Derivation of Statistical Mechanics

[PeterDonis]

It's true for the delta state, but not true for most other fine grained states.

In classical physics, isn't any fine-grained state a delta function in phase space (i.e., a single phase space point)?

 

[Nullstein]

It's true for the delta state, but not true for most other fine grained states.

First of all, the delta state is the physically most relevant case, since arguably, a true physical system is always in such a pure state, independent of what we know about it. Thus, the argument that the thermal state arises as a coarse graining of the true physical state in the way you imagine, is already falsified.

But moreover, you are just making a claim without proof here. How do you know that the situation is better for other states? Gibbs' H-theorem certainly won't help here. Let's just recapitulate what Gibbs' H-theorem says and what is doesn't say:

1. Gibbs' H-theorem says the following: The entropy of the coarse grained version of a time evolved state is greater than the entropy of the time evolved state itself. This is the content of equation (51.20), Gibbs' H-theorem.
2. Gibbs' H-theorem does not claim that the entropy of the coarse grained state at later time is greater than the entropy of the coarsed grained state at earlier time. That's just not the content of the theorem.

If you apply Gibbs' H-theorem once, it only allows you to say that the coarse grained state has higher entropy than the fine grained state. You can't use the theorem to conclude that the entropy of the coarse grained version of the fine grained state increases over time. And one application of the theorem is not enough either, because it doesn't state that the coarse grained state has maximum entropy.

In order to use Gibbs' H-theorem to argue for an approach of equilibrium, you must argue the theorem consecutively, i.e. there must be a time evolution, then a coarse graining, then a time evolution, then a coarse graining and so on. Only in this situation, you get an increase of entropy at every step and you will get a convergence towards the highest entropy state, which is a thermal state.

Anyway, I have checked also the book by Jancel, "Foundations of Classical and Quantum Statistical Mechanics", Sec. "Discussion of the generalised H-theorem" starting at page 169. He starts with "It can be seen from the foregoing developments that this theorem is obtained without any special assumptions, except the fundamental assumption of statistical mechanics, which is necessary for the statistical description of macroscopic phenomena and which cannot contradict the reversibility of the laws of mechanics." Likewise, the section ends with: "... the evolution takes place according to the deterministic laws of mechanics, the irreversibility arising from the gross nature of our observations. Thus, it is not possible to have any kind of contradiction between the statistical conclusions of the theory and the reversible behaviour of mechanical systems." I think it contradicts your view.

This is just prose. The book contains the exact same math as Tolman, so the above arguments apply here as well. A single level of coarse graining is not enough to conclude an increase of entropy to $S_{max}$. The only thing that you can conlcude is that at all times, the entropy of the coarse grained state is higher than the entropy of the fine grained state. No relation between the entropies of coarse grained states at different times is implied.

 

[Demystifier]

In classical physics, isn't any fine-grained state a delta function in phase space (i.e., a single phase space point)?

Not in classical statistical physics. That's why we have phase-space distributions. Fine grained distribution is defined on a continuous phase space, while in the coarse grained distribution the phase space is divided into finite cells.

 

[Demystifier]

@Nullstein, is there a peer-reviewed paper, or a book, that explicitly agrees with you in saying that Gibbs H-theorem does not work for the reasons you are explaining here? If not, do you consider the possibility to write a paper by yourself?

 

[Nullstein]

Not in classical statistical physics. That's why we have phase-space distributions. Fine grained distribution is defined on a continuous phase space, while in the coarse grained distribution the phase space is divided into finite cells.

Also in statistical physics, the fine grained distribution is a delta distribution. The idea of statistical physics is to derive macroscopic properties from the microscopic details. The microscopic theory doesn't suddenly change just because we want to compute macroscopic observables. The ontology of classical mechanics as crystal clear: Every particle has one position and one position only (similarly for momentum). This must be the starting point for every proper derivation of statistical physics.

