That is, sticking to classical ideas, I don’t see why that “quantity with dimensions of action” can’t have an arbitrary value.

Yes, it can have any value you want. But its value affects the free energy you compute for the classical ideal gas. If you want to get the “right answer”—the answer we now believe to be close to the answer for an actual gas—you need to pick this quantity with dimensions of action to be Planck’s constant.

But suppose we didn’t have quantum mechanics! Would we be in serious trouble?

Probably not. Since only energy *differences* are measurable, one can argue—as Tobias did earlier in this long conversation, that an additive constant ambiguity in our definition of free energy doesn’t affect any physical predictions. That sounds correct to me.

But there’s more. The ambiguity also affects the *entropy* you compute for the classical ideal gas! This is more surprising, because—at least nowadays—we’re less likely to think of entropy as being defined only up to an additive constant.

But in fact, if you try to measure the entropy of a substance (as opposed to computing it), you’ll see that we typically fix this constant using the Third Law of Thermodynamics:

The entropy of a perfect crystal at absolute zero is exactly equal to zero.

So, we measure the entropy of a gas at a given temperature by first chilling it down to absolute zero, or close, and then keeping careful track of how much heat energy it takes to warm it up to the given temperature. If we can’t afford to do this experiment, entropy is defined only up to a constant.

(And indeed, we can never afford to get all the way down to absolute zero: we have to hope that close is good enough.)

But a classical ideal gas never freezes and forms a crystal! I believe its entropy just keeps dropping more and more as we go closer and closer to absolute zero! More precisely, it goes to : at low temperatures it goes roughly like , up to some fudge factors.

In short, the free energy and energy of a classical ideal gas are ambiguous up to additive constants, and the latter behaves in a rather annoying way. If there were no quantum mechanics we could learn to live with this, but in fact physicists were irritated by this problem until quantum mechanics came along.

I found this article quite helpful:

• S. M. Tan, *Statistical Physics*, Chapter 4: the Classical Ideal Gas.

though they should have come out and said what I just said.

]]>to compute the free energy of a classical ideal gas ‘on the nose’, not just up to a constant, we’re forced to introduce a quantity with dimensions of action.

But in the absence of any desire to go beyond the classical situation, I don’t get this part:

This quantity later turns out to equal Planck’s constant.

That is, sticking to classical ideas, I don’t see why that “quantity with dimensions of action” can’t have an arbitrary value.

]]>This has the effect … of introducing a fundamental length scale, which allows us to make the partition function dimensionless and also choose a specific value for it.

Bruce wrote:

By “it” I think you mean the length scale.

No, I meant the partition function.

It might be helpful to recall the more familiar but completely analogous situation in statistical mechanics, where we need to choose a quantity with dimensions of action to make the partition function dimensionless, and also give the partition function a specific value.

This is important, for example, when we try to compute the free energy of a classical ideal gas, which is

If we only know the partition function up to a constant multiple, we only know the free energy up to an additive constant. And if a quantity isn’t dimensionless, it’s usually considered bad to take its logarithm (though Tobias has argued above that it can be okay).

Earlier I wrote:

]]>When people compute the free energy of a classical ideal gas. they actually want to know the answer ‘on the nose’, not just up to a constant. The reason, perhaps, is that in this subject everyone assumes that the free energy of a box full of vacuum is zero. So a zero of energy has already been fixed.

You can see one version here:

• S. M. Tan,

Statistical Physics, Chapter 4: the Classical Ideal Gas.where the final answer is in equation (4.31). One strange thing about this particular version of the calculation is that it starts as a quantum calculation and then takes the classical limit. You will see that the volume of the box divided by the ‘thermal DeBroglie wavelength’ of the gas molecules shows up in the answer—see eq. (4.36) for an explanation of what I mean.

Thus,

the answer involves Planck’s constant, even in the classical limit!You can also try to compute the free energy of the classical ideal gas purely using classical mechanics. People must have tried this before quantum mechanics was invented. This is closer to what we’re talking about here. In this you naively start by computing

where describes the momentum of particles in a 3-dimensional box describes the position of those particles, and is the kinetic energy of those particles, a quadratic function of .

However, this formula for is ‘wrong’, because is not dimensionless! It has units of momentum times position to the

nth power: that is, action to thenth power. To make it dimensionless we need to divide the measureby something with units of action to the

nth power… and this is where Planck’s constant, or more precisely shows up! In this approach, it enters in a somewhat ad hoc way.In other words, we’re seeing that to compute the free energy of a classical ideal gas ‘on the nose’, not just up to a constant, we’re

forcedto introduce a quantity with dimensions of action. This quantity later turns out to equal Planck’s constant.So, we’re seeing a way that quantum mechanics pushes its nose under the door even when you didn’t invite it, like a camel that you wish would stay outside your tent.

This has the effect … of introducing a fundamental length scale, which allows us to make the partition function dimensionless and also choose a specific value for it.

By “it” I think you mean the length scale. Can you elaborate on how this allows you to choose a specific value for it? I don’t see why, in the pure classical case, it’s insufficient to just choose an arbitrary value for it.

]]>For Markov processes where the equilibrium distribution is some other distribution the Kullback-Leibler divergence of relative to decreases. This is the simplest version of the Second Law that applies to all Markov processes with a finite set of states.

But in fact there’s a more powerful generalization, which is what Blake uses. For any Markov process, if we take two probability distributions and and evolve them forward in time, the Kullback-Leibler divergence of relative to decreases!

This subsumes all the results I mentioned previously. It has the great advantage that we don’t need to know an equilibrium distribution to apply it. This, I claim, is the right way to think about the Second Law for Markov processes.

]]>Just a suggestion for an application!

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