# How to prove Caratheodory's Concept of temperature

Lenker T D . Caratheodory's concept of temperature[J]. Synthese, 1979

This is a discussion in the article：

let $$S_i$$ be the state space of the thermodynamic system $$\sum i = 1, 2, 3$$, and assume there exists a set of sufficiently smooth relations $$F_{ij}: S_i×S_j→R$$ such that the systems $$∑ i$$ and $$∑ j$$ are in thermal equilibrium while in states $$s_i$$ and $$s_j$$ respectively if and only if $$F_{ij}(s_i, s_j)=0$$ for $$1≤i$$, $$j≤3$$ and $$i≠j$$. These relations satisfy the conditions:

(1.1) $$F_{ij}(s_i, si_j)=0$$ if and only if $$F_{ji}(s_j, s_i)= 0$$.

(1.2) If $$F_{ij}(s_i, s_j)=0$$ and $$F_{jk}(s_j, s_k) = 0$$, then $$F_{ik}(s_i, s_k)=0$$.

Does the existence of these relations imply that there are continuous functions

$$t_i$$:$$s_i→R, i = 1, 2, 3$$ such that $$t_1(s_1)=t_2(s_2)=t_3(s_3)$$ exactly when

$$F_{12}(S_1, s_2) = F_{13}(s_l, s_3) = F_{23}(S_2, s_3) = 0$$?

The following example due to Whaples [8] shows that without more restrictive assumptions, such functions need not exist.

EXAMPLE: Let $$s_1 = s_2 = s_3 = R$$ and let $$F_{ij}(s_i, s_j) = sin [π(s_i - s_j)]$$ for $$i, j = 1, 2, 3$$.

Assume there exist continuous functions

$$t_i$$: $$Si→R$$, $$i = 1, 2, 3$$ such that $$t_1(s_1)=t_2(s_2)=t_3(s_3)$$ if and only if

$$F_{12}(s_1, s_2)= F_{13}(s_1,s_3) = F_{23}(s_2, s_3)=0$$.

Thus $$t_1(s_1)=t_2(s_2)=t_3(s_3)$$ if and only if

$$s_1 ≡s_2 mod 1 ≡ s_3 mod 1$$.

Consider the triples $$(0, 2, 3)$$ and $$(1, 2, 3)$$ belonging to $$s_1 × s_2 × s_3$$. This implies $$t_1(0) = t_1(1)$$ and $$t_1(s) ≠ t_1(0)$$ for $$s ∈ (0, 1)$$. Thus ti is not continuous.

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I have some questions:

①why choose $$F_{ij}: S_i×S_j→R$$ as a smooth relations,is multiplication has some special meaning?

②why $$t_1(S_1)= t_2(s_2)= t_3(s_3)$$ if and only if $$s_1 ≡s_2 mod 1 ≡ s_3 mod 1$$ but not $$s_1 ≡s_2 ≡ s_3$$

③why the triples $$(0, 2, 3)$$ and $$(1, 2, 3)$$ belonging to $$s_1 × s_2 × s_3$$,What is the role of these triples?

• We do have MathJax here that makes your equations look better. You can see the notation page in help center for details, if you aren't familiar with it. – Kyle Kanos Apr 3 at 17:00
• Thank you, your advice help me a lot. – 地山谦 Apr 7 at 9:56