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Continuous ideal-gas bulk limit and finite-block bridge

Statement

Assume the Axiom of Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )). Fix dimension $d \geq 1$ , particle mass $m > 0$ , action reference $h > 0$ , inverse energy $β > 0$ , and positive cube volumes $V_{N}$ with $N / V_{N} \to ρ \in (0, \infty)$ . For $N$ indistinguishable classical noninteracting particles in a cube of volume $V_{N}$ , use the phase-space reference $d q d p / (N! h^{d N})$ and Hamiltonian $H_{N} (q, p) = \sum_{i = 1}^{N} ∣ p_{i} ∣^{2} / (2 m)$ . Define $λ_{β} = h \sqrt{β / (2 π m)}$ . Then the dimensionless partition integral and free-energy limits are $Z_{N} = \frac{V_{N}^{N}}{N! λ_{β}^{d N}}, \frac{- β^{- 1} \log Z_{N}}{N} ⟶ β^{- 1} [\log (ρ λ_{β}^{d}) - 1], \frac{- β^{- 1} \log Z_{N}}{V_{N}} ⟶ β^{- 1} ρ [\log (ρ λ_{β}^{d}) - 1] .$ There is also the following finite-momentum-block limit. Let $M = d N$ , let $G_{1}, \dots, G_{M}$ be independent standard normal variables, set $A_{N} = M^{- 1} \sum_{a = 1}^{M} G_{a}^{2}$ , and define canonical momenta $P_{a}^{c a n} = \sqrt{m / β} G_{a}$ and shell-prepared momenta $P_{a}^{s h} = \sqrt{m / β} G_{a} / \sqrt{A_{N}}$ . On the null event $A_{N} = 0$ , set the shell-prepared vector to $(\sqrt{m M / β}, 0, \dots, 0)$ . The shell-prepared vector has exactly energy $d N / (2 β)$ , and for every fixed $r$ and bounded continuous $ψ : R^{r} \to R$ , $E ψ (P_{1}^{s h}, \dots, P_{r}^{s h}) ⟶ E ψ (P_{1}^{c a n}, \dots, P_{r}^{c a n}) .$ This is weak convergence for each fixed block in this free-particle model. The shell preparation is the stated normalized-Gaussian construction, and no finite-width shell or interacting-gas equivalence is asserted.

Facts & Assumptions

Given: The free-particle Hamiltonian, phase-space normalization, and normalized-Gaussian shell preparation.

[A1]

Assume the Axiom of Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )), as required by the Lebesgue-measure construction and change-of-variables suppliers below. $d, N \geq 1$ , $m, h, β, V_{N} > 0$ , $N / V_{N} \to ρ \in (0, \infty)$ , and the $N$ -particle Hamiltonian is additive kinetic energy in a cubic position box.

[A2]

For the block limit, the $G_{a}$ are independent with density $(2 π)^{- 1 / 2} e^{- x^{2} / 2}$ and $r$ is fixed while $N \to \infty$ .

[F1]

Under [A1], Assuming countable choice, $L (R^{n})$ is a sigma-algebra containing every elementary set and $λ_{n}$ is a complete measure extending elementary volume makes the set function of Lebesgue measurable sets, the family $L (R^{n})$ , and the restricted set function $λ_{n}$ a complete measure, and Lebesgue measure is sigma-finite, and every metrically bounded subset of $R^{n}$ has finite outer measure makes it sigma-finite. Assuming countable choice, every Borel subset of $R^{n}$ is Lebesgue measurable makes continuous densities and their Borel level sets measurable; A box in $R^{n}$ with parameters $a_{i} \leq b_{i}$ is Lebesgue measurable of measure $\prod_{i < n} (b_{i} - a_{i})$ , whichever of its faces are included gives the position cube its stated volume and gives singletons measure zero. Integrable real and complex functions, and their integrals defines the signed integrals used below; The Lebesgue integral is linear on $L^{1} (μ)$ makes them linear, and Monotonicity and nonnegative homogeneity of the nonnegative integral gives the nonnegative monotonicity and homogeneity used in moment and tail bounds.

[F2]

The product measure of two sigma-finite measure spaces and For sigma-finite factors, the product measure exists, has the rectangle formula, is sigma-finite, and is unique: finite products of the sigma-finite Lebesgue factors carry the rectangular product measure. The Euclidean Lebesgue measure is the completion of the product of the factor Lebesgue measures identifies its completion with Euclidean Lebesgue measure, so the displayed Borel densities have the same integrals in either convention.

