Does equilibrium imply max. entropy in statistical mechanics?

[JesseM]

Does "equilibrium" imply max. entropy in statistical mechanics?

I've gotten myself confused thinking about the meaning of "equilibrium" in statistical mechanics. I thought I remembered that an isolated system at equilibrium is equally likely to be in any possible microstate, which means there would be some very tiny probability it would be in a low-entropy state far from the maximum entropy possible for the system. But when physicists talk about "non-equilibrium thermodynamics", aren't they talking about systems which are far from maximum entropy? Perhaps the difference is that non-equilibrium thermodynamics is dealing with systems that are not isolated, but are receiving energy from the outside or something? Or maybe my definition of equilibrium is wrong, and an isolated system at equilibrium is not equally likely to be in every possible microstates, but just those microstates corresponding to a maximum-entropy macrostate?

 

[dextercioby]

I've gotten myself confused thinking about the meaning of "equilibrium" in statistical mechanics. I thought I remembered that an isolated system at equilibrium is equally likely to be in any possible microstate, which means there would be some very tiny probability it would be in a low-entropy state far from the maximum entropy possible for the system.

That's absolutely true.The highlighted part could be rigurously proven using the axiomatical formulation of equilibrium SM.

But when physicists talk about "non-equilibrium thermodynamics", aren't they talking about systems which are far from maximum entropy?

What do you mean by far away?How do you measure "far"...?

Perhaps the difference is that non-equilibrium thermodynamics is dealing with systems that are not isolated, but are receiving energy from the outside or something?

Well,of course it's dealing with nonisolated systems,else,if the systems would be isolated,how would the non-equilibrium states appear...?

Or maybe my definition of equilibrium is wrong, and an isolated system at equilibrium is not equally likely to be in every possible microstates, but just those microstates corresponding to a maximum-entropy macrostate?

Yep,it's true that if one imposes the maximum condition on the entropy functional,he gets equally probable microstates.I haven't been taught the other way around.You might search for it in textbooks on SM.

Daniel.

P.S.I hope u know the condition of equilibrium for a statistical system,right...?
Time-independent hamiltonian in the Schrödinger picture of classical and quantum mechanics+Time independent probability density/density operator for the statistical system.

 

[JesseM]

What do you mean by far away?How do you measure "far"...?

It shouldn't really matter--no matter what cutoff you use for how low entropy must be before it's "far from maximum entropy", there should be some nonzero probability the system will have entropy this low, right? (assuming you restrict yourself to entropies which can be found in states somewhere in the system's phase space, of course). Just as an example, if all the molecules in a box filled with gas collected in one corner of the box, then I think most people would call that far from maximum entropy--would you still say the gas was at equilibrium if it spontaneously went into this state?

Well,of course it's dealing with nonisolated systems,else,if the systems would be isolated,how would the non-equilibrium states appear...?

Like I said, I was confused about whether "equilibrium" was a statement about the entropy or about the probability of different states. If you wait some sufficiently huge amount of time, random fluctuations in a system at equilibrium will produce states whose entropy is far lower than the maximum, no? And yet it would still be correct to say the system is at equilibrium at these rare moments, correct?

P.S.I hope u know the condition of equilibrium for a statistical system,right...?
Time-independent hamiltonian in the Schrödinger picture of classical and quantum mechanics+Time independent probability density/density operator for the statistical system.

Thanks, I didn't know that, although I might have learned it at some point when I took a course on this stuff.

 

[dextercioby]

Funny you mention fluctuations.You know that for equilibrium ensembles,they go to zero in the thermodynamical limit for all obervable quantities.So the more or less "sponteous" separations of the molecules (which would violate Boltzmann's H theorem) are extremely improbable.But the probability is nevertheless nonzero.

Daniel.

