physics of janus geometries - cern · 2018. 11. 16. · janus / icft correspondence btz black hole...

Janus / ICFT correspondenceBTZ black hole

Janus black holes+Multi-faced JBH

Physics of Janus Geometries

Dongsu Bak

University of Seoul(CERN-CKC)

December 9, 2014

Dongsu Bak Physics of Janus Geometries



I Janus /ICFT correspondence: deformation of AdS/CFTcorrespodence [Bak-Gutperle-Hirano, 03, 07]

I Interface CFTThe coupling costant dual to a marginal operator Od−1(x)jumps across the interface while keeping SO(d − 2, 2) out ofSO(d − 1, 2) conformal symmetry.∫

dd−1x [LCFT + γ ε(x1)Od−1(x) ]

Below we take O to be theLagrange density operator:

Od−1(x) = L(x) ↔ φ(x) : dilaton




Janus geometry

I AdS Einstein scalar system

S =1

16πG

∫ddx√g(R + (d − 1)(d − 2)− gab∂aφ∂bφ

)I Einstein scalar equations: d = 3

Rab + 2gab = ∂aφ∂bφ

∂a(√ggab∂bφ) = 0

I This system can be embedded into IIB SUGRA in a consistentmanner in the 3 or 5 dimensions. Here we shall be mainlyconcerned with the three dimensional case.




Janus Solution

I AdS2 slicing ansatz

ds2 = f (µ)[dµ2 + ds2AdS2

]= f (µ)

[dµ2 +

−dt2 + dξ2

ξ2

], φ = φ(µ)

I This ansatz leads to ordinary differential equations,

f ′f ′ = 4f 3 − 4f 2 + 4γ2f , φ′ =γ√f

where γ is the deformation parameter related to the jump ofthe coupling constant.




The case without deformation: φ turned off

I We start with the Poincare patch coordinates

ds2 =1

z2[−dt2 + dx2 + dz2] =

1

cos2 µ

[dµ2 +

−dt2 + dξ2

ξ2

]Make the coordinatetransformation

z = ξ cosµ , x = ξ sinµ

where −π/2 ≤ µ ≤ π/2.

I The scale function fads(µ) = 1cos2 µ

in this Poincare casewithout deformation.




I Janus deformation: the ode can be solved exactly.

fads(µ) =1

cos2 µ→ f (µ) =

κ2+sn2(κ+(µ+ µ0), k2)

where

κ2± ≡1

2(1±

√1− 2γ2)

k2 ≡ κ2−/κ2+ =γ2

2+ O(γ4)

µ0 ≡ K (k2)/κ+=π

2

(1 +

3

8γ2 + O(γ4)

)I The slicing coordinate µ is now ranged over

−µ0 ≤ µ ≤ µ0 , µ0 >π

2

I The range of the bulk angular coordinate is elongated leadingto the wedge shape deformations. But nothing singular!




I Let us turn to the dilaton part

φ(µ) =1√2

log

(1 + ksn2(κ+µ, k

2)

1− ksn2(κ+µ, k2)

)The dilaton starts from −φ0 to +φ0

−φ0 ≤ φ ≤ φ0

where

γ =1√2

tanh√

2φ0

I Hence the corresponding value of dilaton jumps through theinterface from the view point of the boundary conformal fieldtheory, which is the characteristic of the ICFT.




I The other useful form of the Janus solution

ds2 = dr2 + f (r)−dt2 + dξ2

ξ2

f (r) =1

2

(1 +

√1− 2γ2 cosh(2r)

)φ(r) =

1√2

log

(1 +

√1− 2γ2 +

√2γ tanh(r)

1 +√

1− 2γ2 −√

2γ tanh(r)

)

where we introduced the coordinate r by

dr = dµ√

f (µ) , −∞ ≤ r ≤ +∞

I As I explained, one can see that there is nothing singular asthe coordinate r changes.




BTZ black holes: the case without deformation

I Let us begin with the description of usual BTZ solution,

ds2 = −(R2 − 1)dt2 +dR2

R2 − 1+ R2dx2

I x ∼ x + L, if one wishes to consider the system of size L.




I The Euclidean solution is given by

ds2 = (R2 − 1)dτ2 +dR2

R2 − 1+ R2dx2

I To see the behavior near R = 1, we use the coordinate

cosh q = R

which leads to the metric

dq2 + sinh2qdτ2

I Therefore the euclidean time τ has to be τ ∼ τ + 2π to avoidany singularity → 2π = β → Tbtz = 1/2π

I One can get an arbitrary temperature T by an appropriaterescaling of BTZ coordinates. Namely whenever we needtemperature scale, the rule is to replace 1 by 2πT .




