole system. So we lose quantum theory ...

In string theory, we must make black holes from the objects present in string theory.

A simple example: 2-charge holes

(Susskind, Sen, Vafa ’94-’95) Let us compactify spacetime as

∼ M9,1 → M4,1 × T 4 × S1

momentum mode P winding mode NS1

Winding charge n1 Mass = Charge Momentum charge Mass = Charge

np

A black hole with winding charge only

Smicro = ln[256] ∼ 0

(Does not grow with )

n1 √ n1 √

Horizon is singular Bekenstein entropy vanishes

A = 0

Smicro = Sbek = 0

A black hole with momentum charge only

Smicro = ln[256] ∼ 0

(Does not grow with ) Horizon is singular Bekenstein entropy vanishes

A = 0

Smicro = Sbek = 0

∼ np

• ∼

np

LT = n1L

• e2π

√ 2√n1np

Each quantum of harmonic k carries momentum So we have to count ‘partitions’ of states

= n1np

(Susskind ’93, Sen ’94)

8 bosonic + 8 fermionic degrees of freedom P = 2πnp L = 2π(n1np) LT

Total momentum Computing the entropy Smicro = 2π √ 2√n1np = √

Smicro = 4π√n1np

T 4 × S1 K3 × S1

− ∼

• k nk = n1np
• ×

2πk LT

Now let us make a black hole with these charges ... n1 √

np

• For compactification,

geometry gives a Bekenstein - Wald entropy

K3 × S1

× × Sbek = A 2G = 4π√n1np = Smicro

(Dabholkar ’04)

So the 2-charge hole (‘Sen-Vafa hole’) gives a complete story for black hole entropy in string theory ......

The 2-charge black hole is called the ‘small black hole’ since R corrections to the action affect its horizon area To get a black hole whose area is given by just the usual Einstein action R, we need 3 charges ...

Recall that we had compactified spacetime as

∼ M9,1 → M4,1 × T 4 × S1

+

T 4

S1

+ 5-branes

Make a bound state

• f 3 kinds of charges

+

T 4 S1

n1

n5

n1n5

‘Effective string’ with total winding number

+

Vibrations of effective string are partitioned into harmonics in usual way

Smicro = 2π√n1n5np

Make a black hole with momentum, winding, 5-brane charges ....

Sbek = A 4G = 2π√n1n5np = Smicro

(Strominger and Vafa, 1996)

Thus at least for extremal black holes (mass=charge) we understand something about the entropy from a microscopic viewpoint ...

Near-extremal holes have also been understood ... (Callan and Maldacena 1996)

L

+ n1 √

np

• LT = n1L

Recall the way we made the 2-charge black hole ... This allowed us to count the states of the black hole, so we solve the entropy problem, but what about the information puzzle?

A key point The elementary string (NS1) does not have any LONGITUDINAL vibration modes This is because it is not made up of ‘more elementary particles’ Not a mode for the elementary string Thus only transverse oscillations are permitted This causes the string to spread over a nonzero transverse area Momentum is carried by transverse

• scillations

LT = n1L L L

‘Naive geometry’ An ‘actual geometry’

Making the geometry We know the metric of one straight strand

• f string

We know the metric of a string carrying a wave -- ‘Vachaspati transform’ We get the metric for many strands by superposing harmonic functions from each strand

(Dabholkar, Gauntlett,Harvey, Waldram ’95, Callan,Maldacena,Peet ’95)

In our present case, we have a large number of strands, so we ‘smear over them to make a continuous ‘strip’ (Lunin+SDM ’01)

ds2

string

= H[−dudv + Kdv2 + 2Aidxidv] +

4

• i=1

dxidxi +

4

• a=1

dzadza Buv = 1 2[H − 1], Bvi = HAi e2φ = H

H−1 = 1 + Q1 LT LT dv | x − F(v)|2 K = Q1 LT LT dv( ˙ F(v))2 | x − F(v)|2 Ai = −Q1 LT LT dv ˙ Fi(v) | x − F(v)|2

LT = n1L

• F(x − t)

‘Effective string’ with total winding number

A final step: Use dualities to map to D1

• D5

+

T 4 S1

n′

1 = np

D1 branes D5 branes n′

5 = n1

n′

1n′ 5 = n1np

....

