Searching Physics Stackexchange for entropy I have found several posts regarding entropy, lately most of the questions why entropy is a state variable.

This got me thinking. I have understood so far that from the following two posts

Is entropy $S$ a fundamental quantity like Temperature?

Proof for $\oint \frac{dQ}{T}=0 $ in a reversible process

that I can proof that there is such thing as a state variable in a reversible process defined by

$$ d S = \frac{d Q_{rev}}{T} $$

While for an irreversible process $$ S \geq 0$$ holds. The commonly found argument with two bodies of different temperature brought together and as each has its own temperature, the changes in entropy are different resulting in an overall change of entropy

$$ \Delta S = \Delta S_A + \Delta S_B = - \frac{d Q_A}{T_A} + \frac{d Q_B}{T_B} $$

The problem I have with the equation is the following:

We add two terms that assume reversible processes $\Delta S_A$ and $\Delta S_B$, add them up and a new quantity that clearly is non-zero but should still be a state variable in this irreversible case. How can I be sure that in such an irreversible case the introduction of entropy is meaningful and that such a state variable even exists? So far the aforementioned posts have only proven its existence for the reversible case! And why do I even have to take the temperature of the corresponding boundaries $T_A$ and $T_B$ and not something like a temperature $\frac{T_A + T_B}{2}$ for both. In such a case the entropy would still be $0$ for an irreversible process.

I had also a look at Entropy as a state function - Is it just a postulate of the second principle? and Is "entropy" not a state variable for irreversible process? but none of them gives a logical answer to that and makes it seem as if the entropy of an irreversible system is just a postulate, is it?


3 Answers 3


As it was taught to me, that at least from the phenomenological point of view, the entropy is more fundamental than temperature.

The first law of thermodynamics postulates the existance a state function $U$ that we call internal energy. Then the second law of thermodynamics postulates the existance a state function $S$ that we call entropy. They satisfy their specific laws.

The entropy is function of the internal energy and other state parameremters: $$S = S(U,V,N,\dots)$$ where dots represent other possible state parameters, like magnetizaion for example. We have $$dS = \frac{\partial S}{\partial U} dU + \frac{\partial S}{\partial V} dV + \frac{\partial S}{\partial N} dN + \dots $$

We can define new state functions: $$\frac{1}{T}:= \frac{\partial S}{\partial U}$$ $$p := T \frac{\partial S}{\partial V}$$ $$\mu := -T \frac{\partial S}{\partial N}$$ So the temperature, pressure, chemical potential etc. are calaculated from entropy, not the other way around. (In particular, we can read the defintion of temperature as the amount of energy required to increase the entropy by a single unit, while keepeing the other state parametrs constant.)

We have then $$dS = \frac{1}{T} dU + \frac{p}{T} dV - \frac{\mu}{T} dN + \dots $$ or $$dU = T dS - p dV + \mu dN + \dots $$

The mechanical work performed on the system is calculated from the formula $$W = -\int p dV$$ so if there is no matter transfer or changes in other state parameters we have $$\Delta U = W + \int T dS$$ We call the part equal to $\int T dS$ the heat transfer: $$ Q : = \int T dS$$

We then proceed to prove other, non-fundamental, laws of thermodynamics.

  • $\begingroup$ That sounds interesting and consistent... This means the second law postulates the existence of entropy? Can you tell me then how this derivation for a reversible path and an ideal gas gets along without it. Where is it hidden? In the assumption of an internal energy $E_{in}(V,T)$ (which corresponds to your $U$)? $\endgroup$
    – Giuseppe
    Commented Apr 4, 2020 at 12:28
  • $\begingroup$ It looks as if the symmetry in that case was enough to derive that such a quantity exists for a reversible process - without assuming beforehand the existence of such a state variable. $\endgroup$
    – Giuseppe
    Commented Apr 4, 2020 at 12:35
  • 1
    $\begingroup$ @Giuseppe if you define $\Delta S := \int \frac{\delta Q}{T}$ in reversible processes, you need to know what what are $\delta Q$ and $T$. While $\delta Q$ is simple (you just define it as $\delta Q = dU + p dV$), but $T$ is more complex. Whatever way you do it, at some point you need to postulate that it has some property that will guarantee that $S$ is a state function, that is that the integral that defines it does not depend on the path (on which irreversible process is considered), only on end points. In the end, it is equivalent to postulating the existence of the state function $S$. $\endgroup$ Commented Apr 4, 2020 at 12:47
  • $\begingroup$ That sounds plausible! I will have a look at in more detail. Any literature you recommend on the topic? $\endgroup$
    – Giuseppe
    Commented Apr 4, 2020 at 12:50
  • $\begingroup$ @Giuseppe Landau Lifshitz statistical physics, second chapter. In the first chapter the existence of the entropy state function is even connected to more fundamental principles. But be aware, this text is nothing for beginners. $\endgroup$
    – user224659
    Commented Apr 4, 2020 at 13:10

