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.
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?
Assuming that you plunge your hand into the water at a random point in time, yes you will get scalded with probability ~1. This means that the water is “hot” in the same sense that the lottery is “fair” even if you know what the winning numbers will be—if you don’t use that winning knowledge and instead just pick a series of random numbers, as you
would if you didn’t know the winning numbers, then of course you will still lose. I suppose if you are willing to call such a lottery “fair”, then by that same criterion, the water is hot. However, if you use this criterion, I suspect a large number of people would disagree with you on what exactly it means for a lottery to be “fair”. If, on the other hand, you would call a lottery in which you know the winning numbers “unfair”, you should be equally willing to call water about which you know everything “cold”.
This means that the water is “hot” in the same sense that the lottery is “fair”
Well, if I know the winning numbers but Alice doesn’t, the lottery is “fair” for Alice. If I know everything about that cup of water, but Alice doesn’t, is the water at zero Kelvin for me but still hot for Alice?
Temperature in the thermodynamic sense (which is the same as the information-theoretic sense if you have only ordinary macroscopic information) is the same as average energy per molecule, which has a lot to do with phase changes for the obvious reason.
In exotic cases where the information-theoretic and thermodynamic temperatures diverge, thermodynamic temperature still tells you about phase changes but information-theoretic temperature doesn’t. (The thermodynamic temperature is still useful in these cases; I hope no one is claiming otherwise.)
You probably know this, but average energy per molecule is not temperature at low temperatures. Quantum kicks in and that definition fails. dS/dE never lets you down.
Actually, no! There have been kinda-parallel discussions of entropy, information, probability, etc., here and in the Open Thread, and I hadn’t been paying much attention to which one this was.
Anyway, same post or no, it’s as good a place as any to point someone to for a clarification of what notion of temperature I had in mind.
If, on the other hand, you would call a lottery in which you know the winning numbers “unfair”, you should be equally willing to call water about which you know everything “cold”.
In the lottery, there is something I can do with foreknowledge of the numbers: bet on them. And with perfect knowledge of the microstate I can play Maxwell’s demon to separate hot from cold. But still, I can predict from the microstate all of the phenomena of thermodynamics, and assign temperatures to all microstates that are close to equipartition (which I am guessing to be almost all of them). These temperatures will be the same as the temperatures assigned by someone ignorant of the microstate. This assignation of temperature is independent of the observer’s knowledge of the microstate.
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.
Assuming that you plunge your hand into the water at a random point in time, yes you will get scalded with probability ~1. This means that the water is “hot” in the same sense that the lottery is “fair” even if you know what the winning numbers will be—if you don’t use that winning knowledge and instead just pick a series of random numbers, as you would if you didn’t know the winning numbers, then of course you will still lose. I suppose if you are willing to call such a lottery “fair”, then by that same criterion, the water is hot. However, if you use this criterion, I suspect a large number of people would disagree with you on what exactly it means for a lottery to be “fair”. If, on the other hand, you would call a lottery in which you know the winning numbers “unfair”, you should be equally willing to call water about which you know everything “cold”.
Well, if I know the winning numbers but Alice doesn’t, the lottery is “fair” for Alice. If I know everything about that cup of water, but Alice doesn’t, is the water at zero Kelvin for me but still hot for Alice?
And will we both predict the same result when someone puts their hand in it?
Probably yes, but then I will have to say things like “Be careful about dipping your finger into that zero-Kelvin block of ice, it will scald you” X-)
It won’t be ice. Ice has a regular crystal structure, and if you know the microstate you know that the water molecules aren’t in that structure.
So then temperature has nothing to do with phase changes?
Temperature in the thermodynamic sense (which is the same as the information-theoretic sense if you have only ordinary macroscopic information) is the same as average energy per molecule, which has a lot to do with phase changes for the obvious reason.
In exotic cases where the information-theoretic and thermodynamic temperatures diverge, thermodynamic temperature still tells you about phase changes but information-theoretic temperature doesn’t. (The thermodynamic temperature is still useful in these cases; I hope no one is claiming otherwise.)
You probably know this, but average energy per molecule is not temperature at low temperatures. Quantum kicks in and that definition fails. dS/dE never lets you down.
Whoops! Thanks for the correction.
Aha, thanks. Is information-theoretic temperature observer-specific?
In the sense I have in mind, yes.
I am somewhat amused that you linked to the same post on which we are currently commenting. Was that intentional?
Actually, no! There have been kinda-parallel discussions of entropy, information, probability, etc., here and in the Open Thread, and I hadn’t been paying much attention to which one this was.
Anyway, same post or no, it’s as good a place as any to point someone to for a clarification of what notion of temperature I had in mind.
In the lottery, there is something I can do with foreknowledge of the numbers: bet on them. And with perfect knowledge of the microstate I can play Maxwell’s demon to separate hot from cold. But still, I can predict from the microstate all of the phenomena of thermodynamics, and assign temperatures to all microstates that are close to equipartition (which I am guessing to be almost all of them). These temperatures will be the same as the temperatures assigned by someone ignorant of the microstate. This assignation of temperature is independent of the observer’s knowledge of the microstate.