> 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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