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The state of thermodynamic equilibrium is typically defined referring to the system's behaviour in the future. Can the definition be formulated in terms of measurable quantities?

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  • $\begingroup$ Isn't temperature a measurable quantity? $\endgroup$ – Kyle Kanos Feb 16 '15 at 19:26
  • $\begingroup$ All thermodynamic quantities, including temperature, are defined in the equilibrium state. Therefore, there must be a way to establish the latter using measurements. $\endgroup$ – Enoch Arden Feb 16 '15 at 19:29
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Technically, a system is in thermodynamic equilibrium if:

  • Every part of the system has a well-defined temperature. For example:

    • If there's an ideal gas, the velocities satisfy the Maxwell-Boltzmann distribution (for some temperature);
    • If there are photons, their intensity and spectrum satisfies the blackbody radiation formula (for some temperature)
    • If there are chemical bonds breaking and re-forming, the higher-energy configurations occur less often, exactly following the Boltzmann distribution (for some temperature)
    • etc. etc.
  • All of those temperatures of all of those components are exactly the same.

I guess I could have just said "It satisfies the Boltzmann distribution, for some fixed temperature". The above is just a roundabout way to say that. But I wanted to be more specific. :-D

**Footnote: In practice, people refer to a system as being in thermodynamic equilibrium even if some degrees of freedom (i.e., ways to store energy) do not satisfy the Boltzmann distribution ... as long as those degrees of freedom can only change so very very slowly that we can treat it as being stuck in its configuration "forever". A stupid example is nuclear degrees of freedom: According to the Boltzmann distribution, hydrogen atoms should fuse together into helium. But that will only happen if you wait 100 grillion gazillion years, or if we're talking about stars or nuclear weapons. So we freely refer to systems as being in thermodynamic equilibrium even if they have hydrogen atoms in them. That's an extreme example. There are many more murky examples. Is my refrigerator in thermodynamic equilibrium, even though the cheese is gradually going bad (undergoing slow chemical reactions)? Not really but yes for some purposes.

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  • $\begingroup$ This definition is circular: temperature is defined for a system in equilibrium. The latter has to be established before the temperature can be measured. And we are talking about thermodynamics and macroscopic measurements, the Boltzmann distribution is irrelevant. $\endgroup$ – Enoch Arden Feb 16 '15 at 19:57
  • $\begingroup$ @EnochArden - The velocity distribution of a gas is macroscopically measurable - look at doppler shifts of spectral lines for example - as is whether light satisfies a blackbody distribution etc. I take the Boltzmann distribution as a starting point: P~exp(-E/kT). There's a letter "T" in that formula, it's a parameter, and everybody refers to this parameter as "temperature". That's all I mean by "temperature", there's nothing circular there. $\endgroup$ – Steve Byrnes Feb 16 '15 at 20:09
  • $\begingroup$ @EnochArden - The laws of thermodynamics are not axioms handed down by God. They are consequences of statistical mechanics, just as all macroscopic laws of physics can be derived from microscopic starting points. See any statistical mechanics textbook for detailed derivations. $\endgroup$ – Steve Byrnes Feb 16 '15 at 20:11
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    $\begingroup$ It is exactly the opposite. The laws of thermodynamics are empirical observations. And they are fundamental principles which cannot be reduced to the laws of mechanics. Statistical mechanics is a conjecture used to interpret the laws of thermodynamics based on some a priori probabilistic assumptions. $\endgroup$ – Enoch Arden Feb 16 '15 at 21:16

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