Time dependent Interface in AdS Black Hole Spacetime

Koichi Nagasaki1

School of Physics, University of Electronic Science and Technology of China,

Address: No.4, Section 2, North Jianshe Road, Chengdu, Sichuan 610054, China

Abstract

We consider a D5-brane solution in AdS black hole spacetime. This is a defectsolution moving in subspace of AdS5 × S5. This non-local object is realized by theprobe D5-brane moving in black hole spacetime. We found this probe brane does notpenetrate the black hole horizon. We also found the solution does not depend on themotion on S5 subspace.

Contents

1 Introduction 1

2 Flat spacetime 32.1 Equations of motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42.2 Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6

3 Black hole spacetime 63.1 Equations of motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . 83.2 Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10

4 Discussion 10

1 Introduction

In the AdS/CFT correspondence [1, 2, 3] non-local operators are a useful tool forstudying. For example, in [4], we considered a kind of non-local operator called ”aninterface” and found the agreement in the results of gauge theory and gravity theory inthe leading order of the power series of λ/k2. Here λ is the ’t Hooft coupling λ = g2YMNand k is a parameter which specifies the gauge flux on the probe brane. This non-localoperator is realized in the D3/D5 brane system. The multiple D3-brane create a curvedspace-time and one D5-brane is treated as a probe on the AdS5×S5 spacetime. Thesebrane systems are related to nonequilibrium systems in the AdS/CFT correspondence[5, 6, 7, 8]. A time dependent solution of the D5-brane we consider in this paper hasits application to cosmology in anti de-Sitter space and brane world scenario [9, 10].

The motivation to consider the AdS/CFT correspondence including the D5-braneon black hole spacetimes is related to a conjecture called “complexity - action” (CA).

1koichi.nagasaki24@gmail.com

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In this statement complexity is a quantity which is expected to be related many blackhole problems [11, 12, 13, 14]. Complexity has the origin in computational science andit quantifies how hard to create the final state from the initial state [15, 16, 17, 18, 13,14, 19, 20, 21]. In black hole physics this quantity describes a quantum state of theblack hole interior or Einstein-Rosen Bridge [22]. According to the holographic relationstated in the first paragraph, complexity must have the holographic counterpart. Thatis called holographic complexity. CA conjecture [23, 24] states that this holographicquantity is the action which is calculated in a bulk region called “the Wheeler DeWittpatch.” This is the region bounded by null surfaces anchored at the given time onboundaries. The calculation of the action has a difficulty caused by the divergenceat the AdS boundary where the metric diverges. To compute the growth rate ofthe action, we only need to take into account the bulk region inside the horizon asexplained in [25]. The left panel of Figure 1 depicts Wheeler DeWitt patch on aPenrose diagram. The two diagonal lines represent the black hole horizon. As thetime passes, this region changes as shown in the right of Figure 1. The time definedon a boundary theory is denoted by tL. The separated regions (2), (3) and (4) do notcontribute the development. We only have to find the contribution of region (1) tomeasure the time development of the WDW action.

Figure 1: Left: Wheeler DeWitt patch. Right: Development of the region

We found the static defect solutions in the AdS black hole spacetime in [26, 27].These describes the embedding of the D5-brane in AdS black hole spacetime. Theresults tell us that the D5-brane can only extend outside of the horizon if there is anon-zero gauge flux on the D5-brane. In CA conjecture holographic complexity —the action is calculated by the integral over the Wheeler-DeWitt patch. Especially, inorder to find the growth rate of it, we only have to integrate inside the horizon. Thenthe DBI action does not contribute the growth of complexity in these cases.

A time dependent system is interesting for studying the AdS/CFT correspondence.Its behavior is different from the static cases. For example, the extremal surfaces donot penetrate the horizon in a static case, whereas it can penetrate the horizon in thetime dependent case as stated in [28]. Then it will be worth to consider a movingdefect [29] in AdS black hole spacetime. These time dependent cases have never beenstudied for the AdS/CFT correspondence and CA conjecture. If such a solution isfound, this can be a good way to test CA relation. As stated in the first, a newparameter is useful to compare the holographic quantities.

