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Consider particles interacting only by long-range (inverse square law) forces, either attractive or repulsive. I am comfortable with the idea that their behavior may be described by the collsionless Boltzmann equation, and that in that case the entropy, defined by the phase space integral $-\int f \log f \, d^3x \, d^3v$, will not increase with time. All the information about the initial configuration of the particles is retained as the system evolves with time, even though it becomes increasingly harder for an observer to make measurements to probe that information (Landau damping).

But after a long enough time most physical systems relax to a Maxwellian velocity distribution. The entropy of the system will increase for this relaxation to occur. Textbooks tend to explain this relaxation through a collisional term in the Boltzmann equation ('collisions increase the entropy'). A comment is made in passing that an assumption of 'molecular chaos' is being made, or sometimes 'one-sided molecular chaos.' My question is, how do the collisions that underlie the added term in the Boltzmann equation differ from any collision under an inverse square law, and why do these collisions increase entropy when it is clear that interactions with an inverse square law force do not generally increase entropy (at least on the time scale of Landau damping?) And finally, how valid is the so-called molecular chaos assumption?

EDIT: I should clarify that, if entropy is to increase, then it is probably necessary to invoke additional short-range forces in addition to the long range inverse square law forces. I suppose I could rephrase my question as 'what sort of short-range forces are necessary to explain the collisional term in the Boltzmann equation, and how do they increase entropy when inverse-square law collisions do not?' If the question is too abstract as written, then feel free to pick a concrete physical system such as a plasma or a galaxy and answer the question in terms of what happens there.

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  • $\begingroup$ hmm... this question is interesting but it seems a bit too general. possibly if you could define your system more precisely we could get a better understanding. I don't think the inverse square force law comes into play at all in the maxwell-boltzmann distributions. they arise purely from statistical considerations and momentum conservation. $\endgroup$
    – Timtam
    Commented Mar 11, 2012 at 9:26
  • $\begingroup$ @Timtam I do want to keep the discussion rather general, in case the same answer can apply to different systems such as plasmas and galaxies. For the sake of concreteness, I suppose one could focus on either one of those two systems, and make as many additional assumptions about them as necessary to answer the question. Also, I've made an edit so as to allow the explanation to rely on other (short range) particle interactions, in case that's necessary to answer the question. $\endgroup$
    – kleingordon
    Commented Mar 11, 2012 at 9:31

2 Answers 2

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The statement that the entropy increases because of collisions is incorrect. The conservation of phase space volume is a theorem of Hamiltonian mechanics, and therefore applies to all known physical systems, regardless of whether they contain nonlinear forces, collisions or anything else.

What actually happens is that although the phase space volume doesn't change as you integrate the trajectories forward, it does get distorted and squished and folded in on itself until the system becomes experimentally indistinguishable from one with a bigger phase space volume. The information that was originally in the particles' velocity distribution ends up in subtle correlations between the particles' motions, and if you ignore those correlations, that's when you get the Maxwell distribution. The increase in entropy is not something that happens on the level of the system's microscopic dynamics; instead it occurs because some of the information we have about the system's initial conditions becomes irrelevant for making future predictions, so we choose to ignore it.

There is an excellent passage about this (in a slightly different context) in this paper by Edwin Jaynes, which gives a thorough criticism of the kind of textbook explanation that you mention. (See sections 4, 5 and 6.) It explains the issues involved in this much more eloquently than I can, so I highly recommend you give it a look.

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  • $\begingroup$ Thanks, that helps. I'm still confused as to how the definition of entropy can exist without reference to a physical scale on which correlations are ignored. $\endgroup$
    – kleingordon
    Commented Mar 11, 2012 at 11:31
  • $\begingroup$ The simple answer is that it can't. A scale below which details are ignored is always implicit in any definition of entropy. It's quite possible for two different observers to each assign a different entropy to the same system if they differ in what they can measure about it. See bayes.wustl.edu/etj/articles/gibbs.paradox.pdf for a really nice example. (I should note that this is all within a specific interpretation of statistical mechanics known as MaxEnt. I believe it to be the correct interpretation, but views do differ.) $\endgroup$
    – N. Virgo
    Commented Mar 11, 2012 at 13:49
  • $\begingroup$ But we have the unambiguous definition of entropy as -$\int f \, \log f \, d^3x \, d^3p$. In the context of classical physics, shouldn't this have a unique value regardless of who makes the measurement? Or are you saying that there is ambiguity in the way the integral is computed depending on how one handles the implicit limit that is used to define the integral? $\endgroup$
    – kleingordon
    Commented Mar 12, 2012 at 5:50
  • $\begingroup$ No, I'm saying that there's ambiguity in how $f$ is defined in that equation. Traditionally, $f$ was defined as the fraction of the time that the system spends in a given state, in the limit of infinite time. But this only makes sense if you assume the system is already in equilibrium, because how can something that's defined in terms of an infinite time period change over time? In the more modern interpretation, $f$ represents an experimenter's knowledge of the system's microstate - it's a probability distribution because that knowledge is incomplete. Thus it depends on what you can measure. $\endgroup$
    – N. Virgo
    Commented Mar 12, 2012 at 9:30
  • $\begingroup$ Hmm, okay, I'm starting to get it. At some point in the not-too-distant future I might want to chat to work out some of the remaining things that are niggling me, if you're willing. Thanks for the help. $\endgroup$
    – kleingordon
    Commented Mar 12, 2012 at 9:39
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The entropy increase comes from the assumption that you can close the system on the kinetic level, thereby (i) making the dynamics tractable and getting a transport equation, and (ii) disregarding extremely high frequency contributions and paying for this with an entropy increase.

Any interaction leads to collision terms; the details only matter for the particular form of the collision integral but not for its existence.

There are different ways to obtain the Boltzmann equation, but all share the above features. The molecular chaos assumption works only for classical ideal gases. For a modern derivation of kinetic equations and in particular the Boltzmann equation from fundamental principles (i.e., quantum field theory), see

Yu. B. Ivanov, J. Knoll, and D. N. Voskresensky, Self-Consistent Approximations to Non-Equilibrium Many-Body Theory, Nucl. Phys. A 657 (1999), 413--445. hep-ph/9807351

and related papers. See also Good reading on the Keldysh formalism

Edit: In an operator-based formalism, the kinetic approximation forces the density matrix to take the form $e^{-S/k_B}$, where $S$ is a 1-particle operator. This eliminates lots of (not all) high frequency contributions, as the exact dynamics destroys this form, so the approximation must project it back to it instantaneously. For understanding how the projection works see the book by Grabert on Projection operator techniques.

Calzetta did some work on kinetic theory in curved spaces (search the arXiv: http://lanl.arxiv.org); maybe this is more directly related to your question.

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  • $\begingroup$ Thanks for the response. I haven't yet consulted the references but I do have follow-up questions based on what you've posted. If the entropy increase comes from disregarding high frequency contributions to the distribution function, it would seem that some sort of cut-off scale would be required. But such a scale does not appear in the definition of the entropy. How can this be? Also, when you say that any interaction leads to collision terms, how does this work in a specific case like gravity? That is, how do some gravitational interactions increase entropy but others don't? $\endgroup$
    – kleingordon
    Commented Mar 11, 2012 at 11:26
  • $\begingroup$ Any interaction produces collision terms. You need to work out the corresponding microscopic expression to the collision integral. The details are always messy, so I won't try to give a sample calculation. Look at work by Calzetta (scholar.googel.com: author:Calzetta kinetic) for work in thisdirection. $\endgroup$
    – Arnold Neumaier
    Commented Mar 11, 2012 at 11:38

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