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I’ve been trying to gain an understanding of non-equilibrium thermodynamics. I’ve been told that out of thermodynamic equilibrium, macroscopic state variables, such as temperature and pressure, are not well-defined. As I understand it, in equilibrium, on a microscopic level, the energy eigenstates of the system will follow a distribution proportional to $e^{-\beta H}$, where $\beta =(k_{B}T)^{-1}$. The temperature is then the parameter characterising the distribution of possible energies that particles can have in equilibrium. It can loosely be interpreting as being proportional to the average kinetic energy of the particles in the system. Out of thermodynamic equilibrium, the distribution is not as simple as this (in particular, it cannot be parametrised by a single “temperature” parameter). However, I’m a bit unsure about the details.

My question is: why is temperature, and for that matter all other state variables, not well-defined out of thermodynamic equilibrium?

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  • $\begingroup$ I'm also curious about this. But not only about temperature. About all thermodynamics quantities (such as entropy, pressure, volume, magnetization, etc.) $\endgroup$ – thermomagnetic condensed boson Feb 3 '18 at 15:12
  • $\begingroup$ @no_choice99 Good point. I’ll update my question. $\endgroup$ – user35305 Feb 3 '18 at 15:14
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In a non-technical sense, you are already quite familiar with at least one system that is far from thermodynamic equilibrium: The Earth's atmosphere. The Earth's atmosphere does not have a single temperature or pressure. They instead vary from place to place and from time to time. ("If you don't like the weather in New England now, just wait a few minutes.") The solution for intensive variables such as temperature and pressure is simple: Make those intensive variables local functions of position and time.

This approach can't be applied directly to extensive variables such as mass, entropy, and energy. What can be done is to divide one extensive variable by another, thereby yielding an intensive variable such as density (mass divided by volume) or specific energy (energy divided by mass).

These local intensive variables can be related to other local intensive variables by various thermodynamics equations, recast to use only local intensive variables. As a simple example, the ideal gas law ($PV=NRT$) becomes $P = \rho RT/\mu$, where $\mu$ is the molar mass of the gas in question. As a more complex example, Reynold's Transport Theorem can be used to relate a number of local thermodynamic variables.

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  • $\begingroup$ Thanks for the answer. I’ve heard though, that the temperature, for example, in a plasma of particles created during reheating, does not have a well-defined temperature until it thermalises. Is this simply because it does not have a single temperature, and the distribution of energies is not given by the Boltzmann distribution?! $\endgroup$ – user35305 Feb 3 '18 at 20:10
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Temperature is an intensive quantity characterizing a system only in equilibrium (thermodynamics). Outside equilibrium it is , in general, no defined. The concept has been extended to non-equilibrium thermodynamics assuming a local equilibrium, so that phenomena like Fourier's law of heat conduction can be described. Following Boltzmann, a (classical) system in non-equilibrium, e.g. a gas, can be described by a distribution function in phase space $$f(\vec r, \vec p, t)$$ which gives the time dependent probability of finding a molecule at time $t$ at location $\vec r$ and with momentum $\vec p$. This distribution function can be found by solving the Boltzmann transport equation, which usually includes a scattering term. Sometimes an "effective temperature" is defined for out of equilibrium distributions, e.g., for "hot electrons" in semiconductors in high electric fields, or in describing population inversion in lasers. But this "temperature" is not a thermodynamic quantity, it is just an auxiliary parameter for the approximate description of a system.

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  • $\begingroup$ Why exactly is temperature not defined out of equilibrium? Is it because it is a macroscopic quantity defined in terms of an equilibrium distribution of energies? $\endgroup$ – user35305 Feb 3 '18 at 21:17

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