Why is $\dfrac{1}{T}$ chosen as integrating factor for heat differential $\delta Q$ in Caratheodory's formulation of the 2nd law of thermodynamics? It's clear to me why it is that way in Clausius approach but I am not exactly sure how mathematics imposes $\dfrac{1}{T}$ instead of just $T$, for example.
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asked Oct 13 at 16:34
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1$\begingroup$ T. Frankel speaks briefly about this in his book The Geometry of Physics on p. 187 (which contains a very good exposition of Charatheodory's theory) but ultimately refers to the paper Boyling, J.B. An axiomatic approach to classical thermodynamics, Proc. R. Soc., London, A 329 (1972), pp. 35–70. $\endgroup$– Kurt G.Commented Oct 13 at 17:08
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2$\begingroup$ If you define the integrating factor $\theta = \frac{1}{T}$ as a multiplier and not as a divider you get what is called a coldness scale not a hotness scale. That we do not use coldness is a historical accident in the sense that in everyday speech we associate higher temperature is what we call warmer (having more entropy). "Coldness" would be mathematically better in the sense that it is more natural to say that infinite coldness is now inaccessible rather than the absolute zero. $\endgroup$– hyportnexCommented Oct 13 at 17:21
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1$\begingroup$ See Transport Phenomena, Bird et al, Chapter 11, Homework Problem 11.d.1. This shows that, when one takes the overall energy balance equation, subtracts the mechanical energy balance equation, and divides by T, you obtain an equation for the entropy variation with time and station position which reduces to a state junction when viscous dissipation and entropy generation from heat conduction are negligible. $\endgroup$– Chet MillerCommented Oct 13 at 20:49
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2$\begingroup$ At that level, it's just a conventional choice, made to be compatible with preexisting definitions. $\endgroup$– knzhouCommented Oct 15 at 17:34
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$\begingroup$ The axiomatic paper by Boyling I mentioned the other day works entirely in Caratheodory's theory and devotes an entire section to derive $dS=\delta Q/T$ from other principles. This paper can be read freely online in jstor.org. Therefore I am wondering why you put up that 500pt bounty. I guess you should perhaps ask a more specific question. $\endgroup$– Kurt G.Commented Oct 18 at 10:01
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2 Answers
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If the expression : $$pdx+qdy\;\;\;\;\;\;\;\;\;\;\;\;\;(1)$$ is not a total differential, i.e.$$pdx+qdy=0\;\;\;\;\;\;\;\;(2)$$ equation (2) always has, according to the existing theorem, a general integral which is written in the form:$$f(x,y)=C$$ the function $f(x,y)$ must verify the relationship $$\frac{\partial f(x,y)}{\partial x}+\frac{\partial f(x,y)}{ \partial y}\frac{dy}{dx}=0$$ where, according to (2), we must replace$\frac{dy}{dx}$ by $-\frac{p}{q}$, that is to say we must have identically $$\frac{\frac{\partial f}{\partial x}}{p}=\frac{\frac{\partial f}{\partial y}}{q}$$
let $\mu$ be the common value of these two ratios, we have $$\frac{\partial f}{\partial x}=\mu p\;,\;\frac{\partial f}{ \partial y}=\mu q$$ that is to say that $\mu$ is the integrating factor of expression (1).
This reasoning shows that every expression $pdx+qdy$ has an integrating factor.
For the function $\mu$ to be an integrating factor of (1), it is necessary and sufficient that we have the relation (Schwarz's theorem): $$\frac{\partial (\mu p)}{\partial y}=\frac{\partial (\mu q)}{ \partial x}\;\;\;\;\;\;\;\;\;\;\;\;\;(3)$$
For one mole of ideal gas, we have the following equations $$PV=RT\;\;,\;\; \delta Q=C_{v}dT+PdV$$ according to these relations, there exists a state function $ S=C $ such that $$ dS=\mu \delta Q $$ $$dS=\mu \left(C_{v}dT+PdV \right)=\mu \left(C_{v}dT+\frac{RT}{V}dV \right)$$
if we assume $\mu=\frac{1}{T}$ the previous equation becomes $$ dS=\frac{C_{v}}{T}dT+\frac{R}{V}dV$$ and condition (3) is verified, i.e. $$\frac{\partial( \frac{C_v}{T})}{\partial V} \Bigg\vert_{T} =\frac{\partial( \frac{R}{V})}{\partial T} \Bigg\vert_{V} $$ $$\frac{1}{T}\frac{\partial C_v}{\partial V} \Bigg\vert_{T} =\frac{1}{V}\frac{\partial R}{\partial T} \Bigg\vert_{V}=0$$ so $$ dS= \frac{ \delta Q}{T} $$
ps: I've never heard of this formulation, but I tried :-)
-The mathematical part see: higher mathematics volume II by V.Smirnov.
(he introduced the relation (3) in the form of Green's formula: $\int \int_{\sigma}\left(\frac{\partial Q }{ \partial x }-\frac{\partial P}{dy}\right)d{\sigma}= \int_{l}Pdx+Qdy=0 =>\frac{\partial P}{dy}=\frac{\partial Q}{ \partial x }\;$ in our case ,$P=\mu p\; , Q=\mu q$ )
-The thermodynamic part (formulas) see (in French) Thermodynamics solved problems Higher and special mathematics 1st university cycle - Hubert Lumbroso.
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$\begingroup$ condition (3) is the same as proposition (8) of this document. math.univ-lyon1.fr/~pujo/COURS-EDO-2013.pdf or the definition of the integrating factor (definition (19 )) $\endgroup$ Commented Oct 20 at 8:24
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$\begingroup$ Resolution of the Pfaff equation $ Pdx+Qdy+Rdz=0$ see: Theory of ordinary differential equations and integral equations, I.Pétrovsky or a little pushed by Élie Cartan: neo-classical-physics.info/uploads/3/4/3/6/34363841/… $\endgroup$ Commented Oct 20 at 8:32
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$\begingroup$ Errata: definition (19) of the document: $\frac{ \partial \mu a}{\partial x}=\frac{ \partial \mu b}{\partial t}$ same form as theorem (4). $\endgroup$ Commented Oct 20 at 8:50
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You can denote the factor as $\beta$, and express everything just as well. However, by comparing the mathematical results with thermodynamics, we can find that integrating factor $\beta$ has to be inversely proportional to ideal gas temperature, or absolute temperature $T$. So sometimes it is denoted $1/T$ right from the start, anticipating the result and avoiding introducing a quantity (inverse temperature) which does not appear in thermodynamics of Clausius and Kelvin (they used temperature $T$).
