Where is the PdV term in the first law of black-hole thermodynamics?

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1 Where is the PdV term in the first law of black-hole thermodynamics? Brian P Dolan National University of Ireland, Maynooth, Ireland and Dublin Institute for Advanced Studies, Ireland 9th Vienna Central European Seminar on Particle Physics and Quantum Field Theory, 30th November 2012

2 Outline Review of black hole thermodynamics Temperature and Entropy Laws of Thermodynamics Hawking radiation Pressure and Enthalpy Where is the PdV term? What is V? Critical behaviour Conclusions and Outlook

3 Temperature and Entropy Schwarzschild line element: ds 2 = (r)dt (r) dr2 +r 2 (dθ 2 +sin 2 θdφ 2 ), ( (r) = 1 2G NM r ) r h = 2G N M Area: A = 16πG 2 N M 2 Entropy: S A (l 2 l 2 Pl = G N, c = 1) Bekenstein (1972) Pl Temperature, T = κ 2π : κ =surface gravity Hawking (1974) Schwarzschild, κ = 1 4G N M T = 8πG N M Solar mass black hole: T = K, S 10 78

4 Temperature and Entropy Schwarzschild line element: ds 2 = (r)dt (r) dr2 +r 2 (dθ 2 +sin 2 θdφ 2 ), ( (r) = 1 2G NM r ) r h = 2G N M Area: A = 16πG 2 N M 2 Entropy: S A (l 2 l 2 Pl = G N, c = 1) Bekenstein (1972) Pl Temperature, T = κ 2π : κ =surface gravity Hawking (1974) Schwarzschild, κ = 1 4G N M T = 8πG N M Solar mass black hole: T = K, S 10 78

5 Temperature and Entropy Schwarzschild line element: ds 2 = (r)dt (r) dr2 +r 2 (dθ 2 +sin 2 θdφ 2 ), ( (r) = 1 2G NM r ) r h = 2G N M Area: A = 16πG 2 N M 2 Entropy: S A (l 2 l 2 Pl = G N, c = 1) Bekenstein (1972) Pl Temperature, T = κ 2π : κ =surface gravity Hawking (1974) Schwarzschild, κ = 1 4G N M T = 8πG N M Solar mass black hole: T = K, S 10 78

6 Temperature and Entropy Schwarzschild line element: ds 2 = (r)dt (r) dr2 +r 2 (dθ 2 +sin 2 θdφ 2 ), ( (r) = 1 2G NM r ) r h = 2G N M Area: A = 16πG 2 N M 2 Entropy: S A (l 2 l 2 Pl = G N, c = 1) Bekenstein (1972) Pl Temperature, T = κ 2π : κ =surface gravity Hawking (1974) Schwarzschild, κ = 1 4G N M T = 8πG N M Solar mass black hole: T = K, S 10 78

7 Temperature and Entropy Schwarzschild line element: ds 2 = (r)dt (r) dr2 +r 2 (dθ 2 +sin 2 θdφ 2 ), ( (r) = 1 2G NM r ) r h = 2G N M Area: A = 16πG 2 N M 2 Entropy: S A (l 2 l 2 Pl = G N, c = 1) Bekenstein (1972) Pl Temperature, T = κ 2π : κ =surface gravity Hawking (1974) Schwarzschild, κ = 1 4G N M T = 8πG N M Solar mass black hole: T = K, S 10 78

8 Temperature and Entropy Schwarzschild line element: ds 2 = (r)dt (r) dr2 +r 2 (dθ 2 +sin 2 θdφ 2 ), ( (r) = 1 2G NM r ) r h = 2G N M Area: A = 16πG 2 N M 2 Entropy: S A (l 2 l 2 Pl = G N, c = 1) Bekenstein (1972) Pl Temperature, T = κ 2π : κ =surface gravity Hawking (1974) Schwarzschild, κ = 1 4G N M T = 8πG N M Solar mass black hole: T = K, S 10 78

