Alphabeta Math

1 · Prerequisites

2 · Summary

These worked examples display the translation zero-mode divergence of an unpinned harmonic chain, evaluate a Maxwell speed law and a Poisson particle-count case including the vacuum sector, specialize the exact Ising-ring correlation length, check a hard-rod virial remainder at half packing, and compare finite shell support with canonical Gaussian tails. The shell marginal comparison is conditional on Countable Choice through its Lebesgue change-of-variables supplier. These are exact calculations under the named models, not measurements or evidence that any system realizes a preparation.

3 · Logical flowchart

4 · Definitions, postulates, theorems and proofs

None yet.

5 · Examples, experiments and counterexamples

ExampleConstruction: AI-adaptedVerification: AI-adaptedprecheck passjudge pass (gpt-6-luna)Open item page →

Unpinned translation zero-mode divergence

Physical scope

Finite periodic harmonic chain with no pinning; conditional on ACω for the declared Euclidean Lebesgue reference, the continuous translation zero mode makes the canonical partition function diverge.

Example

Assume ${A C}_{ω}$ as defined in The Axiom of Countable Choice ( ${A C}_{ω}$ ) for the Euclidean Lebesgue measure of the phase-space boxes below. Take $N \geq 3$ resolved displacement coordinates $q_{i} \in R$ in metres and momenta $p_{i} \in R$ in kg m s $^{- 1}$ on a periodic chain, with $q_{N + 1} = q_{1}$ , mass $m > 0$ kg, spring stiffness $κ > 0$ in J m $^{- 2}$ , and $H (q, p) = \sum_{i = 1}^{N} \frac{p_{i}^{2}}{2 m} + \frac{κ}{2} \sum_{i = 1}^{N} (q_{i + 1} - q_{i})^{2} .$ Adopt inverse temperature $β > 0$ in J $^{- 1}$ and fixed action reference $h_{0} = a_{0} > 0$ in J s under Declared example preparations and thermodynamic interpretation. Because the coordinates are resolved, use the reference $d q d p / h_{0}^{N}$ with no particle-exchange factorial. Define the partition integral $Z : = \int_{R^{N} \times R^{N}} \exp (- β H (q, p)) \frac{d q d p}{h_{0}^{N}},$ where $\exp$ is the real exponential of The real exponential function and the number $e$ by a power series. The unpinned position integral is infinite, so this model has no normalized canonical probability on the full position space.

Facts & Assumptions

Given: ${A C}_{ω}$ , the periodic-chain Hamiltonian, $m > 0$ kg, $κ > 0$ J m $^{- 2}$ , $β > 0$ J $^{- 1}$ , $h_{0} = a_{0} > 0$ J s, and the resolved-coordinate reference above.

[F1]

Declared example preparations and thermodynamic interpretation: when $Z$ is finite and positive, the canonical density is $Z^{- 1} \exp (- β H)$ relative to the chosen reference measure; the fixed action reference is part of the preparation, and resolved harmonic coordinates retain their declared labels without an added particle-exchange factorial.

[F2]

Under ${A C}_{ω}$ , Lebesgue measure of a box is its product volume, and a bounded Jordan measurable set has Lebesgue measure equal to its Jordan content (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, Lebesgue outer measure is at most Jordan outer content, and a bounded Jordan measurable set is Lebesgue measurable with Lebesgue measure equal to its Jordan content).

[F3]

A $C^{1}$ diffeomorphism with nonzero Jacobian transforms Riemann integrals on compact Jordan sets by the absolute determinant (Change of variables for an injective $C^{1}$ map on a compact Jordan set).

[F4]

Every continuous real function on a nonempty compact metric space is bounded and attains a finite maximum (A continuous real-valued function on a nonempty compact metric space is bounded and attains a greatest and a least value).

[F5]

The nonnegative Lebesgue integral is monotone under pointwise ordering (Monotonicity and nonnegative homogeneity of the nonnegative integral).

[A1]

The example adopts the displayed unpinned periodic-chain Hamiltonian and full $R^{N}$ displacement domain.

Verification

1.1givenA1F3algebra

Put $r_{1} : = 0$ and use coordinates $u : = q_{1}$ , $r_{i} : = q_{i} - q_{1}$ for $2 \leq i \leq N$ . The linear bijection $(u, r_{2}, \dots, r_{N}) \mapsto (q_{1}, \dots, q_{N}) = (u, u + r_{2}, \dots, u + r_{N})$ has determinant $1$ . With $r_{N + 1} : = r_{1}$ , every periodic difference satisfies $q_{i + 1} - q_{i} = r_{i + 1} - r_{i}$ , so the potential is independent of $u$ ; the momentum term is also independent of $u$ . On each compact coordinate box the change-of-variables supplier [F3] therefore gives the phase-volume Jacobian $1$ .

