Definition of Statistical/Thermodynamic Equilibrium

Question

Suppose we have some system $A$ coupled with a reservoir $R$ where $A$ can exchange energy with $R$ but not particles or volume. The system $A$ will have some (time) average energy $U= \langle E \rangle.$ Given this $U,$ one can define the temperature: $$\frac{1}{kT} = \beta = \left(\frac{\partial (\ln \Omega(E))}{\partial E} \right)_{E= U; N \& V \text{fixed}}$$

Then the probability of $A$ being in a particular microstate $e_i$, with the microstate having energy $E$, "when $A$ is in equilibrium," will be given by the Boltzmann distribution: $$P(e') \propto \exp(-\beta E).$$

The probability of $A$ being in any microstate with energy $E$, denoted $P(E),$ is then given by:

$$P(E) \propto \rho(E) \exp(-\beta E)$$ Where $\rho(E)$ is the density of states. The implication of this definition seems strange; however, because it implies that any state of $A$ is possible in equilibrium. In other words, the definition of "A is in equlibrium" has nothing to do with the actual physical state $A$ is in at any given moment in time. What is the difference between the following:

1. $A$ is in equilibrium at time $t_{-1}$ and happened to fluctuate to some extremely low energy $E_{\text{cold}}$ at time $t_0.$
2. $A$ was cooled down to be made extremely cold, and placed into the reservoir at time $t_0$ with energy $E_{\text{cold}}.$ Non-equilibrium mechanics can then be used to calculate the rate at which it's temperature should approach $T$.

In both cases the physical situation at time $t_0$ is exactly the same, but in the first case we would say that it was in equilibrium, and in the second case we would not. My guess is that this has to do with information of the observer. In other words, the canonical ensemble refers to the situation where the system $A$ has been placed into the reservoir $R$, and no further information is known. In that case, as that observer, you couldn't say that $A$ was at $E_{\text{cold}}$ at time $t_0;$ you could only ever give a (time independent) probability distribution for the energies of $A$.

But this isn't quite satisfactory, because presumably the observer could do a quick measurement with a thermometer of $A$ and determine that it's energy had fluctuated wildly.

Answer 1

I just think you are overcomplicating this, by taking words a bit too literally.

Before bringing modern statistical mechanics into it, we can say that there is some idea of isolated thermodynamic equilibrium when energy and other conserved quantities are not entering or leaving that system and the system thus has a constant set of degrees of freedom. What makes it equilibrium is that all hold the same average energy, the average density of particles is uniform, and so on for any other observable conserved quantities we can measure: the local averages of observables should be constant across the system, and constant over time. So we can kinda base our whole idea on the equipartition theorem... Everything isolated and stable and constant and boring. (And “temperature, definition 1” is the average energy in a degree of freedom.)

In isolated equilibrium the total energy is a constant, not a variable. Not every state is possible, and indeed several states are precluded because they spontaneously change in observable ways.

The word “spontaneous” is also important here, it means “changing toward that most boring state” or so, you fill half the container with red gas and half with blue gas and it spontaneously mixes. You could start by defining spontaneity and then your isolated equilibrium is when the system is isolated and allowed to finish all of its spontaneous changes. (And “temperature, definition 2” is “hook up a small thermometer and wait until it stops spontaneously changing.”)

The thing is, the rest of the equilibrium definition is not too much more complicated, it is that a collection of systems can be in mutual equilibrium with each other, if the supersystem they constitute is in isolated thermodynamic equilibrium.

So I think you don't have any questions with regard to isolated thermal equilibrium. Your entire question comes when we move to statistical mechanics and say “these changes are spontaneous because the system is transitioning into a larger macrostate” and this implies some fluctuations where the system transitions away from that larger macrostate as noise, and is noise still an equilibrium process if it is a large enough jump. Now at some level this was always possible, even when we're looking at equidistibution, that says the average energy in a degree of freedom, it doesn't say what happens when so many degrees of freedom drop below average that the results is macroscopically observable.

What the statistical mechanic says, is “I don't care because the largest macrostate is generally exponentially larger than nearby other macrostates.” In other words, yes a bunch of different energies are possible but the really high energies that you could notice like 5kT, are exponentially suppressed. It is true that those large fluctuations take you out of equilibrium, but for the fluctuations to be large enough that we see them with our naked eyes, that just means you're transitioning back to a regime where you need to actually look at the detailed dynamics because the notion of “I can't tell the difference between these microstates” is starting to evaporate; your microscopy is too good.

(If you want a phrase for this to Google, I have heard it called the “coarse graining” of phase space.)

