The black hole information paradox · Black hole entropy Gas with entropy S Suppose we throw a box...

The black hole information paradox

Samir D. Mathur

Gravity

Thermodynamics Quantum Theory

BlackHoles

Black hole entropy

Gas withentropy S

Suppose we throw a box of gas into a black hole

The gas disappears, and we seem to have reduced theentropy of the Universe

Have we violated the second law of thermodynamics ?

The twist operator

October 24, 2009

E = M TdS = dE S =A

4G=

4π(2GM)2

4G= 4πGM2 (1)

1T

=dS

dM= 8πGM (2)

λ ∼ 1T∼ GM (3)

∆t ∼ GM ω ∼ T ∼ 1GM

(4)

tevap ∼M

ω∆t ∼ G2M3 (5)

V = −Gm1m2

r(6)

e+ e− (7)

M m E ∼ mc2 − GMm

rE ∼ 0 r =

GM

c2→ 2GM

c2(8)

r ∼ 3 Km r ∼ 108 Km 108 (9)

F = maGMm

r2= ma → a =

GM

r2(10)

dSmatter

dt+

dShole

dt≥ 0 (11)

1

So we save second law of thermodynamics ...

Suppose we assume that the hole has an entropy

Then

λ ∼ R (15)

∆t ∼ R

c(16)

tevap ∼G2M3

�c2(17)

T =1

8πGM=

dE

dS(18)

tevap ∼ 1063 years (19)

eS (20)

S = ln(# states) (21)

101077(22)

S =c3

�A

4G→ A

4G(c = � = 1) (23)

S = ln 1 = 0 (24)

Ψ ρ (25)

A ∼ e−Sgrav , Sgrav ∼1

16πG

�Rd4x ∼ GM2 (26)

# states ∼ eS , S ∼ GM2 (27)

A = 0 Sbek =A

4G= 0 Smicro = ln(1) = 0 (28)

�F (y − ct) (29)

n1 np e4π√

n1np (30)

Smicro = 4π√n1np Sbek−wald = 4π√

n1np Smicro = Sbek (31)

2

(Bekenstein ’72, Hawking ’74)

The entropy puzzle

The black hole has entropy

Statistical Mechanics says that

λ ∼ R (15)

∆t ∼ R

c(16)

tevap ∼G2M3

�c2(17)

T =1

8πGM=

dE

dS(18)


eS (20)


2

Thus the black hole should have states

λ ∼ R (15)

∆t ∼ R

c(16)

tevap ∼G2M3

�c2(17)

T =1

8πGM=

dE

dS(18)


eS (20)


2

Solar mass : states

λ ∼ R (15)

∆t ∼ R

c(16)

tevap ∼G2M3

�c2(17)

T =1

8πGM=

dE

dS(18)


eS (20)


101077(22)

2

Where are these states ?

λ ∼ R (15)

∆t ∼ R

c(16)

tevap ∼G2M3

�c2(17)

T =1

8πGM=

dE

dS(18)


eS (20)


101077(22)

S =c3

�A

4G→ A

4G(c = � = 1) (23)

S = ln 1 = 0 (24)

Ψ ρ (25)


16πG

�Rd4x ∼ GM2 (26)

# states ∼ eS , S ∼ GM2 (27)

A = 0 Sbek =A

4G= 0 Smicro = ln(1) = 0 (28)

�F (y − ct) (29)

n1 np e4π√

n1np (30)



2

Fix M, look for small distortions of the hole ...

Find NO allowed distortions:

Black hole geometry completely fixed by its conserved quantum numbers: mass, charge, angular momentum

This would mean

λ ∼ R (15)

∆t ∼ R

c(16)

tevap ∼G2M3

�c2(17)

T =1

8πGM=

dE

dS(18)


eS (20)


101077(22)

S =c3

�A

G→ A

4G(c = � = 1) (23)

S = ln 1 = 0 (24)

2

Where are the states of the black hole ?

‘Black holes have no hair’ (John Wheeler)

A more serious (but related) problem: The information paradox

+_

e+e-

Schwinger pair production process(interesting, but not a problem !)

State of created quanta is entangled

↑↓ − ↓↑

Sent = ln 2

Entanglement entropy

After steps

Sent = N ln 2

N

The information problem

⊗ |0�2|0�2� + |1�2|1�2�

⊗ |0�n|0�n� + |1�n|1�n�

. . .

⊗ |0�1|0�1� + |1�1|1�1�

ΨM

Schwinger process in the gravitational field

Possibilities

Sent = N ln 2

To have this entanglement, the remnant should have at least internal states 2N

But how can we have an unbounded degeneracy for objects with a given mass ?

Planck mass remnant

Sent = N ln 2

Complete evaporation

The radiated quanta are in an entangled state, but there is nothing that they are entangled with !

They cannot be described by any wavefunction, but only by a density matrix

failure of quantum mechanics

The two problems are related ...

A normal body emits radiation from its surface, so that the radiation depends on the microstate. In the black hole the radiation is pulled from the vacuum

A normal body has visiblemicrostates, so the entropycan be found by counting them

What do string theorists say ?

The entropy problem :

Does represent a count of states ? Sbek =A

4G

(A) There are 10 - 4 = 6 compact directions in string theory

(B) The coupling constant is a free modulus that we can varyg

(C) If we look at states with mass = charge, then their number does not change with (supersymmetric protection of BPS states)

g

g → 0 g large

All charges are obtained by wrapping strings, branes etc along compact directions

Smicro = 2π√n1n2n3

Entropy is given through number of intersection points

Expect that gravitational field will extend around branes to make a black hole

Find

Sbek ≡ A

4G= 2π

√n1n2n3

(Susskind 93, Sen 95, Strominger-Vafa 96 ...)

