Black holes, the Van der Waals gas, compressibility and the speed - PowerPoint PPT Presentation

Black holes, the Van der Waals gas, compressibility and the speed of sound Brian P . Dolan Dept. of Mathematics, Heriot-Watt University, Edinburgh and Maxwell Institute for Mathematical Sciences, Edinburgh EMPG, 25th Sept. 2013 Brian Dolan Black hole compressibility 0/20

Outline Review of black hole thermodynamics Entropy and Temperature 1st and 2nd laws Hawking radiation Smarr relation Pressure and Enthalpy Enthalpy and the 1st law Volume Equation of state Critical behaviour Compressibility Speed of sound Conclusions Brian Dolan Black hole compressibility 1/20

Entropy and Temperature A Pl ( ℓ 2 Pl = � G / c 3 , G = c = 1). • Entropy: S ∝ ℓ 2 Bekenstein (1972) • Temperature, T = κ � 2 π : κ = surface gravity. Hawking (1974) 1 Schwarzschild black-hole: κ = 4 M � T = 8 π M . Solar mass black hole: T = 6 × 10 − 8 K , S ≈ 10 78 . • Internal energy U ( S ) : T = ∂ U ∂ S . Identify M = U ( S ) ⇒ dM = TdS . Schwarzschild: r h = 2 M , A = 4 π r 2 h = 16 π M 2 , dA = 32 π MdM , TdA Pl = 4 dM . ℓ 2 Brian Dolan Black hole compressibility 2/20

Entropy and Temperature A Pl ( ℓ 2 Pl = � G / c 3 , G = c = 1). • Entropy: S ∝ ℓ 2 Bekenstein (1972) • Temperature, T = κ � 2 π : κ = surface gravity. Hawking (1974) 1 Schwarzschild black-hole: κ = 4 M � T = 8 π M . Solar mass black hole: T = 6 × 10 − 8 K , S ≈ 10 78 . • Internal energy U ( S ) : T = ∂ U ∂ S . Identify M = U ( S ) ⇒ dM = TdS . Schwarzschild: r h = 2 M , A = 4 π r 2 h = 16 π M 2 , dA = 32 π MdM , TdA Pl = 4 dM . ℓ 2 Brian Dolan Black hole compressibility 2/20

Entropy and Temperature A Pl ( ℓ 2 Pl = � G / c 3 , G = c = 1). • Entropy: S ∝ ℓ 2 Bekenstein (1972) • Temperature, T = κ � 2 π : κ = surface gravity. Hawking (1974) 1 Schwarzschild black-hole: κ = 4 M � T = 8 π M . Solar mass black hole: T = 6 × 10 − 8 K , S ≈ 10 78 . • Internal energy U ( S ) : T = ∂ U ∂ S . Identify M = U ( S ) ⇒ dM = TdS . Schwarzschild: r h = 2 M , A = 4 π r 2 h = 16 π M 2 , dA = 32 π MdM , TdA Pl = 4 dM . ℓ 2 Brian Dolan Black hole compressibility 2/20

Entropy and Temperature A Pl ( ℓ 2 Pl = � G / c 3 , G = c = 1). • Entropy: S ∝ ℓ 2 Bekenstein (1972) • Temperature, T = κ � 2 π : κ = surface gravity. Hawking (1974) 1 Schwarzschild black-hole: κ = 4 M � T = 8 π M . Solar mass black hole: T = 6 × 10 − 8 K , S ≈ 10 78 . • Internal energy U ( S ) : T = ∂ U ∂ S . Identify M = U ( S ) ⇒ dM = TdS . Schwarzschild: r h = 2 M , A = 4 π r 2 h = 16 π M 2 , dA = 32 π MdM , TdA Pl = 4 dM . ℓ 2 Brian Dolan Black hole compressibility 2/20

Entropy and Temperature A Pl ( ℓ 2 Pl = � G / c 3 , G = c = 1). • Entropy: S ∝ ℓ 2 Bekenstein (1972) • Temperature, T = κ � 2 π : κ = surface gravity. Hawking (1974) 1 Schwarzschild black-hole: κ = 4 M � T = 8 π M . Solar mass black hole: T = 6 × 10 − 8 K , S ≈ 10 78 . • Internal energy U ( S ) : T = ∂ U ∂ S . Identify M = U ( S ) ⇒ dM = TdS . Schwarzschild: r h = 2 M , A = 4 π r 2 h = 16 π M 2 , dA = 32 π MdM , TdA Pl = 4 dM . ℓ 2 Brian Dolan Black hole compressibility 2/20

