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A particularly weak formulation of thermodynamics is

  1. Two bodies are said to be in thermal equilibrium if no heat transfer occurs when they are put in diathermal contact. Thermal equilibrium is an equivalence relation.
  2. Energy is conserved in thermodynamic processes.
  3. There's no cycle that only extracts heat from a single system and performs the equivalent work.
  4. All energy transfer between thermodynamic systems is bounded.

This is enough to define temperature as "the variable that shows up in the Carnot efficiency equation",

$$\eta_\text{rev}=1-\frac{T_C}{T_H}$$

and by definition it's the same for two bodies in thermal equilibrium. However, is there any construction we can do to prove that the converse also holds? That two bodies with the same temperature must be in thermal equilibrium?

EDIT: If we're allowed to assume entropy $S(U)$ is strictly concave for any physically meaningful system (i.e. anything that's not a thermal reservoir) we can argue as follows: assume systems $S_1$ and $S_2$ are at the same temperature $T$. If they have energies $U_1$ and $U_2$ and were to transfer heat $Q$, the entropy after this process would be $$S_1(U_1+Q)+S_2(U_2-Q)<S_1(U_1)+\frac{Q}{T}+S_2(U_2)-\frac{Q}{T}=S_1(U_1)+S_2(U_2)$$ and therefore is impossible. Note that only heat transfer is considered so the other extensive variables are held fixed for each system. Derivations for other fixed variable constraints are probably possible given suitable thermodynamic potentials.

How to argue for this strict concavity?

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  • $\begingroup$ Comments have been moved to chat; please do not continue the discussion here. Before posting a comment below this one, please review the purposes of comments. Comments that do not request clarification or suggest improvements usually belong as an answer, on Physics Meta, or in Physics Chat. Comments continuing discussion may be removed. $\endgroup$
    – Buzz
    Commented Jul 30, 2023 at 19:01

1 Answer 1

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This is enough to define temperature as "the variable that shows up in the Carnot efficiency equation",

$$\eta_\text{rev}=1-\frac{T_C}{T_H}$$

and by definition it's the same for two bodies in thermal equilibrium.

Yes...

However, is there any construction we can do to prove that the converse also holds? That two bodies with the same temperature must be in thermal equilibrium?

No. Just because two bodies in thermal contact have the same temperature, it does not mean they have to be in thermal equilibrium, that is, it does not mean there is no heat flow. There can be heat flow.

Your definition implies that same temperatures of two reservoirs allow only zero or lower efficiency of an hypothetical cyclic engine working between the two reservoirs. But this can be achieved in various ways: maybe some heat is transferred from one reservoir to the engine, but then equal or higher heat is dumped by the engine into the other reservoir, and thus net work done is zero or negative. This sounds weird, but we actually have to allow for this in theory in order for the reversible engine to be defined at all, which we need so the Carnot limit can be derived.

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  • $\begingroup$ Agreed. Of course you can transfer heat between objects at the same temperature: use an engine to take heat from the first while raising a weight, then dump the same heat onto the second by lowering the weight back down. However, thermal equilibrium is all about spontaneity in my mind. Might just be a bad way to look at it, but we really don't expect heat to flow when we just put two things at the same temperature into diathermal contact. In fact, for any finite transfer between finite systems such a transfer would violate the second law. $\endgroup$ Commented Jul 28, 2023 at 5:53
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    $\begingroup$ "Just because two bodies in thermal contact have the same temperature, it does not mean they have to be in thermal equilibrium, that is, it does not mean there is no heat flow. There can be heat flow." How can that be. Heat is defined as energy transfer due solely to temperature difference. Zero temperature difference, zero heat. $\endgroup$
    – Bob D
    Commented Jul 28, 2023 at 12:46
  • $\begingroup$ @GabrielGolfetti we don't expect the heat to flow in a macroscopic experiment because we actually assume thermal equilibrium. But thermodynamics as a theory is based on the idea that heat can be taken off one body and put into another, at the same temperature (from reservoir to working medium in a reversible engine). That's why two bodies at the same temperature need not be in equilibrium. I don't see how this could possibly violate 2nd law - entropy just changes is location, it does not decrease. $\endgroup$ Commented Jul 28, 2023 at 17:02
  • $\begingroup$ @BobD That is a practical understanding of heat based on modern interpretation of Newton's law of cooling, which is often valid. But that law is not universal, and it is not how heat is defined in thermodynamics. Heat transfer between solid bodies is energy transfer that cannot be expressed as product of intensity and extension and thus cannot be interpreted as some kind of work. $\endgroup$ Commented Jul 28, 2023 at 17:05
  • $\begingroup$ @JánLalinský Sorry Jan, but I respectfully disagree. $\endgroup$
    – Bob D
    Commented Jul 28, 2023 at 17:08

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