Canonical ensemble can be used to derive probability distribution for the internal energy of the closed system at constant volume $V$ and number of particles $N$ in thermal contact with the reservoir.

Also, it is stated that the temperature of both system and reservoir is the same, i.e. they are in the thermal equilibrium.

In statistical thermodynamics, the thermal equilibrium arises naturally from following consideration:

If we consider closed system of interest in thermal contact with the reservoir such that these two form an isolated system than total energy $E$ is defined as: $E = E_S + E_R$

Total number of microstates of the isolated system with internal energy $E$ is given by: $$ \Omega(E) = \sum_{E_S}\Omega_S(E_S)\Omega_R(E - E_S) = \sum_{E_S}f(E_S) $$

If we wanted to find the internal energy of the system $E_S$ for which the number of microstates of the isolated system $f(E_S)$ has a maximum value, we do the following: $$\frac {df(E_S)} {dE_S} = 0$$

This leads us to: $$ \Omega_R\frac {d\Omega_S} {dE_S} + \Omega_S \frac {d\Omega_R} {dE_S} = 0 $$

Because the total system is isolated: $$dE_S = - dE_R$$

Which leads us towards: $$ \frac {dln\Omega_S} {dE_S} = \frac {dln\Omega_R} {dE_R} $$

Taking into account Boltzmann definition of entropy, we know that the previous expression defines temperature equality of the closed system and the reservoir or thermal equilibrium condition.

Important thing to note is that thermal equilibrium is defined when the closed system of interest has the internal energy which corresponds to the maximum number of microstates of the isolated system.

Such state is the most probable and in the thermodynamic limit we are basically certain to find the system's internal energy very close to the value which maximizes the number of microstates. This means that the concept of thermal equilibrium only has sense in the case of thermodynamic limit.

However, canonical ensemble can be used, as it is mentioned earlier, to derive the probability distribution of the internal energy of the system in the more general case when the thermodynamic limit isn't supposed.

It is from the canonical ensemble that we know that the variance of internal energy for the closed system tends to zero when the number of particles tends to infinity.

This means that if the thermodynamic limit isn't assumed, thermal equilibrium condition losses its significance as system can be found in many different energy states with considerable probability not only in the energy state which maximizes the number of microstates as it is the case in the thermodynamic limit.

This is what doesn't really make sense because the canonical ensemble should be applicable in more general case than thermodynamic limit and yet it uses the concept of thermal equilibrium as its condition which only has sense in the thermodynamic limit.

I think solution to this question may be that the thermal equilibrium isn't the general requirement for the canonical ensemble and that it is only valid in the thermodynamic limit which covers almost all cases in reality. Although, if you look on the Wikipedia: https://en.m.wikipedia.org/wiki/Canonical_ensemble you'll find that thermal equilibrium is included in the definition of the canonical ensemble.


3 Answers 3


One needs to distinguish between a closed system (microcanonical ensemble), and a system in a contact with a heat bath (canonical ensemble and canonical distribution). In the microcanonical ensemble the energy is fixed, and the equilibrium corresponds to the maximum of entropy. In the case of the canonical ensemble, one maximizes the entropy of the joint system (system+heat bath), and speaks of the equilibrium between the two, which means that the probability distribution of states is unchanged and depends only on the energy of states.

  • $\begingroup$ This distinction is clear. I don't mention microcanonical ensemble here. Misunderstanding may arise with the term $closed$ $system$ as I sometimes use different terms for the closed system of interest. I always used term $isolated$ $system$ to refer to the joint system, though. $\endgroup$ Commented Jul 26, 2022 at 17:40
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    $\begingroup$ Good answer, see also this nice one: physics.stackexchange.com/a/369412/226902 $\endgroup$
    – Quillo
    Commented Apr 2, 2023 at 11:15

What you are discussing here is not "the canonical ensemble" but "the equivalence of the canonical and microcanonical ensembles". You start by assuming that a system and environment have a fixed total energy, that is that they are together in the microcanonical ensemble, and deduce, using the assumption of taking the thermodynamic limit, that this will give the same results as the canonical ensemble.

If we do not assume the thermodynamic limit, then this equivalence no longer holds. This does not invalidate the canonical ensemble (which can be justified independent of the microcanonical ensemble) or call the concept of thermodynamic equilibrium simply means the choice of ensemble matters for our results. In other words if our system is small enough that fluctuations are not negligible, we have to pay attention to which quantities are actually held fixed and which are allowed to fluctuate.

  • $\begingroup$ Yes, thanks for the answer. However, problem is that how can we define that canonical ensemble holds for system and reservoir being in thermal equilibrium when concept of thermal equilibrium only has sense in the thermodynamic limit? If this is the case than how can we use the canonical ensemble for example to derive the probability distribution of internal energy of the system when such distribution should hold in the more general case than thermodynamic limit? $\endgroup$ Commented Jul 26, 2022 at 20:13
  • $\begingroup$ You have chosen a very narrow definition of thermodynamic equilibrium, but we can definethe concept more generally. Two objects are in thermodynamic equilibrium if they are in thermal contact but there is no flow of heat between them i.e. they bodies can exchange energy but on average they don't. There may be fluctuations around this average, but this does not change the fact that the objects are in equilibrium. This definition does not depend on the thermodynamic limit, although predictions based on it may become less useful if fluctuations are large $\endgroup$ Commented Jul 26, 2022 at 20:39
  • $\begingroup$ I wouldn't agree that my thermal equilibrium definition is narrow. In fact, how you defined it, is what exactly holds in the thermodynamic limit; energy isn't exchanged on average because if that isn't the case there is a considerable probability to find the system with the energy which deviates more from its expectation value. $\endgroup$ Commented Jul 27, 2022 at 10:20
  • $\begingroup$ I think I've go it in the meantime. I think answer to my question may be that when it is said that temperature is fixed in the canonical ensemble, that doesn't neccesarily imply thermal equilibrium, but rather it just specifies temperature as a variable which defines the canonical ensemble along with the volume and number of particles. $\endgroup$ Commented Jul 27, 2022 at 16:20

It is unclear what you mean by

the canonical ensemble should be applicable in more general case than thermodynamic limit

The relationship $$\Omega(E) = \sum_{E_S}\Omega_S(E_S)\Omega_R(E-E_S) \tag{1}$$ does not require the thermodynamic limit. However, in order to identify the equation

$$\frac{\ln\Omega_S}{dE_S}=\frac{\ln\Omega_R}{dE_r} \tag{2}$$

as the condition of thermodynamic equilibrium we must make use of the thermodynamic limit because to derive (2) it from (1) we must make the additional assumption

$$ \ln\sum_{E_S}\Omega_S(E_S)\Omega_R(E-E_S) \sim \ln\Big( \Omega_S(E_S)\Omega_R(E-E_S) \Big)_\text{max} \tag{3} $$

which says that the log of the summation can be replaced by the log of the maximum term. We then identify the maximum term by applying Eq (2).

To summarize, the summation in Eq (1) has a maximum term regardless of the thermodynamic limit and we can identify it using Eq (2). In the thermodynamic limit the log of the maximum term is so much bigger than the rest that all other terms in the summation can be ignored. Only in this limit we obtain the familiar relationships of thermodynamics, for example, that the probability of canonical microstate is $p_i=e^{-\beta E_i}/Q$.

If we apply Eq (2) to a small system we will identify the maximum term in Eq (1) and nothing more. Terms of the summation that are comparable in magnitude will be important and the system will be seen to fluctuate over a wide range of energies, unlike thermodynamic systems whose equilibrium energy is essentially constant.


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