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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 ACω as required by Exact canonical and grand ideal-gas preparations. In a canonical sector N≥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 2/(mβ) and its ensemble mean speed is 8/(πmβ). In the separate grand model, which includes the vacuum sector, choose the chemical potential μchem=β−1log⁡(λ3/V), so the mean activity parameter ν=exp⁡(βμchem)V/λ3 equals 1, where V is in m3, λ in metres, and μchem in joules. The particle count then has the Poisson(1) law, including P(N=0)=P(N=1)=exp⁡(−1).

Facts & Assumptions

Given: ACω, 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 Z0=1.

[F1]

Exact canonical and grand ideal-gas preparations: in d=3, the uniformly selected particle's speed mode and mean are 2/(mβ) and 8/(πmβ); with the defined vacuum sector, the grand count has Poisson mass exp⁡(−ν)νj/j! for every j≥0, where ν=exp⁡(βμchem)V/λ3.

[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,∞), so exp⁡(log⁡x)=x for x>0.

[A1]

Choose the grand model parameter μchem=β−1log⁡(λ3/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⁡(βμchem)=λ3/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.

Depends on

Dependency tree · 2 levels

Sources