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Samir D. Mathur The Ohio State University The fuzzball paradigm Lecture III
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Page 1: Lecture III Samir D. Mathur The Ohio State University · 2018. 10. 22. · Samir D. Mathur The Ohio State University ... p ⇠ S micro D ⇠ R h The states are not spherically symmetric

Samir D. Mathur

The Ohio State University

The fuzzball paradigm

Lecture III

Page 2: Lecture III Samir D. Mathur The Ohio State University · 2018. 10. 22. · Samir D. Mathur The Ohio State University ... p ⇠ S micro D ⇠ R h The states are not spherically symmetric

Previous lecture

Page 3: Lecture III Samir D. Mathur The Ohio State University · 2018. 10. 22. · Samir D. Mathur The Ohio State University ... p ⇠ S micro D ⇠ R h The states are not spherically symmetric

|n⇧total = (J�,total�(2n�2))

n1n5(J�,total�(2n�4))

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

A

4G= S = 2⇥

⌥n1n2n3

�E =1

nR+

1

nR=

2

nR

�E =2

nR

S = ln(1) = 0 (6)

S = 2⌥

2⇥⌥

n1n2 (7)

S = 2⇥⌥

n1n2n3 (8)

S = 2⇥⌥

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

e2�⇥

2⇥

n1np

1 +Q1

r2

1 +Qp

r2

e2�⇥

2⇥

n1n5

w = e�i(t+y)�ikz w(r, �, ⇤) (10)

B(2)MN = e�i(t+y)�ikz B(2)

MN(r, �, ⇤) , (11)

2

D ⇠ RH

Horizon willnot form

The size of bound states grows with the number of branes

Fractionation generates low tension objects that can stretch far

Page 4: Lecture III Samir D. Mathur The Ohio State University · 2018. 10. 22. · Samir D. Mathur The Ohio State University ... p ⇠ S micro D ⇠ R h The states are not spherically symmetric

L =⇥

dx[�14F a

µ⇥Fµ⇥a +i

2⇥⌅⇥ + . . .]

P =2�np

L=

2�(n1np)LT

p =2�k

LT

k

knk = n1np

e2⇤�

2�

n1np

S = 2�⇤

2⇤n1np

LT = n1L

L

1

S ⇥ E ⇥⇤

E⇤

E (81)

n1 n1 np np (82)

S = 2⇥⇤

2(⇤

n1 +⇤

n1)(⇤

np +⇤

np) ⇥⇤

E⇤

E ⇥ E (83)

S = 2⇥(⇤

n1 +⇤

n1)(⇤

n5 +⇤

n5)(⇤

np +⇤

np) ⇥ E32 (84)

S = 2⇥(⇤

n1 +⇤

n1)(⇤

n2 +⇤

n2)(⇤

n3 +⇤

n3)(⇤

n4 +⇤

n4) ⇥ E2 (85)

S = AN

N⇥

i=1

(⇤

ni +⇤

ni) ⇥ EN2 (86)

ds2 = �dt2 +�

i

a2i (t)dxidxi (87)

S = 2⇥(⇤

n1 +⇤

n1)(⇤

n2 +⇤

n2)(⇤

n3 +⇤

n3)(⇤

n4 +⇤

n4) (88)

S = 2⇥(⇤

n1 +⇤

n1)(⇤

n2 +⇤

n2)(⇤

n3 +⇤

n3) (89)

n4 = n4 � 1 (90)

Smicro = 2⇥⇤

2⇤

n1np = Sbek (91)

Smicro = 2⇥⇤

n1n5np = Sbek (92)

Smicro = 2⇥⇤

n1n5npnkk = Sbek (93)

Smicro = 2⇥⇤

n1n5(⇤

np +⇤

np) = Sbek (94)

Smicro = 2⇥⇤

n5(⇤

n1 +⇤

n1)(⇤

np +⇤

np) = Sbek (95)

Smicro = 2⇥(⇤

n5 +⇤

n5)(⇤

n1 +⇤

n1)(⇤

np +⇤

np) (96)

Smicro = 2⇥(⇤

n1 +⇤

n1)(⇤

n2 +⇤

n2)(⇤

n3 +⇤

n3)(⇤

n4 +⇤

n4) (97)

ni = ni � ni (98)

E =�

i

(ni + ni) mi (99)

S = CN⇥

i=1

(⇤

ni +⇤

ni) (100)

Pa =�

i

(ni + ni) pia (101)

R [n1, n5, np, ��, g, LS1 , VT 4 ] (102)

⇥ Rs (103)

P =2⇥np

L(104)

6

S ⇥ E ⇥⇤

E⇤

E (81)

n1 n1 np np (82)

S = 2⇥⇤

2(⇤

n1 +⇤

n1)(⇤

np +⇤

np) ⇥⇤

E⇤

E ⇥ E (83)

S = 2⇥(⇤

n1 +⇤

n1)(⇤

n5 +⇤

n5)(⇤

np +⇤

np) ⇥ E32 (84)

S = 2⇥(⇤

n1 +⇤

n1)(⇤

n2 +⇤

n2)(⇤

n3 +⇤

n3)(⇤

n4 +⇤

n4) ⇥ E2 (85)

S = AN

N⇥

i=1

(⇤

ni +⇤

ni) ⇥ EN2 (86)

ds2 = �dt2 +�

i

a2i (t)dxidxi (87)

S = 2⇥(⇤

n1 +⇤

n1)(⇤

n2 +⇤

n2)(⇤

n3 +⇤

n3)(⇤

n4 +⇤

n4) (88)

S = 2⇥(⇤

n1 +⇤

n1)(⇤

n2 +⇤

n2)(⇤

n3 +⇤

n3) (89)

n4 = n4 � 1 (90)

Smicro = 2⇥⇤

2⇤

n1np = Sbek (91)

Smicro = 2⇥⇤

n1n5np = Sbek (92)

Smicro = 2⇥⇤

n1n5npnkk = Sbek (93)

Smicro = 2⇥⇤

n1n5(⇤

np +⇤

np) = Sbek (94)

Smicro = 2⇥⇤

n5(⇤

n1 +⇤

n1)(⇤

np +⇤

np) = Sbek (95)

Smicro = 2⇥(⇤

n5 +⇤

n5)(⇤

n1 +⇤

n1)(⇤

np +⇤

np) (96)

Smicro = 2⇥(⇤

n1 +⇤

n1)(⇤

n2 +⇤

n2)(⇤

n3 +⇤

n3)(⇤

n4 +⇤

n4) (97)

ni = ni � ni (98)

E =�

i

(ni + ni) mi (99)

S = CN⇥

i=1

(⇤

ni +⇤

ni) (100)

Pa =�

i

(ni + ni) pia (101)

R [n1, n5, np, ��, g, LS1 , VT 4 ] (102)

⇥ Rs (103)

P =2⇥np

L(104)

6

L

2-charge NS1-P extremal hole

M9,1 ! M4,1 ⇥ S1 ⇥ T 4IIB:

covering space

A NS1 string carrying the momentum P in the form of travelling waves

Page 5: Lecture III Samir D. Mathur The Ohio State University · 2018. 10. 22. · Samir D. Mathur The Ohio State University ... p ⇠ S micro D ⇠ R h The states are not spherically symmetric

L =⇥

dx[�14F a

µ⇥Fµ⇥a +i

2⇥⌅⇥ + . . .]

P =2�np

L=

2�(n1np)LT

p =2�k

LT

k

knk = n1np

e2⇤�

2�

n1np

S = 2�⇤

2⇤n1np

LT = n1L

L

1

L =⇥

dx[�14F a

µ⇥Fµ⇥a +i

2⇥⌅⇥ + . . .]

P =2�np

L=

2�(n1np)LT

p =2�k

LT

k

knk = n1np

e2⇤�

2�

n1np

S = 2�⇤

2⇤n1np

LT = n1L

L

1

L =⇥

dx[�14F a

µ⇥Fµ⇥a +i

2⇥⌅⇥ + . . .]

P =2�np

L=

2�(n1np)LT

p =2�k

LT

k

knk = n1np

e2⇤�

2�

n1np

S = 2�⇤

2⇤n1np

LT = n1L

L

1

‘Naivegeometry’

An ‘actual geometry’

‘singularhorizon’

Page 6: Lecture III Samir D. Mathur The Ohio State University · 2018. 10. 22. · Samir D. Mathur The Ohio State University ... p ⇠ S micro D ⇠ R h The states are not spherically symmetric

Generic states will have structure at the string scale, but we can estimatethe size of the region over which the metric deformations are nontrivial

A

G⇠ p

n1np ⇠ Smicro

D ⇠ Rh

The states are not spherically symmetric

We do not find a horizon

Page 7: Lecture III Samir D. Mathur The Ohio State University · 2018. 10. 22. · Samir D. Mathur The Ohio State University ... p ⇠ S micro D ⇠ R h The states are not spherically symmetric

r = 0

Horizon

Traditional picture Simple fuzzballstate

More complicated fuzzball state

Modes evolve like in a piece of coal, so there is no information problem

Page 8: Lecture III Samir D. Mathur The Ohio State University · 2018. 10. 22. · Samir D. Mathur The Ohio State University ... p ⇠ S micro D ⇠ R h The states are not spherically symmetric

D1-D5-P states

Avery, Balasubramanian, Bena, Carson, Chowdhury, de Boer, Gimon, Giusto, Guo, Hampton, Keski-Vakkuri, Levi, Lunin, Maldacena, Maoz, Martinec, Niehoff, Park, Peet, Potvin, Puhm, Ross, Ruef, Russo, Saxena, Shigemori, Simon, Skenderis, Srivastava, Taylor, Turton, Vasilakis, Virmani, Warner ...

