# Reversible cycle approximated by Carnot cycles

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)?

• Should the integrand be noted as $\frac{Q_{C,k,n}}{T_{C,k,n}}+\frac{Q_{H,k,n}}{T_{H,k,n}}$? – hyportnex Aug 17 '15 at 0:27

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.
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 ${\displaystyle\oint} _\textrm{rev}\frac{\delta Q}{T}=0$.
• 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! – Self-teaching worker Aug 18 '15 at 10:46
• 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.) – hyportnex Aug 18 '15 at 13:39
• 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. – hyportnex Aug 18 '15 at 14:10