 

[JesseM]

http://biophysics.asu.edu/main/thorpe/classes/phy541/ensembles.html says that for an isolated system, "equilibrium" means maximum entropy, while the idea that each microstate should be equally probable is called the "principle of ergodicity":

Relation of statistical mechanics to thermodynamics

Statistical mechanics is the microscopic justification of thermodynamics. Using the principle of ergodicity, we ignore dynamics (F=ma) and assume that all dynamically accessible microscopic states are equally likely. There are some macroscopic constraints on the accessible states, such as fixed volume (V), energy (E), or number of particles (N). A distribution of different states subject to a set of constraints is known as a thermodynamic or statistical ensemble. We will calculate sums over these statistical ensembles, known as partition functions. The thermodynamic properties follow from the partition function.

Common thermodynamic ensembles

* Microcanonical has fixed number, volume, and energy. Equilibrium macrostate has maximum entropy.
* Canonical ensemble has fixed number and volume, and exchanges energy with a reservoir at temperature T. Equilibrium macrostate has minimum Helmholtz free energy (F=E-TS).
* Grand canonical ensemble has fixed volume and exchanges energy and particles with a reservoir at temperature T and chemical potential mu. The equilibrium macrostate maximizes the pressure, p=-(E-TS-mu N)/V.
* Other ensembles exist for special situations, such as an external magnetic field.

So would you say that the definitions on this page are wrong, or does "equilibrium macrostate" mean something different from "equilibrium ensemble", or did I misunderstand your earlier answer?

 

[dextercioby]

Nope,i said it.It all depends on the approach.There are 2 ways of doing SM.The traditional way,or the axiomatical way.

I've explained what is an equilibrium macrostate for a statistical system.That and the equal probability for accessible microstates in the microcanonical ensemble induces a maximum entropy.

Daniel.

P.S.The principle of ergodicity (the ergodicity hypothesis initially formulated by Boltzmann) is something else.

 

[dextercioby]

The whole point is that the traditional approach to SM (see for example is sections 6.1 pp.6.3 from [1]) is rather confusing.It's the reason for which theorists go for axiomatical theories.

Daniel.

---------------------------------------------------------
[1]K.Huang,"Statistical Mechanics",Wiley,2-nd edition,1987.

 

[JesseM]

I've explained what is an equilibrium macrostate for a statistical system.

Which section of your earlier posts are you referring to?

That and the equal probability for accessible microstates in the microcanonical ensemble induces a maximum entropy.

Does "accessible" mean all possible microstates the system could be in, or only those microstates compatible with the maximum-entropy macrostate?

http://wug.physics.uiuc.edu/courses/phys113/fall02/Lectures/113_4_fa02/sld014.htm defines the "equilibrium macrostate" as the macrostate whose value of nL has the largest number of microstates. For example, if the number of particles N was 1000, the equilibrium macrostate would have nL=500. So, would the "accessible microstates" which are given equal probability in the microcanonical ensemble mean the set of all microstates with nL=500, or would it mean all possible microstates, with nL taking any value between 0 and 1000?

 

[dextercioby]

The post-scriptum of post #2.

Well,this "accessible microstates" is the nastiest and most unintuitive part of the microcanonical ensemble.Read Huang.

Daniel.

 

[JesseM]

The amazon reviews say that Huang gives a very abstract and mathematically advanced presentation, so given the fact that my knowledge of thermodynamics is very elementary (only took one course in college, and I've forgotten most of it, although some would probably come back to me if I reviewed my old textbook) I'm not too confident I'd be able to follow it. The reviews also say Huang focuses on the "kinetic" approach to thermodynamics rather than the "equilibrium" approach, and I think the equilibrium approach is what I studied. But I'll look for it at the library and see if it's any help.

In the meantime, is it possible to say what "accessible states" would mean in the specific case I mentioned where the only macro-parameter is nL, the number of particles on the left side of the chamber? Or is there no simple answer to this question?

