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Finite canonical partition, energy, variance, entropy and free-energy identities

Physical scope

Exact finite-sum predictions for the adopted canonical law on a fixed finite microstate list with fixed internal energies; the statistical free-energy value is not automatically identified with a macroscopic Legendre potential or an empirical thermodynamic measurement.

Statement

Let $(Ω, ϵ)$ be a finite microstate model with energies $E_{x} = ϵ (x) J$ from Finite statistical microstate model. Let $T > 0$ , $β = (k_{B} T)^{- 1}$ , and $P_{β} (x) = e^{- β E_{x}} / Z (β)$ be the adopted preparation of Canonical sampling assumption for finite and normalized statistical models. With ${\overset{ˉ}{E}}_{β} = \sum_{x} P_{β} (x) E_{x}$ , ${Var}_{β} (E) = \sum_{x} P_{β} (x) (E_{x} - {\overset{ˉ}{E}}_{β})^{2}$ , and $S_{β} = - k_{B} \sum_{x} P_{β} (x) \log P_{β} (x)$ , the finite partition sum is positive and $- \partial_{β} \log Z = {\overset{ˉ}{E}}_{β}, \partial_{β}^{2} \log Z = {Var}_{β} (E), S_{β} = k_{B} (\log Z + β {\overset{ˉ}{E}}_{β}) .$ Consequently the statistical free-energy value $F_{s t a t} : = - k_{B} T \log Z$ satisfies $F_{s t a t} = {\overset{ˉ}{E}}_{β} - T S_{β}$ . The derivatives hold the finite state set and its energy list fixed. No equivalence with a macroscopic entropy or Helmholtz potential is asserted.

Facts & Assumptions

Given: A finite nonempty state set, a fixed energy list in joules, $T > 0$ , the fixed SI value of $k_{B}$ , and the adopted canonical weights.

[F1]

Finite statistical microstate model: $Ω$ is finite and nonempty and each $E_{x}$ is a fixed energy in joules.

[F2]

Canonical sampling assumption for finite and normalized statistical models: $P_{β} (x) = e^{- β E_{x}} / Z$ with $Z = \sum_{x} e^{- β E_{x}}$ and $β = 1 / (k_{B} T) > 0$ in $J^{- 1}$ .

[F3]

The exponential function is smooth and $(\exp)^{'} = \exp$ : $\frac{d}{d β} e^{- β E_{x}} = - E_{x} e^{- β E_{x}}$ for each fixed $E_{x}$ .

[F4]

The natural logarithm has derivative 1/x and equals the integral from 1 to x of 1/t and The natural logarithm as the inverse of the exponential function: for differentiable positive $Z$ , $(\log Z)^{'} = Z^{'} / Z$ .

Proof

Given: The finite canonical model and its fixed energy values.

1.1F1F2algebra

Since $Ω$ is finite and nonempty and every $e^{- β E_{x}} > 0$ , the finite sum $Z = \sum_{x} e^{- β E_{x}}$ is strictly positive and finite. The weights $P_{β} (x)$ therefore sum to one by their definition.

2.1F2F3F4step 1.1

Differentiating the finite sum term by term using [F3] gives $Z^{'} = - \sum_{x} E_{x} e^{- β E_{x}}$ and $Z^{''} = \sum_{x} E_{x}^{2} e^{- β E_{x}}$ . Applying [F4] yields $- \partial_{β} \log Z = - (Z^{'} / Z) = {\overset{ˉ}{E}}_{β}$ and $\partial_{β}^{2} \log Z = Z^{''} / Z - (Z^{'} / Z)^{2} = \sum_{x} P_{β} (x) E_{x}^{2} - {\overset{ˉ}{E}}_{β}^{2} = {Var}_{β} (E)$ .

3.1F2step 1.1algebra∎

For each $x$ , $\log P_{β} (x) = - β E_{x} - \log Z$ . Substituting this into the finite entropy sum gives $S_{β} = - k_{B} \sum_{x} P_{β} (x) \log P_{β} (x) = k_{B} (β {\overset{ˉ}{E}}_{β} + \log Z)$ . Since $β k_{B} = 1 / T$ , rearrangement gives ${\overset{ˉ}{E}}_{β} - T S_{β} = - k_{B} T \log Z = F_{s t a t}$ .

Finite canonical partition, energy, variance, entropy and free-energy identities

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources