## Main Question or Discussion Point

Hi guys, I have a question regarding the Gibbs paradox. Let's say we take a container A of helium gas and place it next to another container B also containing helium gas at the same temperature and pressure. When we remove the partition (wall) between them, what happens?

If we use classical physics, then the entropy increases by 2Nk ln(2), since we treat every atom as distinguishable. The mixing causes an increase in entropy.

If we use quantum physics, however, the atoms are indistinguishable. So, if we take our partition function and divide it by N!, then after using Stirling's approximation we get a new value for S, where S(container A) + S(container B) = S(complete container after partition removed). According to the math, the entropy does not increase.

But I cannot believe this: before the partition is removed, there is a definite amount of atoms in the left half of the complete container and a definite amount of atoms in the right half of the complete container. After, there is no longer a definite amount of atoms in each half, and so I believe that even in a quantum mechanical description, the entropy should increase.

So, does it increase or does it not? If yes, then where is the math that gives us the corrected value for S incorrect? If no, then where did I go wrong conceptually?

Thanks.

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Mapes
Homework Helper
Gold Member
Perceptive question. If I understand it correctly, you're asking why the entropy doesn't increase when the two gas containers are combined, since, after all, we're removing the constraint of N atoms in each. Normally, when we remove constraints, entropy increases as a system moves to a new equilibrium. But not in this case: the containers are already at equilibrium upon connection, each with entropy S, and there is no possible way for the total entropy to increase above 2S.

Another way of looking at it is to imagine closing a shutter at the midpoint some time after the containers are connected. As you mentioned, the lack of a midpoint wall constraint allows the sides to become unbalanced; after the shutter is closed, one side will typically have slightly <N atoms while the other will have slightly >N atoms. But there's no contradiction; this just means that one side will have entropy slightly <S, the other slightly >S and the total entropy will not exceed 2S.

Thanks Mapes for the response.

Let's say that system A has a definite energy (or, more precisely, an energy between E and E + dE). Same goes for system B. Then the total energy of our combined system, whether there is a partition or not, is definite. Whenever an isolated system with definite energy is in equilibrium, we know, according to the fundamental postulate of statistical mechanics, that it is equally likely to be in its accessible states. This situation is known as a microcanonical ensemble.

Instantly after we remove the partion, the system is not equally likely to be in its accessible states. This is because right after it is removed, we know exactly how many atoms are in each half. However, a state where there are slightly more atoms on the left half is a state accessible to the system. Therefore the system cannot be in equilibrium after the partition is removed. The entropy should increase.

My thinking is that just because A and B have the same temperature and pressure, that does not mean that they will be in equilibrium when the partition is removed.

Mapes
Homework Helper
Gold Member
Instantly after we remove the partion, the system is not equally likely to be in its accessible states. This is because right after it is removed, we know exactly how many atoms are in each half.
I'm with you until this point. How is the microstate with an equal number of atoms on each side not a valid microstate for the given macrostate variables T, P, and 2E?

My thinking is that just because A and B have the same temperature and pressure, that does not mean that they will be in equilibrium when the partition is removed.
But what does equilibrium mean to you, if not an equivalence in pressure, temperature, and energy (assuming negligible other sources of system energy)?

jambaugh
Gold Member
this just means that one side will have entropy slightly <S, the other slightly >S and the total entropy will not exceed 2S.
That is not quite right. The Sterling approximation gives additivity (entropy becomes extensive) but the exact form will not be additive and the entropy of double the particles in double the volume will not quite equal twice the original entropy but be slightly more. The amount is insignificant especially at larger scales but qualititively different from zero.

The information that exactly N particles are in each box is lost and one rather has a very sharp probability distribution. So Sam your intuition is right but remember you applied Sterling's approximation which equates to ignoring this tiny increase in entropy.

What I found more fascinating is when you look at the entropy of two entangled quantum systems. The entropy of the composite can be zero (sharply defined entangled pair) while the entropy of either piece is non-zero due to the entanglement.

Fundamentally entropy is not extensive.

So I make the outragious claim:
The entropy of the universe is zero! The entropy of specific systems is due to their entanglment with the rest of the universe!

Mapes
Homework Helper
Gold Member
Ahh, interesting. I must say my studies of large systems of typical matter have not covered these details. Thanks!

atyy
How about something like this? In quantum mechanics, the distributions are the Bose-Einstein or Fermi-Dirac distributions. These only become the classical distribution divided by N! in the high temp limit. Since we already took a classical limit before removing the partition, we can also ignore quantum effects after removing the partition.

Jambaugh,

Thanks for pointing out the math error for Stirling's approximation. It turns out that if one makes the following better approximation:

ln(N!) = N ln(N) - N + (1/2) ln(2 pi N)

then there is a very slight increase in entropy when the partition is removed for the first time. However, when we place the partition back, the entropy stays the same. Why? It is because we do not know exactly how many atoms are going into the left or right half. So the entropy can not change when we place the partition back, or when we remove it for a second time.

Now when the partition has been put back, the true number of atoms in the left half does not change in time. When the partition is gone, the true number of atoms in the left half will change in time. But regardless, the entropy is the same with or without partition.

So here is how I believe I would summarize my thoughts: if the gases are composed of the same type of atom, then when the partition is removed for the first time, there is a slight (very slight!) entropy increase. From there, the entropy does not change no matter how many times one reinserts or removes the partition.

Is this it? If so, thanks guys for the help.

