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Stream: deprecated: thermodynamics

Topic: stat mech of the Carnot engine

Owen Lynch (Nov 10 2021 at 23:57):

There's some weird stuff going on with the Carnot Engine if we think about it from a statistical mechanical perspective.

While the gas is thermally connected to a heat bath, the ensemble for its state is a canonical ensemble. Thus, when we close the connection with the heat bath, the ensemble should still be a canonical ensemble, because we haven't observed that it is at a specific energy. But now we are in a very weird state, we know that energy can't flow in or out (which is typically accompanied by a microcanonical ensemble), but we are in a canonical ensemble!!

Owen Lynch (Nov 10 2021 at 23:58):

I now am going to see what happens as we slowly expand the chamber by slowly decreasing the external pressure, I expect what will happen is that we evolve to some weird non-canonical ensemble!!

John Baez (Nov 11 2021 at 00:44):

This is an excellent thing to think about! Someone must have studied the stat mech of the Carnot engine, but I haven't seen it.

It'll be interesting to see what happens when you start with a canonical ensemble and then expand that chamber without heat flowing in or out. It's supposed to be changing "isentropically", right?

John Baez (Nov 11 2021 at 00:46):

I wrote a blog article about the Carnot cycle which I can use as cheat sheet:

• The ideal monatomic gas.

Nathaniel Virgo (Nov 11 2021 at 03:21):

This is a handwavy physics argument, but I'd naively expect that when you take that microcanonical-ensemble-that-looks-like-a-canonical-ensemble and slowly decrease the external pressure, you'll end up in another microcanonical-ensemble-that-that-looks-like-a-canonical-ensemble at a different temperature. My non-rigorous reasoning is that if that didn't happen then in the next step in the Carnot cycle, when you connect the system to another heat bath at a lower temperature, the system would not be in equilibrium with the new heat bath. So there would be a loss of free energy / gain in the universe's entropy as they equilibriate, and the Carnot cycle wouldn't be reversible.

Or maybe there really are losses of that kind and they vanish in the thermodynamic limit somehow? I don't really know.

Owen Lynch (Nov 11 2021 at 16:16):

That makes a lot of sense to me Nathaniel!

Owen Lynch (Nov 11 2021 at 16:20):

I'm having trouble doing the rigorous mathematical argument behind it though. I.e., if I have one particle whose energy is known, I can compute the decrease in that energy as pressure decreases and the chamber expands. But in order for the pressure to be constant when the energy is random I need to have many particles so I can apply the law of large numbers. Then each of these particles is giving up some kinetic energy when they hit the boundary of the chamber, but figuring out exactly how this works is not obvious to me.

Owen Lynch (Nov 11 2021 at 16:20):

Oh, maybe I have a solution.

Owen Lynch (Nov 11 2021 at 16:22):

We can assume that each particle transfers momentum into the barrier continuously.

Owen Lynch (Nov 11 2021 at 16:25):

There's something else going on that's weird here.

Owen Lynch (Nov 11 2021 at 16:32):

While we are connected to the heat bath, the canonical ensemble assumption is borne out by time-averaging, which means that we can simplify by just working with one particle. However, once we are no longer connected to the heat bath, we can't rely on time-averaging, we can only do particle-averaging. I.e., if we have many particles, each of which are independent, then averaging across all the particles gives us estimates according to the canonical distribution. But averaging a single particle over time no longer does that.

Owen Lynch (Nov 11 2021 at 16:33):

Which means that we have a make a LLN argument to say that there is in fact a well-defined pressure

Owen Lynch (Nov 11 2021 at 16:34):

For any finite number of particles, as soon as we close off the heat bath, pressure becomes a random variable, that approaches a constant as $N \to \infty$.

Owen Lynch (Nov 11 2021 at 16:38):

I think actually what's going on here is that the canonical ensemble and the microcanonical ensemble end up being the same as $N \to \infty$, as the energy of the canonical ensemble goes to a constant.

Owen Lynch (Nov 11 2021 at 16:39):

And the distribution energies for an individual particle in the microcanonical ensemble becomes approximately canonical.

Owen Lynch (Nov 11 2021 at 16:41):

So basically, as $N \to \infty$, we can approximate the Carnot cycle by actually saying that we're in a canonical ensemble when we are in contact with the heat bath, and in a microcanonical ensemble with energy equal to the expectation of energy of the canonical ensemble when we are not in contact with the heat bath.

Owen Lynch (Nov 11 2021 at 16:44):

It then makes sense to say that we shift along microcanonical ensembles as we decrease pressure, as we are giving up energy by doing work on the barrier, but that energy could come from any of the particles, so we know nothing about the new ensembles except for their new amount of energy.

Owen Lynch (Nov 11 2021 at 16:46):

So in practice @Nathaniel Virgo, I think there are losses of free energy somehow, but as $N \to \infty$, the losses go to $0$ because there's no difference between the canonical and microcanonical ensembles.

