Alphabeta Math
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Hard-core spheres in the explicit dilute domain

Example

Assume Countable Choice. Fix d∈N with d≥1, a hard-core diameter a∈R with a>0, and β∈R with β>0. For x=(xj)j<d∈Rd, write ∥x∥2:=∑j<dxj2, and let B(0,a):={x∈Rd:∥x∥2<a}. Let λd denote Lebesgue measure on Rd. Define the hard-core pair potential va:Rd→R∪{+∞} by va(x)=+∞ when ∥x∥2<a and va(x)=0 when ∥x∥2≥a. Its stability constant may be chosen as B=0, the function x↦e−βva(x)−1 is −1 on B(0,a) and 0 elsewhere, and its absolute integral is Cβ=λd(B(0,a)). For a complex activity z∈C, with modulus ∣z∣ as defined by ∣x+iy∣=x2+y2, the sufficient analytic domain is e∣z∣λd(B(0,a))<1.

Facts & Assumptions

Given: Countable Choice, d∈N with d≥1, a∈R with a>0, β∈R with β>0, and the hard-core potential in the Example.

[A1]
[F1]

Stability, superstability and regular tails: stability means that every N-particle pair-energy sum is at least −BN; the potential must be even Borel.

[F2]

The stable continuum Mayer pressure series converges in an explicit dilute disk: under Countable Choice, for a stable even Borel potential with finite Cβ, the centered-cube pressure series is absolutely convergent whenever ∣z∣e2βB+1Cβ<1.

[F3]

Euclidean balls have positive finite Lebesgue measure: under Countable Choice, the Euclidean ball B(0,a) is Lebesgue measurable and has positive finite measure.

[F5]

The Euclidean inner product ⟨x,y⟩=∑k<nxkyk on Rn: ∥x∥2=∑j<dxj2, so ∥−x∥2=∥x∥2.

[F6]

The nonnegative integral agrees with the simple integral on simple functions and The integral of a nonnegative simple function: the integral of an indicator of a measurable set equals that set's measure.

[F7]

Real and imaginary parts, complex conjugation, and modulus: if z=x+iy with x,y∈R, its complex modulus is ∣z∣=x2+y2.

[F8]

Absolute value in an ordered field, Basic properties of the absolute value: real absolute value is defined by sign, is nonnegative, and satisfies ∣−1∣=1 and ∣0∣=0.

Proof

Given: The model and assumptions in the Example, [A1], [F1], [F2], [F3], [F4], [F5], [F6], [F7], and [F8].

1.1A1F1F4F5algebra

The Euclidean norm is even by [F5], so va(−x)=va(x). The ball B(0,a) is open, hence Borel, by [F4]; its complement is Borel as well. The target singleton {+∞} is Borel by [F4], and {0}=⋂m≥1(−1/m,1/m) is Borel by the extended-real topology and sigma-algebra closure in [F4]. Since va takes the constant value +∞ on the ball and 0 on its complement, the preimage of any target Borel set is among the empty set, the ball, its complement, and all of Rd; hence va is Borel. Every pair energy is either 0 or +∞, so every finite total energy is at least zero; therefore the stability inequality holds with B=0.

1.2A1F3F6F8givenalgebra

Under the stated convention, e−βva(x)−1=−1 on B(0,a) and 0 elsewhere, so ∣fβ∣=1B(0,a) by [F8]. The ball lemma [F3] gives measurability and finite measure, and [F6] gives Cβ=∫1B(0,a)dx=λd(B(0,a))∈(0,∞).

2.1A1F2F7step 1.1step 1.2algebra∎

The hypotheses of [F2] hold by steps 1.1–1.2 and [A1]. Substituting B=0 and Cβ=λd(B(0,a)) into its sufficient disk gives e∣z∣λd(B(0,a))<1. This is a sufficient analytic domain; it is not asserted to be necessary and implies no freezing or coexistence point.

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