#### Holographic micro thermofield geometries of BTZ black holes

Received: April
Published for SISSA by Springer
Open Access 5 6 7
c The Authors. 5 6 7
0 Department of Chemistry, Brown University
1 Center for Theoretical Physics of the Universe
2 Department of Physics, Ewha Womans University
3 Department of Physics, Sejong University
4 Physics Department, University of Seoul
5 nd that their Penrose diagrams are no
6 Institute for Basic Science , Seoul 08826 , Korea
7 Providence , Rhode Island 02912 , U.S.A
We nd general deformations of BTZ spacetime and identify the corresponding thermo eld initial states of the dual CFT. We deform the geometry by introducing bulk elds dual to primary operators and nd the back-reacted gravity solutions to the quadratic order of the deformation parameter. The dual thermo eld initial states can be deformed by inserting arbitrary linear combination of operators at the mid-point of the Euclidean time evolution that appears in the construction of the thermo eld initial states. The deformed geometries are dual to thermo eld states without deforming the boundary Hamiltonians in the CFT side. We explicitly demonstrate that the AdS/CFT correspondence is not a linear correspondence in the sense that the linear structure of Hilbert space of the underlying CFT is realized nonlinearly in the gravity side. We also longer a square but elongated horizontally due to deformation. These geometries describe a relaxation of generic initial perturbation of thermal system while xing the total energy of the system. The coarse-grained entropy grows and the relaxation time scale is of order =2 . We clarify that the gravity description involves coarse-graining inevitably missing some information of nonperturbative degrees.
AdS-CFT Correspondence; Black Holes; Conformal Field Theory
Holographic micro thermo eld geometries of BTZ black holes
Contents
1 Introduction 2 3 4
Einstein scalar system
Linearized perturbation
Linearized solution including back reaction
Boundary stress tensor and horizon area
Convenient form of coordinates
Field theory construction
Other examples of micro-geometries
Bulk dynamics
Conclusions
B Other perturbation with m2 6= 0
Introduction
one may study various aspects of gravity and
eld theories in a rather precisely de ned
horizon and gravitational singularities.
is the inverse of
PI CI OI e 4 H0 with H0 denoting
generic perturbation of thermal system can be achieved.
Namely such states are still
which we take as .
present work.
Penrose diagram and horizon area. In section 4, we present the
eld theory description
various scalar perturbations.
partially overlap with ours in this paper.
We begin with the three dimensional Einstein scalar system
S =
There are also in general interactions between these bulk
elds, which we shall ignore in
this note. The dimension
of the corresponding dual operator is related to the mass by
d) = `2m2
and recover it whenever it is necessary. The Einstein equation reads
and the scalar equation of motion is given by
m2 2 gab = @a @b
= 0
known AdS3
M4 background where M4 may be either T 4 or K3 [10]. Thus our
The BTZ black hole in three dimensions can be written as
ds2 =
where the coordinate ' is circle compacti ed with '
' + 2 . Of course here we turn
is ensured if the Euclidean time coordinate tE has a period
Gibbons-Hawking temperature is then
2
= 2 `R . The corresponding
The mass of the black hole can be identi ed as
The boundary system is de ned on a cylinder
T =
M =
ds2B =
dt2 + `2d'2
theory is related to the Newton constant by
Thus the entropy of the system becomes
while the energy of the system can be expressed as
in terms of the quantities of CFT.
Linearized perturbation
Introducing new coordinates ( ; ; x) de ned by
c =
2 R 4G 3` 2G
S =
M =
x =
tanh `2 =
should be xed further.
the BTZ black hole metric (2.5) can be rewritten as
ds2 =
d 2 + d 2 + cos2 dx2
A( ; ; x)
B( ; ; x)
= ( ; ; x)
equations of motion (2.3) and (2.4) reduce to
(@~A)2 +
(@xA)2
A @~2A = 2A
A2 (@~ )2 + AB (@x )
3(@~B)2
2B @~2B +
@~B @~
@xA@xB + 2B@x2A
= 0;
As a power series in , the scalar eld may be expanded as
( ; ; ') =
in the leading order becomes
tan @ h + tan @ h + @~2h
R2 cos2
@'2 h = 0
. Here for simplicity, we shall consider
h = cos2 sin
a =
b =
0( ) + 1( ) cos 2
0( ) + 1( ) cos 2
tan @ h + tan @ h + @~2h
h = 0
corresponding to integral dimensions.