I notice that this page says:

Suppose we have a case where the system has a set of states (called the “accessible” states) that are equiprobable, i.e. Pi = 1/W for some constant W. The remaining (“inaccessible”) states are unoccupied, i.e. they all have Pi = 0. The constant W is called the multiplicity. It necessarily equals the number of accessible states, since (in accordance with the usual definition of “probability”) we want our probabilities to be normalized: <sum>Pi = 1.

In this less-than-general case, the entropy (as defined by equation 3) reduces to
S = logW (18)

...

By way of example, equation 18 is normally applied to microcanonical systems. Microcanonical means that the system is isolated, i.e. constrained to have a definite, constant energy.

So, equation 18 indicates the W microstates they are referring to as "accessible" are only the ones with entropy S, suggesting that states with different entropies would not be considered "accessible". And it is only these states they are calling equiprobable, while all other states are defined to have probability zero--does the last line mean this would be the probability distribution of the microcanonical ensemble, if S is taken as the maximum entropy?

 

[dextercioby]

Yes,i've told u.The probability density has a typical theta-Heaviside dependence and is zero for "unaccessible" states,and 1/#of accessible microstates for these accessible microstates.

Then Landau & Lifschitz.I've also told u about Greiner,Neise & Stocker.

Daniel.

 

[Juan R.]

JesseM

Random questions.

Equilibrium in usual sense of macroscopic bodies with short range forces, without ligadures, and isolation with stability implies a maximum in entropy.

Yes, in non-equilbrium thermodynamics one deals with systems outside of state of maximum of entropy corresponding to eq. One deal with states of non-maximum entropy in the usual sense.

Effectively, an isolated system at equilibrium is equally likely to be in every possible microstates compatible with ligadures at the eq. state. The rest of possible nonequilibrium microstates are not visited at the equilibrium state, but can be instantaneous visited by means of thermal fluctuations.

You can that when the number of particle at left and right is not the same the macroscopic chemical potential at both sides is not the same, violating the macroscopic condition for chemical (or matterial) equilibrium that can find in any standard textbook.

E.g. Chemical Thermodynamics: Basic Theory and Methods, 6th Edition -- by Irving M. Klotz, Robert M. Rosenberg.

dextercioby

let me a small comment, your

“That's absolutely true.The highlighted part could be rigurously proven using the axiomatical formulation of equilibrium SM.”

Would better read like “is derived from equilibrium SM using some ad hoc unproven arguments”


JesseM

If you force the molecules to remain in the corner, e.g. with a wall, it is equilibrium (forced). Other case is not equilbrium and system spontaneously evolutions to right equilibrium filling the whole volume

Equilibrium by def. is the no variation of magnitudes on isolated systems. At the usual level of standard SM, that is compatible with maximum entropy. But the inverse is not always true.

On those “rare” moments the system is outside of equilibrium because the fluctuation put it outside of equilibrium.

dextercioby

I am sorry to say this but regarding fluctuations you are rather wrong (even assuming that the TL had some rigorous mathematical basis).

Fluctuations do not violate Boltzmann's H theorem. This is a very usual misunderstanding of Bolztmann H-theorem that one find in literature.

For a more rigorous treatment of this topic, you can see my previous preprint (search in Google physchem/0109003), my analysis of the situation and references cited therein (specially references 11, 12, and 13), or you can also consult an improved version that will be freely available on my web in some days (www.canonicalscience.com)

You said that “are extremely improbable.But the probability is nevertheless nonzero.” This is totally false; fluctuations are totally common and verifiable experimentally with basic laboratory material. One can compute the size of fluctuation in temperature using Einstein well-known formula. A simple two decimal digits thermometer or a commercial electronic ph-meter are sufficient for detecting fluctuation in Temperature or in Concentrations even in macroscopic matter.

Fluctuations in gas systems of around 100 molecules are still more abrupt.


JesseM

Only a note for you the “principle of ergodicity” (sometime called ergodic hypothesis or other names) is only applicable to equilibrium states. In other approaches, it is a theorem derived just at equilibrium states.

