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A volume-filling field changes ideal-gas pressure

Physical scope

A classical ideal positional gas in dimension d at least two with a constant per-particle external energy throughout the volume; a nonzero shift cannot be represented by a uniformly tempered exterior, while the zero shift agrees with the empty exterior. No empirical observation is asserted.

Example

For the ideal positional gas $φ = 0$ in dimension $d \geq 2$ and a bounded measurable region of dimensionless volume $0 \leq V < \infty$ , take $z > 0$ and $T > 0$ , with $β = (k_{B} T)^{- 1} > 0$ in $J^{- 1}$ . The grand partition is $Ξ_{V} (z, β) = e^{z V}$ . Now impose a constant external energy $c \in R$ joules per particle at every point of the volume, adding $c N$ to the configuration energy and using the same grand weighting convention. The new grand partition is $Ξ_{V}^{(c)} = e^{z e^{- β c} V}$ . For $V > 0$ , the dimensionless per-volume log-partition rate changes from $z$ to $z e^{- β c}$ when $c \neq 0$ . At $V = 0$ , both partitions are $1$ and no per-volume rate is assigned. For $c \neq 0$ , this energy shift cannot be represented on the nonempty cubes $Λ_{n}$ by the deterministic exterior interaction used by Qualified van Hove and tempered-exterior thermodynamic limits. For $c = 0$ , it vanishes and agrees with the empty exterior $\emptyset \in Ω (0)$ of Van Hove regions and uniformly tempered exterior preparations. For every $c$ , the ideal potential itself fails the cited theorem's stronger interaction hypotheses.

Facts & Assumptions

[F1]

Stable regularly tempered classical positional preparation: $V$ is dimensionless reference volume, $β$ is inverse energy, and the ideal model has $H_{N} = 0$ .

[A1]

Finite-volume equilibrium ensemble sampling: $Ξ_{V} = \sum_{N \geq 0} z^{N} Z_{V, N}$ with the indistinguishable positional factor $1 / N!$ .

[F2]

Van Hove regions and uniformly tempered exterior preparations: the deterministic exterior class $Ω (t)$ has one fixed square-count bound $t \geq 0$ , and $W_{n} (ω, ζ) = \sum_{x \in ω} \sum_{y \in ζ ∖ Λ_{n}} φ (x - y)$ . Thus $\emptyset \in Ω (0)$ and, for the ideal potential $φ = 0$ , $W_{n} = 0$ for every allowed exterior.

[L1]

Qualified van Hove and tempered-exterior thermodynamic limits: its current conditional statement treats exterior energy only as $W_{n} (ω, ζ)$ for a fixed $ζ \in Ω (t)$ and assumes a superstable regular pair potential with an additional positive divergent core for the exterior conclusion. The ideal potential has no such positive core. The theorem is draft and remains conditional on its mathematical suppliers.

Verification

Given: The finite-volume grand weight, an ideal gas, and the uniform per-particle shift $c$ over the entire region.

1.1givenF1A1algebra

Since $H_{N} = 0$ , the finite-volume canonical positional integral is $Z_{V, N} = V^{N} / N!$ . Summing the adopted grand weights gives $Ξ_{V} = \sum_{N \geq 0} (z V)^{N} / N! = e^{z V}$ .

2.1givenA1step 1.1algebra

Adding $c N$ to the energy multiplies the $N$ -particle weight by $e^{- β c N}$ , so $Ξ_{V}^{(c)} = \sum_{N \geq 0} (z e^{- β c} V)^{N} / N! = e^{z e^{- β c} V}$ . When $V > 0$ , division of the two logarithms by $V$ gives $z$ and $z e^{- β c}$ ; these differ for $c \neq 0$ because $β > 0$ and $z > 0$ . When $V = 0$ , only the vacuum contributes, both partitions equal $1$ , and division by $V$ is undefined.

3.1givenF2L1step 2.1∎

For $φ = 0$ , [F2] gives $W_{n} (ω, ζ) = 0$ for every allowed exterior. If $c \neq 0$ , a singleton $ω = {x}$ in any $Λ_{n}$ has $c N = c \neq 0$ , so no such exterior represents this energy shift. If $c = 0$ , the shift is zero and agrees with $ζ = \emptyset \in Ω (0)$ . In either case the ideal potential lacks the positive divergent core required by [L1], so the calculation in step 2.1 contradicts no qualified exterior conclusion.

A volume-filling field changes ideal-gas pressure

Example

Facts & Assumptions

Verification

Depends on

Dependency tree · 2 levels