only. Let us organize the series expansions of the metric variables by
A = A0 1 +
B = B0 1 +
The leading order equations for the metric part then become
A0 = cos2 ;
B0 =
of motion up to some extra homogeneous solutions.
Using the remaining components
= 2
with the simplest solution of (3.7).
Linearized solution including back reaction
(1 + 6 cos 2 + 5 cos 4 ) + c1 tan +
(1 + tan )
(5 + cos 4 + 6 (2 + cos 2 ) tan ) + c3 cos2
+ c4(2 + cos 2 ) tan
(13 + 16c3) cos2
cos sin
in (3.8) can be written in more convenient forms
(1 + 6 cos 2 + 5 cos 4 ) +
(1 + tan )
(5 + cos 4 + 6 (2 + cos 2 ) tan ) + c3 cos2
(13 + 16c3) cos2
cos sin +
0 =
1 =
0 = c2
1 =
0 =
1 =
0 = c2
1 =
where we introduce
One then nds
( 0 + 1 cos 2 ) =
( 0 + 1 cos 2 ) =
( ; ) = 1
( ; ) = 1
0 =
1 =
0 = c2
1 =
(5 + cos 4
(1 + 6 cos 2 + 5 cos 4 )
cos 4 +
( 0 + 1 cos 2 ) + O( 4)
( 0 + 1 cos 2 ) + O( 4)
(1 + 2 cos2 ) cos 2
6(2 + cos 2 )) + c3 cos2
(25 + 16c3) cos2
We now require that A and B have expansions
( 0 + 1 cos 2 ) + O( 4) = (
x c2 = 2116 and c3 =
one has
9 . This choice xes the freedom of coordinate scaling. Therefore
0)2, one may
0 =
1 =
0 =
1 =
0( ) =
10 cos2 ) +
(1 + tan )
(1 + 2 cos2 )(1 + tan )
(21 + 12 cos2 )(1 + tan )
(1 + tan )
One further nds
0 =
1 =
0 =
1 =
In this coordinate system, the (orbifold) singularity is still located at
. Hence the
directional coordinate is ranged over [
other hand, the spatial in nity is at
0( ) so that the
coordinate is ranged over
nds the Penrose diagram is elongated horizontally in a
-dependent manner. We
nd that
Boundary stress tensor and horizon area
may be identi ed as
hO(t; ')i =
8 G`3 cosh2 t`R2 tanh `2 =
expanded as
0)4. Let us de ne ( ) by
( ) =
0( ). The functions A and B can be
hO(0; ')i = 0
@t hO(t; ')ijt=0 =
td =
A =
B =
q cos 2 +
where q = 3
transformation,
the metric becomes
denotes higher order terms in
By the coordinate
q cos 2~ sin2 ~ +
q sin 2~ sin 2~ +
d~2 + d~2 + cos2 ~ dx2
= ~ +
= ~ +
which are time independent.
( ) =
27 9 cos2 +22 cos4
24 cos6
changed. The mass and pressure are then given by
M = 2 ` p =
The horizon area (length) becomes
A( ) = 2 R 1
In the region near
of A( ), from which one nds A(
2 , our small
approximation breaks down since the coe cient of
2 , our small
approximation breaks down since the coe cient of 2 term becomes too large.
A(0) is given by
A(0) = 2 R 1
Convenient form of coordinates
we make the following coordinate transformation
cos 2 cos 2 +
Then the metric turns into the form
a =
a =
bx =
sin 2 + cos2
cos2 (11 + 4 cos2 ) cos 2
10 cos2 ) +
4 cos2 ) cos 2
3 + 4 cos2 (2
coordinate ranges ;
=2; =2].