Answer 2

This question revolves around the concepts of macrostate and microstate in Statistical Mechanics and the role played by microstates in ensemble theory for equilibrium systems.

An important concept to consider is that almost always, there are many microstates corresponding to a given macrostate. In particular, in the canonical ensemble, all the microstates must be considered due to the possibility of an arbitrarily large exchange of energy with the reservoir. However (and this is the key point), their contribution to the averages is weighted by their probability. The probability of all the microstates having energy between $E$ and $E+dE$ depends on the macrostate through the temperature of the reservoir and on the energy $E$ according to the formula $$ P(E,dE) \propto \rho(E) \exp (-\beta E) dE. $$ The product $\rho(E) \exp (-\beta E)$ is usually strongly peaked around the average energy $$ \bar E= \frac{\int E \rho(E) \exp (-\beta E) dE}{\int \rho(E) \exp (-\beta E) dE}, $$ thus implying that all the sets of microstates with energies not coinciding with $\bar E$, although possible, have an extremely low probability. It is usually difficult to grasp that, for macroscopic systems, a tiny deviation from $\bar E$ is already very improbable; large deviations have such a small probability that their contribution to the averages can be safely ignored.

Therefore, although all the microstates are present in a canonical ensemble at temperature $T=\frac{1}{k\beta}$, only those with energy very close to $\bar E$ can be considered as typical microstates for that temperature. Microstates with energy significantly different from $\bar E$ are not typical; their probability is negligible, and their presence in a physical realization of the system can only be justified as a transient in a non-equilibrium ensemble approaching the equilibrium.

The consequence of such considerations is that just looking at a microstate does not allow us to assign it to a specific reservoir temperature, just as we cannot assign the probability of the event it belongs to from a single outcome. The presence of an observable of the system whose average coincides with the fixed reservoir temperature should not confuse us.

Answer 3

My guess is that this has to do with information of the observer. In other words, the canonical ensemble refers to the situation where the system $A$ has been placed into the reservoir $R$, and no further information is known. In that case, as that observer, you couldn't say that $A$ was at $E_{\text{cold}}$ at time $t_0;$ you could only ever give a (time independent) probability distribution for the energies of $A$.

Jaynes' Maximum entropy approach to deriving (macro)canonical distribution is a valid point of view, but the "traditional" reasoning in statistical physics is somewhat different. (See, e.g., Microcanonical ensemble through Maximum Entropy method.)

Thermodynamic equilibrium
In this traditional approach the equilibrium is usually defined by formulation like "all the fast relaxation processes have ended, and all the slow relaxation processes are too slow on the relevant time scale". In other words, for practical purposes thermodynamic equilibrium is approximate. E.g., very slow molecules may never hit the walls of the container during the observation time, but this doe snot prevent us from extending integral to zero when calculating pressure - the error is negligible.

Thinking of the size of the error leads us to draw a distinction between situations (1) and (2) mentioned in the Q.: by fluctuations in statistical physics we usually mean deviations of the order of $1/\sqrt{N}$ - i.e., the error in our approximation of the distribution by equilibrium one, which vanishes in thermodynamic limit. These fluctuations can be very large, but such large fluctuations are very rare/improbable - e.g., the probability that the Red Sea opened in front of Moses is finite, but so small, that we can neglect it for all the practical purposes, and confidently say that it has never happened. (Remember, that the relevant scale here is $e^{N_A}$, where $N_A\approx 10^{24}$.)

There are also situations that one often describes as equilibrium, but which are really stationary states, because they are characterized by permanent energy/matter fluxes. E.g., the surface of the Sun can be considered to be in equilibrium, characterized by temperature, usual statistical distributions, etc. In fact, this surface constantly loses energy via radiation, and this energy is resupplied by the thermonuclear reactions within the Sun, but for many practical purposes it is irrelevant. (See How does radiation become black-body radiation?.)

In the context of deriving hydrodynamic equations from Boltzmann equation, one also speaks o local equilibrium - a Boltzmann-like distribution with parameters (temperature and potentials) slowly varying through space, which however turns to zero the collision integral.

Ergodicity
Another factor to consider is the assumption of ergodicity, which underpins the "traditional" derivations. It means that in practice we average the state of the system over a rather long period of time, rather than over an ensemble, assuming that the two averages are equal. This directly addresses situation (1) in the Q.: fluctuations take place all the time, but they average out during the observation time.

On the other hand, a system genuinely driven out of equilibrium (situation 2) would show evolution, even with the averaging. In essence, we deal with quantities defined as: $$ U(t)=\frac{1}{T}\int_{t-T/2}^{t+T/2}u(t) $$ In equilibrium $u(t)$ - the instantaneous value - is changing all the time, but $U(t)$ remains nearly constant (to required precision).

