9
$\begingroup$

I am trying to understand the difference between thermodynamic and thermal equilibriums. According to wikipedia, they really are distinct; you can have one without the other. However, I am having trouble visualizing a somewhat less abstract and more physically realizable system where you can have one without the other. Is there some "non-pathological" system where we can see that it is in thermodynamic but not thermal equilibrium and another where we have thermal but not thermodynamic equilibrium?

I'm motivated by this question by a statement I heard regarding local and global temperature. In some system which can be partitioned into subsystems where local temperature is a well-defined quantity, the claim was made that the system can be in thermodynamic equilibrium and hence have a well-defined global temperature, yet the local temperature can fluctuate and differ from the global temperature. This seems quite paradoxical to me and hard to wrap my mind around, though in principle it seems possible.

$\endgroup$
2
  • $\begingroup$ I've always used the two terms interchangeably. It appears that, at least with the view of Adkins, that the point being made is that a given system may have subsystems and that while some of the subsystems may reach thermal equilibrium with an external body, other subsystems may not. Hence, (partial) thermal equilibrium without thermodynamic equilibrium. A real-world example is how when a material is shock-compressed, the electronic subsystem may quickly reach thermal equilibrium with itself and thus have a well defined temperature, even though it is still not in equilibrium with the lattice. $\endgroup$
    – user93237
    Dec 28, 2016 at 20:50
  • 1
    $\begingroup$ Thermodynamic equilibrium consists of thermal + radiative + chemical + mechanical equilibrium. $\endgroup$
    – Poutnik
    Feb 10, 2022 at 8:21

2 Answers 2

3
$\begingroup$

This is a question of meaning of words - semantics. I did not make any research into how the terms are used in the classic physics literature, but I think the following view is quite reasonable:

  • system is in thermal equilibrium when any two parts of it are in direct or indirect thermal contact (capable of exchanging heat) but no appreciable heat is exchanged from one to another even after a long time (concern with heat flow)

  • system is in thermodynamic equilibrium when all macroscopic processes in it subsided(concern with any macroscopic flow, not only heat).

Is there some "non-pathological" system where:

  • we can see that it is in thermodynamic but not thermal equilibrium:

Yes, for example two gases of different temperature trapped in a cylindrical vessel separated by movable but thermally insulating piston. If the pressure of the gas is the same in both compartments, the system will remain without change indefinitely, so it is in thermodynamic equilibrium (under given constraint of perfect insulation). However, it is clearly not in thermal equilibrium, because there is no possibility of thermal exchange between the two parts.

Admittedly, this is an idealized situation, as there are no perfect thermal insulators in the real world. In practice such a system would not be in thermodynamic equilibrium either, as some heat would flow from the hotter to the colder gas. But it is possible to make walls that would maintain the state of different temperatures for a very long time without appreciable change in the macroscopic state, so this would be a system in thermodynamic equilibrium for all practical purposes.

  • another where we have thermal but not thermodynamic equilibrium:

Yes, for example a gas that is being compressed by a moving wall. The temperature and pressure increases everywhere as work is being performed on the gas elements; but there is no heat flux anywhere. The system is in thermal equilibrium, but not in thermodynamic equilibrium, as macroscopic changes are happening.

Of course, in practice the compression would not occur uniformly in the whole space, but there would be compression waves and heat exchange between elements of the gas. The system is thus not even in thermal equilibrium. But if the compression is done slowly enough, heat exchanges will be negligible and most energy transfer will be via work. Thus the system would be in thermal equilibrium for all practical purposes.

$\endgroup$
4
  • $\begingroup$ This classic problem is the so-called "adiabatic piston". Here you have a cylinder filled with ideal gas but is separated with a frictionless movable wall. Assume that the walls of the cylinder and also the partition are impermeable and adiabatic, and starting from equilibrium (equal pressure) you move the partition towards one side (where because of the compression you higher pressure and temperature than on the side) and let it go freely. What happens, will the system ever settle and if yes, how? See Gruber: Thermodynamics of systems with internal adiabatic constraints, E JP 20 (1999) $\endgroup$
    – hyportnex
    Dec 29, 2016 at 16:46
  • $\begingroup$ @hyportnex, your question is far from the original one. If you genuinely want to ask it, please create separate question here: physics.stackexchange.com/questions/ask $\endgroup$ Dec 29, 2016 at 21:48
  • $\begingroup$ Thanks, that's helpful. As an extension, in the case with thermodynamic but not thermal equilibrium, is temperature of the total system a well-defined quantity? We would simply define global temperature in the normal way with as in a microcanonical ensemble, if I'm not mistaken. $\endgroup$
    – Aaron
    Dec 30, 2016 at 21:10
  • $\begingroup$ @Aaron, I think that is not possible, since the system consists of two parts at different temperatures. The ratio $\Delta U/\Delta S$ would depend on which subsystem got the energy. In other words, the internal energy of such constrained system is not a function of total entropy. This kind of system is weird example far from applications of thermodynamics, so perhaps it is not useful to say it is in thermodynamical equilibrium. In any case thermodynamical equilibrium is intuitively more general concept than equality of temperatures. $\endgroup$ Dec 30, 2016 at 21:34
0
$\begingroup$

Thermal equilibrium is only concerned with one form of internal energy namely Kinetic Energy (KE) of the molecules. In a gas the temperature is proportional to the mean KE of the molecules.

In contrast, thermodynamic equilibrium is the state of maximum entropy. Now, contrary to what some think, entropy is affected by all forms of internal energy including phase change energy (sometimes referred to as latent heat) and chemical and potential energy. I will focus on potential energy (PE) which can be due to location within a force field like gravity or centrifugal force, or it could be PE relating to things like wound up clock springs or pistons pushed down a cylinder creating extra pressure that would push them back when released.

Considering the troposphere (lowest layer of the atmosphere) where there is a non-zero temperature gradient we need firstly to understand that maximum entropy is the state when there are no longer any unbalanced energy potentials because, in accord with the Second Law of Thermodynamics, as entropy increases there will be unbalanced energy potentials that are diminishing. These potentials must take into account all forms of internal energy, but, if we assume there are no phase changes currently taking place in a column of the troposphere and no chemical reactions, then what we need to consider is PE and KE only.

It is the SUM of this PE and KE which must be considered. At maximum entropy there will be no difference in the sum (PE + KE) between the top and bottom of that column, or anywhere in between. However, clearly PE is greater at the top, and so KE must be less at the top. Since it is only KE that affects temperature we have a temperature gradient that was formed by gravity acting on these molecules. This was first explained by Josef Loschmidt in the 1870's.

So, this state of thermodynamic equilibrium has a temperature gradient and so we would not say that the column of air was in thermal equilibrium.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.