Miami 2006 AdS/CFT Correspondence with Heat Conduction ...

Miami 2006

AdS/CFT Correspondence with Heat Conduction

GEORGE SIOPSIS

Department of Physics and AstronomyThe University of Tennessee, Knoxville

December 14, 2006

⇒

J. Alsup, C. Middleton and G. S., hep-th/0607139

AdS/CFT Correspondence with Heat Conduction 1

STRONG FORCE: Quantum Chromodynamics (QCD)

u u u

p ndd

d

nucleons consist of quarks.

Quarks and gluons have colorsI cannot be free (directly detected)Only color neutral objects can be “seen”(nucleons, mesons, etc)Scale:

ΛQCD = 0.236 GeV

g

du

u d

interaction via gluons.F asymptotic freedom

At high energies: coupling constant decreasesI spin-1 charged bosons (gluons) make vacuum paramagnetic

αs(E) =2π

(11− 23Nf) ln(E/ΛQCD)

[Politzer; Gross and Wilczek - Nobel Prize 2004] Nf : number of quarks.

George Siopsis Miami - 12/14/06


“A second unexpected connection comes from studies carried out using the Relativistic HeavyIon Collider, a particle accelerator at Brookhaven National Laboratory. This machine smashestogether nuclei at high energy to produce a hot, strongly interacting plasma. Physicists havefound that some of the properties of this plasma are better modeled (via duality) as a tiny blackhole in a space with extra dimensions than as the expected clump of elementary particles inthe usual four dimensions of spacetime. The prediction here is again not a sharp one, as thestring model works much better than expected. String-theory skeptics could take the point ofview that it is just a mathematical spinoff. However, one of the repeated lessons of physics isunity - nature uses a small number of principles in diverse ways. And so the quantum gravitythat is manifesting itself in dual form at Brookhaven is likely to be the same one that operateseverywhere else in the universe.”

– Joe Polchinski

AuAufireball



AdS/CFT correspondence and hydrodynamics[Policastro, Son and Starinets]

correspondence between N = 4 SYM in the large N limit and type-IIB stringtheory in AdS5 × S5.

I in strong coupling limit of field theory, string theory is reduced to classi-cal supergravity, which allows one to calculate all field-theory correlationfunctions.→ nontrivial prediction of gauge theory/gravity correspondence

entropy of N = 4 SYM theory in the limit of large ’t Hooft coupling is precisely3/4 the value in zero coupling limit.long-distance, low-frequency behavior of any interacting theory at finite tem-perature must be described by fluid mechanics (hydrodynamics).universality: hydrodynamics implies very precise constraints on correlation func-tions of conserved currents and stress-energy tensor:I correlators fixed once a few transport coefficients are known.



Hydrodynamicsconserved current: jµ

chemical potential µ = 0, so in thermal equilibrium

〈j0〉 = 0

retarded thermal Green function

GRµν(ω, q) = −i

∫d4x e−iq·x θ(t)〈[jµ(x), jν(0)]〉 ,

where q = (ω, q), x = (t, x)

I determines response to a small external source coupled to the current.ω and q small:

• external perturbation varies slowly in space and time

• macroscopic hydrodynamic description for its evolution is possible.

diffusion equation

∂0j0 = D∇2j0 ,

where D is a diffusion constant with dimension of length.



⇒ overdamped mode, dispersion relation

ω = −iDq2 ,

pole at ω = −iDq2 in the complex ω-plane, in the retarded correlation func-tions of j0

stress-energy tensor Tµν

∂0T00 + ∂iT0i = 0 ,

∂0T0i + ∂jTij = 0 ,

where

T00 = T00 − ρ, ρ = 〈T00〉 ,T ij = T ij − pδij = − 1

ρ + p

[η

(∂iT

0j + ∂jT0i − 2

3δij∂kT0k

)+ζδij∂kT0k

],

ρ (p): energy density (pressure)η (ζ): shear (bulk) viscosity.two types of eigenmodes:

• the shear modes - transverse fluctuations of momentum density T0i, with



a purely imaginary eigenvalue

ω = −iDq2 , D =η

ρ + p,

• sound wave - simultaneous fluctuation of energy density T00 and longitu-dinal component of momentum density T0i, with dispersion relation

ω = usq − i

2

1

ρ + p

(ζ +

4

3η

)q2 , u2

s =∂p

∂ρ.

conformal theory ⇒ stress-energy tensor is traceless, so

ρ = 3p , ζ = 0 , us =1√3



GravityThe non-extremal 3-brane background is a solution of type-IIB low energyequations of motion.In the near-horizon limit r ¿ R, the metric becomes

ds210 =(πTR)2

u

(−f(u)dt2 + dx2 + dy2 + dz2

)+

R2

4u2f(u)du2+R2dΩ2

5 ,

where T = r0πR2 is Hawking temperature, u =

r20r2

, f(u) = 1− u2.

