Use Boltzmann factors to derive the exponential formula

Step 1 of 2

Consider a system with a single air molecule, let \(s_{1}\) be the state when the molecule at the sea level and \(s_{2}\) be the state when the molecule at height of z, assume that the energy is only potential energy therefore the difference in the energy between the states \(s_{1}\) and \(s_{2}\) is the potential energy which is \(\Delta E=m g z\), the ratio of \(s_{2}\) state probability to state \(s_{1}\) probability is:

\(\begin{array}{c}
\frac{\mathcal{P}\left(s_{2}\right)}{\mathcal{P}\left(s_{1}\right)}=\frac{e^{-E_{2} / k T}}{e^{-E_{1} / k T}}=e^{-\Delta E / k T} \\
\frac{\mathcal{P}\left(s_{2}\right)}{\mathcal{P}\left(s_{1}\right)}=e^{-m g z / k T}
\end{array}\)