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I have seen two definitions of thermodynamic limit.

Definition 1. This is the common definition in statistical physics textbooks, $$N \to \infty ,V \to \infty ,N/V = {\rm{constant}}$$ There are many reasons to consider this thermodynamic limit one of which is that surface effects should be negligible as one is only interested in bulk properties. However, I can’t see any evidence that the above mathematical statement can leads to negligible surface effect. If $V \to \infty $, then the surface area of the system will also tends to infinity. Then I found the second definition of thermodynamic limit,

Definition 2. I found this definition in chapter 3 of "Statistical Mechanics of Lattice Systems" by Sacha Friedli and Yvan Velenik. In 3.2.1 (page 83), a sequence ${\Lambda _n}$ converges to $ {Z^d} $ ($ {Z^d} $ is an infinite lattice in d dimension) in the sense of van Hove is defined as,

$$\mathop {\lim }\limits_{n \to \infty } \frac{{\left| {{\partial ^{{\rm{in}}}}{\Lambda _n}} \right|}}{{\left| {{\Lambda _n}} \right|}} = 0$$

The meaning of the symbols can be seen in the book. It seems that this definition was used as the definition of thermodynamic limit in this book.

This definition of thermodynamic limit clearly shows that the surface effects are negligible. So my question is are these two definitions of thermodynamic limit equivalent?


As said by @GiorgioP, the meaning of the symbols related are listed below,

$ {Z^d} $ is d dimensional cubic lattice (infinite); ${\Lambda _n}$ is a sequence ($n \ge 1$) of finite volume subsets of $ {Z^d} $, that is, ${\Lambda _n} \subset {Z^d},\left| {{\Lambda _n}} \right| < \infty$ (PS: i didnt find the defination of $\left| \Lambda \right|$ in the book, through the context i think it might be the number of vertexes in $\Lambda$)
${\partial ^{{\rm{in}}}}{\Lambda _n}$ is the surface of ${\Lambda _n}$ which is defined by $${\partial ^{\ln }}\Lambda \mathop = \limits^{{\rm{ def }}} \{ i \in \Lambda :\exists j \notin \Lambda ,j\sim i\}$$
here ${j \sim i}$ means vertex j and i and bonded by a single edge in Ising model.

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  • $\begingroup$ I think math.SE would have a better chance at this. I tried doing some constructions and in the end I found that the two definitions are equal (in lattices) iff van Hove convergence is equivalent to normal convergence (which probably is not the case). I can post an answer that shows where I got (which isn't that far) but it is not enough to say a lot. $\endgroup$
    – acarturk
    Dec 31, 2019 at 11:03
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    $\begingroup$ @acarturk I do not agree that math.SE would be a better site for this question. Thermodynamic limit is a mathematical procedure strongly based on the underlying physics. Moreover, at least one of the authors of the cited book is an active contributor to SE.physics. What a better place to get an authoritative answer? $\endgroup$
    – GiorgioP
    Dec 31, 2019 at 11:16

2 Answers 2

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Missing a definition of the symbols you are reporting from the book cited in Definition 2, it is difficult to make more than some guess about the two definitions' mutual relations. It is highly probable that Definition 2 is about the thermodynamic limit in a system on a lattice.

However, there is a general observation, which is a partial answer to your question. The thermodynamic limit is a formal tool for recovering all the classical thermodynamic systems' properties from the ensemble formulation of statistical mechanics. The underlying physical idea is to look for the limit of some intensive version of thermodynamic potentials when the system's size grows. As such, its precise definition varies with the system and with the ensemble.

For example, the thermodynamic limit in the usual microcanonical, canonical, and grand canonical ensembles are:

  • (microcanonical) $$ N \rightarrow \infty, V \rightarrow \infty, E \rightarrow \infty, $$ keeping constant $E/V$ and $N/V$.
  • (canonical) $$ N \rightarrow \infty, V \rightarrow \infty $$ keeping constant $N/V$ at a fixed temperature.
  • (grand canonical) $$ V \rightarrow \infty $$ at a fixed temperature and chemical potential.

Moreover, if additional or different extensive variables are required to describe a macrostate, they should be included as well.

The basic assumption which justifies the disappearance of finite size effects at the thermodynamic limit is that they would not scale like the volume $V$ but at least as $V^{\alpha}$, with $\alpha<1$. For a finite 3D system where particles interact via short-range forces, one would expect surface effects scaling like $V^{2/3}$. The difficult part is to show rigorously that such an expectation is actually satisfied by realistic interaction models.