Thus is a count of states Sbek

The information problem : What happens to the entanglement ?

⊗ |0�2|0�2� + |1�2|1�2�

⊗ |0�n|0�n� + |1�n|1�n�

. . .

⊗ |0�1|0�1� + |1�1|1�1�

ΨM

String theorists: String theory is unitary, so thereshould be no problem, really

GR folks: But what is the resolution to the problem? Hawking has an explicit computation, and you have not shown us what is wrong with it !!

⊗ |0�2|0�2� + |1�2|1�2�

⊗ |0�n|0�n� + |1�n|1�n�

. . .

⊗ |0�1|0�1� + |1�1|1�1�

ΨM

+ δψ1

+ δψ2

+ δψn

There will of course be small corrections to the leading order Hawking computation

Small correction to a large number of pairs will (hopefully) disentangle the inner and outer quanta

Stringtheorists

You must think we are such fools.

We know there will always be small corrections from quantum gravity.

If that could remove the entanglement, why wouldwe be worrying about the information paradox for35 years ?

⊗ |0�2|0�2� + |1�2|1�2�

⊗ |0�n|0�n� + |1�n|1�n�

. . .

⊗ |0�1|0�1� + |1�1|1�1�

ΨM

+ δψ1

+ δψ2

+ δψn

GR folks

Kip Thorne

John Preskill

Stephen Hawking

In 2005, Stephen Hawking surrendered his bet to John Preskill,based on such an argument of ‘small corrections’ ...

(Subleading saddle points in a Euclidean path integral give exponentially small corrections to the leading order evaporation process)

But Kip Thorne did not agree to surrender the bet ...

So who is right ?

+_

e+e-

entangled pairs

1√2(↑↓ − ↓↑)⊗ 1√

2(↑↓ − ↓↑) . . .

+ corrections

Sent ≈ N ln 2

Schwinger process

For the surface is spacelike

From (??) we are given that

||ψ2||2 = �ψ2|ψ2� ≡ �21 < �2 (65)

|�ψ1|ψ2�| ≡ �2 < � (66)

S(p) = (�21 − �22) lne

(�21 − �22)+O(�3) < � (67)

SN+1 > SN + ln 2− 2� (68)

ds2 = − (1− 2M

r)dt2 +

dr2

(1− 2Mr )

+ r2(dθ2 + sin2 θdφ2) (69)

r > 2M r < 2M t = constant r = constant (70)

4

The black hole is described by the Schwarzschild metric

The black hole

For the surface is spacelike


||ψ2||2 = �ψ2|ψ2� ≡ �21 < �2 (65)

|�ψ1|ψ2�| ≡ �2 < � (66)

S(p) = (�21 − �22) lne

(�21 − �22)+O(�3) < � (67)

SN+1 > SN + ln 2− 2� (68)

ds2 = − (1− 2M

r)dt2 +

dr2

(1− 2Mr )

+ r2(dθ2 + sin2 θdφ2) (69)


4

Crucial point about the black hole:


||ψ2||2 = �ψ2|ψ2� ≡ �21 < �2 (65)

|�ψ1|ψ2�| ≡ �2 < � (66)

S(p) = (�21 − �22) lne

(�21 − �22)+O(�3) < � (67)

SN+1 > SN + ln 2− 2� (68)

ds2 = − (1− 2M

r)dt2 +

dr2

(1− 2Mr )

+ r2(dθ2 + sin2 θdφ2) (69)


4


||ψ2||2 = �ψ2|ψ2� ≡ �21 < �2 (65)

|�ψ1|ψ2�| ≡ �2 < � (66)

S(p) = (�21 − �22) lne

(�21 − �22)+O(�3) < � (67)

SN+1 > SN + ln 2− 2� (68)

ds2 = − (1− 2M

r)dt2 +

dr2

(1− 2Mr )

+ r2(dθ2 + sin2 θdφ2) (69)


4


||ψ2||2 = �ψ2|ψ2� ≡ �21 < �2 (65)

|�ψ1|ψ2�| ≡ �2 < � (66)

S(p) = (�21 − �22) lne

(�21 − �22)+O(�3) < � (67)

SN+1 > SN + ln 2− 2� (68)

ds2 = − (1− 2M

r)dt2 +

dr2

(1− 2Mr )

+ r2(dθ2 + sin2 θdφ2) (69)


4

r=0 horizon

t=constant

r=constant

We have to draw spacelike slices to foliate spacetime

(no time-independent slicing possible)

Entangled pairs

The Hawking process

Stretching of spacetime causes field modes to get excited

r=0 horizon

correlated pairs

Older quanta move apart

initial matter

Hawking state

E = hν =hc

λ(38)

|�ψ�|H|ψ� − �ψ�|Hs.c.|ψ�| ≤ � (39)

lp � λ � Rs (40)

2� (41)

|ξ1� =1√2

�|0�|0�+ |1�|1�

�(42)

|ξ2� = √2

�|0�|0� − |1�|1�

�(43)

SN+1 = SN + ln 2 (44)

S = Ntotal ln 2 (45)

|ψ� → |ψ1�|ξ1�+ |ψ2�|ξ2� (46)

||ψ2|| < � (47)

SN+1 < SN (48)

SN+1 > SN + ln 2− 2� (49)

S(A) = −Tr[ρA ln ρA] (50)

bN+1 cN+1 {b} {c} p = {cN+1 bN+1} (51)

S({b}+ p) > SN − � (52)

S(p) < � (53)

S(cN+1) > ln 2− � (54)

S({b}+ bN+1) + S(p) > S({b}) + S(cN+1) (55)

S(A+B) + S(B + C) ≥ S(A) + S(C) (56)

S({b}+ bN+1) > SN + ln 2− 2� (57)

