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Finite spherical-shell gas and its exact marginal

Physical scope

Finite-N classical ideal kinetic energy for one operationally indistinguishable species, conditional on Countable Choice for the exact shell marginal, with a specifically adopted rotational Gaussian-direction preparation; no dynamical ergodicity or general coarea claim.

Statement

Assume ${A C}_{ω}$ as defined in The Axiom of Countable Choice ( ${A C}_{ω}$ ), the mathematical hypothesis required by the spherical-marginal supplier. Let $N, d \geq 1$ , $D = d N \geq 2$ , and let $B = \prod_{j = 1}^{d} [0, L_{j}] \subset R^{d}$ have side lengths $L_{j} > 0$ in metres and volume $V = \prod_{j} L_{j}$ in $m^{d}$ . Let $m > 0$ kg and $E > 0$ joules, and set $R = \sqrt{2 m E}$ in kg m s $^{- 1}$ . Take one operationally indistinguishable species of $N$ particles as specified by Classical particle counting and physical observables. On the auxiliary labelled cover, take independent uniform positions $q_{i} \in B$ with density $V^{- 1}$ and, independently, the momentum law $P = R Z / ∣ Z ∣,$ where $Z \in R^{D}$ has the independent standard-normal coordinate law defined in Exact spherical-shell coordinate marginal and fixed-block Gaussian limit and $P = 0$ is assigned on the null event $Z = 0$ . This product law is invariant under particle permutations, so it defines the stated model preparation on the quotient; a one-particle marginal means the law of a uniformly selected particle. This is a specified spherical-shell preparation: $H (P) = ∣ P ∣^{2} / (2 m) = E$ almost surely, with no assumption about a trajectory or equilibration. For any momentum block of $k$ coordinates with $1 \leq k < D$ , its density is $f_{D, k}$ from Exact spherical-shell coordinate marginal and fixed-block Gaussian limit. In particular the total-energy variance is zero J $^{2}$ .

Compare the separate canonical ideal-gas preparation of Exact canonical and grand ideal-gas preparations with the same $N, d, m, B$ and one-species convention, at $β = D / (2 E) > 0$ . Its mean total energy is also $E$ , but its variance is $D / (2 β^{2}) > 0$ J $^{2}$ . If $N > 1$ , the uniformly selected particle's $d$ -component shell marginal has compact support because $d < D$ , whereas its canonical Gaussian momentum has positive probability outside every finite-radius ball. The two finite preparations therefore differ despite their matched mean energy. For $N = 1$ , the full momentum vector is singular on its sphere and the density formula for $k = D$ is not asserted.

Facts & Assumptions

Given: ${A C}_{ω}$ , the rectangular box with side lengths in metres and volume in m $^{d}$ , the one-species operational indistinguishability assumption, particle number and dimension, positive mass and energy, shell radius $R$ , and the explicitly adopted independent position/spherical-momentum preparation in the Statement.

[F1]

Declared example preparations and thermodynamic interpretation: each shell or product law is an explicitly adopted probability preparation, separate from a canonical law and not inferred from Hamiltonian dynamics.

[F2]

Exact spherical-shell coordinate marginal and fixed-block Gaussian limit: assuming ${A C}_{ω}$ , for $D \geq 2$ , $1 \leq k < D$ , and $R > 0$ , the independent standard-normal coordinate law has an orthogonally invariant direction; $R Z / ∣ Z ∣$ is concentrated on $∣ P ∣ = R$ , its $k$ -coordinate marginal has the stated exact density, and its coordinate second moment is $R^{2} / D$ .

[F3]

Exact canonical and grand ideal-gas preparations: for the massive $d$ -dimensional canonical ideal gas with $β > 0$ , total energy has mean $d N / (2 β)$ and variance $d N / (2 β^{2})$ , and each component of the uniformly selected particle's momentum is Gaussian with variance $m / β$ .

[F4]

Classical particle counting and physical observables: for one operationally indistinguishable species, the labelled lift carries the symmetric permutation action and a permutation-invariant one-particle marginal is interpreted as the law of a uniformly selected particle.

[A1]

This permutation-invariant spherical-shell law is a chosen finite classical distribution on one operationally indistinguishable species; it is not a claim that an isolated gas dynamically reaches it or that it is the only microcanonical convention.

Proof

1.1givenF1A1algebra

Under the shell preparation [F1] and its explicit law [A1], $∣ P ∣ = R$ almost surely by the definition of $P$ ; hence $H (P) = R^{2} / (2 m) = E$ on every realization outside the null event $Z = 0$ . A constant random variable has variance zero.

2.1F2step 1.1

For $1 \leq k < D$ , the first $k$ momentum coordinates are the projection in [F2], so their exact Lebesgue density is the displayed $f_{D, k}$ , which vanishes outside the closed radius- $R$ ball. The event that this block has norm $R$ requires the remaining $D - k$ Gaussian coordinates to vanish; their joint density and $D - k \geq 1$ imply that this event has probability zero.

2.2givenF3step 1.1algebra

Choose $β = D / (2 E)$ . By [F3], the canonical mean energy is $D / (2 β) = E$ and its variance is $D / (2 β^{2}) > 0$ . Thus it cannot have the shell law of step 1.1, whose energy variance is zero.

3.1A1F2F3F4step 2.1∎

If $N > 1$ , then $d < D$ . Particle permutations act by orthogonal coordinate permutations on $R^{D}$ , so [F2] and [A1] make every particle block marginal identical; by the selected-particle convention [F4], their uniform mixture is the law of a uniformly selected particle. It has a density supported in the closed radius- $R$ ball, with no boundary atom by step 2.1. By [F3], the canonical Gaussian of a uniformly selected particle has strictly positive density at every point of $R^{d}$ , so its integral over the open annulus $R < ∣ p ∣ < 2 R$ is positive, while the shell probability of that annulus is zero. Therefore the one-particle marginals differ as well. For $N = 1$ , $d = D$ and the full shell law is singular, so no $k = D$ density assertion has been used.

Remarks

This result compares two stated finite probability laws. It proves no equivalence or inequivalence theorem for general thermodynamic ensembles and no dynamical equilibration claim.

Finite spherical-shell gas and its exact marginal

Statement

Facts & Assumptions

Proof

Remarks

Depends on

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