Entropy and Temperature A Pl ( ℓ 2 Pl = � G / c 3 , G = c = 1). • Entropy: S ∝ ℓ 2 Bekenstein (1972) • Temperature, T = κ � 2 π : κ = surface gravity. Hawking (1974) 1 Schwarzschild black-hole: κ = 4 M � T = 8 π M . Solar mass black hole: T = 6 × 10 − 8 K , S ≈ 10 78 . • Internal energy U ( S ) : T = ∂ U ∂ S . Identify M = U ( S ) ⇒ dM = TdS . Schwarzschild: r h = 2 M , A = 4 π r 2 h = 16 π M 2 , dA = 32 π MdM , TdA Pl = 4 dM . ℓ 2 Brian Dolan Black hole compressibility 2/20

Entropy and Temperature A Pl ( ℓ 2 Pl = � G / c 3 , G = c = 1). • Entropy: S ∝ ℓ 2 Bekenstein (1972) • Temperature, T = κ � 2 π : κ = surface gravity. Hawking (1974) 1 Schwarzschild black-hole: κ = 4 M � T = 8 π M . Solar mass black hole: T = 6 × 10 − 8 K , S ≈ 10 78 . • Internal energy U ( S ) : T = ∂ U ∂ S . Identify M = U ( S ) ⇒ dM = TdS . Schwarzschild: r h = 2 M , A = 4 π r 2 h = 16 π M 2 , dA = 32 π MdM , TdA Pl = 4 dM . ℓ 2 Brian Dolan Black hole compressibility 2/20

Entropy and Temperature • Entropy: S = 1 A Pl ( ℓ 2 Pl = � G / c 3 , G = c = 1). 4 ℓ 2 Bekenstein (1972) • Temperature, T = κ � 2 π : κ = surface gravity. Hawking (1974) 1 Schwarzschild black-hole: κ = 4 M � T = 8 π M . Solar mass black hole: T = 6 × 10 − 8 K , S ≈ 10 78 . • Internal energy U ( S ) : T = ∂ U ∂ S . Identify M = U ( S ) ⇒ dM = TdS . Schwarzschild: r h = 2 M , A = 4 π r 2 h = 16 π M 2 , dA = 32 π MdM , TdA Pl = 4 dM . ℓ 2 Brian Dolan Black hole compressibility 2/20

First and Second laws • More generally M = U ( S , J , Q ) , (angular momentum, electric charge), First Law of Black Hole Thermodynamics dM = T dS + Ω dJ + Φ dQ . • Can extract useful work in a Penrose process. • Kerr black hole ( Q = 0): 1 − 4 π 2 J 2 1 � � T = . S 2 8 π M S • Extremal: J max = 2 π ⇒ T = 0. • Reduce J ⇒ extract energy. 1 • Maximum efficiency for fixed S : η = 1 − 2 ≈ 29 % . √ Second Law of Black Hole Thermodynamics ∆ S ≥ 0 . Brian Dolan Black hole compressibility 3/20

First and Second laws • More generally M = U ( S , J , Q ) , (angular momentum, electric charge), First Law of Black Hole Thermodynamics dM = T dS + Ω dJ + Φ dQ . • Can extract useful work in a Penrose process. • Kerr black hole ( Q = 0): 1 − 4 π 2 J 2 1 � � T = . S 2 8 π M S • Extremal: J max = 2 π ⇒ T = 0. • Reduce J ⇒ extract energy. 1 • Maximum efficiency for fixed S : η = 1 − 2 ≈ 29 % . √ Second Law of Black Hole Thermodynamics ∆ S ≥ 0 . Brian Dolan Black hole compressibility 3/20

First and Second laws • More generally M = U ( S , J , Q ) , (angular momentum, electric charge), First Law of Black Hole Thermodynamics dM = T dS + Ω dJ + Φ dQ . • Can extract useful work in a Penrose process. • Kerr black hole ( Q = 0): 1 − 4 π 2 J 2 1 � � T = . S 2 8 π M S • Extremal: J max = 2 π ⇒ T = 0. • Reduce J ⇒ extract energy. 1 • Maximum efficiency for fixed S : η = 1 − 2 ≈ 29 % . √ Second Law of Black Hole Thermodynamics ∆ S ≥ 0 . Brian Dolan Black hole compressibility 3/20

