Microcanonical Ensemble

Transcription

1 Entropy for Department of Physics, Chungbuk National University October 4, 2018

2 Entropy for A measure for the lack of information (ignorance): s i = log P i = log 1 P i. An average ignorance: S = k B i P is i = k B i P i log P i = k B log P i. We call S as entropy (Shannon s Entropy). k B : Boltzmann s constant.

3 Entropy for Maximum Entropy The prior probability distribution maximizes entropy (the average ignorance) while respecting macroscopic constraints. A natural constraint of normalization is i P i = 1. Applying the Lagrange multiplier method, [S + λ( i P i 1)] = 0, where stands for the derivative with respect to P. To be specific, [ S + λ( ] [ P i 1) = P i log P i + λ( ] P i 1) P P i i i = [ ] d log P i log P i P i + λ dp i i = i [ log P i 1 + λ] = 0.

4 Entropy for Maximum Entropy It leads log P i 1 + λ = 0. Finally, we get P i = e λ 1 = const. 1 Ω.

5 Entropy for Boltzmann s Entropy In an equilibrium state, the probability of i state is given by p i = 1 Ω. Then the entropy can be expressed as S = log P i = log 1 Ω = log Ω. We will see the meaning of entropy later.

6 Entropy for Boltzmann and Shannon Ludwig Boltzmann and Claude Shannon

7 Entropy for Fixed energy and the number of particles. We define Ω is the accessible volume in phase space with E H E + E. Ω = dp dq, E H E+ E where P = (p 1, p 2,, p 3N ) and Q = (q 1, q 2,, q 3N ). The probability and the expectation value is then P = 1 Ω, O = 1 Ω E H E+ E O(P, Q)dP dq. Conceptually the microcanonical ensemble approach is extremely simple, in practice it is not so easy.

8 Entropy for Ideal Gas in a box The Hamiltonian of N-ideal gas molecules in a box with volume V (i.e, helium atoms at high temperatures and low densities): H = 3N i p 2 i 2m + V (x i). (1) where the potential V (x i ) is given by 0 if x V and otherwise V =. The number (or volume) of states is Ω = dp dq = E H E+ E dq dp E H E+ E = Ω Q Ω P = V N Ω(P ), since there is V = 0 in a box.

9 Entropy for Momentum Space The constant energy surface is a sphere in 3N-dimensional space, 3N i=1 p 2 i = 2mE = R 2, where radius R = 2mE. If we define Σ(E) as the volume of region H E, Ω P = dp E H E+ E = Σ(E + E) Σ(E). Then, the volume Σ can be computed as Σ(E) = dp H E = 3N i=1 p2 i 2mE dp.

10 Entropy for The volume of n-dimensional sphere with radius R is V N (R) = dx 1 dx 2 dx N x2 i R2 = R N dy 1 dy 2 dy N Ni y2i 1 = R N C N, where C N = N i y2 i 1 dy 1dy 2 dy N. N i

11 Entropy for [ N N π = e dx] x2 = = dv N (R)e (x2 1 +x x2 N ) dx 1 dx 2 dx N e (x2 1 +x x2 N ) where dv N (R) = dx 1 dx 2 dx N = NR N 1 C N dr. In n-dimensional polar coordinates, N π = NR N 1 C N e R2 dr 0 = C N N 2 0 ( ) X N 2 1 e X N N dx = C N 2 Γ 2 = C N Γ (N/2 + 1) = C N (N/2)!, where X = R 2 and dx = 2RdR. Therefore, C N = πn/2 (N/2)!, V N(R) = R N πn/2 (N/2)!.

12 Entropy for Momentum Space The volume of region H E where Ω P = Σ(E + E) Σ(E) Σ(E + E) Σ(E) = E E E dσ(e) de, dσ(e) de = d π 3N/2 (2mE) 3N/2 de (3N/2)! = 3N 2 π 3N/2 (2mE) 3N/2 1 (3N/2)! = (2πm)3N/2 E 3N/2 1. (3N/2 1)! (2)

13 Entropy for Number of States & Entropy The total number (volume) of states is Ω = V N E (2πm)3N/2 E 3N/2 1. (3N/2 1)! (3) The entropy is then k B log Ω, and log Ω = log V N + log E + 3N 2 log(2πm) ( ) 3N log E log(3n/2 1)! N log V + 3N 2 log(2πme) (3N/2) log(3n/2) + (3N/2).

14 Entropy for Entropy for Ideal Gas The entropy is S Nk B log V + 3Nk B 2 1 T = log(2πme) (3Nk B /2) log(3n/2) + (3Nk B /2). P T = ( ) S E ( ) S V V,N E,N = 3Nk B 2E, = Nk B V. Therefore, we can derive equipartition theorem and the equation of state for the ideal gas, E = 3 2 Nk BT, P V = Nk B T.

15 Entropy for Gibbs Paradox Compare 2N particles in two boxes with volume V and 2N particles in a box with volume 2V : S 1 = 2k B log V N, S 2 = k B log(2v ) 2N.

16 Entropy for Gibbs Paradox & Gibbs factor All particles are perfectly identical or indistinsuishable: S 1 = 2k B log V N /N!, S 2 = k B log(2v ) 2N /(2N)!.

17 Entropy for Planck Constant h has the unit of momentum distance. The uncertainty principle: δxδp h. Applying two factors, we find the Sackur-Tetrode formula: Ω = Ω crude /(N!h 3n ). [ ( V 4πmE S(N, V, E) = Nk B log N 3Nh 2 ) 3/2 ] + Nk B 5 2. (4)