Thermodynamics quantities

I The horizon is located at R = 1 and the horizon length isgiven by the system size L and the corresponding BH entropyis identified with

S =L

4G→ c

6(2πT ) L ← lads

4G=

c

6I The mass of the BTZ BH can be computed using the

standard holographic method leading to

M = pL =L

16πG=

c

6

L

4π(2πT )2 =

c

6πT 2L




Main topic: Janus black holes [ Bak-Gutperle-Janik 11]

I The black hole version of the Janus geometry is dual to thefinite temperature version of ICFT.

I Due to the Janus deformation, we do not have thetranslational symmetry of the BTZ any more. Hence thegoverning equations become nonlinear partial differentialequations, which depend on two coordinates.

I The construction of the solution should be highly nontrivialbut, based on ads/cft, its existence is clear.




Interface entropy [Azeyanagi, Karch, Takayanagi, Thompson 07]

I Entanglement entropy of boundary theoryA reduced density matrix can be defined by tracing over allstates in B:

SA = −trHAρA log ρA

with ρA = trHBρ

I Holographic entanglement entropy

SA =Area(Γ∂A)

4G




I The minimal length curve in Janus geometry: geodesic isgiven by the curve of constant ξ = ξ0 and parametrized by r .

Area(Γ) = RAdS3

∫dr = RAdS3

(r∞(Γ)− r−∞(Γ)

)I Holographic cut-off regularization:

r±∞ = ∓(

log ε+1

2log√

1− 2γ2 − log(2ξ0))

Area(Γ) = RAdS3

(2 log

2ξ0ε− log(

√1− 2γ2)

)




Interface degrees

I Entropy

SA =c

6log

L

ε+

1

4Gln 1/

√1− 2γ2

I Interface Hilbert space

gI =[1/√

1− 2γ2] 1

4G=(

cosh√

2φ0) c

6

The exponential of the interface entropy can be interpreted asan extra Hilbert space dimensions due to the presence of theinterface.




Construction

I Recall the BTZ black hole solution

ds2 = (R2 − 1)dτ2 +dR2

R2 − 1+ R2dx2

I Introducing sin y = 1/R, the BTZ metric can be written as

ds2 =1

sin2 y

[cos2 y dτ2 + dx2 + dy2

]I y ∈ [0, π/2]. y = 0: boundary and y = π/2: horizon

I We take the following ansatz:

ds2 =dx2 + dy2

A(x , y)+

dτ2

B(x , y), φ = φ(x , y)




I The Einstein dilaton equations become

(~∂A)2 − A ~∂2A = 2A− A2 (~∂φ)2

3(~∂B)2 − 2B ~∂2B = 8B2/A~∂B · ~∂φ− 2B ~∂2φ = 0

I The 0th order: A0 = sin2 y and B0 = tan2 y : original BTZ

I Dilaton perturbation: φ = ϕγ + O(γ3) + O(γ5)

tan y∂yϕ− sin2 y ~∂2ϕ = 0

Linearized dilaton:

φ = γsinh x√

sinh2 x + sin2 y+ O(γ3)

Boundary condition: φ(x , y = 0) = γε(x) + O(γ3)




I The leading order perturbation of the metric part is of ordergamma square: Use the expansionsA = A0(1 + γ2a + O(γ4)), B = B0(1 + γ2b + O(γ4))

−→ −2a + sin2 y ~∂2a = 4sin2y(~∂ϕ)2

−2 tan y∂yb + sin2 y ~∂2b = 4a

I Metric part of the linearized Janus BH

ds2 =f (µ)

sinh2 x + sin2 y


]+ O(γ4)

where the coordinate µ is related by

tanµ =sinh x

sin y




Troubles in numerical approach

I One has to deal with nonlinear diff equations with twovariables.

I Even more serious problem in our case is the range ofcoordinates or the validity of a single coordinate chart sincethe geometry may not be covered by a single coordinate chartin general.

I In the Janus solution, the ranges of µ and r coordinates!There is no way to fix the range of the coordinates in a priorimanner.

I A main problem of numerical relativity! Some fundamentalprogress is needed in this respect of numerical relativity.