9 First and Second laws Internal energy U(S), T = U S : identify M = U(S) First Law of Black Hole Thermodynamics With S = a A G N dm = du = T ds r h = 2G N M, A = 16πGN 2M2, ds = 32πaG N TdS = 4adM a = 1 4 Bekenstein-Hawking Entropy S = 1 4 A l 2 Pl MdM, T = 8πG N M,

10 First and Second laws Internal energy U(S), T = U S : identify M = U(S) First Law of Black Hole Thermodynamics With S = a A G N dm = du = T ds r h = 2G N M, A = 16πGN 2M2, ds = 32πaG N TdS = 4adM a = 1 4 Bekenstein-Hawking Entropy S = 1 4 A l 2 Pl MdM, T = 8πG N M,

11 First and Second laws Internal energy U(S), T = U S : identify M = U(S) First Law of Black Hole Thermodynamics With S = a A G N dm = du = T ds r h = 2G N M, A = 16πGN 2M2, ds = 32πaG N TdS = 4adM a = 1 4 Bekenstein-Hawking Entropy S = 1 4 A l 2 Pl MdM, T = 8πG N M,

12 First and Second laws Internal energy U(S), T = U S : identify M = U(S) First Law of Black Hole Thermodynamics With S = a A G N dm = du = T ds r h = 2G N M, A = 16πGN 2M2, ds = 32πaG N TdS = 4adM a = 1 4 Bekenstein-Hawking Entropy S = 1 4 A l 2 Pl MdM, T = 8πG N M,

13 First and Second laws Internal energy U(S), T = U S : identify M = U(S) First Law of Black Hole Thermodynamics With S = a A G N dm = du = T ds r h = 2G N M, A = 16πGN 2M2, ds = 32πaG N TdS = 4adM a = 1 4 Bekenstein-Hawking Entropy S = 1 4 A l 2 Pl MdM, T = 8πG N M,

14 First and Second laws Internal energy U(S), T = U S : identify M = U(S) First Law of Black Hole Thermodynamics With S = a A G N dm = du = T ds r h = 2G N M, A = 16πGN 2M2, ds = 32πaG N TdS = 4adM a = 1 4 Bekenstein-Hawking Entropy S = 1 4 A l 2 Pl MdM, T = 8πG N M,

15 Hawking radiation Free energy, F(T), is Legendre transform of U(S) F = U TS = M κa 8πG N Schwarzschild: F = M 2 = 16πGT Heat capacity: C = U T = T 2 F T 2 = 8πG NM 2 = 2S < 0 Negative Heat capacity! Radiates with power P AT 4 3 G 2 n M2 Lifetime τ M P G2 N M3, M 1012 kg τ years If the black hole is shrinking, is the heat capacity C V or C P?

16 Hawking radiation Free energy, F(T), is Legendre transform of U(S) F = U TS = M κa 8πG N Schwarzschild: F = M 2 = 16πGT Heat capacity: C = U T = T 2 F T 2 = 8πG NM 2 = 2S < 0 Negative Heat capacity! Radiates with power P AT 4 3 G 2 n M2 Lifetime τ M P G2 N M3, M 1012 kg τ years If the black hole is shrinking, is the heat capacity C V or C P?

17 Hawking radiation Free energy, F(T), is Legendre transform of U(S) F = U TS = M κa 8πG N Schwarzschild: F = M 2 = 16πGT Heat capacity: C = U T = T 2 F T 2 = 8πG NM 2 = 2S < 0 Negative Heat capacity! Radiates with power P AT 4 3 G 2 n M2 Lifetime τ M P G2 N M3, M 1012 kg τ years If the black hole is shrinking, is the heat capacity C V or C P?

18 Hawking radiation Free energy, F(T), is Legendre transform of U(S) F = U TS = M κa 8πG N Schwarzschild: F = M 2 = 16πGT Heat capacity: C = U T = T 2 F T 2 = 8πG NM 2 = 2S < 0 Negative Heat capacity! Radiates with power P AT 4 3 G 2 n M2 Lifetime τ M P G2 N M3, M 1012 kg τ years If the black hole is shrinking, is the heat capacity C V or C P?

