# How is the Maxwell-Boltzmann curve obtained?

Is it purely qualitative and determined by experiment?

Or is there some function which defines the familiar graph?

(Image source: Wikipedia)

• – Ivan Neretin Oct 4 '16 at 11:31
• There is a theoretical derivation and it is actually quite long. I wouldn't mind typing it up, but I'm not quite free today :( You can probably find it in any physical chemistry textbook. – orthocresol Oct 4 '16 at 11:33
• There is a lovely experiment showing that the maxwell-Boltzmann curve is correct. The experiment observing how a narrow, horizontal beam Cs atoms fall under gravity . The original paper is Esterman, Simpson & Stern, Phys. Rev. 19487, v71 p238 but is behind a paywall, the nobel lecture by stern is at nobelprize.org/nobel_prizes/physics/laureates/1943/… – porphyrin Oct 4 '16 at 12:54
• @porphyrin - the archive for Physical Review is most decidedly not behind a paywall. Go to journals.aps.org/pr/issues/71/4 to find the paper. – Jon Custer Oct 4 '16 at 22:02
• @porphyrin - Crud. I take that back - you need an institutional, personal, or US public or highs school library. I remember the archive going up originally intending to be wide open. Sigh - must have been too much mass-downloading abuse. Sadness. – Jon Custer Oct 4 '16 at 22:21

There are two common types of graph, one with kinetic energy as the $x$-axis, and one with speed. For the most part, they are qualitatively the same, although there is a slight difference in curvature near the origin. The speed function is much more commonly used, because one can more easily extract useful information from it, such as mean velocities, or mean free paths. It is:
$$f(v) = (4\pi v^2) \cdot \left(\frac{m}{2\pi kT}\right)^{3/2} \cdot \exp{\left(-\frac{mv^2}{2kT}\right)}$$
where $m$ refers to the mass of the particle, $k$ Boltzmann's constant, and $T$ the temperature.