Maxwell speed distribution

The Maxwell velocity distribution [1] [2] [3] [4] provides probability that the speed of a molecule of mass m lies in the range v to v+dv is given by

$P(v)dv = 4 \pi v^2 dv \left( \frac{m}{2 \pi k_B T} \right)^{3/2} \exp (-mv^2/2k_B T)$

where T is the temperature and $k_B$ is the Boltzmann constant. The maximum of this distribution is located at

$v_{\rm max} = \sqrt{\frac{2k_BT}{m}}$

The mean speed is given by

$\overline{v} = \frac{2}{\sqrt \pi} v_{\rm max}$

and the root-mean-square speed by

$\sqrt{\overline{v^2}} = \sqrt \frac{3}{2} v_{\rm max}$

Derivation

Consider an ideal gas with particles of unity mass in the three-dimensional ($3D$) space. As long as there is not a privileged direction in the equilibrium, we can take any direction in the space and to study the discrete time evolution of the velocity distribution in that direction. Let us call this axis $U$. We can complete a Cartesian system with two additional orthogonal axis $V,W$. If $p_n(u){\mathrm d}u$ represents the probability of finding a particle of the gas with velocity component in the direction $U$ comprised between $u$ and $u + {\mathrm d}u$ at time $n$, then the probability to have at this time $n$ a particle with a $3D$ velocity $(u,v,w)$ will be $p_n(u)p_n(v)p_n(w)$. The particles of the gas collide between them, and after a number of interactions of the order of system size, a new velocity distribution is attained at time $n+1$. Concerning the interaction of particles with the bulk of the gas, we make two simplistic and realistic assumptions in order to obtain the probability of having a velocity $x$ in the direction $U$ at time $n+1$: (1) Only those particles with an energy bigger than $x^2$ at time $n$ can contribute to this velocity $x$ in the direction $U$, that is, all those particles whose velocities $(u,v,w)$ verify $u^2+v^2+w^2\ge x^2$; (2) The new velocities after collisions are equally distributed in their permitted ranges, that is, particles with velocity $(u,v,w)$ can generate maximal velocities $\pm U_{max}=\pm\sqrt{u^2+v^2+w^2}$, then the allowed range of velocities $[-U_{max},U_{max}]$ measures $2|U_{max}|$, and the contributing probability of these particles to the velocity $x$ will be $p_n(u)p_n(v)p_n(w)/(2|U_{max}|)$. Taking all together we finally get the expression for the evolution operator $T$. This is:

$p_{n+1}(x)=Tp_n(x) = \int\int\int_{u^2+v^2+w^2\ge x^2}\,{p_n(u)p_n(v)p_n(w)\over 2\sqrt{u^2+v^2+w^2}} \; {\mathrm d}u{\mathrm d}v{\mathrm d}w\,.$

Let us remark that we have not made any supposition about the type of interactions or collisions between the particles and, in some way, the equivalent of the Boltzmann hypothesis of molecular chaos would be the two simplistic assumptions we have stated on the interaction of particles with the bulk of the gas. In fact, the operator $T$ conserves in time the energy and the null momentum of the gas. Moreover, for any initial velocity distribution, the system tends towards its equilibrium, i.e. towards the Maxwellian Velocity Distribution (MVD). This means that

$\lim_{n\rightarrow\infty} T^n \left(p_0(x)\right) \rightarrow p_f(x)=MVD\;(1D\;case)\,.$

Let us sketch now all these properties.

First, we introduce the norm $||\cdot||$ of positive functions (one-dimensional velocity distributions) in the real axis as

$\vert\vert p\vert\vert=\int_{-\infty}^{+\infty} p(x) dx.$

Then we have the following exact results:

Theorem 1

For any $p$ with $||p||=1$, we have $||Tp||=||p||$.

This can be interpreted as the conservation of the number of particles or in an equivalet way of the total mass of the gas.

Theorem 2

The mean value of the velocity in the recursion $p_n=T^np_0$ is conserved in time. In fact, it is null for all $n$:

$===\cdots==\cdots=0\,,$

where

$=\int_{-\infty}^{+\infty}f(x)g(x){\mathrm d}x\,.$

It means that the zero total momentum of the gas is conserved in its time evolution under the action of $T$.

Theorem 3

For every $p$ with $||p||=1$, we have

$====\cdots==\cdots \,.$

It means that the mean energy per particle is conserved and in consequence, by Theorem 1, the total energy of the gas is conserved in time.

Theorem 4

The one-parametric family of normalized gaussian functions $p_{\alpha}(x)=\sqrt{\alpha\over\pi}e^{-\alpha x^2}$, $\alpha\ge 0$, $||p_{\alpha}||=1$, are fixed points of the operator $T$. In other words, $Tp_{\alpha}=p_{\alpha}$.

Conjecture

As a consequence of the former theorems and by simulation of many exmaples, it can be claimed the following conjecture:

For any $p$ with $||p||=1$, with finite $$ and verifying $\lim_{n\rightarrow\infty} ||T^np(x)-\mu(x)||=0$, the limit $\mu(x)$ is the fixed point $p_{\alpha}(x)=\sqrt{\alpha\over\pi}\,e^{-\alpha x^2}$, with $\alpha=(2\,)^{-1}$.

Conclusion

In physical terms, it means that for any initial velocity distribution of the gas, it decays to the Maxwellian distribution, which is just the fixed point of the dynamics. Recalling that $=k\tau$, with $k$ the Boltzmann constant and $\tau$ the temperature of the gas, and introducing the mass $m$ of the particles, let us observe that the MVD (above presented) is recovered in its $3D$ format:

$MVD = p_{\alpha}(u)p_{\alpha}(v)p_{\alpha}(w)=\left({m\alpha\over\pi}\right)^{3\over 2}\,e^{-m\alpha (u^2+v^2+w^2)} \;\;\; with \;\;\; \alpha=(2k\tau)^{-1}.$