First and Second laws • More generally M = U ( S , J , Q ) , (angular momentum, electric charge), First Law of Black Hole Thermodynamics dM = T dS + Ω dJ + Φ dQ . • Can extract useful work in a Penrose process. • Kerr black hole ( Q = 0): 1 − 4 π 2 J 2 1 � � T = . S 2 8 π M S • Extremal: J max = 2 π ⇒ T = 0. • Reduce J ⇒ extract energy. 1 • Maximum efficiency for fixed S : η = 1 − 2 ≈ 29 % . √ Second Law of Black Hole Thermodynamics ∆ S ≥ 0 . Brian Dolan Black hole compressibility 3/20

First and Second laws • More generally M = U ( S , J , Q ) , (angular momentum, electric charge), First Law of Black Hole Thermodynamics dM = T dS + Ω dJ + Φ dQ . • Can extract useful work in a Penrose process. • Kerr black hole ( Q = 0): 1 − 4 π 2 J 2 1 � � T = . S 2 8 π M S • Extremal: J max = 2 π ⇒ T = 0. • Reduce J ⇒ extract energy. 1 • Maximum efficiency for fixed S : η = 1 − 2 ≈ 29 % . √ Second Law of Black Hole Thermodynamics ∆ S ≥ 0 . Brian Dolan Black hole compressibility 3/20

First and Second laws • More generally M = U ( S , J , Q ) , (angular momentum, electric charge), First Law of Black Hole Thermodynamics dM = T dS + Ω dJ + Φ dQ . • Can extract useful work in a Penrose process. • Kerr black hole ( Q = 0): 1 − 4 π 2 J 2 1 � � T = . S 2 8 π M S • Extremal: J max = 2 π ⇒ T = 0. • Reduce J ⇒ extract energy. 1 • Maximum efficiency for fixed S : η = 1 − 2 ≈ 29 % . √ Second Law of Black Hole Thermodynamics ∆ S ≥ 0 . Brian Dolan Black hole compressibility 3/20

First and Second laws • More generally M = U ( S , J , Q ) , (angular momentum, electric charge), First Law of Black Hole Thermodynamics dM = T dS + Ω dJ + Φ dQ . • Can extract useful work in a Penrose process. • Kerr black hole ( Q = 0): 1 − 4 π 2 J 2 1 � � T = . S 2 8 π M S • Extremal: J max = 2 π ⇒ T = 0. • Reduce J ⇒ extract energy. 1 • Maximum efficiency for fixed S : η = 1 − 2 ≈ 29 % . √ Second Law of Black Hole Thermodynamics ∆ S ≥ 0 . Brian Dolan Black hole compressibility 3/20

First and Second laws • More generally M = U ( S , J , Q ) , (angular momentum, electric charge), First Law of Black Hole Thermodynamics dM = T dS + Ω dJ + Φ dQ . • Can extract useful work in a Penrose process. • Kerr black hole ( Q = 0): 1 − 4 π 2 J 2 1 � � T = . S 2 8 π M S • Extremal: J max = 2 π ⇒ T = 0. • Reduce J ⇒ extract energy. 1 • Maximum efficiency for fixed S : η = 1 − 2 ≈ 29 % . √ Second Law of Black Hole Thermodynamics ∆ S ≥ 0 . Brian Dolan Black hole compressibility 3/20

Hawking radiation ∂ T = − T ∂ 2 F • Heat capacity: C = ∂ U ∂ T 2 . • Free energy, F ( T ) , is Legendre transform of U ( S ) F = U − TS = M − κ A 8 π . • Schwarzschild: F = M � 2 = 16 π T . • Heat capacity: C = − 8 π M 2 = − 2 S < 0 . Negative! � • Radiates with power P ∼ AT 4 � � 3 ∼ M 2 . � , M ∼ 10 12 kg ⇒ τ ∼ 10 10 years . P ∼ M 3 Lifetime: τ ∼ M • Fixing J or Q � = stabilises the black hole. Brian Dolan Black hole compressibility 4/20