Validity of Maxwellian distribution for interacting particles? I have read in a few (relatively credible) sources (e.g. Cambridge Tripos exam) that the Maxwell-Boltzmann speed distribution can be valid for interacting particles, but I have not been able to find a satisfactory proof/intuitive explanation of this, hence I seek advisement in two areas:
1) What is the intuitive explanation here? Is it possible to argue that the intermolecular forces are independent of velocity (and hence speed), maintaining the isotropy of the distributions? I have also read a brief comment (in another stackexchange post which for some reason I cannot find again) that this is only valid for short-range forces. What distinguishes the long-range forces from the short-range ones (assuming the argument I have mentioned is correct, since it would only be coordinate-dependent not velocity-dependent)?
2) Mathematically, how do we model the exponential term? What does the potential energy look like, would it be quadratic in $r_i$ or $v_i$ of the $i^{\text th}$ particle?
Thanks!
 A: Intuitive explanations may be misleading and, more important, they heavily depend on the previous knowledge on a subject matter. Without a good training and a formal proof I  would hardly believe that, even in a gas, the velocity distribution is a Maxwellian.
Certainly, as you noticed, the fact that interaction potentials do not depend on velocities plays an important role to ensure that interacting and not interacting systems behave similarly. However, I would observe that the consequence of this fact on velocity distribution is not trivially obvious. In particular, I do not see a really convincing way to prove the isotropy of the distribution without using formulae.
Within statistical mechanics the result can be obtained in a particularly simple way in the canonical ensemble. There, the probability distribution in phase space can be directly written as the Boltzmann's factor of the Hamiltonian $H$:
$$
\rho( \{ {\bf r_i,p_i}  \})= \frac{e^{-\beta H( \{ {\bf r_i,p_i}  \})}}{Z}
$$
which factorizes into a function of the positions times a function of the momenta as soon as the Hamiltonian can be written in the separated form $H( \{ {\bf r_i,p_i}  \}) = K( \{ {\bf p_i}  \})+P( \{ {\bf r_i}  \})$. Since the one-particle velocity distribution can be obtained as marginal distribution, by integrating $\rho$ over all coordinates and all momenta but one, it is clear, in this case, that the presence of a potential energy $P$ does not modifies the velocity distribution. Maxwell distribution is recovered for classical systems, since the kinetic energy term corresponding to the motion of the centers of mass of the molecules is a sum of quadratic terms.
However, already for classical particles, the general validity of such a result can be established only for macroscopic systems. Indeed, for finite, small size systems, this nice factorization does not hold in the micro canonical o in the gran canonical ensembles. Therefore the one-particle velocity distribution is not exactly a maxwellian, even for a perfect gas. One needs to take the limit for very large systems to recover both, the equivalence of results in different ensembles and the Maxwell's distribution. 
It is worth noticing that since the simple result in the canonical ensemble depends only on the separation of the Hamiltonian into a (quadratic) kinetic and a potential energy term, its validity cannot depend on the range of the potential, provided that the thermodynamic limit exists.