− ∼

• k nk = n1np
• k nk = n′

1n′ 5

ds2 =

• H

1 + K [−(dt − Aidxi)2 + (dy + Bidxi)2] +

• 1 + K

H dxidxi +

• H(1 + K)dzadza

H−1 = 1 + Q LT LT dv | x − F(v)|2 K = Q LT LT dv( ˙ F(v))2 | x − F(v)|2 Ai = − Q LT LT dv ˙ Fi(v) | x − F(v)|2 dB = − ∗ dA

• | −

dB = − ∗4 dA

(Lunin+SDM ’01, also ‘Supergravity supertubes’ (Emparan+Mateos+T

• wnsend ’01)

Geometry for D1

• D5

4G A 4G ∼ √n1n5 ∼ Smicro = 2π √ 2√n1n5

• √

4G ∼ ∼ A 4G ∼

• n1n5 − J ∼ Smicro = 2π

√ 2

• n1n5 − J

Scale of the ‘fuzzball’

(Lunin+SDM ’02)

D1-D5 CFT state

D1-D5 gravity dual

• F(y − ct) =

NS1-P geometry NS1-P state

− (αi1

−1)n1(αi2 −2)n2 . . . |0

S,T dualities

A simple example

NS1- P : one turn

• f a uniform helix

D1-D5: CFT state has all loops ‘singly wound’, and all spins aligned Make metric from profile function F

− AdS3 × S3

going over to flat space at infinity

|ktotal = (J−,total

−(2k−2))n1n5(J−,total −(2k−4))n1n5 . . . (J−,total −2

)n1n5 |1total Piece of global

AdS3 × S3 × T 4

Field theory representation

• f brane state

Geometry created by this state many ‘bits’ The horizon of 2-charge holes needed higher derivative corrections Consider D1-D5-P , which does not need such corrections at leading order

ds2 = −1 h(dt2 − dy2) + Qp hf (dt − dy)2 + hf

• dr2

N

r2

N + a2η + dθ2

• +

h

• r2

N − na2η + (2n + 1)a2ηQ1Q5 cos2 θ

h2f 2

• cos2 θdψ2

+ h

• r2

N + (n + 1)a2η − (2n + 1)a2ηQ1Q5 sin2 θ

h2f 2

• sin2 θdφ2

+ a2η2Qp hf

• cos2 θdψ + sin2 θdφ

2 + 2a√Q1Q5 hf

• n cos2 θdψ − (n + 1) sin2 θdφ
• (dt − dy)

− 2aη√Q1Q5 hf

• cos2 θdψ + sin2 θdφ
• dy +
• H1

H5

4

• i=1

dz2

i

√

f = r2

N − a2η n sin2 θ + a2η (n + 1) cos2 θ

h =

• H1H5, H1 = 1 + Q1

f , H5 = 1 + Q5 f

η ≡ Q1Q5 Q1Q5 + Q1Qp + Q5Qp

Non-extremal hole

(Jejalla, Madden, Ross Titchener ’05)

Geometry has a classical ‘ergoregion instability’

(Cardoso, Dias, Jordan, Hovdebo, Myers, ’06)

The same process that gives Hawking radiation from the black hole now gives us the exact frequency of the instability (Chowdhury + SDM (2007) One (non-typical) microstate

• f a non-extremal hole

∼

1 4 lp

N α lp ∼

1 4 lp

small g large g Conventional picture Fractionation ‘Fuzzball’ Summary of the fuzzball picture

The structure of extremal holes

Infinite throat horizon singularity

‘fuzzball

cap’

F(y − ct)

‘Fuzzball’ Classical states and quantum fuzzballs Fix Total Energy Put energy in a few harmonics, large

• ccupation number

for each harmonic Coherent state

Energy in many harmonics,

• ccupation number
• rder unity in each

Generic quantum state Size depends on mean harmonic order, fluctuations depend on occupation number

Summary

(A) Fuzzball picture offers a resolution of the information paradox ...

lp Nα lp

(B) Lessons must be very general, with applications to other places ...e.g. Cosmology Can we (should we) ask the question: What is the most entropic state in string theory when the energy density is taken to infinity?