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boltz07 - ln Pr E 1 = ln Ω E-E 1/kT Pr E 1 = Ze-E 1/kT...

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Phys. 123B May 21, 2007 A small system in contact with a heat reservoir Find the probability of the small system being in energy state E 1 . The small system is in state E 1 . E * is the total energy of the combined system and E 1 E * The reservoir has temperature T assumed to be fixed. When small system is in state E 1 the reservoir energy must be E = E * - E 1 The probability (Pr) of small system having energy E 1 is proportional to the multiplicity of the reservoir at energy E . The small system has multiplicity = 1. Pr( E 1 ) Ω( E ) = Ω( E * - E 1 ) Now expand in Taylor series the log of the probability around E * . ln Ω( E * - E 1 ) ln Ω( E * ) - βE 1 where β = ln Ω /∂E | E * β = 1 k ∂S/∂E = 1 kT where T is the reservoir temperature.
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Unformatted text preview: ln Pr( E 1 ) = ln Ω( E * )-E 1 /kT Pr( E 1 ) = Ze-E 1 /kT where Z is a constant independent of the energy of the small system. Since the probabilities sum to one Z = Σ i e-βE i Entropy change of a Reservoir. Reservoir energy E , Temperature T . The reservoir absorbs an amount of heat Q but temperature is constant. Expand log of multiplicity in Taylor series. Δ ln Ω = ln Ω( E + Q )-ln Ω( E ) = Q ∂ ln Ω( E ) ∂E | E Since S = k ln Ω and 1 /T = dS dE de±ne β = 1 kT Then Δ ln Ω = Qβ = Q kT The important result for a reservoir is Δ S = Q T...
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