Derivation of Statistical Mechanics

[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.