# Confusion regarding temperature and entropy

In my thermodynamics/statistical mechanics book ("Concepts in thermal physics", Blundell), the temperature of a system A has been defined as being one over the derivative of the entropy of A with respect to the total (average) internal energy, evaluated at constant volume. This seems to imply that as long as a system has a well defined entropy and internal energy, then a temperature can be assigned by simply using this formula. But what if a system has a non-uniform temperature (meaning that it has no well-defined total temperature), but yet a well-defined entropy and internal energy? Doesn't this definition of temperature apply anymore in that case? This also get me to my second question. Can an entropy be assigned to a system with a non-uniform temperature? My guess is yes, since you can add up (or integrate, in the continuous case) all the entropies of various subsystems of A to get the total entropy. But at the same time, entropy is meant to be a state function which is only defined when the system is in a well-defined macroscopic state (which includes temperature??).

• You would recommend you to complement the Blundell with some other book. Although it is a nice introductory book, I had exactly the same kind of doubts after the course in which I used it (I have to admit that I am not the best student, so others might have a different opinion).
– JGBM
Jan 9, 2021 at 11:50
• @Jonhy Yeah, I have another book as well, but I actually find Blundell's book to be better. I feel like it has a really great structure. But of course, it would have been even better if it cleared things like this up as well. Jan 9, 2021 at 12:05
• That structure is its strongest point and I really appreciate it to. But sometimes, things are just messy and complicated xD.
– JGBM
Jan 9, 2021 at 12:15

If the temperature is non-uniform, then the system is not at equilibrium and none of this applies.

The entropy (from the perspective of statistical mechanics - the number of microstates) is always well-defined (since is fundamentally combinatoric and does not depend on a given configuration of the system), but the temperature you obtain by this process is the temperature of the system, with a given energy $$E$$, when it finally reaches equilibrium.

You can eventually do this locally, if the system takes a long time to reach equilibrium, but it will only be an approximation.

Although the subject is called thermodynamics, it is more like thermostatics, since there is no dynamic (there is no time). The study of non-equilibrium thermodynamics (which is actually thermodynamics) and the definition of quantities, like the temperature, in these systems is a field of research.

Actually, you can easily notice that you need to be working in the context of equilibrium. The laws of thermodynamics imply that a system with non-uniform temperature eventually thermalize. This would introduce some notion of time, as you would need to label your quantities. The equilibrium properties are the only ones that do not depend on time.

• Thank you for the answer. But one of the reasons why I wondered whether entropy is defined for non-equilibrium states, is that my book has several examples where, say, they calculate the total entropy increase of systems A and B as they come into equilibrium (i.e. as their temperatures apporach one another) to show that it satisfies the second law. But since the two systems don't have equal temperatures during the process, it seems like such a total entropy for the combined system shouldn't be well defined, according to your assertions. So what's going on here? Jan 9, 2021 at 12:13
• I like the comment on the "dynamics* part of the word "thermodynamics". Thermostatics would be wrong as well, though. In the sense that flows and transformations are part of thermodynamics, it is just that enough time is given. Jan 9, 2021 at 12:59
• I agree with @Alchimista. A large part of thermal physics is indeed the study of equilibrium states. However, two other large parts are phase changes, or the changes from one equilibrium state to another, and transport properties. Both of these, especially the latter, have dynamical considerations.
– CGS
Jan 9, 2021 at 13:09
• I know that is a bit more complicated than that. I just wanted to point out a point of view that is not emphasized.
– JGBM
Jan 9, 2021 at 18:51
• @FelisSuper The total entropy is always well-defined, but the results you obtain are for the new system at equilibrium. The entropy does not capture non-equilibrium properties of your system. The most you might try is to use the fact that you know the energy that each system has at a given time (you know you much energy is transferred by the interface), and use that to calculate the dependence of the total entropy in time. However, the temperature obtained does not make sense. It would be like talking about the combined temperature of two isolated systems.
– JGBM
Jan 9, 2021 at 19:16

Some people feel that it is valid to evaluate thermodynamic state properties locally (i.e., per unit mass), and I am one of them. In the case of temperature, for example, the partial derivative of specific internal energy with respect to specific entropy at constant specific volume would be the local temperature. Without assuming such local validity, it would be impossible to get accurate solutions to transient systems and steady flow systems experiencing transport processes. So we could never design heat exchangers, cooling towers, boilers, chemical reactors, distillation columns, absorption columns, piping systems, or any other industrial scale chemical processing equipment.

To add some references to the previous answers:

The very meaningful questions you are asking are addressed in the field of continuum thermomechanics, which studies the interplay of mechanical and thermal properties of generic bodies (systems) in generic dynamical processes.

Two beautiful books that introduce this theory are: