Ads/CFT Correspondence

AdS/CFT Correspondence

GEORGE SIOPSIS

Department of Physics and Astronomy The University of Tennessee Knoxville, TN 37996-1200 U.S.A. e-mail: [email protected]

Notes by James Kettner

Fall 2010 ii Contents

1 Path Integral 1 1.1 Quantum Mechanics ...... 1 1.2 Statistical Mechanics ...... 3 1.3 Thermodynamics ...... 4 1.3.1 Canonical ensemble ...... 4 1.3.2 Microcanonical ensemble ...... 4 1.4 Field Theory ...... 4

2 Schwarzschild Black Hole 7

3 Reissner-Nordstrom¨ Black Holes 11 3.1 The holes ...... 11 3.2 Extremal limit ...... 13 3.3 Generalizations ...... 15 3.3.1 Multi-center solution ...... 15 3.3.2 Arbitrary dimension ...... 16 3.3.3 Kaluza-Klein reduction ...... 17

4 Black Branes from String Theory 19

5 Microscopic calculation of Entropy and Hawking Radiation 23 5.1 The hole ...... 23 5.2 Entropy ...... 25 5.3 Finite temperature ...... 27 5.4 Scattering and Hawking radiation ...... 29

iii iv CONTENTS LECTURE 1

Path Integral

1.1 Quantum Mechanics

Set ~ = 1. Given a particle described by coordinates (q, t) with initial and ﬁnal coordinates of (qi, ti) and (qf , tf ), the amplitude of the transition is

∗ hqi, ti | qf , tf i = hψ |qf , tf i = ψ (qf , tf ) . where ψ satisﬁes the Schrodinger¨ equation

∂ψ 1 ∂2ψ i = − . ∂t 2m ∂q2

Feynman introduced the path integral formalism Z −iS hqi, ti | qf , tf i = [dq] e q(ti)=qi,q(tf )=qf R where S = dtL is the action and L = L(q, q˙) is the Lagrangian, e.g.,

1 L = mq˙2 − V (q) 2 We convert to imaginary time via the Wick rotation τ = it so

1 L → − mq˙2 − V (q) ≡ −L 2 E where the derivative is with respect to τ, LE is the total energy (which is bounded below),R and the subscript E stands for Euclidean. The action becomes iS → SE = dτLE so the path integral becomes Z [dq] e−SE . 2 LECTURE 1. PATH INTEGRAL

The major contribution to the path integral’s value comes from minimum SE, i.e., δS = 0

i.e., from the classical trajectory q = qcl(τ). Consider a perturbation of the classical trajectory

q(τ) = qcl (τ) + δq (τ) S = S + ... Z cl Z [dq] e−SE = e−Scl [dq] e− ...

−Scl hqi, ti | qf , tf i ≈ e

The correlators are Z −SE hqi,τ i |T [q (τ 1) q (τ 2) ··· ]| qf,τ f i = [dq] q (τ 1) q (τ 2) ··· e q(τ i)=qi,q(τ f )=qf (1.1.1) where T denotes a time ordered product. Notice that the path integral automatically takes care of time ordering. This can easily be seen by splitting the path integral into time intervals bounded by τ 1, τ 2, ... . We are interested in the vacuum expectation values

G(τ 1, τ 2,... ) = h0 |T [q (τ 1) q (τ 2) ··· ]| 0i . (1.1.2) To write them in terms of a path integral, notice that by using the time evolution operator, we have −τ f H |qf , τ f i = e |qf , 0i (1.1.3)

we see that in the limit τ f → ∞, all terms but one vanish. Therefore,

|qf , τ f i → |0i h0 | qf , 0i

Similarly, hqi, τ i| → hqi, 0 | 0i h0| as τ i → −∞. Substituting these limiting expres- sions into equation (1.1.1) gives R −SE h0 |T [q (τ 1) q (τ 2) ··· ]| 0i hqi, 0 | 0i h0 | qf , 0i = [dq] q (τ 1) q (τ 2) ... e R h0 | 0i hq, 0 | 0i h0 | q, 0i = [dq] e−SE R [dq]q(τ )q(τ ) ... e−SE G(τ , τ ,... ) = R 1 2 1 2 [dq]e−SE where we integrate over all trajectories q(τ) with τ ∈ (−∞, +∞) and no speciﬁed end points. 1.2 Statistical Mechanics 3

1.2 Statistical Mechanics

Set the Boltzmann constant kB = 1. Denote temperature by T ≥ 0; At T = 0 the system settles in the ground state. Vacuum expectation values correspond to T = 0. To study ﬁnite T , introduce the partition function X Z = e−En/T states D ¯ ¯ E X ¯ ¯ = n ¯e−H/T ¯ n states X ³ ´ = Tr e−H/T |ni hn| states³ ´ = Tr e−H/T Z D ¯ ¯ E ¯ ¯ = dq q, 0 ¯e−H/T ¯ q, 0 Z = dq hq, 0 | q, 1/T i Z Z = dq [dq0] e−SE 0 0 Z q (0)=q (1/T )=q = [dq] e−SE q(0)=q(1/T )

Therefore, the partition function is given by a path integral over periodic orbits of period 1/T .

1 2 Example 1 Quantum mechanics: consider a free particle V = 0 and L = 2 mq˙ . Then using path integrals r m 2 −m(qf −qi) /2(τ f −τ i) hqi, ti | qf , tf i = e 2π(τ f − τ i)

Example 2 Statistical mechanics: Set τ i = 0, τ f = 1/T , and qi = qf = q. Z r mT Z = dq which is infinite so put thesystem in a box 2π r mT = V where V is the volume (length). 2π which is the same as Z ∞ dp 2 Z = V e−p /2m 2π r−∞ mT = V . 2π 4 LECTURE 1. PATH INTEGRAL

1.3 Thermodynamics

1.3.1 Canonical ensemble Expand the partition function around the classical trajectory to get

Z = e−Scl+ ... = e−F/T

where F is the Helmholtz free energy so

F = SclT.

1 −En/T The probability of a state |ni is pn = Z e . The internal energy is U = hEi X = pnEn states ∂ (ln Z) = T 2 ∂T The entropy is

X ∂ (ln Z) U F S = p ln p = T − ln Z = − n n ∂T T T states

which means F = U − TS.