S(A+B) ≥ |S(A)− S(B)| (58)

lp Nαlp (59)

SN = S({b}) (60)

A = {b} B = p = {bN+1 + cN+1} (61)

A = {b} B = bN+1 C = cN+1 (62)

S({b}+ bN+1) + S(p) ≥ S({b}) + S(cN+1) (63)

3

10 light years77

(We will use a discretized picture for simplicity; for full state see e.g. Giddings-Nelson)

Hawking’s argument

E = hν =hc

λ(38)

�ψ�|H|ψ� ≈ �ψ�|Hs.c.|ψ�+O(�) (39)

lp � λ � Rs (40)

2� (41)

S1 = |0�|0�+ |1�|1� (42)

SN+1 = SN + ln 2 (43)

3

E = hν =hc

λ(38)

�ψ�|H|ψ� ≈ �ψ�|Hs.c.|ψ�+O(�) (39)

lp � λ � Rs (40)

2� (41)

|ξ1� = |0�|0�+ |1�|1� (42)

SN+1 = SN + ln 2 (43)

3

: Entanglement entropy after N pairs have been created

The radiation state (green quanta) are highly entangled with the infalling members of the Hawking pairs (red quanta)

E = hν =hc

λ(38)

�ψ�|H|ψ� ≈ �ψ�|Hs.c.|ψ�+O(�) (39)

lp � λ � Rs (40)

2� (41)

|ξ1� = |0�|0�+ |1�|1� (42)

SN+1 = SN + ln 2 (43)


3

E = hν =hc

λ(38)

|�ψ�|H|ψ� − �ψ�|Hs.c.|ψ�| ≤ � (39)

lp � λ � Rs (40)

2� (41)

|ξ1� =1√2

�|0�|0�+ |1�|1�

�(42)

|ξ2� = √2

�|0�|0� − |1�|1�

�(43)

SN+1 = SN + ln 2 (44)


|ψ� → |ψ1�|ξ1�+ |ψ2�|ξ2� (46)

||ψ2|| < � (47)

SN+1 < SN (48)

SN+1 > SN + ln 2− 2� (49)


bN+1 cN+1 {b} {c} p = {cN+1 bN+1} (51)

S({b}+ p) > SN − � (52)

S(p) < � (53)

S(cN+1) > ln 2− � (54)

S({b}+ bN+1) + S(p) > S({b}) + S(cN+1) (55)

S(A+B) + S(B + C) ≥ S(A) + S(C) (56)

S({b}+ bN+1) > SN + ln 2− 2� (57)

S(A+B) ≥ |S(A)− S(B)| (58)

lp Nαlp (59)

SN = S({b}) (60)

A = {b} B = p = {bN+1 + cN+1} (61)

A = {b} B = bN+1 C = cN+1 (62)

S({b}+ bN+1) + S(p) ≥ S({b}) + S(cN+1) (63)

3

E = hν =hc

λ(38)

|�ψ�|H|ψ� − �ψ�|Hs.c.|ψ�| ≤ � (39)

lp � λ � Rs (40)

2� (41)

|ξ1� =1√2

�|0�|0�+ |1�|1�

�(42)

|ξ2� = √2

�|0�|0� − |1�|1�

�(43)

SN+1 = SN + ln 2 (44)


|ψ� → |ψ1�|ξ1�+ |ψ2�|ξ2� (46)

||ψ2|| < � (47)

SN+1 < SN (48)

SN+1 > SN + ln 2− 2� (49)


bN+1 cN+1 {b} {c} p = {cN+1 bN+1} (51)

S({b}+ p) > SN − � (52)

S(p) < � (53)

S(cN+1) > ln 2− � (54)

S({b}+ bN+1) + S(p) > S({b}) + S(cN+1) (55)

S(A+B) + S(B + C) ≥ S(A) + S(C) (56)

S({b}+ bN+1) > SN + ln 2− 2� (57)

S(A+B) ≥ |S(A)− S(B)| (58)

lp Nαlp (59)

SN = S({b}) (60)

A = {b} B = p = {bN+1 + cN+1} (61)

A = {b} B = bN+1 C = cN+1 (62)

S({b}+ bN+1) + S(p) ≥ S({b}) + S(cN+1) (63)

3

E = hν =hc

λ(38)

|�ψ�|H|ψ� − �ψ�|Hs.c.|ψ�| ≤ � (39)

lp � λ � Rs (40)

2� (41)

|ξ1� =1√2

�|0�|0�+ |1�|1�

�(42)

|ξ2� = √2

�|0�|0� − |1�|1�

�(43)

SN+1 = SN + ln 2 (44)


|ψ� → |ψ1�|ξ1�+ |ψ2�|ξ2� (46)

||ψ2|| < � (47)

SN+1 < SN (48)

SN+1 > SN + ln 2− 2� (49)


bN+1 cN+1 {b} {c} p = {cN+1 bN+1} (51)

S({b}+ p) > SN − � (52)

S(p) < � (53)

S(cN+1) > ln 2− � (54)

S({b}+ bN+1) + S(p) > S({b}) + S(cN+1) (55)

S(A+B) + S(B + C) ≥ S(A) + S(C) (56)

S({b}+ bN+1) > SN + ln 2− 2� (57)

S(A+B) ≥ |S(A)− S(B)| (58)

lp Nαlp (59)

SN = S({b}) (60)

A = {b} B = p = {bN+1 + cN+1} (61)

A = {b} B = bN+1 C = cN+1 (62)

S({b}+ bN+1) + S(p) ≥ S({b}) + S(cN+1) (63)

3

Entangled state

If the black hole evaporates away,we are left in a configuration which cannot be described by a pure state

(Radiation quanta are entangled, but there is nothing that they are entangled with)

We can get a remnantwith which the radiationis highly entangled

So what can small corrections do ?