Here we choose tE ranged over [
de ned by
g (tE; ') = g( tE; ')
Oy(t; ') = O(t; ')
We associate the interval [ 2
1; 1). This
{ 11 {
initial state is given by
j (0; 0)i = p
hnjU jmi jmiL
U is in general given by
The Lorentzian time evolution is given by the Hamiltonian
where the left-right Hamiltonians are identi ed with
U = T exp
dtEH(tE)
HLT (tL)
1 dtL + 1
HR(tR) dtR
HL(tL) = H
HR(tR) = H itR
is relevant to the initial ket state.
left in nities. Then with
0( ), one has the relations
tR = sin
tL =
tER = sinh E
tLE =
which identi es the boundary times tR and tL. Since
is ranged over [ 2 ; 2 ], one sees
that tR and tL are ranged over (
above becomes
1; 1) as expected. Now by analytic continuation, the
where, from the Euclidean geometry, one
nds that E is ranged over (
nds that tER can be chosen to be ranged over ( 4 4
; ) whereas tLE to be ranged over
Note that the points tE =
4 is not associated with the right nor the left boundaries of the
1; 1). One
without deforming the Hamiltonian.
of g( E ; '). For the Janus deformation in [8], one
nds the analytic continuation indeed
works. We leave further clari cation of this issue to future works.
{ 12 {
Thus the time evolution is given by
j (tL; tR)i = T exp i
dt0LHLT (t0L)
1 T exp
HR(t0R) j (0; 0)i
by the von Neumann de nition
R(tR) = trLj (tL; tR)ih (tL; tR)j
SR(tR) =
trR R(tR) log R(tR)
tary operator U = T exp
h i R0tR dt0RHR(t0R)i. For the undeformed case with Hamiltonian
tion is fully preserved, which implies that the corresponding
ne-grained (von Neumann)
bative objects in string theory. In quantum
eld theory on R
S1, one may prove that
boundary is given by
hO(t; ')i = h (0; 0)j1
O(t; ') j (0; 0)i
can be evaluated perturbatively as
hO(t; ')i = `
trO(t; ')O( i(s
H0 + O( 3) (4.15)
{ 13 {
given by
trO(t; ')O(t0; '0)e
m= 1
approximation. See [8, 13] for the normalization factor.
of ' with
hO(t; ')i =
dx g0(s)
cosh 2 (t + is) + cosh x
t0) + cosh 2 ` ('
'0 + 2 m) + i
g0(s) can be determined by demanding
The function g0(z) is identi ed as
[ cosh(v + iu) + cosh x]2 =
cosh2 v
g0(z) =
where O njc (j
0) and O njs (j
1) are de ned by
operator precisely at tE =
. (There is, however, an example where the Lorentzian
simpli es as
U in (4.6) is modi ed to
O njc =
O njs =
d' cos j'
d' sin j'
O (t; ')jt=0
O (t; ')jt=0
4 creates a deformation of state without
combination of operators
V =
corresponding to such deformation of states described in the above.
Other examples of micro-geometries
The corresponding expectation value can be given by
gn(s) =
hO(t; ')in =
hO(t; ')i0 + O( 3)
larly by
hn( ; ) =
h1( ; ) =
hO(t; ')i1 =
form is di erent from that of the previous solution.