1.3.2 Microcanonical ensemble The microcanonical ensemble is isolated from its environment. The number of states with a given energy E is given by the multiplicity g(E) = eS(E), i.e., the entropy counts the number of different states. X Z = e−E/T g(E) energy levels

−E0/T ≈ g(E0)e = eS(E0)−E0/T

∂S 1 where E0 maximizes S(E) − E/T . We deduce ∂E = T .

1.4 Field Theory

µ Set the speed of light c = 1. Let x = (t, x) be the positionR 4-vector. Consider a real 4 ﬁeld ϕ(x) with dynamics governed by the actionR S = d xL where L (∂µϕ, ϕ) is the Lagrangian density; the Lagrangian is L = d3xL. Use the Wick rotation τ = it so 1.4 Field Theory 5

2 2 2 2 2 2 ds = −dt + dx = dτ + dx = dsE (Euclidean continuation). The correlation functions or Green functions are

G(x , x ,... ) = h0 |T (ϕ(x )ϕ(x ) ··· )| 0i 1 2 R 1 2 [dϕ] ϕ (x ) ϕ (x ) ··· e−SE = R 1 2 [dϕ] e−SE as before. Taking an arbitrary function J as a source current, the generating functional is given by Z ³ ´ R 4 Z [J] = [dϕ] e−SE + d xJϕ and ¯ ¯ δ δ ¯ G(x1, x2.... ) = ··· Z [J] ¯ (1.4.1) δJ(x1) δJ(x2) J=0 Example 3 For a free ﬁeld of mass m, 1 1 L = (∂ φ)2 + m2φ2 + Jφ E 2 µ 2 The classical ﬁeld equation is the Klein-Gordon equation

2 2 −∇ φcl + m φcl = J

Writing a quantum ﬁeld as φ = φcl + δφ we obtain Z (2) 4 SE = Scl + S (δφ) ,Scl = d LE(∂clµ, φcl)

Notice that there is no linear terms, because it is proportional to the ﬁeld equation. Then Z[J] = e−Scl where we omitted a constant which was independent of J. We have Z 1 S = d4xd4x0J(x)D (x, x0)J(x0) cl 2 E where DE is the propagator,

2 2 0 4 0 (−∇ + m )DE(x, x ) = δ (x − x )

Using (1.4.1), we easily deduce

G(x1, x2) = DE(x1, x2)

You are invited to check that this is still valid when interactions are included. 6 LECTURE 1. PATH INTEGRAL LECTURE 2

Schwarzschild Black Hole

Spacetime is provided with a metric tensor gµν so that a line element has length

2 µ ν ds = gµν dx dx

2 2 2 3 In ﬂat spacetime, ds = −dt + dx (x ∈ R ), so gµν = ηµν = diag(−1 1 1 1) as a matrix. We denote the determinant of gµν by g. The Einstein equations are ½ 1 0 , with just gravity not matter R − Rg = . µν 2 µν source , in the presence of matter

ν µ Rµν is the Ricci tensor (the contracted curvature tensor Rµνρ) and R = Rµ (its trace) 1 is the Ricci scalar. Rµν − 2 Rgµν = 0 arises from the action Z 1 √ S = d4x −gR. 16πG If we take the trace of the Einstein equation in empty space, we get 1 Rµ − Rgµ = 0 µ 2 µ which implies R = 0 so Rµν = 0. Schwarzschild found the solution

dr2 ds2 = −f(r)dt2 + + r2dΩ2 f(r) 2

2 2 2 2 2 where dΩ2 = dθ + sin θdϕ is the line element on the sphere S and 2GM f(r) = 1 − r Where do G and M come from? Compare to electromagnetism with Maxwell’s equations without currents

µν ∂µF = 0. 8 LECTURE 2: Schwarzschild Black Hole

2 Q Its simplest symmetric solution is Aµ = (A0, 0) with ∇ A0 = 0 so A = r , i.e. the Coulomb potential. Q turns out to be the charge by Gauss’s law. In General Relativity, we can read off the mass from

gtt ≈ −1 + 2V (r) if V (r) is small (i.e., when r → ∞), where V (r) is the Newtonian potential (V (r) = GM/r). In the Schwarzschild solution f(r) diverges as r → 0 which is a true singularity. There is also a singularity (which is an artifact of the coordinate system) at r = 2GM ≡ r+ which is called the horizon. While not a coordinate singularity, the horizon is significant because inside (r < r+) not even light√ can escape.p Since f(r+) = 0, proper time for an observer approaching the horizon ( −ds2 = f(r)dt) passes quickly while to a distant observer it appears to take an infinite time to reach the horizon. Now let us define τ = it so Minkowski space becomes Euclidean (ds2 = dτ 2 + dx2) and the Schwarzschild metric becomes ³ r ´ dr2 2 + 2 ¡ ¢ 2 2 ds ≈ 1 − dτ + r+ + r dΩ2. r 1 − r

When r = r+ + ε is outside but close to the horizon (ε > 0),

2 ε 2 r+ 2 2 2 ds ≈ dτ + dε + r+dΩ2. r+ ε Note both ε and τ are completely independent of Ω so the space near the horizon neatly 2 separates into a sphere S of radius r+ and a two-dimensional manifold of metric

2 ε 2 r+ 2 ds2 = dτ + dε r+ ε √ τ To understand this manifold, change coordinates to ρ = 2 r+ε and χ = to get 2r+

2 2 2 2 ds2 = ρ dχ + dρ which is similar to polar coordinates but χ is not restricted to be between 0 and 2π. The resulting spacetime is a cone (the circumference of a closed curve with constant ρ is not 2πρ). We want to eliminate the conical singularity. If it is to be a plane, χ must be between 0 and 2π so τ must be between 0 and 4πr+. Recall that periodic imaginary time is an attribute of statistical systems of temperature T which is the iverse period. Thus the black hole has temperature 1 1 T = = 4πr+ 8πGM which is the Hawking temperature. It looks like we have a statistical system, but what are the states? In thermodynamics, dU = T dS where U is the total energy; in this 1 case, it must be the mass. Thus dM = 8πGM dS which means Z 4πr2 A S = 8πGMdM = 4πGM 2 = + = + 4G 4G Schwarzschild Black Hole 9

where A+ is the area of the horizon; this is the Bekenstein-Hawking formula. It is remarkably universal. Normally entropy is proportional to volume and therefore mass (in an ordinary star with N particles and n degrees of freedom, there are nN possible states so entropy S is proportional to N ln n which is proportional to mass M (and volume) so we have two surprises: the entropy is proportional to surface area (the ﬁrst hint of holography) and to the square of the mass. Also as T → 0, the number of states usually goes to one so 1 S → 0, but, in this case, T = 8πGM so as T → 0, M → ∞ : entropy is increasing. The relation between temperature and mass implies that the heat capacity is given by dM C = = −8πGM 2 < 0 dT which is an unstable thermodynamic system. Introduce the partition function with the Euclidean action SE Z Z = [dg] e−SE ≈ e−Scl . which should be quantum gravity. If the Ricci scalar R = 0 then Scl = 0 but we had better be careful. The Ricci tensor is a second derivative and the Lagrangian should be independent of second order and higher derivatives. We can integrate by parts but we must keep the surface terms at R > r+ and let R → ∞ at the end. Z √ 1 3 Ssurface = − d x hK 8πG surface