Q_Goest
Homework Helper
Gold Member
Hi Sam. I deleted my previous post because I missed the finer point you were looking to determine. This isn't really a question about Gibbs paradox, it's a question about Maxwell's demon, isn't it?

Gibbs himself did think of the correction to the entropy that arises from this reasoning. It led him to pioneer the thermodynamics of surfaces (definition fo surface free energy, surface tention etc. etc.).

So, if you have some bulk substance of finite volume, it also has a surface. If you place two bulk volumes containing the same substance at the same temperature and pressure next to each other and remove the separation, the change that happens can be described in terms of removing the surface.

Of course, you can then argue that surfaces are finite too and then go on to compute the contribution of edges, define the so-called line tension etc. etc.

jambaugh
Gold Member
Jambaugh,
...
Now when the partition has been put back, the true number of atoms in the left half does not change in time. When the partition is gone, the true number of atoms in the left half will change in time. But regardless, the entropy is the same with or without partition.

So here is how I believe I would summarize my thoughts: if the gases are composed of the same type of atom, then when the partition is removed for the first time, there is a slight (very slight!) entropy increase. From there, the entropy does not change no matter how many times one reinserts or removes the partition.

Is this it? If so, thanks guys for the help.
That sums it up nicely. Of course we're still approximating here. The order of magnitude error is greater (I think) than this little dabble of entropy increase when we first remove the partition. Its been a while since I played with this.

atyy
I don't really understand why the entropy would or wouldn't increase when you put the partition back. Isn't that how you would prepare the coarse grained system in the first place - so you would get a decrease each time it's inserted, and then an increase after it's removed? Or is the divided contained prepared some other way the first time round? (in the quantum case)

You cannot put the partition back without performing work. You cannot neglect the surface tension if you want to discuss this problem in a consistent way.

atyy
You cannot put the partition back without performing work. You cannot neglect the surface tension if you want to discuss this problem in a consistent way.
Is there any essential difference between classical and quantum ideal gases - I could guess maybe there's no surface term for the classical ideal gas, whereas there will be for quantum ideal gases, since bose or fermi statistics are a kind of "interaction"?

Is there any essential difference between classical and quantum ideal gases - I could guess maybe there's no surface term for the classical ideal gas, whereas there will be for quantum ideal gases, since bose or fermi statistics are a kind of "interaction"?
There is a difference, but in the limit for dilute gasses this difference vanishes, as you woud expect. In that dilute limit, you do have a surface term that has a purely entropic origin.

jambaugh
Gold Member
I'm not sure what the debate is now. You can make the opening between the two half tanks arbitrarily small (and allow it to stay open an arbitrarily long time to compensate for scaling issues). And so you can close the opening with effectively zero work requirements.

The issue is not one of the partitioning mechanism. The initial allowing of the two systems to exchange particles is an irreversible process. You cannot assure that exactly N particles reside in each tank after the fact without exerting some free energy.
You must count particles in each tank and selectively allow the correct number to pass one-way from the tank with a surplus. This selective one-way herding of particles is where the "refrigeration" comes in i.e. the entropy gets lowered back to the original pre-junction value.

Note once the two systems are allowed to join and then are separated you cannot use the original (exact value of N) entropy formula for each. The total entropy will be the same (corrected) combined entropy as when they were joined. Subsequent opening and closing will not further change the entropy.

Ok, but then I would say that after combining the two tanks via the small hole, all you need to do is count the number of molecules in one tank to bring the entropy in the tank back to the lower value (apart from the weak dependence on N within the sqrt(N) fluctuation range).

Then, if you know what the new value for N in one of the tanks is and you again combine it with the other tank, the information gets effectively lost again and the entropy increases again. You need to count yet again to decrease the entropy.

So, counting the number of molecules alone is enough to decrease the entropy. Since the number of molecules typically fluctuates within a range proportional to sqrt(N), you would expect that the entropy decreases by k Log[sqrt(N)] as a result of counting.

This does not contradict the Second Law. You can only count the number of molecules by coupling the system to a computer and then the number of molecules will be stored in the computer memory. One can store the number minus N to save memory space and this number is completely random. So, the entropy of the computer memory increases by exactly the amount by which the entropy of the gas diminishes.

One could still argue that the number stored in the memory should not count toward the entropy, but to be able to continue the repeatred combining and counting, the memory has to be erased and then the irreversible entropy change will occur.

atyy
Subsequent opening and closing will not further change the entropy.
Quantum mechanically the energy eigenstates depend on the boundary conditions, so why wouldn't a change in the boundary conditions matter?

jambaugh
Gold Member
atty,
I was thinking of the semi-classical description. In a fully quantum description yes the geometry of the opening and opening and closing will affect the system. You get a Casimir effect as well I suppose due to as you say altering the available modes.

Count I.
It seems to me that particle number is an extrensic quantity like temperature and pressure in this case. You can e.g. weigh the tanks to determine particle number (in principle since one can take arbitrarily long in making the measurement). So it seems to me you should be able to measure that without affecting the entropies. But something doesn't seem right with my reasoning on this. Hmmm.... I think I'm smelling a demon here.

I'll ponder this further.

So I make the outragious claim:
The entropy of the universe is zero! The entropy of specific systems is due to their entanglment with the rest of the universe!
The fine grained entropy of an ideal gas is always zero. For an unconfined ideal gas, the coarse grained entropy tends to infinity. I don't think you can define the entropy of the universe per se. See Section 2 of the following article:

http://www.mi.ras.ru/~vvkozlov/fulltext/196_e.pdf [Broken]

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