Classical Blackhole Thermodynamics (Con'd)Quantum Blackhole Thermodynamics8.821 F2008 Lecture 24Blackhole Thermodynamics (Con’d)Lecturer: McGreevy Scribe: Wing-Ho KoDecember 2, 20081 Classical Blackhole Thermodynamics (Con’d)Previously we have stated the third law for black hole thermodynamics. To rephrase the result:3rd Law of blackhole thermodynamics. Extremal blackholes are unnatural.In spite of this, extremal blackh oles are beloved by string theorists, because the computations areeasier. In particular, Sextr. BH(T = 0) = ln(# of ground states), and SUSY implies that this isindependent of couplings and hence can be computed in the weak coupling limit.Previously we have also stated the s econd law, but in which we only considered blackholes alone.We now consider a more general version the second law in which both blackholes and normal stuffare included:Generalized 2rd Law (GSL) of blackhole thermodynamics. [Bekenstein PRD 9 3292(1974)]Stotal= Snormal stuff+ζA~GNδStotal≥ 0 for physical processes.where ζ is a number yet to be determined.The idea behind the GSL is that it will be possible to build perpetual machines if it is false.However, there is a problem with the “lowering-the-box” energy extraction scheme described in aprevious lecture. To be more specific, suppose we lower an (infinitesimal) box of entropy from ∞to the event horizon and extract energy along the way as described in the previous lecture. Whenthe box is at the event horizon, we know that Ebox(r = rs) = 0, and if we let go at that point,∆A = 0 for th e blackhole according to the first law . Hence the GSL is violated.The other laws of classical blackhole thermodynamics also contain “flaws”:1• Flaw # 1. Classically a blackhole cannot radiate, so TBH= 0 ?• Flaw # 3. Classically, the area of each individual blackhole must be non-d ecreasing. Butthermodynamics should require only that δStotal≥ 0. Thus the blackhole thermodynamicsand ordinary thermodynamics are not in good parallel.1The apparent contradictions in the blackhole thermodynamic laws can be resolved by realizing th atthe blackhole entropy SSek= A/~GNcontains a factor of ~, hence SBek→ ∞ in the classical limit(~ → 0) when A and GNare fixed. In particular, for the box-lowering scenario, even though thechange in area δA is infi nitesimal, it can still lead to a finite change ∆S in entropy. Hence there isno contradiction with the GSL.Similarly, from the 1st Law dM = TBekdSBek. As ~ → 0, TBek= ~κ/8πζ → 0. Hence flaw # 1 doesnot lead to a contradiction in the classical limit.Also, as ~ → 0, any finite δA for any blackhole will lead to ∆S = ∞, so flaw # 3 does not lead toa contradiction in the classical limit.2 Quantum Blackhole ThermodynamicsOur major result is the Hawking radiation law :Hawking Radiation. A blackhole radiates like a blackbody with T = TBH=~κ2π.Let’s first consider some consequences before considering the explanation:1. This fixes the numerical constant ζ to be 1/4. i.e., SBH=A4~GN.2. The area of individual blackhole can decrease, thu s fixes flaw # 3 of GSL.3. The resu lting dA satisfies GSL for reasonable equation of state (E.O.S.). For example, con-sider radiation in 4d. We h ave E = (1/4)aT4and S = (1/3)aT3. Treating blackhole as ablackbody in fl at space, dSrad=43dETBH, while energy conservation implies that dM = −dEand hence dSBH= −dETBH. Summing up2, dStotal= d(Srad+ SBH) =13dETBH> 0.4. Let’s return to the box-lowering puzzle,(a) Bekenstein bound (Conjecture). Entropy in a box of linear size R must satisfySbox≤ 2πER.(b) Buoyancy force. Consider a box near event horizon, so that it’s top side is at r2and thebottom side is at r1(see Fig. 1). The local temperature T1(T2) on the top (bottom)1Perhaps this is not a big deal, since individual blackholes can merge.2Consequently, this says that the speed of sound is c2sound= 1/3.2side satisfies T2/T1=pgtt(r1)/pgtt(r2). Hence T1> T2. Consequently, there is moreradiation from the bottom than from the top3, which produces a buoyancy force ∝displaced radiation.As a result, there is a critical rcrit> rsfor which E(displaced radiation) = Ebox(rcrit).By dropping the box at rcrit, ∆SBH= E/TBH, but Sbox≤ Sdisplaced radiation= E/TBHevent horizonrr2r1T2T1Figure 1: Buoyancyforce on a box.Now let’s return to the question of why TBHtakes this particular form.For a generic (possibly non-extremal) blackhole, the event horizon is lo-cated at a zero of the emblackening factor f.For example, for Schwarzchild blackh ole,ds2Sch= −fdt2+dr2f+ r2dΩ2≈ κ2R2dt2+ dR2+ r(R)2dΩ2= −R2dη2+ dR2+ . . .≈ −dT2+ dZ2+ dX2+ dY2where in the above we have us ed :4f = 1 − rH/r, rH= 2GM, R(r) =pr(r − 2GM) + . . ., κ = 1/4GM, η = κt, T = R sinh η and Z = R cosh η.Notice that after the transformation, we obtain the Minkowski space R3,1. In other words, theRindler space is a Minkowski space in th e coordinate frame of a uniformly accelerating observer.Graphically, what we get is Fig. 2. Notice that the two lines defined by {r = 2GM, η = ∞ } and{r = 2GM, η = − ∞ } divide th e space into four regions { I, II, III, IV }.The key observation is that region I is self-contained. i.e., region II and III can’t communicatewith it, while information from region IV passes through the line {r = 2GM, η = −∞} and hencecorresponds to initial data.Therefore, at T = 0, the Z < 0 (marked by L on the figure) degree of freedom (d.o.f.) are totallyirrelevant and useless for region I. Thus we should trace over th em when computing stuff in regionI.Now, Claim:ρR= trL|g.s.ihg.s.| =1Ze−2πHRwhere HRis the riddler Hamiltonian:5HR= “∂∂η” =Zconst. T surf aceTab∂∂ηanband Z = trRe−2πHR.3analogous to more water at the bottom of sea4η is called the “Rindler time” and plays the role of rapidity5Note that [HR] = 1 since the riddler time [η] = 1.3r = 2GM, η = ∞r = 2GM , η = −∞r < 2GMfixed Rfixed ηRLZTIIIIIIIVto asymptoticallyflat spaceto singularityFigure 2: The geometry of the Rindler space.Notice that the density matrix we obtained is not a pure state b ut is entangled with the L d.o.f.(hence S = −tr ρRln ρR6= 0). This is a thermal density matrix with temperature Triddler= 1/2π.Proof of Claim [Unruh 1976]. Consider a scalar field φ in Rindler space. A complete set ofcommuting d.o.f’s at T = 0 isφ(x, y, z) =φR(x, y, z) for Z > 0φL(x, y, z) for Z < 0φRand φLcommute because they are at spacelike separated