Also, since we are talking about applications to Bohmian mechanics here: The ability to resolve the measurement problem crucially depends on the fact that Bohmian particles have a single true trajectory. If you don't have this, then you can't explain why only one branch of the wave function is realized.

@Nullstein, is there a peer-reviewed paper, or a book, that explicitly agrees with you in saying that Gibbs H-theorem does not work for the reasons you are explaining here? If not, do you consider the possibility to write a paper by yourself?

The very book you quoted agrees with me:

And you can also read that the statement that I have made: You can only conclude that the coarse grained entropy at later time is higher than the coarse grained entropy at the initial time ($\bar H(t)\leq\bar H(0)$):

You cannot assert that $\bar H(t_3)\leq\bar H(t_2)$ (for $t_3>t_2>t_1=0$), because you would need to assume that $P(t_2) = \rho(t_2)$ for that:

This can't be true in this situation, because

If you want to use the theorem to assert that $\bar H(t_3)\leq\bar H(t_2)$, you must thus perform another coarse graining step in order to obtain $P(t_2) = \rho(t_2)$, so the assumptions of the theorem are fulfilled again.

Therefore, unless there is intermediate coarse graining, the theorem says nothing about convergence to equilibrium.

In fact, the book you cited actually supports my view of how statistical mechanics emerges, namely from the ergodic hypothesis:

 

[Demystifier]

@Nullstein I think that your claims against Gibbs H-theorem are much stronger than those quotes. In other words, I don't think that those quotes confirm those of your claims with which I disagree. But anyway, I think that at this point we can agree to disagree and conclude this discussion.

 

[Nullstein]

@Nullstein I think that your claims against Gibbs H-theorem are much stronger than those quotes. In other words, I don't think that those quotes confirm those of your claims with which I disagree. But anyway, I think that at this point we can agree to disagree and conclude this discussion.

I don't think my claims are stronger. I just agree with the author of that book that Gibbs' H-theorem can't be used to derive the statistical equilibrium distribution. He makes this clear in multiple places throughout the chapter, I only quoted a fraction of them. The author wrote the exact same things as I did. Also my post #56 is an easily understandable proof with a concrete, physically relevant example that shows that it can't work. I don't think there is room to agree to disagree, because it is not possible for someone who has internalized the arguments, to disagree. There is no vagueness in this argument, all of this is exact mathematics. Also, I am convinced that this is really standard knowledge and I'm not making any claim beyond that which is taught in a standard course on statistical physics.

 

[PeterDonis]

Coarse graining in Gibbs' H-theorem means projecting the fine grained state on a lattice version of phase space with fixed sized phase cells. Again, please read the book (specifically eq. (51.3)).

I have gone through the book in some detail now. I see where I was misunderstanding the "coarse graining" terminology as it is used in the book (I am used to seeing that term used differently): as the book uses the term, the size of the phase space cells is fixed, yes.

However, I don't see anything in the book that justifies this claim of yours:

Gibbs' H-theorem really evolves every initial distribution into equilibrium.

The only distributions that are even considered in the book with regard to the Gibbs H-theorem are distributions in which the fine grained state $\rho$ is a uniform distribution in each finite coarse-grained phase space cell (the value of $\rho$ can vary from cell to cell, but it is uniform over each individual cell). The basic argument given in the book is that if we start with an ensemble that is uniformly distributed over a single phase space cell (which is what justifies the initial condition $\rho = P$), the time evolution of that ensemble will not be uniformly distributed over a single phase space cell (so we will have $\rho \neq P$ at the later time, which is what leads to the conclusion that the Gibbs $H$ decreases).

 

[jbergman]

I have gone through the book in some detail now. I see where I was misunderstanding the "coarse graining" terminology as it is used in the book (I am used to seeing that term used differently): as the book uses the term, the size of the phase space cells is fixed, yes.

