Classical thermodynamics always discusses entropy in the light of reversible processes, and it lies at the heart of the definition of entropy. But do these reversible processes exist in Nature, or are they just a gedankenexperiment?

Since if they do not exist, then classical thermodynamics gives a lower bound on the increase of entropy, but says nothing about the difference in entropy between the reversible and irreversible process. Doesn't this make the classical definition useless for a quantitative approach? Or is there proof that the entropy difference between the reversible and irreversible process is small for most everyday cases?

  • $\begingroup$ Most of the processes we are aware of have $T$ symmetry! $\endgroup$
    – Ali
    May 3, 2014 at 16:37
  • $\begingroup$ Some quantum systems can reversible flip between two states, for example Benzene rings. But this is far from the statistical definition of entropy that you are interested. I'm not aware of any physical system or thermodynamic engine which can be made completely reversible (i.e. operate at the Carnot limit), it's probably impossible. Carnot efficiency give the maximum efficiency of a heat energy and therefore the minimum amount if entropy that needs to be generated. $\endgroup$
    – boyfarrell
    May 5, 2014 at 5:02
  • $\begingroup$ No. Reversible process do not exist in nature, but many processes have sufficiently small losses to treat them as reversible in theory. It is the usual "good enough" mentality that allows us to do physics (and science in general) in the first place. $\endgroup$ May 7, 2023 at 16:44
  • $\begingroup$ Particle interactions in QFT described by Feynman diagrams are time reversible. In other words, they look the same when played forward in time and backward in time (i.e. they conserve momentum, lepton number, etc). However, this is not true on large scales where quantum effects are not present. $\endgroup$
    – Tachyon
    May 7, 2023 at 18:31

2 Answers 2


Of course, thermodynamics says things about differences in entropy between reversible and irreversible processes. In fact, we can analyse how irreversible a process was and what to do in order to, for instance, extract more work from it. That's a very important part of thermodynamics.

But, no, there are no truly reversible processes — at least for daily macroscopic, bulk phenomena; I might be unaware of some crazy quantum-relativistic-boson-fermion experiment :). In practice, all real processes are irreversible, even though some can be good approximations of reversible ones (e.g. if you go really slowly).

Also, your statement "classical thermodynamics gives a lower bound on the increase of entropy" is a bit weird. That "lower bound" would be zero, so I'd prefer to say that it states that global entropy never decreases. Apart from that, classical thermodynamics also goes on to explore quantities like availability, irreversibility and efficiency, and it can do a good work of optimizing the hell out of your irreversible processes.

  • 3
    $\begingroup$ "all real processes are irreversible" should probably be revised to "all real thermodynamic processes are irreversible". After all, that's the primary conceptual difficulty with thermodynamics; it imposes a macroscopic time-reversal asymmetry, while microscopically, everything is time-reversal symmetric. (The Standard Model has $CPT$ symmetry, after all.) $\endgroup$ Jun 8, 2018 at 16:15

Hold on everyone! I may have found an example of a reversible process.

Consider a closed system with immovable diabatic walls consisting of ice and water at equilibrium at $T=273.15$, the fusion temperature of water. Since this is a phase-transformation, it occurs reversibly at constant $T$ and $P$ with $\Delta S^\circ = 0$.

Now if we add heat to the system, some of the ice turns into water, and if we draw heat from the system, some of the water turns into ice, both reversibly.

The only requirement is that the heat source and sink must be reversible, which is also not a problem since we can take two similar solid-liquid equilibrium systems at the fusion temperature, one with fusion temperature higher than that of water, which will act as source of heat, and one with fusion temperature lower than that of water, which will act as sink of heat.

Since all processes are reversible phase-transformations, these occur reversibly with $\Delta S^\circ = 0$.

  • $\begingroup$ The only thing is missing in you explanation is how to transfer "heat" rversibly, which of course, cannot be done except in the "limit" asymptoticatcally, approximately, etc., assuming that the heat source/sink has infinite size compared to the system (diabatically enclosed ice water) and the source/sink has no other interaction but reversible exchange of heat (an assumption). $\endgroup$
    – hyportnex
    May 7, 2023 at 17:04
  • $\begingroup$ @hyportnex since the walls are diabatic, all we have to do is to put the source/sink in touch with the system. Further, we do not need to have an infinite size source/sink, to carry out a finite reversible process. Moreover, the heat exchange is reversible because both the source/sink and the system are undergoing phase-transformation, which happens reversibly. $\endgroup$
    – ananta
    May 7, 2023 at 17:10
  • $\begingroup$ If two bodies are of the same temperature and otherwise are in mechanical/electrical etc. equilibrium then then they are also in thermal equilibrium. While maintaining mechanical equilibrium the only way to transport thermal energy from one to another is by a temperature difference that is NOT zero, it maybe small, it maybe insignificant but NOT zero, and then a tiny bit of irreversibility will rear its ugly head... $\endgroup$
    – hyportnex
    May 7, 2023 at 17:18
  • $\begingroup$ @hyportnex the source and sink are at higher and lower temperatures than the system, respectively... i am not using the same ice/water equilibrium system for the source and sink but different materials which have higher and lower fusion temperatures, respectively. $\endgroup$
    – ananta
    May 7, 2023 at 17:20
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    $\begingroup$ In theory, the only way to avoid that irreversible heat evolution is by employing a Carnot engine at the interface between your system and the reservoir, but even a Carnot engine needs some amount of irreversibility to be able absorb thermal energy in its isothermal legs. For a thorough understanding I suggest you read publ.royalacademy.dk/books/83/569?lang=da publ.royalacademy.dk/backend/web/uploads/2019-07-23/AFL%202/… and tandfonline.com/doi/abs/10.1080/… and $\endgroup$
    – hyportnex
    May 7, 2023 at 17:38

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