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What's the most fundamental definition of temperature? Is it the definition concern about average energy, number of micro states, or what?

By "fundamental", I mean "to be applied" in such general cases as Black Hole's Temperature, Accelerated Frame's Radiation,...

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Related: – Qmechanic Jul 6 '14 at 13:10
I have seen physics lectures where temperature is defined as proportional to the mean energy of the constituent particles and this seems to be a common misconception. It works for ideal gasses only. And its approximately true for most multiparticle systems warmer than a few $K$: to get a good feel for how this approximation arises, apply the fundamental definition $\beta = T^{-1} = \partial S/\partial U$ (a definition that also works for black holes), work out the temperature of an ensemble of quantum harmonic oscillators at equilibrium and you find that the mean energy ..... – WetSavannaAnimal aka Rod Vance Sep 5 '15 at 9:55
.... for each constituent oscillator is $\left<E\right> = \frac{\hbar\,\omega}{2}\,\coth\left(\frac{1}{2}\,\beta\,\hbar\,\omega \right)\to k\,T$ when $k\,T\gg \hbar\,\omega$. – WetSavannaAnimal aka Rod Vance Sep 5 '15 at 9:58
up vote 6 down vote accepted

It's the differential relationship between internal energy and entropy: \begin{align} dU &= T\,dS + \cdots \\ \frac{\partial S}{\partial U} &= \frac 1T \end{align} As energy is added to a system, its internal entropy changes. Remember that the (total) entropy is $$ S = k \ln\Omega, $$ where $\Omega$ is the number of available microscopic states that the system has. The second law of thermodynamics is simply probabilistic: entropy tends to increase simply because there are more ways to have a high-entropy system than a low-entropy system. The logarithm matters here. If you double the entropy of a system (by, say, combining two similar but previously-isolated volumes of gas) you have squared $\Omega$.

Consider two systems with different $U,S,T$ that are in contact with each other. One of them has small $\partial S/\partial U$: a little change in internal energy causes a little change in entropy. The other has a larger $\partial S/\partial U$, and so the same change in energy causes a bigger change in entropy. Because they're in contact with each other, random fluctuations will carry tiny amounts of energy $dU$ from one system to the other. But because of the internal differences that lead to different numbers of internal states, it becomes overwhelmingly more likely that energy will flow from the system with small $\partial S/\partial U$ (reducing its entropy by a little) and into the system with larger $\partial S/\partial U$ (increasing its entropy by a lot). So we call the first one "hot" and the second one "cold."

This definition even extends to the case where entropy decreases as energy is added, in which case the absolute temperature is negative. It also explains why those negative temperatures are "hotter" than ordinary positive temperatures: in that case adding energy to the positive-temperature system increases its entropy, and removing energy from the negative-temperature system also increases its entropy.

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It seems like the process "adding energy to the positive-temperature system increases its entropy, and removing energy from the negative-temperature system also increases its entropy." will go on until the negative-temperature runs out of energy? – Mr.T Jul 7 '14 at 6:56
No, because $\partial S/\partial U$ changes with internal energy. The one-way flow of energy goes on until both systems have the same $\partial S/\partial U$, at which point energy flows in both directions are equally likely and we say that the two systems are at the same temperature. – rob Jul 7 '14 at 12:38

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