(York-Gibbons-Hawking action) where K = trKµν is the extrinsic curvature of the surface and 1 K = n gαβ∂ g µν 2 α β µν µ (nµ = √1 δ is the unit vector perpendicular to the surface). α grr r r GM 2GM K = − 1 − ττ R2 R r 2GM K = R 1 − θθ R r 2GM K = R sin2 θ 1 − ϕϕ R 2 2 2 hττ = f(R) hθθ = R hϕϕ = R sin θ

Then Z Z Z 1 1/T 2π π √ Ssurface = − dτ dϕ dθ hK 8πG 0 0 0 2R − 3GM = − 2GT 10 LECTURE 2: Schwarzschild Black Hole

which is the classical action. It diverges as R → ∞. But in order to properly deﬁne it, we need to introduce a ref- erence point. We shall subtract the contribution of empty space. The latter is obtained by setting M = 0. However, there is a complication. We obtained the temperature in the case M 6= 0 by demanding that there be no conical singularity. There is no such constraint in empty space (M = 0). Instead, we shall match the boundaries of the two spacesp (at r = R). The time direction at r = R for the black hole has length 1 √ 1 T gττ = T f(R). If we make time periodic with period 1/T0 for M = 0, then that will be its length at any r. We need to match r 1 1 p 1 2GM = f(R) = 1 − T0 T T R

i.e., choose the temperature of empty space to be T0 (red-shifted). Then 2R − 3GM R S = − + cl 2GT GT 0r 2R − 3GM R 2GM = − + 1 − 2GT GT R 3M M = − + O(1/R) 2T T M = 2T Now

F = U − TS = M − TS 1 πr2 = M − + 8πGM G 1 4πG2M 2 = M − 8πGM G 1 = M 2

since r+ = 2GM. Therefore, F S = cl T as expected. Can we understand the entropy microscopically by counting degrees of freedom? This is hard at ﬁnite temperature. It is easier to ask the question in the limit T → 0 because the system then settles to its ground state. The problem is that in this limit, M → ∞, so this limit is hard to understand. We need a better black hole. LECTURE 3

Reissner-Nordstrom¨ Black Holes

3.1 The holes

Let us combine gravity with electromagnetism to ﬁnd a charged black hole. The action is Z Z Z 1 √ 1 √ 1 √ S = d4x −gR + d3x hK − d4x −gF F µν 16πG 8πG 16π µν so there is now a source (electromagnetic ﬁeld energy corresponds to mass). We have 1 R − g R = 8πT µν 2 µν µν where (from electromagnetism) µ ¶ 1 1 T = gρσF F − g F F ρσ µν 4π µρ νσ 4 µν ρσ which has trace zero so there is no scale: no mass, no distance, no time, just photons. R = 0, just as in the Schwarzschild black hole, but we now have the Maxwell equations with no charges, µν ∇µF = 0 Q The most symmetric solution is the 4-vector with time component A0 = r ; however, we want the potential to be zero at the horizon so we set Q Q A0 = − . r r+ and 2GM GQ2 f(r) = 1 − + r r2 12 LECTURE 3: Reissner-Nordstrom¨ Black Holes

This is the Reissner-Nordstrom¨ black hole. The horizon r+ (0 = f(r+)) is the solution to r2 − 2GMr + GQ2 = 0

so p 2 2 2 r± = GM ± G M − GQ

(r− is inside r+ so we cannot observe it). When Q = 0, we get r+ = 2GM as 2 2 2 expected. The minimal r+ occurs when G M − GQ = 0 so there is a minimal mass

Q2 M = min G Below this mass, we would have a naked singularity, but we stick with dressed singu- larities. The temperature is given by

1 0 T = f (r+) 4π µ ¶ 1 2GM 2GQ2 = − 4π r2 r3 µ + ¶+ 1 1 GQ2 = − 3 4π r+ r+ The entropy is A πr2 S = + = + . 4G G The law of the thermodynamics reads

dM = T dS + µdQ

where µ = Q plays the role of a chemical potential and Q, in effect, counts the Gr+ number of charges. Differentiating, µ ¶ µ ¶ ∂M ∂S = T ∂r+ ∂r+ µ ¶ Q Q ∂M Q = ∂Q Gr r+ +

This seems√ to be nicer than a Schwarzschild black hole. In the extremal case, r+ = GM = GQ, Ã ! 1 1 (GM)2 T = − 3 = 0 4π r+ r+ S = πQ2 6= 0

violating the third law of thermodynamics. 3.2 Extremal limit 13

Look at the classical action M Q2 Scl = − 2 2T 2T r+ and F U = − S T T M πr2 = − + T G which is not equal to the classical action.

2 2 F M πr+ Q Scl − = − + − 2 T 2T G 2T r+ Q2 = − T r+ µQ = − T so TScl = F − µQ is the Gibbs free energy. Two different calculations have given the same answer.

3.2 Extremal limit

Look at the extremal limit more closely

2GM GQ2 f(r) = 1 − + r r2 2r r2 = 1 − + + + r r2 ³ r ´2 = 1 − + r 0 If we change coordinates to r = r − r+,

µ 0 ¶2 r −2 0 f = 0 = H (r ) r + r+

0 0 2 where H(r ) = 1 + r+/r is an harmonic function (∇ H = 0).

