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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.

Wall arrivals are not an unbiased sample of the gas: faster atoms arrive more frequently. The normalized speed distribution among incident atoms isThe angular factor in the arrival rate cancels when calculating the mean kinetic energy. ThusUsing $I_5=2I_3/\alpha$ givesBy contrast, an atom chosen randomly from the bulk has mean kinetic energy $3k_{B}T/2$. The arrival weighting raises the normal-direction contribution to $k_{B}T$ while each tangential direction still contributes $k_{B}T/2$. This is also the mean energy carried by atoms effusing through a thin aperture from an equilibrated reservoir.

For a classical dilute ideal gas in thermal equilibrium, the probability of a single-particle velocity is proportional to the Boltzmann factor for kinetic energy, $e^{-m{v}^2/(2k_{B}T)}$. Velocities with speed between $v$ and $v+dv$ fill a spherical shell of velocity-space volume $4\pi v^{2}dv$. The normalized Maxwell speed distribution is thereforeThe factor $v^2$ counts states, rather than representing an extra Boltzmann energy factor.

Set $\alpha =m/(2k_{B}T)$ and use $I_j={\int }_0^{\infty }{v}^j{e}^{-\alpha v^2}dv$. Normalization cancels in ratios, so the given recurrence givesConsequentlyThe root mean square speed is larger than the mean speed, consistent with the positive variance of the speed distribution.

A speed distribution is the probability distribution of the magnitude of velocity. When it has a probability density function $f$, the fraction of particles with speeds between $v$ and $v+dv$ is $f(v)dv$, and ${\int }_0^{\infty }f(v)dv=1$. The Maxwell speed distribution describes an equilibrium ideal gas; the effusive beam speed distribution weights that population by its rate of arrival.