Page 9: Lecture III Samir D. Mathur The Ohio State University · 2018. 10. 22. · Samir D. Mathur The Ohio State University ... p ⇠ S micro D ⇠ R h The states are not spherically symmetric

NS1

P

D5

D1

NS1-P bound state D1-D5 bound state

fractional D1 branesn01n

05 We can join up these fractional

strings in different ways

Page 10: Lecture III Samir D. Mathur The Ohio State University · 2018. 10. 22. · Samir D. Mathur The Ohio State University ... p ⇠ S micro D ⇠ R h The states are not spherically symmetric

r = 0 (129)

r = 2M (130)

t (131)

V ⇥ V

G

V

2G(132)

�E =2

n1n5R(133)

�h =1

k+ (

l

2�m)

1

k� 2mn (134)

�h =1

k+ (

l

2� m)

1

k(135)

✓F (y � ct) = (136)

(��1)n1(��2)

n2 . . . |0⇤ (137)

8

NS1-Pstate

r = 0 (129)

r = 2M (130)

t (131)

V ⇥ V

G

V

2G(132)

�E =2

n1n5R(133)

�h =1

k+ (

l

2�m)

1

k� 2mn (134)

�h =1

k+ (

l

2� m)

1

k(135)

✓F (y � ct) = (136)

(�i1�1)

n1(�i2�2)

n2 . . . |0⇤ (137)

8

D1-D5state

(b) A mode maps toa loop with winding and spin in the direction

↵i�k

ki

(c) NS1-P: X

k

k nk = n1np

D1-D5:X

k

k nk = n01n

05

n1 ! n05

np ! n01

(a) NS1-P D1-D5

Page 11: Lecture III Samir D. Mathur The Ohio State University · 2018. 10. 22. · Samir D. Mathur The Ohio State University ... p ⇠ S micro D ⇠ R h The states are not spherically symmetric

NS1-P D1-D5

NS1 string source KK monopole tube

Dualize the geometries

(Lunin, Maldacena, Maoz 03)

(Lunin+SDM 01)

Thus these solutions haveno source, just a novel topology

The family of D1D5 geometries can be quantizedand their number gives the correct entropy

(Rhychkov 06)

Page 12: Lecture III Samir D. Mathur The Ohio State University · 2018. 10. 22. · Samir D. Mathur The Ohio State University ... p ⇠ S micro D ⇠ R h The states are not spherically symmetric

Elementary objects in IIB string theory

graviton string (NS1) NS 5-brane

D1, D3, D5, D7, D9branes

Kaluza-Kleinmonopole

Any one of these objects can be mapped to any other by S,T dualities,which are exact symmetries of the theory

Page 13: Lecture III Samir D. Mathur The Ohio State University · 2018. 10. 22. · Samir D. Mathur The Ohio State University ... p ⇠ S micro D ⇠ R h The states are not spherically symmetric

D1-D5-P near-extremalD1-D5-P extremal

General states:

[J+�1�

+2 ](|0iR)n1n5 A supergravity

quantum localizedin the 'cap'

(SDM,Saxena,Srivastava 03)

Page 14: Lecture III Samir D. Mathur The Ohio State University · 2018. 10. 22. · Samir D. Mathur The Ohio State University ... p ⇠ S micro D ⇠ R h The states are not spherically symmetric

[J+�2n+1 . . . J

+�1|0+iR]n1n5

(Giusto, SDM, Saxena 04)

string(Gava+Narian 02.Lunin+SDM 03)

(using the pp-wave formalism)

Page 15: Lecture III Samir D. Mathur The Ohio State University · 2018. 10. 22. · Samir D. Mathur The Ohio State University ... p ⇠ S micro D ⇠ R h The states are not spherically symmetric

|�� = J+�nR

. . . J+�1J

+�nL

. . . J+�1[�k]

Nk |0�R

ergoregion

(Jejalla, Madden, Ross Titchener ’05)

General program pioneered by Bena+Warner: find large families ofsolutions.

Several different innovative techniques used

KK monopoles

spherescarrying fluxes

Page 16: Lecture III Samir D. Mathur The Ohio State University · 2018. 10. 22. · Samir D. Mathur The Ohio State University ... p ⇠ S micro D ⇠ R h The states are not spherically symmetric

A toy model

Page 17: Lecture III Samir D. Mathur The Ohio State University · 2018. 10. 22. · Samir D. Mathur The Ohio State University ... p ⇠ S micro D ⇠ R h The states are not spherically symmetric

Start with the 3+1 dimensional Schwarzschild metric

Make the analytic continuation t ! �i⌧

ds2 = �(1� r0r)dt2 +

dr2

1� r0r

+ r2(d✓2 + sin2 ✓d�2)

ds2 = (1� r0r)d⌧2 +

dr2

1� r0r

+ r2(d✓2 + sin2 ✓d�2)

0 ⌧ < 4⇡r0Let the direction be a circle ⌧

This gives the 4-d Euclidean Schwarzschild geometry

⌧r

✓, �Not part of spacetime

Page 18: Lecture III Samir D. Mathur The Ohio State University · 2018. 10. 22. · Samir D. Mathur The Ohio State University ... p ⇠ S micro D ⇠ R h The states are not spherically symmetric

Dimensional reduction

4+1 dim

3+1 dim

g⌧⌧ = e2p3�

� =

p3

2ln(1� r0

r)

Dimensional reduction of the circle gives a scalar in the 3+1 spacetime

Page 19: Lecture III Samir D. Mathur The Ohio State University · 2018. 10. 22. · Samir D. Mathur The Ohio State University ... p ⇠ S micro D ⇠ R h The states are not spherically symmetric

gEµ⌫ = e1p3�gµ⌫

ds2E = �(1� r0r)

12 dt2 +

dr2

(1� r0r )

12

+ r2(1� r0r)

12 (d✓2 + sin2 ✓d�2)

The 3+1 metric is defined as

The action is

S =1

16⇡G

Zd4x

p�g

✓RE � 1

2�,µ�

⌧r

✓, �

Page 20: Lecture III Samir D. Mathur The Ohio State University · 2018. 10. 22. · Samir D. Mathur The Ohio State University ... p ⇠ S micro D ⇠ R h The states are not spherically symmetric

The stress tensor is the standard one for a scalar field

Tµ⌫ = �,µ�,⌫ � 1

2gEµ⌫�,��

,�

which turns out to be

Tµ⌫ = diag{�⇢, pr, p✓, p�} = diag{�f, f,�f,�f}

f =3r20

8r4(1� r0r )

32

(a) We see that the energy density and radial pressure are positive. The tangential pressures are negative

(b) All these quantities diverge as we reach the tip of the cigar.

gtt never changes sign, so there is no horizon(c) (SDM 16)

Page 21: Lecture III Samir D. Mathur The Ohio State University · 2018. 10. 22. · Samir D. Mathur The Ohio State University ... p ⇠ S micro D ⇠ R h The states are not spherically symmetric

So what happened to Buchdahl’s theorem?

Because the radial pressure diverged, Buchdahl would have discarded this solution as unphysical.

But we see that the problem is with the dimensional reduction: the full spacetime is completely smooth

pressure will diverge somewhere if radius of ball is

p = 0

R <9

4M

R = 2M

Page 22: Lecture III Samir D. Mathur The Ohio State University · 2018. 10. 22. · Samir D. Mathur The Ohio State University ... p ⇠ S micro D ⇠ R h The states are not spherically symmetric

Hawking radiation from fuzzballs

Page 23: Lecture III Samir D. Mathur The Ohio State University · 2018. 10. 22. · Samir D. Mathur The Ohio State University ... p ⇠ S micro D ⇠ R h The states are not spherically symmetric

Recall our difficulty with the information paradox …

(Strong coupling) Entangled pairs;Entanglement keepsgrowing

Radiates like anormal body;no problem of growing entanglement

(weak coupling)

The average rate of radiation is the same in both cases, but the detailed mechanism of radiation is very different

Page 24: Lecture III Samir D. Mathur The Ohio State University · 2018. 10. 22. · Samir D. Mathur The Ohio State University ... p ⇠ S micro D ⇠ R h The states are not spherically symmetric

ergoregion

|�� = J+�nR

. . . J+�1J

+�nL

. . . J+�1[�k]

Nk |0�R

(a) We take a simple CFT state at weak coupling

(b) We know its metric at strong coupling ... it does not have a horizon as naively expected, but a ‘cap’

Page 25: Lecture III Samir D. Mathur The Ohio State University · 2018. 10. 22. · Samir D. Mathur The Ohio State University ... p ⇠ S micro D ⇠ R h The states are not spherically symmetric

2 The non-extremal microstate geometries: Review

In this section we recall the microstate geometries that we wish to study, and explain how asuitable limit can be taken in which the physics can be described by a dual CFT.