In the above reason we would like to consider a moving object in the AdS blackhole spacetime. The whole spacetime consists of the product of AdS5 and S5. We cansuppose some types of motion of the D-branes. For simplicity, in this paper we would

2

like to consider a motion which has a spherical symmetry which has the rotation inS2 subspace of S5.

The organization of this paper is as follows. In Section 2 we start with the studyof the flat AdS case and find its solution. This case corresponds to the flat spacetimesolution [4] added the angular momentum. In Section 3 we generalize the solution tothe black hole spacetime. We conclude this paper in Section 4 with some commentsfor the results.

2 Flat spacetime

In this section we consider the embedding of the D5-brane in AdS spacetime AdS5×S5.The metric of the AdS part with radius RAdS is given by time coordinate t, cartesiancoordinates xi; i = 1, 2, 3 and the radial coordinate 1/y as

ds2AdS =R2

AdS

y2(−dt2 + dy2 +

3∑i=1

dx2i ). (1)

The S5 metric is given by coordinates θ ∈ [0, π/2], ϕ ∈ [0, 2π), ψ ∈ [0, π/2] andφ, χ ∈ [0, 2π) as

ds2S5 = dθ2 + sin2 θdϕ2 + cos2 θdΩ23

= dθ2 + sin2 θdϕ2 + cos2 θ(dψ2 + sin2 ψdφ2 + cos2 ψdχ2). (2)

In the following we put RAdS = 1 for simplicity. The Ramond-Ramond 4-form isdefined as

C4 = − 1

y4dtdx1dx2dx3 + 4α4, (3)

where α4 is a 4-form satisfying dα4 = volume of S5.We consider the D5-brane rotates in subspace of S5. We assume that the D5-brane

extends to the directions

t, x1, x2 ∈ AdS5, ψ, φ ∈ S5, (4)

and one dimensional subspace in other directions. By parameter σ, this embedding isgiven by the following functions

y = y(σ), x3 = x3(σ), θ = θ(σ), ϕ = ωt+ g(σ). (5)

There is a gauge flux on the D5-brane which is expressed by symmetry as

F = F ′(ψ)dψ ∧ dφ, A = F (ψ)dφ. (6)

In this section ′ means the derivative by the variables indicated in eq.(5) and eq.(6).The induced metric is

ds2D5 = −( 1

y2− ω2 sin2 θ

)dt2 +

(y′2 + x′23y2

+ θ′2 + g′2 sin2 θ)dσ2 + 2ωg′ sin2 θdtdσ

+dx21 + dx22

y2+ cos2 θ(dψ2 + sin2 ψdφ2).

Adding the flux, the above metric is in matrix form

Gind + F

=

− 1y2

(1− ω2y2 sin2 θ) ωg′ sin2 θ

ωg′ sin2 θ 1y2

(y′2 + x′23 + y2θ′2 + y2g′2 sin2 θ)

1/y2

1/y2

cos2 θ F ′

−F ′ cos2 θ sin2 ψ

.

(7)

3

Then the DBI action is

SDBI/T5 = −2πT V∫dydψ

√cos4 θ sin2 ψ + F ′2

y4L(σ), (8)

where T is the integral over the time interval and V is the integral on (x1, x2)- plane.The Wess-Zumino action is, by substituting the RR4-form (3),

SWZ/T5 =

∫F ∧ C4 = −2πT V

∫F ′x′3dψdσ

y4. (9)

Summing them, the D5-brane action is

SD5/T5 = −2πT V∫dψdσ

y4

(√cos4 θ sin2 ψ + F ′(ψ)2L(σ) + F ′(ψ)x′3

). (10)

The equation of motion for F (ψ) is

d

dψ

( F ′(ψ)√cos4 θ sin2 ψ + F (ψ)′2

)= 0. (11)

