I read on Wikipedia that the Maxwell-Boltzmann distribution is given by:

$$f(v) = \sqrt{\left(\frac{m}{2 \pi kT}\right)^3}\, 4\pi v^2 e^{- \frac{mv^2}{2kT}}....(1)$$

In Wikipedia's words: "This probability density function gives the probability, per unit speed, of finding the particle with a speed near $v$". Now later on, they define the "distribution for velocity vector" as:

$$f_\mathbf{v} (v_x, v_y, v_z) = \left(\frac{m}{2 \pi kT} \right)^{3/2}\exp \left[-\frac{m(v_x^2 + v_y^2 +v_z^2)}{2kT}\right]....(2)$$ and this gives "The probability of finding a particle with velocity in the infinitesimal element $[dv_x, dv_y, dv_z]$ about velocity $v = [v_x, v_y, v_z]$." which, for one dimension reduces to: $$f_v (v) =\sqrt{\frac{m}{2 \pi kT}}\exp \left[\frac{-mv^2}{2kT}\right].$$ but since the speed just equal to the absolute value of the velocity in one dimension, after making the appropriate change of variables $u=|v|$ in $(1)$, we get: $$f_u (u) =\sqrt{\frac{2m}{ \pi kT}}\exp \left[\frac{-mu^2}{2kT}\right]....(3) $$

where $u$ ranges from $0$ to $\infty$.

  1. But shouldn't $(3)$ correspond to $(1)$ since the interpretaions, as mentioned (by Wikipedia, at least) are the same? why are the equations different?
  2. Also how can the probability of finding a particle in the velocity neighborhood about $v=0$ be the highest but probability of finding a particle in the speed neighborhood of $u=0$ be $0$ ? i.e , physically, why does $(2)$ peak at $0$ when $(1)$ is $0$ at $0$?
  • 1
    $\begingroup$ $(1)$ includes all the velocities with magnitude $v$ distributed over all directions, i.e. you need to take the spherical shell from $v$ to $v+dv$ into accounts. $(2)$ specifies every components. By the way, if the gases are confined in $1$-D space, there's no analogous distribution of velocities at all. Since any colliding particles are just exchanging their velocities due to elastic head-on collision. $\endgroup$ Commented Mar 5, 2017 at 11:11
  • $\begingroup$ Related physics.stackexchange.com/q/99331/226902 and physics.stackexchange.com/q/734935/226902 $\endgroup$
    – Quillo
    Commented Jan 17, 2023 at 0:26

1 Answer 1


You have to go back to see where your equation $(1)$ came from.

For one dimension the probability of particles having a velocity between $\vec v_{\rm x}$ and $\vec v_{\rm x}+ d\vec v_{\rm x}$ is given by

$$f(\vec v_{\rm x})\;d\vec v_{\rm x} =\sqrt{\frac{m}{ 2\pi kT}}\exp \left[\frac{-mv_{\rm x}^2}{2kT}\right]\,d\vec v_{\rm x}$$

The speed distribution is given by $$f(v_{\rm x})\;dv_{\rm x} =2\,\sqrt{\frac{m}{ 2\pi kT}}\exp \left[\frac{-mv_{\rm x}^2}{2kT}\right]\,dv_{\rm x}=\sqrt{\frac{2m}{ \pi kT}}\exp \left[\frac{-mv_{\rm x}^2}{2kT}\right]\,dv_{\rm x}$$

the factor $2$ being there because the speed (magnitude) of $\vec v_{\rm x}$ is the same as that of $-\vec v_{\rm x}$.
This is in agreement with your equation $(3)$.

Equation $(1)$ came from the idea that in three dimensions there is no preferred direction so

$f(\vec v_{\rm x},\vec v_{\rm xy},\vec v_{\rm z})\,d\vec v_{\rm x}d\vec v_{\rm y}d\vec v_{\rm z} = f(\vec v_{\rm x})d\vec v_{\rm x}\,f(\vec v_{\rm y})d\vec v_{\rm y}\,f(\vec v_{\rm z})d\vec v_{\rm z}$

You now have to count all the speeds which are the same ie the magnitude of the velocity $v$ is the same where $v^2 = v^2_{\rm x}+v^2_{\rm y}+v^2_{\rm z}$.

In this three dimensional case the volume of a shell of radius $v$ and thickness $dv$ is $d\vec v_{\rm x}d\vec v_{\rm y}d\vec v_{\rm z}= 4 \pi v^2 dv$ is being considered which results in your equation $(1)$.

$$f(v) \,dv = \sqrt{\left(\frac{m}{2 \pi kT}\right)^3}\, 4\pi v^2 \exp \left[\frac{-mv^2}{2kT}\right] \,dv$$

The equivalent distribution for two dimensions with an area of a ring of radius $v$ and thickness $dv$ is $d\vec v_{\rm x}d\vec v_{\rm y}= 2 \pi v\, dv$ and $v^2 = v^2_{\rm x}+v^2_{\rm y}$ is

$$f(v) \,dv = \left(\frac{m}{2 \pi kT}\right)\, 2\pi v\, \exp \left[\frac{-mv^2}{2kT}\right] \,dv$$


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