= Solution
In the <canonical ensemble>, a one-particle velocity state has kinetic energy $E(\mathbf v)=m|\mathbf v|^2/2$ and hence probability density proportional to $e^{-\beta E}$, where $\beta=(k_BT)^{-1}$. The three Cartesian <Gaussian integrals> normalize the density to the <Maxwell-Boltzmann velocity distribution>
$$
\boxed{
f(\mathbf v)=
\left(\frac{m}{2\pi k_BT}\right)^{3/2}
\exp\left(-\frac{m|\mathbf v|^2}{2k_BT}\right).}
$$
Integrating over a spherical shell of radius $v$ and area $4\pi v^2$ gives the <Maxwell speed distribution>
$$
\boxed{
g(v)=4\pi v^2
\left(\frac{m}{2\pi k_BT}\right)^{3/2}
e^{-mv^2/(2k_BT)},\qquad v\geq0.}
$$
Since
$$
\frac d{dv}\log g(v)=\frac2v-\frac{mv}{k_BT},
$$
its unique positive maximum is the <Most probable speed in the Maxwell distribution>
$$
\boxed{v_{\rm mp}=\sqrt{\frac{2k_BT}{m}}.}
$$
Back to article page
