According to Bogolubov postulate (various texts name it differently) in Non-equilibrium thermodynamics, the number of needed parameters to describe our system is decreasing with time, and finally at infinity, once our system reaches its equilibrium state, we will need only the usual three thermodynamic parameters $T,V,N$ to define the state.

Ok, this seems to be very logical, however, how does entropy fit in there? Generally speaking we can say that entropy describes the disorder in our system, and the higher the entropy, the less order we have, and more disorder means less symmetry in the system. (Or is this statement wrong?)
And when a system has high symmetry, we need less parameters to describe it, which uncovers a contradiction than I can't understand:

A system has reached equilibrium state, when the system's entropy reaches its maximum (we are talking here about isolated systems only), but maximum entropy means maximum disorder, maximum disorder means the need for a lot of parameters to describe it due to luck of order. But that is the opposite of what the postulate above states.

So where is the mistake in this statement? (I asked my professor and he answered that defining entropy in a non-equilibrium state is, for now, an open question but I think that the answer should lie somewhere else...) (in case there is really some relation between Entropy and Symmetry, please provide some math how this is accomplished)

  • $\begingroup$ Maybe one could establish a link between the notion of disorder of a system and a measure of how far from being finitely generated its automorphism group is : the more ordered the system, the more discrete its automorphism group gets. $\endgroup$ – Sylvain JULIEN Nov 8 '17 at 23:10

Just to add another look to the answer of Paul. As you know any thermodynamic potential has its own variables. As the function of its variables it achieves a minimum in equilibrium. Entropy is one of such potentials that can be used on the equal basis with the others. The only difference is that it achieves minimum rather than maximum. It appeared historically this way, that entropy has been defined as it is. Put minus in front of it, and it will be as all the others.

OK, its own variables are the internal energy, E, the volume, V, and the number of particles N. The first of these, the internal energy, is very inconvenient in use. One typically cannot measure this parameter independently. It is not often evident, how to calculate it into measurable parameters to compare with experiment. For this reason the entropy is rarely used.

In fact the philosophy behind is that you first define the set of variables that is adequate to the problem, and then work with the corresponding potential. The set (V, T, N) corresponds to the free energy. Use that!

Second. You write:

[...] and more disorder means less symmetry in the system. (Or is this statement wrong?)

You are definitely wrong. Below I just give you a counter-example that will clarify the situation:

a)The symmetry group of a crystalline solid is a discrete group with (i) translations over a discrete set of vectors, rotation over discrete set of angles, and reflection in a discrete set of planes. Eventually the symmetry might be even more complex, but this does not influence my example.

b) Introducing a disorder one may transform this crystal into amorphous solid. This has the symmetry group with continuous translations and rotations and the infinite set of the mirror planes. This group is continuous, the so-called, Euclidean motion group. Any symmetry group of any crystal is its subgroup. Thus, by increasing the disorder we increased the symmetry.

c) By further increasing the disorder (say, by increasing the temperature) we may bring our solid into the liquid state. Here the symmetry is still higher, since it allows all movements with a constant volume. The Euclidean motion group is the subgroup of this one.

One may also give a number of such examples within the solid state.

Though it is typical that during phase transitions a high-temperature phase has a higher symmetry, there are also opposite examples. They are too specific and I will not give them here. There are also examples for transitions between solid states with the symmetry groups that have no group-subgroup relations between one-another. In this case one cannot decide which symmetry is higher.

To summarize all this one should say that there is no solid rule about the relation between disorder and symmetry, though increasing disorder is often followed by the symmetry increase.