The horizon corresponds to u = 1, spatial infinity to u = 0.gauge theory/gravity correspondence:

• background metric with non-extremality parameter r0 is dual to N = 4SU(N) SYM at finite temperature T in the limit of N →∞, g2

Y MN →∞.



Thermal R-current correlators5d Maxwell action

S = − 1

4g2SG

∫d5x

√−g F aµνFµν a , , g2

SG =16π2R

N2

Fix gauge

Au = 0

residual gauge transformations

Aµ → Aµ + ∂µΛ

where Λ is independent of u.Fourier decomposition (momentum in z-direction)

Aµ =∫

d4q

(2π)4e−iωt+iqzAµ(q, u) .

Dimensionless energy and momentum,

w =ω

2πT, q =

q

2πT,



5d Maxwell equations,

1√−g∂ν[√−ggµρgνσ(∂ρAσ − ∂σAρ)] = 0 ,

reduce to ordinary differential equations

wA′t + qf A′

z = 0 ,

A′′t −

1

uf

(q2At + wqAz

)= 0 ,

A′′z +

f ′

fA′

z +1

uf2

(w2Az + wqAt

)= 0 ,

A′′α +

f ′

fA′

α +1

uf

(w2

f− q2

)Aα = 0 , α = x, y

prime denotes derivative w.r.t. u.I invariant under residual gauge transformations

At → At − ωΛ , Az → Az + qΛ

Deduce

Az =uf

wqA′′

t −qw

At ,



A′′′t +

(uf)′

ufA′′

t +w2 − q2f(u)

uf2A′

t = 0 ,

2nd-order differential equation for A′t. Define

A′t = (1− u)νF (u) , ν = ±iw/2

“incoming wave” boundary condition at horizon singles out ν = −iw/2.

Equation for F (u):

F ′′ +(

1− 3u2

uf+

iw1− u

)F ′ +

iw(1 + 2u)

2ufF +

w2[4− u(1 + u)2]

4uf2F − q2

ufF = 0 .

In long-wavelength, low-frequency limit, w and q are small,

⇒ solution double series in w and q2,

F (u) = F0 + wF1 + q2G1 + w2F2 + wq2H1 + q4G2 + · · · .

F0 = C , F1 =iC

2ln

2u2

1 + u, G1 = C ln

1 + u

2u.



take limit u → 0 assuming boundary conditions

limu→0

At(u) = A0t , lim

u→0Az(u) = A0

z .

C =q2A0

t + wqA0z

iw− q2

C is invariant under the residual gauge transformations.Similarly, we find

Aα =8A0

a(1− u)−iw/2

8− 2iw ln 2 + π2q2

[1 +

iw2

ln1 + u

2

+q2

2

(π2

12+ Li2(−u) + lnu ln(1 + u) + Li2(1− u)

)]+ O

(w2, q4,wq2

).



Green functions

Gabxx = Gab

yy = −iN2Tω δab

16π+ · · · ,

Gabtt =

N2Tq2 δab

16π(iω −Dq2)+ · · · ,

Gabtz = Gab

zt = − N2Tωq δab

16π(iω −Dq2)+ · · · ,

Gabzz =

N2Tω2 δab

16π(iω −Dq2)+ · · · ,

where

D =1

2πT.

correlation functions of j0 and jz contain diffusion poleI expected from hydrodynamics.



compare with weak coupling:I diffusion constant ∝ mean free path

D ∼ 1

(g2YMN)2T ln 1

(g2YMN)

, g2YMN ¿ 1 .

D as a function of the ’t Hooft coupling g2YMN is

D = fD(g2YMN)

1

T,

where fD(x) ∼ 1−x2 lnx

for x ¿ 1 and fD(x) = 12π for x À 1.



Shear Modecompute two-point function of stress-energy tensor in the boundary theoryI consider small perturbation of metric

gµν = g(0)µν + hµν

where g(0)µν is given by

ds25 =π2T2R2

u

(−f(u)dt2 + dx2

)+

R2

4f(u)u2du2 .