h( ; ) =
0 h0( ; ) + 1 h1( ; )
solves the linearized scalar eld equation in (3.7) where
1 are real. From this
problem from the beginning. One nds
nonvanishing n2 implies
f n1n2 ( ; ) =
( ) where
is ranged over [
( ); 0R( )]. One nds that
R=L( ) =
2 ( ; =2)
= 1 + 2GR=L( ) + O( 4)
a( ; ) =
b( ; ) =
12 a210( ; ) + 0 1 a201( ; )
+ sin 3
18 cos 2 + cos 4 )
relation GL( 0; 1) = GR( 0;
function g(s) by
illustrated for various
1 with
The corresponding scalar eld reads
g0(s) =
h0 =
1 + sin
and the vev becomes
hO(t; ')i =
to present. The choice
will also give the scalar solution given by
g(s; ') = gn(s) =
hn( ; ) =
Finally we consider the case of massive scalar whose dual operator O
has a general
the case with `2m2 = 3 (
given by
h = cos
1(sin ) + 2Q
1(sin ))
Note that this reduces to (3.12) or (5.12) for massless case (
h = cos3
construct a rather general state by the insertion
{ 17 {
with V = P
O200, one can choose the linear combination
which leads to
g(s; ') =
0 g0(s) + 0 g0(s)
V = ( 0 + i 0)O200 = C200 O200
h( ; ) =
h =
string theory.
Bulk dynamics
better to use the coordinates ( ; ) 2 [
as introduced in section 3.3. Namely
h( ; ) = he( ; ) + ho( ; )
he( ; ) =
n hn( ; ) ; ho( ; ) =
discuss properties of this solution. First of all, there is a symmetry
) =
) = hn( ; )
{ 18 {
) =
) = he( ; )
h(0; ) = q1( )
@ h( ; )j =0 = q2( )
condition is relaxed. Now we shall give an initial condition at
= 0 by
perturbations can be independent from each other. Note that the set
which leads to the symmetry of the solution
j n = 0; 1; 2;
the Dirichlet boundary condition. The cos
factor here follows from the fact that we are
now the basis is given by
j n = 0; 1; 2;
argued below, our gravity description fails near
the singularities are spacelike. Away from
particular nothing special happens near horizon regions.
{ 19 {
gather lies in the right side of horizons that are colored in red.
description by the boundary
eld theory. In particular
R(tR) contains all those
microis available
obBut he cannot cross the past horizon of the right side, which is the
45 red line in gure 6.
basically from the causality imposed by the horizon.
Note however that the e ect is of order
2 since missing information is mainly due
to the horizon change that is of order 2
. The higher order contributions including
gravthe information regarding the perturbative gravity
uctuation may be restored.
{ 20 {
metric description of micro-geometries.
missing information is stored in such nonperturbative degrees.
Conclusions
vacuum, which is the thermalization of any initial perturbation.
We will report the related study elsewhere.
Acknowledgments
in 2017.
{ 21 {
For n = 1 case, one has
h1 = cos2 sin (1
A = cos2
B =
a =
b =
a = (
b =
the form
0 =
1 =
2 =
0 =
1 =
2 =
= 1
= 1
37 cos2
+ 126 cos4
120 cos6 ) +
(9 + 15 cos2
+ 44 cos4
(1 + 2 cos 2) tan
(3 + 23 cos2
