# When do gas molecules deviate from Maxwell Boltzmann distribution?

The velocity of the gas molecules at room temperature obey Maxwell Boltzmann distribution. When we say velocity, the distribution is true for each component right ? means vx, vy and vz are in MB distribution. At higher temperature, the distribution just widen and in lower temperature it become narrower. Sill,the distribution is MB. So is there any circumstances, the gas deviates from MB distribution? I guess when there is a gas flow, am I right? Is there any other situation the deviation happen?

• Sure. When you can no longer consider all particles are interchangeable, yuo must take into account the simmetrization postulate. According to their symmetry, the'll be bosons or fermions, and you'll have to check statistics of Bose-Einstein or Fermi-Dirac. – FGSUZ May 25 '18 at 22:01
• There is such a thing as nonthermal distributions for cosmic ray particles (usually a power law with exponent near -2). – Kyle Kanos May 25 '18 at 22:31

Somewhat tautologically, the answer is: Whenever the assumptions for the Maxwell-Boltzmann distribution are not met (in good approximation). The Maxwell-Boltzmann distribution assumes a gas of non-interacting, classical (here: non-quantum and non-relativistic) particles in equilibrium.

To enumerate some possibilities, where those assumptions are not met:

• If the gas is composed of non-interacting quantum particles, you will have a Bose or Fermi distribution instead (those are defined in terms of energies, since velocities are not too good a concept deep in the quantum regime). More generally, there are quantum corrections to the Maxwell-Boltzmann distribution.

• If the gas is sufficiently hot that relativistic corrections to the kinetic energy become important.

• If the particles in the gas are interacting and the interaction is velocity dependent there are corrections to the Maxwell-Boltzmann distribution.

• If the gas of classical particles is not in equilibrium the distribution will not be Maxwell-Boltzmann (and a real gas is never in equilibrium, just close to equilibrium on some length scale). Note, that in small parcels of a flowing gas the velocities in a co-moving frame are very close to being Maxwell-Boltzmann distributed.

As to the part about the components of the velocity, you can write the distribution in different ways (in units where $k_B = 1$) and both are often called the Maxwell-Boltzmann distribution:

• The velocities are distributed $p(\vec v) \propto \exp\left(-m\vec v^2/2T\right)$ and this means that $p(v_i) \propto \exp\left(-mv_i^2/2T\right)$ ($i \in \{x, y, z\}$)

• The speed (absolute value of the velocity) follows a distribution $p(v) \propto v^2 \exp\left(-mv^2/2T\right)$. The components of the velocity obviously do not follow this distribution.

As Ján Lalinský pointed out in the comments, the Maxwell-Boltzmann distribution need not be corrected for most interacting classical gases (even the English Wikipedia article on the Maxwell-Boltzmann distribution at the time of this writing states, that the van-der-Waals gas has corrections to the velocity distribution). The German Wikipedia includes a derivation of the Maxwell-Boltzmann distribution under the assumption of an arbitrary, velocity independent, conservative force (which has a potential $U(\vec r_1, \ldots, \vec r_N)$) acting on the particles. I'll repeat the argument here in abbreviated form. The $N$ particles of a classical gas follow the Boltzmann statistics, in equilibrium the phase space distribution is given by: \begin{align*} \rho(\vec p_1, \ldots, \vec p_N, \vec r_1, \ldots, \vec r_N) = \frac 1 Z e^{-\beta H(\vec p_1, \ldots, \vec p_n, \vec r_1, \ldots, \vec r_N)}, \\ Z = \frac{1}{N!(2\pi\hbar)^{3N}}\int d^{3N}r \int d^{3N}p\, e^{-\beta H(\vec p_1, \ldots, \vec p_n, \vec r_1, \ldots, \vec r_N)}. \end{align*} Since the Hamilton function is given by $H = \sum_i \frac{p_i^2}{2m} + U(\vec r_1, \ldots, \vec r_N)$ and $e^{a+b} = e^ae^b$ the partition function factors as $$Z = \frac{1}{N!(2\pi\hbar)^{3N}}\int d^3r e^{-\beta U(\vec r_1, \ldots, \vec r_N)} \int d^3p\, e^{-\beta \sum_i \frac{p_i^2}{2m}}.$$ Now we can derive the probability density for the momentum of some particle $j$: \begin{align*} P(\vec p) &= \left<\delta(\vec p - \vec p_j)\right> = \frac 1 Z \frac{1}{N!(2\pi\hbar)^{3N}}\int d^{3N}p \int d^{3N}r\, \delta(\vec p - \vec p_j) \rho(\vec p_1, \ldots, \vec p_N, \vec r_1, \ldots, \vec r_N) \\ &= \frac{\int d^{3N}r e^{-\beta U(\vec r_1, \ldots, \vec r_N)} \int d^{3N}p\, \delta(\vec p - \vec p_j) e^{-\beta \sum_i \frac{p_i^2}{2m}}}{\int d^{3N}r e^{-\beta U(\vec r_1, \ldots, \vec r_N)} \int d^{3N}p\, e^{-\beta \sum_i \frac{p_i^2}{2m}}} = \frac{e^{-\beta \frac{p^2}{2m}}}{\int d^3p_j e^{-\beta \frac{p_j^2}{2m}}} = \left(\frac{\beta}{2m\pi}\right)^{\frac 3 2} e^{-\beta \frac{p^2}{2m}} \end{align*} Note, how all position integrals and all but one momentum integrals cancel due to the factorization of the exponentials (and therefore the integrals). The only remaining step is to transform the distribution from momentum to velocity as argument: $$P(\vec v) = \left(\frac{m}{2\pi T}\right)^{\frac 3 2} e^{-\beta \frac{mv^2}{2}}.$$
Note, that this derivation has surprising consequences. For example an ensemble of classical harmonic oscillators in thermal equilibrium with a bath will have Maxwell-Boltzmann distributed velocities, since they fit the derivation above with $U = \sum_i \frac{kr_i^2}{2}$.
• By long range I mean those interactions which do not allow expressing total energy as sum over energy attributed to parts of the material system, which is the usual condition under which the Boltzmann distribution is derived (also manifested by the fact that energy is homogeneous function of $N,S,V$). That is possible for interactions whose interparticle potential decays faster than $1/r^2$, such as van der Waals interaction does. But it is not valid for gravitational interaction or Coulomb interaction of same charges. – Ján Lalinský May 27 '18 at 20:55