Maxwell-Boltzmann distribution works even for interacting classical particles?

Question

In many places I see it mentioned that the Maxwell-Boltzmann distribution

$$ n_i \propto \exp(-\beta E_i) $$

(number $n_i$ of particles in state $i$ decreases exponentially with the energy $E_i$ of the state)

holds only if the particles do not interact or the interactions are negligible. For example in the "MB statistics" Wiki article it is mentioned several times:

In statistical mechanics, Maxwell–Boltzmann statistics describes the average distribution of non-interacting material particles over various energy states

(Limits of applicability) Note, however, that all of these statistics [MB, FD, BE] assume that the particles are non-interacting and have static energy states.

(Derivation) Maxwell–Boltzmann statistics can be derived in various statistical mechanical thermodynamic ensembles: (...) In each case it is necessary to assume that the particles are non-interacting

However I've been following Tolman's text and it seems that the MB distribution holds for classical particles even if they interact with one another, so long as the entire dynamical system follows canonical equations of motion (actually I'm not sure what the exact requirement is, but it seems to be pretty liberal).

I've skimmed back portions of chapters 3 and 4 (ideally I would reread Tolman's entire derivation of the MB distribution, but it's so so long-winded) and this seems to indeed be the case, but I'm not 100% sure yet, and so I come here to ask.

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Sketch of derivation of MB distribution for gas of centrally interacting particles

Consider a gas consisting of a large number $n$ of classical point particles, distinguishable but otherwise identical, interacting with one another through central forces (say, each particle is bound to every other particle by an ideal spring):

$$ H = \frac1{2m}\sum_i p_i^2 + \sum_{i,j} V(|\mathbf q_i-\mathbf q_j|) $$

First note that the complete dynamical (micro)state of the gas is specified by the coordinates $x_i, y_i, z_i$ and momenta $p_{xi}, p_{yi}, p_{zi}$ of the particles. Or, the phase space of the gas is a product of $n$ single-particle phase spaces. This parallels Tolman's development; we have this footnote p. 71:

This implies that we can regard the coordinatos and momenta of the molecules themselves as giving a sufficiently precise specification of the intermolecular fiold, and can neglect tho circumstance that, in general, a knowledge of additional variables would really be neeessary to give an exact spocification to tho complieated electromagnetic ffeld existing betweon tho molecules. This approximation is valid when tho velocities of the electric charges involved are small compared with that of light.

Anyway, suppose now that the only thing we know about the gas is that its total energy is $E$. By the principle of equal a priori probabilities of statistical mechanics, it is equally likely to be in any one of the dynamical microstates compatible with this constraint on the energy.

This is the microcanonical distribution $\rho$. Notice that since the particles are interacting, the dynamical microstate of the system is changing over time as particles exchange energy. However, from the canonical equations of motion we have Liouville's theorem. And so $\rho$, which depends only on $E$, a constant of the motion, is constant over time. This means that the system is appropriately represented by the microcanonical ensemble every instant in time.

And so we may apply the usual argument. "What is the probability $n_i/n$ that a given gas particle is in the (single-particle) state $i$?" (we have partitioned the phase space of the particles into discrete units etc…). With the usual argument we arrive at

$$ n_i \propto \exp(-\beta E_i) $$

where $E_i$ is the energy of the $i$th state.

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So, is that right?

This answer and some of its comments seem to agree roughly with this (MB distribution works for interacting classical particles) (though it's unclear exactly what the restrictions on the interaction should be), but they don't go too deep on the subject.

As far as I know, the Boltzmann distribution is valid even for interacting gases, provided the particle interaction is short-range

It seems that within classical statistical mechanics the velocity distribution in thermal equilibrium is always the Maxwell-Boltzmann distribution, no matter what the interactions are (as long as they don't depend on the velocities).

Given how much I see it stated that MB only works for non-interacting particles, I would like more details on the matter. Who's wrong? Where did this idea that it had to be non-interacting come from?

Answer 1

I think it is a matter of definitions: microcanonical, canonical, and grand canonical distributions are quite general and applicable to quantum and classical systems, interacting or not (the assumtpions involved in tehir derivation are of different nature).

Boltzmann distribution is usually a synonym for the canonical distribution, although in some contexts it is probably limited to the classical case. Maxwell distribution is the distribution of velocities under the Boltzmann distribution with no potential. So it is really more narrowly defined and may exclude some interactions, but one often says Maxwell-Boltzmann instead of Boltzmann (which is often the term for Canonical).

Where the absence of interactions does play a role is in deriving some of the more specific results, such as the equations of state or heat capacities.

Answer 2

The one-particle energy distribution $$ n_i \propto \exp{(-\beta \epsilon_i)}, $$ where $E_i$ is a one-particle value of the energy, is usually called Boltzmann distribution (or Boltzmann factor). `The same name is also used for the $N$-body distribution $$ n \propto \exp{(-\beta E)}, $$ where now $E$ is the total energy of an $N$ particle system.

The name Maxwell-Boltzmann distribution is assigned to the distribution of the one-particle velocities in a classical system.

It is always possible to define an n-particle distribution, just averaging over all the degrees of freedom, but those of a single particle. However, only in the case of factorizable probability distributions, it is possible to establish a connection between the one-particle distribution and corresponding one-particle energy (or velocity).

To be more explicit, in the case of a quantum system, the one-particle energy eigenvalues are defined only if the Hamiltonian can be written as a sum of one-particle Hamiltonian. Even the presence of a weak pair-wise interaction prevents introducing a meaningful one-particle Hamiltonian and consequently, there is no way to introduce in general one-particle energies.

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Side remark.

Notice that in some cases the original degrees of freedom can be transformed into new degrees of freedom such that the new Hamiltonian is close to a separable Hamiltonian. In such special cases, it may be meaningful to introduce a one-particle distribution, although the new particles are collective states made of the original degrees of freedom. The case of phonons in a harmonic crystal is prototypical of such a possibility.

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Now, let's examine a little more in detail the Maxwell-Boltzmann velocity distribution function in the case of classical particles. In that case, notwithstanding some quite widespread confusion, the one-particle distribution is well defined even in the case of a strongly interacting system. The reason is that for a classical system it is meaningful to consider separately kinetic and potential energy (classically there is no problem of non-commutation). A formal proof is contained in an addendum to the answer cited in the question. Therefore, since the one-particle velocity distribution depends only on the kinetic energy part of the Hamiltonian, and this is always separable, the MB distribution of velocities is the correct classical distribution for any classical interacting or non-interacting system.