2.1 General nonextremal geometries

Let us recall the setting for the geometries of [13]. Take type IIB string theory, and compactify10-dimensional spacetime as

M9,1 → M4,1 × T 4 × S1 (2.1)

The volume of T 4 is (2π)4V and the length of S1 is (2π)R. The T 4 is described by coordinateszi and the S1 by a coordinate y. The noncompact M4,1 is described by a time coordinate t, aradial coordinate r, and angular S3 coordinates θ,ψ,φ. The solution will have angular momentaalong ψ,φ, captured by two parameters a1, a2. The solutions will carry three kinds of charges.We have n1 units of D1 charge along S1, n5 units of D5 charge wrapped on T 4 × S1, and np

units of momentum charge (P) along S1. These charges will be described in the solution bythree parameters δ1, δ5, δp. We will use the abbreviations

si = sinh δi, ci = cosh δi, (i = 1, 5, p) (2.2)

The metrics are in general non-extremal, so the mass of the system is more than the minimumneeded to carry these charges. The non-extremality is captured by a mass parameter M .

With these preliminaries, we can write down the solutions of interest. The general non-extremal 3-charge metrics with rotation were given in [23]

ds2 = − f√

H1H5

(dt2 − dy2) +M

H1H5

(spdy − cpdt)2

+

H1H5

(r2dr2

(r2 + a21)(r

2 + a22) − Mr2

+ dθ2

)

+

(√

H1H5 − (a22 − a2

1)(H1 + H5 − f) cos2 θ

H1H5

)

cos2 θdψ2

+

(√

H1H5 + (a22 − a2

1)(H1 + H5 − f) sin2 θ

H1H5

)

sin2 θdφ2

+M

H1H5

(a1 cos2 θdψ + a2 sin2 θdφ)2

+2M cos2 θ√

H1H5

[(a1c1c5cp − a2s1s5sp)dt + (a2s1s5cp − a1c1c5sp)dy]dψ

+2M sin2 θ√

H1H5

[(a2c1c5cp − a1s1s5sp)dt + (a1s1s5cp − a2c1c5sp)dy]dφ

+

H1

H5

4∑

i=1

dz2i (2.3)

4

where

Hi = f + M sinh2 δi, f = r2 + a21 sin2 θ + a2

2 cos2 θ, (2.4)

The D1 and D5 charges of the solution produce a RR 2-form gauge field given by [6]

C2 =M cos2 θ

H1[(a2c1s5cp − a1s1c5sp)dt + (a1s1c5cp − a2c1s5sp)dy] ∧ dψ

+M sin2 θ

H1[(a1c1s5cp − a2s1c5sp)dt + (a2s1c5cp − a1c1s5sp)dy] ∧ dφ

−Ms1c1

H1dt ∧ dy − Ms5c5

H1(r2 + a2

2 + Ms21) cos2 θdψ ∧ dφ. (2.5)

The angular momenta are given by

Jψ = − πM

4G(5)(a1c1c5cp − a2s1s5sp) (2.6)

Jφ = − πM

4G(5)(a2c1c5cp − a1s1s5sp) (2.7)

and the mass is given by

MADM =πM

4G(5)(s2

1 + s25 + s2

p +3

2) (2.8)

It is convenient to define

Q1 = M sinh δ1 cosh δ1, Q5 = M sinh δ5 cosh δ5, Qp = M sinh δp cosh δp (2.9)

Extremal solutions are reached in the limit

M → 0, δi → ∞, Qi fixed (2.10)

whereupon we get the BPS relation

Mextremal =π

4G(5)[Q1 + Q5 + Q5] (2.11)

The integer charges of the solution are related to the Qi through

Q1 =gα′3

Vn1 (2.12)

Q5 = gα′n5 (2.13)

Qp =g2α′4

V R2np (2.14)

2.2 Constructing regular microstate geometries

The solutions (2.3) in general have horizons and singularities. One can take careful limits ofthe parameters in the solution and find solutions which have no horizons or singularities. In[24] regular 2-charge extremal geometries were found while in [6, 7] regular 3-charge extremal

5

where

Hi = f + M sinh2 δi, f = r2 + a21 sin2 θ + a2

2 cos2 θ, (2.4)

The D1 and D5 charges of the solution produce a RR 2-form gauge field given by [6]

C2 =M cos2 θ

H1[(a2c1s5cp − a1s1c5sp)dt + (a1s1c5cp − a2c1s5sp)dy] ∧ dψ

+M sin2 θ

H1[(a1c1s5cp − a2s1c5sp)dt + (a2s1c5cp − a1c1s5sp)dy] ∧ dφ

−Ms1c1

H1dt ∧ dy − Ms5c5

H1(r2 + a2

2 + Ms21) cos2 θdψ ∧ dφ. (2.5)

The angular momenta are given by

Jψ = − πM

4G(5)(a1c1c5cp − a2s1s5sp) (2.6)

Jφ = − πM

4G(5)(a2c1c5cp − a1s1s5sp) (2.7)

and the mass is given by

MADM =πM

4G(5)(s2

1 + s25 + s2

p +3

2) (2.8)

It is convenient to define

Q1 = M sinh δ1 cosh δ1, Q5 = M sinh δ5 cosh δ5, Qp = M sinh δp cosh δp (2.9)

Extremal solutions are reached in the limit

M → 0, δi → ∞, Qi fixed (2.10)

whereupon we get the BPS relation

Mextremal =π

4G(5)[Q1 + Q5 + Q5] (2.11)

The integer charges of the solution are related to the Qi through

Q1 =gα′3

Vn1 (2.12)

Q5 = gα′n5 (2.13)

Qp =g2α′4

V R2np (2.14)

2.2 Constructing regular microstate geometries

The solutions (2.3) in general have horizons and singularities. One can take careful limits ofthe parameters in the solution and find solutions which have no horizons or singularities. In[24] regular 2-charge extremal geometries were found while in [6, 7] regular 3-charge extremal

5

where

Hi = f + M sinh2 δi, f = r2 + a21 sin2 θ + a2

2 cos2 θ, (2.4)

The D1 and D5 charges of the solution produce a RR 2-form gauge field given by [6]

C2 =M cos2 θ

H1[(a2c1s5cp − a1s1c5sp)dt + (a1s1c5cp − a2c1s5sp)dy] ∧ dψ

+M sin2 θ

H1[(a1c1s5cp − a2s1c5sp)dt + (a2s1c5cp − a1c1s5sp)dy] ∧ dφ

−Ms1c1

H1dt ∧ dy − Ms5c5

H1(r2 + a2

2 + Ms21) cos2 θdψ ∧ dφ. (2.5)

The angular momenta are given by

Jψ = − πM

4G(5)(a1c1c5cp − a2s1s5sp) (2.6)

Jφ = − πM

4G(5)(a2c1c5cp − a1s1s5sp) (2.7)

and the mass is given by

MADM =πM

4G(5)(s2

1 + s25 + s2

p +3

2) (2.8)

It is convenient to define

Q1 = M sinh δ1 cosh δ1, Q5 = M sinh δ5 cosh δ5, Qp = M sinh δp cosh δp (2.9)

Extremal solutions are reached in the limit

M → 0, δi → ∞, Qi fixed (2.10)

whereupon we get the BPS relation

Mextremal =π

4G(5)[Q1 + Q5 + Q5] (2.11)

The integer charges of the solution are related to the Qi through

Q1 =gα′3

Vn1 (2.12)

Q5 = gα′n5 (2.13)

Qp =g2α′4

V R2np (2.14)

2.2 Constructing regular microstate geometries

The solutions (2.3) in general have horizons and singularities. One can take careful limits ofthe parameters in the solution and find solutions which have no horizons or singularities. In[24] regular 2-charge extremal geometries were found while in [6, 7] regular 3-charge extremal

5

(Jejalla, Madden, Ross Titchener ’05)

Page 26: Lecture III Samir D. Mathur The Ohio State University · 2018. 10. 22. · Samir D. Mathur The Ohio State University ... p ⇠ S micro D ⇠ R h The states are not spherically symmetric

As in any statistical system, each microstate radiates a little differently

Ψ = ψ(x)e−iωt (201)

L =1

2∂µφ∂µφ (202)

τ (203)

|ψ⟩1 =1√2

(1.1|0⟩b1 ⊗ |0⟩c1 + 0.9|1⟩b1 ⊗ |1⟩c1) (204)

E = mc2 E = mc2 − GMm

rE ∼ 0 r ∼ GM

c2(205)

|Ψ⟩ = [|0⟩b1|0⟩c1 + |1⟩b1|1⟩c1]⊗ [|0⟩b2|0⟩c2 + |1⟩b2|1⟩c2]

. . .

⊗ [|0⟩b1|0⟩c1 + |1⟩b1|1⟩c1] (206)

eiθ e−iθ (207)

c, !, G (208)

A

G∼ √

n1np ∼ Smicro (209)

R R + R2 (210)

Sbw =A

2G= 4π

√n1n2 = Smicro (211)

K3 × T 2 (212)

S = 2√

2π√

n1n5 (213)

AdS3 × S3 × T 4 (214)

∼ (n1n5)1

6 lp (215)

ΓCFT = V ρL ρR (216)

ΓCFT = V ρL ρR (217)

11

Ψ = ψ(x)e−iωt (201)

L =1

2∂µφ∂µφ (202)

τ (203)

|ψ⟩1 =1√2

(1.1|0⟩b1 ⊗ |0⟩c1 + 0.9|1⟩b1 ⊗ |1⟩c1) (204)

E = mc2 E = mc2 − GMm

rE ∼ 0 r ∼ GM

c2(205)

|Ψ⟩ = [|0⟩b1|0⟩c1 + |1⟩b1|1⟩c1]⊗ [|0⟩b2|0⟩c2 + |1⟩b2|1⟩c2]

. . .

⊗ [|0⟩b1|0⟩c1 + |1⟩b1|1⟩c1] (206)

eiθ e−iθ (207)

c, !, G (208)

A

G∼ √

n1np ∼ Smicro (209)

R R + R2 (210)

Sbw =A

2G= 4π

√n1n2 = Smicro (211)

K3 × T 2 (212)

S = 2√

2π√

n1n5 (213)

AdS3 × S3 × T 4 (214)

∼ (n1n5)1

6 lp (215)

ΓCFT = V ρL ρR (216)

ΓCFT = V ρL ρR (217)

11

Occupation numbersof left, right excitationsBose, Fermi distributionsfor generic state

Emissionvertex

Occupation numbersfor this particularmicrostate

Emission from the special microstate is peaked at definite frequenciesand grows exponentially, like a laser .....

Page 27: Lecture III Samir D. Mathur The Ohio State University · 2018. 10. 22. · Samir D. Mathur The Ohio State University ... p ⇠ S micro D ⇠ R h The states are not spherically symmetric

Radiation from the special microstate’

(B) Emission rate grows exponentially with time because after n de-excited strings have been created, the probability for creating the next one is Bose enhanced by (n+1)

(A) The emitted frequencies are peaked at

Emission rate grows as

Ψ = ψ(x)e−iωt (201)

L =1

2∂µφ∂µφ (202)

τ (203)

|ψ⟩1 =1√2

(1.1|0⟩b1 ⊗ |0⟩c1 + 0.9|1⟩b1 ⊗ |1⟩c1) (204)

E = mc2 E = mc2 − GMm

rE ∼ 0 r ∼ GM

c2(205)

|Ψ⟩ = [|0⟩b1|0⟩c1 + |1⟩b1|1⟩c1]⊗ [|0⟩b2|0⟩c2 + |1⟩b2|1⟩c2]

. . .

⊗ [|0⟩b1|0⟩c1 + |1⟩b1|1⟩c1] (206)

eiθ e−iθ (207)

c, !, G (208)

A

G∼ √

n1np ∼ Smicro (209)

R R + R2 (210)

Sbw =A

2G= 4π

√n1n2 = Smicro (211)

K3 × T 2 (212)

S = 2√

2π√

n1n5 (213)

AdS3 × S3 × T 4 (214)

∼ (n1n5)1

6 lp (215)

ΓCFT = V ρL ρR (216)

ΓCFT = V ρL ρR (217)

R S1 l, mφ, mψ λ (218)

ωCFTR = ωgravity

R ωCFTI = ωgravity

I (219)

ω = ωgravityR + iωgravity

I (220)

ωCFTR =

1

R[−l − 2 − mψm + mφn] (221)

m = nL = n + R + 1, n = nL − nR Exp[ωCFTI t] (222)

11

a†|ni =pn+ 1|n+ 1i

where there is no incoming wave, but we still have an outgoing wave carrying energy out toinfinity. These instability frequencies are given by solutions to the transcendental equation

−e−iνπΓ(1 − ν)

Γ(1 + ν)

2

)2ν=

Γ(ν)

Γ(−ν)

Γ(12(1 + |ζ| + ξ − ν))Γ(1

2 (1 + |ζ|− ξ − ν))

Γ(12(1 + |ζ| + ξ + ν))Γ(1

2 (1 + |ζ|− ξ + ν))(3.50)

We reproduce the solution to this equation, found in [14], in appendix B. In the large R limit(2.27) the instability frequencies are real to leading order

ω ≃ ωR =1

R(−l − mψm + mφn − |− λ − mψn + mφm|− 2(N + 1)) (3.51)

where N ≥ 0 is an integer. The imaginary part of the frequency is found by iterating to ahigher order; the result is

ωI =1

R

(

[l!]2

[

(ω2 − λ2

R2)Q1Q5

4R2

]l+1l+1+NCl+1

l+1+N+|ζ|Cl+1

)

(3.52)

Note that ωI > 0, so we have an exponentially growing perturbation. Our task will be toreproduce (3.51),(3.52) from the microscopic computation.

4 The Microscopic Model: the D1-D5 CFT

In this section we discuss the CFT duals of the geometries of [13]. Recall that we are workingwith IIB string theory compactified to M4,1×S1×T 4. The S1 is parameterized by a coordinatey with

0 ≤ y < 2πR (4.53)

The T 4 is described by 4 coordinates z1, z2, z3, z4. Let the M4,1 be spanned by t, x1, x2, x3, x4.We have n1 D1 branes on S1, and n5 D5 branes on S1 × T 4. The bound state of these branesis described by a 1+1 dimensional sigma model, with base space (y, t) and target space adeformation of the orbifold (T 4)n1n5/Sn1n5

(the symmetric product of n1n5 copies of T 4). TheCFT has N = 4 supersymmetry, and a moduli space which preserves this supersymmetry. Itis conjectured that in this moduli space we have an ‘orbifold point’ where the target space isjust the orbifold (T 4)n1n5/Sn1n5

[28].The CFT with target space just one copy of T 4 is described by 4 real bosons X1, X2, X3,

X4 (which arise from the 4 directions z1, z2, z3, z4), 4 real left moving fermions ψ1,ψ2,ψ3,ψ4

and 4 real right moving fermions ψ1, ψ2, ψ3, ψ4. The central charge is c = 6. The completetheory with target space (T 4)n1n5/Sn1n5

has n1n5 copies of this c = 6 CFT, with states thatare symmetrized between the n1n5 copies. The orbifolding also generates ‘twist’ sectors, whichare created by twist operators σk. A detailed construction of the twist operators is given in[19, 20], but we summarize here the properties that will be relevant to us.

The twist operator of order k links together k copies of the c = 6 CFT so that the Xi,ψi, ψi

act as free fields living on a circle of length k(2πR). Thus we end up with a c = 6 CFT on acircle of length k(2πR). We term each separate c = 6 CFT a component string. Thus if we arein the completely untwisted sector, then we have n1n5 component strings, each giving a c = 6CFT living on a circle of length 2πR. If we twist k of these component strings together by atwist operator, then they turn into one component string of length k(2πR). In a generic CFTstate there will be component strings of many different twist orders ki with

i ki = n1n5.

10

Page 28: Lecture III Samir D. Mathur The Ohio State University · 2018. 10. 22. · Samir D. Mathur The Ohio State University ... p ⇠ S micro D ⇠ R h The states are not spherically symmetric

spinning star

light cones tilt so much thatevery object must rotate

The gravity solution

Page 29: Lecture III Samir D. Mathur The Ohio State University · 2018. 10. 22. · Samir D. Mathur The Ohio State University ... p ⇠ S micro D ⇠ R h The states are not spherically symmetric

Ergoregions produce particle pairs

A quantum rotating oppositeto the ergoregion slows down rotation, and so decreases energy: Net negative energy as seen from infinity

Ergoregion instability: the produced waves grow exponentially

Page 30: Lecture III Samir D. Mathur The Ohio State University · 2018. 10. 22. · Samir D. Mathur The Ohio State University ... p ⇠ S micro D ⇠ R h The states are not spherically symmetric

Radiation: The gravity calculation

S = 2⇥⇧

n5(⇧

n1 +⇧

n1)(⇧

np +�

np) (57)

= 2⇥⇧

n5(E

⇧m1mp

) (58)

S = 2⇥⇧

n1n5npnkk (59)

S = 2⇥⇧

n1n5nkk(⇧

np +�

np) (60)

= 2⇥⇧

n1n5(E

⇧mpmkk

) (61)

S = 2⇥⇧

n1n5(⇧

np +�

np)(⇧

nkk +⇧

nkk) (62)

⇤ lp (63)

⇤ n16 lp (64)

M9,1 ⌅M4,1 ⇥ T 4 ⇥ S1 (65)

E/(2mkk) = 0.5 (66)

E/(2mkk) = 1.2 (67)

Lz ⇤ [g2��4⇧n1n5np

V R]13 ⇤ Rs (68)

�S (69)

eS (70)

eS+�S (71)

S = 2⇥⇥

n1n5np(1� f) + 2⇥�

n1n5npf(⇧

nk +⇧

nk) (72)

nk = nk =1

2

�E

mk=

1

2Dmk(73)

D ⇤ [

⇧n1n5npg2��4

V Ry]13 ⇤ RS (74)

�S = S � 2⇥⇧

n1n5np = 1 (75)

S =A

4G(76)

mk ⇤G5

G24

⇤ D2

G5(77)

D ⇤ G135 (n1n5np)

16 ⇤ RS (78)

N� lp (79)

eS (80)

5

Graviton with indices on the torus is a scalar in 6-d

Smicro = 2π√

n1n5np (152)

Sbek =A

4G= 2π

√n1n5np = Smicro (153)

ni = ni (154)

X i N pi = wi ρ ρ Ni (155)

X1 X2 X3 w = {1, .5, .5} N = 2 (156)

a1 a2 a3 (157)

L1 L2 (158)

P =∑

i

(ni + ni) pi (159)

S = S − λ(Ebranes − E) = AN

N∏

i=1

(√

ni +√

ni) − λ(2mini − E) (160)

[email protected] [email protected]

S ∼ ED−1

D (161)

S ∼ E (162)

S = A(√

n1 +√

n1)(√

n2 +√

n2)(√

n3 +√

n3) . . . (√

nN +√

nN ) (163)

∼ EN2 (164)

SO(4) ≈ SU(2) × SU(2) (165)

(1

2,1

2) (166)

X1, X2, X3, X4 ψ+, ψ+,ψ−, ψ− ψ+, ¯ψ+, ψ−, ¯ψ

−(167)

(0, 0) (0,1

2) (

1

2, 0) (168)

nL nR { (169)

h12 ≡ Ψ !Ψ = 0 (170)

M4,1 → t, r, θ,ψ,φ (171)

S1 → y y : (0, 2πR) (172)

ClV [l] V (173)

N = n1n5 (174)√

N − n√

n ≈√

N√

ndn

dt∝ n (175)

9

Smicro = 2π√

n1n5np (152)

Sbek =A

4G= 2π

√n1n5np = Smicro (153)

ni = ni (154)

X i N pi = wi ρ ρ Ni (155)

X1 X2 X3 w = {1, .5, .5} N = 2 (156)

a1 a2 a3 (157)

L1 L2 (158)

P =∑

i

(ni + ni) pi (159)

S = S − λ(Ebranes − E) = AN

N∏

i=1

(√

ni +√

ni) − λ(2mini − E) (160)

[email protected] [email protected]

S ∼ ED−1

D (161)

S ∼ E (162)

S = A(√

n1 +√

n1)(√

n2 +√

n2)(√

n3 +√

n3) . . . (√

nN +√

nN ) (163)

∼ EN2 (164)

SO(4) ≈ SU(2) × SU(2) (165)

(1

2,1

2) (166)

X1, X2, X3, X4 ψ+, ψ+,ψ−, ψ− ψ+, ¯ψ+, ψ−, ¯ψ

−(167)

(0, 0) (0,1

2) (

1

2, 0) (168)

nL nR { (169)

h12 ≡ Ψ !Ψ = 0 (170)

M4,1 → t, r, θ,ψ,φ (171)

S1 → y y : (0, 2πR) (172)

ClV [l] V (173)

N = n1n5 (174)√

N − n√

n ≈√

N√

ndn

dt∝ n (175)

9

The mass of the extremal D1-D5 system is

Mextremal =πM

4G(5)(s2

1 + s25 + 1) (2.37)

From (2.8) we see that the energy of the system above the energy of the extremal D1-D5 systemis

∆MADM ≃ πM

8G(5)(1 + 2s2

p) ≃ π

8G(5)

Q1Q5

R2nm(s−2 + s2)

8G(5)

Q1Q5

R2(m2 + n2 − 1)

=1

2R(m2 + n2 − 1)n1n5 (2.38)

where we used (2.32),(2.12),(2.13) and (2.25). Note that this result is consistent with our initialobservation (2.28) that M becomes small for large R.

In the large R limit that we have taken we also have, using (2.21) and (2.36)

r2+ ≈ −Q1Q5

R2

s2

s−2 − s2

r2− ≈ −Q1Q5

R2

s−2

s−2 − s2(2.39)

which gives

r2+ − r2

− ≈ Q1Q5

R2(2.40)

3 The instability of the geometries: Review

Shortly after the construction of the above 3-charge regular geometries it was shown in [14]that these geometries suffered from an instability. This was a classical ergoregion instabilitywhich is a generic feature of rotating non-extremal geometries. In this section we will reproducethe computations of [14] to find the complex eigenfrequencies for this instability in the large Rlimit.

3.1 The wave equation for minimally coupled scalars

We consider a minimally coupled scalar field in the 6-dimensional geometry obtained by dimen-sional reduction on the T 4. Such a scalar arises for instance from hij , which is the gravitonwith both indices along the T 4. The wave equation for the scalar is

✷Ψ = 0 (3.41)

We can separate variables with the ansatz [27, 13, 14]5

Ψ = exp(−iωt + iλy

R+ imψψ + imφφ)χ(θ)h(r) (3.42)

5Our conventions are slightly different from those in [14]: we have the opposite sign of λ, for us positive ω

will correspond to positive energy quanta, and for us ω has dimensions of inverse length.

8

Smicro = 2π√

n1n5np (152)

Sbek =A

4G= 2π

√n1n5np = Smicro (153)

ni = ni (154)

X i N pi = wi ρ ρ Ni (155)

X1 X2 X3 w = {1, .5, .5} N = 2 (156)

a1 a2 a3 (157)

L1 L2 (158)

P =∑

i

(ni + ni) pi (159)

S = S − λ(Ebranes − E) = AN

N∏

i=1

(√

ni +√

ni) − λ(2mini − E) (160)

[email protected] [email protected]

S ∼ ED−1

D (161)

S ∼ E (162)

S = A(√

n1 +√

n1)(√

n2 +√

n2)(√

n3 +√

n3) . . . (√

nN +√

nN ) (163)

∼ EN2 (164)

SO(4) ≈ SU(2) × SU(2) (165)

(1

2,1

2) (166)

X1, X2, X3, X4 ψ+, ψ+, ψ−, ψ− ψ+, ¯ψ+, ψ−, ¯ψ

−(167)

(0, 0) (0,1

2) (

1

2, 0) (168)

nL nR { (169)

h12 ≡ φ !φ = 0 (170)

M4,1 → t, r, θ, ψ, φ (171)

S1 → y y : (0, 2πR) (172)

9

Smicro = 2π√

n1n5np (152)

Sbek =A

4G= 2π

√n1n5np = Smicro (153)

ni = ni (154)

X i N pi = wi ρ ρ Ni (155)

X1 X2 X3 w = {1, .5, .5} N = 2 (156)

a1 a2 a3 (157)

L1 L2 (158)

P =∑

i

(ni + ni) pi (159)

S = S − λ(Ebranes − E) = AN

N∏

i=1

(√

ni +√

ni) − λ(2mini − E) (160)

[email protected] [email protected]

S ∼ ED−1

D (161)

S ∼ E (162)

S = A(√

n1 +√

n1)(√

n2 +√

n2)(√

n3 +√

n3) . . . (√

nN +√

nN ) (163)

∼ EN2 (164)

SO(4) ≈ SU(2) × SU(2) (165)

(1

2,1

2) (166)

X1, X2, X3, X4 ψ+, ψ+, ψ−, ψ− ψ+, ¯ψ+, ψ−, ¯ψ

−(167)

(0, 0) (0,1

2) (

1

2, 0) (168)

nL nR { (169)

h12 ≡ φ !φ = 0 (170)

M4,1 → t, r, θ, ψ, φ (171)

S1 → y y : (0, 2πR) (172)

9(Solve by matching inner and outer region solutions)

Page 31: Lecture III Samir D. Mathur The Ohio State University · 2018. 10. 22. · Samir D. Mathur The Ohio State University ... p ⇠ S micro D ⇠ R h The states are not spherically symmetric

Ψ = ψ(x)e−iωt (201)

L =1

2∂µφ∂µφ (202)

τ (203)

|ψ⟩1 =1√2

(1.1|0⟩b1 ⊗ |0⟩c1 + 0.9|1⟩b1 ⊗ |1⟩c1) (204)

E = mc2 E = mc2 − GMm

rE ∼ 0 r ∼ GM

c2(205)

|Ψ⟩ = [|0⟩b1|0⟩c1 + |1⟩b1|1⟩c1]⊗ [|0⟩b2|0⟩c2 + |1⟩b2|1⟩c2]

. . .

⊗ [|0⟩b1|0⟩c1 + |1⟩b1|1⟩c1] (206)

eiθ e−iθ (207)

c, !, G (208)

A

G∼ √

n1np ∼ Smicro (209)

R R + R2 (210)

Sbw =A

2G= 4π

√n1n2 = Smicro (211)

K3 × T 2 (212)

S = 2√

2π√

n1n5 (213)

AdS3 × S3 × T 4 (214)

∼ (n1n5)1

6 lp (215)

ΓCFT = V ρL ρR (216)

ΓCFT = V ρL ρR (217)

R S1 l, mφ, mψ λ (218)

ωCFTR = ωgravity

R ωCFTI = ωgravity

I (219)

ω = ωgravityR + iωgravity

I (220)

11

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

where there is no incoming wave, but we still have an outgoing wave carrying energy out toinfinity. These instability frequencies are given by solutions to the transcendental equation

−e−iνπΓ(1 − ν)

Γ(1 + ν)

2

)2ν=

Γ(ν)

Γ(−ν)

Γ(12(1 + |ζ| + ξ − ν))Γ(1

2 (1 + |ζ|− ξ − ν))

Γ(12(1 + |ζ| + ξ + ν))Γ(1

2 (1 + |ζ|− ξ + ν))(3.50)

We reproduce the solution to this equation, found in [14], in appendix B. In the large R limit(2.27) the instability frequencies are real to leading order

ω ≃ ωR =1

R(−l − mψm + mφn − |− λ − mψn + mφm|− 2(N + 1)) (3.51)

where N ≥ 0 is an integer. The imaginary part of the frequency is found by iterating to ahigher order; the result is

ωI =1

R

(

[l!]2

[

(ω2 − λ2

R2)Q1Q5

4R2

]l+1l+1+NCl+1

l+1+N+|ζ|Cl+1

)

(3.52)

Note that ωI > 0, so we have an exponentially growing perturbation. Our task will be toreproduce (3.51),(3.52) from the microscopic computation.