By using c(σ), which is a function of σ,

F ′2 =c2 cos4 θ

1− c2 sin2 ψ, F ′ = −κ sinψ, (12)

where we denote the constant factor in the front of sinψ as κ since we know F ′ isindependent of y:

c(σ)2 cos4 θ(σ)

1− c(σ)2=: κ2. (13)

This constant represents the strength of the gauge flux. Therefore we found the ansatzfor the gauge flux to be proportional to the volume form of S2 subspace:

F = −κ sinψdψ ∧ dφ. (14)

Summarizing the above discussion, the action is

SD5/T5 = −2πTV

∫dσ

y4

(√cos4 θ + κ2L(σ)− κx′3

), (15)

L(σ) :=√

(y′2 + x′23 + y2θ′2)(1− ω2y2 sin2 θ) + y2g′2 sin2 θ. (16)

For simplicity we use the following notation in the next section.

Θ :=√

cos4 θ + κ2, S := y′2 + x′23 + y2θ′2, Ω := 1− ω2y2 sin2 θ. (17)

By this notation the Lagrangian is

L =ΘL(σ)− κx′3

y4, L(σ) =

√SΩ + y2g′2 sin2 θ. (18)

2.1 Equations of motion

The equations of motion are

d

dσ

(y′ΘΩ

y4L

)+

4

y5(ΘL− κx′3)−

Θ

y3θ′2Ω + (g′2 − ω2S) sin2 θ

L= 0, (19a)

d

dτ

(θ′

ΩΘ

y2L

)+

2 sin θ cos3 θ

y4ΘL− (g′2 − ω2S) sin θ cos θ

y2Θ

L= 0, (19b)

d

dτ

(x′3

ΩΘ

y4L

)+

4κy′

y5= 0, (19c)

d

dτ

(g′

sin2 θ ·Θy2L

)= 0. (19d)

4

There is a gauge degree of freedom due to reparametrization invariance in σ. We fixΘ/L = 1. Then the equations (19) are

d

dσ

(y′Ωy4

)+

4

y5(Θ2 − κv)− θ′2Ω + (w2 − ω2S) sin2 θ

y3= 0, (20a)

d

dσ

(θ′

Ω

y2

)+

2 sin θ cos3 θ

y4− (g′2 − ω2S) sin θ cos θ

y2= 0, (20b)

d

dσ

(x′3

Ω

y4

)+

4κy′

y5= 0, (20c)

d

dσ

(g′

sin2 θ

y2

)= 0. (20d)

We define

A := − d

dσlog Ω =

2ω2y sin θ(y′ sin θ + yθ′ cos θ)

Ω, (21a)

B :=4

yΩ(κx′3 −Θ2), C :=

(g′2 − ω2S)

Ω, D := −2 sin θ cos3 θ

y2Ω. (21b)

The equations are solved for the second derivative terms as

y′′ = yθ′2 +4y′2

y+ y′A+B + Cy sin2 θ, (22a)

θ′′ =2y′θ′

y+ θ′A+ C sin θ cos θ +D, (22b)

x′′3 =4y′x′3y

+ x′3A−4κy′

yΩ, (22c)

g′′ = −2θ′g′ cot θ +2y′g′

y. (22d)

Static case In the static case, the solution of the above equations must be x3 = κyas obtained in [4]. Let us confirm it. For ω = 0 and θ = 0, the equations of motion(19) are simplified by setting σ = y as

d

dy

(κ

√1 + κ2

y4√

1 + x′23

)+

4κ

y5= 0. (23)

Then in this case, the following is the solution:

x3 = κy, θ(y) = 0, g(y) = 0. (24)

This is surely the stationary solution [4].

Boundary condition y = 0 corresponds to the AdS boundary. The boundarycondition is, since we fixed Θ/L =: 1, in y → 0 limit

1 =Θ

L=

√cos4 θ0 + κ2

(y′2 + x′23 + y2θ′2)(1− ω2y2 sin2 θ0) + g′(0)2y2 sin2 θ0→√

cos4 θ0 + κ2

y′2 + x′23.