See also: What maximizes entropy?

Remark
A probability of a single microstate under Boltzmann distribution is $p_i\propto e^{-\beta E_i}$. The probability of any microstate of energy $E$ is $p(E)\propto\sum_i\delta (E-E_i)e^{- \beta E_i}=\rho(E)e^{- \beta E_i}$ (I admit that the proportionality sign makes this distinction mathematically irrelevant, but it comes back into force when calculating the partition function or discussing the DOS.)

Appendix
In response to the comments, I have reviewed about a dozen introductory texts on statistical mechanics, in search for a clear definition of statistical equilibrium. I stress here that I am talking about statistical mechanics and not thermodynamics texts. I gave found that some of them forgo defining equilibrium completely, assuming perhaps that it is self-evident. Of the rest, I found no a single satisfactory definition, although the main outlines are clear - I add the quotes below.

In general, there are several types of definitions:
Empirical
You know it is an equilibrium when you see it. These definitions rely on our everyday experience that systems evolve towards an equilibrium state. This is a middle ground between the intuitive understanding and the thermodynamic definition below.

Thermodynamic definition
Equilibrium is a state stable against spontaneous changes. This is readily tied to entropy, and sometimes recast as a state in which entropy has its maximum (or more generally, specific thermodynamic potentials have extremum.) I think this definition belongs to thermodynamics rather than statistical physics, since it deals purely with macroscopic quantities and grounded in phenomenological observation.

Traditional definition
Finally, there is what I called above traditional definition, which comprises the following characteristics (although no single book below cites all of them):

• The system experience no macroscopic changes (processes), although it still evolves microscopically (fluctuations)
• The system reaches this state after an infinitely long wait (a more refined version of this statement is that the time is sufficiently long for "fast" relaxation processes to run out, but not too long for the "slow" processes to manifest themselves. This allows to incorporate, e.g., applying stat physics to elementary particles with long but finite lifetime or neglect slow chemical reactions - I heard this many times, but I couldn't find the exact reference.)
• There are no macroscopic fluxes of energy or matter

Quotes
Landau&Livshitz "Statistical physics, Part 1"

If a closed macroscopic system is in a state such that in any macroscopic subsystem the "macroscopic" physical quantities are to a high degree of accuracy equal to their mean values, the system is said to be in a state of statistical equilibrium (or thermodynamic or thermal equilibrium). It is seen from the foregoing that, if a closed macroscopic system is observed for a sufficiently long period of time, it will be in a state of statistical equilibrium for much the greater part of this period. If, at any initial instant, a closed macroscopic system was not in a state of statistical equilibrium (if, for example, it was artificially disturbed from such a state by means of an external interaction and then left to itself, becoming again a closed system), it will necessarily enter an equilibrium state. The time within which it will reach statistical equilibrium is called the relaxation time. In using the term "sufficiently long" intervals of time, we have meant essentially times long compared with the relaxation time.

Reif "Statistical Physics"

A system of many particles (such as our gas) whose macroscopic state does not tend to change in time is said to be in equilibrium.

Tanaka "Methods of Statistical Physics"

The thermodynamic process is a process in which some of the macroscopic properties of the system change in the course of time, such as the flow of matter or heat and/or the change in the volume of the system. It is stated that the system is in thermal equilibrium if there is no thermodynamic process going on in the system, even though there would always be microscopic molecular motions taking place. The system in thermal equilibrium must be uniform in density, temperature, and other macroscopic properties.

Smirnov "Principles of statistical physics"

In the course of the evolution of the system an individual particle can change its state, but the average number of particles in each state is conserved with some accuracy. Such behavior in a closed system is called thermodynamic equilibrium.

Sadovskii "Statistical physics"

When a macroscopic system is in (thermodynamic) equilibrium, its macroscopic characteristics (temperature, volume, pressure etc.) remain constant in time, though its microscopic state continuously changes and we do not know it at all (i.e. where precisely is its phase point on the ergodic surface at the given moment in time)

Reichl "A modern course in statistical physics"

Since the equilibrium state is, by definition, a state which is stable against spontaneous changes, Eq. (2.49) tells us that the equilibrium state is the state of maximum entropy.