Troubles in numerical approach

I Try the numerical ansatz:

ds2 =1

sin2 y

[eV (x ,y)+W (x ,y) cos2 y dτ2 + dx2 + eV (x ,y)−W (x ,y)dy2

]where the coordinate y is ranged over [0, π/2].

I On the horizon side, we have to impose ∂yV (π2 , s) = 0,W (π2 , s) = 0 and Janus BC for y = 0.

I We shall use a new compact coordinate X defined over−1 ≤ X ≡ tanh x ≤ 1 where −∞ < x <∞Troubles! Jump of dilaton atX = 0 near boundary y = 0.Bulk coordinate range of X :−Xmax(y) ≤ X ≤ Xmax(y).




Dilaton as a coordinate

I The dilaton changes monotonically and ranged over

−φ0 ≤ φ ≤ φ0I Naive ansatz:

ds2 =dτ2

tan2 y+ A(y , s)dy2 + 2B(y , s)dyds + C (y , s)ds2

φ = φ0 s




I Let us first represent the undeformed BTZ metric in the form

ds2BTZ =1

sin2 y


]=

dτ2

tan2 y+

1

1− s2 cos2 y

[dy2

sin2 y+

2s cos y dsdy

sin y(1− s2)+

ds2

(1− s2)2

]where we introduced coordinate s using the leading ordersolution of dilaton

φ

φ0= s =

sinh2 x

sinh2 x + sin2 y

I Based on this the following ansatz seems natural.

dτ2

tan2 y+

1

1− s2 cos2 y

[K̃ (y , s)dy2

sin2 y+

2L̃(y , s) dsdy

sin y(1− s2)+

M̃(y , s)ds2

(1− s2)2

]




I Note that both y = 0 and s = 1 corresponds to the boundaryof spacetime and if one approaches the corner along y = 0direction, one finds troubles.

I A little investigation shows the following interpolatingfunction takes care of the all of the singular behavior.

1

1− s2 cos2 y−→ F (s, y) =

α2 + (1− α2)(1− s2)

1− s2(1− α2 sin2 y)

where

γ =tanh

√2φ0√

2, α2 =

tanh√

2φ0√2φ0




Numerical ansatz

I Final form of the ansatz:

ds2 =dτ2

tan2 y+ F (y , s)

[eK(y ,s)dy2

sin2 y+

2L(y , s) dsdy

sin y(1− s2)+

eM(y ,s)ds2

(1− s2)2

]I horizon (y = π/2) BC

K (π

2, s) = 0 L(

π

2, s) = 0 ∂yM(

π

2, s) = 0

I s = 0 BC

L(y , 0) = 0 ∂sK (y , 0) = 0 ∂sM(y , 0) = 0

which is from the Z2 symmetry of our black hole, s ↔ −s.

I y = 0 and s = 1 BC: K = M = 0 and L = s cos y




Numerical result

eK(s,y) with φ0 = 10.The upper left is the dangrouscorner of y = 0 and s = 1.[Romuald Janik]




I Exact Janus BH solution:

ds2 =dτ2

tan2 y+ Fe (y, s)

[dy2

sin2 y+

2f ′(s) dsdy

tan y f (s)+ G(y, s)ds2

]

Fe (y, s) =

[sin2 y +

cos2 y

f (s)

]−1

, f (s) =γ2

1− cosh√

2φ0s

cosh√

2φ0

G(y, s) =φ20f

2(s)

γ4

[γ2 sin2 y + cos2 y

(1−

cosh√2φ0s

cosh√2φ0

+sinh2

√2φ0s

2 cosh2√2φ0

)]

I The solution is basically obtained by the guess work from ourperturbative analysis and using some intuition from the numericalanalysis.

I One can check this is the exact solution using our original Einsteinscalar equation in a direct manner.




Errors ∼ 1/105

We have presented here the error of our numerical analysis in theright side of the figure for the function Exp[K ] and one can seethat numerical error is indeed quite small.




Exact Janus BH

I By the coordinate transformation,

cot2 y = f (s) sinh2 p

one is led to

ds2 = f (s)[sinh2 p dτ2 + dp2 +

φ20γ2

f (s)ds2]

Reintroducing µ by

γ/√

f (s) = dφ/dµ = φ0ds/dµ

one finds a remarkably simple form, which is consistent withthe Janus ansatz

ds2 = f (µ)[sinh2 p dτ2 + dp2 + dµ2

], φ = φ(µ)

I p = 0 corresponds to the horizon. p =∞ corresponds toboundary.




Boundary-horizon map

I First is the discussion of geometric entropy.

The horizon length is defined inthe horizon side whereas thedual FT system is defined atthe boundary side.

Since, in our case, the translational inv. along x is lost, it isnot clear how to map the boundary point to the horizon point.

I Methods 1: Based on the current Qa = εabc ∇bξc [Wald-Iyer]

where ξa is the Killing vector of the time translation symmetry∇aξb +∇bξa = 0.

I This current is automatically conserved ∇aQa = 0.

I In the horizon side it gives us something proportional to thehorizon length which can be used as a definition of entropy.




Method 1

I By the conservation law, the first law of thermodynamics isguaranteed.

TdS = dE + pdLThen the contribution alongthe AA1 and BB1 lines shouldbe zero in order for the firstlaw to work.

This gives us the zero charge condition

dxaQa = 0

along the lines of the boundary horizon map. This gives us aunique map of the boundary point to the horizon point andcorresponding horizon length can be computed.




Method 2

I The second method is based on the null line emanatedorthogonally from a boundary point to the horizon point.

This is related to the concept oflightsheet which is used to show the holographic principle ofgeneral spacetimes. [Bousso]

I The results of the two mappings for the finite L are not thesame but in the large L limit the result is the same leading tothe interface entropy

SI =1

4Glog

1√1− 2γ2




Thermodynamics

I One comment here is that there is no way to define theentropy density in the field theory side in a gauge invariantmanner. There is no such local operator O(x).

I Entropy

S =πTL

4G+ SI

I Energy and pressure

E =πT 2L

4Gp =

πT 2

4G

I The first law of thermodynamics

TdS = dE + pdL




Multi-faced Janus black holes [ Bak-Min 13]

I Multi-faced Janus systems

I This time I do not how to do the numerical analysis due to thecoordinate chart problem.




I The ansatz is the same as before which leads to the sameequations and the leading order of the scalar part is given by

tan y∂yϕ− sin2 y ~∂2ϕ = 0 , φ = ϕγ + O(γ3)

I The multi faced boundary condition is given by

ϕ(x , 0) =∑n

αnε(x − ln)

where αn is the nth interface coefficient.I The solution satisfying this boundary condition is given by

ϕ(x , y) =∑n

αnϕ0(x − ln, y)

where ϕ0 is the single interface solution

ϕ0(x , y) =sinh x√

sinh2 x + sin2 y




I The leading order correction to the metric part of the metricpart is of order γ2, whose source term is the square of theleading scalar part: Remember the expansionsA = A0(1 + γ2a + O(γ4)), B = B0(1 + γ2b + O(γ4))

−→ −2a + sin2 y ~∂2a = 4sin2y(~∂ϕ)2

−2 tan y∂yb + sin2 y ~∂2b = 4a

I Solving this part of the solution is essential to get the leadingorder corrections to the thermodynamic quantities. Thereseems no systematic ways. But we were able to findcompletely general analytic solution which is not illuminatingat all.




Thermodynamic quantities of multi-faced black Janus

Let me illustrate the contributionfor the double interfaces withinterface coefficient α− and α+

respectively located at x = 0 andx = l .

I Casimir energy

E = E0 + Ec , E0 =c

6πT 2L

Ec =c

6

2πT 2l

sinh2 2πTl(2α−α+)γ2 + O(γ4)

I The pressure is found as

pL =c

6πT 2

pl =c

6

2πT 2

sinh2 2πTl(2α−α+)γ2 + O(γ4)




Entropy correlation of multi-faced black Janus

I The entropy can also be identified as

S = S0 + SI , S0 =c

3πTL

SI =c

3

[α2− + α2

+

+2α−α+

(1− coth2πTl +

2πTl

sinh2 2πTl

)]γ2 + O(γ4)

Here we have illustrate thecross term contribution of theinterface entropy as a functionof Tl .

I On the other hand small Tl limit, the interface entropy becomes

SI =c

3(α− + α+)2γ2 + O(γ4)




Final remarks

I Of course this result can be generalized to an arbitrarynumber of interfaces.

I One can check the above from the FT computations.

I Any attempt of numerical analysis of the system will beextremely interesting.

Thanks a lot!


physics of janus geometries - cern · 2018. 11. 16. · janus / icft correspondence btz black hole...

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