For an isolated system, can the entropy decrease or increase?

In any sizable system, the number of equilibrium states are much, much greater then the number of non-equilibrium states. Since each accessible micro state is equally probably, it is overwhelmingly likely that we will find the system in an equilibrium state.

However, for a closed system, one that does not interact with any external system, the number of micro states is fixed. Therefore the entropy is fixed.

At any given time, the system will be in one of its micro states. Even if the system is in a micro state that does not look like equilibrium from a macroscopic point of view, its entropy will remain the same, since entropy is a property of all the micro states, not just a given micro state.

So, are the number of accessible micro states (and hence entropy) of an isolated system constant?

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Isolated system: Since the matter, energy, and momentum is fixed, the total number of microstates available that satisfy these constraints is fixed/constant. So is the entropy constant? Yes, if the system is in equilibrium. No, if the system is not in equilibrium. What does that mean in terms of microstates?

If the system is in equilibrium, all these microstates are equally probable and the system visits each of these microstates over the course of time (also known as ergodic hypothesis). Therefore the each micostate has equal probability and $S=k ln\Omega$. In such a state there is no more increase in entropy possible.

If the system is in non-equilibrium the system doesn't have equal probability of being in every microstate. In fact if the system is stuck in non-equilibrium ( e.g., a hot part and cold part in the box separated by a thermally insulating wall) it cannot access some microstates at all. Hence the system is restricted to fewer microstates. Technically, there is no unique global thermodyanmic state for the whole system and you cannot define entropy. But you can calculate entropy by summing up entropy of different local equilibrium parts (e.g., entropy of the hold part and cold part separately). Since the total number of microstates you can access is small, entropy is lesser than what it could be if you remove that thermally insulating wall letting the system equilibrate. Thus, when you equilibrate more microstates become accessible. Think of the system spreading in the phase space. Thus entropy increases. Once the system has reached complete thermodynamic equilibrium no more entropy increase in possible. All allowable microstates have been made accessible and equally probable!

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Your first sentence, answers, I think, the question. The rest is a subject for another question. –  yalis Feb 2 '13 at 2:20
+1 I just had a long conversation with a grad student friend of mine, and we came to essentially the same conclusion as this answer. This is a much better answer than mine. –  joshphysics Feb 2 '13 at 2:23
Glad to be of help. –  Sankaran Feb 2 '13 at 16:15

Since entropy is

$$S=k_b \ln \Omega,$$

it is synonymous to a certain phase space volume $\Omega$ which we have to fix somehow. Phase space volume is something like a counter of possible states of the system (in the most simple case, the positions and momenta of all particles). This is (normally) done by setting some macroscopic parameters, like temperature, pressure, number of particles etc.

So, if you have a closed system (i.e. a box where nothing can get in or out), you can e.g. fix volume, energy and the number of particles. You end up with $S(E,V,N)=k_b \ln \Omega(E,V,N)$.

This can be understood as the number of all possible configurations inside the box which have the total energy $E$, the total possible volume $V$ and consist of $N$ particles - including all the stuff in the box on the left side, on the right side, evenly distributed, half of it moving and the other half standing still etc. $S$ is a constant number because $E$, $V$ and $N$ are fixed in a closed system.

The problem is that you can make no statement about the system whatsoever. But - instead of $S(E,V,N)$, you can look at other dependencies of $S$. Let's say you divide up the box into two equally sized parts, a left part and a right part. $N_1$ is the number of particles in the left part, and $N_2$ is the number of particles in the right part (also $N_1+N_2=N$).

Now, you can calculate $S(E,V,N_1,N_2)$ and choose $N_1$ and $N_2$. That way, $S$ can change because you can choose $N_1$ and $N_2$ (particles may go from the left to the right part and vice versa). The important point is that $S$ is much greater for $N_1\approx N_2$ than for any other values.