Fitzpatrick "Thermodynamics and Statistical Mechanics"

Let us consider an isolated system in equilibrium. In this situation, we would expect the prob- ability of the system being found in one of its accessible states to be independent of time. This implies that the statistical ensemble does not evolve with time. Individual systems in the ensemble will constantly change state, but the average number of systems in any given state should remain constant. Thus, all macroscopic parameters describing the system, such as the energy and the volume, should also remain constant. There is nothing in the laws of mechanics that would lead us to suppose that the system will be found more often in one of its accessible states than in another. We assume, therefore, that the system is equally likely to be found in any of its accessible states. This is called the assumption of equal a priori probabilities, and lies at the heart of statistical mechanics. In fact, we use assumptions like this all of the time without really thinking about them.

It follows that if a stable equilibrium state has been attained [i.e., one in which no further spontaneous processes (other than random fluctuations) can take place] then this state is such that S is maximized. In other words, it is the most probable state of the system, subject to the given constraints.

Chandler "Introduction to modern statistical mechanics"

Experimentally we know that isolated systems tend to evolve spontaneously toward simple terminal states. These states are called equilibrium states. By simple we mean that macroscopically they can be characterized by a small number of variables.

Virtually no system of physical interest is rigorously in equi- librium. However, many are in a metastable equilibrium that usually can be treated with equilibrium thermodynamics. Generally, if in the course of observing the system, it appears that the system is independent of time, independent of history, and there are no flows of energy or matter, then the system can be treated as one which is at equilibrium, and the properties of the system can be characterized by E V, «!,..., nr alone. Ultimately, however, one is never sure that the equilibrium characterization is truly correct, and one relies on the internal consistency of equilibrium thermodynamics as a guide to the correctness of this description. An internal inconsistency is the signature of nan-equilibrium behavior or the need for additional macroscopic variables and not a failure of thermodynamics.

Huang "Statistical mechanics"

Thermodynamic equilibrium prevails when the thermodynamic state of the system does not change in time.

Mandl "Statistical Physics" More generally, let us consider an isolated system. This system may be in a state containing all sorts of pressure differences, temperature gradients, ihomogeneities of density, concentrations, etc. A system in such a state is of course not in equilibrium. It will change with time as such processes as pressure equalization, thermal conduction, diffusion, etc., occur. Left to itself, the system eventually reaches a state in which all these pressure gradients, etc., have disappeared and the system undergoes no further macroscopically observable changes. We call such a state an equilibrium state.

Wikipedia Thermodynamic equilibrium

Thermodynamic equilibrium is an axiomatic concept of thermodynamics. It is an internal state of a single thermodynamic system, or a relation between several thermodynamic systems connected by more or less permeable or impermeable walls. In thermodynamic equilibrium, there are no net macroscopic flows of matter nor of energy within a system or between systems. In a system that is in its own state of internal thermodynamic equilibrium, not only is there an absence of macroscopic change, but there is an “absence of any tendency toward change on a macroscopic scale.”1

A collection of matter may be entirely isolated from its surroundings. If it has been left undisturbed for an indefinitely long time, classical thermodynamics postulates that it is in a state in which no changes occur within it, and there are no flows within it. This is a thermodynamic state of internal equilibrium.[5][6] (This postulate is sometimes, but not often, called the "minus first" law of thermodynamics.[7] One textbook[8] calls it the "zeroth law", remarking that the authors think this more befitting that title than its more customary definition, which apparently was suggested by Fowler.)

Answer 4

I appreciate very much the answers which have been given to my question, but unfortunately they do not seem (to me) to answer the question in my OP. I cannot tell if this is because I have failed to express my question clearly enough, if I am fundamentally missing a point that others are making, or if the answers provided so far have been inadequate.

After thinking about the issue some more though I have come to an understanding that I think is to my satisfaction. I propose the following definition of "statistical equilibrium:"

A physical system $A$ is in statistical equilibrium, to observer O, with reservoir $R,$ if $O$ has no knowledge to suggest that the combination $A + R$ should be in any particular microstate. That is, $O$ should assume that $A+R$ has maximal entropy. If $R$ is much larger than $A$, and if $A$ and $R$ exchange only energy, then one can show that observer $O$'s best guess that the state $A$ is in will given by the canonical ensemble distribution.

As for the question that I raised in the OP, "what happens if a fluctuation is large enough that observer $O$ can notice it?" the above defintion should imply that the system is no longer in equilibrium. This shouldn't come as a surprise though, because the observer has updated their knowledge. On the other hand, such updates in knowledge will for all practical purposes never occur because changes which would be noticeable by typical observers with ordinary equipment will occur so rarely.

The situation was confusing because strictly speaking the definition I have written above implies that an observer should not be able to make any measurements of $A$ -- because this might give them new knowledge. However, "ordinary" measurements will almost never be precise enough to actually meaningfully give new knowledge, and so for all intents and purposes even if an observer is allowed to make measurements the system will still (to them) be in equilibrium.