A microcanonical ensemble is one that represents an isolated system with fixed number of particles, volume, and energy. In other words, it's an $(N,V,E)$ ensemble.
If the energy is fixed, it will result in a fixed number of microstates. The total number of microstates has no variation with respect to energy, since the energy is constant. If $\Omega(N, V, E)$ is a constant, so is entropy, $k_B \ln \Omega$.
What does it mean to say that a microcanonical ensemble is one that maximizes entropy? With respect to what variable?
Edit: Responding to the first answer, i.e., by @flaudemus, with two well-known arguments in support of the claim, both of which are completely wrong, and do not work (and this is not about that answer; that's how a microcanonical ensemble is taught in the majority of classes in my opinion).
Wrong argument 1: The total number of microstates $\Omega(E)$ is a function of energy $E$ and also varies with energy.
Reasoning why it's wrong: $S$ varies with $E$ in general, not for a specific $(N,V,E)$ ensemble under study. Consider two microcanonical ensembles, $(N,V,E_1)$, and $(N,V,E_2)$. Those are two completely different ensembles, studied independently. They have no relation to each other. It is meaningless to say that if the second ensemble has higher entropy compared to the first one, entropy increased with energy. Well it increased by changing the ensemble you are studying, so you essentially changed the problem. Alternatively, if you change the energy within the same problem, it is no longer a microcanonical ensemble, since the energy is no longer fixed.
Wrong argument 2: In order to see, how entropy is maximized in such a system, consider splitting your system of total energy $E$ into two parts ${\cal S}$ and ${\cal S}'$, and let the energy of ${\cal S}$ be $\epsilon$, and that of ${\cal S}'$ be $E-\epsilon$. Then the total entropy is
$$ S(\epsilon) = S_{\cal S}(\epsilon) + S_{{\cal S}'}(E-\epsilon).$$
Maximum entropy means
$$ \frac{dS(\epsilon)}{d\epsilon} = \frac{dS_{\cal S}(\epsilon)}{d\epsilon} + \frac{dS_{{\cal S}'}(E-\epsilon)}{d\epsilon} = \frac{dS_{\cal S}(\epsilon)}{d\epsilon} – \frac{dS_{{\cal S}'}(E-\epsilon)}{d(E-\epsilon)} = 0.$$
According to our definition of temperature, this is equivalent to
$$ \frac{1}{T_{\cal S}} = \frac{1}{T_{{\cal S}'}},$$
i.e., maximizing entropy means that the two subsystems have the same temperature. In other words, the temperatures of all thermodynamic subsystems of your NVE-system are the same.
Reasoning why it's wrong: All this derivation establishes is the maximization of entropy of a subsystem of that of the microcanonical ensemble, not the system itself. The system as a whole, was always, and will always will be at a fixed entropy $S$, corresponding to fixed total energy $E$. (edit: This is not entirely correct. The maximization is of the conditional entropy as explained in edit 3 below).
Edit 2: Would it be correct to say the following:
- A microcanonical ensemble does not maximize total entropy of the system, since total energy and total entropy is fixed.
- However, if we consider a large subsystem of the total system, that subsystem would have its entropy maximized with respect to its energy, which is fluctuating and not a consant.
- By large subsystem, I mean the subsystem size could be one-half, one-third, but not one-hundredth, or one-thousandth of that of the total system. If that subsystem is too small, it becomes a canoncial ensemble, with the rest of the system becoming the heat bath, and then entropy might not maximize. (Would this be correct reasoning?)
Edit 3: I have read edits by @flaudemus, as well as some texts. I agree that we're dealing with two kinds of total entropies: $S(N, V, E)$ and $S_{c,\epsilon}(N, V, E, \epsilon)$, where (edit: correct expression in Edit 5):
$$\require{enclose} \enclose{horizontalstrike}{S(N, V, E) = \int_0^E S_{c,\epsilon}(N,V,E,\epsilon)\,d\epsilon}$$
$$S_{c,\epsilon}(N, V, E, \epsilon) = S_{\mathcal{S}}(\epsilon) + S_{\mathcal{S'}}(E-\epsilon)$$
However, this is still a serious problem. No stat-mech text clarifies that it's not the first kind of total entropy, but the second kind (let's call it conditional entropy: total entropy of the system at total energy $E$, given that part of the system has energy $\epsilon$).
Not only that, it raises other issues, like why can't we condition the total entropy on something else, like part of the system having n out of N particles, or part of the system having v out of V volume.
But then if $S(N, V, E) \neq S_{c,\epsilon}(N, V, E,\epsilon)$, what about the following (at equilibrium):
$$S_{c,\epsilon}(N, V, E,\epsilon) \stackrel{?}{=} S_{c,n}(N, V, E, n)$$
$$S_{c,\epsilon}(N, V, E,\epsilon) \stackrel{?}{=} S_{c,v}(N, V, E, v)$$
If they are not equal, there is nothing special about $S_{c,\epsilon}(N, V, E,\epsilon)$ compared to other conditional entropies.
Very unfortunate how such critical details are glossed over in 99% of classes and texts.
Finally, that still means total (unconditional) entropy of a microcanonical ensemble stays a constant, but total conditional entropy (conditioned on susbystem energy) is the one that gets maximized.
Edit 4: Some texts I've looked at:
- Schroeder, An Introduction to Thermal Physics
- Kardar, Statistical Physics of Particles
- Sethna, Statistical Mechanics: Entropy, Order Parameters, and Complexity
I only went through short sections, mainly looking for discussion on microcanonical ensemble. As far as I can tell, all of them fall short in clarifying this issue. However, I was able to come up with a relationship between the two total entropies while looking at Fig. 3.1 in Schroeder.
Edit 5: I'm sorry my expression in Edit 3 is incorrect, as pointed out by @flaudemus, but with $d\epsilon$ not $dE$. The correct integral relation applies to microstates, not entropy:
$$\Omega(N, V, E) = \int_0^E \Omega_{c,\epsilon}(N,V,E,\epsilon)\,d\epsilon$$
This gives the following relation between entropies:
$$S(N, V, E) = k_B \ln \int_0^E \exp \frac{S_{c,\epsilon}(N,V,E,\epsilon)}{k_B}\,d\epsilon$$
Best Answer
There are different entropies involved: thermodynamic(function of $E,V,N$) and information-theoretical (function of probabilities, also sometimes called the Gibbs entropy). These are not the same thing.
The general rule is: the thermodynamic entropy for some macroscopic state X is the maximum possible value of information entropy functional for all probability distributions that obey given constraints implied by state X.
The concept of microcanonical ensemble can be understood as "the best" probability description of a system for these constraints: known and fixed energy, volume and number of particles: $E,V,N$.
Information entropy is a function of set of probabilities for all possible states. If the space of states is discrete and has $N$ distinct states, the information entropy functional can be defined as
$$ I[{p_k}] = - \sum_{k=1}^N p_k \ln p_k. $$
(this function was studied by Claude Shannon and Edwin Jaynes who gave some arguments why this function and not some other).
For a system with known and fixed $E,V,N$, all the states in the sum above must be possible, in other words, all states $k$ have the same energy, volume, number of particles. Other states are not included in the sum.
Different probability assignments for those states, while all compatible with constraints implied by $E,V,N$ , may give different values for $I$. The distribution $p_k^* = 1/N$ gives, for large $N$, the maximum possible value for $I$. This maximum possible value then gives thermodynamic entropy of the system in macroscopic state $E,V,N$, according to formula
$$ S(E,V,N) = k_B I [ \{ p_k^*\} ] = k_B \ln N. $$