# What are the Boltzmann factors?

Sep 14, 2017

A general ratio of the population of states can be written in statistical mechanics as:

${N}_{i} / N = \frac{{g}_{i} {e}^{- \beta {\epsilon}_{i}}}{q} = \frac{{g}_{i} {e}^{- \beta {\epsilon}_{i}}}{{\sum}_{i} {g}_{i} {e}^{- \beta {\epsilon}_{i}}}$

where:

• ${g}_{i}$ is the degeneracy of state $i$ with energy ${\epsilon}_{i}$.
• $\beta = \frac{1}{{k}_{B} T}$ is a constant containing the Boltzmann constant and temperature.
• ${N}_{i}$ is the number of particles in state $i$ and $N$ is the total number of particles.

If we then consider a single state relative to energy zero, we have two states such that:

${N}_{1} / {N}_{0} = {N}_{1} / N \cdot \frac{N}{N} _ 0$

$= \frac{{g}_{1} {e}^{- \beta {\epsilon}_{1}}}{\cancel{{g}_{0} {e}^{- \beta {\epsilon}_{0}} + {g}_{1} {e}^{- \beta {\epsilon}_{1}}}} \cdot \frac{\cancel{{g}_{0} {e}^{- \beta {\epsilon}_{0}} + {g}_{1} {e}^{- \beta {\epsilon}_{1}}}}{{g}_{0} {e}^{- \beta {\epsilon}_{0}}}$

Since the $N$'s cancel out, you can see that this can be extended to any number of states. If we then let energy zero be ${\epsilon}_{0} = 0$, then:

$\frac{{N}_{i}}{{N}_{0}} = \frac{{g}_{i}}{{g}_{0}} {e}^{- \beta {\epsilon}_{i}}$

Thus, the population of state $\boldsymbol{i}$ with some energy higher than energy zero is given by ${e}^{- \beta {\epsilon}_{i}} = {e}^{- {\epsilon}_{1} / {k}_{B} T}$, weighted by the ratio of the degeneracies ${g}_{i}$ and ${g}_{0}$.

We call ${e}^{- {\epsilon}_{i} / {k}_{B} T}$ the Boltzmann factor for state $i$.