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How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
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These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Finite shell support versus canonical Gaussian tails

Physical scope

Two operationally indistinguishable particles in a three-dimensional spherical-shell preparation compared with the matched-mean canonical ideal gas, conditional on Countable Choice for the marginal theorem; exact distributions only.

Example

Assume ${A C}_{ω}$ as required by Finite spherical-shell gas and its exact marginal, and use its one-species rectangular-box model. Take $d = 3$ , $N = 2$ , $D = 6$ , mass $m > 0$ in kg, and energy $E > 0$ in joules, so the momentum radius is $R = \sqrt{2 m E}$ in kg m s $^{- 1}$ . A uniformly selected particle's three-momentum block ( $k = 3$ ) under the shell preparation has density $f_{6, 3} (p) = \frac{4}{π^{2} R^{3}} \sqrt{1 - \frac{∣ p ∣^{2}}{R^{2}}}, ∣ p ∣ < R,$ and zero for $∣ p ∣ \geq R$ . A canonical ideal gas with $β = D / (2 E) = 3 / E$ in J $^{- 1}$ has the same mean total energy but assigns positive probability to the annulus $R < ∣ p_{⋆} ∣ < 2 R$ for the uniformly selected particle's momentum $p_{⋆}$ .

Facts & Assumptions

Given: ${A C}_{ω}$ , the one-species rectangular-box shell theorem Finite spherical-shell gas and its exact marginal with $D = 6, k = 3, E > 0$ , and the matched-mean canonical ideal-gas theorem Exact canonical and grand ideal-gas preparations.

[F1]

Finite spherical-shell gas and its exact marginal: assuming ${A C}_{ω}$ , the uniformly selected particle's $k$ -block shell marginal has density $f_{D, k}$ , concentrated in the closed radius- $R$ ball with no boundary atom for $k < D$ , and $β = D / (2 E)$ matches the canonical mean total energy.

[F2]

Exact canonical and grand ideal-gas preparations: in the massive canonical model each momentum component has a Gaussian density with positive variance $m / β$ , hence the vector density is strictly positive on all of $R^{d}$ .

[F3]

Gaussian, gamma and beta product integrals: $Γ (s + 1) = s Γ (s)$ and $Γ (1 / 2) = \sqrt{π}$ , so $Γ (3) = 2$ and $Γ (3 / 2) = \sqrt{π} / 2$ .

[A1]

Choose $d = 3$ , $N = 2$ , and $k = 3$ in the one-species model.

Verification

1.1A1F1F3algebra

Substituting $D = 6, k = 3$ into [F1] gives exponent $(6 - 3 - 2) / 2 = 1 / 2$ and coefficient $Γ (3) / [Γ (3 / 2) (π R^{2})^{3 / 2}] = 4 / (π^{2} R^{3})$ using [F3]. This is the displayed density.

2.1F1F2step 1.1∎

The shell block has probability zero outside the radius- $R$ ball by [F1]. The canonical Gaussian vector density from [F2] is positive throughout the nonempty annulus $R < ∣ p_{⋆} ∣ < 2 R$ , which has positive volume, so that event has positive canonical probability for a uniformly selected particle. Therefore the two finite momentum marginals differ.

Remarks

All distributions here are model predictions under their stated preparations. No observed counts or fitted parameters are involved.

Finite shell support versus canonical Gaussian tails

Example

Facts & Assumptions

Verification

Remarks

Depends on

Dependency tree · 2 levels

Sources