Einstein equations:

Rµν = R(0)µν +R(1)

µν + · · · = 2Λ

3gµν ,Λ =

−6

R2

To first order in hµν,

R(1)µν = − 4

R2hµν .

assume perturbation ∼ e−iωt+iqz



fix gauge

huµ = 0

3 classes of perturbations:

• hxy 6= 0, or hxx = −hyy 6= 0;

• hxt and hxz 6= 0, or hyt and hyz 6= 0;

• htz and all diagonal elements of hµν are nonzero, and hxx = hyy

(sound wave).



Perturbation with hxy 6= 0

h′′xy +1− 3u2

ufh′xy +

1

(2πT )2f2u

(f

∂2hxy

∂z2− ∂2hxy

∂t2

)− 1 + u2

fu2hxy = 0

Introduce φ =uhxy

(πTR)2= hx

y

Fourier component φk(u) satisfies eq. for minimally coupled scalar

φ′′k −1 + u2

ufφ′k +

w2 − q2f

uf2φk = 0 .

For incoming wave at horizon,

φk(u) = (1− u)−iw/2Fk(u) ,

where Fk(u) is regular at u = 1 and can be written as a series

Fk(u) = 1− iw2

ln1 + u

2+

w2

8

[(ln

1 + u

2+ 8(1− q2

w2)

)ln

1 + u

2−4 Li2

1− u

2

]+O(w3) ,

retarded Green function

Gµν,λρ(ω, q) = −i∫

d4x e−iq·x θ(t)〈[Tµν(x), Tλρ(0)]〉 .



Deduce

Gxy,xy(ω, q) = −N2T2

16

(i2πTω + q2

).

shear viscosity of strongly coupled N = 4 SYM plasma (Kubo formula)

η = limω→0

1

2ω

∫dt dx eiωt 〈[Txy(x), Txy(0)]〉 =

π

8N2T3 .

NEXT: Calculation involving different components leads to same result!



Perturbation with htx 6= 0, hxz 6= 0

In this channel correlation functions have diffusion pole.

Einstein equations for Fourier components of Ht = uhtx(πTR)2

, Hz = uhzx(πTR)2

:

H ′t +

qfw

H ′z = 0 ,

H ′′t −

1

uH ′

t −wquf

Hz − q2

ufHt = 0 ,

H ′′z −

1 + u2

ufH ′

z +w2

uf2Hz +

wquf2

Ht = 0 .

Deduce

Hz =uf

wqH ′′

t −f

wqH ′

t −qw

Ht .

H ′′′t −

2u

fH ′′

t +2uf − q2f + w2

uf2H ′

t = 0 .

In the low-frequency, long-wavelength limit, solve as before...



H ′t(u) = (1− u)−iw/2G(u)

G′′ −(

2u

f− iw

1− u

)G′ +

1

f

(2 +

iw2− q2

u+

w2[4− u(1 + u)2]

4uf

)G = 0 .

G(u) = C

[u− iw

(1− u− u

2ln

1 + u

2

)+

q2(1− u)

2

]+ O

(w2,wq2, q4

).

C =q2H0

t + qwH0z

iw− q2

2

.

Deduce correlators

Gtx,tx(ω, q) =N2πT 3q2

8(iω −Dq2)+O(w2,wq2, q4) ,

Gtx,xz(ω, q) = − N2πT 3ωq

8(iω −Dq2)+O(w2,wq2, q4) ,

Gxz,xz(ω, q) =N2πT 3ω2

8(iω −Dq2)+O(w2,wq2, q4) ,

where D = 14πT

I half the diffusion constant for the R-charge.



Deduce η:I recall from hydrodynamics D = η

ρ+p.

Entropy:

s =3

4s0 =

π2

2N2T3 ,

where s0 is entropy at zero coupling.

From s = ∂P∂T , ρ = 3p, deduce ρ + p = π2

2 N2T4, ∴

η =π

8N2T3 ,

η

s=

1

4π

I agrees with Kubo formula.

I no agreement unless s = 34s0.



behavior of η as a function of the ’t Hooft coupling

η = fη(g2YMN)N2T3

where fη(x) ∼ 1−x2 lnx

for x ¿ 1 and fη(x) = π8 for x À 1.

