Maxwell derived the shape of the probability distribution of velocity of gas particles by starting with just two assumptions.

These are:

  1. The probability distribution is rotation invariant.

  2. The components (of velocity of a gas particle) in the direction of the coordinate axes are statistically independent.

And the rest is lovely deduction, but I found that as a layman I don't have any physical intuition as to why the second assumption is plausible. Is there an intuitive explanation behind the second assumption? If not, is there a way to derive the second assumption from a set of more plausible-looking assumptions?

  • $\begingroup$ This assumption was actually the weakest point in Maxwell's derivations, as he fully realized himself. Indeed, in his subsequent paper "On the dynamical theory of heat" he found an alternative derivation avoiding this assumption. Later, Boltzmann further improved on these aspects in his paper "Studien über das Gleichgewicht der lebendigen Kraft zwischen bewegten materiellen Punkten". $\endgroup$ – Yvan Velenik Dec 4 '13 at 18:07
  • $\begingroup$ It's worth pointing out that the velocity components are only statistically independent for a gas of classical, massive particles. In contrast the velocity components have correlations with a gas of relativistic particles. $\endgroup$ – Nanite Jan 3 '14 at 14:53
  • $\begingroup$ I thought the most arbitrary assumption was they were distributed as Gaussian in velocity (or was it energy?) Either way it seems a lucky guess. $\endgroup$ – DWin Feb 2 '14 at 4:32
  • $\begingroup$ @DWin: A vector of independent random variables which has a joint distribution that is invariant under rotations must be a Gaussian. This is in fact a characterization of Gaussian distributions $\endgroup$ – Alex R. Mar 4 '14 at 6:16

If I have a velocity which has some component $v_x$ in the $x$-direction, then is there any reason for you to assume you know anything anything about the component of my velocity which might be in a perpendicular direction, $v_y$?

No. So you can see that it is reasonable to assume that, if you know my $v_x$, my $v_y$ is still unconstrained, i.e. you have no information about it. The same holds for $v_z$, and there you have your statistically independent velocities!

  • 1
    $\begingroup$ I'm not convinced this is indeed the case. If I know that $v_x$ has some extreme value rather greater than $\sqrt{{kT}/{m}}$, is it really evident that $v_y$ is equally likely to be nonzero as when $v_x=0$? $\endgroup$ – Emilio Pisanty Dec 3 '13 at 23:02
  • $\begingroup$ Now that I thought about it a bit more, I think the real point might be the symmetry. Do you have a more explicit argument? I'd be interested because it might help me refine my vague ideas. $\endgroup$ – Danu Dec 3 '13 at 23:16

I'm just going to quote Wikipedia here:

For the case of two colliding bodies in two dimensions, the overall velocity of each body must be split into two perpendicular velocities: one tangent to the common normal surfaces of the colliding bodies at the point of contact, the other along the line of collision. Since the collision only imparts force along the line of collision, the velocities that are tangent to the point of collision do not change. The velocities along the line of collision can then be used in the same equations as a one-dimensional collision.

Key point: During a collision, the components of velocity perpendicular to the axis of contact (which is random) do not, for example, decrease when the other increases. Each collision selects an axis at random and perturbs the velocity in that direction.


This came to my mind after reading some introduction to maximum entropy probability distributions.

Independence can be derived from the following four assumptions:

(1) average momentum of particles inside the box is fixed at 0

(2) average kinetic energy of particles inside the box is fixed

(3) the gas velocity distribution must be maximum entropy probability distribution under the above two constraints (1) and (2).

(4) kinetic energy of a particle can be expressed as $f(v_x) + f(v_y) + f(v_z)$ for some function $f$. (not true for relativistic particles)

(Hopefully (2) and (3) are somewhat more plausible-looking than independence. (2) seems related to equipartition theorem or is it enough to say that the average kinetic energy is fixed simply because of law of large numbers? And (3) is mostly asserting that there is no other hidden constraints (other than (1) and (2)) to be discovered.)

To derive independence, suppose $p(v_x,v_y,v_z)$ is a probability density function (on velocity) that does NOT have independence between x, y, z components of speed. We want to compare the entropy of $p$ to that of $p'$ where $p'$ is defined as $p'(v_x, v_y, v_z) = p_1(v_x) p_2(v_y) p_3(v_z)$ where $p_1$ is the marginal distribution of x component of velocity from $p$ and similarly for $p_2, p_3$. If $p$ shows average momentum of 0, and average kinetic energy of 1, then the same is true for $p'$. ($p'$ and $p$ having same average kinetic energy is derived from (4)). But it is a result in information theory that $h(p) < h(p_1) + h(p_2) + h(p_3) = h(p')$ (where $h(p)$ denotes entropy of $p$). This means that $p$ CANNOT be maximum entropy probability distribution under constraints (1) and (2).


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