For an irreversible process path, the change in entropy is not the integral of dQ/T. It is only dQ/T for an alternate reversible path that you have devised separately between the same two thermodynamic end states. So, for an irreversible process, the first thing you need to do is totally forget about the actual irreversible path; this is no longer relevant. Instead, you need to devise a reversible path between the same two end states and calculate the integral of dQ/T for that path. For more on this, including worked examples for irreversible paths, see the following: https://www.physicsforums.com/insights/grandpa-chets-entropy-recipe/

  • $\begingroup$ Thanks for your answer but I am looking for an explanation to why entropy is a state variable for the irreversible case. How do I know such a variable even exists? I will have a look at your post on tht platform in the meanwhile! $\endgroup$
    – Giuseppe
    Commented Apr 4, 2020 at 11:55
  • 1
    $\begingroup$ The link you mentioned gives a very neat way of finding the entropy change, however it does not justify why that works. Why does the entropy change for a reversible and for an irreversible path stay the same? $\endgroup$
    – user258881
    Commented Apr 4, 2020 at 12:34
  • $\begingroup$ Because entropy is a physical property of the system (state function), so it depends only on the two end states. In the irreversible process, in addition to entropy being exchanged across the boundary of the system by $dQ/T_{boundary}$, it is also generated within the system. $\endgroup$ Commented Apr 4, 2020 at 13:43

For general processes – reversible or irreversible – there are two complications with respect to thermostatics:

1 – The notion of "state" is generalized. A state at time $t$ may be given not just by the values of some set of thermodynamics (or thermomechanics) variables at that time – say, $\bigl(V(t), T(t)\bigr)$ – but by the full history of those values for some previous interval of time $\Delta t$. For example, informally, $$\text{state}(t) = \{V(t'), T(t') \,\vert\; t-\Delta t < t' \le t\} $$ (or more precisely an equivalence class of such histories).

This takes care of materials with memory (eg subject to hysteresis).

This means that the entropy becomes a functional rather than a function; that is, a function of a function (since the state is effectively a function).

In some cases the memory is "differential", that is, it only concerns very short time intervals. Then the history can be approximated by the time-derivative of the thermodynamic quantities. For example we can have, informally, $$\text{state}(t) = \{V(t), T(t), \dot{V}(t), \dot{T}(t)\} .$$

2 – The entropy itself can be a non-unique function(al). That is, we have a (convex) set of possible entropy-function(al)s, different from one another – and not just by a constant. Each of them does its job and leads to identical experimental consequences, so it doesn't matter which we choose. Pedagogical examples are given by Owen (1984) for simplified elastic-plastic materials.

This "causes nonmeasureability of entropy and nonexistence of entropometers" (Samohýl & Pekar 2014, p. 52). This latter fact was discovered in the 1970s. I think one of the first to point it out was W. A. Day (1977).

Despite these two complications, the concept of entropy and its mathematical use has still proved to be fundamental, also in irreversible thermodynamics. Its role changed a little though.

Earlier on, entropy and its related inequality was used, roughly speaking, for checking which processes were possible and which impossible. Its modern use, instead, is for checking which consitutive equations are possible and which impossible. This is a much bigger role, because it decides upon whole physical models, not just processes.

Very instructive simple examples of how this happens, for toy systems described eg by $\{V(t), T(t), \dot{V}(t), \dot{T}(t)\}$ variables, are given in chapter 2 of Samohýl & Pekar (2014).

Here are some references, from different perspectives (applied, mathematical) about the points above:

  • Astarita: Thermodynamics: An Advanced Textbook for Chemical Engineers (Springer 1990).
    – A truly wonderful book! It discusses all points above.

  • Owen: A First Course in the Mathematical Foundations of Thermodynamics (Springer 1984).
    – It gives examples of the non-uniqueness of entropy, and studies the structure of the set of entropy-functions.

  • Samohýl, Pekar: The Thermodynamics of Linear Fluids and Fluid Mixtures (Springer 2014).
    – I warmly recommend chapter 2 of this book for an overview of all points above with simple examples.

  • Truesdell (ed.): Rational Thermodynamics (Springer 1984).

  • Day: An objection to using entropy as a primitive concept in continuum thermodynamics (1977) https://doi.org/10.1007/BF01180089


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