Distinguishability and Statistical Mechanics
I have actually thought about this issue when I was teaching undergraduate statistical mechanics, even though no one has asked me anything about it. Indeed, this was what I meant when I said that, teaching that class allowed me to think about things in a more relaxed fashion, e.g. I had plenty of time to contemplate whatever I wanted for as long as I wanted without the pressure of studying for the next midterm.
Anyway, what I realized is that, if particles are actually distinguishable but you did not know this fact, then you are just inadvertently simplifying a problem. If you did not design any experiments that could detect the differences, then the simplification you have inadvertently made is valid for you experiments. If you knew this ahead of time, it would not change anything, except for the fact that you have knowingly made an approximation.
Let’s consider two thought experiments.
First, let’s imagine humans are technologically backward. What human knew as Nitrogen 14 (14N) actually have two distinct colors, green and red, but of course you didn’t know that. I bring a bottle of green 14N and a bottle of red 14N. I put these two different 14N’s into one big box. I then ask you to calculate their entropy, internal energy, pressure, etc., and then directly measure these properties.
You calculate the thermodynamic properties, and then you measure these properties. Ha! They match perfectly, except for entropy, because you cannot directly measure entropy. They are 14N’s, so you know how they behave.
Then, I turn on the magic color-detector, and all of the suddenly you see green particles and red particles in the process of mixing. The mixing has not yet reached equilibrium, and the entropy is not yet maximized. However, every other thermodynamic property you have calculated is still valid, because every other thermodynamic property is independent of the color.
You have simplified the original problem. Instead of calculating a system of half green 14N and half red 14N, you have considered a problem with just 14N. Since everything you have calculated is independent of the color, your calculations are valid and accurate.
With this extra information, however, now you can calculate their mixing kinetics.
Then, let’s consider another experiment. I am a lair. I have one bottle of 14N and one bottle of 15N, but I tell you that both are 14N’s. You tested for one bottle using mass spectrometry, and confirmed that it is indeed 14N. I put them into a big box and tell you to do the same thing. Since some of the thermodynamic property, e.g. the pressure, depends on the size of the particles, you will find that what your measurement is in fact a little different from your calculation. You will know that I have lied, that the other bottle must not be 14N. Thus, the two populations must be distinguishable. Your simplification was invalid.
Another example is to consider the simple Ising model. In addition to spins up and down, let's we assign two different colors to each spin: green and red. However, if the Ising program is not set up to distinguish the color, i.e. there is no energy associated with either color, then at absolute zero temperature, the system will be all spin up or all spin down, but their color will be randomly distributed. As far as the "Ising universe" is concerned, their colors are indistinguishable, thus the system has the lowest possible energy and zero entropy. Only when different values of energy is assigned to each color, will the Ising model gain the ability to "realize" its "error," and go to a single color configuration at absolute zero temperature.
Your knowledge of the distinguishability of the system will not change how the system behaves. You will know the particles are distinguishable if and only if you perform experiments that can tell them apart.