[F3]

Tonelli's theorem for nonnegative measurable functions on a sigma-finite product and Fubini's theorem for L^1 functions on a sigma-finite product: nonnegative product integrals factor by Tonelli; integrable products may be evaluated iteratively.

[F4]

The Gaussian integral $\int_{- \infty}^{\infty} e^{- x^{2}} d x = \sqrt{π}$ : the improper Riemann Gaussian integral is $\int_{- \infty}^{\infty} e^{- x^{2}} d x = \sqrt{π}$ . A nonnegative improper Riemann integral on a half-line agrees with the Lebesgue integral identifies each convergent nonnegative improper integral on $[0, \infty)$ with its Lebesgue integral. Apply it to the continuous even Gaussian and its even polynomial moments; reflection gives the negative half-line. Thus the improper integral evaluations below also evaluate their Lebesgue integrals.

[F5]

A C^1 diffeomorphism satisfies the change-of-variables formula for nonnegative Lebesgue measurable functions and Square roots exist: a unique $\sqrt{a} \geq 0$ with $(\sqrt{a})^{2} = a$ ; the positives are ${x^{2} : x \neq 0}$ : the scaling substitution $u = \sqrt{c} x$ is valid for $c > 0$ , so $\int_{R} e^{- c x^{2}} d x = \sqrt{π / c}$ .

[F6]

The natural logarithm as the inverse of the exponential function, The natural logarithm has derivative 1/x and equals the integral from 1 to x of 1/t, and The integral logarithm is continuous and strictly increasing on $(0, \infty)$ : the library natural logarithm equals the integral logarithm and is strictly increasing on positive reals.

[F7]

If $u, v$ are differentiable on $[a, b]$ with $u^{'}, v^{'}$ integrable, then $\int_{a}^{b} u v^{'} = u (b) v (b) - u (a) v (a) - \int_{a}^{b} u^{'} v$ : on each finite interval, differentiable factors with integrable derivatives obey the ordinary integration-by-parts identity.

[F8]

The exponential dominates every fixed nonnegative integer power at $+ \infty$ : $R^{k} e^{- R^{2} / 2} \to 0$ for each fixed nonnegative integer $k$ , by setting $x = R^{2} / 2$ and bounding $x^{k / 2}$ by a larger integer power.

[F9]

Heine-Borel in $R^{n}$ : with the Euclidean metric a subset of $R^{n}$ is compact if and only if it is closed and bounded, and the proof by bisection uses no choice principle; the same holds on the real line makes each closed bounded Euclidean ball compact, and Heine-Cantor: a continuous map from a compact metric space to any metric space is uniformly continuous makes a continuous test function uniformly continuous on that ball.

Proof

Given: [A1] and, for the second conclusion, [A2].

1.1A1F1F2F3F4F5

The integrand is nonnegative, so [F3] factors the position and momentum integrals: $Z_{N} = \frac{1}{N! h^{d N}} \int_{Λ_{N}^{N}} \int_{R^{d N}} e^{- β \sum_{i} ∣ p_{i} ∣^{2} / (2 m)} d p d q = \frac{V_{N}^{N}}{N! h^{d N}} {(\int_{R} e^{- β p^{2} / (2 m)} d p)}^{d N} .$ By [F4]–[F5] the one-coordinate integral is $\sqrt{2 π m / β}$ , proving the stated exact formula with $λ_{β} = h \sqrt{β / (2 π m)}$ . The dimensions are consistent: $λ_{β}$ is a length, so $V_{N}^{N} / λ_{β}^{d N}$ and $Z_{N}$ are dimensionless.

1.2A2F1F2F3F4F5F7F8algebra

Let $G$ have the density in [A2]. By [F4]–[F5], it is normalized. The polynomial and Gaussian factors have continuous derivatives on every finite interval, so apply [F7] on $[- R, R]$ first with $u (x) = x$ , $v^{'} (x) = x e^{- x^{2} / 2}$ , and then with $u (x) = x^{3}$ , $v^{'} (x) = x e^{- x^{2} / 2}$ . In each case $v (x) = - e^{- x^{2} / 2}$ . Letting $R \to \infty$ , the endpoint terms vanish by [F8], giving $E G^{2} = 1$ and $E G^{4} = 3$ . Thus $E (G^{2} - 1) = 0$ and $E (G^{2} - 1)^{2} = 2$ . The independence in [A2] and [F3] factor the cross integrals, so expanding the square gives $E (A_{N} - 1)^{2} = M^{- 2} \sum_{a = 1}^{M} 2 = 2 / M .$ For any $η > 0$ , $η^{2} 1_{{∣ A_{N} - 1 ∣ > η}} \leq (A_{N} - 1)^{2}$ , so integration proves $P (∣ A_{N} - 1 ∣ > η) \leq 2 / (M η^{2}) \to 0$ . Hence $A_{N} \to 1$ in probability.