19 Hawking radiation Free energy, F(T), is Legendre transform of U(S) F = U TS = M κa 8πG N Schwarzschild: F = M 2 = 16πGT Heat capacity: C = U T = T 2 F T 2 = 8πG NM 2 = 2S < 0 Negative Heat capacity! Radiates with power P AT 4 3 G 2 n M2 Lifetime τ M P G2 N M3, M 1012 kg τ years If the black hole is shrinking, is the heat capacity C V or C P?

20 Pressure and enthalpy First Law of thermodynamics du = T ds PdV Where is the P dv term in the first law? Include cosmological constant Λ, contributes pressure P and energy density ǫ = P = Λ 8πG N Henneaux+Teitelboim (1984); (1989); Teitelboim (1985) Thermal energy Enthalpy U = U(S,V) U = M +ǫv = M PV M = U +PV M = U +PV = H(S,P) Kastor, Ray+Traschen [ ]

21 Pressure and enthalpy First Law of thermodynamics du = T ds PdV Where is the P dv term in the first law? Include cosmological constant Λ, contributes pressure P and energy density ǫ = P = Λ 8πG N Henneaux+Teitelboim (1984); (1989); Teitelboim (1985) Thermal energy Enthalpy U = U(S,V) U = M +ǫv = M PV M = U +PV M = U +PV = H(S,P) Kastor, Ray+Traschen [ ]

22 Pressure and enthalpy First Law of thermodynamics du = T ds PdV Where is the P dv term in the first law? Include cosmological constant Λ, contributes pressure P and energy density ǫ = P = Λ 8πG N Henneaux+Teitelboim (1984); (1989); Teitelboim (1985) Thermal energy Enthalpy U = U(S,V) U = M +ǫv = M PV M = U +PV M = U +PV = H(S,P) Kastor, Ray+Traschen [ ]

23 Pressure and enthalpy First Law of thermodynamics du = T ds PdV Where is the P dv term in the first law? Include cosmological constant Λ, contributes pressure P and energy density ǫ = P = Λ 8πG N Henneaux+Teitelboim (1984); (1989); Teitelboim (1985) Thermal energy Enthalpy U = U(S,V) U = M +ǫv = M PV M = U +PV M = U +PV = H(S,P) Kastor, Ray+Traschen [ ]

24 Pressure and enthalpy First Law of thermodynamics du = T ds PdV Where is the P dv term in the first law? Include cosmological constant Λ, contributes pressure P and energy density ǫ = P = Λ 8πG N Henneaux+Teitelboim (1984); (1989); Teitelboim (1985) Thermal energy Enthalpy U = U(S,V) U = M +ǫv = M PV M = U +PV M = U +PV = H(S,P) Kastor, Ray+Traschen [ ]

25 What is V? Include Λ, (r) = 1 2G NM r Λ 3 r2 (Λ < 0), (r h ) = 0 M(r h,λ) = H(S,P), Thermodynamic definition of V (BPD [ ]): V = ( ) H P S S = πr2 h, P = Λ l 2 8πG Pl N V = 4 (l Pl S) 3 2 = 4πr3 h 3 π 3 Problem S = πr2 h and V = 4πr3 l 2 h 3 are not independent! Pl Resolved by introducing rotation, J

26 What is V? Include Λ, (r) = 1 2G NM r Λ 3 r2 (Λ < 0), (r h ) = 0 M(r h,λ) = H(S,P), Thermodynamic definition of V (BPD [ ]): V = ( ) H P S S = πr2 h, P = Λ l 2 8πG Pl N V = 4 (l Pl S) 3 2 = 4πr3 h 3 π 3 Problem S = πr2 h and V = 4πr3 l 2 h 3 are not independent! Pl Resolved by introducing rotation, J

27 What is V? Include Λ, (r) = 1 2G NM r Λ 3 r2 (Λ < 0), (r h ) = 0 M(r h,λ) = H(S,P), Thermodynamic definition of V (BPD [ ]): V = ( ) H P S S = πr2 h, P = Λ l 2 8πG Pl N V = 4 (l Pl S) 3 2 = 4πr3 h 3 π 3 Problem S = πr2 h and V = 4πr3 l 2 h 3 are not independent! Pl Resolved by introducing rotation, J