I At weak coupling,

η

sÀ 1

4π



Chemical potential3-Charged black hole:

ds2 = −H−2/3f(r)dt2 + H1/3(

dr2

f(r)+ r2dx2

)

where

f(r) = −m

r2+ r2H , H = (1 + q21/r2)(1 + q22/r2)(1 + q23/r2)

[Behrnd, Cvetic and Sabra]

I interesting thermodynamic properties (Cardy-Verlinde entropy formula, etc)[Klemm, Petkou, G.S. and Zanon]

Entropy density:

s =N2r3H2π

√H(rH)

Chemical potentials (i = 1,2,3)

µi =

√2qi

1 + qi/r2H

√H(rH)



Kubo formula for shear viscosity

η =N2r3H8π2

√H(rH)

so η/s = 14π !

I also agrees with η from diffusion pole![Son and Starinets]

For single charge (q2 = q3 = 0), Kubo formula for thermal conductivity:

κT = πN2T2H

(1 + q1/r2H)2

(2 + q1/r2H)q1/r2H

Wiedemann-Franz law:

κTµ2

ηTH= 8π2



AdS/CFT Correspondence with Heat ConductionGOAL: Understand flowI static black holes not sufficientI try Bjorken flowPerfect Fluid [Janik and Peschanski]

introduce proper time τ and rapidity y on longitudinal plane,

x0 = τ cosh y , x1 = τ sinh y

transverse coordinates: x⊥ = (x2, x3).

Let z = R2/r denote the fifth dimension.Einstein equations in the bulk

RAB −1

2gABR− 6gAB = 0

where the cosmological constant is Λ = −6.I metric ansatz

ds2 =1

z2

[−ea(τ,z)dτ2 + τ2eb(τ,z)dy2 + ec(τ,z)dx⊥2 + dz2

]



most general bulk metric obeying

• boost invariance

• symmetry under reflection in the longitudinal direction (y → −y)

• translational and rotational invariance.

introduce coordinate

v =z

τη

leading behavior of solution in the τ →∞ limit

a(v) = A(v)− 2m(v)b(v) = A(v) + 2(4η − 1)m(v)c(v) = A(v) + 2(1− 2η)m(v)

where

A(v) =1

2

[ln(1 + ∆(η)v4) + ln(1−∆(η)v4)

]

m(v) =1

4∆(η)

[ln(1 + ∆(η)v4)− ln(1−∆(η)v4)

]



with

∆(η) =

√1

3(6η2 − 4η + 1)

Demand regularity in the bulk,

I R2 = RµναβRµναβ non-singular

I unique value

η =1

3Holographic Renormalizationrelate bulk metric to the vacuum expectation value of the stress-energy tensorof the corresponding gauge theory on the boundaryI metric is of the general asymptotically AdS form

ds2 =gµνdxµdxν + dz2

z2

Near the boundary at z = 0 expand

gµν = g(0)µν + z2g

(2)µν + z4g

(4)µν + . . .



where g(0)µν = ηµν, g

(2)µν = 0,

〈Tµν〉 ∝ g(4)µν

from explicit form of g(4)µν , for η = 1/3,

⇒ perfect fluid

Tαβ = (ρ + p)uαuβ + pηαβ

equation of state p = 13ρ

∵ tracelessness due to conformal invariance

mass density and temperature fall off, respectively, as

ρ ∼ 1

τ4/3, T ∼ 1

τ1/3



An extension of the bulk metric [Alsup, Middleton, G.S.]

I relax requirement of invariance under reflection in longitudinal direction.I add appropriate off-diagonal terms in the bulk metric

ds2 = ds2perfect fluid+2τλh||(v)dτdy + 2τσh⊥(v)dτdx⊥

• exponents λ, σ < 0 of τ

• 3 functions: 1 longitudinal h||(v), 2 transverse h⊥(v) = (h2(v), h3(v))

to be determined by the Einstein equations:

¦ mixed transverse and two yx⊥ components are, respectively,

(vh′2 + (2− vc′)h2)(vh′3 + (2− vc′)h3) = 0

(vh′|| + (2− vb′)h||)(vh′⊥ + (2− vc′)h⊥) = 0

exact (no higher-order corrections)! solution:

h⊥(v) =ec(v)

v2

up to overall arbitrary multiplicative constant→ singular function (h⊥(v) ∼ v−2 as v → 0)



→ discard

h⊥(v) = 0

¦ two x⊥τ components yield

(2vh′′2 + (2 + vb′ − vc′)h′2)h3 − (2 ↔ 3) = 0

→ trivially solved.→ unnecessary to invoke regularity; condition of isotropy (h2 = h3) suf-

fices.