2 −2 2 2 02 0 2 2 ds = −H dt + H dr + (r + r+) dΩ2 = −H−2dt2 + H2dx2 , x ∈ R3

The Coulomb potential Q Q Q A = 0 − = − r r+ Hr+ 14 LECTURE 3: Reissner-Nordstrom¨ Black Holes

As we get near the horizon r0 → 0 and µ ¶ 0 2 ³ ´2 2 r 2 r+ 02 2 2 ds ≈ − dt + 0 dr + r+dΩ r+ r so the black hole is the product of a 2-dimensional manifold and the 2-sphere; we shall show that the manifold is anti-de Sitter space (AdS). Recall that in the case of a Schwarzschild black hole, after setting τ = it, we obtained 1 a cone. To get rid of the singularity, we required periodicity under τ → τ + T which 2 determined the temperature T . Here, if we set τ = it/r+, we obtain µ ¶ 1 ds2 = r2 r0 2dτ 2 + dr0 2 . 2 + r0 2 To see what manifold this is, in 3-dimensional Minkowski space with metric

2 0 2 1 2 2 2 ds3 = −(dx ) + (dx ) + (dx ) consider a hyperboloid surface deﬁned by

¡ 0¢2 ¡ 1¢2 ¡ 2¢2 2 x − x − x = r+

0 If we change coordinates to (r+, r , τ) with µ ¶ 1 r0 ¡ ¢ x0 = r + 1 + τ 2 + 2r0 2 µ ¶ 1 r0 ¡ ¢ x1 = r − 1 − τ 2 + 2r0 2 2 0 x = r+r τ

for a ﬁxed r+, we obtain the metric on the hyperboloid, µ ¶ 1 ds2 = r2 r0 2dτ 2 + dr0 2 , 2 + r0 2

which is of the desired form (Euclidean anti-de Sitter space (EAdS2)). There is no restriction on τ, which is consistent with the fact that T = 0. This surface has three isometries:

¡ 0 1 ¢ M1 = i x ∂1 − x ∂0 , a boost along the x-axis ¡ 0 2 ¢ M2 = i x ∂2 − x ∂0 , a boost along the x-axis ¡ 1 2 ¢ L = i x ∂2 − x ∂1 , angular momentum. satisfying the Lorentz algebra

[L, M1] = −iM2

[L, M2] = iM1

[M1,M2] = iL. 3.3 Generalizations 15

In terms of (r0, τ),

0 M = i (−r ∂ 0 + τ∂ ) 1 µ r µτ ¶ ¶ 0 1 1 2 M = i r τ∂ 0 + 1 + − τ ∂ 2 r 2 r0 2 τ µ µ ¶ ¶ 0 1 1 2 L = i r τ∂ 0 + −1 + − τ ∂ r 2 r0 2 τ

0 As we approach the boundary (r0 → ∞) staying on a r = const. slice (so that ∂r0 → 0) we have

M1 → iτ∂τ i ¡ ¢ M → 1 − τ 2 ∂ 2 2 τ i ¡ ¢ L → − 1 + τ 2 ∂ 2 τ Deﬁne

2 D = M1 = iτ∂τ ,H = M2 − L = i∂τ ,K = −M2 − L = iτ ∂τ Evidently, [D,H] = −iH , [D,K] = iK , [H,K] = 2iD which is the conformal algebra in one dimension. D generates dilations, H is the Hamiltonian (generates translations in τ) and K generates special conformal transfor- mations. Exercise: Show that for an operator O(τ),

e−iaH O(τ)eiaH = O(τ + a) , e−iaDO(τ)eiaD = O(eaτ) and ﬁnd the corresponding transformation for K. Thus the isometries of AdS space turned into the generators of the conformal group. Starting with a black hole, we have arrived at a conformal quantum mechanical system; this is the ﬁrst hint at an AdS/CFT correspondence.

3.3 Generalizations 3.3.1 Multi-center solution Recall that in the extremal limit,

2 −2 2 2 2 3 −1 ds = −H dt + H dx (x ∈ R ) ,At = −H

r+ and H = 1 + |x| , an harmonic function. You may guess that the Einstein-Maxwell equations lead to ∇2H = 0, so H can be any harmonic function, in particular a multi- center solution, q q H = 1 + 1 + 2 + ... |x − x1| |x − x2| 16 LECTURE 3: Reissner-Nordstrom¨ Black Holes

consisting of the Coulomb potentials due to charges qi at positions xi (i = 1, 2,... ). The ﬁst term equal to 1 is there for correct asymptotics. All these black holes are in equilibrium because the electric and gravitational forces√ cancel each other (due to 2 G MiMj = GQiQj, where Mi is the mass and Qi = qi/ G is the charge of the ith black hole).

3.3.2 Arbitrary dimension In dimensions d ≥ 4, a Reissner-Nordstrom¨ black hole has metric

dr2 2µ q ds2 = −f(r)dt2 + + r2dΩ2 , f(r) = 1 − + f(r) d−2 rd−3 r2(d−3)

where 8πGM 8πGQ2 2π µ = and , q = , Ω = ¡ ¢ (d − 2)Ω (d − 3)(d − 2)Ω d−2 d−1 d−2 d−2 Γ 2

Ω + d − 2 is the volume of a sphere in dimension d − 2 (Sd−2). The horizon is at r+, where p d−3 2 2 r± = µ ± µ − q, µ ≥ q

the temperature is Ã µ ¶ ! f 0(r ) d − 3 r d−3 T = + = 1 − − . 4π 4πr+ r+

and the entropy is A Ω rd−2 S = + = d−2 + 4G 4G

2 d−2 At extremality, µ = q, so r+ = r− and T = 0. The entropy is S ∝ Q d−3 . For the metric, notice that · ¸ ³r ´d−3 2 f(r) = 1 − + r Changing coordinates to 0d−3 d−3 d−3 r = r − r+ so that the horizon is at r0 = 0, we obtain

2 −2 2 2 2 d−1 ds = −H dt + H d−3 dx , x ∈ R µ ¶ r d−3 H = 1 + + |x|

H is an harmonic function. 3.3 Generalizations 17

3.3.3 Kaluza-Klein reduction Interestingly, the four-dimensional Reissner-Nordstrom¨ black hole (in fact, all of electromagnetism) can also be derived from pure gravity in ﬁve dimensions. The action is Z 1 p S = d5x −g(5)R(5) 16πG5 The space is spanned by coordinates xM (M = 0, 1, 2, 3, 4). If x4 is a loop of length L (compactiﬁed) and there is no dependence on x4 (which is true if we are interested in distances large compared to L), then Z 1 p 1 L S = d4x −g(5)R(5) , = 16πG G G5 where G is Newton’s constant in our four-dimensional world. Let us parametrize the metric as

