Comment on Entropy and Temperature

Fri, 19 Dec 2014 19:29:57 GMT

We don't lose those things.

Suppose that you boil some water in a pot. You take the pot off the stove, and then take a can of beer out of the cooler (which is filled with ice) and put it in the water. The place where you're confusing your friends by putting cans of beer in pots of hot water is by the ocean, so when you read the thermometer that's in the water, it reads 373 K. The can of beer, which was in equilibrium with the ice at a measured 273 K, had some bits of ice stuck to it when you put it in. They melt. Next, you pull out your fancy laser-doppler-shift-based water molecule momentum spread measurer. The result jives with 373 K liquid water. After a short time, you read the thermometer as 360 K (the control pot with no beer reads 371 K). There is no ice left in the pot. You take out the beer, open it, and measure it's temperature to be 293 K and its momentum width to be smaller than that of the boiling water.

What we observed was:

Heat flowed from 373 K water to 273 K beer

The momentum distribution is wider for water at 373 K than at 293 K

Ice placed in 373 K water melts

Our thermometer reads 373 K for boiling water and 273 K for water-ice equilibrium

Now, suppose we do exactly the same thing, but just after putting the beer in the water, Omega tells us the state of every water molecule in the pot, but not the beer. Now we know the temperature of the water is exactly 0 K. We still anticipate the same outcome (perhaps more precisely), and observe the same outcome for all of our measurements, but we describe it differently:

Heat flowed from 0 K water to 273 K beer

The momentum distribution is wider for water at 0 K (or recently at 0 K) than at 293 K

Ice placed in 0 K water melts

Our thermometer reads 373 K for water boiling at 0 K, and 273 K for water-ice equilibrium

So the only difference is in the map, not the territory, and it seems to be only in how we're labeling the map, since we anticipate the same outcome using the same model (assuming you didn't use the specific molecular states in your prediction).

Remember, this isn't my definition. This is the actual definition of temperature used by statistical physicists.

I agree that temperature should be defined so that 1/T = dS/dE . This is the definition that, as far as I can tell, all physicists use. But nearly every result that uses temperature is derived using the assumption that all microstates are equally probable (your second law example being the only exception that I am aware of). In fact, this is often given as a fundamental assumption of statistical mechanics, and I think this is what makes the "glass of water at absolute zero" comment confusing. (Moreover, many physicists, such as plasma physicists, will often say that the temperature is not well-defined unless certain statistical conditions are met, like the energy and momentum distributions having the correct form, or the system being locally in thermal equilibrium with itself.)

I'm having trouble with brevity here, but what I'm getting at is that if you want to show that we can drop the fundamental postulate of statistical mechanics, and still recover the second law of thermodynamics, then I'm, happy to call it a feature rather than a bug. But it seems like bringing in temperature confuses the issue rather than clarifying it.

Discuss