The equilbrium ensembles that you cited are the “microscopic representations” of the state of macroscopic equilibrium. The ensemble corresponding to usual macroscopic equilibrium state of equilibrium thermodynamics that appears in textbooks is the microcanonical one. The macrocanonical is for non-isolated systems interacting with a heat bath and the grand canonical for open systems. The three are for equilibrium situations but only the first is for closed systems and have direct link with usual thermodynamics for closed systems.

The slides are rather confusing and perhaps wrong. The second slide, e.g it appears to say that in equilibrium one can find also microstates with small number of microstates (i.e. nonequilibrium ones). This is not true, since that those states would be not equilibrium states. At equilibrium one find just eq.

For example macrostate 1 and 2 in slide 1 are not equilibrium states (assuming that one can define rigorously eq. for a system of 4 particles). In that case macrostates 1 and 2 are accessible for the system but would be seen like fluctuations outside of the macroscopic equilibrium. I am preparing an article in the topic from a new formulation more general (valid also at nonequilibrium states or mesoscopic regimes), and rather rigorous omitting that class of misunderstanding.

If the number of particles N was 1000, the equilibrium macrostate would have nL=500. Rigorously the equilibrium microstates are those setting nL=500. any other value (e.g. nL=200) would be a spontaneous fluctuation moving the system outside of eq. state.

 

[JesseM]

Yes,i've told u.The probability density has a typical theta-Heaviside dependence and is zero for "unaccessible" states,and 1/#of accessible microstates for these accessible microstates.

Then Landau & Lifschitz.I've also told u about Greiner,Neise & Stocker.

Daniel.

Thanks, I'll look into these. One other quick question--you said the probability density function would be time-independent, which would mean the probability of finding the system in any microstate would be constant over time. So suppose we only have three microstates a, b, and c, and P(a) is the time-independent probability of finding the system in microstate a, P(b) the probability of finding it in b, P(c) of finding it in c. Also say that if you examine it and find it in a, and then after some time t examine it again, the probability it will then be found in b is P(a -> b), and so forth for the other transitions. Given this, would the time-independent condition imply the following must be true:

P(a)*P(a -> a) + P(b)*P(b -> a) + P(c)*P(c -> a) = P(a)

and likewise

P(a)*P(a -> b) + P(b)*P(b -> b) + P(c)*P(c -> b) = P(b)

and

P(a)*P(a -> c) + P(b)*P(b -> c) + P(c)*P(c -> c) = P(c)

?

 

[dextercioby]

"So the more or less "sponteous" separations of the molecules (which would violate Boltzmann's H theorem) are extremely improbable.But the probability is nevertheless nonzero."

Regarding fluctuations you are rather wrong (even assuming that the TL had some mathematical basis).

Fluctuations do not violate Boltzmann's H theorem. This is a very usual misunderstanding of Bolztmann H-theorem that one find in literature.

For a more rigorous treatment of this topic, you can see my previous preprint (search in Google physchem/0109003), my analysis of the situation and references cited therein (specially references 11, 12, and 13), or you can also consult an improved version that will be freely available on my web in some days (www.canonicalscience.com)

You said that “are extremely improbable.But the probability is nevertheless nonzero.” This is totally false; fluctuations are totally common and verifiable experimentally with basic laboratory material. One can compute the size of fluctuation in temperature using Einstein well-known formula. A simple two decimal digits thermometer or a commercial electronic ph-meter are sufficient for detecting fluctuation in Temperature or in Concentrations even in macroscopic matter.

Fluctuations in gas systems of around 100 molecules are still more abrupt.

”

Nope,there's my quote above yours.I didn't say what u claim I've said,so,please,don't put words into my mouth.I might choke...

You've obviously haven't read my post

Here's my quote

"Funny you mention fluctuations.You know that for equilibrium ensembles,they go to zero in the thermodynamical limit for all obervable quantities."