28 What is V? Include Λ, (r) = 1 2G NM r Λ 3 r2 (Λ < 0), (r h ) = 0 M(r h,λ) = H(S,P), Thermodynamic definition of V (BPD [ ]): V = ( ) H P S S = πr2 h, P = Λ l 2 8πG Pl N V = 4 (l Pl S) 3 2 = 4πr3 h 3 π 3 Problem S = πr2 h and V = 4πr3 l 2 h 3 are not independent! Pl Resolved by introducing rotation, J

29 What is V? Include Λ, (r) = 1 2G NM r Λ 3 r2 (Λ < 0), (r h ) = 0 M(r h,λ) = H(S,P), Thermodynamic definition of V (BPD [ ]): V = ( ) H P S S = πr2 h, P = Λ l 2 8πG Pl N V = 4 (l Pl S) 3 2 = 4πr3 h 3 π 3 Problem S = πr2 h and V = 4πr3 l 2 h 3 are not independent! Pl Resolved by introducing rotation, J

30 AdS-Kerr Thermodynamics Rotating black-hole in asymptotically AdS (Kerr-AdS) Angular momentum J = am (G N = = 1,Λ = 3 L 2 ) Area of event horizon: A = 4π(r2 h +a2 ) = 4S 1 a2 L 2 Mass and enthalpy (1+ M = 1 ){ 8PS ( 3 S PS ) 3 +4π 2 J 2} := H(S,P,J) 2 πs Sorkin (1982); Caldarelli, Gognola+Klemm [hep-th/ ] Thermodynamic volume: V = H P = 2 ( {S PS } )+2π 2 J 2 S,J 3πH 3 Cvetic, Gibbons+Kubiz nák [ ]; BPD [ ] ( ) 3V 1/3 4π Equality only for J = 0 ( ) S 1/2 π

31 AdS-Kerr Thermodynamics Rotating black-hole in asymptotically AdS (Kerr-AdS) Angular momentum J = am (G N = = 1,Λ = 3 L 2 ) Area of event horizon: A = 4π(r2 h +a2 ) = 4S 1 a2 L 2 Mass and enthalpy (1+ M = 1 ){ 8PS ( 3 S PS ) 3 +4π 2 J 2} := H(S,P,J) 2 πs Sorkin (1982); Caldarelli, Gognola+Klemm [hep-th/ ] Thermodynamic volume: V = H P = 2 ( {S PS } )+2π 2 J 2 S,J 3πH 3 Cvetic, Gibbons+Kubiz nák [ ]; BPD [ ] ( ) 3V 1/3 4π Equality only for J = 0 ( ) S 1/2 π

32 AdS-Kerr Thermodynamics Rotating black-hole in asymptotically AdS (Kerr-AdS) Angular momentum J = am (G N = = 1,Λ = 3 L 2 ) Area of event horizon: A = 4π(r2 h +a2 ) = 4S 1 a2 L 2 Mass and enthalpy (1+ M = 1 ){ 8PS ( 3 S PS ) 3 +4π 2 J 2} := H(S,P,J) 2 πs Sorkin (1982); Caldarelli, Gognola+Klemm [hep-th/ ] Thermodynamic volume: V = H P = 2 ( {S PS } )+2π 2 J 2 S,J 3πH 3 Cvetic, Gibbons+Kubiz nák [ ]; BPD [ ] ( ) 3V 1/3 4π Equality only for J = 0 ( ) S 1/2 π

33 AdS-Kerr Thermodynamics Rotating black-hole in asymptotically AdS (Kerr-AdS) Angular momentum J = am (G N = = 1,Λ = 3 L 2 ) Area of event horizon: A = 4π(r2 h +a2 ) = 4S 1 a2 L 2 Mass and enthalpy (1+ M = 1 ){ 8PS ( 3 S PS ) 3 +4π 2 J 2} := H(S,P,J) 2 πs Sorkin (1982); Caldarelli, Gognola+Klemm [hep-th/ ] Thermodynamic volume: V = H P = 2 ( {S PS } )+2π 2 J 2 S,J 3πH 3 Cvetic, Gibbons+Kubiz nák [ ]; BPD [ ] ( ) 3V 1/3 4π Equality only for J = 0 ( ) S 1/2 π

