I'm very confused by the account on fluctuations near the equilibrium in chapter 12 of Landau's book on statistical physics. To be brief, the kernel of my doubt is that he states that if you have a observable $X$ such that equilibrium is attained at $X=0,$ then the probability of having a fluctuation of size $x$ is given by the exponential of the entropy evaluated at $x.$ I wonder, what is the mathematical definition of this notion of entropy? I don't even understand that. Below is my attempt at a possible formalization of those notions, but feel free to completely ignore it if you can answer my question right away!

Possible interpretation (surely wrong, but I'm sure that the "solution" to my doubt has to be something similar to this) :

Let $M$ be the space of configurations and $TM$ the phase space. Define a classical observable $X : TM \rightarrow \mathbb{R}. $ If $\mathcal{C}(TM)$ is a suitable space of probability density functions (the details not important at all for the argument here), define an entropy functional $I : \mathcal{C}(TM) \rightarrow \mathbb{R}$ in some way. I am guessing most times the appropiate deifinition would be $I[f]=-\displaystyle \int f(q,p) \text{log}f(q,p) dq dp,$ and I would like to focus specifically on this entropy, but of course other possibilities could/should be considered! We denote by $\overline{X}^f$ the average value of $X$ with respect to the probability given by $f,$ that is, $\overline{X}^f= \displaystyle \int X(q,p) f(q,p) dq dp$

Define an entropy function $S(x) : \mathbb{R} \rightarrow \mathbb{R}$ as: $$S(x) := \text{max} \left( I[f] \ : \ \overline{X}^f=x \right),$$ that is, $S(x)$ is the value of the entropy functional over the function maximizing it while satisfying the restriction that the average value of $X$ is $x.$ Of course, the existence and uniqueness of such a maximizer is not granted and is an important problem, but for the time being I will assume this is the case, since in the boltzmannian/maxwellian and closely related relevant situations existence and uniqueness actually hold. Therefore, the fact that $S$ is a well-defined function is highly-non trivial, but we won't care about this now. Call also $f_x : TM \rightarrow \mathbb{R}$ to each maximizer: $$f_x(q,p) = \text{argmax} \left( I[f] \ : \ \overline{X}^f=x \right). $$

In this context, the identity that I would like want to prove should be: $$\int_{ \{ (q,p) : X(q,p)=x \} } f_0(q,p) dq dp= c \text{e}^{S(x)},$$ where $c$ is some (normalizing) constant.


I suggest you take a look at: The Large Deviation approach to Statistical Mechanics (https://arxiv.org/abs/0804.0327)

This is a beautifully written review article which explains how a sub-branch of Probability Theory (called Large Deviation Theory) can be used as a language to express Statistical Physics. Section V of this article deals with equilibrium fluctuations. (However I recommended you read the preceding sections as well)

The exponential decay away from the mean value is nothing but a generalization of the central limit theorem (CLT) which holds far away from the mean as well (unlike CLT!).

The article motivates, in an intuitive manner, how entropy is the "rate function" that determines how fast the probabilities of these fluctuations decay.

| cite | improve this answer | |
  • $\begingroup$ Thank you for your contribution, it looks very interesting. It's is a very long work, so I cannot accept the answer as for now, but I will try to read it this week and share further doubts! :) $\endgroup$ – Qwertuy Apr 2 '18 at 16:24
  • $\begingroup$ Happy to help :) $\endgroup$ – StatisticalMechanic Apr 2 '18 at 17:30

In principle there does not seem to be that much of a mystery. Entropy is defined on Boltzmann's grave stone (and in Landau) as $$ S = \log\Omega $$ where $\Omega$ is the number of micro states corresponding to some macro state. Now we ask for the probablity that in a sub-volume $V$ of some system $A$ a thermodynamic variable fluctuates from its equilibrium value. The probability is just the ratio of the corresponding number of microstates, so $$ p\sim \exp(\Delta S) $$

| cite | improve this answer | |
  • $\begingroup$ Well, the fluctuation formula is indeed easy from that definition of entropy. But I was using the Gibbs-Jaynes-Shannon entropy functional, not the Boltzmann entropy. They are quite different objects, in principle. One of basic doubts is precisely whether the Boltzmann entropy can be directly related to the entropy functional (via some maximum principle) $\endgroup$ – Qwertuy Apr 6 '18 at 13:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.