My textbook, W.E. Gettys, F.J. Keller, M.J. Skove, Physics 1, gives the definition of a reversible transformation as a transformation that can be inverted by effectuating only infinitesimal changes in the surroundings. I admit that I have no idea of what infinitesimal means in a rigourous mathematical language (if we exclude non-standard analysis, but I do not think that it is used in elementary thermodynamics), therefore the definition is quite confusing to me.

Then the book proves that the entropy of a Carnot cycle is zero and, since a reversible cycle can be approximated by the sum of many Carnot cycles, the entropy of a reversible cycle is zero too.

I suppose that we can mathematically formalize the meaning of such an approximation by saying that for any reversible cycle there is a sequence $\{\{C_{1,n},\ldots,C_{n,n}\}\}_n$ of sets of Carnot cycles $C_{1,n},\ldots,C_{n,n}$ (say that the $n$-th set of Carnot cycles used to approximate is composed by $n$ cycles) such that $$\oint\frac{\delta Q}{T}=\lim_{n\to\infty} \sum_{k=1}^n \frac{Q_{C,k,n}}{T_{C,k,n}}+\frac{Q_{H,k,n}}{T_{C,k,n}}.$$As a side note, I realise that, for any Carnot cycle $C_{k,n}$, we have $\frac{Q_{C,k,n}}{T_{C,k,n}}+\frac{Q_{H,k,n}}{T_{C,k,n}}=0$ and therefore, provided that the approximation holds, ${\displaystyle\oint}\frac{\delta Q}{T}=0$.

But, how can we see that a reversible cycle can be arbitrarily approximated by Carnot cycles (i.e. that ${\displaystyle\oint}\frac{\delta Q}{T}=\lim_{n\to\infty} \sum_{k=1}^n \frac{Q_{C,k,n}}{T_{C,k,n}}+\frac{Q_{H,k,n}}{T_{C,k,n}}$, if we use the notation I have introduced and if I have correctly understood the meaning of the approximation)?

  • $\begingroup$ Should the integrand be noted as $\frac{Q_{C,k,n}}{T_{C,k,n}}+\frac{Q_{H,k,n}}{T_{H,k,n}}$? $\endgroup$
    – hyportnex
    Aug 17, 2015 at 0:27

1 Answer 1


I am not familiar with the textbook you mentioned but there is a conventional "proof" of the equality ${\displaystyle\oint} _\textrm{rev}\frac{\delta Q}{T}=0$ for states that can be described by a thermal parameter, usually temperature $T$ and by a mechanical parameter, such as $V$. The "proof" goes by approximating the area enclosed by the $T,V$ cycle with Carnot cycles, that is consisting of isothermal-adiabatic-isothermal-adiabatic segments. The reason why I used the quotation marks around the word proof because this is not really a proof in a mathematical sense. There are many loose ends, among them the funny way of rectifying the cycle with isothermal pieces that do not join.

It was never obvious to me that the process converges. We have a lot more complicated case when there are several mechanical/deformation variables. The rectification of an arbitrary curve in higher dimensions with non-joining segments must be even more tenuous than in 2D. Note too that the existence of adiabatic surfaces that are explicitly postulated here is equivalent mathematically with the very existence of the entropy function that we wish to prove, so this not really a proof at all.

This whole problem can be side-stepped by a trick that goes back at least to Fermi (see his book), namely the use of an auxiliary Carnot cycle engine tied to a reservoir whose temperature $T_0$ is higher than any one in the system whose cycle you are interested in, and then the system exchanges heat only through this auxiliary Carnot engine at all temperatures. One now only postulates the existence of isothermal heat exchanges, not adiabatics. Using Kelvin's principle it follows that $ T_0 {\displaystyle\oint} \frac{\delta Q}{T} \le 0$ with $T_0 >T$, and from $T_0 >0$ we get ${\displaystyle\oint} \frac{\delta Q}{T}\le 0$ for all cycles. Now if the process is reversible then reversing it we get both ${\displaystyle\oint}_\textrm{rev} \frac{\delta Q}{T}\ge 0$ and ${\displaystyle\oint} _\textrm{rev}\frac{\delta Q}{T}\ge0$ so then ${\displaystyle\oint} _\textrm{rev}\frac{\delta Q}{T}=0$.

  • $\begingroup$ My book exactly gives the conventional proof that you talk about, where it isn't clear to me that the process converges. My book also says that, for an irreversible process the inequality is strict: $\oint\frac{\delta Q}{T}<0$: can we modify Fermi's proof to demonstrate that? I heartily thank you! $\endgroup$ Aug 18, 2015 at 10:46
  • $\begingroup$ not trying to avoid to answer your question but in practice any path $\mathcal C$ for which the contour integral $\oint_\mathcal C \frac{\delta Q}{T} < 0$ is irreversible, and one for which $\oint _{\mathcal C}\frac{\delta Q}{T} = 0$ is reversible. (Probably here we have to include the path $\mathcal C$ and its "immediate" neighborhood, as well.) $\endgroup$
    – hyportnex
    Aug 18, 2015 at 13:39
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    $\begingroup$ I don't know if you can "prove" this way that if the process is not reversible then $\oint_{\mathcal C} \frac{\delta Q}{T} < 0$, i.e., strict inequality. I am not even sure that it is "provable" at all, instead of being a separate postulate. If you want mathematical rigor then you need other formulations of thermodynamics. There are at least 3 that I know of, and are associated with the names of (1) Gibbs, (2) Caratheodory, (3) Coleman-Noll-Truesdell. It is questionable though that the rigor these bring leads to any better physics. $\endgroup$
    – hyportnex
    Aug 18, 2015 at 14:10
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    $\begingroup$ I have checked Bryan:Thermodynamics and it already has this proof of the Clausius inequality on page 60, section 73; the next section 74 has the funny integration... Bryan's book was published in 1907, thirty years before Fermi whose book I wholeheartedly recommend as being one of the clearest written classic on anything. Both books are on the internet free to all. $\endgroup$
    – hyportnex
    Aug 31, 2015 at 22:39
  • $\begingroup$ @hyportnex: Can Caratheodory's axiomatic approach be used to derive the OP's result? I've googled a lot but couldn't get such source where Caratheodory's approach has been used to prove it; instead all the books I'am accustomed to gave more or less the same 'conventional' proof. so, could you please advice on this? $\endgroup$
    – user36790
    Jul 5, 2016 at 19:02

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