4 The Microscopic Model: the D1-D5 CFT

In this section we discuss the CFT duals of the geometries of [13]. Recall that we are workingwith IIB string theory compactified to M4,1×S1×T 4. The S1 is parameterized by a coordinatey with

0 ≤ y < 2πR (4.53)

The T 4 is described by 4 coordinates z1, z2, z3, z4. Let the M4,1 be spanned by t, x1, x2, x3, x4.We have n1 D1 branes on S1, and n5 D5 branes on S1 × T 4. The bound state of these branesis described by a 1+1 dimensional sigma model, with base space (y, t) and target space adeformation of the orbifold (T 4)n1n5/Sn1n5

(the symmetric product of n1n5 copies of T 4). TheCFT has N = 4 supersymmetry, and a moduli space which preserves this supersymmetry. Itis conjectured that in this moduli space we have an ‘orbifold point’ where the target space isjust the orbifold (T 4)n1n5/Sn1n5

[28].The CFT with target space just one copy of T 4 is described by 4 real bosons X1, X2, X3,

X4 (which arise from the 4 directions z1, z2, z3, z4), 4 real left moving fermions ψ1,ψ2,ψ3,ψ4

and 4 real right moving fermions ψ1, ψ2, ψ3, ψ4. The central charge is c = 6. The completetheory with target space (T 4)n1n5/Sn1n5

has n1n5 copies of this c = 6 CFT, with states thatare symmetrized between the n1n5 copies. The orbifolding also generates ‘twist’ sectors, whichare created by twist operators σk. A detailed construction of the twist operators is given in[19, 20], but we summarize here the properties that will be relevant to us.

The twist operator of order k links together k copies of the c = 6 CFT so that the Xi,ψi, ψi

act as free fields living on a circle of length k(2πR). Thus we end up with a c = 6 CFT on acircle of length k(2πR). We term each separate c = 6 CFT a component string. Thus if we arein the completely untwisted sector, then we have n1n5 component strings, each giving a c = 6CFT living on a circle of length 2πR. If we twist k of these component strings together by atwist operator, then they turn into one component string of length k(2πR). In a generic CFTstate there will be component strings of many different twist orders ki with

i ki = n1n5.

10

where there is no incoming wave, but we still have an outgoing wave carrying energy out toinfinity. These instability frequencies are given by solutions to the transcendental equation

−e−iνπΓ(1 − ν)

Γ(1 + ν)

2

)2ν=

Γ(ν)

Γ(−ν)

Γ(12(1 + |ζ| + ξ − ν))Γ(1

2 (1 + |ζ|− ξ − ν))

Γ(12(1 + |ζ| + ξ + ν))Γ(1

2 (1 + |ζ|− ξ + ν))(3.50)

We reproduce the solution to this equation, found in [14], in appendix B. In the large R limit(2.27) the instability frequencies are real to leading order

ω ≃ ωR =1

R(−l − mψm + mφn − |− λ − mψn + mφm|− 2(N + 1)) (3.51)

where N ≥ 0 is an integer. The imaginary part of the frequency is found by iterating to ahigher order; the result is

ωI =1

R

(

[l!]2

[

(ω2 − λ2

R2)Q1Q5

4R2

]l+1l+1+NCl+1

l+1+N+|ζ|Cl+1

)

(3.52)

Note that ωI > 0, so we have an exponentially growing perturbation. Our task will be toreproduce (3.51),(3.52) from the microscopic computation.

4 The Microscopic Model: the D1-D5 CFT

In this section we discuss the CFT duals of the geometries of [13]. Recall that we are workingwith IIB string theory compactified to M4,1×S1×T 4. The S1 is parameterized by a coordinatey with

0 ≤ y < 2πR (4.53)

The T 4 is described by 4 coordinates z1, z2, z3, z4. Let the M4,1 be spanned by t, x1, x2, x3, x4.We have n1 D1 branes on S1, and n5 D5 branes on S1 × T 4. The bound state of these branesis described by a 1+1 dimensional sigma model, with base space (y, t) and target space adeformation of the orbifold (T 4)n1n5/Sn1n5

(the symmetric product of n1n5 copies of T 4). TheCFT has N = 4 supersymmetry, and a moduli space which preserves this supersymmetry. Itis conjectured that in this moduli space we have an ‘orbifold point’ where the target space isjust the orbifold (T 4)n1n5/Sn1n5

[28].The CFT with target space just one copy of T 4 is described by 4 real bosons X1, X2, X3,

X4 (which arise from the 4 directions z1, z2, z3, z4), 4 real left moving fermions ψ1,ψ2,ψ3,ψ4

and 4 real right moving fermions ψ1, ψ2, ψ3, ψ4. The central charge is c = 6. The completetheory with target space (T 4)n1n5/Sn1n5

has n1n5 copies of this c = 6 CFT, with states thatare symmetrized between the n1n5 copies. The orbifolding also generates ‘twist’ sectors, whichare created by twist operators σk. A detailed construction of the twist operators is given in[19, 20], but we summarize here the properties that will be relevant to us.

The twist operator of order k links together k copies of the c = 6 CFT so that the Xi,ψi, ψi

act as free fields living on a circle of length k(2πR). Thus we end up with a c = 6 CFT on acircle of length k(2πR). We term each separate c = 6 CFT a component string. Thus if we arein the completely untwisted sector, then we have n1n5 component strings, each giving a c = 6CFT living on a circle of length 2πR. If we twist k of these component strings together by atwist operator, then they turn into one component string of length k(2πR). In a generic CFTstate there will be component strings of many different twist orders ki with

i ki = n1n5.

10

These exact match the CFT values(Chowdhury+SDM '07)

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Thus the microstate radiates like a piece of coal; there is noinformation problem

Page 34: Lecture III Samir D. Mathur The Ohio State University · 2018. 10. 22. · Samir D. Mathur The Ohio State University ... p ⇠ S micro D ⇠ R h The states are not spherically symmetric

How is the semiclassical approximation violated at the horizon?

Page 35: Lecture III Samir D. Mathur The Ohio State University · 2018. 10. 22. · Samir D. Mathur The Ohio State University ... p ⇠ S micro D ⇠ R h The states are not spherically symmetric

In 1972, Bekenstein taught us that black holes have an entropy

S = c3

~A4G ⇠ A

l2p

This means that a solar mass black hole has states⇠ 1010144

This is far larger than the number of states of normal matter with the same energy

Page 36: Lecture III Samir D. Mathur The Ohio State University · 2018. 10. 22. · Samir D. Mathur The Ohio State University ... p ⇠ S micro D ⇠ R h The states are not spherically symmetric

As it approaches the horizon radius, there is a small amplitude for it to tunnel into a fuzzball state

Consider a collapsing shell

Page 37: Lecture III Samir D. Mathur The Ohio State University · 2018. 10. 22. · Samir D. Mathur The Ohio State University ... p ⇠ S micro D ⇠ R h The states are not spherically symmetric

Amplitude for tunneling

Stunnel ⇠1

16⇡G

Zd4x

p�gR

Let us set all length scales to L ⇠ GM

Zd4x

p�g ⇠ (GM)4 R ⇠ 1

(GM)2

Stunnel ⇠ GM2

Page 38: Lecture III Samir D. Mathur The Ohio State University · 2018. 10. 22. · Samir D. Mathur The Ohio State University ... p ⇠ S micro D ⇠ R h The states are not spherically symmetric

The probability of tunneling into the chosen fuzzball state is

Ptunnel = |Atunnel|2 ⇠ e�2Stunnel

Using our estimate

We should now multiply thus by the number of fuzzball states we can tunnel to

Ptunnel ⇠ e�↵GM2

N ⇠ eSbek ⇠ eA4G ⇠ e

4⇡(GM)2

G ⇠ e4⇡GM2

We thus see that it is possible for the total probability for tunneling into fuzzballs can be order unity

P N ⇠ 1(SDM 0805.3716)

Page 39: Lecture III Samir D. Mathur The Ohio State University · 2018. 10. 22. · Samir D. Mathur The Ohio State University ... p ⇠ S micro D ⇠ R h The states are not spherically symmetric

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

Toy model

Page 40: Lecture III Samir D. Mathur The Ohio State University · 2018. 10. 22. · Samir D. Mathur The Ohio State University ... p ⇠ S micro D ⇠ R h The states are not spherically symmetric

(Bena, Mayerson, Puhm, Vernocke 15)

(Kraus+SDM 15)

An argument can be made using Kraus-Wilczek tunneling from black holes that the exponentials exactly cancel

For simple families of fuzzball microstates the entropy enhanced tunneling has been explicitly calculated

We call this phenomenon 'Entropy enhanced tunneling'

Page 41: Lecture III Samir D. Mathur The Ohio State University · 2018. 10. 22. · Samir D. Mathur The Ohio State University ... p ⇠ S micro D ⇠ R h The states are not spherically symmetric

⇥ ⇥ R (15)

�t ⇥ R

c(16)

tevap ⇥G2M3

�c2(17)

T =1

8⇤GM=

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)

⇥ ⌅ (25)

A ⇥ e�Sgrav , Sgrav ⇥1

16⇤G

�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 e4�⇥

n1np (30)

Smicro = 4⇤⌅n1np Sbek�wald = 4⇤⌅

n1np Smicro = Sbek (31)

⇥f = e�iHt⇥i ⇥i = eiHt⇥f (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

⇥ ⇥ R (15)

�t ⇥ R

c(16)

tevap ⇥G2M3

�c2(17)