Owen Lynch (Nov 11 2021 at 16:46):

I'm now going to actually do the calculation for the microcanonical ensemble of why entropy stays constant.

Owen Lynch (Nov 11 2021 at 16:47):

I expect it's because the velocities of the particles lose entropy, because we know that they are getting smaller, but the positions of the particles gain entropy, because they could be over a larger space.

Owen Lynch (Nov 12 2021 at 14:34):

OK, I've done the calculations, and it works out. I now want to know why!!

Owen Lynch (Nov 12 2021 at 14:36):

It feels very quantum actually. As we increase the volume of the box, we know less about the positions of the particles, but more about their momentum

Owen Lynch (Nov 12 2021 at 14:37):

Isentropicness feels very much like the uncertainty principle

Owen Lynch (Nov 12 2021 at 14:37):

I.e., whatever we do, the sum of the entropies of the position and momentums remains constant

Owen Lynch (Nov 12 2021 at 14:38):

@John Baez thoughts about this?

Owen Lynch (Nov 12 2021 at 15:03):

Also, I'll put the derivation here for anyone interested.

Suppose that there are $N$ particles in a 1-dimensional box $[0,L]$. The total energy of all the particles is $N U$. Then if we take the microcanonical distributions, we approximately have $N$ independent particles, each of which has position uniformly distributed in $[0,L]$ and momentum normally distributed with mean $0$ and variance $\sigma^2 = 2mU$, where $m$ is the mass of a single particle.

The average velocity of a particle is $v = \sqrt{\frac{2U}{m}}$. This means that each particle impacts the right-hand wall once every $\frac{2L}{v}$ seconds, and transfers a momentum of $2mv$ into the wall. Thus, the force on the wall from all the particles is

$F = N\frac{2mv}{\frac{2L}{v}} = \frac{mv^2}{L} = N \frac{2 U}{L}$

This gives a differential equation for energy as a function of length that we can solve, because the work done on the wall is equal to the decrease in energy.

$d (N U) = F d L = -N\frac{2 U}{L} d L$

Solving this gives

$U = U_0 \frac{L_0^{2}}{L^{2}}$

We can now compute the total entropy as a function of $L$. The entropy of the positions of the particles is simply $N \log(L)$. The entropy of the momenta of the particles is $\frac{N}{2} \log(2 \pi e \sigma^2)$. Recall that $\sigma^2 = 2 m U = 2 m U_0 \frac{L_0^{2}}{L^{2}}$, so this reduces to

$\frac{N}{2} \log(4 \pi e m U_0 L_0^2) - N\log(L)$

Thus, the sum of the entropies of position and momentum are

$N \log(L) + \frac{N}{2} \log(4 \pi e m U_0 L_0^2) - N\log(L) = \frac{N}{2} \log(4 \pi e m U_0 L_0^2)$

which is a constant.

Owen Lynch (Nov 12 2021 at 15:05):

I feel that this doesn't really answer why the entropy is a constant, it just sort of happened to work out that way.

Owen Lynch (Nov 12 2021 at 15:07):

Of course, the units don't work out, but that's because I haven't bothered to add in various constants for entropy that properly speaking should be in there

Owen Lynch (Nov 12 2021 at 15:14):

I'm going to refresh my memory on the quantum analogue of this system to see if I can dredge any insight from that.

Fabrizio Genovese (Nov 12 2021 at 15:20):

are you sure that $v = \sqrt{\frac{U}{2m}}$?

Fabrizio Genovese (Nov 12 2021 at 15:21):

It feels like the 2 should be on the numerator, if I understand what you are doing

Fabrizio Genovese (Nov 12 2021 at 15:21):

(This seems also what you did when you derived $F$)

Owen Lynch (Nov 12 2021 at 15:22):

No, I'm not sure, let me check

Owen Lynch (Nov 12 2021 at 15:23):

Yes, you are right

Owen Lynch (Nov 12 2021 at 15:23):

Yeah, I did the correct thing later on though

Nathaniel Virgo (Nov 12 2021 at 15:54):

It feels like this should be related to the general results about the equivalence between ensembles in the thermodynamic limit. (See this paper by Touchette, for example.) I wonder if one could start from those kinds of result and show that the change in entropy always has to be zero, for any system where they hold? (This is just a guess.)

Owen Lynch (Nov 12 2021 at 18:29):

Somehow I don't think so, i.e. I believe that this is relevant to the other issue of the transition between stages of the carnot cycle, but I don't think that has much bearing on the change in entropy within an isentropic stage.

Owen Lynch (Nov 12 2021 at 18:30):

I think what I really need is a statistical-mechanical understanding of $dQ$, the classical heat transfer.

Owen Lynch (Nov 12 2021 at 18:30):

(which is a 1-form that is not the differential of a 0-form)

Owen Lynch (Nov 12 2021 at 18:31):

(I.e., $Q$ is not a state function)