34 AdS-Kerr Thermodynamics Rotating black-hole in asymptotically AdS (Kerr-AdS) Angular momentum J = am (G N = = 1,Λ = 3 L 2 ) Area of event horizon: A = 4π(r2 h +a2 ) = 4S 1 a2 L 2 Mass and enthalpy (1+ M = 1 ){ 8PS ( 3 S PS ) 3 +4π 2 J 2} := H(S,P,J) 2 πs Sorkin (1982); Caldarelli, Gognola+Klemm [hep-th/ ] Thermodynamic volume: V = H P = 2 ( {S PS } )+2π 2 J 2 S,J 3πH 3 Cvetic, Gibbons+Kubiz nák [ ]; BPD [ ] ( ) 3V 1/3 4π Equality only for J = 0 ( ) S 1/2 π

35 AdS-Kerr Thermodynamics Rotating black-hole in asymptotically AdS (Kerr-AdS) Angular momentum J = am (G N = = 1,Λ = 3 L 2 ) Area of event horizon: A = 4π(r2 h +a2 ) = 4S 1 a2 L 2 Mass and enthalpy (1+ M = 1 ){ 8PS ( 3 S PS ) 3 +4π 2 J 2} := H(S,P,J) 2 πs Sorkin (1982); Caldarelli, Gognola+Klemm [hep-th/ ] Thermodynamic volume: V = H P = 2 ( {S PS } )+2π 2 J 2 S,J 3πH 3 Cvetic, Gibbons+Kubiz nák [ ]; BPD [ ] ( ) 3V 1/3 4π Equality only for J = 0 ( ) S 1/2 π

36 Critical behaviour JP T > 0 below green curve, locally stable above red curve C P on red curve: two values of SP for given JP ( small and large black holes), merge to one at critical point (JP) crit, (SP) crit (Caldarelli, Gognola+Klemm [hep-th/ ]) (SP) crit , (JP) crit , (VP 3 2) crit SP

37 Critical behaviour JP T > 0 below green curve, locally stable above red curve C P on red curve: two values of SP for given JP ( small and large black holes), merge to one at critical point (JP) crit, (SP) crit (Caldarelli, Gognola+Klemm [hep-th/ ]) (SP) crit , (JP) crit , (VP 3 2) crit SP

38 Equation of state and critical exponents Define t := T T c T c, v := V V c V c, p := P P c P c Expand the equation of state about the critical point: p = 242t 081tv 021v 3 +o(t 2,tv 2,v 4 ) cf van der Waals gas: p = 4t 6tv 3 2 v3 +o(t 2,tv 2,v 4 ) C V = T/ T S V,J t α ; At fixed p < 0, v > v < t β ; Isothermal compressibility, κ T = 1 ( V ) V P T,J t γ ; At t = 0, p v δ ; Mean Field Exponents α = 0, β = 1, γ = 1, δ = 3 2

39 Equation of state and critical exponents Define t := T T c T c, v := V V c V c, p := P P c P c Expand the equation of state about the critical point: p = 242t 081tv 021v 3 +o(t 2,tv 2,v 4 ) cf van der Waals gas: p = 4t 6tv 3 2 v3 +o(t 2,tv 2,v 4 ) C V = T/ T S V,J t α ; At fixed p < 0, v > v < t β ; Isothermal compressibility, κ T = 1 ( V ) V P T,J t γ ; At t = 0, p v δ ; Mean Field Exponents α = 0, β = 1, γ = 1, δ = 3 2