2 (5) M N µ ν 2Φ ¡ 4 µ¢2 ds ≡ gMN dx dx = gµν dx dx + e dx + Aµdx Then the Ricci scalar reads 1 R(5) = R − 2e−Φ∇2eΦ − e2ΦF F µν 4 µν where R is the Ricci scalar in our four-dimensional world. The action becomes Z µ ¶ 1 √ 1 S = d4x −g R − ∂µΦ∂ Φ − e2ΦF F µν . 16πG µ 4 µν consisting of a gravitational field gµν , the dilaton Φ and a photon Aµ. Notice that all three fields are massless. 2 M Consider a particle of the dilaton field. SInce it is massless, p = p pM = 0 . Because 4 4 ip4x ip4(x +L) the fifth dimension has a finite length L, p4 must be quantized: e = e so p4L = 2πn for some integer n. Then in the four dimensions we can observe, the particle appears to have mass m with µ ¶ 2π 2 m2 = p pµ = p2 = n2 µ 4 L so m = 0, 2π/L, 4π/L, . . . , the Kaluza-Klein tower. These masses are, in principle, observable, but we have not observed them. If L ¿ 1, these masses would be very large and, therefore, may not be observable at present (or ever). To get the Reissner-Nordstrom¨ black hole, set Φ = 0. Then the action reduces to one with gravity and electromagnetism which is exactly where we got the Reissner- Nordstrom¨ black hole earlier. Additionally, there exists a solution with Φ 6= 0. We obtain µ ¶ f(r) dr2 r ρ ds2 = − dt2 + H7/8 + r2dΩ2 , f(r) = 1 − + ,H = 1 + H1/8 f(r) 2 r r 18 LECTURE 3: Reissner-Nordstrom¨ Black Holes

−2Φ 3/2 −1 e = H , and At = H . In the extremal limit r+ → 0

ds2 = −H−1/8dt2 + H7/8dx2 , x ∈ R3 .

The singularity at r = |x| = 0 has vanishing area, therefore the entropy S = 0. LECTURE 4

Black Branes from String Theory

The Kaluza-Klein picture we have presented above is classical and, just as we need quantum mechanics to understand the electron (which is a point particle, i.e. a singularity), we need quantum mechanics to understand a black hole. Unfortunately (because it is very complicated) the only quantum theory of gravity we possess is string theory. To avoid inﬁnities in the construction of quantum gravity, string theory complicates things:

• Strings live in ten dimensions

• Strings vibrate. The greater the frequency, the higher the mass. Also an inﬁnite number of vibrational modes should correspond to an inﬁnite number of particles. The massless modes must be the particles we see. The electron’s mass is 0.5 MeV , the proton’s is 1 GeV , and the W particle has a mass of 80 GeV , but the vibrational modes have mass/energy on the order of the Planck mass/energy 1019 GeV . Presumably these particles appeared only at the beginning of the universe.

• We need supersymmetry: every particle must have a super partner.

Example 1. Type IIA. Low energy (massless strings). In ten dimensions specify a metric gµν , a scalar ϕ, an anti-symmetric tensor Bµν , a vector potential Aµ, and an anti-symmetric tensor Aµνρ. As usual Fµν = ∂µAν − ∂ν Aµ; we can do a similar thing for Bµν and Aµνρ to get anti-symmetric tensors

Hµνρ = ∂µBνρ + ∂ρBµν + ∂ν Bρµ

Fµνρσ = ∂µAνρσ. 20 LECTURE 4: Black Branes from String Theory

Also we can anti-symmetrize Fµνρσ + HµνρAσ to get a Gµνρσ. The action is Z µ ¶ 1 √ R − 1 (∂ ϕ)2 − 1 eϕ/2H Hµνρ− 10 2 ¡ µ 12 µνρ ¢ S10 = d x −g 1 5ϕ/2 µν 1 µνρσ 16πG10 e Fµν F − GµνρσG Z 4 12 1 µ1...µ10 − ² Bµ1µ2 Fµ3...µ6 Fµ7...µ10 32πG10 where ² is the Levi-Civita tensor (zero if any indices are repeated, ±1 if the permutation of the indices is even (odd)). In eleven dimensions the situation is simpler. We only require an anti-symmetric tensor AABC with anti-symmetric FABCD as above. The action is Z p µ ¶ 1 11 (11) (11) 1 ABCD S11 = d x −g R + FABCDF . 16πG11 18 (11) If the tenth dimension is tightly wrapped, we can partition the matrix gAB into a 9 by (10) 9 matrix gµν , a 9 by 1 column Aµ (and the same 1 by 9 row) and a scalar ϕ:

µ (10) ¶ (11) gµν Aµ gAB = . Aµ ϕ

If we restrict the indices of AABC to be from 0 to 9, we get a 10-dimensional tensor Aµνρ and setting Bµν = Aµν(10), we obtain all the ﬁelds for the 10-dimensional case and S(11) gives S(10). Example 2. Type IIB. p-brane solutions. We are given an anti-symmetric tensor

Aµ1...µp+1 and a metric µ ¶ ¡ ¢ 1 ds2 = −H(p−7)/8 −f(r)dt2 + dy2 + H(p+1)/8 dr2 + r2dΩ2 f(r) 8−p where ρ7−p r7−p H = 1 + , e2ϕ = H(3−p)/2, f(r) = 1 − + , and y ∈ Rp. r7−p r7−p This would be a black hole except for the presence of dy. It appears as if a black hole is stretched through other dimensions and is called a black brane around hyi. In the extremal limit r+ → 0, ¡ ¢ ds2 = −H(p−7)/8 −dt2 + dy2 + H(p+1)/8dx2 where x ∈ R9−p. As r → 0, H has a singularity and ³ρ´−(7−p)2/8 ¡ ¢ ³ρ´(7−p)(p+1)/8 ds2 ≈ −dt2 + dy2 + dx2. r r When p = 3, we get something special, the dilaton ϕ is constant and r2 ¡ ¢ ρ2 ds2 ≈ −dt2 + dy2 + dx2 ρ2 r2 r2 ¡ ¢ ρ2 = −dt2 + dy2 + dr2 + ρ2dΩ2. ρ2 r2 5 Black Branes from String Theory 21

The last term is just the ﬁve sphere S5 and r would not fall out if p 6= 3. The other terms are the anti-deSitter space AdS5 (3 spatial and 1 time dimension). The situation is similar to two dimensional conformal theory on the boundary giving quantum mechanics. 22 LECTURE 4: Black Branes from String Theory LECTURE 5