Rules of the game:

(a) The region where pairs are being produced has ‘normal evolution’; i.e. vacuum modes evolve as expected upto corrections of order

(b) The stuff inside the hole can be reshuffled in any way we want, but the quanta that have left cannot be altered

�

� � 1Making rigorous the statement that ‘Nothing happens at the horizon’

1√2

�|0�1|0�1� + |1�1|1�1�

�

+ �1√2

�|0�1|0�1� − |1�1|1�1�

�

Theorem: Small corrections to Hawking’s leading order computation do NOT remove the entanglement

(SDM 09)δSent

Sent< 2�

Bound does not depend on the number of pairs N

A

B C

D

E = hν =hc

λ(38)

�ψ�|H|ψ� ≈ �ψ�|Hs.c.|ψ�+O(�) (39)

lp � λ � Rs (40)

2� (41)

|ξ1� = |0�|0�+ |1�|1� (42)

|ξ2� = |0�|0� − |1�|1� (43)

SN+1 = SN + ln 2 (44)


|ψ� → |ψ1�|ξ1�+ |ψ2�|ξ2� (46)

||ψ2|| < � (47)

SN+1 < SN (48)

SN+1 > SN + ln 2− 2� (49)


bN+1 cN+1 {b} {c} p = {cN+1 bN+1} (51)

S({b}+ p) > SN − � (52)

S(p) < � (53)

S(cN+1) > ln 2− � (54)

S({b}+ bN+1) + S(p) > S({b}) + S(cN+1) (55)

S(A+B) + S(B + C) ≥ S(A) + S(C) (56)

S({b}+ bN+1) > S) + ln 2− 2� (57)

S(A+B) ≥ |S(A)− S(B)| (58)

3

E = hν =hc

λ(38)

�ψ�|H|ψ� ≈ �ψ�|Hs.c.|ψ�+O(�) (39)

lp � λ � Rs (40)

2� (41)

|ξ1� = |0�|0�+ |1�|1� (42)

|ξ2� = |0�|0� − |1�|1� (43)

SN+1 = SN + ln 2 (44)


|ψ� → |ψ1�|ξ1�+ |ψ2�|ξ2� (46)

||ψ2|| < � (47)

SN+1 < SN (48)

SN+1 > SN + ln 2− 2� (49)


3

etc.

Basic tool : Strong Subadditivity (Lieb + Ruskai ’73)

_In Schwinger processor in black hole: Entanglement rises witheach emission

Cannot resolve theproblem as long ascorrections to low energy dynamics are small at the horizon

Conclusion:

An order unity correction at the horizon means that we need ‘hair’ ..

Thus we have a conflict between two ‘theorems’:

(a) The Hawking argument, supplements by the inequality that shows its robustness to small corrections

(b) The ‘no hair theorem’, which encodes our failure to find any alternative to the black hole geometry

So, what is the resolution?

(Avery, Balasubramanian, Bena, Chowdhury, de Boer, Gimon, Giusto, Keski-Vakkuri, Levi, Maldacena, Maoz, Park, Peet, Potvin, Ross, Ruef, Saxena, Skenderis, Srivastava, Taylor, Turton, Warner ...)

The traditional expectation ...

|n�total = (J−,total−(2n−2))

n1n5(J−,total−(2n−4))

n1n5 . . . (J−,total−2 )n1n5 |1�total (5)

A

4G= S = 2π

√n1n2n3

∆E =1

nR+

1

nR=

2

nR

∆E =2

nR

S = ln(1) = 0 (6)

S = 2√

2π√

n1n2 (7)

S = 2π√

n1n2n3 (8)

S = 2π√

n1n2n3n4 (9)

n1 ∼ n5 ∼ n

∼ n14 lp

∼ n12 lp

∼ n lp

M9,1 → M4,1 ×K3× S1

A

4G∼

�n1n5 − J ∼ S

A

4G∼√

n1n5 ∼ S

e2π√

2√

n1np

1 +Q1

r2

1 +Qp

r2

e2π√

2√

n1n5

w = e−i(t+y)−ikz w(r, θ, φ) (10)

B(2)MN = e−i(t+y)−ikz B(2)

MN(r, θ, φ) , (11)

2

= 2π√

n1n5(2

�E

2mp) (54)

S = 2π√

n5(√

n1 +√

n1)(√

np +�

np) (55)

= 2π√

n5(E

√m1mp

) (56)

S = 2π√

n1n5npnkk (57)

S = 2π√

n1n5nkk(√

np +�

np) (58)

= 2π√

n1n5(E

√mpmkk

) (59)

S = 2π√

n1n5(√

np +�

np)(√

nkk +√

nkk) (60)

∼ lp (61)

∼ n16 lp (62)

M9,1 →M4,1 × T 4 × S1 (63)

E/(2mkk) = 0.5 (64)

E/(2mkk) = 1.2 (65)

Lz ∼ [g2α�4√n1n5np

V R]13 ∼ Rs (66)

∆S (67)

eS (68)

eS+∆S (69)

S = 2π�

n1n5np(1− f) + 2π�

n1n5npf(√

nk +√

nk) (70)

nk = nk =1

2

∆E

mk=

1

2Dmk(71)

D ∼ [

√n1n5npg2α�4

V Ry]13 ∼ RS (72)

∆S = S − 2π√

n1n5np = 1 (73)

S =A

4G(74)

mk ∼G5

G24

∼ D2

G5(75)

D ∼ G135 (n1n5np)

16 ∼ RS (76)

Nα lp (77)

5

weak coupling

strongcoupling

But it seems in string theory the opposite happens ...




