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The Fundamental Postulate says:

In equilibrium, all accessible microstates are equally likely.

Accessible means having same energy.(right?)

Let a container is taken full of gas having number of particles $N_,$ volume $V$ and energy $E\:_;$ the system is isolated.

At equilibrium, the system would be in that macrostate which would have the maximum multiplicity or the largest number of microstates; that would correspond to gas totally dispersed over the whole volume $V\;.$

However, consider the case when the gas gets confined to the left half of the container that is in volume $V/2\;_;$ this macrostate would have a much lesser multiplicity than the former one that is $$\Omega(N,V,E)\gt \Omega\left(N,\frac{V}{2}, E\right) $$

However, as the microstates corresponding to $(N,V/2,E)$ have the same energy $E\;_,$ all the microstates of $(N,V,E)$ and $(N,V/2,E)$ are equally likely in equilibrium.

But, since, the microstate corresponding to $N,V/2,E$ doesn't represent the microstate corresponding to the maximum entropic macrostate, all the microstates $\Omega\left(N,\frac{V}{2}, E\right)$ must be called fluctuation.

But fluctuation, in equilibrium?

At first, I couldn't believe this; how fluctuation happens in equilibrium.

But according to the Fundamental Postulate, all microstates having the same energy are equally likely in equilibrium. This means the microstates of the fluctuation can be exhibited by the system in equilibrium as they are equally likely along the other microstates having the same energy $E.$

How can this happen? How can fluctuation microstates occur in equilibrium, inspite of having the same energy $E\,?$

I thought I was mistaking; only the microstates corresponding to $(N,V,E)$ can occur in equilibrium. But also, I can't deny the fact the the microstates having the same energy $E$ are equally likely.

Also, as written here:

We will first consider an isolated system, typically a gas enclosed in a box, which is thermally insulated. So, any time evolution of the system will be subject to the constraint that the total energy remains constant. Left for a long time, it is believed to be in equilibrium. We further assume that given an isolated system in equilibrium, it is found with equal probability in each of its accessible microstates. This is the postulate of equal a priory probability. ... Now each macrostate comprises of numerous microstates. For example, all the gas confined to only one half of the box, is a macrostate. There is a huge number of ways this can happen, by various arrangements of particles and their momenta. The gas uniformly occupying the whole volume of the box, is another macrostate. And again, there are a huge lot of microstates associated with this macrostate. Now each microstate is equally probable, but we never actually see a gas occupying only one half of its container. Why does that happen? It happens because the number of microstates associated with the gas occupying the whole volume are overwhelmingly large, compared to the microstates associated with the gas occupying only one half of the box.

Notice the words, 'equilibrium', 'each' and the follow-up question 'Why does that happen'?; the author clearly means that at equilibrium, since all the microstates are equally likely, not only the macrostate having the gas uniformly spread over the entire volume has the greatest probability to appear; but also the macrostate corresponding to the gas confined to the left-half of the container can appear in equilibrium; it is only that the former is exhibited mostly.

So, as the Fundamental Postulate permits all the microstates having the same energy $E$ to occur equally likely in equilibrium, it is inferred from that the fluctuation occurs in equilibrium since the microstates $\Omega(N,V/2,E)$ are equally likely in equilibrium :(

So, my question is:

  • Does fluctuation occur really in equilibrium as its microstates are permitted to occur equally likely in the system at equilibrium? Or is it wrong?

What am I actually missing in utilising the Fundamental Postulate and fluctuation? Can anyone please help me clear out my confusion?

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  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$ – David Z Feb 11 '16 at 13:31
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At equilibrium, the system would be in that macrostate which would have the maximum multiplicity or the largest number of microstates; that would correspond to gas totally dispersed over the whole volume $V\;.$

This is wrong, and based on a misunderstanding of terminology. You have microstates, and macrostates. A macrostate assigns a probability to each microstate. Some macrostates are in thermal equilibrium. Those states assign zero to many microstates, and they assign the same non-zero number to all the other microstates. That's it. That fundamental terminology needs to be understood.

However, as the microstates corresponding to $(N,V/2,E)$ have the same energy $E\;_,$ all the microstates of $(N,V,E)$ and $(N,V/2,E)$ are equally likely in equilibrium.

If only the volume $V/2$ were available, then you would assign the probability $1/\Omega\left(N,\frac{V}{2}, E\right)$ to the microstates of energy $E$ where they are all on the left (and zero to all the other microstates). If the full volume $V$ were available, then you would assign the probability $1/\Omega\left(N,V, E\right)$ to all the microstates of energy $E$ (and zero to all the other microstates).

In the latter case, since all of the microstates of energy $E$ are available (including the ones where they are all on the left) you might get one of those where they are all on the left. But the probability of getting one of them is $\Omega\left(N,\frac{V}{2}, E\right)/\Omega\left(N,V, E\right)\ll 1$ because there are $\Omega\left(N,\frac{V}{2}, E\right)$ many of them and each has a probability of $1/\Omega\left(N,V, E\right).$

But, since, the microstate corresponding to $N,V/2,E$

That's not meaningful. There were $\Omega\left(N,\frac{V}{2}, E\right)$ many states of energy $E$ where they were are on the left. You haven't specified which one you are talking about.

$\Omega\left(N,\frac{V}{2}, E\right)$ must be called fluctuation.

No. $\Omega\left(N,\frac{V}{2}, E\right)$ is a number, it's like 5 or 10 except it is much larger.