2.1A1F6step 1.1algebra

Monotonicity of $\log$ from [F6] gives $\int_{1}^{N} \log x d x \leq \log N! \leq \int_{1}^{N + 1} \log x d x$ , hence $\log N! = N \log N - N + O (\log (N + 1))$ . Again by [F6], $0 \leq \log x = \int_{1}^{x} d t / t \leq 2 \sqrt{x}$ for $x \geq 1$ , so $\log (N + 1) / N \leq 2 \sqrt{N + 1} / N \to 0$ . Substituting the exact $Z_{N}$ yields $N^{- 1} \log Z_{N} = \log (V_{N} / N) + 1 - d \log λ_{β} + o (1) \to - \log ρ + 1 - d \log λ_{β} .$ Multiplication by $- β^{- 1}$ proves the per-particle limit; multiplying by $N / V_{N} \to ρ$ proves the per-volume limit.

2.2A2step 1.2

Since $A_{N} > 0$ almost surely, the shell-prepared formulas define random variables on a probability-one set. On the null event $A_{N} = 0$ , assign the fixed shell vector $(\sqrt{m M / β}, 0, \dots, 0)$ ; this measurable extension leaves every expectation unchanged and has the same stated energy. On $A_{N} > 0$ , algebra gives $\sum_{a = 1}^{d N} \frac{(P_{a}^{s h})^{2}}{2 m} = \frac{1}{2 β A_{N}} \sum_{a} G_{a}^{2} = \frac{M}{2 β} = \frac{d N}{2 β} .$ This exact energy constraint holds on every outcome where $A_{N} > 0$ , an event of probability one since the joint Gaussian density assigns zero mass to the origin.

2.3A2F1F3F9step 1.2

For fixed $r$ , the scalar multiplier $A_{N}^{- 1 / 2}$ tends to $1$ in probability by continuity at $1$ and step 1.2. Therefore $(P_{1}^{s h}, \dots, P_{r}^{s h}) - (P_{1}^{c a n}, \dots, P_{r}^{c a n})$ tends to zero in probability: restrict first to the event $\sum_{a = 1}^{r} G_{a}^{2} \leq R^{2}$ , where its norm is at most $\sqrt{m / β} R ∣ A_{N}^{- 1 / 2} - 1 ∣$ , and then bound the complement by $P (\sum_{a = 1}^{r} G_{a}^{2} > R^{2}) \leq r / R^{2}$ using the second moment. For bounded continuous $ψ$ , choose $R$ so this tail is small and additionally restrict to $∣ A_{N}^{- 1 / 2} - 1 ∣ \leq 1$ , whose complement has probability tending to zero. On this event both blocks lie in the ball of radius $2 \sqrt{m / β} R$ . Use uniform continuity of $ψ$ on the compact ball of radius $2 \sqrt{m / β} R$ , and then use convergence in probability of the difference. For any $ϵ > 0$ , choose $δ > 0$ by that uniform continuity. The expectation difference is at most $ϵ + 2 ∥ ψ ∥_{\infty} [r / R^{2} + P (∣ A_{N}^{- 1 / 2} - 1 ∣ > 1) + P (∥ P_{1 : r}^{s h} - P_{1 : r}^{c a n} ∥ > δ)]$ , with the multiplier assigned arbitrarily on its null exceptional event. Sending first $N \to \infty$ and then $R \to \infty$ proves the claimed expectation limit.

3.1A1A2F1step 2.3∎

If positions are included, choose them independently and uniformly in the same cube under both preparations. After coordinatewise rescaling by the side length, every fixed position block already has the same uniform law in both models and is independent of the momentum vector. The same bounded-continuous test argument as in 2.3, first on compact sets and then on their small tails, proves the joint fixed-block limit. This does not change the energy computation because the Hamiltonian has no position dependence.

Continuous ideal-gas bulk limit and finite-block bridge

Statement

Facts & Assumptions

Proof

Depends on

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