$$\left( \text{probability of a molecule having velocity }\vec{v}\right) \propto e^{-mv^{2}/2kT}$$ So the most likely velocity vector for a molecule in an ideal gas is zero. Given what we know about Boltzmann factors, this result should hardly be surprising.

Schroeder's Thermal Physics, page-232.

First of all I don't understand how is that true. I can see that average velocity should be zero. But why should most likely velocity vector should be zero?
And if that is true then, after that we derive the speed distribution. And it goes to zero for $v=0$. So that means that we won't find any molecule with zero speed (at rest in container's frame). But that also means that there is no molecule with zero velocity!
SO then how could zero velocity be the most likely velocity if there is no molecule with zero velocity??

  • $\begingroup$ The probability is measured as the area below the Boltzmann distribution between two values, probability here is not discrete so you won't find any particle with exact zero velocity, but you can calculate the probability of finding a particle with velocity between 0 and 0.0001 $\endgroup$ May 5, 2022 at 8:45
  • $\begingroup$ Do not mess-up velocity vector magnitude with it's direction. Average velocity direction is zero, but magnitude - is not. And Maxwell–Boltzmann law speaks about velocity magnitudes distribution. $\endgroup$ May 5, 2022 at 8:59
  • 1
    $\begingroup$ The most likely vector in $\Bbb R^3$ is the location of the probability density's global maximum; this is separate from the question of what value in $[0,\,\infty)$ is most likely for $v:=|\vec{v}|$. In fact $v=0$ has probability density $0$ (not just probabillity $0$), because there's only one $v=0$ point, whereas each $v>0$ comprises a surface in $\vec{v}$-space. $\endgroup$
    – J.G.
    May 5, 2022 at 9:03

1 Answer 1


The distribution is $$ w(v_x,v_y, v_z)=Ae^{-\frac{m(v_x^2+v_y^2+v_z^2)}{2k_BT}}, $$ where A is the normalization constant that can be easily found by integration (see Gaussian integral): $$ \int dv_xdv_ydv_z w(v_x,v_y, v_z) = 1. $$ The average velocity along any direction is zero, e.g., $$ \langle v_x\rangle = \int dv_xdv_ydv_z v_x w(v_x,v_y, v_z)=0 $$ This can be shown either by direct integration by parts or from the symmetry reasons: the odd function is integrated in symmetric limits (from $-\infty$ to $\infty$).

On the other hand, the speed, i.e., the magnitude of this vector, is not zero: $$ v=\sqrt{v_x^2+v_y^2+v_z^2}, \langle v\rangle =\int dv_xdv_ydv_z v w(v_x,v_y, v_z)>0, $$ since the integrand is positive everywhere. However, usually one prefers to calculate $\langle v^2\rangle$, since in this case integration is easier.

But why should most likely velocity vector should be zero?

The most likely velocity corresponds (by definition) to the peak of the distribution, which here is at $v_x=v_y=v_z=0$, so it is zero as well.

  • $\begingroup$ Okay , thanks, so I understand that avergre velcity ought to be zero and average speed will be greater than zero. And I understand the math. Suppose if I were to pick a molecule at random and check its velocity. Then its most likely that it would have velcity close to zero. It just sounds very conterintuitive to me. 😝 $\endgroup$
    – mum
    May 7, 2022 at 8:08
  • $\begingroup$ @mum 1) most likely means that the probability of this even is higher than probability of any other event, but it is not higher than the probability of all other events. The probability of finding a molecule with any non-zero velocity is higher than the probability of hitting exactly $v_x=v_y=v_z=0$ (strictly speaking, the probability of such an event is zero, since we have a continuous distribution). 2) we average over all particles: if half of them have velocity in one direction and the other half in the opposite, the average velocity is zero, but the average speed is not. $\endgroup$
    – Roger V.
    May 7, 2022 at 8:21

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