  • $\begingroup$ although I marked it as an answer, I think your example of minus sign for entropy is not correct because this sign will not change at all the fact that entropy max/min= High disorder whatever the sign is. $\endgroup$ – TMS Nov 7 '12 at 16:06
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    $\begingroup$ Let me stress this again. There is NO necessary connection between entropy and order except in the case of a totally isolated system. $\endgroup$ – Paul J. Gans Nov 7 '12 at 22:16
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    $\begingroup$ Two comments: (i) entropy is always a thermodynamic potential, whereas the other potentials, Gibbs, free energy, Enthalpy... are only thermodynamic potential under precise constraints. E.g. Gibbs energy G is a thermodynamic potential only for constant temperature, constant pressure, and constant composition. Otherwise G is not a thermodynamic potential and cannot be used to predict evolution or stability. $\endgroup$ – juanrga Nov 8 '12 at 11:57
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    $\begingroup$ And (ii) this general character of entropy is the reason which entropy is the central concept, and used almost exclusively, in modern formulations for the study of evolution, stability, and fluctuations: modern thermodynamics, extended thermodynamics... You cannot develop the general theory of thermodynamic stability using G, H, or A and precisely the modern formalism of extended thermodynamics is based in extending the functional expression for the entropy and then using this extended entropy for the study of phenomena. $\endgroup$ – juanrga Nov 8 '12 at 12:05
  • $\begingroup$ First to answer to TMS. If you have the entropy, S, that has an absolute minimum, the value -S has an absolute maximum in the same point. Just try to plot them. $\endgroup$ – Alexei Boulbitch Nov 12 '12 at 9:19

First, using $(P,T,V)$ at equilibrium is both redundant and insufficient. It is redundant because you are using two conjugate variables $P$ and $V$ and you only need one of them. It is insufficient, because you are lacking composition $N$ and maybe other variables (it depends of the specific system) to describe the equilibrium state.

Second, whereas entropy is a well-defined quantity (both at equilibrium and outside of it) disorder is a subjective term. Moreover, we know situations of maximum entropy which are not of the highest symmetry and vice versa.

Effectively, entropy is a maximum at an equilibrium state, but equilibrium means that a lot of parameters are unneeded to describe it. This is because during the approach to equilibrium those parameters become redundant. Imagine a fluid in non-equilibrium regime. You need to know lots of parameters such as density, pressure, temperature at each point $r$ of the fluid: $\rho(r)$, $P(r)$, and $T(r)$. But at equilibrium, the condition of equilibrium says that temperature, pressure and density have to be the same at each point $r$ of the fluid. Instead a list of hundred of values you can describe the whole fluid at equilibrium with only three values $\rho$, $P$, and $T$. You could still give a list of $\rho(r)$, $P(r)$, and $T(r)$, but this is completely redundant.

The basic idea is that the more far from equilibrium, the more complex the system is, and you need more parameters to describe the system.

  • $\begingroup$ Regarding P,V,N you right, it's my mistake and I corrected it. Also as I described it's clear for me why there is few parameters when we approach equilibrium, also I used the term "order" just as a link to the "symmetry", and that was the main point of my question, also can you please mention that situations when max. entropy doesn't implements highest symmetry? $\endgroup$ – TMS Nov 5 '12 at 23:48
  • $\begingroup$ An immiscible mixture of two components has the highest symmetry and becomes homogeneous when is not at equilibrium. When equilibrium is reached the mixture is more 'ordered' (less symmetrical) with two well-separated phases. $\endgroup$ – juanrga Nov 6 '12 at 11:49

You have two different questions here. The first, about extra parameters away from equilibrium is easy to answer. In a totally non-equilibrium system you'd need to specify the locations and momentum of every single particle in the system. That's $6N$ for an $N$-particle system. At equilibrium that number is reduced to three.

But the real confusion comes in the second question. The entropy of a system does NOT go to a maximum at equilibrium other than in the case that the system is totally isolated from the surroundings. In other cases the direction of equilibrium is determined by one or another of the system free energies.

For example a system in a container of fixed volume that lets heat in and out (we assume the system sits in a constant temperature bath at a temperature of $T$) it is the Helmholtz free energy which chemists call $A$ that determines equilibrium. The entropy in such a system is NOT a maximum.

