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dxu comments on Entropy and Temperature - Less Wrong
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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?
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.