ph2b_final_soln

ph2b_final_soln - PH2b[15 pts Problem#1 Black Hole...

This preview shows pages 1–5. Sign up to view the full content.

PH2b [15 pts] Problem #1: Black Hole Thermodynamics (a) The energy of a photon is E γ = hc/λ . The total energy of N photons is E = Nhc/λ . The energy of an equivalent mass is E = Mc 2 . We just set them equal and Fnd this mass such that 2 = Nhc/λ (1) M = Nh/λc (2) (b) S k B N (3) N = Mλc/h (4) But we want the photons with the smallest possible energy, which would be giving it the largest possible wavelength, which we are told is radius of the black hole, R . N = MRc/h (5) and S k B MRc/h (6) (c) A black hole has no features other than M , so it has no features (i.e.) bumps and hills on it, so its surface area is that of a perfect sphere. A =4 πR 2 (7) R =( A 4 π ) 1 / 2 (8) M = Rc 2 2 G = c 2 2 G ( A 4 π ) 1 / 2 (9) S k B MRc/h = k B c 2 2 G A 4 π c h A (10) (d) R is a function of mass, and mass is a function of Energy so we can write the entropy E 2 . N = ER hc (11) = E hc 2 GM c 2 (12) = 2 G hc 5 E 2 (13) giving, S = k B N (14) = k B ± 2 G hc 5 E 2 ² (15) then ∂S ∂E = 1 T = k B ± 4 G hc 5 E ² (16) 1

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
PH2b T = hc 5 4 k B GE = hc 3 4 k B GM (17) (e) P = AσT 4 =4 πR 2 σ ± hc 3 4 k B GM ² 4 = c 8 h 4 πσ 16 G 2 k 4 B M 2 (18) (f) dE dt = c 2 dM dt = P = c 8 h 4 16 G 2 k 4 B M 2 where E = Mc 2 is the energy radiated M 2 dM = c 6 h 4 16 G 2 k 4 B dt Now we integrate both sides and get 1 3 M 3 = c 6 h 4 16 G 2 k 4 B t + C if we assume that at t = 0 the black hole has not radiated energy yet (therefore C=0) and at time t evapourate the mass is M o then t evapourate = 16 G 2 k 4 B M 3 o 3 c 6 h 4 (19) Then t sun 2 . 7 × 10 68 seconds and t sun 5 . 4 × 10 - 91 seconds. 2
PH2b [15 pts] Problem #2: Thermodynamic Potentials (a) (4 pts) To show that Φ is minimized with respect to particle and energy variations, we must frst show that its diFerential vanishes at equilibrium. At equilibrium, τ and μ are constants, so the diFerential o± Φ is given by d Φ = dU - τdσ - μdN =0 . (20) We must also show that Φ is indeed a minimum at equilibrium. Knowing that entropy is a maximum at equilibrium, we seek to write entropy in terms o± Φ. Assuming a system S ±ree to exchange energy and particles with a bath R , we write the total entropy as σ = σ R + σ S = σ R ( U - U S ,N - N S )+ σ S ( U S S ) ± σ R ( U,N ) - U S ± ∂σ R ∂U R ² V,N - N S ± R ∂N R ² V,U + σ S ( U S S ) . (21) Using ± R R ² V,N 1 τ , (22) and ± R R ² V,U ≡- μ τ , (23) we fnd σ = σ R ( ) - U S τ + μN S τ + σ S = σ R ( ) - Φ S τ . (24) Now σ R ( ) is a constant since U and N are constants, and σ is maximized at equilibrium. There±ore, Φ must be minimized at equilibrium. (b) (3 pts) The Gibbs sum Z is given by Z = ³ U,N g ( ) e ( - U ) , (25) where the multiplicity g ( ) gives the number o± states with energy U and particle number N . Writing the multiplicity as exp( σ ), we can write Z = ³ U,N e ( τσ + μN - U ) . (26) By substituting the expression ±or Φ, we fnd Z = ³ U,N e - Φ( τ,μ,V ; U,N ) . (27) The constant ±actor is unity! (Pretty trivial.) 3

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
PH2b (c) (4 pts) Φ is minimized at equilibrium; therefore, exp( - Φ ) is maximized at equilibrium, and the term in the Gibbs sum which dominates is the equilibrium term. Thus, we approximate Z ± e - Φ( τ,μ,V ) , (28) which gives us Φ( τ, μ, V )= - τ ln Z . (29)
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 01/08/2011 for the course PH 2b taught by Professor Martin during the Winter '08 term at Caltech.

Page1 / 11

ph2b_final_soln - PH2b[15 pts Problem#1 Black Hole...

This preview shows document pages 1 - 5. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online