Microscopic calculation of Entropy and Hawking Radiation

5.1 The hole

We want to understand why entropy is so large. Can it be explained from the action in this example? In ﬁve dimensions, the Reissner-Nordstrom¨ black hole has the metric ds2 = H−2dt2 + H2dx2 where H = 1 + Q2/r2 (r2 = x2). The 5-brane has metric ¡ ¢ ds2 = H−1/4 dt2 + dy2 where H is the same as in the Reissner-Nordstrom¨ black hole and e2ϕ = H−1 is singular. Strominger and Vafa looked at

2 −3/4 −1/4 2 2 2 ds = H1 H5 − dt + dy5 + (Hn − 1) (dt + dy) µ ¶1/4 H1 2 1/4 3/4 2 + dy + H1 H5 dx H5 where 2 ρn −2ϕ H5 Hn = 1 + 2 and e = , r H1 y and x are in R4 and all the y-dimensions are very small. Consider some special cases. Case 1: When ρ1 = ρn = 0, the metric becomes

2 −1/4 ¡ 2 2¢ −1/4 2 3/4 2 ds = H5 −dt + dy5 + (H5) dy + H5 dx −1/4 ¡ 2 2 2¢ 3/4 2 = H5 −dt + dy5 + dy + H5 dx 24 LECTURE 5: Microscopic calculation of Entropy and Hawking Radiation

where the y-terms form a 5-dimensional Euclidean space or, if small, a torus (5-brane).

Case 2: When ρ5 = ρn = 0, the metric becomes

2 −3/4 ¡ 2 2¢ 1/4 ¡ 2 2¢ ds = H1 −dt + dy5 + H1 dy + dx

where the x and y terms contribute 8 dimensions and the y5 term is a 1-brane.

Case 3: When ρ1 = ρ5 = 0, the metric becomes

2 2 2 2 2 2 ds = −dt + dy5 + (Hn − 1) (dt + dy5) + dy + dx

with no brane.

?? The complete solution is a bound state of a 5-brane from the survival of H5 and a 1- brane from the survival of H1.A 5-dimensional observer should perceive a black hole; the y5-direction will have a momentum proportional to n/L which will be perceived as mass.

µ µ ¶ ¶2 2 −1 2 1 2 2 ds = −Hn dt + Hn dy5 + − 1 dt + dy + dx Hn looks like the Kaluza-Klein reduction where

2 ¡ 4 µ¢2 ds = ...stuff dx + Aµdx

and, here we can set µ ¶ 1 At = − 1 . Hn ?? Now take the complete metric

µ ¶1/4 2 −3/4 −1/4 2 H1 2 −3/4 −1/4 −1 2 1/4 3/4 2 ds = H1 H5 Hn (dy5 + Atdt) + dy −H1 H5 Hn dt +H1 H5 dx H5

5 where y (y5) is a small, closed loop in R (R). A 5-dimensional observer sees the last two terms which equal

−3/4 −1/4 −1 ¡ 2 2¢ H1 H5 Hn −dt + H1H5Hndx

which is similar to a Reissner-Nordstrom¨ black hole: 1 1 ¡ ¢ ds2 = − dt2 + Hdx2 = −dt2 + H3dx2 H2 H2

1/3 and, if ρ1 = ρ5 = ρn, it is exactly the same. We could set H = (H1H5Hn) so the difference is that a RN black hole as one charge while the general case has three. 5.2 Entropy 25

5.2 Entropy

To get the entropy, we need the area A = 2π2r3 where, ?? in the RN case (Hdx2 = √ + 2 2 2 Hr dΩ ), r+ = limr→0 Hr . ?? Therefore

Ã !3 µ 2 2 2 ¶1/6 2 ρ1 ρ5 ρn 2 A = 2π lim r = 2π (ρ1ρ5ρn) r→0 r2 r2 r2

¡ ¢ 2GM Mass for a Schrodinger¨ black hole, is deduced from gtt = − 1 − r . In the general case of d dimensions gtt ≈ −1 + 2V as r → ∞ where V is the potential energy; in fact µ 8πGM 2π(d−2)/2 gtt = −1 + d−3 , µ = , and Ωd−1 = . r (d − 2)Ωd−2 Γ((d − 2)/2)

−2/3 For 5 dimensions, G5M = 3πµ/4 and gtt = − (H1H5Hn) ; expanding in 1/r, we get 2 ρ2 + ρ2 + ρ2 g = −1 + 1 5 n + ... tt 3 r2 so π ¡ ¢ G M = ρ2 + ρ2 + ρ2 . 5 2 1 5 n Mass splits into three pieces (with no binding energy) due to the 1-brane, the 5-brane, and the momentum.

Let ρn = 0 and look at the metric near the horizon ?? r → 0 ?? r p r2 ¡ ¢ ρ ρ ρ3 ¡ ¢ ds2 ≈ p −dt2 + dy2 + 1 dy2 + 1 5 dr2 + r2dΩ2 3 5 ρ r2 3 ρ1ρ5 5 Ã q ! r q r2 ¡ ¢ dr2 ρ = p −dt2 + dy2 + ρ ρ3 + 1 dy2 + ρ ρ3dΩ2 3 5 1 5 r2 ρ 1 5 3 ρ1ρ5 5

4 where the terms in parenthesis represent AdS3, the next term is a 4-torus, T , and the last term is just the 3-sphere, S3. ?? Can we understand entropy in this model? The 1-brane can be taken to be wrapped tightly N1 times, the 5-brane N5 times, no momentum, and only one state so S = 0. This suggests entropy comes from momentum. String theory tells us that excitations of branes arise from the movement of strings attached to branes. Branes have no binding energy and they are points (no deeper structure). We can set the coupling constant to one. The only parameter that is available to determine energy is the length of the string `s; thus the unit of energy is 1/`s. If the 1-brane wraps N1 times around a length L1, 2 2 it must have N1L1/`s elementary units so its energy is N1L1/`s. M1 = N1L5/2π`s. 2 8 10 6 8 In 4-dimensions, G ∝ ` so we expect G10 ∝ `s, in fact G = 16π `s. The equation 26 LECTURE 5: Microscopic calculation of Entropy and Hawking Radiation

¡ ¢ π 2 2 2 G5M = 2 ρ1 + ρ5 + ρn results from gravity alone, not string theory. Now

π G M = ρ2 5 1 2 1 6 8 16π `s N1L5 π 2 2 = ρ1 L1...L5 2π`s 2 4 2 (2π) 6 ρ1 = `sN1. L1...L4

Similarly the 5-brane

N5L1...L5 1 5 2 = 16π M5 `s `s 2 ρ5 = N5`s.

