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This is a question that was rised when we were discussing "what is melting actually". How many particles you need to form a liquid or solid. I have some remarks to point out what I want to know.

Q: How many particles is needed to observe a phase transition?

A:

  1. just one, phase transition is not a collective phenomena.
  2. 1 million, Ising model in 1000x1000 lattice produces a phase transition.
  3. $N_a$, phase transition is a thermodynamic concept and you always need to be in TD limit.
  4. The question is dependent on the process. For instance, in phenomena A you need N particles, but in B you need M particles.
  5. The question is stupid, you should ask X instead.

I would like to know which arguments are true and which are false (and why). Also, how many particles you need to form a liquid from solid (one particle with Gibbs energy above some limit X is a liquid?).

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3 Answers 3

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Option 3. An equilibrium phase transition is a non-analytic point of the thermodynamic free-energy. For a finite number of particles, the free-energy is always analytic. So you cant get a phase transition. Kardar discusses this point.

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I would say you should first understand how precise your instruments are. While Vijay Murthy is correct that, technically, there are no phase transitions for a finite number of particles, the discontinuous functions of the thermodynamic limit can be typically approximated with analytical functions with arbitrary precision, so, as your instruments' precision cannot be arbitrarily high, sometimes you just cannot tell an analytic function from a discontinuous one. I would say, the typical relative difference between intensive thermodynamic parameters for an infinite system and an N-particle system is $N^{-\frac{1}{2}}$, so, if $N\sim 10^{23}$ for a typical macroscopic system, it's difficult to tell, say, its specific heat from that of an infinite system.

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  • $\begingroup$ I thought this was an in-principle question since the OP asked "This is a question that was rised when we were discussing "what is melting actually"." You are right in practice. I agree. $\endgroup$
    – Vijay Murthy
    Commented May 12, 2012 at 17:37
  • $\begingroup$ Neither do I disagree with you. But if we treat this question as an "in principle" one, another issue arises: how precise is the theory you're basing your reply on? You see, standard statistical physics does not take into account relativistic effects, or the fact that "in principle" we should take into account the infinite number of soft photons, which is crucial, if we are talking about finite/infinite number of particles. So the theory is not arbitrarily precise either. $\endgroup$
    – akhmeteli
    Commented May 12, 2012 at 19:00
  • $\begingroup$ I only meant classical finite temperature phase transitions. For this case, the theory is quite well developed since the late 1970s (the Domb-Green-Lebowtiz series on phase transitions and critical phenomena cover these aspects). I am not sure if the OP wanted to consider quantum and relativistic effects. And my knowledge of quantum phase transitions is pretty poor. $\endgroup$
    – Vijay Murthy
    Commented May 12, 2012 at 19:09
  • $\begingroup$ Neither do I know what exactly OP wanted to consider. I am just trying to say that the theory you use does not claim arbitrarily high precision, so my note on precision may be relevant as well. $\endgroup$
    – akhmeteli
    Commented May 12, 2012 at 19:14
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Glad to see that your questioning title coincides with the seminar title of mine, ``How Many is Different? Ideal Bose gas revisited", which I have repeated given at various places last two years, including the most recent one (refs. linked).

Whether finite systems may feature a mathematical singularity or not is an old controversial issue. In fact, at the Van der Waals memorial meeting in 1937, the audience could not agree on the question, whether partition function for a finite system could or could not explain a sharp phase transition. So the chairman of the session, Kramers, put it to a vote!

At quantum level, for sure, a statistical partition function for a finite system is an analytic function of (typically) temperature and volume. This implies, e.g. the specific heat at constant volume, Cv, is finite, never diverges. However, if we consider alternatively the specific heat at constant `pressure', this quantity, Cp, may become singular, as shown first here.

According to them, the ideal Bose gas confined in a cubic box, (the standard textbook quantum system), may undergo a liquid-gas-type phase ``transition" under constant pressure condition, even though it consists of a finite number of particles.

The punchline is to choose an alternative section condition on the domain of the analytic function (temperature-volume plane), such as the constant pressure condition, and to realize a singularity.

J-H.

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