(25)In order to avoid the singularity at the boundary, the righthand side of the thirdequation (22c) is

4y′

yΩ(x′3(1− ω2y2 sin2 θ)− κ)→ 4y′

yΩ(x′3(0)− κ). (26)

5

Then, from eq.(25) and eq.(26) we impose the following condition:

y′(0) = cos2 θ0, x′3(0) = κ. (27)

We impose the Neumann boundary condition for the angular direction θ′(0) = 0. Therighthand side of the forth equation of motion (22d) approaches

2g′

y sin θ(y′ sin θ − yθ′ cos θ)→ 2g′ cos θ0

y sin θ0(cos θ0 sin θ0 − yθ′), ∴ g′(0) = 0. (28)

From the above discussion we obtain the appropriate boundary condition

y(0) = 0, θ(0) = θ0, x3(0) = κσ, g(0) = 0, (29a)

y′(0) = cos2 θ0, θ′(0) = 0, x′3(0) = κ, g′(0) = 0. (29b)

2.2 Solution

We solve the equations of motion (22) under the initial condition (29). The resultsare summarized in the following seven figures.

The function y increases linearly for the value of σ in the case θ0 = 0 as shown inFigure 2.

The next three figures show the gauge flux dependence. Figure 3 and Figure 5show the behavior of θ and g as functions of y. In these figures, neglecting numericalerror, we read θ and g do not grow from zero. Figure 4 shows linear relationshipx3 = κy as the same in [4].

The next three figures show the angular momentum dependence (see Figure 6,Figure 7 and Figure 8). From these results, we find that there is no dependence onthe angular momentum. It can also be seen from the θ plot (Figure 6) and the factthat the ω dependence in eqs.(20) exists only in the combination with sin θ.

3 Black hole spacetime

In this section we generalize the solution obtained in the previous section to the blackhole spacetime. The metric is

ds2AdS5×S5 =r2

R2AdS

(−hdt2 + d~x2) +R2

AdS

r2dr2

h+R2

AdSds2S5 , (30)

h(r) = 1− r4Hr4, (31)

where RAdS is the radius of the AdS and the S5 metric is, by coordinates θ, ψ ∈ [0, π/2]and ϕ, φ, χ ∈ [0, 2π),

ds2S5 = dθ2 + sin2 θdϕ2 + cos2 θ(dψ2 + sin2 ψdφ2 + cos2 ψdχ2). (32)

In the following, we put RAdS = 1. By changing variables, y = 1/r, this metricbecomes

ds2AdS5×S5 =1

y2(−hdt2 +

dy2

h+ d~x2) + ds2S5 . (33)

In the same way as before, we consider the case where the D5-brane extends tothe directions

t, x1, x2 ∈ AdS5, ψ, φ ∈ S5. (34)

By parameter σ the embedding of the D5-brane is

y = y(σ), x3 = x3(σ), θ = θ(σ), ϕ = ωt+ g(σ). (35)

6

0 20 40 60 80 1000

20

40

60

80

100y(

)

= 0.0= 0.1= 0.2= 0.3= 0.4

Figure 2: Increase of y for θ0 = 0

0 20 40 60 80 100y

7

8

9

10

1e 11+6.2831853071= 0.0= 0.1= 0.2= 0.3= 0.4

Figure 3: κ dependence of θ

0 20 40 60 80 100y

0

5

10

15

20

25

30

35

40

x 3

= 0.0= 0.1= 0.2= 0.3= 0.4

Figure 4: κ dependence of x3

0 20 40 60 80 100y

0.04

0.02

0.00

0.02

0.04

g

= 0.0= 0.1= 0.2= 0.3= 0.4

Figure 5: κ dependence of g

The gauge flux is obtained in the same way as the flat case,

F = −κ sinψdψ ∧ dφ. (36)

The RR4-from is

C4 = − 1

y4dtdx1dx2dx3 + 4α4, (37)

where α4 is a 4-form satisfying dα4 = volume of S5. The induced metric becomes

ds2D5 = −( hy2− ω2 sin2 θ

)dt2 +

(y′2/h+ x′23y2

+ θ′2 + g′2 sin2 θ)dσ2 + 2ωg′ sin2 θdtdσ

+dx21 + dx22

y2+ cos2 θ(dψ2 + sin2 ψdφ2).

