The entropy $S$ of a system is defined as $$S = k\ln \Omega.$$ What precisely is $\Omega$? It refers to "the number of microstates" of the system, but is this the number of all accessible microstates or just the number of microstates corresponding to the systems current macrostate? Or is it something else that eludes me?
Statistical-Mechanics – Ambiguity in the Definition of Entropy
definitionentropystatistical mechanics
Related Solutions
A macrostate is a set of microstates. Some microstates are thermal, others are not.
Without the assumption of being in thermal equilibrium you can't assume anything since any possible microstate is possible. And lots of possibilities macrostates could be picked.
Usually you want to group your macrostates according to a state variables such as pressure, volume, total energy or something like that.
And when you break your 30 microstates into three groups: A, B, and C you can ask yourself if each group is classified according to a state variable such as pressure, volume, total energy or something like that.
And even if it is, then all you might know is the state variable, and even that maybe not precisely. For instance the volume isn't precisely known since the exact locations of all the many parts is not known.
Now even when the microstate is a particular microstate, and that microstate is assigned specifically to a particular macrostate, that doesn't tell you how that macrostate assigns probabilities to the microstates in it
And you can assume that each microstate in the macrostate is equally likely. But that is just an assumption. If energy is conserved, then the dynamics will always keep the total system at a configuration with that fixed initial energy, so it doesn't change from any state to just any state.
Doesn't thermal equilibrium mean the macrostate having the greatest multiplicity?
It means so much more. Firstly, it requires that a macrostate is a probability distribution on microstates. Secondly, the macrostate is specified by some (macro) state variables. Thirdly, the particular distribution specified by the hypothesis requires that the space of all microstates be partitioned (partitioned by different values of the state variables) and each partition has an equal probability assigned to every microstate in that part of the partition. I always imagine different floors of a mansion, where your variables constrain you to a different floor and each room in a floor is equally likely.
Now, thermal equilibrium doesn't mean having the greatest multiplicity. You could have $N$ particles of some gas at a certain pressure $P_0$ and volume $V$ and that could be one macrostate, and there might be macrostates with a larger multiplicity with the same $N$ and $V$ and larger $P_+\gt P_0$ but there isn't enough energy for those $N$ particles to have that pressure, there just isn't enough kinetic energy to spread around to get them to the $P_+\gt P_0$ macrostate variable.
If you want to go to the mansion example. Imagine that you have two mansions and one person can go up a floor if (and only if) the other person goes down a floor. When they are both by some stairs then one can go up and the other can go down. But if that places one of them into a floor with trillions more rooms available than the other one had, then they are way way more likely to stay in the configuration with the one stuck in the floor with way more rooms.
So energy can be exchanged between the two people, but the additional energy spends most of it's time with one of them having more energy if they can exchange it. Eventually they could get to a level where one gains as many rooms as the other one loses. And that joint collection roughly is the macrostate of the combined system.
And when that happens we say they are thermal equilibrium. And the thing they have in common, temperature, is how many additional rooms they gain per bit of energy. Maybe one has stairs that are longer, so going up/down one flight for it is going up/down 5 flights for the other. But maybe the rate at which the floor have more rooms changes with floors at a different rate. There could still be a $\textrm dS/\textrm dE$ in common.
Can thermal equilibrium have fluctuation?
The macrostate could, in principle change from thermal equilibrium and more to have to two subsystems be at different temperature instead of the same temperature. But that would require bouncing around until you are by some stairs going down instead of towards any of the many more options on staying on the same floor, and then continuing to do the improbable floor after floor until the temperature of the two subsystems are very out of line.
And even if it happened, it could just go back to equilibrium. The idea is that for a large enough system, the time to wait for such a thing to happen is just really really long.
The density $\rho$ would count the number of microstates within the volume $d^{3N}p\,d^{3N}q$ that satisfies the energy constraint $E<\mathcal{H}<E+\Delta$. So you'd actually have:
\begin{align} \Gamma(E) &= \int_{E<\mathcal{H}<E+\Delta} d^{3N}p\,d^{3N}q \\ &= \int \rho\,d^{3N}p\,d^{3N}q \end{align}
Best Answer
Entropy is a property of a macrostate, not a system. So $\Omega$ is the number of microstates that correspond to the macrostate in question.
Putting aside quantization, it might appear that there are an infinite number of microstates, and thus the entropy is infinite, but for any level of resolution, the number is finite. And changing the level of resolution simply multiplies the number of microstates by a constant amount. Since it is almost always the change in entropy, not the absolute entropy, that is considered, and we're taking the log of $\Omega$, it actually doesn't matter if the definition of S is ambiguous up to a constant multiplicative factor, as that will cancel out when we take dS. So with a little hand waving (aka "normalization"), we can ignore the apparent infinity of entropy.