It's correct.

let me a small comment, your

“That's absolutely true.The highlighted part could be rigurously proven using the axiomatical formulation of equilibrium SM.”

Would better read like “is derived from equilibrium SM using some ad hoc unproven arguments”

Great,i think the traditional approach to equilibrium SM is crap. The axiomatical one rules.


Daniel.

 

[dextercioby]

More or less spontaneous separations of molecules in a closed system would definitely lower the entropy and enter conflict with H theorem.Howevernthey are possible solutions to Poincaré's theorem.But,i said,highly improbable.

These separations (which in a way remind of Maxwell's demon) are not fluctuations.I know very well what fluctuations are.

Daniel.

 

[Juan R.]

I think that i quote correctly

Funny you mention fluctuations.You know that for equilibrium ensembles,they go to zero in the thermodynamical limit for all obervable quantities.So the more or less "sponteous" separations of the molecules (which would violate Boltzmann's H theorem) are extremely improbable.But the probability is nevertheless nonzero.

Daniel.

I'm soory but I thought that I had quote to you correctly in my previous post.

1) There is no violation of H theorem (as already explained).

2) In the "thermodynamic limit", that is, constant concentration and large systems ignoring superfittial effects of the order of (1/N), the fluctuations are not extremely improbable. It are very probable, in fact we can measure it without problems even in 1000 L of water. For small systems (e.g. 1000 molecules) fluctuations are still more heavy and easily detectable.

That was my point.

 

[Juan R.]

regarding the "axiomatic" formulation, it does not work better than traditional SM approach

 

[Juan R.]

More or less spontaneous separations of molecules in a closed system would definitely lower the entropy and enter conflict with H theorem.Howevernthey are possible solutions to Poincaré's theorem.But,i said,highly improbable.

These separations (which in a way remind of Maxwell's demon) are not fluctuations.I know very well what fluctuations are.

Daniel.

The lower of entropy by "more or less" spontaneous separation of molecules in closed system does not violate H-theorem. They are not highly improbable, can be computed from ratio of trajectories and measured in laboratory.

Moreover, could you say that is a fluctuation for eviting possible misunderstanding for me part of your words.

 

[dextercioby]

I'm soory but I thought that I had quote to you correctly in my previous post.

1) There is no violation of H theorem (as already explained).

If in an isolated closed system the entropy decreases by the formation of "clusters" of molecules,then H theorem's violated.

2) In the "thermodynamic limit", that is, constant concentration ignoring superfitial (1/N) effects, the fluctuations are not extremely improbable.

I didn't say the reverse.

regarding the "axiomatic" formulation, it does not work better than traditional SM approach

It doesn't,but definitely it's more elegant.Incidentally,as F.Schwabl[1] shows,it's more correct to use Boltzmann's formula instead of Gibbs' for nonequilibirum system.But for equilibrium SM,the two formulas describe the same physics,but on different ways.

Daniel.

----------------------------------------------------------
[1]F.Schwabl:"Statistical Mechanics",Springer Verlag.

 

[Juan R.]

If in an isolated closed system the entropy decreases by the formation of "clusters" of molecules,then H theorem's violated.

Daniel.

On any isolated closed system, any entropy decreasing does not violate the H-theorem. See references quoted above.

Perhaps is a bit more elegant but I do not see that was more correct to use Boltzmann's formula instead of Gibbs' for nonequilibirum system. Both agree, if you refer to entropy formulas.

 

[dextercioby]

Schwabl explains why.

I agree,H theorem is basically a statistical formulation for the second principle of thermodynamics.

Daniel.

 

[Juan R.]

Schwabl explains why.

I agree,H theorem is basically a statistical formulation for the second principle of thermodynamics.

Daniel.

Thanks to Schwabl then.

