Effusive beam speed distribution ($g(v)=vf(v)/⟨v⟩$)

Which speed distribution does an arrival counter measure? Suppose an equilibrium, nonrelativistic ideal gas escapes by effusion through a thin opening into vacuum. The opening is small compared with the mean free path; the reservoir remains at temperature $T$. Let $f(v)$ denote its normalized Maxwell speed distribution, with mass $m$ and $a=m/(2k_{B}T)$:At a given direction toward the opening, the number crossing per unit time is proportional to the normal component $v\cos \theta$. Integrating over a fixed accepted cone contributes an angular factor independent of $v$. Thus the normalized probability density function for speeds among arriving particles isIts integral is one because ${\int }_0^{\infty }{v}^3{e}^{-a{v}^2}dv=1/(2a^2)$. The reservoir mode is $v=\sqrt{2k_{B}T/m}$; the arrival mode is $\sqrt{3k_{B}T/m}$. Faster particles get more opportunities to cross the opening. This is a selection effect, not heating after emission.

Specify the observable before applying the formula. An instantaneous count in a fixed segment of the beam weights the arrival flux by the time spent there, proportional to $1/v$; in this ideal model its normalized speed law returns to $f(v)$. A detector's speed-dependent efficiency, source collisions or a velocity selector changes the arriving law again. Counting arrivals and counting particles currently present are different measurements.