T =1

8⇤GM=

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)

⇥ ⌅ (25)

A ⇥ e�Sgrav , Sgrav ⇥1

16⇤G

�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 e4�⇥

n1np (30)

Smicro = 4⇤⌅n1np Sbek�wald = 4⇤⌅

n1np Smicro = Sbek (31)

⇥f = e�iHt⇥i ⇥i = eiHt⇥f (32)

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 …

Thus the black hole is not a semiclassical object

Page 42: Lecture III Samir D. Mathur The Ohio State University · 2018. 10. 22. · Samir D. Mathur The Ohio State University ... p ⇠ S micro D ⇠ R h The states are not spherically symmetric

A pictorial description of ‘entropy enhanced tunneling’

Page 43: Lecture III Samir D. Mathur The Ohio State University · 2018. 10. 22. · Samir D. Mathur The Ohio State University ... p ⇠ S micro D ⇠ R h The states are not spherically symmetric

(1) Shell far outside horizon, semiclassical collapse

(2) As shell approaches its horizon, there is a nucleation of Euclidean Schwarzschild ‘bubbles’just outside the shell

A pictorial description of ‘entropy enhanced tunneling’

Page 44: Lecture III Samir D. Mathur The Ohio State University · 2018. 10. 22. · Samir D. Mathur The Ohio State University ... p ⇠ S micro D ⇠ R h The states are not spherically symmetric

(3) The bubbles cost energy, which is drawn from the energy of the shell. The shell now has a lower energy, which corresponds to a horizon radius that is smaller. The shell thus moves inwards without forming a horizon

(4) As the shell reaches closeto its new horizon, more bubbles nucleate, and so on.

Page 45: Lecture III Samir D. Mathur The Ohio State University · 2018. 10. 22. · Samir D. Mathur The Ohio State University ... p ⇠ S micro D ⇠ R h The states are not spherically symmetric

(5) Instead of a black hole with horizon, we end up with a horizon sized structure which has no horizon or singularity

Page 46: Lecture III Samir D. Mathur The Ohio State University · 2018. 10. 22. · Samir D. Mathur The Ohio State University ... p ⇠ S micro D ⇠ R h The states are not spherically symmetric

The causality paradox

Page 47: Lecture III Samir D. Mathur The Ohio State University · 2018. 10. 22. · Samir D. Mathur The Ohio State University ... p ⇠ S micro D ⇠ R h The states are not spherically symmetric

The causality problem

Light cones point inwards

So information cannot come out

When the hole evaporates away, what happens to the information in the shell?

Page 48: Lecture III Samir D. Mathur The Ohio State University · 2018. 10. 22. · Samir D. Mathur The Ohio State University ... p ⇠ S micro D ⇠ R h The states are not spherically symmetric

In quantum gravity do we have to stay inside the light cone?

(A) Perturbative quantum gravity

Light cones fluctuate underLight cones in flat spacetime ⌘µ⌫⌘µ⌫ + hµ⌫

Can we have small violations of causality?

No: We can quantize and we will find that its causal propagators vanish outside the light cone

hµ⌫

Page 49: Lecture III Samir D. Mathur The Ohio State University · 2018. 10. 22. · Samir D. Mathur The Ohio State University ... p ⇠ S micro D ⇠ R h The states are not spherically symmetric

(B) Nonperturbative quantum gravity:

Bubble nucleation: false vacuum (red) changes to true vacuum (brown)

bubble surface does not move faster than light

Page 50: Lecture III Samir D. Mathur The Ohio State University · 2018. 10. 22. · Samir D. Mathur The Ohio State University ... p ⇠ S micro D ⇠ R h The states are not spherically symmetric

A general state is a superposition of many 3-manifolds

So which light cones should we choose to determine causality ?

[(3)g]

For a general state there is no well defined notion of causality

General states

Page 51: Lecture III Samir D. Mathur The Ohio State University · 2018. 10. 22. · Samir D. Mathur The Ohio State University ... p ⇠ S micro D ⇠ R h The states are not spherically symmetric

Assume that the quantum vacuum state also satisfies these symmetries

Then we conjecture that in our full quantum gravity theory there is a definition of local operators, such that causality is maintained using the light cones of the maximally symmetric background

But consider a space with a maximal symmetry group like Minkowki space, de-Sitter space etc.

|0i

We also assume that for gently curved space, we get 'approximate locality'

Page 52: Lecture III Samir D. Mathur The Ohio State University · 2018. 10. 22. · Samir D. Mathur The Ohio State University ... p ⇠ S micro D ⇠ R h The states are not spherically symmetric

The region used in the Hawking argument is gently curved

Thus the leakage of the wavefunction outside the light cone must be small

The small corrections theorem then tells us that these small violations of nonlocality will not help

If we assume order unity effect of nonlocal physics, then there was no black hole puzzle in the first place ...

We can just take the stuff inside the hole and place it outside, and then there is no paradox

Page 53: Lecture III Samir D. Mathur The Ohio State University · 2018. 10. 22. · Samir D. Mathur The Ohio State University ... p ⇠ S micro D ⇠ R h The states are not spherically symmetric

Many of the alternatives to fuzzballs invoke some kind of nonlocal physics:

(A) Nonlocality on scale for low energy modes (Giddings)M

(B) Nonlocal effects on scales M3

Wormholes between the hole and its radiation (Maldacena and Susskind 2013)

Bits describing the radiation are not independent of bits describing theremaining hole (Papadodimas and Raju)

(C) Nonlocal effects on infinitely long length scales

Gauge modes arising from diffeomorphisms at infinity (Hawking,Perry Strominger 2015)

Page 54: Lecture III Samir D. Mathur The Ohio State University · 2018. 10. 22. · Samir D. Mathur The Ohio State University ... p ⇠ S micro D ⇠ R h The states are not spherically symmetric

Q: Can we show a concrete computation in string theory which gives effects outside the light cone?

We will assume that there is no significant leakage outside the light cone ...

Page 55: Lecture III Samir D. Mathur The Ohio State University · 2018. 10. 22. · Samir D. Mathur The Ohio State University ... p ⇠ S micro D ⇠ R h The states are not spherically symmetric

Resolving the causality paradox

Page 56: Lecture III Samir D. Mathur The Ohio State University · 2018. 10. 22. · Samir D. Mathur The Ohio State University ... p ⇠ S micro D ⇠ R h The states are not spherically symmetric

We can ask how the causality problem is avoided in the fuzzball paradigm …

(1) M M 0

r = 2M

Shell falls in at speed of light

Sees only normalphysics when far

(2) M

r = 2M

r = 2(M +M 0)

When shell approaches horizon, it tunnels into fuzzballs …

Page 57: Lecture III Samir D. Mathur The Ohio State University · 2018. 10. 22. · Samir D. Mathur The Ohio State University ... p ⇠ S micro D ⇠ R h The states are not spherically symmetric

We have seen that at this location a tunneling into fuzzballs becomes possible …

But we can still ask: What local property tells the infalling shell that it should start tunneling into fuzzballs at this location?

If the local spacetime is the vacuum (or close to the vacuum), then one might try to use the equivalence principle and say that the shell can feel nothing as it passes this location

M

r = 2M

r = 2(M +M 0)

Page 58: Lecture III Samir D. Mathur The Ohio State University · 2018. 10. 22. · Samir D. Mathur The Ohio State University ... p ⇠ S micro D ⇠ R h The states are not spherically symmetric

Is there a picture of spacetime where low energy matter sees nothing special, but matter with energy more than a given threshold sees significantly altered physics ?

Toy model: The Matrix modelc = 1

The essential point is that spacetime is not just a manifold, but has an additional property that we can call the “depth” or “thickness”

Page 59: Lecture III Samir D. Mathur The Ohio State University · 2018. 10. 22. · Samir D. Mathur The Ohio State University ... p ⇠ S micro D ⇠ R h The states are not spherically symmetric

Toy model: The Matrix model (Das+Jevicki ….)c = 1

L = Tr⇣12M(t)2 � V (M)

⌘M : N ⇥N matrix

Z =R Q

i,j dMij dM⇤ij eiL

Eigenvalues behave like fermions, so the lowest energy state has energy levels filled upto a fermi surface …

Is there a picture of spacetime where low energy matter sees nothing special, but matter with energy more than a given threshold sees significantly altered physics ?

Page 60: Lecture III Samir D. Mathur The Ohio State University · 2018. 10. 22. · Samir D. Mathur The Ohio State University ... p ⇠ S micro D ⇠ R h The states are not spherically symmetric

Small deformation on the fermi sea travel as massless bosons

But large deformation on the fermi sea suffer a distortion

The waveform can ‘fold’ over, after which it is no longer described by a classical scalar field … (Das+SDM 95)

Page 61: Lecture III Samir D. Mathur The Ohio State University · 2018. 10. 22. · Samir D. Mathur The Ohio State University ... p ⇠ S micro D ⇠ R h The states are not spherically symmetric

In fact the matrix model gives a fermi sea of varying depth

wave does not notice depth of sea

wave is distorted bytouching bottom of sea

r

The matrix model does not actually have a black hole …

Also, in our actual fuzzball paradigm, what effect provides theanalogue of the varying depth fermi sea?