The momentum in the y5 direction p5 = 2πn/L5, n = 0, 1, 2, ... is the last piece of the puzzle :

6 8 2 (2π) `s ρn = 2 n L1...L4L5

and, putting it all together,

S = 2π2ρ ρ ρ p1 5 n = 2π N1N2n.

This is one success of string theory. ?? We now intend to count the degrees of freedom of this system. Note that strings can begin and end on any brane, but only strings between the 1-brane and the 5-brane contribute. For instance, when N1 = 2 and N5 = 2, the string will return to its starting point after going around twice, but, if N1 = 2 and N5 = 3, the string will return to its starting point after going around six times. In general, if N1 and N5 are relatively prime, the string goes around N1N5 times. If both numbers are very large, we can ignore the relatively prime requirement since they will be close to a relatively prime pair. Then Leff = N1N5L5. Then

2πn 2πn Leff 2π p5 = = = N1N5n L5 Leff L5 Leff

and 2π/Leff can be considered the quantum of momentum. Given a momentum, we want to count all of the ways of getting that momentum to get the number of states. We do so in a physical way. Suppose the quantum of energy is E0 and ni particles have energy iE0 at temperature T so the total energy is E = n1E0 + n22E0 + n33E0 + .... 5.3 Finite temperature 27

Deﬁne the partition function X Z = e−E/T E X = e−(n1E0+n22E0+n33E0+ ...)/T n Ãi !Ã ! X∞ X∞ = e−n1E0/T e−n22E0/T ...

n1=0 n2=0 1 1 = ... 1 − e−E0/T 1 − e−2E0/T P ¡ ¢ −NE0/T The free energy is given by F = −T ln Z = T N ln 1 − e . Take the high temperature limit T → ∞ and let x = NE0/T and ∆x = E0/T → 0. Then X T ¡ −x¢ F = E0 ln 1 − e E0 T 2 E X ¡ ¢ = 0 ln 1 − e−x E0 T T 2 X ¡ ¢ = ∆x ln 1 − e−x E 0 Z T 2 ¡ ¢ ≈ ln 1 − e−x dx E0 T 2 π2 = . E0 6

2 ¡ ¢ The entropy is S = − ∂F = π T and the internal energy is E = −T 2 ∂ F = ∂T 3 E0 ∂T T π2 T 2 T 2 6n = nE0 so = 2 which we substitute into the entropy 6 E0 E0 π √ r r 6n π2 n N N n S = = 2π = 2π 1 5 . π 3 6 6 This is not the whole story because string theory says there are four different strings so we should multiply by 4 under the radical; furthermore, supersymmetry gives fermions p n for which ni must be either 0 or 1 so Sfermion = 2π 12 . Therefore s µ ¶ N N n 4 p S = 2π 1 5 4 + = 2π N N n. 6 2 1 5 We have come to the same conclusion independently of the calculation using the horizon area.

5.3 Finite temperature

Suppose T is slightly greater than zero. To heat the system, we could introduce an anti- brane or let the strings run in both directions. Let the number of left (right) moving 28 LECTURE 5: Microscopic calculation of Entropy and Hawking Radiation

strings be n+ (n−). A 5-dimensional observer would see a charge Q ∝ n+ − n− (Mn ∝ n+ + n−). The metric is

µ ¶ µ ¶1/4 µ 2 ¶ 2 −3/4 −1/4 2 2 1 ¡ ¢2 H1 2 1/4 3/4 dr 2 2 ds = H1 H5 −dt + dy5 + 2 ρ+dt + ρ−dy5 + dy +H1 H5 + r dΩ3 r H5 f(r)

2 r+ where ρ splits into ρ+ and ρ+ and f(r) = 1 − r2 so there are two horizons at 0 and 2 2 2 r+. ρ+ − ρ− = r+ so, when ρ+ = ρ−, r+ = 0 and we are back to the case T = 0. Let us consider physical properties where µ ¶ dr2 ds2 = − (H H H )−2/3 fdt2 + (H H H )1/3 + r2dΩ2 5 1 5 n 1 5 n f 3

2 2 with Hn = 1 + ρ−/r . The area of the horizon is p ¯ 2 3 ¯ A+ = 2π r+ H1H5Hn r=r+ .

Assume ρ1, ρ5 >> r+, then s 2 2 2 2 2 3 ρ1ρ5 r+ + ρ− A+ = 2π r+ 4 2 = 2πρ1ρ5ρ+. r+ r+

Expand the metric to get

2 ρ2 + ρ2 + ρ2 r2 g = − (H H H )−2/3 f = −1 + 1 5 − + + + .... tt 1 5 n 3 r2 r2

2 2 3π The 1/r comprise µ/r where in classical gravity G5M = 4 µ; therefore µ ¶ π 3 G M = ρ2 + ρ2 + ρ2 + r2 5 2 1 5 − 2 +

= G5 (M1 + M5 + Mn) .

and µ ¶ π 1 π G M = ρ2 + r2 ≈ ρ2 5 1 2 1 2 + 2 1 µ ¶ π 1 π G M = ρ2 + r2 ≈ ρ2 5 5 2 5 2 + 2 5 µ ¶ π 1 π ¡ ¢ G M = ρ2 + r2 ≈ ρ2 + ρ2 . 5 n 2 − 2 + 4 + −

