The one-dimensional Maxwell Boltzmann probability speed distribution is given by

\begin{equation} f(u)du=2\sqrt{\frac{m}{2\pi k_BT}} e^{-\frac{mu^2}{2k_BT}}du, \end{equation} where $u$ is the speed of the particle, $m$ the mass of the particle, $T$ the temperature, and $k_B$ Boltzmann's constant. And the factor of 2 comes from the $\pm$ velocity contributions.

The 2D and 3D speed distributions are proportional to $u$ and $u^2$, respectively. And so have a zero probability of a particle having zero speed. Why does the 1D speed distribution have a greater than zero probability of the speed being zero?


1 Answer 1


First, don't confuse probability and probability density, of which the former is the integral of the latter over a range of interest. The probability of having zero speed in any number of dimensions is exactly zero. But probability density is non-zero in 1D case.

The difference in zero or non-zero probability density comes from the difference between the concepts of speed and velocity. For a speed $u$ in 1D case there are two velocities: $+u$ and $-u$. In 2D there's a circle of them:


which has zero circumference when $u=0$. Similarly for 3D case, where there's a sphere of velocities that result in the same speed, and the area of this sphere is zero when the speed $u=0$.

This makes probability density scale proportionally to these circumferences/areas, which reflects that there are more ways (different velocities) to get higher speed than to get a lower speed in 2D and 3D spaces.


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