¦ vy and yτ components are, respectively,

a0(v)h||(v) + a1(v)h′||(v) + h′′||(v) +O(τ−2+4η+2λ) = 0

b0(v)h||(v) + b1(v)h′||(v) + h′′||(v) +O(τ−2+4η+2λ) = 0



where

a0 = −2(1 + λ)

ηv2+−a′ + 1+λ

ηb′ + 2c′

v+

1

2(a′ − b′ − 2c′)b′ − b′′

a1 =3η − λ− 1

ηv− 1

2(a′ + b′ − 2c′)

b0 = − 4

v2+

2(a′ + b′ + 4c′)v

− 1

2(a′2 − a′b′ + b′2)− a′c′ − b′c′ − 3

2c′2 − a′′ − b′′ − 2c′′

b1 =1

v2− 1

2(a′ + b′ − 2c′)

For consistency (a1 = b1), choose

λ = 2η − 1

for consistency of the expansion, we need

η <1

2

[for which τ has negative exponent, −2 + 4η + 2λ = −4 + 8η < 0]We need to ensure a0 = b0, as well.I To this end, we need to determine a, b, c from:



¦ diagonal components

6b′ − vb′2 + 12c′ − 2vb′c′ − 3vc′2 − 2vb′′ − 4vc′′ +O(τ−2+2η, τ−4+8η) = 0−6a′ + va′2 − 12c′ + 2va′c′ + 3vc′2 + 2va′′ + 4vc′′ +O(τ−2+2η, τ−4+8η) = 0

−6a′ + va′2 − 6b′ + va′b′ + vb′2 − 6c′ + va′c′ + vb′c′ + vc′2 + 2va′′ + 2vb′′ + 2vc′′

+O(τ−2+2η, τ−4+8η) = 0−6a′ − 6b′ + va′b′ − 12c′ + 2va′c′ + 2vb′c′ + vc′2 +O(τ−2+2η, τ−4+8η) = 0

¦ and vτ component,

2a′−6ηa′−2b′−4ηb′+vηb′2−8ηc′+2vηb′c′+3vηc′2+2vηb′′+4vηc′′+O(τ−2+2η, τ−4+8η) = 0

I To leading order, identical to perfect fluid case⇒ same a, b, c (satisfy a0 = b0).

¦ rest of eqs. coalesce, yielding an ODE which uniquely determines h||(v) atleading order,

h||(v) =1

v2

[1−∆v4

1 + ∆v4

](1

2−η)/∆ (

1−∆2v8)1

2− η

∆

[(1−∆v4)2η/∆ − (1 + ∆v4)2η/∆

]

we discarded singular solution (behaving ∼ v−2 as v → 0).expand

h||(v) ∼ v2 + (2η − 1)v6 + . . .



up to an irrelevant overall constant factor.For η = 1/3, reduces to

h||(v) =v2

1 + 13v4

special value of η derived by demanding regularity of R2 = RµναβRµναβ

- unchangedFinally, our metric reads

ds2 =1

z2

[−

(1− e

3z4

τ 4/3

)2

1 + e3

z4

τ 4/3

dτ2 +(1 + e

3z4

τ 4/3

)(τ2dy2 + dx2

⊥) + dz2 +2Az4

1 + e3

z4

τ 4/3

dydτ

τ

]

A is an arbitrary constant characterizing the departure from the symmetricstate giving rise to perfect fluid hyrdodynamics on the boundary.



HydrodynamicsFrom holographic renormalization

〈Tyτ〉 ∼ g(4)yτ ∼ 1

τ

In order to understand how our solution relates to the gauge theory fluid,choose stress-energy tensor with arbitrary energy flux,

Tµν =

T ττ T τy T τ2 T τ3

T τy T yy 0 0T τ2 0 T22 0T τ3 0 0 T33

.

Minkowski space metric

ds24 = −dτ2 + τ2dy2 + (dx⊥)2

From local conservation law

∇αTαβ = ∂αTαβ + ΓααλTλβ + Γβ

αλTαλ = 0

where Γyyτ = 1

τ = Γyτy and Γτ

yy = τ ,



♥ for β = τ ,

∂τT ττ + ∂yT τy +1

τT ττ + τT yy + ∂⊥T τ⊥ = 0

♥ for β = y,

∂τT τy + ∂yT yy +3

τT τy = 0

♥ for transverse components, β = i (i = 2,3),

∂τT τi +1

τT τi + ∂jT

ij = 0

♥ from conformal invariance (tracelessness)

−T ττ + τ2T yy + T22 + T33 = 0

match with bulk metric through holographic renormalization:

→ stress-energy tensor to the order we are considering should not dependon rapidity (y) or transverse coordinates (x⊥).