However, I don't see anything in the book that justifies this claim of yours:

The only distributions that are even considered in the book with regard to the Gibbs H-theorem are distributions in which the fine grained state $\rho$ is a uniform distribution in each finite coarse-grained phase space cell (the value of $\rho$ can vary from cell to cell, but it is uniform over each individual cell). The basic argument given in the book is that if we start with an ensemble that is uniformly distributed over a single phase space cell (which is what justifies the initial condition $\rho = P$), the time evolution of that ensemble will not be uniformly distributed over a single phase space cell (so we will have $\rho \neq P$ at the later time, which is what leads to the conclusion that the Gibbs $H$ decreases).

I don't think that's true. Can you cite where in the book that the distribution is uniform in each cell?

What I read was is that $P$ is equal to the average of $\rho$ over a cell.

However as $\rho$ evolves that original volume changes shape and no longer matches our coarse-graining grid so that at a later time $P$ is no longer equal to the average of $\rho$ over a coarse grained cell.

 

[PeterDonis]

Can you cite where in the book that the distribution is uniform in each cell?

Page 170, in the paragraph above equation 51.14:

"Since the fine-grained density $\rho$ will have a constant value inside each such region..."

Note carefully that I did not say that all possible distributions have $\rho$ constant in each phase space cell (which is what "each such region" in the above quote refers to). I only said that, in the proof of the Gibbs H-theorem, Tolman only considers such distributions. And that is what he is doing in the quote above.

 

[Nullstein]

You have a misunderstanding here. It's not that Gibbs' H-theorem is inapplicable. The idea of this route to thermal equilibrium is that you artificially coarse grain your state in order to apply Gibbs' H-theorem (without any physical justification). This is the unphysical step that I have been pointing out all along. The actual state of the system is always a delta state, but you have to introduce some artifical coarse graining in order to apply the theorem. And it doesn't even suffice to coarse grain once in the beginning. You have to do it over and over, because a single application of Gibbs' H-theorem doesn't suffice to reach $S_\text{max}$. Gibbs' H-theorem only shows that $S(t_1) > S(t_0)$ and $S(t_2) > S(t_0)$, but from this it doesn't follow that $S(t_2) > S(t_1)$ (from $2 > 0$ and $1 > 0$, it doesn't follow that $1 > 2$). In order to show that, you must again perform a coarse graining at $t_1$, so you can apply Gibbs' H-theorem. Unless there is a physical source for these coarse graining steps (e.g. coupling to some heat bath), this route to thermal equilibrium is therefore excluded.

 

[PeterDonis]

You have a misunderstanding here. It's not that Gibbs' H-theorem is inapplicable.

What specifically are you responding to here? Whatever it is, it might help to quote it.

 

[PeterDonis]

Gibbs' H-theorem only shows that $S(t_1) > S(t_0)$

No, that's not quite what it shows. What it shows is that $S(t_1) > S(t_0)$ if the system was not in the thermal equilibrium state (whose $H$ value corresponds to $S_\text{max}$) at $t_0$. If the system was in the thermal equilibrium state at $t_0$, then we cannot conclude that $S(t_1) > S(t_0)$.

 

[PeterDonis]

The actual state of the system is always a delta state

In classical physics, yes. (Note, however, that this is not true in quantum physics. But here I take it that we're only discussing classical physics.)

but you have to introduce some artifical coarse graining in order to apply the theorem.

The coarse-graining is not "artificial". It's a reflection of the fact that we don't know which particular point in phase space represents the actual microscopic state of the system. That's the whole reason for using ensembles in the first place. If we knew with infinite precision the exact point in phase space that represented the actual microscopic state of the system, we would have no need for thermodynamics or statistical mechanics at all.

 

[PeterDonis]

a single application of Gibbs' H-theorem doesn't suffice to reach $S_\text{max}$

You are misunderstanding the argument. The argument is not that we apply the H-theorem to explain how $S_\text{max}$ is "reached". The argument is that the H-theorem tell us that, unless the system is in a state where the entropy is already $S_\text{max}$, the entropy will increase with time. We don't need to know specifically how the system "reaches" $S_\text{max}$.

