Entropy and Temperature

This is a good article making a valuable point. But this —

Temperature is sometimes taught as, "a measure of the average kinetic energy of the particles," because for an ideal gas U/N = (3/2) kBT. This is wrong, for the same reason that the ideal gas entropy isn't the definition of entropy.

— is a confusing way to speak. There is such a thing as "the average kinetic energy of the particles", and one measure of this thing is called "temperature" in some contexts. There is nothing wrong with this as long as you are clear about what context you are in.

If you fall into the sun, your atoms will be strewn far and wide, and it won't be because of something "in the mind". There is a long and perfectly valid convention of calling the relevant feature of the sun its "temperature".

if you know the states of all the molecules in a glass of hot water, it is cold in a genuinely thermodynamic sense: you can take electricity out of it and leave behind an ice cube.

I am not sure this is true as stated. An omniscient Maxwell demon that would only allow hot molecules out runs into a number of problems, and an experimentally constructed Maxwell's demon works by converting coherent light (low entropy) into incoherent (high entropy).

[Spoiler alert: I can't find any 'spoiler' mode for comments, so I'm just going to give the answers here, after a break, so collapse the comment if you don't want to see that]

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For the entropy (in natural units), I get

%20=%20N%20\ln%20N%20-%20\frac{E}{\epsilon}%20\ln%20\frac{E}{\epsilon}%20-%20\left(%20N%20-%20\frac{E}{\epsilon}%20\right)%20\ln%20\left(%20N%20-%20\frac{E}{\epsilon}%20\right))

and for the energy, I get

%20=%20\frac{\epsilon%20N}{e%5E{\epsilon%20/%20T}%20-%201})

Is this right? (upon reflection and upon consulting graphs, it seems right to me, but I don't trust my intuition for statistical mechanics)

This is related to the physics of computation: ultimate physical computers coincides with temperatures approaching need to 0 K - (reversible computing, Landauer Principle) . Heat/entropy is computational stupidity.

Incidentally, this also explains the fermi paradox: post singularity civilizations migrate out away from their hot stars into the cold interstellar spaces, becoming dark matter. (which however, does not imply that all cold dark matter is intelligent)

Temperature is then defined as the thermodynamic quantity that is the shared by systems in equilibrium.

I think I've figured out what's bothering me about this. If we think of temperature in terms of our uncertainty about where the system is in phase space, rather than how large a region of phase space fits the macroscopic state, then we gain a little in using the second law, but give up a lot everywhere else. Unless I am mistaken, we lose the following:

• Heat flows from hot to cold
• Momentum distribution can be predicted from temperature
• Phase changes can be predicted from temperature
• The reading on a thermometer can be predicted from temperature

I'm sure there are others. I realize that if we know the full microscopic state of a system, then we don't need to use temperature for these things, but then we wouldn't need to use temperature at all.

if you know the states of all the molecules in a glass of hot water, it is cold in a genuinely thermodynamic sense: you can take electricity out of it and leave behind an ice cube.

If you're able to do this, I don't see why you'd be using temperature at all, unless you want to talk about how hot the water is to begin with (as you did), in which case you're referring to the temperature that the water would be if we had no microscopic information.

Probability is in the mind.

I am not sure what this means. In what sense is probability in the mind, but energy isn't? Or if energy is in the mind, as well, what physical characteristic is not and why?

Thanks for this. I am definitely going to use the Gibbs paradox (page 3 of the Jaynes paper) to nerd-snipe my physics-literate friends.

I'l follow suit with the previous spoiler warning.

SPOILER ALERT .

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I took a bit different approach from the others that have solved this, or maybe you'd just say I quit early once I thought I'd shown the thing I thought you were trying to show:

If we write entropy in terms of the number of particles, N and the fraction of them that are excited: α ≡ E/(Nε) , and take the derivative with respect to α, we get:

dS/dα = N log [(1-α)/α]

Or if that N is bothering you (since temperature is usually an intensive property), we can just write:

T = 1/(dS/dE) = E / log[(1-α)/α]

This will give us zero temperature for all excited or no excited particles (which makes sense, because you know exactly where you are in phase space), and it blows up at half particles are excited. This means that there is no reservoir hot enough to get from α < .5 to α = .5 .

so temperature is in the mind

I am not quite sure in which way this statement is useful.

"..and for an encore goes on to prove that black is white and gets himself killed on the next zebra crossing." -- Douglas Adams

Probability is in the mind. Entropy is a function of probabilities, so entropy is in the mind. Temperature is a derivative of entropy, so temperature is in the mind.

If I plunge my hand into boiling water, I will get scalded. Will I still get scalded if I know the position and momentum of every particle involved? If so, what causes it? If not, where does this stop -- is everything in the mind?

ETA: I should have reread the discussion first, because there has been a substantial amount about this very question. However, I'm not sure it has come to a conclusion that resolves the question. Also, no-one has taken on Shalizi's conundrum that someone cited, that Bayesian reasoners should see entropy decrease with time.

ETA2: One response would be that the detailed knowledge allows one to predict the same injury, by predicting the detailed properties of all of the particles through time. But I find this unsatisfying, because the easiest way to get that prediction is to start by throwing away almost all of the information you started with. Find the temperature you would attribute to this microstate if you didn't know the microstate, and make the prediction you would have made knowing only the temperature. This will almost invariably give the right answer: it will be right as often as you would actually get scalded from boiling water. If the simplest way to make some objective prediction is in terms of a supposedly subjective quantity that is not being experienced by any actual subject, just how subjective is that quantity?

ETA 3: Consider the configuration space of the whole system, a manifold in some gigantic number of dimensions, somewhat more than Avogadro's number. I am guessing that with respect to any sensible measure on that manifold, almost all of it is in states whose temporal evolution almost exactly satisfies equipartition. Equipartition gives you an objective definition of temperature.

Equipartition is what you will see virtually all of the time, when you do not know the microstate. But it is also what you will see when you do know the microstate, for virtually every microstate. The only way to come upon a non-equipartitioned pan of hot water is by specifically preparing it in such a state.

But what is a sensible measure, if we do not deliberately define it in such a way as to make the above true? You could invent a measure that gave most of its mass to the non-equipartitioned cases. But to do that you would have to already know that you wanted to do that, in order to devote the mass to them. I think there's a connection here to the matter of (alleged) "Bayesian brittleness", and to the game of "follow the improbability". But I do not quite see a resolution of this point yet. Entsophy/Joseph Wilson seemed to be working towards this on his blog last year, until he had some sort of meltdown and vanished from the net. I had intended to ask him how he would define measures on continuous manifolds (he had up to then only considered the discrete case and promised he would get to continuous ones), but I never did.

There is a peculiar consequence of this, pointed out by Cosma Shalizi. Suppose we have a deterministic physical system S, and we observe this system carefully over time. We are steadily gaining information about its microstates, and therefore by this definition, its entropy should be decreasing.

You might say, "the system isn't closed, because it is being observed." But consider the system "S plus the observer." Saying that entropy is nondecreasing over time seems to require that the observer is in doubt about its own microstates. What does that mean?