2.1F1F2F4F5step 1.1algebra∎

Under the declared reference convention [F1], choose any compact coordinate box $C$ of positive volume in $(r_{2}, \dots, r_{N}, p)$ . The Hamiltonian is continuous and hence bounded above on $C$ by a finite $H_{C}$ by [F4]. By [F2], $C$ has Lebesgue measure $vol (C)$ . On $u \in [- R, R]$ and $(r, p) \in C$ , the integrand is at least $e^{- β H_{C}}$ ; the determinant-one map in step 1.1 and [F2] show this slab contributes at least $2 R e^{- β H_{C}} vol (C) / h_{0}^{N}$ to the partition integral by [F5]. This lower bound is unbounded as $R \to \infty$ , so $Z = \infty$ and no normalization exists.

Remarks

This calculation identifies a translation zero mode. It makes no claim about separately confined or pinned models.

ExampleConstruction: AI-adaptedVerification: AI-adaptedprecheck passOpen item page →

Maxwell speed marginal and Poisson number example

Physical scope

Conditional on ACω for the Euclidean Lebesgue product reference; exact d=3 massive ideal-gas canonical speed law and one grand-canonical activity choice for one operationally indistinguishable species, including the vacuum sector; no measured gas data.

Example

Assume ${A C}_{ω}$ as required by Exact canonical and grand ideal-gas preparations. In a canonical sector $N \geq 1$ of the massive $d = 3$ ideal-gas model of one operationally indistinguishable species, choose any mass $m > 0$ in kg and inverse temperature $β > 0$ in J $^{- 1}$ . The most probable speed of a uniformly selected particle is $\sqrt{2 / (m β)}$ and its ensemble mean speed is $\sqrt{8 / (π m β)}$ . In the separate grand model, which includes the vacuum sector, choose the chemical potential $μ_{c h e m} = β^{- 1} \log (λ^{3} / V),$ so the mean activity parameter $ν = \exp (β μ_{c h e m}) V / λ^{3}$ equals $1$ , where $V$ is in m $^{3}$ , $λ$ in metres, and $μ_{c h e m}$ in joules. The particle count then has the Poisson $(1)$ law, including $P (N = 0) = P (N = 1) = \exp (- 1)$ .

Facts & Assumptions

Given: ${A C}_{ω}$ , the massive ideal-gas canonical and grand preparations of Exact canonical and grand ideal-gas preparations for one operationally indistinguishable species, with $d = 3$ , $m, β, V, λ > 0$ and the vacuum convention $Z_{0} = 1$ .

[F1]

Exact canonical and grand ideal-gas preparations: in $d = 3$ , the uniformly selected particle's speed mode and mean are $\sqrt{2 / (m β)}$ and $\sqrt{8 / (π m β)}$ ; with the defined vacuum sector, the grand count has Poisson mass $\exp (- ν) ν^{j} / j!$ for every $j \geq 0$ , where $ν = \exp (β μ_{c h e m}) V / λ^{3}$ .

[F2]

The real exponential function and the number $e$ by a power series, The natural logarithm as the inverse of the exponential function: $\exp$ and $\log$ are inverse functions on $R$ and $(0, \infty)$ , so $\exp (\log x) = x$ for $x > 0$ .

[A1]

Choose the grand model parameter $μ_{c h e m} = β^{- 1} \log (λ^{3} / V)$ ; this is a parameter selection, not an observation.

Verification

1.1F1

Specializing the $d = 3$ clauses of [F1] gives the stated mode and mean of the model speed distribution; these are distributional functionals, not values assigned to every individual particle.

2.1A1F1F2algebra∎

Substituting [A1] and applying [F2] gives $\exp (β μ_{c h e m}) = λ^{3} / V$ , hence $ν = 1$ in [F1]. The Poisson formula then gives $P (N = j) = \exp (- 1) / j!$ , so the $j = 0$ and $j = 1$ probabilities both equal $\exp (- 1)$ .

Remarks

The displayed values are exact consequences of the specified distributions. They are not measured speeds or particle counts.

ExampleConstruction: AI-adaptedVerification: AI-adaptedprecheck passOpen item page →

Exact finite-ring correlations and limiting correlation length at a chosen coupling

Physical scope

Zero-field finite periodic nearest-neighbor Ising ring at the exact parameter choice beta J=(log 2)/2; no observed magnetic data.

Example

For any finite $N \geq 2$ at zero field, choose coupling $J > 0$ in joules and inverse temperature $β = (\log 2) / (2 J)$ in J $^{- 1}$ . Then $t = \tanh (β J) = 1 / 3$ and the exact ring correlation at separation $1 \leq r < N$ is $E_{N} [s_{i} s_{i + r}] = \frac{3^{- r} + 3^{- (N - r)}}{1 + 3^{- N}} .$ At fixed $r$ , the limit is $3^{- r} = e^{- r \log 3}$ , so the correlation length is $1 / \log 3$ lattice spacings.

