8.821 F2008 Lecture 23: Black Hole ThermodynamicsLecturer: McGreevy Scribe: Tongyan LinDecember 2, 2008In today’s lecture we’ll discuss the laws of black hole thermodynamics and how Ad S black holesare related to finite temperature CFTs, and Koushik will give a related presentation.1 Laws of ThermodynamicsRecall from last time that for a black holeArea ∼ Entropyκ ∼ T (1)where κ was the surface gravity. The near-horizon metric isds2∼ −κ2ρ2dt2+ dρ2+ ... = κ2ρ2dτ2+ dρ2+ ... (2)when we go to Euclidean time τ ≡ it. If τ has periodicity τ ∼ τ +2π/κ then the euclidean geometryis regular.Recall the canonical ensemble thermal partition function isZth= tre−H/T(3)where e−H/Tpropagates the system with imaginary time t = 1/iT . Thermal equilibrium is equiv-alent to periodic euclidean time with period 1/T , so we identify κ with temperature T .The laws of (stationary) black hole thermodynamics, analogous to the usual laws of therm odynam-ics, are:• 0th (thermal equilibrium): κ is constant over the event horizon. This means temperatureis constant in space and time. Thu s stationary black holes are in thermal equilibrium withconstant temperature. John thinks the pr oof of the 0th law doesn’t depend on the shape ofthe black hole, as long as its a station ary solution.1• 1st (conservation of energy):dE = dM = ΩdJ + ΦdQ +κ8πGdA (+P dV ) (4)ΩdJ is the change in rotational energy, ΦdQ is the electrical energy, andκ8πGdA = T dS isheat exchange. This law relates th e change in the energy (or equivalently mass) to changesin various properties of the black hole.The last term describing mechanical work P dV isn’t present for black holes bu t IS for blackbranes...• 2nd (entropy increases): This is the area theorem for a black hole we proved last lecture,˙A ≥ 0, since S =A4~G. (Proof of the exact relation between S and A in a later lecture.)• 3rd (absolute zero entropy): κ (or rather T ) cannot taken to zero in a finite number ofsteps. This doesn ’t m ean that S(T = 0) = 0, but it does probably mean at T = 0 th ere is aminimum in entropy.These laws follow from Einstein’s Equation, the energy condition we discussed last class, andassuming we have stationary b lack holes.1.1 3rd lawSince we discussed the 2nd law last time, and the 0th and 1st laws are pretty convincing, we nowprovide some evidence for the validity of the 3rd law.First, why isn’t it tru e that S(T = 0) = 0? Counterexamples are everywhere if you just open youreyes to them:• It is well known to some people that there exist supersymmetric theories with LARGE groundstate degeneracies, ∝ eQαwhere Q is the charge and α is some power. So, S(T = 0) =ln(degeneracies) ∼ Qα.• The Kerr-Newman black hole is another counterexample. Here are some facts about the KNblack h ole that you can easily derive or look up:A = 4π(2M(M + µ) − Q2)µ =pM2− Q2− J2/M2κ = 4πµ/A (5)The black hole is extremal when µ = 0. (This is also the BPS bou nd when the black holeis supersymmetric.) If µ < 0 then there is a naked singularity. Note that that when µ = 0,κ ∝ T = 0 but S ∝ A 6= 0.As for the claims of the 3rd law, we have some anecdotal evidence. Let’s consider a non-extremalKN black hole with J = 0, in other words a non-extremal RN black h ole, so Q < M. (EverywhereQ is really |Q|).2What can we do to try to make this black hole extremal? We need to throw on some charge q andmass m, such that the black h ole becomes extremal, namelyM + m = Q + q (6)How, the mass m is attracted to the beautiful black hole by a force F ∼ Mm/r2but the charge q isrepulsed by a force F ∼ Qq/r2. Thus for the matter to fall in freely, M m > Qq. According to somemysterious algebra, this relation along with Q < M actually implies Q + q < M + m. Thereforeyou have to force the matter onto the b lack h ole, w hich somehow adds heat and prevents you fromco oling the black hole. Or you have to throw in infinitesimal little bits which takes FOREVER.2 CFT at finite temperatureWe’re going to use the power of AdSCFT to describe CFTs at finite temperature with black holes.In particular we mean a 3 + 1 dimensional relativistic CFT. The partition function isZ(τ ) = tre−H/T= e−F/T(7)with free energy F , on a space with geometry S1th× Σ3where the S1has radius 1/T, τ ∼ τ + 1/Tand Σ3is some 3 m an ifold. We can give Σ3finite volume as an IR regulator.This is a deformation of the IR physics (modes with ω ≫ T = EKKdon’t notice).For large V3= V ol(Σ3), then F = cV3T4which is clear from extensivity of F and dimensionalanalysis.3 AdS black holesThis object goes by many names, such as planar black hole, Poincare black hole, black br ane...This is a black hole in AdSD+1, but probably many of the equations below mean D = 4. Themetric isds2=L2z2−fdt2+ d~x2+dz2ff = 1 −z4z4m(8)We again p ut the ~x coordinates on a finite volume space, for example in box of volume V3, x ∼x + V1/33, periodic BCs. Notice that if f = 1 we get the Poincare AdS metric, and in fact f onlydeviates from 1 at larger z representing the fact that this is an IR deformation.Whence:3• It solves Ein s tein’s equations with a cosmological constant Λ =(D+1)(D+2)2L2and asymptotesto Poincare AdS, differing only in the IR region with a horizon at z = zm, fixed t.• It’s the double Wick rotation of the confining solution with t = iythere, y = −itthere.• Analogous to how we got the AdS solution from the near-horizon limit of D3-branes, it’s thenear-horizon limit of black 3-branes in R9,1, in particular the near-extremal RR soliton withgeometry:ds2=−fdt2+ d~x2pH(r)+pH(r)dr2f(r)+ r2dΩ25H(r) = 1 +˜L4r4f(r) = 1 −r4Hr4(9)(Again f = 1 gives the usual RR soliton.) Note that there also exists a black hole whichasymptotes to GLOBAL AdS (with boundary S1×S3), which is known as AdS-Schwarzchildwhich describes a CFT on S3at finite temperature T.Let’s check out the horizon p roperties so we can find the usual thermodynamic quantities we’reinterested in. The near-horizon metric isds2∼ κ2ρ2dτ2+ dρ2+L2z2md~x2(10)where κ = 2/zmand the temperature is T = κ/(2π) = 1/(πzm). Meanwhile the area of horizon isA =Zz=zm,fixedt√gd3x =Lzm3V3(11)Therefore the entropy is (in the deconfined phase of the gauge theory):S =A4G5=L34G5V3z3m=N22π(πT )3V3=π22N2V3T3(12)(Recall from long ago thatL34G5=N22π.)Again, we want some anecdotal evidence (at least) to support the claim that this describes a CFTin thermal