The sum of the induced metric and the gauge flux is

GD5 + F

=

− 1y2

(h− ω2y2 sin2 θ) ωg′ sin2 θ

ωg′ sin2 θ 1y2

(y′2

h + x′23 + y2θ′2) + g′2 sin2 θ

1/y2

1/y2

cos2 θ F ′

−F ′ cos2 θ sin2 ψ

.

(38)

Then the DBI action is

SDBI/T5 = −2πTV

∫dydψ

√cos4 θ sin2 ψ + F ′2

y4L(σ). (39)

7

0 20 40 60 80 100y

5

6

7

8

9

10

1e 11+6.2831853071= 0.0= 0.1= 0.2= 0.3= 0.4

Figure 6: ω dependence of θ

0 20 40 60 80 100y

0

2

4

6

8

10

x 3

= 0.0= 0.1= 0.2= 0.3= 0.4

Figure 7: ω dependence of x3

0 20 40 60 80 100y

0.04

0.02

0.00

0.02

0.04

g

= 0.0= 0.1= 0.2= 0.3= 0.4

Figure 8: ω dependence of g

The WZ term is

SWZ/T5 = −2πTV

∫dψdy

F ′(ψ)x′3y4

. (40)

Summing them the action is

SD5/T5 = −2πTV

∫dψdy

y4

[√cos4 θ sin2 ψ + F ′2L(y) + x′3F

′]. (41)

We can determine the gauge flux in the same way as found in eq.(14). Substitutingthe gauge flux (36), the action is

SD5/T5 = −2πTV

∫dy

y4(√

cos4 θ + κ2L(σ)− κx′3), (42)

L(σ) =

√(y′2h

+ x′23 + y2θ′2)

(h− ω2y2 sin2 θ) + hy2g′2 sin2 θ.

3.1 Equations of motion

For notational convenience, we define

Θ :=√

cos4 θ + κ2, S :=y′2

h+ x′23 + y2θ′2, Ω := h− ω2y2 sin2 θ. (43)

In this notation the action is

SD5 ∼∫dσ

ΘL− κvy4

, L =

√SΩ + hg′2y2 sin2 θ. (44)

8

The equations of motion for y,θ,x3 and g are

d

dσ

(Θ

y4(y′/h)Ω

L

)+

4

y5(ΘL− κv)− Θ

y4L(yθ′2Ω + y(hg′2 − ωS) sin2 θ)

− Θ

y4L

(− y′2Ω

h2+ S + y2g′2 sin2 θ

)∂h = 0, (45a)

d

dσ

(Θ

y4θ′y2Ω

L

)+

2 sin θ cos3 θ

Θ

L

y4− Θ

y4y2(hu2 − ω2S) sin θ cos θ

L= 0, (45b)

d

dσ

(Θ

y4x′3Ω

L− κ

y4

)= 0, (45c)

d

dσ

(Θ

y4hy2 sin2 θ

Lg′)

= 0. (45d)

We choose the gauge L/Θ = 1, The equations of motion become

d

dσ

(y′h

Ω

y4

)+

4

y5(Θ2 − κv)− 1

y3(θ′2Ω + (hg′2 − ωS) sin2 θ)