A

4G= S = 2π

√n1n2n3

∆E =1

nR+

1

nR=

2

nR

∆E =2

nR

S = ln(1) = 0 (6)

S = 2√

2π√

n1n2 (7)

S = 2π√

n1n2n3 (8)

S = 2π√

n1n2n3n4 (9)

n1 ∼ n5 ∼ n

∼ n14 lp

∼ n12 lp

∼ n lp

M9,1 → M4,1 ×K3× S1

A

4G∼

�n1n5 − J ∼ S

A

4G∼√

n1n5 ∼ S

e2π√

2√

n1np

1 +Q1

r2

1 +Qp

r2

e2π√

2√

n1n5

w = e−i(t+y)−ikz w(r, θ, φ) (10)


MN(r, θ, φ) , (11)

2

weak coupling

strongcoupling

Size of bound states grows with coupling, number of branes ...

(SDM 97)

Start with the simplest microstates:

weak coupling

strongcoupling

Recall that weak coupling states are described by intersecting branes wrapped on compact directions

No horizon, no singularity ... so we do not get the traditional hole ...

Many such constructions have been done ...

General lesson: Eigenstates in string theory do not have a traditional horizon

That is, there is no smooth region straddling the horizon where low energy pair modes evolve as expected on gently curved spacetime

Geometry can be different from the traditional hole, or more generally, there is no geometry, just a quantum ‘fuzz’ fuzzballs

Thus we have finally found the ‘hair’ for black holes ...

Nature of the hair:

Compact directions makelocally nontrivial fibrations over the noncompact directions

small compact direction circle Angular sphere of

noncompact directions

eSbek states upon quantization

Thus the hair are fundamentally a nonperturbative construct involving the extra dimensions ...

and for k ≥ 0

|− k�total = (J+,total−(2k−2))

n1n5(J+,total−(2k−4))

n1n5 . . . (J+,total−2 )n1n5(J+,total

0 )n1n5 |1�total (2.15)

Similarly, for k > 1

|k�total = (J−,total−(2k−2))

n1n5(J−,total−(2k−4))

n1n5 . . . (J−,total−2 )n1n5 |1�total (2.16)

3 Constructing the gravity duals

In [?] the 2-charge D1-D5 solutions were found by dualizing to the FP system, which hasa fundamental string (F) wrapped on S1 carrying momentum (P) along S1. Metrics forthe vibrating string were constructed, and dualized back to get D1-D5 geometries. Thegeneral geometry was thus parametrized by the vibration profile �F (v) of the F string.But a 1-parameter subfamily of these D1-D5 geometries had been found earlier [?, ?], bylooking at extremal limits of the general axially symmetric D1-D5 geometry found in [?].

We do not have an analogue of the procedure of [?] for 3-charge systems. We willfollow instead the analogue of [?, ?] and take an extremal limit of the general 3-chargesolution to obtain solutions with D1, D5 and P charges. Taking the limit needs somecare, and it will be important to know in advance the properties of the CFT states forwhich we will be finding the duals. The procedure will give us the duals of the states|n�total which were discussed in the last section. We will find that the dual geometriesare completely smooth, with no horizon and no singularity.

3.1 Spectral flow in the gravity description

In [?, ?] the following 2-charge D1-D5 solution was found (setting Q1 = Q5 = Q forsimplicity)

ds2 = −1

h(dt2 − dy2) + hf

�dθ2 +

dr2

r2 + a2

�− 2aQ

hf(cos2 θdydψ + sin2 θdtdφ)

+ h

��r2 +

a2Q2 cos2 θ

h2f 2

�cos2 θdψ2 +

�r2 + a2 − a2Q2 sin2 θ

h2f 2

�sin2 θdφ2

�+ dzidzi

(3.1)

where

a =Q

R, f = r2 + a2 cos2 θ, h = 1 +

Q

f(3.2)

Let R >>√

Q. In the region r <<√

Q the geometry (??) becomes

ds2 = −(r2 + a2 cos2 θ)

Q(dt2 − dy2) + Q

�dθ2 +

dr2

r2 + a2

�

− 2a(cos2 θdydψ + sin2 θdtdφ) + Q(cos2 θdψ2 + sin2 θdφ2) (3.3)

5

√α�

α� <g2α�3n1np

V4LM

∼ ls ∼ lp ∼ (n1n5)αlp

L, V4, g

AdS3 × S3 × T 4

4

Geometry for simplestate (winding =1)

Generic D1D5P CFT stateSimple states: all components the same,excitations fermionic, spin aligned

The numerator is r2dr

2 = r2Ndr

2N , and we get a cancellation of the factors r

2N . We will

see below that in the extremal metric the point rN = 0 acts like an origin of polarcoordinates, so the choice (??) is the correct one to define a coordinate rN with range(0,∞).

We also find that other terms in the metric and gauge field are finite in the extremallimit; this can be verified using (??),(??). We get the extremal solution (in the stringframe)

ds2 = −1

h(dt

2 − dy2) +

Qp

hf(dt− dy)2 + hf

�dr

2N

r2N + a2η

+ dθ2

�

+ h

�r2N − na

2η +(2n + 1)a2ηQ1Q5 cos2 θ

h2f 2

�cos2 θdψ2

+ h

�r2N + (n + 1)a2η − (2n + 1)a2ηQ1Q5 sin2 θ

h2f 2

�sin2 θdφ2

+a

2η2Qp

hf

�cos2 θdψ + sin2 θdφ

�2

+2a√

Q1Q5

hf

�n cos2 θdψ − (n + 1) sin2 θdφ

�(dt− dy)

− 2aη√

Q1Q5

hf


�dy +

�H1

H5

4�

i=1

dz2i (4.21)

C2 =a√

Q1Q5 cos2 θ

H1f(−(n + 1)dt + ndy) ∧ dψ

+a√

Q1Q5 sin2 θ

H1f(ndt− (n + 1)dy) ∧ dφ

+aηQp√

Q1Q5H1f(Q1dt + Q5dy) ∧


�

− Q1

H1fdt ∧ dy − Q5 cos2 θ

H1f

�r2N + (n + 1)a2η + Q1

�dψ ∧ dφ (4.22)

e2Φ =

H1

H5(4.23)

f = r2N − a

2η n sin2 θ + a2η (n + 1) cos2 θ

h =�

H1H5, H1 = 1 +Q1

f, H5 = 1 +

Q5

f(4.24)

11

The numerator is r2dr

2 = r2Ndr

2N , and we get a cancellation of the factors r

2N . We will

see below that in the extremal metric the point rN = 0 acts like an origin of polarcoordinates, so the choice (??) is the correct one to define a coordinate rN with range(0,∞).