40 Equation of state and critical exponents Define t := T T c T c, v := V V c V c, p := P P c P c Expand the equation of state about the critical point: p = 242t 081tv 021v 3 +o(t 2,tv 2,v 4 ) cf van der Waals gas: p = 4t 6tv 3 2 v3 +o(t 2,tv 2,v 4 ) C V = T/ T S V,J t α ; At fixed p < 0, v > v < t β ; Isothermal compressibility, κ T = 1 ( V ) V P T,J t γ ; At t = 0, p v δ ; Mean Field Exponents α = 0, β = 1, γ = 1, δ = 3 2

41 Equation of state and critical exponents Define t := T T c T c, v := V V c V c, p := P P c P c Expand the equation of state about the critical point: p = 242t 081tv 021v 3 +o(t 2,tv 2,v 4 ) cf van der Waals gas: p = 4t 6tv 3 2 v3 +o(t 2,tv 2,v 4 ) C V = T/ T S V,J t α ; At fixed p < 0, v > v < t β ; Isothermal compressibility, κ T = 1 ( V ) V P T,J t γ ; At t = 0, p v δ ; Mean Field Exponents α = 0, β = 1, γ = 1, δ = 3 2

42 Equation of state and critical exponents Define t := T T c T c, v := V V c V c, p := P P c P c Expand the equation of state about the critical point: p = 242t 081tv 021v 3 +o(t 2,tv 2,v 4 ) cf van der Waals gas: p = 4t 6tv 3 2 v3 +o(t 2,tv 2,v 4 ) C V = T/ T S V,J t α ; At fixed p < 0, v > v < t β ; Isothermal compressibility, κ T = 1 ( V ) V P T,J t γ ; At t = 0, p v δ ; Mean Field Exponents α = 0, β = 1, γ = 1, δ = 3 2

43 Equation of state and critical exponents Define t := T T c T c, v := V V c V c, p := P P c P c Expand the equation of state about the critical point: p = 242t 081tv 021v 3 +o(t 2,tv 2,v 4 ) cf van der Waals gas: p = 4t 6tv 3 2 v3 +o(t 2,tv 2,v 4 ) C V = T/ T S V,J t α ; At fixed p < 0, v > v < t β ; Isothermal compressibility, κ T = 1 ( V ) V P T,J t γ ; At t = 0, p v δ ; Mean Field Exponents α = 0, β = 1, γ = 1, δ = 3 2

44 Gibbs Free Energy Gibbs Free Energy, G(T,P,J) = H(S,P,J) TS: (J = 1) P G T

45 Kerr-Reissner-Nordström-AdS Similar story for J = 0,Q 0 (Reissner-Nordström AdS) Gibbs free energy is a swallowtail, same as van der Waals Champlin, Emparan, Johnson+Myers [hep-th/ ; ] Exponents are mean field, Kubiz nák+mann [arxiv: ] J 0, Q 0 Caldarelli, Gognola+Klemm [hep-th/ ]

46 Kerr-Reissner-Nordström-AdS Similar story for J = 0,Q 0 (Reissner-Nordström AdS) Gibbs free energy is a swallowtail, same as van der Waals Champlin, Emparan, Johnson+Myers [hep-th/ ; ] Exponents are mean field, Kubiz nák+mann [arxiv: ] J 0, Q 0 Caldarelli, Gognola+Klemm [hep-th/ ]

47 Kerr-Reissner-Nordström-AdS Similar story for J = 0,Q 0 (Reissner-Nordström AdS) Gibbs free energy is a swallowtail, same as van der Waals Champlin, Emparan, Johnson+Myers [hep-th/ ; ] Exponents are mean field, Kubiz nák+mann [arxiv: ] J 0, Q 0 Caldarelli, Gognola+Klemm [hep-th/ ]

48 Q Kerr-Reissner-Nordström-AdS Similar story for J = 0,Q 0 (Reissner-Nordström AdS) Gibbs free energy is a swallowtail, same as van der Waals Champlin, Emparan, Johnson+Myers [hep-th/ ; ] Exponents are mean field, Kubiz nák+mann [arxiv: ] J 0, Q 0 Caldarelli, Gognola+Klemm [hep-th/ ] J Line of second order critical points (mean field) Unique black hole Large and small black holes