→ Solution:



T τy =Cτ3

, Tτy =−Cτ

T τ⊥ =C⊥τ

, Tτ⊥ =−C⊥

τ

C, C⊥ = (C2, C3) arbitrary constants.I matches prediction of gravity dual for T τy.I stress-energy tensor in terms of ρ = T ττ (energy density) and the arbitraryconstants C, C⊥ (determining energy flux),

Tµν =

ρ Cτ3

C2τ

C3τC

τ3 −1τ ∂τρ− 1

τ2ρ 0 0C2τ 0 ρ + 1

2τ∂τρ 0C3τ 0 0 ρ + 1

2τ∂τρ

.

demanding isotropy, τ2T yy = T22 = T33,

ρ ∼ 1

τ4/3

matching perfect fluid behavior.

I equation of state also unchanged, p = 13ρ.



For the energy flux, we obtain velocity components

uy ∼ τT τy

T ττ∼ 1

τ2/3, u⊥ ∼ T τ⊥

T ττ∼ τ1/3

transverse components grow with time (unstable solution). ∴ ought to set

C⊥ = 0

matches expected behavior from the gravity dual

I corresponding off-diagonal components of the metric were set to zero bydemanding regularity of the solution to the Einstein equations.

Summarizing,

T ττ =B

τ4/3, Tττ =

Bτ4/3

T yy =B

3τ10/3, Tyy =

B3

τ2/3

T ii =B

3τ4/3, Tii =

B3τ4/3

(i = 2,3)

T τy =Cτ3

, Tτy =−Cτ



To gain further insight into the nature of this fluid, consider the general caseof a dissipative relativistic fluid with stress-energy tensor and particle current,respectively,

Tαβ = (ρ + p)uαuβ + pηαβ + tαβ , Jα = nuα + jα

n is the particle density (for a perfect fluid, Jα = nuα)dissipative pieces satisfy

uαtαβ = uαjα = 0

Standard thermodynamic arguments using the relations

ρ + p = Ts + µn , dρ = Tds + µdn

lead to the general expressions

tαβ = −η[∇αuβ + uαuγ∇γuβ + (α ↔ β)]−(ζ − 2

3η

)[ηαβ + uαuβ]∇γuγ

jα = −κ(∂α + uαuβ∂β)µ

T

in terms of the coefficient of thermal conductivity κ and shear and bulk viscosi-ties, η and ζ, respectively.



Assuming no particle transport, ~J = ~0 (vectors in 3d),

n~u = κ(~∇+ ~uuβ∂β)µ

T

This can be solved as an expansion in the derivatives of µ/T ,

~u =κn

~∇µ

T+ . . .

justified because local velocities are small (|~u| ¿ 1).To match the desired behavior of the stress-energy tensor, assume no viscosity,

η = ζ = 0

energy flow in the longitudinal direction, ∴

u⊥ = 0

energy flux in the longitudinal direction

T τy = (ρ + p)uτuy

to leading order,

T τy =4

3ρuy + · · · = 4κρ

3nτ2∂y

µ

T+ . . .



gradient of chemical potential to leading order,

∂yµ

T=

3nC4κBτ1/3 + . . .

from the Gibbs-Duhem relation

dp = sdT + ndµ

we obtain the temperature gradient

∂yµ

T=

ρ + p

n∂y

1

T=

4B3nτ4/3

∂y1

T

∴ the temperature gradient is

∂y1

T=

9n2C16κB2

τ5/3 + . . .

energy flux

Tτy = κT∂yT

where κT is the standard definition of thermal conductivity in non-relativisticmechanics.



We obtain

κT = κ(

ρ + p

nT

)2

To leading order,

κT =16κB2

9n2T2τ8/3

¦ ¦ ¦I to add viscosity consider next perturbative order.

[Nakamura and Sin; Janik]

I Elliptic flow is also possible through modified metric ansatz.[Sin, Nakamura and Kim]



CONCLUSIONS

• We are beginning to understand non-perturbative QCD effects using stringtheory

• RHIC’s fireball can be described by a dual black hole

• RHIC and LHC may probe black holes and provide information on stringtheory as well as non-perturbative QCD effects.


Miami 2006 AdS/CFT Correspondence with Heat Conduction ...

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