Derivation of Boltzmann distribution for 1-dimensional potential

p(h):pressure n(h):number of particles in unit volume F(h):force applied to the each particle U(h):potential energy of the particle From equation of state, we obtain: p(h)V=NRT→p(h)=n(h)kbT Considering equilibrium of forces in the gas between h and h+dh, we obtain: Sp(h+dh)+n(h)F(h)Sdh=p(h)S→p(h+dh)+n(h)F(h)dh=p(h)→p(h+dh)−p(h)dh=−n(h)F(h)→dpdh=−n(h)F(h) By substituting (1) to (2),we can derive differential equation (3). kbTdndh=−n(h)F(h) ODE (3) can be solved by dividing both sides by n. 1ndndh=−F(h)kbT Integrating both sides by h results in following. ∫1ndn=−1kbT∫F(h)dh \begin{equation} \leftrightarrow \int \frac{1}{n}dn=-\frac{1}{k_b...