Page 62: Lecture III Samir D. Mathur The Ohio State University · 2018. 10. 22. · Samir D. Mathur The Ohio State University ... p ⇠ S micro D ⇠ R h The states are not spherically symmetric

Conjecture: The Rindler region outside the hole has a different set of quantum fluctuations from those in a patch of empty Minkowski space (‘pseudo-Rindler’)

Quantum fluctuations will be different near the surface of the fuzzball since there is a nontrivial structure there …

(a) What is the nature of these fluctuations?

(b) Why should they be important ?

Page 63: Lecture III Samir D. Mathur The Ohio State University · 2018. 10. 22. · Samir D. Mathur The Ohio State University ... p ⇠ S micro D ⇠ R h The states are not spherically symmetric

(a) The fluctuations are the fluctuations to larger fuzzballs

M ! M +�M

Our energy is still so this is a virtual excitations (vacuum fluctuation)

M

Page 64: Lecture III Samir D. Mathur The Ohio State University · 2018. 10. 22. · Samir D. Mathur The Ohio State University ... p ⇠ S micro D ⇠ R h The states are not spherically symmetric

(b) The reason these fluctuations are important is because they are ‘entropy enhanced’ (there are a lot of fuzzballs with that larger mass)

Exp [Sbek(M +�M)] states

Page 65: Lecture III Samir D. Mathur The Ohio State University · 2018. 10. 22. · Samir D. Mathur The Ohio State University ... p ⇠ S micro D ⇠ R h The states are not spherically symmetric

Rindler space

pseudo-Rindler space

(Quantum fluctuations are different from empty space)

At a location depending on the energyof the quantum, there will be a tunnelinginto fuzzballs …

This resolves the causality paradox(SDM 17)

Page 66: Lecture III Samir D. Mathur The Ohio State University · 2018. 10. 22. · Samir D. Mathur The Ohio State University ... p ⇠ S micro D ⇠ R h The states are not spherically symmetric

Complementarity and fuzzball complementarity

Page 67: Lecture III Samir D. Mathur The Ohio State University · 2018. 10. 22. · Samir D. Mathur The Ohio State University ... p ⇠ S micro D ⇠ R h The states are not spherically symmetric

't Hooft, Susskind ....

Is it possible that the interior of a black hole is a manifestationof some new physics (different from the physics outside) ?

infallingobject

information reflects from stretched horizonfor the purpose of the outside observer

A second copy of the information continues to fall in

Normally we cannot do 'quantum cloning'.But here it is allowed since we cannot compare the two copies easily

Page 68: Lecture III Samir D. Mathur The Ohio State University · 2018. 10. 22. · Samir D. Mathur The Ohio State University ... p ⇠ S micro D ⇠ R h The states are not spherically symmetric

(a) But what reflects the information off the horizon?

(b) Also, if we look at good slices, what special physics separates the interior from the exterior?

r = 0

r = 1

The Firewall argument tries to gives a rigorous proof that this idea does not work

Page 69: Lecture III Samir D. Mathur The Ohio State University · 2018. 10. 22. · Samir D. Mathur The Ohio State University ... p ⇠ S micro D ⇠ R h The states are not spherically symmetric

But with fuzzballs things are quite different

We have a surface rather than a smooth horizon,so there is no difficulty in radiating the informationback

There is no interior, so can we have any notion of smooth infall?

Should we not already say that the surface of a fuzzball will have to behave like a firewall?

No, because there is a possibility which we will call Fuzzball complementarity

Page 70: Lecture III Samir D. Mathur The Ohio State University · 2018. 10. 22. · Samir D. Mathur The Ohio State University ... p ⇠ S micro D ⇠ R h The states are not spherically symmetric

There is a vast space of fuzzball states

N ⇠ eSbek(M)

When the collapsing shell tunnels into fuzzballs, then its state keeps evolving in this large space of fuzzball states …

….

A collapsing shell tunnels into a linear combination of fuzzball states

Page 71: Lecture III Samir D. Mathur The Ohio State University · 2018. 10. 22. · Samir D. Mathur The Ohio State University ... p ⇠ S micro D ⇠ R h The states are not spherically symmetric

Thus we must study dynamics is SUPERSPACE, the space of all gravity configurations

Superspace, the space of all fuzzball configurations

The full quantum gravity state is a wavefunctional over superspace

Page 72: Lecture III Samir D. Mathur The Ohio State University · 2018. 10. 22. · Samir D. Mathur The Ohio State University ... p ⇠ S micro D ⇠ R h The states are not spherically symmetric

Conjecture: This evolution in superspace can be approximately mapped to infall in the classical black hole

Fuzzballcomplementarity

E � T

Page 73: Lecture III Samir D. Mathur The Ohio State University · 2018. 10. 22. · Samir D. Mathur The Ohio State University ... p ⇠ S micro D ⇠ R h The states are not spherically symmetric

This may seem strange, but something like this happened with AdS/CFT duality (Maldacena 1997) …

Create random excitations A complicated set of

gluon excitations spreads on the D-branes

We can map this complicated evolution to free infall of the graviton

The difference is that AdS/CFT duality is exact, while fuzzball complementarity is an approximate map

Page 74: Lecture III Samir D. Mathur The Ohio State University · 2018. 10. 22. · Samir D. Mathur The Ohio State University ... p ⇠ S micro D ⇠ R h The states are not spherically symmetric

Fuzzball complementarity

Free infall onto the fuzzball is a hard impact process with

E � T

For these hard impact processes the evolution in the space of fuzzballs map to the ‘vibrations’ of empty space

Different fuzzballs radiate different at energies E ⇠ T

Page 75: Lecture III Samir D. Mathur The Ohio State University · 2018. 10. 22. · Samir D. Mathur The Ohio State University ... p ⇠ S micro D ⇠ R h The states are not spherically symmetric

Causality and the firewall argument

Page 76: Lecture III Samir D. Mathur The Ohio State University · 2018. 10. 22. · Samir D. Mathur The Ohio State University ... p ⇠ S micro D ⇠ R h The states are not spherically symmetric

What is the firewall argument?

Hawking (1975) showed that if we have the vacuum at the horizon(no hair) then there will be a problem with growing entanglement

This is equivalent to the statement: If we assume that

Ass:1 The entanglement does not keep rising, but instead drops down after some point like a normal body

Then the horizon cannot be a vacuum region.

AMPS use the same argument of bits and strong subadditivity that we used to make the Hawking argument rigorous (the small corrections theorem)

So what is the difference between the Hawking paradox and the firewall argument?

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AMPS tried to make the a stronger statement, by adding an extra assumption

Ass2: Outside the stretched horizon, we have ‘normal physics’ (Effective field theory). In particular, a shell comingin at the speed of light will encounter no new physics till it hits the stretched horizon (causality)

AMPS claim: Given assumptions Ass:1 and Ass:2, an infalling object will encounter quanta with energies reaching planck scale as it approaches the horizon (Firewall)

black hole

stretched horizon

infalling shell only seeseffective field theory(no new physics) outside the hole

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The intuition behind the AMPS argument is simple:

In the Hawking computation with vacuum horizon, the particles do not actually materialize until they are long wavelength (low energy) excitations

If we replace the black hole by a normal hot body, then we will haveno entanglement problem (by definition)

But we can now follow the radiation quanta back to the emitting surface, where they will be high energy real quanta

high energy quanta

Page 79: Lecture III Samir D. Mathur The Ohio State University · 2018. 10. 22. · Samir D. Mathur The Ohio State University ... p ⇠ S micro D ⇠ R h The states are not spherically symmetric

But there is a problem with the AMPS argument:

The two assumptions Amp:1 and Amp:2 are in conflict with each other …

(1) M M 0

r = 2M

Shell falls in at speed of light

Sees only normalphysics (Ass:2)

(2) M

r = 2M

r = 2(M +M 0)

Shell passes throughits own horizon without drama, since by causality it has not seen the hole

Page 80: Lecture III Samir D. Mathur The Ohio State University · 2018. 10. 22. · Samir D. Mathur The Ohio State University ... p ⇠ S micro D ⇠ R h The states are not spherically symmetric

(3) r = 2(M +M 0) Light cones point inwards inside

the new horizon, so the informationof the shell cannot be sent to infinity without violating causality.

This contradicts assumption Ass:1

Thus there is a conflict between Assumptions Ass:1 and Ass:2 made in the firewall argument.

If we drop the new assumption Ass:2, then we cannot argue that there is a firewall: in fact we can construct a bit model where high energy quanta feel ‘no drama’ at the horizon (fuzzball complementarity)

(SDM+Turton 2013, SDM 2015)

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Suppose an object of energy E>>kT falls in

Now there are possible states of the holeeS(M+E)

So most of the new states created after impact are not entangled with the radiation at infinity

(This is just like the entanglement before the halfway evaporation point)

Complementarity is the dynamics of these newly created degrees of freedom, and says that this dynamics is captured by the physics of the black hole interior

AMPS worry only about experiments with Hawking modes b, c, but these have E~kT

Nf

Ni= eS(M+E)

eS(M) = eS(M)+�S

eS(M) = e�S ⇡ eEkT � 1

Page 82: Lecture III Samir D. Mathur The Ohio State University · 2018. 10. 22. · Samir D. Mathur The Ohio State University ... p ⇠ S micro D ⇠ R h The states are not spherically symmetric

SUMMARY

In string theory we find the fuzzball paradigm, where black holesdo not have the traditional structure, and radiate like normal bodies.

The lesson: The scale of quantum gravity excitations increases with the number of particles involved, and always prevents horizon formation

lp ! N↵lp


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