## Appendix A Boltzmann Equation  
[PERSON] uses statistical mechanics [25; 28] to develop the path between the microscopic and macroscopic scale. To account for the micro-state, in [PERSON]'s statistics, the particles must be identic and distinguishable [29]. In this case, this assumption necessary for atoms or molecules. For all the microstates [27; 29]:  
\[Q\approx\frac{N!}{\prod_{i=0}^{I}n_{i}!} \tag{1}\]  
Here, \(Q\) represents the number of configurations, \(N\) is the total number of molecules and \(n_{i}\) the number of molecules in the energetic state \(i\). With [PERSON]'s approximation (\(lnx!\approx xlnx-x\)), the solution to Equation (1) gives (by using \(\sum\limits_{i}n_{i}=N\)):  
\[\ln(Q)=N\ln(N)-\sum\limits_{i}n_{i}\ln(n_{i}) \tag{2}\]  
The domination configuration is when \(Q\) is at a maxim, \(d\ln Q=0\), then, when \(Q\) is a maximum:  
\[d\ln(Q)=\sum\limits_{i}\frac{\partial\ln(Q)}{\partial n_{i}}dn_{i}=0 \tag{3}\]  
To solve Equation (3), we need to use the Lagrange multiplier and the two further constraints. The first concerns energy (\(U=\sum\limits_{i=0}^{I}E_{i}n_{i}\) see Equation (3)) as expressed by \(\sum\limits_{i}E_{i}dn_{i}=0\) and the second concerns species (\(N=\sum\limits_{i}n_{i}\) see Equation (4)) as expressed by \(\sum\limits_{i}dn_{i}=0\). With the Equation (3) and the constraints, the Lagrange multiplier gives:  
\[\sum\limits_{i}\frac{\partial\ln(Q)}{\partial n_{i}}dn_{i}+\alpha\sum \limits_{i}dn_{i}-\beta\sum\limits_{i}E_{i}dn_{i}=0 \tag{4}\]  
where \(\alpha\) and \(\beta\) are constant. All the \(dn_{i}\) are treated as independent. Then, to satisfy \(d\ln Q=0\), the following equation is obtained:  
\[\frac{\partial\ln(Q)}{\partial n_{i}}+\alpha-\beta E_{i}=0 \tag{5}\]The first term of this equation can be transformed by considering the Equation (A2) under the following form:  
\[\frac{\partial\ln(Q)}{\partial n_{i}}=\frac{\partial N\ln(N)}{\partial n_{i}}- \sum\limits_{j}\frac{\partial n_{j}\ln(n_{j})}{\partial n_{i}}\] (A6)  
In this equation, \(\frac{\partial N\ln N}{\partial n_{i}}\) can be derived and gives \(\frac{\partial N\ln N}{\partial n_{i}}=\ln(N)+1\) and \(\sum\limits_{j}\frac{\partial n_{j}\ln(n_{j})}{\partial n_{i}}\) can be derived and gives to \(\sum\limits_{j}\frac{\partial n_{j}\ln(n_{j})}{\partial n_{i}}=\ln(n_{i})+1\). To do this derivation, the following equation is needed: \(\frac{\partial N}{\partial n_{i}}=\frac{\partial\sum\limits_{i}n_{i}}{ \partial n_{i}}=1\). Then, the Equation (A6) gives:  
\[\frac{\partial\ln(Q)}{\partial n_{i}}=-\ln\Bigl{(}\frac{n_{i}}{N}\Bigr{)}\] (A7)  
Finally, with the Equation (A7), the Equation (A5) gives:  
\[-\ln\Bigl{(}\frac{n_{i}}{N}\Bigr{)}+\alpha-\beta E_{i}=0\] (A8)  
Therefore:  
\[\frac{n_{i}}{N}=e^{\alpha}e^{-\beta E_{i}}\] (A9)  
The determination of \(\alpha\) gives (with Equation (A9) and \(N=\sum\limits_{i}n_{i}\)):  
\[e^{\alpha}=\frac{1}{\sum_{i}e^{-\beta E_{i}}}\] (A10)  
With the Equations (A9) and (A10):  
\[\ln(Q)=N\ln(N)-\sum\limits_{i}n_{i}\ln(n_{i}) \tag{2}\]0 (A11)  
\(Z\) is the partition function. \[\ln(Q)=N\ln(N)-\sum\limits_{i}n_{i}\ln(n_{i}) \tag{2}\]1 (A12)  
To find \(\beta\), the Equation (A11) is injected into the Equation (A2) to give:  
\[\ln(Q)=N\ln(N)-\sum\limits_{i}n_{i}\ln(n_{i}) \tag{2}\]2 (A13)  
The development of the Equation (A13) gives:  
\[\ln(Q)=N\ln(N)-\sum\limits_{i}n_{i}\ln(n_{i}) \tag{2}\]3 (A14)  
With Equation (3), \(U=\sum\limits_{i=0}^{I}E_{i}n_{i}=\sum\limits_{i=0}^{I}E_{i}\frac{N}{Z}e^{- \beta E_{i}}\) and Equation (4), \(N=\sum\limits_{i}n_{i}=\sum\limits_{i}\frac{N}{Z}e^{-\beta E_{i}}\). Then, the Equation (A14) gives:  
\[\ln(Q)=N\ln(N)-\sum\limits_{i}n_{i}\ln(n_{i}) \tag{2}\]4 (A15)  
Which allows determining \(\beta\) by using Equation (A15) (\(N\ln Z\) is a constant).