= Effusive beam speed distribution
{title2=$g(v)=vf(v)/\langle v\rangle$}
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 $a=m/(2k_BT)$:
$$
f(v)=\frac{4a^{3/2}}{\sqrt\pi}v^2e^{-av^2},\qquad v>0.
$$
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 is
$$
\boxed{g(v)=\frac{vf(v)}{\langle v\rangle}=2a^2v^3e^{-av^2}}.
$$
Its <integral> is one because $\int_0^\infty v^3e^{-av^2}\,dv=1/(2a^2)$. The reservoir <mode> is $v=\sqrt{2k_BT/m}$; the arrival <mode> is $\sqrt{3k_BT/m}$. 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. \b[Counting arrivals and counting particles currently present are different measurements.]
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