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I've read that equilibrium is the state at which entropy has a maximum. However I found this definition paradoxical because on books entropy is usually defined only for equilibrium states. How is it possible?

A possible explanation: I think that it could be that entropy is actually defined for quasi equilibrium states (meaning states at which thermodynamics property like temperature have a meaning) and the quasi equilibrium state at which entropy has a maximum is a truly equilibrium state.

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  • $\begingroup$ Please provide a source in which context this is stated. Are you familiar with the concept of information entropy? $\endgroup$
    – Jakob
    Oct 14 at 7:19
  • $\begingroup$ There are different ways to define entropy, e.g., see this answer: physics.stackexchange.com/a/669707/247642 $\endgroup$ Oct 14 at 9:31
  • $\begingroup$ Do you want an answer from the perspective of thermodynamics, or statistical mechanics. $\endgroup$
    – Bob D
    Oct 15 at 16:30
  • $\begingroup$ Thermodynamics please $\endgroup$
    – SimoBartz
    Oct 15 at 17:15
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There are many thermodynamic quantities that can be well-defined for out-of-equilibrium as well as equilibrium states. Internal energy, volume and entropy are a few.

We define entropy in general by asserting that it is the sum of the entropies of all the parts of a system. Then you can take the parts small enough that each is in internal equilibrium, yet large enough that thermodynamics still applies (I am giving a thermodynamic answer here). The thermodynamic limit allows both requirements to be satisfied simultaneously.

Note that when each part is in internal equilibrium we don't necessarily have an equilibrium state overall because the different parts may not be in equilibrium with each other.

There is a difficulty with cases where a thermodynamic limit may be unclear or too great an approximation. For example in a highly irregular case such as turbulent flow it may not be possible to find parts that are large enough to be treated accurately in a thermodynamic limit (the limit of large numbers of microstates). In such cases one may adopt the statistical definition $S = -k_{\rm B} \sum p_i \ln p_i$. The standard approach is to assert that this is a definition, and one then gives arguments to show that such a definition agrees with the thermodynamic statement $dS = dQ_{\rm rev}/T$ in the thermodynamic limit.

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The following is from a thermodynamics perspective, rather than statistical mechanics.

I've read that equilibrium is the state at which entropy has a maximum.

I believe you are referring to the maximum entropy principle. If so, from a thermodynamics perspective, the maximum entropy principle states that for an isolated system (a system for which there is no energy or mass transfer with the surroundings) the entropy is maximized at equilibrium. (See https://en.wikipedia.org/wiki/Principle_of_minimum_energy#:~:text=The%20maximum%20entropy%20principle%3A%20For,energy%20is%20minimized%20at%20equilibrium.)

However I found this definition paradoxical because on books entropy is usually defined only for equilibrium states

Yes, entropy is a (equilibrium) thermodynamics state property. But I don't believe there is a paradox.

For an isolated system if the system is not in internal equilibrium, entropy is not maximized. Only when it comes to internal equilibrium is entropy maximized. And if it is not in internal equilibrium it can only be brought to internal equilibrium as a result of an irreversible process, which by definition generates (increases) entropy.

Take the often used example of the free expansion of an ideal gas into a vacuum. The system consists of an ideal gas and a vacuum separated by a partition bounded by a rigid, thermally insulated vessel. So we have an isolated system (no work, heat, or mass transfer with the surroundings). The gas (one part of the system) is initially in internal equilibrium and the vacuum (the other part of the system) is in internal equilibrium, but the gas and the vacuum are not in equilibrium with each other due to the pressure and temperature difference (temperature of a perfect vacuum being zero).

In order to bring the two parts of the system into equilibrium, an opening in the partition is created and the gas allowed to freely expand into the vacuum until equilibrium is established. The process is irreversible and entropy is generated (entropy increases). For the process, $\Delta U=0$ and for an ideal gas $\Delta T=0$. The change in entropy can be calculated by assuming a reversible isothermal path connecting the initial and final states of the system. The entropy of the new state of the system is now maximized.

Hope this helps.

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  • $\begingroup$ You say "entropy is a (equilibrium) thermodynamics state property" and "if the system is not in internal equilibrium, entropy is not maximized". But if system is not in internal equilibrium entropy is not defined. I see a paradox in saying entropy is both not at max and not defined $\endgroup$
    – SimoBartz
    Oct 18 at 9:32
  • $\begingroup$ I was referring to different parts of a system not being in equilibrium with one another but individually in internal equilibrium, like the example I gave. See the third paragraph of Andrew Steene's answer. In that example the gas was initially in internal equilibrium with some initial entropy. $\endgroup$
    – Bob D
    Oct 18 at 10:48
  • $\begingroup$ The total system doesn't have an entropy if it' not at equilibrium, and if it has it then is always maximized because it's in equilibrium. Moreover the system made by two parts as you described I think it's in equilibrium, indeed you need to modify the system (removing the separating wall) to let the system evolve in a new equilibrium. $\endgroup$
    – SimoBartz
    Oct 18 at 13:05
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    $\begingroup$ If the two parts of the system were in equilibrium with one another, nothing would happen if the partition is removed. It appears no matter what I (or Andrew Steen said in his 3rd paragraph) say, it appears you will remain unconvinced. That's OK, but I see no point in continuing the discussion. Hope you find the answer you are looking for. $\endgroup$
    – Bob D
    Oct 18 at 13:50
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From the paper Entropy and Time by Arieh Ben-Naim:

Unfortunately, there is no general definition of equilibrium which applies to all systems. Callen [35*,36**], introduced the existence of the equilibrium state as a postulate. He also emphasized that any definition of an equilibrium state is necessarily circular.

I don't know if that view is universal, but certainly you can talk about the entropy of a nonequilibrium system in the way that Andrew Steane said (summing up the entropies of the parts, with the assumption that each test volume is well-described by whatever distribution and with the assumption that the entropy is extensive). So I think your answer is correct and that there are many assumptions baked into the word "equilibrium".


*35. Callen H.B. Thermodynamics. John Wiley and Sons; New York, NY, USA: 1960

**36. Callen H.B. Thermodynamics and an Introduction to Thermostatistics. 2nd ed. John Wiley and Sons; Hoboken, NJ, USA: 1985.

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Boltezmans entropy equation is what is used for equilibrium situations, thats what the OP has no doubt read in a book, i.e. like this.

However, Boltezeman cannot be used in non-equilibrium systems, for that you need to use Shannons entropy equation. That the resolution to the OP's 'paradox'.

For more on the connection between entropy definitions, see my answer

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