We also find that other terms in the metric and gauge field are finite in the extremallimit; this can be verified using (??),(??). We get the extremal solution (in the stringframe)

ds2 = −1

h(dt

2 − dy2) +

Qp

hf(dt− dy)2 + hf

�dr

2N

r2N + a2η

+ dθ2

�

+ h

�r2N − na

2η +(2n + 1)a2ηQ1Q5 cos2 θ

h2f 2

�cos2 θdψ2

+ h

�r2N + (n + 1)a2η − (2n + 1)a2ηQ1Q5 sin2 θ

h2f 2

�sin2 θdφ2

+a

2η2Qp

hf


�2

+2a√

Q1Q5

hf

�n cos2 θdψ − (n + 1) sin2 θdφ

�(dt− dy)

− 2aη√

Q1Q5

hf


�dy +

�H1

H5

4�

i=1

dz2i (4.21)

C2 =a√

Q1Q5 cos2 θ

H1f(−(n + 1)dt + ndy) ∧ dψ

+a√

Q1Q5 sin2 θ

H1f(ndt− (n + 1)dy) ∧ dφ

+aηQp√

Q1Q5H1f(Q1dt + Q5dy) ∧


�

− Q1

H1fdt ∧ dy − Q5 cos2 θ

H1f

�r2N + (n + 1)a2η + Q1

�dψ ∧ dφ (4.22)

e2Φ =

H1

H5(4.23)

f = r2N − a

2η n sin2 θ + a2η (n + 1) cos2 θ

h =�

H1H5, H1 = 1 +Q1

f, H5 = 1 +

Q5

f(4.24)

11

4.2 Taking the extremal limit

To get the extremal limit we must take

M → 0, δi →∞ (i = 1, 5, p) (4.11)

keeping the Qi fixed. This gives

cosh2 δi =Qi

M+

1

2+ O(M)

sinh2 δi =Qi

M− 1

2+ O(M) (4.12)

We must also take suitable limits of a1, a2 so that the angular momenta are held fixed.It is useful to invert (??):

a1 = −√

Q1Q5

M

γ1 cosh δ1 cosh δ5 cosh δp + γ2 sinh δ1 sinh δ5 sinh δp

cosh2 δ1 cosh2 δ5 cosh2 δp − sinh2 δ1 sinh2 δ5 sinh2 δp

a2 = −√

Q1Q5

M

γ2 cosh δ1 cosh δ5 cosh δp + γ1 sinh δ1 sinh δ5 sinh δp

cosh2 δ1 cosh2 δ5 cosh2 δp − sinh2 δ1 sinh2 δ5 sinh2 δp

(4.13)

Using (??) we find

a1 = −(γ1 + γ2) η

�Qp

M− γ1 − γ2

4

�M

Qp+ O(M3/2)

= −a η

�Qp

M+ a

2n + 1

4

�M

Qp+ O(M3/2)

a2 = −(γ1 + γ2) η

�Qp

M+

γ1 − γ2

4

�M

Qp+ O(M3/2)

= −a η

�Qp

M− a

2n + 1

4

�M

Qp+ O(M3/2) (4.14)

where we have defined the dimensionless combination

η ≡ Q1Q5

Q1Q5 + Q1Qp + Q5Qp(4.15)

and in the second equalities we have used the specific values for γ1 and γ2 given in (??).We thus see that for generic values of γ1, γ2 and Qp the parameters a1 and a2 diverge

when M → 0. There are two exceptions:(a) Qp = 0, which is the case considered in [?, ?]; in this case a1 and a2 go to finite valueswhen M → 0.

9

(Giusto SDM Saxena 04)

How does Hawking radiation arise ?

The Black Hole

Spinning starcauses ‘frame-dragging’

Far away light conesare close to normal

Close by, light cones tilt so much that every object MUST rotate ERGOREGION

A curiousity: Ergoregions

Hawking radiation

Traditional picture

Ergoregion

Actually radiation comes out just like from any other object, not from the vacuum

Why does semiclassical intuition fail ?

??

Shell collapses to make a black hole ...

Consider the amplitude for the shell to tunnel to a fuzzball state

Amplitude to tunnel is very small

But the number of states that one can tunnel to is very large !