The confusion arises because the TOTAL entropy of system and surroundings does go to a maximum IF and ONLY IF we assume that the surroundings are surrounded (sorry about that) by rigid impenetrable walls. The problem here is that the entropy change in the surroundings is very hard to measure. This is why the free energies were invented. Using them one does not have to attempt to compute the entropy change in the surroundings.

The entropy is associated with "order" (whatever that means) again only in the totally isolated case. In all others the association is far more complex.

  • $\begingroup$ Obviously i was talking about isolated systems, I clarified that now, also my question about linking entropy to symmetry not parameters count... $\endgroup$ – TMS Nov 6 '12 at 8:32
  • $\begingroup$ There are many issues with this answer. First, the 6N variables are not enough for a full quantum description. Second, the entropy of a non-isolated system is also maximal at eq. Precisely the canonical distribution for a system in thermal contact with surrounds is found by maximizing the entropy of the system (Lagrange multipliers method). Third, any chemists textbook explains that Helmholtz free energy is a thermodynamic potential only for T,V,N constant. For T,P,N constant chemists use Gibbs free energy... $\endgroup$ – juanrga Nov 6 '12 at 11:59
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    $\begingroup$ I am clearly talking about a classical description and $6N$ is correct. The entropy of a non-isolated system does NOT go to a maximum and is not found the way you say if I understand you correctly. And I'll add to your comment that the entropy is a thermodynamic potential only for E,V,N constant systems. Other than these I agree with what you wrote. $\endgroup$ – Paul J. Gans Jan 4 '13 at 1:54

more disorder means less symmetry in the system

This isn't true. Here is a straightforward counterexample:

Consider a room full of a gas, with no external potentials like gravity, so that the state of the gas at a given time is completely determined by its density, pressure, and temperature.

Now, let's say that initially, the gas was all bunched up in one corner of the room. This setup is out of equilibrium, with a low entropy, but the density does not have either rotational or translational symmetry.

The gas eventually drifts into equilibrium, wherein the entropy is at a maximum, and the density and pressure of the gas in the room are constant throughout: this is rotationally and translationally symmetric.

So, entropy has increased, and so has the number of symmetries of the density of the gas.

  • $\begingroup$ Even so I see what you talking about, but I believe your specific example is wrong, because its quit possible to have density of such symmetrical properties even in non-equilibrium! nothing prevents that. $\endgroup$ – TMS Jun 28 '13 at 14:55
  • $\begingroup$ Of course, but in those cases, when entropy increases the symmetries stay the same, and they do not have less symmetry in equilibrium. The physics is simpler than in Alexei Boulbitch's examples but the lesson is the same. $\endgroup$ – Chay Paterson Jun 28 '13 at 15:02

This may not be a great answer. I was considering this exact question (Is high symmetry the same as high entropy?) when listening to Brian Greene, "The Fabric of the Cosmos". This is the answer I plan to give my high school physics class: Our intuitive "feel" for symmetry leads us astray. Looking at liquid and solid molecules of water, we may see the beautiful crystalline structure as high symmetry. This is where our intuition leads us astray. Consider the classic system of two separated gasses. If I try to "fold" the system over the divider, no particles "match up". When the divider is lifted and the gasses mix, there will be many more "matches" upon folding. The previous answers also tell us that there is also more to comparing symmetry and entropy, but this gives me a mental image that basically shows how high entropy is comparable to high symmetry.

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    $\begingroup$ DO NOT referr people to these kinds of popular-science books. $\endgroup$ – Abhimanyu Pallavi Sudhir Jun 28 '13 at 13:08
  • $\begingroup$ The abstract concepts of entropy and symmetry are both measures of how many ways you can change some aspects of a system while leaving other aspects unchanged. However, great confusion is easy to generate by talking about either/both without carefully defining the system/properties under discussion. So, why should thermodynamic entropy of a gas have any simple relation to density symmetry of same gas? By switching to different properties, we effectively are talking about two different systems as far as entropy/symmetry is concerned. $\endgroup$ – Ron Burk Apr 29 '14 at 3:44

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