Does fluctuation occur really in equilibrium as its microstates are permitted to occur equally likely in the system at equilibrium? Or is it wrong?

When you are in the larger microstate you have a non-zero (but super tiny) probability of getting a microstate where they are all on the left side. Just like if you flip four coins then each outcome $HHHH,$ $HHHT,$ $HHTH,$ $HHTT,$ $HTHH,$ $HTHT,$ $HTTH,$ $HTTT,$ $THHH,$ $THHT,$ $THTH,$ $THTT,$ $TTHH,$ $TTHT,$ $TTTH,$ $TTTT$ are equally likely, but there is only a $1/16$ chance of getting a state like $HHHH$ and if you had a thousand coins then there would only be a $2^{-1000}$ chance of getting all heads. That's pretty small. If you had $10^{24}$ coins, you realistically wouldn't see them all come out heads, the probability is nonzero, but $2^{-10^{24}}$ is quite small.

I've read To each microstate there corresponds exactly one macrostate.

That only holds sometimes. It definitely holds when two macrostates have different total energy or different numbers of particles. But the passage you quoted literally is talking about a case where all being on the left is in these two different macrostates. Accessibility isn't determined solely by energy, it is determined by whether it is consistent with the macroscopic state variables such as energy, volume, pressure, temperature.

Have you saw the article I quoted in my question above? It says all the gas confined to only one half of the box, is a macrostate and then says The gas uniformly occupying the whole volume of the box, is another macrostate. And again, there are a huge lot of microstates associated with this macrostate. Now each microstate is equally probable, but we never actually see a gas occupying only one half of its container.- they are two different macrostates aren't they?

Let's say you had a big box of volume $V$ and you placed a thin barrier down the middle and filled the left side (of volume $V/2$) with gas of total energy $E$. Then there would be a macrostate describing that, and there would be a probability of a given microstate (out of many many possible). Now if you very quickly moved the barrier out of the way, so fast that no gas was touching it while you moved it then you could argue that the microstate changed or it didn't (it happened so fast no particle had time to move). And you could argue that the macrostate changed or it didn't. And in both cases the argument would be purely semantic. You know zero particles moved.

Now if the first macrostate was in thermal equilibrium then each of the many microstates was equally likely. And whichever one it was in at that moment, that microstate is one of the vastly many more microstates available to the volume $V$ system. But it is one of them. And there are way way way more microstates in the volume $V$ system. So if you tried to look again later and hoped to find it all on the left side again, the chance would be $\Omega\left(N,\frac{V}{2}, E\right)/\Omega\left(N,V, E\right)\ll 100\%.$ It is a nonzero chance. But you aren't going to see it.

Make sure you understand that example and the math, then reread your passage again. Whether something is "the same" microstate or "the same" macrostate doesn't affect what the probabilities are. It is what it is. When someone says the particles can be anywhere in the large box, they could be anywhere, they could even be positioned so all are on the left side. But its so so so unlikely when you have $10^{24}$ particles. So the chance is small.

Wouldn't the occurrence of microstates corresponding to $E,V/2, N$ be fluctuation from the macrostate having the greatest multiplicity?

If you describe your macrostate as just one number, $V$ (the size of the box) and another number, $E$ the total energy, then if you have all your particles on the left with a particular velocity each, then that collection of positions and velocities is the same for the two macrostates. But for the macrostate with more volume available such a microstate is highly improbable in equilibrium since so few states are like that and all the states are equally likely.

Now, you could if you wanted describe your macrostate with more numbers. You could divide your box into regions and have a density of particles in each region and an average energy per particle in that region. And then you could ask which macrostates of that are equilibrium ones. And for that, it would not be equilibrium to have totally different densities and average energies in different parts of a box. And you could achieve that from a rare transition. But that is a different question (though it might address what you want to know).

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  • $\begingroup$ First of all, thanks for the answer. In the latter case, since all of the microstates of energy $E$ are available (including the ones where they are all on the left)- do you want to say the microstates where the gas is on the left-half is both under the macrostate $(N,V,E)\;?$ However, I've read To each microstate there corresponds exactly one macrostate. You are saying otherwise- the microstates under $N,V/2,E$ can't be under $N,V,E\;.$ $\endgroup$ – user36790 Feb 8 '16 at 4:47
  • $\begingroup$ @user36790 Edited $\endgroup$ – Timaeus Feb 8 '16 at 5:02
  • $\begingroup$ I asked this in chat; David Z answered like you; when I asked about the same, ACuriousMind told both are correct on their own assumptions. My assumption is correct provided the system is in equilibrium. And I'm talking about Equilibrium Statistical Mechanics. $\endgroup$ – user36790 Feb 8 '16 at 5:05
  • $\begingroup$ Check this discussion: chat.stackexchange.com/rooms/71/conversation/… and check this pdf: ls.poly.edu/~jbain/physinfocomp/lectures/… where it is written. $\endgroup$ – user36790 Feb 8 '16 at 5:09
  • $\begingroup$ Have you saw the article I quoted in my question above? It says all the gas confined to only one half of the box, is a macrostate and then says The gas uniformly occupying the whole volume of the box, is another macrostate. And again, there are a huge lot of microstates associated with this macrostate. Now each microstate is equally probable, but we never actually see a gas occupying only one half of its container.- they are two different macrostates aren't they? $\endgroup$ – user36790 Feb 8 '16 at 5:20

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