+ 150 cos4
144 cos6 ) +
(3 + 24 cos 2
72 cos4 ) tan
(111 + 23 cos 2 + 94 cos4
(27 + 3 cos 2 + 122 cos4
60 cos6 ) +
120 cos6 )
(37 + 20 cos 2
4 cos4 ) tan
8 cos4 ) tan
13 cos 2 + 234 cos4
240 cos6 ) +
4 cos 2) tan
18(1 + 2 cos2 ) cos 2 + (1 + 8 cos2
24 cos4 ) cos 4 ] + O( 4)
2[74 + 40 cos2
+ 2( 9
+ 8 cos4 ) cos 2
4 cos2 ) cos 4 ] + O( 4)
{ 22 {
cos2 (37
126 cos2
+ 120 cos4 )
44 cos2
+ 36 cos4 )
222 cos2
+ 144 cos4 )
cos2 (37
106 cos2
+ 60 cos4 )
146 cos2
+ 120 cos4 )
234 cos2
+ 240 cos4 )
Along the future horizon
the horizon length is a monotonically increasing function
sin2 ~(9 cos 2~
2 cos 4~) +
(9 sin 2~
sin 4~ cos 2~) sin 2~ +
0 =
1 =
2 =
0 =
1 =
2 =
= ~ +
= ~ +
0( ) =
2 ( =2; )
18 cos 2 + cos 4 ) + O( 4)
Penrose diagram in
gure 7. As in section 3.2, the metric can be transformed to the
standard BTZ metric (3.29) by the coordinate transformation,
279 + 93 cos2
782 cos4
+ 3064 cos6
4272 cos8
+ 1920 cos10
A(0) = 2 R 1
A coordinate transformation
[36 sin 2 (cos 2 + 2 sin 2 ) + sin 4 (cos 4 + 4 sin 4 )]
(74 + 18 cos 2 cos 2 + cos 4 cos 4 )
gives the metric of the form (3.35) with
a =
74 cos2
+ 252 cos4
252 cos2
+ 240 cos4
+ 6 cos 2 ( 18 + 3 cos2
44 cos4
+ 36 cos6 )
cos 4 ( 6 + cos2
174 cos4
+ 144 cos6 )]
+ 6 cos 2 (33 + 44 cos2
+ cos 4 ( 47 + 18 cos2 ( 7 + 8 cos2 )]
97 cos2
+ 60 cos4 )
a =
bx =
and (A.3) with
0 =
1 =
2 =
explicit form of the solution is given by
h = cos3
cos 2 (330
16 cos2 (15
64 cos2
+ 60 cos4 ))
cos 4 ( 15 + 76 cos2
960 cos4
+ 960 cos6 )]
45 cos2
18 cos4
+ 40 cos6 ) +
(165 + 275 cos2
132 cos4
+ 108 cos6 )
(1 + 2 cos 2) tan
(15 + 115 cos2
402 cos4
144 cos6 )
24 cos4 ) tan
{ 24 {
(27 + 20 cos2
+ 4 cos4 ) tan
115 cos2
+ 118 cos4
+ 120 cos6 )
+ 8 cos4 ) tan
65 cos 2 + 18 cos4
240 cos6 )
4 cos 2) tan
22(1 + 2 cos2 ) cos 2
(1 + 8 cos2
24 cos4 ) cos 4 ] + O( 4)
2[54 + 40 cos2
+ 8 cos4
+ 16 cos4 ) cos 2
4 cos2 ) cos 4 ] + O( 4)
0 =
1 =
2 =
0 =
1 =
2 =
cos2 (45 + 18 cos2
cos2 (55 + 132 cos2
108 cos4 )
cos2 (5 + 42 cos2
+ 144 cos4 )
cos2 (135 + 154 cos2
cos2 (55 + 2 cos2
120 cos4 )
18 cos2
+ 240 cos4 )
1 =
2 =
= 1
= 1
0( ) =
2 ( =2; )
22 cos 2
cos 4 ) + O( 4)
= ~ +
= ~ +
sin2 ~(11 cos 2~ + 2 cos 4~) +
(11 sin 2~ + sin 4~ cos 2~) sin 2~ +
{ 25 {
A( ) = 2 R 1 +
375 + 125 cos2
+ 50 cos4
264 cos6
+ 848 cos8
640 cos10
A(0) = 2 R 1
A coordinate transformation
gives the metric of the form (3.35) with
a =
a =
bx =
[44 sin 2 (cos 2 + 2 sin 2 )
sin 4 (cos 4 + 4 sin 4 )]
(54 + 22 cos 2 cos 2
cos 4 cos 4 )
45 cos2
18 cos4
+ cos 2 ( 660 + 110 cos2
+ 264 cos4
216 cos6 )
+ cos 4 ( 30 + 5 cos2
+ 282 cos4
+ 144 cos6 )]
cos2 [6(45 + 18 cos2
+ 2 cos 2 (605
+ cos 4 (235
132 cos2
+ 108 cos4 )
18 cos2 (29 + 8 cos2 )]
[3885 + 120 cos2
872 cos4
+ 480 cos6
2 cos 2 (975 + 8 cos2 ( 125 + 44 cos2
+ 60 cos4 ))
cos 4 (75
380 cos2
+ 192 cos4
960 cos6 )]
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