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Residual entropy in a finite canonical model

Physical scope

A finite discrete Gibbs preparation with a specified energy gap and entropy convention, using a positive Kelvin-valued canonical parameter; no identification with an entropy-derived macroscopic temperature, macroscopic entropy postulate, or empirical material is assumed.

Example

Let $X = {1, \dots, N}$ be a finite set of microstates with energy values $E_{i}$ in joules. Let $E_{0} = \min_{i} E_{i}$ , let exactly $g \geq 1$ states have energy $E_{0}$ , and suppose every other state has energy at least $E_{0} + Δ$ for one fixed gap $Δ > 0$ joules. For $T > 0$ kelvins, use $T$ as the positive parameter in this finite canonical model and adopt the probabilities $w_{i} (T) = \exp ⁣ (- \frac{E_{i} - E_{0}}{k_{B} T}), Z (T) = \sum_{i = 1}^{N} w_{i} (T), p_{i} (T) = \frac{w_{i} (T)}{Z (T)}$ and the model entropy $S_{G} (T) = - k_{B} \sum_{i = 1}^{N} p_{i} (T) \log p_{i} (T) .$ Then $\lim_{T ↓ 0} S_{G} (T) = k_{B} \log g .$ This is a calculation for the adopted finite statistical model, not a deduction of the canonical preparation from macroscopic thermodynamics.

Facts & Assumptions

Given: The finite state set, its energy values, the positive gap, $T > 0$ , and the specified canonical probabilities and entropy in the Example.

[F1]

SI quantities and unit conventions: $k_{B} > 0$ has units $J K^{- 1}$ , so each exponent is dimensionless and $S_{G}$ is in $J K^{- 1}$ .

[F2]

Entropic temperature, pressure, and molar chemical potentials: The macroscopic thermodynamic temperature $T_{c}$ is the entropy-derived state function $1 / S_{c, U}$ on a designated open chart. The canonical parameter $T$ here is not identified with that function; the Gibbs probabilities are adopted as a separate finite model.

[F3]

The natural logarithm as the inverse of the exponential function: $\log$ is the natural logarithm with domain $(0, \infty)$ .

[F4]

Order, continuity, range, and the product, quotient, and reciprocal laws for the natural logarithm: $\log$ is continuous and strictly increasing, and $\log (1 / x) = - \log x$ for $x > 0$ .

[F5]

The exponential tends to $+ \infty$ at $+ \infty$ and to $0$ at $- \infty$ : $\exp (x) \to 0$ as $x \to - \infty$ .

[F6]

The logarithm grows more slowly than every positive real power: $\log x / x \to 0$ as $x \to + \infty$ .

Verification

1.1givenF1F2F5algebra

In the limit of the stipulated canonical parameter $T ↓ 0$ [F2], each excited state $i \notin G : = {i : E_{i} = E_{0}}$ has $(E_{i} - E_{0}) / (k_{B} T) \to + \infty$ , so $w_{i} (T) \to 0$ by [F5]. Ground-state weights are $1$ . Since there are finitely many states, $Z (T) \to g$ , each ground-state probability tends to $1 / g$ , and each excited-state probability tends to $0$ .

2.1F1F3F4step 1.1algebra

By continuity of $\log$ at $1 / g > 0$ from [F4], the total ground-state contribution to entropy tends to $- k_{B} g \frac{1}{g} \log (1 / g) = k_{B} \log g .$

2.2F3F4F6step 1.1algebra

For an excited state with $p_{i} (T) \to 0$ , set $x = 1 / p_{i} (T) \to + \infty$ . By [F4], $p_{i} \log p_{i} = - \log (x) / x$ , which tends to $0$ by [F6]. The finite sum of excited-state contributions therefore tends to $0$ .

3.1F1F4step 2.1step 2.2algebra∎

Combining the ground and excited contributions in the finite sum gives $S_{G} (T) \to k_{B} \log g$ . In the $g = 1$ case this limit is $0$ ; for $g > 1$ it is positive because $\log$ is strictly increasing and $\log 1 = 0$ .

Residual entropy in a finite canonical model

Example

Facts & Assumptions

Verification

Depends on

Dependency tree · 2 levels

Sources