 

[Nullstein]

No, that's not quite what it shows. What it shows is that $S(t_1) > S(t_0)$ if the system was not in the thermal equilibrium state (whose $H$ value corresponds to $S_\text{max}$) at $t_0$. If the system was in the thermal equilibrium state at $t_0$, then we cannot conclude that $S(t_1) > S(t_0)$.

Yes, I don't mention such trivial details for the sake of brevity.

In classical physics, yes. (Note, however, that this is not true in quantum physics. But here I take it that we're only discussing classical physics.)

Gibbs' H-theorem is classical only, so I don't see why you bring up quantum theory.

The coarse-graining is not "artificial". It's a reflection of the fact that we don't know which particular point in phase space represents the actual microscopic state of the system.

The laws of physics don't care about what we know or don't know. The challenge is to derive the ensemble distribution from the from the true microscopic details. If a derivation can't achieve that, it has failed (as also acknowledged in the book).

That's the whole reason for using ensembles in the first place. If we knew with infinite precision the exact point in phase space that represented the actual microscopic state of the system, we would have no need for thermodynamics or statistical mechanics at all.

Sure, but we need to derive the ensemble from the microscopic details. We can't just make up assumptions that are in contradiction to the axioms. And if we assume that the microscopic distribution isn't a delta, we are in contradiction with the axioms. Once we have derived the ensemble, we can work with that. That's the whole point.

You are misunderstanding the argument. The argument is not that we apply the H-theorem to explain how $S_\text{max}$ is "reached". The argument is that the H-theorem tell us that, unless the system is in a state where the entropy is already $S_\text{max}$, the entropy will increase with time.

No, that's not what the theorem says. Read again what is written in the book. I have also explained it multiple times already. The theorem says that $S(t_1) > S(t_0)$. It doesn't say that $S(t_2) > S(t_1)$. You can't concluce this without the additional assumption of an intermediate coarse graining step at $t_1$.

We don't need to know specifically how the system "reaches" $S_\text{max}$.

Nobody claimed that. You need to know though that the entropy keeps increasing and the theorem doesn't imply this without intermediate coarse graining.

 

[Nullstein]

Let me try to explain it one last time. The statement of Gibbs' H-theorem is:

If $\rho(t_0) = P(t_0)$, where $\rho$ is the fine grained distribution and $P$ is the coarse grained distribution, then for $t_1 > t_0$, we have $S(t_1) \geq S(t_0)$ and $\rho(t_1) \neq P(t_1)$. ($S$ is computed with respect to $P$).

Now if we have $t_2>t_1$, the theorem explicitely says that $\rho(t_1) \neq P(t_1)$, so the theorem proves that we can't apply it again and therefore we are not allowed to conclude that $S(t_2) \geq S(t_1)$.

So we don't know whether the entropy of $P$ keeps increasing after $t_1$. In order to conlcude that, we need to coarse grain again by setting $\rho(t_1):=P(t_1)$ and apply the theorem again. The relevant quotations were given in post #65. If you disagree, you should be able to point to the exact sentence you disagree with.

 

[PeterDonis]

if we assume that the microscopic distribution isn't a delta

Nowhere is that assumed. The ensemble is not the same as the actual delta-function microscopic state, nor is it claimed to be. As I said, the ensemble is used because we don't know the actual delta-function microscopic state. If we did, we would have no need for thermodynamics or statistical mechanics at all.

Read again what is written in the book.

I have, multiple times. It doesn't say what you claim it says.

Let me try to explain it one last time.

You have already said the same thing multiple times. It didn't convince me then and it doesn't convince me now.

I think we're done.

 

[jedishrfu]

This thread has run its course and is now deadlocked in a let's agree to disagree state. Consequently it's a good time to conclude, call it a draw, move on and close this thread.

I'd like to thank all who have posted here with the hope that some insight will come out of this discussion to resolve the remaining issues of disagreement.