# Root Mean Square Velocity of Particles in a Gas with a Maxwellian Velocity Distribution

### R.M.S. Velocity in a Nonrelativistic Gas

The root mean square velocity of particles in a gas (atoms, molecules, or electrons, etc.) can be calculated from the velocity distribution of the particles. In the case that this velocity distribution is the Maxwellian velocity distribution, we have the following formulas for a particle of mass $m$:

$v({\rm r.m.s.}) = \sqrt{\left(\frac{3kT}{m}\right)}$

or

$v \sim 23,000 \left(\frac{kT}{{\rm 1 \ keV}}\right)^{\frac{1}{2}} \left(\frac{m_{e}}{m}\right)^{\frac{1}{2}} \ \ {\rm km \ s^{-1}}$

where $m_{e}$ is the electron rest mass. Recall that the ratio of the mass of a proton to an electron is $\sim 1836.152$ (source: NIST). We can also express the r.m.s. velocity or speed as a fraction of the speed of light. This form is particularly useful if we know the particle rest mass in the same units as the temperature (or more correctly, $kT$). For example, if know $kT$ in keV, we can use 511~keV for the electron rest mass, or 938 GeV for the proton rest mass.

$\frac{v}{c} = \sqrt{\left(\frac{3kT}{mc^{2}}\right)}$

These formulae are only applicable in the nonrelativistic regime.

Note that the peak velocity of a Maxwellian distribution is

$v({\rm peak}) = \sqrt{\left(\frac{2kT}{m}\right)}$

In general, for any arbitrary velocity distribution $f(v)$ with an arbitrary normalization, and with velocities going from $-\infty$ to $+\infty$, the r.m.s. velocity is simply the square root of the integral of $f(v)$ weighted by $v^{2}$ (and then the whole thing is normalized by the total area under $f(v)$):

$v({\rm r.m.s.}) \ = \ \left[ \frac{ \int_{-\infty}^{+\infty}{ v^{2} \ f(v) \ dv}}{ \int_{-\infty}^{+\infty}{ f(v) \ dv}} \right]^{\frac{1}{2}}$