We now have

2 2 n+ + n− ∝ MN ∝ ρ+ + ρ−

n+ − n− ∝ Q ∝ 2ρ+ρ− 5.4 Scattering and Hawking radiation 29 so ¡ ¢2 n+ ∝ ρ+ + ρ− ¡ ¢2 n− ∝ ρ+ − ρ− . √ √ In terms of n±, S ∝ ρ+ ∝ n+ + n− or p ¡√ √ ¢ S = 2π N1N5 n+ + n− which is expectedp since the rightp and left moving modes are completely independent: S+ = 2π N1N5n+, S− = 2π N1N5n−, S = S+ + S−. With two independent quantum systems, our counting agrees with classical mechanics. To count the degrees of freedom, we want to look near the horizon r+, but carefully because r+ is very small. When r is small the H functions can be approximated by ρ2/r2. As before the metric becomes p 2 µ ¶ r 3 µ 2 ¶ r 1 ¡ ¢2 ρ ρ ρ dr ds2 = p −dt2 + dy2 + ρ dt + ρ dy + 1 dy2 + 1 5 + r2dΩ2 3 5 r2 + − 5 ρ r2 f(r) 3 ρ1ρ5 5 2 q 2 q r r ¡ ¢ 1 ¡ ¢2 dr ρ = p −dt2 + dy2 + p ρ dt + ρ dy + ρ ρ3 + ρ ρ3dΩ2 + 1 dy2. 3 5 3 + − 5 1 5 r2 − r2 1 5 3 ρ ρ1ρ5 ρ1ρ5 + 5 The last two terms are the sphere S3 and the torus T 4. The other terms form a three 2 dimensional manifold M which was AdS3 when T was zero. To study the metric ds3 02 2 2 of M, set r = r + ρ− to get ?? Ã ¡ ¢ ¡ ¢ µ ¶ ! 1 r02 − ρ2 r02 − ρ2 r02dr02 ρ + ρ 2 ds2 = p − − + dt2 + ρ2ρ2 ¡ ¢ ¡ ¢ + r02 dy + + − dt . 3 3 r02 1 5 02 2 02 2 5 r02 ρ1ρ5 r − ρ− r − ρ+ ?? This is Banados-Teitelboim-Zanelli or BTZ black hole. If r0 → ∞, the large 02 2 2 2 02 02 02 2 parenthetical term reduces to −r dt + ρ1ρ5dr /r + r dy5 which is the metric of AdS3. This is unusual because, far from the horizon, we do not get ﬂat space; in this case going to inﬁnity means going as far away as possible without leaving the horizon. Being near the horizon introduces a cosmological constant to get negative curvature.

5.4 Scattering and Hawking radiation

What do we mean by going to infinity and not leaving the horizon? Look at radiation thrown at the black hole. Introduce a scalar field ϕ which is massless and has minimal coupling (no term coupling with curvature); the action Z 1 √ S = d5x −g∂ ϕ∂ ϕgµν 2 µ ν give the Klein-Gordon field equation ¤ϕ = 0 which we want to solve for scattering. Use a spherically symmetric 5-dimensional wave not involving angles

−iωt ϕ(r, t) = Φω(r)e . 30 LECTURE 5: Microscopic calculation of Entropy and Hawking Radiation

The Laplacian gives 1 ¡ √ ¢0 √ grr −gΦ0 + ω2Φ = 0 −g ω ω which is 1 f(r) ¡ 3 0 ¢0 2 3 f(r)r Φω + ω Φω = 0. (4.1) H1H5Hn r 3/2 3/2 In four or ﬁve dimensions behaves as 1/r : Φω(r) = ψ(r)/r and we obtain µ ¶ ¡ ¢0 3 −f(r) f(r)ψ0 + ω2H H H − f(r)(4 − 3f(r)) ψ = 0 (4.2) 1 5 n 4r2

which is actually a Schrodinger¨ equation. Introduce tortoise coordinates dr∗ = dr/f(r) so Z µ ¶ r+ r − r+ r∗ = dr/f(r) = r + ln 2 r + r+

where r > r+. Equation 4.2 becomes

d2ψ − 2 + V ψ = 0 dr∗ where V is the coefﬁcient of ψ in Equation 4.2. As r → 0, V → −∞ and, as r → ∞, the H functions go to 1 and V → −ω2. Calculating the transmission coefﬁcient for a wave coming from +∞ is a classical problem, not a quantum mechanical one, but we cannot possibly solve it exactly. With the assumption ρ1, ρ5 >> r+, we will look for a solution near the horizon and one away from the horizon and patch them together. The √ intermediate region is given by the harmonic mean ρm = r+ρ1: r r r r ρ r + = + << 1 and m = + << 1. ρm ρ1 ρ1 ρ1

2 r+ r+ r+ Far away, space is ﬂat and r & ρ , < << 1 so f(r) = 1 − 2 ≈ 1 and r∗ m r ρm r becomes r so 3 V = −ω2 + 4r2 which can be solved to give a Bessel function √ √ ψ = A rJ1 (ωr) + B rN1 (ωr) A B Φ = J (ωr) + N (ωr) . ω r 1 r 1

If we now assume low frequency ω << 1/ρm, then, as r → ρm, ωr gets very small, J1 (ωr) → ωr/2, and N1 (ωr) → ∞ so we must set B = 0 and we have

A Φ = ω. ω 2 5.4 Scattering and Hawking radiation 31

2 2 2 2 Near the horizon, r . ρm, use equation 4.1 with H1 ≈ ρ1/r , H5 ≈ ρ5/r to get 2 2 µ 2 ¶ f ¡ ¢0 ρ ρ ρ fr3Φ0 + ω2 1 5 1 + − r3 ω r4 r2 which can be solved to yield

0 −i(a+b)/2 Φω = A f(r) F (−ia, −ib; 1 − ia − ib; f(r)) ¡ ¢ ¡ ¢ where F is a hypergeometric function, a = ωρ1ρ5/ ρ+ + ρ− , and b = ωρ1ρ5/ ρ+ − ρ− . To match, let r → ρm so f(r) ≈ 1 and Γ (1 − ia − ib) A A0 must match ω. Γ (1 − ia) Γ (1 − ib) 2 We now calculate the incoming ﬂux and transmitted ﬂux. 1 ¡ ¢ F = f(r)r3Φ∗ Φ + c.c. in 2i ω ω

dFin iωr and dr = 0. Far away we have an incoming wave proportional to e and a reﬂected wave proportional to e−iωr; as r → ∞,

µ iωr ¶ A −3πi/4 e Φω = √ e + c.c. 2πω r3/2 so ¯ ¯ 1 ¯A¯2 F = ¯ ¯ . in 4π ¯ 2 ¯ 0 −i(a+b)/2 We now want the absorbed ﬂux. Near the horizon (r → r+) Φω → A f(r) and 2 0 2 Fabs = −r+(a + b) |A | . Then F T = abs Fin ¯ ¯ ¯A0 ¯2 = −r2 (a + b)16π ¯ ¯ + ¯ A ¯ ¯ ¯ ω2 ¯Γ (1 − ia) Γ (1 − ib)¯2 = −16πr2 (a + b) ¯ ¯ + 4 ¯ Γ (1 − ia − ib) ¯ abe2π(a+b) → −(2πω)2 . (e2πa − 1)(e2πb − 1) The factor 1 ³ ´−1 = e2πωρ1ρ5/(ρ++ρ−) − 1 e2πa − 1 looks like the Planck (black body) distribution ³ ´−1 e~ω/kT − 1 .