Toy model: Small amplitude to tunnel to a neighboring well, but there are a correspondingly large number of adjacent wells

In a time of order unity, the wavefunction in the central well becomes a linear combination of states in all wells (SDM 07)

λ ∼ R (15)

∆t ∼ R

c(16)

tevap ∼G2M3

�c2(17)

T =1

8πGM=

dE

dS(18)


eS (20)


101077(22)

S =c3

�A

4G→ A

4G(c = � = 1) (23)

S = ln 1 = 0 (24)

Ψ ρ (25)


16πG

�Rd4x ∼ GM2 (26)

# states ∼ eS , S ∼ GM2 (27)

A = 0 Sbek =A

4G= 0 Smicro = ln(1) = 0 (28)

�F (y − ct) (29)

n1 np e4π√

n1np (30)



Ψf = e−iHtΨi Ψi = eiHtΨf (32)

S ∼ E34 S ∼ E S ∼ E2 S ∼ E

92 (33)

H = Hvacuum + O(�) � (34)

Z =�

D[g]e−1� S[g] (35)

2

Measure has degeneracy of states

Action determines classical trajectory

For traditional macroscopic objects the measure is order while theaction is order unity

λ ∼ R (15)

∆t ∼ R

c(16)

tevap ∼G2M3

�c2(17)

T =1

8πGM=

dE

dS(18)


eS (20)


101077(22)

S =c3

�A

4G→ A

4G(c = � = 1) (23)

S = ln 1 = 0 (24)

Ψ ρ (25)


16πG

�Rd4x ∼ GM2 (26)

# states ∼ eS , S ∼ GM2 (27)

A = 0 Sbek =A

4G= 0 Smicro = ln(1) = 0 (28)

�F (y − ct) (29)

n1 np e4π√

n1np (30)




S ∼ E34 S ∼ E S ∼ E2 S ∼ E

92 (33)

H = Hvacuum + O(�) � (34)

Z =�

D[g]e−1� S[g] (35)

� (36)

2

Path integral

But for black holes the entropy is so large that the two are comparable ...

We have a failure of the semiclassical approximation ...

What happens if someone falls into a black hole ?

(SDM+Pumberg 2011)

(a) Low energy dynamics (E ∼ kT )

No horizon, radiation from ergoregions, so radiation like that from any warm body

no information loss since radiation depends on choice of microstate

(b) Correlators in high energy infalling frame (E � kT )

ψk

ψk

Collective oscillations of fuzzball gives Green’s functions thatequal Green’s functions in empty space ....

Summary

(A) The black hole information paradox is a serious problem: It does not allow GR to be consistent with quantum unitarity

⊗ |0�2|0�2� + |1�2|1�2�

⊗ |0�n|0�n� + |1�n|1�n�

. . .

⊗ |0�1|0�1� + |1�1|1�1�

ΨM

Inequality shows that there is no way around this problem unless we have an order unity change to low energy evolution at the horizon

(B) The problem then is the ‘no hair theorem’ which prevented us from finding any significant change to the evolution at the horizon

(C) In String theory we can now make explicit constructions of the microstates of the black hole. It turns out that they do have ‘hair’; in fact a horizon never forms

= 2π√

n1n5(2

�E

2mp) (54)

S = 2π√

n5(√

n1 +√

n1)(√

np +�

np) (55)

= 2π√

n5(E

√m1mp

) (56)

S = 2π√

n1n5npnkk (57)

S = 2π√

n1n5nkk(√

np +�

np) (58)

= 2π√

n1n5(E

√mpmkk

) (59)

S = 2π√

n1n5(√

np +�

np)(√

nkk +√

nkk) (60)

∼ lp (61)

∼ n16 lp (62)

M9,1 →M4,1 × T 4 × S1 (63)

E/(2mkk) = 0.5 (64)

E/(2mkk) = 1.2 (65)

Lz ∼ [g2α�4√n1n5np

V R]13 ∼ Rs (66)

∆S (67)

eS (68)

eS+∆S (69)

S = 2π�

n1n5np(1− f) + 2π�

n1n5npf(√

nk +√

nk) (70)

nk = nk =1

2

∆E

mk=

1

2Dmk(71)

D ∼ [

√n1n5npg2α�4

V Ry]13 ∼ RS (72)

∆S = S − 2π√

n1n5np = 1 (73)

S =A

4G(74)

mk ∼G5

G24

∼ D2

G5(75)

D ∼ G135 (n1n5np)

16 ∼ RS (76)

Nα lp (77)

5




A

4G= S = 2π

√n1n2n3

∆E =1

nR+

1

nR=

2

nR

∆E =2

nR

S = ln(1) = 0 (6)

S = 2√

2π√

n1n2 (7)

S = 2π√

n1n2n3 (8)

S = 2π√

n1n2n3n4 (9)

n1 ∼ n5 ∼ n

∼ n14 lp

∼ n12 lp

∼ n lp

M9,1 → M4,1 ×K3× S1

A

4G∼

�n1n5 − J ∼ S

A

4G∼√

n1n5 ∼ S

e2π√

2√

n1np

1 +Q1

r2

1 +Qp

r2

e2π√

2√

n1n5

w = e−i(t+y)−ikz w(r, θ, φ) (10)


MN(r, θ, φ) , (11)

2

weak coupling

strongcoupling

(D) One can then ask how semiclassical intuition failed. The reason can be traced to the large entropy of gravitational states of the black hole, which made the measure in the path integral compete with the classical action

λ ∼ R (15)

∆t ∼ R

c(16)

tevap ∼G2M3

�c2(17)

T =1

8πGM=

dE

dS(18)


eS (20)


101077(22)

S =c3

�A

4G→ A

4G(c = � = 1) (23)

S = ln 1 = 0 (24)

Ψ ρ (25)


16πG

�Rd4x ∼ GM2 (26)

# states ∼ eS , S ∼ GM2 (27)

A = 0 Sbek =A

4G= 0 Smicro = ln(1) = 0 (28)

�F (y − ct) (29)

n1 np e4π√

n1np (30)




S ∼ E34 S ∼ E S ∼ E2 S ∼ E

92 (33)

H = Hvacuum + O(�) � (34)

Z =�

D[g]e−1� S[g] (35)

2This should be a basic lesson for the behavior of quantum gravity in general, in all situations where we have a sufficiently large density of quanta ...