This means that most of the possible configurations have approximately the same number of particles on the left and on the right side. Because every configuration is equally probable, most of the time, the system will look like this. But not all the time, though - even when $S(E,V,N_1,N_2)=0$, there is exactly one microscopic configuration which fulfils the macroscopic parameters, and it will occur at some time (if the system is ergodic, but that is a different story). In the given example, the entropy will increase and decrease all the time when particles go from the left half to the right half.

But by definition, the percentage of time when the system is in a specific configuration $N_1$,$N_2$ is proportional to $\Omega(E,V,N_1,N_2)=\exp\left( \frac {S(E,V,N_1,N_2)} {k_b}\right)$ (I just thought of this last formula, please correct me if I'm wrong).

In conclusion: The term entropy is meaningless without the parameters one wants to fix/vary.

Funny example: Let's calculate $S(N_{\textrm{people alive on earth}})$ for the whole universe which describes the number of possible configurations of the universe for a given number of living people on earth. $S(0)$ will probably be much bigger than for any other $N_{paoe}$, which means that, most of the time, there will be no people on earth and the entropy will be huge (this is my interpretation of the the heat death of the universe). Then, they might come back and the entropy will be very small.

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ok, but in your example, you are dividing the isolated system into two non-isolated, interacting subsystems (left and right). The entropy of either system can increase or decrease but on the whole we are just sampling different micro states. Entropy should not be dependent on the particular micro state that the system is currently sampling. It only depends on the total number of micro states. In a way that seems to contradict intuition. The universe is closed (?) but it seems that its entropy will increase until heat death is reached. Yet, the number of micro states of the universe is fixed? –  yalis Feb 2 '13 at 1:17
The division is necessary for the entropy to be able to change. If the entropy only depends on constant parameters (e.g. E, V, N), it is constant. This also applies to a closed universe. For other parameters (e.g. $N_1$, rotation speed of mars, number of posts on physics.SE), the entropy changes (see my last example). –  Rafael Reiter Feb 2 '13 at 10:01

Yes the entropy of an isolated system can increase. Here's an example. Take a system consisting of an ideal gas in a thermally insulated box. Let the initial condition of the system be such that the gas occupies the left half of the box with the right side being vacuum. After the initial time, the gas will expand freely to fill the box. Thermal insulation and free expansion guarantee that $Q = 0$ and $W = 0$, so by the first law $\Delta U = 0$. The entropy of the system will be larger in this final state because of this equation (which gives ideal gas entropy) and the fact that $\Delta U = 0$ and $\Delta N$ = 0 while $\Delta V > 0$.

Hope that helps!

Cheers!

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I actually wrote up this example in the initial draft of the question, but deleted it. The thing here is this: Are we not considering the left half of the system as a sub-system of the closed the system. This subsystem interacts with the whole system by expanding. You are "adding volume" to the subsystem and thus increasing the number of micro states. Thus, I would think that a truly isolated system cannot expand. –  yalis Feb 2 '13 at 0:49
Another angle. Say gas particles uniformly distributed at some initial time. It's quite possible, though extremely unlikely, that at at some point in the future, all the particles were on one side of the box. Has the entropy of the system decreased? Not sure how it can, since entropy is associated with the number of micro states, which has not changed. –  yalis Feb 2 '13 at 0:50
Yea you make a very good point that I was actually thinking about as well when I was writing up this example. To put it another way, viewed as a quantum system, the Von-Neumann entropy cannot change since isolated systems undergo unitary time evolution and the Von-Neumann entropy remains invariant under such an evolution of the density matrix of the system. I've heard that a resolution of such issues has to do with quantum "decoherence," but I'm by no means an expert on such things. –  joshphysics Feb 2 '13 at 1:26
I'm almost tempted to start another thread asking about this specific example of the gas in the box and how it can be reconciled with the quantum/microscopic considerations we are both concerned about. I'm not entirely sure such questions have even been resolved to be honest. –  joshphysics Feb 2 '13 at 1:36