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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 ACω 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≥1, rod length a>0 in metres, mass m>0 in kg, inverse temperature β>0 in J−1, and fixed action reference h0>0 in J s. Set box length L=2Na in metres. Then L>Na and number density ρ=N/L=1/(2a) in m−1, so aρ=1/2. The exact force-conjugate pressure is βp1D=2ρ. Truncating its virial series after the first two terms gives ρ+aρ2=3ρ/2, with exact remainder ρ/2.

Facts & Assumptions

Given: ACω and the one-dimensional hard-rod model from Exact Tonks hard-rod pressure and convergent virial series with N≥1, a>0 metres, L>Na metres, m>0 kg, β>0 J−1, h0>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, βp1D=ρ/(1−aρ) and the remainder after M terms is ρ(aρ)M/(1−aρ).

[A1]

Choose L=2Na and truncate after M=2 terms.

Verification

1.1A1algebra

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

2.1F1step 1.1A1algebra∎

The exact equation in [F1] gives βp1D=ρ/(1−1/2)=2ρ. The first two series terms are ρ+aρ2=ρ+ρ/2=3ρ/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.

Depends on

Dependency tree · 2 levels

Sources