Let me an interesting point for your knowledge of "fluctuations"

Not only the H-theorem or the second law are not violated. The deviations of the observed behavior from the "predicted" by the H-theorem (or the second law) are perfectly compatible with thermodynamics (or SM) and computed from Einstein formula (in the thermodynamics case) or from any other valid formula in the molecular approach.

 

[JesseM]

One other quick question--you said the probability density function would be time-independent, which would mean the probability of finding the system in any microstate would be constant over time. So suppose we only have three microstates a, b, and c, and P(a) is the time-independent probability of finding the system in microstate a, P(b) the probability of finding it in b, P(c) of finding it in c. Also say that if you examine it and find it in a, and then after some time t examine it again, the probability it will then be found in b is P(a -> b), and so forth for the other transitions. Given this, would the time-independent condition imply the following must be true:

P(a)*P(a -> a) + P(b)*P(b -> a) + P(c)*P(c -> a) = P(a)

and likewise

P(a)*P(a -> b) + P(b)*P(b -> b) + P(c)*P(c -> b) = P(b)

and

P(a)*P(a -> c) + P(b)*P(b -> c) + P(c)*P(c -> c) = P(c)

?

Anyone, is this correct? To find a time-independent probability distribution, do you have to look for the eigenvector with eigenvalue 1 of the matrix of transition probabilities for different microstates?

 

[Juan R.]

I think that are not focusing correctly the molecular mechanism of the system. It is difficult to obtain the correct molecular view from usual SM literature .

For your system, assuming, closed, equilibrium, etc, etc, the only reactions possible are (the = indicates double arrow)

A = B

A = C

B = C

The reaction A = A is not defined and, formally, it is redundant

Take the first reaction, the rate of P(A) in it is computed from

dP(A)/dt = P(B->A)P(B) - P(A->B)P(A).

At equilibrium the rate is zero. Then

P(B->A)P(B) = P(A->B)P(A)

Due to a basic property of matter (some textbooks call it "microscopic reversiblity" which is incorrect. It is best expresed from the canonical omega coefficient related to time symmetry and CPT), one know that

P(B->A) = P(A->B)

therefore

P(B) = P(A)

This is the principle of ergodicity or the equality a priori of probabilities. Note that here I derived like a theorem valid on determined situations.

Repeating for other reactions one finds P(C) = P(A).

By conservation of total probability

P(A) + P(B) + P(C) = 1 = 3 P(A).

therefore P(A) = 1/3.

In general for W accesible microstates j in closed systems at microcanonical eq. P(j) = 1/W.


Details in my "new" paper (will be accesible at my web).

Extension of thermodynamics to new levels of molecular description

 

[JesseM]

For your system, assuming, closed, equilibrium, etc, etc, the only reactions possible are (the = indicates double arrow)

A = B

A = C

B = C

The reaction A = A is not defined and, formally, it is redundant

They aren't meant to be "reactions", just transitions from one microstate to another during a given time interval. (the density matrix gives probabilities for each individual microstate, correct?) Surely there can be a nonzero probability that if you measure a quantum system to be in a certain state at a certain time, and then you measure it again some time later, you will find it to be in the same state?

 

[Juan R.]

The concept of reaction is dependent of the formalism used. An interaction between two electrons is not usually considered a reaction in chemistry, but it is in the formalism of particle physics. In other approaches, the above processes are reactions, a generalized concept of reactions.

In your previously posted slides with atoms in two cavities, the atoms can experience a basic reaction.

Atom left = Atom right

The detailed atomic-molecular mechanics for the above reaction is exactly equal to the atomic-molecular approach for any other more typical reaction in the usual chemical sense: collisions.


In the microcanonical ensemble, the diagonal part of the density matrix, only when the system is at one diagonal state, that is when the quantum state is not represented by a wavefunction.

If your measurement process does not interfer in the system heavily, and after of isoalte the system again it can achieve the same initial state, the reply is yes, it can be again in the same state.