λ ∼ R (15)

∆t ∼ R

c(16)

tevap ∼G2M3

�c2(17)

T =1

8πGM=

dE

dS(18)


eS (20)


101077(22)

S =c3

�A

4G→ A

4G(c = � = 1) (23)

S = ln 1 = 0 (24)

Ψ ρ (25)


16πG

�Rd4x ∼ GM2 (26)

# states ∼ eS , S ∼ GM2 (27)

A = 0 Sbek =A

4G= 0 Smicro = ln(1) = 0 (28)

�F (y − ct) (29)

n1 np e4π√

n1np (30)




S ∼ E34 S ∼ E S ∼ E2 S ∼ E

92 (33)

2

λ ∼ R (15)

∆t ∼ R

c(16)

tevap ∼G2M3

�c2(17)

T =1

8πGM=

dE

dS(18)


eS (20)


101077(22)

S =c3

�A

4G→ A

4G(c = � = 1) (23)

S = ln 1 = 0 (24)

Ψ ρ (25)


16πG

�Rd4x ∼ GM2 (26)

# states ∼ eS , S ∼ GM2 (27)

A = 0 Sbek =A

4G= 0 Smicro = ln(1) = 0 (28)

�F (y − ct) (29)

n1 np e4π√

n1np (30)




S ∼ E34 S ∼ E S ∼ E2 S ∼ E

92 (33)

2

λ ∼ R (15)

∆t ∼ R

c(16)

tevap ∼G2M3

�c2(17)

T =1

8πGM=

dE

dS(18)


eS (20)


101077(22)

S =c3

�A

4G→ A

4G(c = � = 1) (23)

S = ln 1 = 0 (24)

Ψ ρ (25)


16πG

�Rd4x ∼ GM2 (26)

# states ∼ eS , S ∼ GM2 (27)

A = 0 Sbek =A

4G= 0 Smicro = ln(1) = 0 (28)

�F (y − ct) (29)

n1 np e4π√

n1np (30)




S ∼ E34 S ∼ E S ∼ E2 S ∼ E

92 (33)

2

λ ∼ R (15)

∆t ∼ R

c(16)

tevap ∼G2M3

�c2(17)

T =1

8πGM=

dE

dS(18)


eS (20)


101077(22)

S =c3

�A

4G→ A

4G(c = � = 1) (23)

S = ln 1 = 0 (24)

Ψ ρ (25)


16πG

�Rd4x ∼ GM2 (26)

# states ∼ eS , S ∼ GM2 (27)

A = 0 Sbek =A

4G= 0 Smicro = ln(1) = 0 (28)

�F (y − ct) (29)

n1 np e4π√

n1np (30)




S ∼ E34 S ∼ E S ∼ E2 S ∼ E

92 (33)

2

?

?

?

Radiation

String gas/brane gas

Fractional brane gas:The stuff insideblack holes

Matter is also crushed to high densities in the early Universe ...

So these lessons on phase space may radically change our picture of the dynamics there ...

(E) Cosmology

The infall problem

What happens to an object (E >> kT) that falls into the black hole? What does an infalling observer ‘feel’ ?

??

Low energy radiationmodes are corrected by order unity, no information loss in process of creation

Is it possible that the dynamics of highenergy infalling objects can approximated by the traditional black hole geometry in some way?

Now we know that black hole microstates are fuzzballs. let us see if we can do any better ...

Central part of eternal black hole diagram looks like a piece of Minkowski spacetime, Horizons look like Rindler horizons

So complementarity looks as strange as asking that we get destroyed at a Rindler horizon, and in a dual description we continue past the horizon

Rindler space: Accelerated observers see a thermal bath

t = R sinh τ

x = R cosh τ

Minkowski spacetime

Right RindlerWedge

Left RindlerWedge

R = R0

τ = τ0

t

x

R = R0An observer moving along sees a temperature

T =1

2πR0

The Minkowski vacuum can be written an an entangled sum of Rindler states

|0�M =�

k

e−Ek2π |Ek�L R�Ek|

An observation

If there is a scalar field ,then the Rindler states willhave a bath of scalar quanta

φ If has a interaction, then this bath of scalar quantawill be interacting

φ φ3

The graviton is a field that isalways present, so we will havea bath of (interacting) gravitons

Thus expect fully nonlinear quantum gravity near Rindler horizon

Rindler coordinates ‘end’ at the boundary of the wedge

Thus it is logical to expect that the gravity solution for Rindler states should also ‘end’

But this is exactly what fuzzball microstates do !

|0�M =�

k

e−Ek2π |Ek�L R�Ek|

=� ⊗

Thus we expect :

Black Holes : Israel (1976): The two sides of the eternal black hole are the two entangled copies of a thermal system in thermo-field-dynamics

Re[t]Im[t]

Maldacena (2001): This implies that the dual to the eternal black hole is two entangled copies of a CFT

Van Raamsdonk (2009): CFT states are dual to gravity solutions ... so we should be able to write an entangled sum of CFT states as an entangled sum of gravity states ...

�⊗ =

Thus we can expect that summing over fuzzball microstates will generate the eternal black hole spacetime

(SDM + Plumberg 2011)

Is it reasonable to expect that sums over (disconnected) gravitational solutions can be a different (connected) gravitational solution ?

Something like this happens in 2-d Euclidean CFT ...

The fuzzball microstates do not have horizons, but the eternal black hole spacetime does ...

ψk ψk

�

k

e−τhk−τ hk ⊗ =

‘Sewing’ process in CFT

(a) Low energy dynamics (E ∼ kT )

No horizon, radiation from ergoregions, so radiation like that from any warm body

no information loss since radiation depends on choice of microstate

(b) Correlators in high energy infalling frame (E � kT )

ψk

ψk

for generic states ψk

�⊗

mψm ψm

�ψk|O1O2|ψk� ≈�

m

e−βEm�ψm|O1O2|ψm�

≈

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