49 Conclusions Λ 0 P dv term in black hole 1st law First Law du = TdS +ΩdJ +ΦdQ PdV Black hole mass is identified with enthalpy, H(S,P): dm = dh = T ds +ΩdJ +ΦdQ +V dp Thermodynamic volume: V = ( ) H P S When J 0, V and S are not independent, ( ) 3V 2 ( ) S 3, 4π π reverse iso-perimetric inequality Gibbs free energy: G(T,P) = T lnz, (Z = e I E) Line of second order critical points in J Q plane with mean field exponents and van der Waals type equation of state

50 Outlook Heat capacity: C P = T T ; C V = S T T P Condensed matter applications? S V Higher dimensional rotating black holes; other backgrounds? de Sitter, Λ > 0, P < 0? Beyond mean field?

51 Smarr relation Ordinary thermodynamics: U(S,V,n) (n = number of moles) is a function of extensive variables U is also extensive λ d U(S,V,n) = U(λ d S,λ d V,λ d n) U = S U U +V S V +n U n Euler equation U = ST VP +nµ (µ = chemical potential) G = U +VP ST = nµ Gibbs-Duhem relation Black hole: S λ 2 S, P λ 2 P, J λ 2 J, M λm λh(s,p,j) = H(λ 2 S,λ 2 P,λ 2 J) M = H(S,P,J) H = 2S H S 2P H P +2J H J H = 2ST 2VP +2J Ω Smarr relation Smarr (1973); Kastor, Ray+Traschen [ ]

52 Penrose Processes Mechanical energy: dw = du = TdS ΩdJ ΦdQ +PdV Most efficient is S = const and decrease J and Q Maximum work starting from an extremal black-hole Start from J = J max, Q = 0 and keep P constant maximum efficiency: η max = 518%, for PS = 1837 BPD [ ] (compare η max = = 29%, eg Wald (1984)) Charged-Kerr-AdS: η max = 75% (compare η max = 50%)

53 Compressibility BPD [arxiv: ] V P Adiabatic compressibility: κ = 1 V κ J=0 = 0 Maximum for J max (T = 0): κ max = eg P = 0, S,J 0 2S(1+8PS) (3+8PS) 2 (1+4PS) ( ) 2 κ max = 2S 9 = 4πM2 9 = M M ms 2 kg 1 cf neutron star, M M, R 10km, degenerate Fermi gas κ ms 2 kg 1 Very stiff equation of state! ρ = M V, speed of sound v 2 s = ρ P, S,J Speed of Sound vs 2 (2πJ) 4 = 1+ ) 2 (2S 2 +(2πJ) v2 s 1, with v s = 1 for J = 0

54 Internal Energy U = H +PV U(S,V,J) = ( π ) 3 ( )( ) 3V S 2 (3V ) 2 S 4π 2π 2 +J2 J 2 4π ( ) S 3 π, BPD [ ] As J 0, ( 3V 2 ( 4π) S 3 π) T = U J S V,J finite lim 2 J 0 is finite ( 3V 4π) 2 ( π) S 3 But lim 2 U J 0 C S 2 V = U T V = T S T V,J 0 as J 0 P = U V S,J Virial expansion at fixed T (large V), P = T 2 ( 4π 3V ) 1/3 1 ( ) 4π 2/3 + 4π J 2 ( ) J 4 8π 3V 3 V 2 +o V 11/3

55 P V diagram P as a function of ( 3V 4π) 1/3, curves of constant T for J = 1 Critical point at T c = 00418J 1/2, P c = /J and V C = 1158J 3/2 (Caldarelli, Gognola+Klemm [hep-th/ ])

56 Equation of state Equation of state T(V,P) = 4π { (3V 4π ) πGP ( 3V 4π } )1 3 P V

57 BTZ black hole d 2 s = (r)dt (r)dr 2 +r 2 dφ 2, (r) = 8G N M + r2 L 2 (Λ = 1 L 2 ) r h = 8G N ML, S = πr h 2l Pl (l Pl = G N ) Enthalpy H = M: H(S,P) = 4l Pl π S2 P Equation of state: Thermodynamic volume: PV 1 2 = π 4l Pl T V = πr 2 h Gibbs free energy: G = H TS = M

