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Finite-size mixing and distinguishability conventions

Physical scope

Exact finite-N calculation for two equal ideal-gas compartments at equal temperature; distinguishes adopted indistinguishable-label and fixed-species conventions and reports no empirical observation.

Example

Take two initially separate equal-volume boxes, each containing $N \geq 1$ particles of the same mass $m > 0$ at the same $T > 0$ , with ideal Hamiltonian $K = \sum p_{j}^{2} / (2 m)$ . Adopt the classical reference with $N!$ counting for indistinguishable labels and fixed action reference $a_{0} > 0$ . Removing the partition while fixing total particle number $2 N$ gives $\frac{Δ S_{i d e n t i c a l}}{k_{B}} = \log ⁣ (\frac{4^{N} (N!)^{2}}{(2 N)!}) > 0,$ whereas for two distinguishable species with fixed $N$ of each, $\frac{Δ S_{s p e c i e s}}{k_{B}} = 2 N \log 2.$ The probability under the final identical-particle positional law of finding exactly $N$ particles in each half is $(\binom{2 N}{N}) / 4^{N}$ , the reciprocal factor inside the logarithm above. The identical-particle entropy change is $O (k_{B} \log (N + 1))$ , hence sublinear in $N$ ; no exact zero at finite $N$ is claimed. For the distinguishable-species comparison, initially confine species $A$ (exactly $N$ particles) to the first box and species $B$ (exactly $N$ ) to the second; after removing the partition, each species can occupy the full volume $2 V$ .

Facts & Assumptions

Given: The two ideal-gas preparations, fixed action reference, equal temperature and the stated particle-counting convention.

[A1]

Each ideal gas uses the adopted classical $1 / N!$ quotient-counting reference when labels are operationally indistinguishable (Classical particle counting and physical observables).

[A2]

Both states are canonical at the same $β = (k_{B} T)^{- 1} > 0$ with the SI temperature convention SI quantities and unit conventions, and the position volumes are $V$ initially per side and $2 V$ after partition removal.

[F1]

The Gaussian integral $\int_{- \infty}^{\infty} e^{- x^{2}} d x = \sqrt{π}$ proves $\int_{R} e^{- x^{2}} d x = \sqrt{π}$ . For each $a > 0$ , $y = \sqrt{a} x$ is a linear diffeomorphism with Jacobian $\sqrt{a}$ , so $\int_{R} e^{- a x^{2}} d x = a^{- 1 / 2} \int_{R} e^{- y^{2}} d y = \sqrt{π / a}$ . Applying this scaling separately in each momentum coordinate gives the stated product of Gaussian factors.

[F2]

For a canonical Gibbs density, the entropy identity is $S = k_{B} (\log Z + β ⟨ H ⟩)$ , the equality case of the reference-relative Gibbs identity (Reference-relative entropy and Gibbs variational inequality).

Verification

Given: [A1]–[A2] and the counting, Gaussian and canonical entropy interfaces [F1]–[F2].

1.1A1A2F1F2algebra

For $r$ particles in position volume $W$ , separate the $r$ position integrals and use [F1] in all $d r$ momentum coordinates. For one coordinate put $a = β / (2 m)$ ; integrating $\frac{d}{d p} (p e^{- a p^{2}}) = e^{- a p^{2}} - 2 a p^{2} e^{- a p^{2}}$ over $R$ gives $⟨ p^{2} ⟩ = m / β$ , as the boundary term vanishes by Gaussian decay. Thus the mean kinetic energy is $d r / (2 β)$ . The position law factorizes into $r$ uniform position laws. With $λ_{T} = a_{0} \sqrt{β / (2 π m)}$ , the partition function is $Z_{r} (W) = \frac{W^{r}}{r! a_{0}^{d r}} {(\frac{2 π m}{β})}^{d r / 2} = \frac{(W / λ_{T}^{d})^{r}}{r!} .$ For equal total particle count $2 N$ before and after, the $β U$ terms in [F2] cancel, so the entropy change is $k_{B}$ times the logarithm of the partition-function ratio.

2.1A1A2step 1.1algebra

Initially the ratio is $Z_{2 N} (2 V) / Z_{N} (V)^{2}$ $= 4^{N} (N!)^{2} / (2 N)!$ ; for two fixed distinguishable species it is $[Z_{N} (2 V) / Z_{N} (V)]^{2} = 4^{N}$ . The first ratio exceeds one because $(2 N)! / (N!)^{2} = (\binom{2 N}{N}) < \sum_{j = 0}^{2 N} (\binom{2 N}{j}) = 4^{N}$ . Under the final positional law, the $2 N$ particles independently fall in either half with probability $1 / 2$ , so the central binomial probability is $(\binom{2 N}{N}) / 4^{N}$ , the reciprocal of the identical-particle ratio.

3.1A1step 2.1algebra∎

For each integer $n \geq 1$ , monotonicity of $\log x$ on unit intervals gives $n \log n - n + 1 \leq \log (n!) \leq n \log n - n + 1 + \log n .$ Indeed the lower bound integrates $\log x \leq \log j$ on $[j - 1, j]$ for $j = 2, \dots, n$ , and the upper bound integrates $\log x \geq \log j$ on $[j, j + 1]$ for $j = 1, \dots, n - 1$ , then adds $\log n$ . Substitution for $n = N$ and $n = 2 N$ cancels the terms linear in $N$ in $N \log 4 + 2 \log (N!) - \log ((2 N)!)$ , leaving a quantity bounded in absolute value by a constant multiple of $\log (N + 1)$ . Therefore $Δ S_{i d e n t i c a l} / N \to 0$ as $N \to \infty$ .

Remarks

This is a finite mathematical consequence of the two declared counting conventions. It is not a mixing experiment, and the macroscopically sublinear correction does not make the finite- $N$ entropy change vanish.

Finite-size mixing and distinguishability conventions

Example

Facts & Assumptions

Verification

Remarks

Depends on

Dependency tree · 2 levels

Sources