− 1

y4

(− y′2Ω

h2+ S + y2g′2 sin2 θ

)∂h = 0, (46a)

d

dσ

(θ′

Ω

y2

)+

2 sin θ cos3 θ

y4− (hu2 − ω2S) sin θ cos θ

y2= 0, (46b)

d

dσ

(x′3Ωy4− κ

y4

)= 0, (46c)

d

dσ

(h sin2 θ

y2g′)

= 0. (46d)

We define the following:

A := − d

dσlog Ω =

2ω2y sin θ(y′ sin θ + yθ′ cos θ) + 4r4Hy3y′

Ω, (47a)

B :=4h

yΩ(κv −Θ2), C :=

hg′2 − ωSΩ

, D := −2 sin θ cos3 θ

y2Ω, (47b)

E :=h(S + y2g′2 sin2 θ)

Ω∂h. (47c)

By solving the equations for the second derivative terms,

y′′ =4y′2

y+ hyθ′2 + y′A+B + Chy sin2 θ + E, (48a)

θ′′ =2θ′y′

y+ θ′A+D + C sin θ cos θ, (48b)

x′′3 =4x′3y

′

y− 4κy′

yΩ+ x′3A, (48c)

g′′ = −2θ′g′ cot θ +2y′g′

y− y′g′∂h

h. (48d)

Boundary condition The boundary condition is, since we fixed Θ/L = 1, iny → 0 limit,√

cos4 θ0 + κ2

(y′2

h + x′23 + y2θ′2)(h− ω2y2 sin2 θ) + hy2g′2 sin2 θ=

√cos4 θ0 + κ2

y′2 + x′23= 1. (49)

From the third equation of motion (48c)

4y′

yΩ(x′3(h− ω2y2 sin2 θ)− κ) ≈ 4y′

yΩ(x′3 − κ). (50)

9

From equations (49) and (50) we find the conditions,

x′3(0) = κ, y′ = cos2 θ0. (51)

We impose the Neumann condition θ′(0) = 0 at the AdS boundary. The forth equation(48d) gives

2g′

y sin θ(y′ sin θ − yθ′ cos θ) ≈ 2g′ cos θ0

y sin θ(cos θ0 sin θ0 − yθ′), g′(0) = 0. (52)

Summarizing the appropriate boundary condition is

y(0) = 0, θ(0) = θ0, x3(0) ≈ κσ, g(0) = 0, (53a)

y′(0) = cos2 θ0, θ′(0) = 0, x′3(0) = κ, g′(0) = 0. (53b)

3.2 Solution

We solve the equations of motion (48) under the boundary condition (53). The resultsof the numerical calculation are shown in the next seven figures.

Figure 9 shows the behavior of x3 as a function of y for different gauge fluxes. Inthis plot the angular momentum is ω = 0.5 and the mass is m = 1 as one can see thereflection point at the horizon y = yh. For large values of κ the curve becomes smoothat the peak of y.

Figure 10 shows the behavior of x3 for different masses. For m = 0 the relation ofy and x3 is linear. For non-zero mass the curve bends in the vicinity of the horizon.Since the equations of motion are not explicitly dependent on x3, there are similarsolutions which are parallel translated along x3 direction.

As we can see in Figure 11 and Figure 12, θ and g do not grow as the same for theprevious section.

Since θ = 0 is the solution, in the same way as before, there is no angular mo-mentum dependence as shown in the next three figures (see Figure 13, Figure 14 andFigure 15).

In κ = 0 case we choose the gauge σ = y. Since identically θ = 0, each factorbecome L =

√1 + x′23 h, Θ =

√1 + κ2 and Ω = h. The equations (45) are simplified

asd

dy

(x′3h√

1 + κ2

y4L

)= 0. (54)

We can see that x′3 = 0 is the solution. This is the unique solution which can penetratethe horizon. In this case the action integrated over the Wheeler DeWitt patch is simply,SD5 ∼

∫∞yhdy/y4. The form of the D5-brane in (x3, y)-plane is depicted in Figure 16.

4 Discussion

In the plots of the solution we found the behavior of a moving interface in subspaceS5 of AdS5×S5 black holes. In this paper, we found two things: The first is nontrans-parency of the horizon and the second is the independence of the angular momentumin S5 subspace. Let us discuss them in detail.

The D5-brane does not penetrate the horizon. In complexity - volume [14, 30, 31]and complexity - action [24, 23] conjecture, we need to consider the surface or regionwhich extends to the interior of the horizon. Especially the growth rate of the actionis calculated from the integration inside the horizon. We want to find the effect of thenon-local moving object such as [29] to black hole complexity. For this reason, we werelooking for the case the probe brane penetrates the horizon. In [26, 27] we found thatthe static probe brane does not penetrate the horizon. Then we tried to find a time

10

0.0 0.2 0.4 0.6 0.8 1.0y

0.0

0.2

0.4

0.6

0.8

1.0

x 3= 0.2= 0.3= 0.4= 0.5= 0.6

Figure 9: κ dependence of x3

0.0 0.2 0.4 0.6 0.8 1.0y

0.0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

x 3

m = 0.0m = 1.0m = 2.0m = 3.0m = 4.0

Figure 10: Mass dependence of x3

0.00 0.25 0.50 0.75 1.00 1.25 1.50 1.75 2.00y

7.4

7.6

7.8

8.0

8.2

8.4

1e 11+6.2831853071

m = 0.0m = 1.0m = 2.0m = 3.0m = 4.0

Figure 11: Mass dependence of θ

0.00 0.25 0.50 0.75 1.00 1.25 1.50 1.75 2.00y

0.04

0.02

0.00

0.02

0.04

g

m = 0.0m = 1.0m = 2.0m = 3.0m = 4.0

Figure 12: Mass dependence of g

dependent solution in this paper. According to the analysis, the probe D5-brane doesnot extend the horizon in the same way as the static case. We restricted the motionof the D5-brane in S5 subspace. Then we found that the motion in S5 part does notaffect the embedding in AdS bulk. In order to find the probe brane which affects CAconjecture, we may need to consider a system which has the motion in AdS bulk.

The second accomplishment is the discovery of the independence of the angularmomentum in S5. That was due to the trivial solution for longitude coordinate θ = 0.This result says the behavior of the probe brane in the AdS5 part is decoupled fromthe motion in the S5 subspace.

For this result, the interpretation in CFT side has another meaning because theaction in AdS spacetime leads the drag force in the boundary theory [3]. In our casethere are two interfaces on the boundary. The gauge theory on the each side of theinterface has gauge group SU(N) and SU(N − k).

In this work we focus on the motion restricted on a subspace of S5. For future work,we are interested in time dependent solutions moving on the AdS subspace. We expectto find a solution which can exist in the horizon for non-zero gauge flux. This casegives non-trivial action which includes the gauge flux. This will be a generalization of(4.13) of [32] which includes the nonzero gauge flux. This gauge flux give us a usefulway to compare the holographic quantities as stated in introduction. Then, if we findthe such time dependent solution, this give a good example to test CA conjecture forAdS black holes.

11

0.0 0.2 0.4 0.6 0.8 1.0y

0.0

0.2

0.4

0.6

0.8

1.0x 3

= 0.0= 0.1= 0.2= 0.3= 0.4

Figure 13: ω dependence of x3

0.0 0.2 0.4 0.6 0.8 1.0y

7.4

7.6

7.8

8.0

8.2

8.4

1e 11+6.2831853071

= 0.0= 0.1= 0.2= 0.3= 0.4

Figure 14: ω dependence of θ

0.0 0.2 0.4 0.6 0.8 1.0y

0.04

0.02

0.00

0.02

0.04

g

= 0.0= 0.1= 0.2= 0.3= 0.4

Figure 15: ω dependence of g

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Horizon

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Figure 16: Form of the D5-brane

Acknnowledgments

I would like to thank Sung-Soo Kim and Satoshi Yamaguchi for helpful discussion.

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