In my professors script for Thermodynamics stated that for different systems there can only be one global affine transformation the reasoning is the following:
The additivity of the entropy of different systems strongly constrains 𝑆(𝑋).
As 𝑆(𝑋) encodes an order, any$$\tilde{𝑆}(𝑋) := 𝛼𝑆(𝑋) +𝛽, 𝛼 > 0$$ is suitable as well for an individual system.
Introducing different 𝛼, 𝛽 for different systems, where$$ 𝑆(𝑋) ≤ 𝑆(𝑋′), 𝑆(𝑌) ≤ 𝑆(𝑌′) \text{ therefore } 𝑆(𝑋)+𝑆(𝑌) ≤ 𝑆(𝑋′)+𝑆(𝑌′)$$would allow for$$\tilde{𝑆}(𝑋) ≤ \tilde{𝑆}(𝑋′), \tilde{𝑆}(𝑌) ≤ \tilde{𝑆}(𝑌'), \text{ but } \tilde{𝑆}(𝑋)+\tilde{𝑆}(𝑌) > \tilde{𝑆}(𝑋′)+\tilde{𝑆}(𝑌′)$$violating eq (6.15) which is:$$ 𝑋 ⪯ 𝑋′ ⇔ 𝑆(𝑋) ≤ 𝑆(𝑋′)$$[find suitable numbers to show this (was ex to the reader)].
Additivity therefore implies the global calibration of entropy across all types of systems:
There can be only one global affine transformation 𝑆(𝑋) → 𝑆˜(𝑋) := 𝛼𝑆(𝑋) + 𝛽, 𝛼 >$0$.
𝛽 will be fixed by the third law of thermodynamics; 𝛼 fixes the unit of entropy.
I am pretty confused because lets say we chose different affine transformations for S(X) and S(Y) such that:
𝑆˜(𝑋) :=$\alpha _1 $S(X) +$\beta _1$ and 𝑆˜(𝑌) :=$\alpha _2 $S(Y) +$\beta _2$ with$\alpha _1 ,\alpha _2 >0$
than:
\begin{align}\tilde S(X)+\tilde S(Y) &> \tilde S(X')+\tilde S(Y') \\[6pt]\Longleftrightarrow\quad\alpha_1 S(X)+\beta_1+\alpha_2 S(Y)+\beta_2&>\alpha_1 \tilde S(X')+\beta_1+\alpha_2 \tilde S(Y')+\beta_2 \\[6pt]\Longleftrightarrow\quad\alpha_1\bigl(S(X)-S(X')\bigr)&>\alpha_2\bigl(S(Y')-S(Y)\bigr)\\[6pt]\end{align}which is contradicted by:$S(X)\le S(X'),\qquad S(Y)\le S(Y'),\qquad \alpha_1,\alpha_2>0.$
- $\begingroup$Please use Mathjax properly.$\endgroup$Tobias Fünke– Tobias Fünke2025-11-14 11:18:10 +00:00CommentedNov 14 at 11:18
- $\begingroup$Is there a tutorial on that somewhere this is my first question?$\endgroup$ColBi– ColBi2025-11-14 11:21:59 +00:00CommentedNov 14 at 11:21
- 2$\begingroup$Hi, sorry I forgot to link:MathJax Tutorial.$\endgroup$Tobias Fünke– Tobias Fünke2025-11-14 11:24:53 +00:00CommentedNov 14 at 11:24
- $\begingroup$You are getting the logical inference wrong. You started out with the last-most line, and you want to derive a contradiction. The way to do it is as your professor did it. You, however, went around in a loop. No wonder you get nowhere.$\endgroup$naturallyInconsistent– naturallyInconsistent2025-11-14 18:32:56 +00:00CommentedNov 14 at 18:32