As the title implies, I had some issues understanding how Clausius theorem

$$\int_{Closed}\frac{dQ}{T} = 0$$ can be derived for an arbitrary cycle starting by slicing it into multiple Carnot cycles. Since posting the question I managed to solve my misunderstanding starting from Roger Vadim's comment, but the question has been closed due to it not being related to physics according to the help center suggestions. For this reason I have edited the question since maybe others might have the same problem as me.

The reasoning I followed is from a physics book that I was going through and got to a part where starting from

$$\frac{Q_1}{T_1} - \frac{Q_2}{T_2} = 0$$

for $T_1 > T_2$, the authors say that we can rewrite this as

$$\sum \frac{Q}{T} = 0$$ keeping the sign convention for $Q$, then

$$\sum \frac{dQ}{T} = 0$$ for infinitely small cycles, thus an infinitely small quantity of heat are exchanged and finally

$$\int_{Closed}\frac{dQ}{T} = 0$$

on a reversible cycle that is decomposed in multiple infinitesimal Carnot cycles. This is the reasoning in the book.

What I couldn't manage to visualize is how this is done in a greater detail. For instance, assume that we have a circle like cycle in $pV$ and I want to apply the reasoning from the book. I assume that I have to slice the circle/disk using isotherms and near the borders of the cycle I should add adiabatic transitions from one isotherm to the next one in order to create the Carnot cycles. This should be like trying to represent the circle/disk on a discrete grid, but the pixels are long Carnots in-between very close isotherms.

Additionally I assume that I can split thos long Carnot cycles with multiple adiabatic curves so that inside the circular disk a small Carnot cycle takes its entire input heat from a Carnot engine placed above it in terms of temperature, and gives all of its output heat to another small Carnot cycle below it in terms of temperature. On the border however I assume that I get something like in the image below.

hand drawn schematic of the image

For a better understanding of what I tried to draw in the image, the circular cycle is drawn in the top right corner and then I zoom in on one region of it marked by the small rectangle. That "zoom" is redrawn in bottom left region where the arc $AB$ represents a small part of the cycle and with dotted lines I tried to fit a Carnot cycle such that the points $A$ and $B$ are at some well chosen positions on this small Carnot cycle.

Based on my choice of the region on the large cycle, heat should come through the upper isotherm $dQ_{isoT}$, and some of it $dQ_{AB}$ should pass through $AB$ in order to get to the lower isotherm. In general $dQ_{isoT} \neq dQ_{AB}$.

I assume that the $dQ$ from the integral should be the heat that gets into/out of the system, namely $dQ_{AB}$ but based on the reasoning of getting to the integral the heat should be from a Carnot cycle, namely $dQ_{isoT}$.

So either $dQ_{isoT}=dQ_{AB}$ for reasons that I cannot see (and I refer to mathematically getting this approximation) or there is something else that I missed entirely?

EDIT: The solution to my misunderstanding consists of two parts, one solved by Roger Vadim's answer, and the second one where I assumed that I have to slice the arbitrary cycle with isotherms and approximate the edges as being adiabatic, when I actually I had to slice the cycle with adiabatic curves and assume that the heat transfer is done at near constant temperature.

  • 1
    $\begingroup$ I didn't fully follow your reasoning, but the idea is to fill the cycle with Carnot cycles in the same way as it is filled with squares of the chequered paper in your image. Except that instead of straight lines we now have isotherms and adiabatas. Perhaps googling up figures of Curvilinear coordinates might help $\endgroup$
    – Roger V.
    Jun 23, 2022 at 8:48
  • $\begingroup$ Ok, but if that is so, then how do we take into account the cycles that are near the border of the cycle where the Carnot cycles cannot be used to describe the shape of the curve without having errors? Do we just ignore them in a hand-wavy manner cause physics or is there some mathematical reasoning that can be applied here? $\endgroup$ Jun 23, 2022 at 9:52
  • $\begingroup$ There is an error, of course, but it becomes smaller and smaller as the cycles become smaller, and ultimately vanish. If you had a good course on differential calculus, this should not appear particularly striking... though generalizations to curvilinear coordinates are usually studied elsewhere (in advanced elasticity or relativity courses). $\endgroup$
    – Roger V.
    Jun 23, 2022 at 9:57
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    $\begingroup$ I think my main issue was related to the curvilinear coordinates because I find it harder to accept that the surface can be split in this types of shapes when compared with rectangles. I think I got it now. $\endgroup$ Jun 23, 2022 at 10:00
  • $\begingroup$ What does all this have to do with thermodynamics? $\endgroup$
    – Bob D
    Jun 23, 2022 at 18:33


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