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  • $\begingroup$ Sir, thank you very much for your kind reply! your answer is very enlightening. one question is that can we treat the fluctuations as a consequece of finite size of the system. $\endgroup$
    – FaDA
    Dec 31, 2019 at 13:06
  • $\begingroup$ Be careful. Infinite size systems do not suppress fluctuations in general. Only relative fluctuations of extensive quantities, provided one is not at criticality. $\endgroup$
    – GiorgioP
    Dec 31, 2019 at 13:12
  • $\begingroup$ Sir do you mean that in infinite size systems, fluctuations can also exists. $\endgroup$
    – FaDA
    Dec 31, 2019 at 13:24
  • $\begingroup$ @FaDA Sure. It is the relative fluctuation which goes to zero with increasing the size of the system. $\endgroup$
    – GiorgioP
    Dec 31, 2019 at 14:29
  • $\begingroup$ Does the statement "in infinite size systems, fluctuations can also exists" corresponds to extensive properties. I mean it seems to me that the intensive properties tends to some constant value independent of the system size when approaching the thermodynamic limit. $\endgroup$
    – FaDA
    Dec 31, 2019 at 15:10
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I have nothing to add to @GiorgioP 's excellent answer, so I'll focus on issues specific to the book you address. (From your question, it is clear that several aspects I discuss below are clear for you; I nevertheless address them in case they might be useful to other readers.)

First, if $A\subset\mathbb{Z}^d$, then $|A|$ indeed denotes the number of vertices in $A$ (see the list of conventions, page xv of the free version of the book).

Spin systems vs fluids

The way you look at the thermodynamic limit applies better to a fluid, in which the number of molecules and the volume of the system are distinct quantities. The definition you state from our book applies to the Ising model, in which there is one spin per vertex, so the volume $V$ and the number of constituents $N$ would be equal in this case.

You'll find statements closer to those you are used to in the lattice-gas interpretation of this model. In this interpretation, a $+$ spin corresponds to a vertex occupied by a "particle", while a $-$ spin corresponds to an empty vertex. In this case, the number of $N$ of particles and the volume $V$ of the system are again distinct quantities and the thermodynamic limit takes the more usual form.

As @GiorgioP mentioned, you then have various notions of thermodynamic limits according to the ensemble you are considering. This is discussed in detail in Chapter 4 of the book. There is also an informal discussion in the introductory chapter (Section 1.3).

Convergence in the sense of van Hove

Finally, the precise condition you state (convergence in the sense of van Hove) is required if you want to prove convergence of the pressure (or of the thermodynamic potential relevant to the ensemble you consider) of finite-volume systems to a universal limit (that is, a limit that does not depend on the "shape" of the boxes in which your finite systems are contained or on the boundary condition used). You must then be able to show that boundary effects induce only higher-order corrections that disappear in the limit. For systems with short-range interactions, it turns out that these corrections scale like the size of the boundary of the system and will thus be negligible (remember that we are interested in densities) provided that the size of the boundary of the system grows slower than its volume, which is precisely what van Hove's condition guarantees.

Why is this usually not stated explicitly in Physics?

The answer is very simple: usually, in Physics, on works with nice sequences of "boxes" (cubes, for example), so this problem does not arise (or, better, the condition is automatically satisfied). This is also what is done in Chapter 4 of the book.

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  • $\begingroup$ Sir, thank you very much for your answer and your gorgeous book. I followed your book on the youtube so I skipedd Chapter 4, I will check it later. One question I am still not very sure is that, after above discussion, I wonder if the statement $\psi (\beta ,h) = \mathop {\lim }\limits_{{\rm{thermodynamiclimit}}} \psi _\Lambda ^{\mathop \# \limits^ }(\beta ,h)$ means there is no fluctuation for intensive quantity $\psi $ in the thermodynamic limit. $\endgroup$
    – FaDA
    Jan 1, 2020 at 12:12
  • $\begingroup$ Sir, thank you very much for your answer and your gorgeous book. I followed your book on the youtube so I skipedd Chapter 4, I will check it later. One question I am still not very sure is that, after above discussion, I wonder if the statement $\psi (\beta ,h) = \mathop {\lim }\limits_{\Lambda \Uparrow {Z^d}} \psi _\Lambda ^\# (\beta ,h)$ means there is no fluctuation for intensive quantity $\psi $ in the thermodynamic limit. $\endgroup$
    – FaDA
    Jan 1, 2020 at 12:22
  • $\begingroup$ @FaDA : in the thermodynamic limit, the density associated to extensive observables (for instance the magnetization density in an Ising model) do not fluctuate (except, possibly, at first-order phase transitions). Concerning thermodynamic potentials, these quantities do not fluctuate even in finite systems. Note that these quantities are not observables: they are defined at the level of ensembles, not individual microscopic configurations. $\endgroup$ Jan 1, 2020 at 12:22
  • $\begingroup$ (And thanks for your nice words about our book.) $\endgroup$ Jan 1, 2020 at 12:23
  • $\begingroup$ Note that, as @GiorgioP mentions in his comment, this does not mean that there are no fluctuations in the thermodynamic limit. Even in an infinite Ising system, say, the value of the spin at a particular vertex fluctuates! This is true of any local observables. Only averages of local observables over the whole lattice take deterministic values. $\endgroup$ Jan 1, 2020 at 12:25

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