What are the requirements for a system to be described by the Boltzmann distribution in equilibrium?

For example, should all the particles be identical? No attractors in the phase space? Intuitionally, what are those systems? I know it is applied outside of ideal gases, so where do we draw the line?


3 Answers 3


I will be blunt. As fas I know, nobody knows a priori for which systems equilibrium statistical mechanics will work or not.

Part of the current effort to determine which systems are fine being described by equilibrium statistical mechanics focuses on various proofs of ergodicity for such systems. For now, they are somewhat limited to either a restrictive definition of ergodicity (Birkhoff's theorem for hard sphere systems) or for a broader definition of ergodicity with constraints on the observables to be ergodic for very short range interaction. I would say that the latter attempt was pioneered by Khinchin in his book of the mathematical foundations of statistical mechanics.

Others take on a very different path and consider as fundamental a form of Bayesian inference to justify the use of such a formalism. This line of work was strongly advocated by E.T. Jaynes and more recently by the likes of R. Balian in his book from microphysics to macrophysics.

The existence of attractors is not to be ruled out from the outset. For instance, one can study the thermodynamics of a table and figure out its thermal expansion coefficient, specific heat capacity, Young modulus, electrical properties and so forth. All of this can be done by using equilibrium statistical mechanics (and the canonical ensemble most of the time). Yet, the centre of mass of the table itself does not wonder about (on Earth at least) with some Boltzmann probability in my lounge. This means that the Boltzmann weight is to be used only for some aspects of the motion of the particles comprising the table but not for all. In that sense, most of the statistical thermodynamics we do in practice has to deal with the state of a system in a particular non-ergodic attractor; and that's fine.

Finally, it also depends what you consider as being given to justify the use of the canonical ensemble. For instance, even when given the rule of equally likely micro states for an isolated system it does not always imply that a system made of two parts (one small and one big) will reduce to a Boltzmann weight for the small sub-system. In particular that does not hold if the particles in each subsystem have long range interactions (like gravitational ones). General theorems on this particular assumption (microcanonical implies canonical) have been derived by using large deviation theory.


There is unfortunately no necessary conditions on which every physicist agree to retrieve equilibrium statistical mechanics. However, recent advances have permitted to show what does not constitute a necessary condition for statistical mechanics to work. I would mention:

  • Chaos is neither enough nor sufficient to claim that a system will equilibrate according to equilibrium statistical mechanics (contrary to what has been thought and taught for a long time). This has been illustrated in the general result provided by the KAM theorem, by the famous Pasta-Fermi-Ulam problem and by the work on this very topic by Vulpiani for instance.

  • Large system size. Although there are many results that show that many statistical mechanics results actually hold for large system sizes, there are also instances where this is not necessary. One can mention the proof of the Birkhoff's ergodic theorem for various billiards and also some issues that become even bigger as system size actually increases (like in the Pasta-Fermi-Ulam problem).

  • $\begingroup$ I really love this answer because you reference a ton of related material. You clarify a lot of concepts for me. I just have one question. It is clear that the topic of which systems are described by the Boltzmann distribution is a topic of ongoing research, but could you give the necessary conditions that have been established thus far? $\endgroup$ May 1, 2016 at 11:03
  • $\begingroup$ @DionysiosGerogiadis: If the microcanonical ensemble is assumed true for an isolated system, then the Boltzmann distribution follows for systems with comprising particles that have short range interactions. $\endgroup$
    – gatsu
    May 1, 2016 at 17:00
  • $\begingroup$ @DionysiosGerogiadis: If the equiprobability of microstates for isolated systems is not postulated, then arguments along the line of Khinchin's say that equilibrium statistical mechanics probably holds for large system sizes to calculate expectation values of some additive variables (like kinetic energy) of short ranged interacting particles. Jaynes' (very different) line of argument would be to redefine the ensemble itself if it is found that we have chosen originally a too broad or too narrow ensemble; it can be thought of as an iterative epistemological approach. $\endgroup$
    – gatsu
    May 1, 2016 at 17:11

There are several ways to characterize a system of particles. You can see them as whether they are isolated or in thermal equilibrium or in mechanical or chemical equilibrium. Similarly, you can ask whether a system is made up of identical bosons, identical fermions or distinguishable particle. In each of this cases, different statistics prevails. Simply because the count of microstates change drastically from one case to another.

The Maxwell-Boltzmann statistics and the Boltzmann distribution function are different things. The Boltzmann distribution where $F = $1/z$ e^{-\beta E }$ is true for system which is in contact with a thermal reservoir(i.e. it's temperature is fixed). This is also known as cannonical distribution. Here $z = \sum e^{-\beta E }$ for all E. There must be no other interaction of any kind. Only the thermal interaction with a reservoir. Of course for quantum systems(identical particles) these ideas need to be extended(see Pathria for example) but the idea is same.

On the other hand Maxwell-Boltzmann Statistics is not extendable to fermions or bosons. It is the statistics of distinguishable particles. It has nothing to do with Boltzmann distribution. By that I mean that if we allow particle interaction with environment and so on, this statistics gives some other distribution. e.g. For isolated system, it assumes equal probability to all states.

  • $\begingroup$ "Simply because the count of microstates change drastically from one case to another." This sentece really clarifies things for me. So is this what distinguishes Fermi, Bose-Einstein and Boltzmann statistics? $\endgroup$ May 1, 2016 at 11:05
  • $\begingroup$ Yes. This is the distinction between different statistics. $\endgroup$
    – Ari
    May 1, 2016 at 13:10

The boundary lines are, 1. isolated system 2. equilibrium 3. no composition change such as chemical reaction

  • $\begingroup$ Could you please give a more intuitional interpretation of these three things? For example, does isolated system mean constant energy? Equilibrium is very straightforward, but I do not get what no composition change means. Also, could you give a small hint to the happens if one of those things break down? (Is there a different distribution for systems that are not isolated?) $\endgroup$ May 1, 2016 at 4:59
  • $\begingroup$ Isolated system has no mass and energy exchange with the surrounding. If it gets a single piece of energy and reaches to another state of equilibrium, Boltzmann distribution is still valid but with different partition function. If the system continuously get energy, then Boltzmann distribution cannot describe this change and one has to introduce another parameter into the distribution. $\endgroup$
    – user115350
    May 1, 2016 at 5:36
  • $\begingroup$ When a system has mass exchange, for dense system, the distribution will be Fermi-Dirac distribution or Bose-Einstein distribution depending on whether it is fermions or bosons. $\endgroup$
    – user115350
    May 1, 2016 at 5:43
  • $\begingroup$ Thank you for touching up on fermions and bosons! The classification of such systems is becoming more and more clear through such comments. $\endgroup$ May 1, 2016 at 10:58

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