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I know that the canonical ensemble can be derived by maximizing the entropy under the constraint that the mean of the energy is fixed. But is there an intuitive reason for that?

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    $\begingroup$ I don't understand. By definition, this is what the canonical ensemble means. $\endgroup$ Commented Dec 9, 2022 at 17:32
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    $\begingroup$ Canonical ensemble is (also?/usually?) defined by means of a heat reservoir, which is free to exchange energy with the system of interest. The two are at the same temperature and energy is exchanged. OP seems to be asking how this can be seen as equivalent to maximizing the entropy subject to constraints. $\endgroup$
    – hft
    Commented Dec 9, 2022 at 19:11
  • $\begingroup$ (Sorry, I should have mentioned, the above comment was directed at @TobiasFünke) $\endgroup$
    – hft
    Commented Dec 9, 2022 at 19:28
  • $\begingroup$ @hft I think Jaynes addresses this in his "Information theory and statistical mechanics I". My first comment was addressing the title, btw. $\endgroup$ Commented Dec 9, 2022 at 19:31
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    $\begingroup$ Does this answer your question? Why do we only use energy in the micro canonical ensemble? (Not the same ensemble, but the same reason) $\endgroup$
    – Roger V.
    Commented Dec 9, 2022 at 19:33

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If there isn't a limit to mean energy, even states with absurdly high energy could be as probable as states with low energy. In absence of other similar limit, maximizing entropy would lead to uniform distribution over all states (all states equally probable). That would also imply infinite temperature, which is not appropriate for real systems in thermodynamic equilibrium at some finite temperature.

A limit to mean energy implies that the maximizing distribution is such that the higher the energy of microstate is, the less probable that state is. This is more appropriate for real systems at finite temperature.

This only means one has to limit mean energy, not fix it. But it is easier to find the maximizing probability distribution for the condition of fixed mean energy, rather than for the condition of energy being limited from above by some maximum limit.

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    $\begingroup$ You can maximize the entropy subject to only the constraint $\sum_i p_i = 1$ and this results in a probability distribution appropriate for the microcanonical ensemble, not canonical. $\endgroup$
    – hft
    Commented Dec 9, 2022 at 19:13
  • $\begingroup$ @hft You're right, so I've tried to fix my answer. $\endgroup$ Commented Dec 9, 2022 at 19:23
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I know that the canonical ensemble can be derived by maximizing the entropy under the constraint that the mean of the energy is fixed. But is there an intuitive reason for that?

A few definitions/preliminaries to start:

  1. In the canonical (or "Gibbs") ensemble, we consider the temperature to be fixed. This is often described as considering the system of interest to be in contact with a "large" "heat bath/reservoir," with which the system of interest can exchange energy.
  2. In the microcanonical ensemble, we consider the energy (and other integrals of the motion, which we ignore here) to be fixed. This is often described as considering the system of interest to be completely isolated.

In the microcanonical ensemble, the distribution function is $$ \rho_{micro}(E) = A\delta(E-E_0)\;, $$ where $E_0$ is the fixed energy of the isolated system. Any microstate that satisfies the condition is equally probable.

The above microcanonical ensemble can be compared with the result of minimizing the entropy subject to only the condition of conservation of probability. Both result in a constant probability for every (accessible) microstate.

The canonical ensemble can be derived by considering the microcanonical ensemble of a system (1) in contact with a heat bath (2). The total distribution is: $$ A\delta(E_1 + E_2 - E_0)d\Gamma_1 d\Gamma_2 $$

The probability of a microstate of system (1) being in state $n$ is: $$ p_n = A\int\delta(E_n + E_2 - E_0)d\Gamma_2\;, $$ which can be shown to equal: $$ w_n = \tilde A e^{-E_n/T}\;, $$ where $T$ is the temperature.

The above probability expression also results from maximizing the entropy subject to two constraints: (1) $\sum_i p_i = 1$; and (2) $\sum_i p_i E_i = U$, where $U$ is some constant, which we will identify as the thermodynamic energy of the macrostate.

So, in the canonical ensemble, we see that states with arbitrary energies can be excited, but their probability is low if the energies are very high. This is because the system is free to exchange energy with the heat bath.

Similarly, when we maximize the entropy subject to the constraint $\sum_i p_i E_i = U$, we are saying that the microstates of any energy are possible, but nevertheless if we are going to characterize a system in equilibrium we know that the macrostate will be characterizable in terms of some thermodynamic energy $U$, which is why we need the constraint.

This is the best intuitive explanation I can give, and I note it is not really "intuitive," but rather just an explanation of how various ways of looking at ensembles are related. Not sure if there is a better "intuitive" explanation, but if so I look forward to reading about it.

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