Facts & Assumptions

Given: The finite zero-field ring theorem Finite periodic Ising-ring correlations and exact decay with $J > 0$ , $β > 0$ , and its indexed-bond convention.

[F1]

Finite periodic Ising-ring correlations and exact decay: with $t = \tanh (β J)$ , $C_{N} (r) = (t^{r} + t^{N - r}) / (1 + t^{N})$ and, for $0 < t < 1$ , $ξ = - 1 / \log t$ .

[A1]

Set $β J = (\log 2) / 2$ in the finite model.

Verification

1.1A1algebra

Since $\tanh (x) = (e^{2 x} - 1) / (e^{2 x} + 1)$ , the choice [A1] gives $t = (2 - 1) / (2 + 1) = 1 / 3$ .

2.1F1step 1.1algebra∎

Substituting $t = 1 / 3$ in [F1] gives the displayed finite- $N$ correlation. Its fixed- $r$ limit is $3^{- r}$ by the theorem's error bound, and $3^{- r} = e^{- r \log 3}$ gives $ξ = 1 / \log 3$ .

Remarks

This is a finite-model calculation. The displayed coupling is a chosen parameter and is not an experimentally fitted value.

ExampleConstruction: AI-adaptedVerification: AI-adaptedprecheck passjudge pass (gpt-6-luna)Open item page →

Packing boundary and exact truncated virial error

Physical scope

Conditional on ACω for the Euclidean Lebesgue product reference; finite one-dimensional Tonks gas with common positive rod length, positive mass, inverse temperature and action reference at half the close-packing density; exact force-conjugate pressure and geometric virial remainder only.

Example

Assume ${A C}_{ω}$ as required by Exact Tonks hard-rod pressure and convergent virial series. Use the one-species hard-rod model of Exact Tonks hard-rod pressure and convergent virial series with $N \geq 1$ , rod length $a > 0$ in metres, mass $m > 0$ in kg, inverse temperature $β > 0$ in J $^{- 1}$ , and fixed action reference $h_{0} > 0$ in J s. Set box length $L = 2 N a$ in metres. Then $L > N a$ and number density $ρ = N / L = 1 / (2 a)$ in m $^{- 1}$ , so $a ρ = 1 / 2$ . The exact force-conjugate pressure is $β p_{1 D} = 2 ρ$ . Truncating its virial series after the first two terms gives $ρ + a ρ^{2} = 3 ρ / 2$ , with exact remainder $ρ / 2$ .

Facts & Assumptions

Given: ${A C}_{ω}$ and the one-dimensional hard-rod model from Exact Tonks hard-rod pressure and convergent virial series with $N \geq 1$ , $a > 0$ metres, $L > N a$ metres, $m > 0$ kg, $β > 0$ J $^{- 1}$ , $h_{0} > 0$ J s, and $ρ = N / L$ in m $^{- 1}$ .

[F1]

Exact Tonks hard-rod pressure and convergent virial series: under those full hard-rod hypotheses, $β p_{1 D} = ρ / (1 - a ρ)$ and the remainder after $M$ terms is $ρ (a ρ)^{M} / (1 - a ρ)$ .

[A1]

Choose $L = 2 N a$ and truncate after $M = 2$ terms.

Verification

1.1A1algebra

The choices give $ρ = N / (2 N a) = 1 / (2 a)$ and therefore $a ρ = 1 / 2$ .

2.1F1step 1.1A1algebra∎

The exact equation in [F1] gives $β p_{1 D} = ρ / (1 - 1 / 2) = 2 ρ$ . The first two series terms are $ρ + a ρ^{2} = ρ + ρ / 2 = 3 ρ / 2$ . Applying the remainder formula in [F1] with $M = 2$ gives $ρ (1 / 2)^{2} / (1 - 1 / 2) = ρ / 2$ , exactly the difference.

Remarks

This is a finite-parameter illustration of the exact one-dimensional model. It is not a numerical measurement or an all-fluid virial expansion.

ExampleConstruction: AI-adaptedVerification: AI-adaptedprecheck passOpen item page →

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.

Gaussian and gamma product tools — Examples

1 · Prerequisites

2 · Summary

3 · Logical flowchart

4 · Definitions, postulates, theorems and proofs

5 · Examples, experiments and counterexamples

Unpinned translation zero-mode divergence

Example

Facts & Assumptions

Verification

Remarks

Maxwell speed marginal and Poisson number example

Example

Facts & Assumptions

Verification

Remarks

Exact finite-ring correlations and limiting correlation length at a chosen coupling

Example

Facts & Assumptions

Verification

Remarks

Packing boundary and exact truncated virial error

Example

Facts & Assumptions

Verification

Remarks

Finite shell support versus canonical Gaussian tails

Example

Facts & Assumptions

Verification

Remarks

Sources