58 Heat capacity: C P = T = S > 0 P T S Phase transition Compare pure AdS 3 with (r) = 1+ r2 L 2 (Equivalent to setting M = 1 8G N in BTZ) Enthalpy, H = 1 8G N = constant, T = 0 Gibbs free energy: G = H = 1 8G N If M < 1 8G N in BTZ, pure AdS 3 has lower free energy Equivalent to temperature: T = 2G N P π

59 Quantum corrections to BTZ equation of state (Brown+Henneaux (1984); Maloney+Witten [ ]) Rotating BTZ black hole: (angular momentum J) ds 2 = (r)dt ( (r) dr2 +r 2 dφ 4G ) 2 NJ r 2 dt (r) = ) ( 8G N M + r2 L G2 N J2 r 2 Inner and outer event horizons: [ r± 2 = 4G N ML 2 1± 1 Hawking temperature: ( J ML ]1 ) 2 2 T = (r h ) 4π = (r2 h r2 ) 2πL 2 r h

60 Partition function Wick rotate to Euclidean time: t it E, J ij E Also r h r E,+ and r ir E, where [ ]1 ( ) 2 2 re,± 2 = 4G NML 2 JE 1+ ±1 ML Define τ = r E, +ir E,+ L Inverse Hawking temperature: 1 2πT = (Im(τ) > 0) r E,+ L 2 (re,+ 2 +r2 E, ) = L Im Define q = e 2πiτ in terms of which Z BTZ = (q q) L 16 G N to all orders in perturbation theory n=2 ( 1 ) τ 1 q n 2

61 Entropy corrections Set J = 0: T = r h 2π, τ = i r L 2 h L and q = e 4π2 LT The J = 0 partition function: Z BTZ = e π 2 TL G N n=2 ) (1 e 4π2 n 2 TL Defining x = TL = r h 2πL, the Gibbs free energy is G(T,P) = T lnz BTZ = π2 x 2 Entropy, S < 1 4 (area) 2G N +2T n=2 ln (1 e ) 4π2 nx

62 Equation of state (quantum volume): V(T,P) = G [ P = πrh 2 1 8πG N T L (BPD [arxiv: ]) n=2 ] n e 4π2nx 1 P classical quantum

63 Higher dimensional black holes Line element: d 2 s = (r)dt (r)dr 2 +r d d 2 Ω (d) Ω (d) = 2π d 2 : volume of d-dimensional unit sphere Γ( d 2) Generalisation: event horizon any d-dimensional Einstein space R ij = R d g ij with constant Ricci curvature R and unit radius volume Ω (d) (eg flat torus, or CP d 2 for even d) (r) = R d(d 1) 16πG N Ω (d) d Planck length l d Pl = G N : S = Ω (d) 4 r d h l d Pl M r d 1, P = Λ 8πG N 2Λ d(d +1) r2

64 Identify H(S,P) = M H(S,P) = S R 4π d 1 ( 4l d Pl S Ω (d) gives thermodynamic volume V = (BPD [arxiv: ]) Equation of state ( ) H P S ) 1 d + 16πG N P d +1 = Ω (d)r d+1 h d +1 ( 4l d Pl S Ω (d) { T = ( ) (d +1)V 1 ( ) 1 } d+1 (d +1)V d+1 R +16πGN P 4πd Ω (d) Ω (d) ) 1 d,

65 Heat capacity 16πG N P C P = Sd 16πG N P ( ) 2 4l d Pl S d Ω (d) ( ) 2 4l d Pl S d Ω (d) For R > 0, phase transition at T = 2 R(d) G N P d π For flat event horizon, R = 0, C P = Sd +R R,

Where is the PdV term in the first law of black hole thermodynamics?

arxiv:109.17v3 [gr-qc] 8 Nov 016 Where is the PdV term in the first law of black hole thermodynamics? Brian P. Dolan Department of Mathematical Physics, National University of Ireland, Maynooth, Ireland