# Boltzmann Distribution

## 1 Definition

The Boltzmann distribution gives the relative fraction of atoms in two states in thermal equilibrium at a certain temperature, taking into account the degeneracies of these states and the energy difference between states. This distribution applies to large ensembles of atoms such that statistical arguments are valid. The formula for the Boltzmann distribution is given by the following:

${\displaystyle {\frac {n_{2}}{n_{1}}}={\frac {g_{2}}{g_{1}}}e^{-{\frac {\Delta E}{kT}}}\,\!}$

where ${\displaystyle n_{1},n_{2}}$ are the number (or number density) of atoms in energy states 1 and 2, respectively, ${\displaystyle g_{1},g_{2}}$ are the respective degeneracies of those energy states (i.e., how many distinct configurations of the atom have that same energy state), ${\displaystyle \Delta E=E_{2}-E_{1}}$ is the energy difference between the two states, ${\displaystyle k}$ is Boltzmann’s constant, and ${\displaystyle T}$ is the temperature describing the distribution of states in the system.

Note the distinction between the Boltzmann distribution, which describes the distribution of energy states in the system at a given temperature, and the Maxwell-Boltzmann distribution, which instead describes a distribution of velocities at a given temperature.

The Boltzmann distribution tells us which transitions to expect from a population, since, for a transition to occur, we must begin with particles in the starting state such that they may transition to the end state. Together with the Saha equation, it can also tell us the ratio of element abundances in stellar astrophysics, allowing for the interpretation of stellar spectra.

## 2 Derivation

Consider an isolated statistical ensemble of N particles with energies given by

${\displaystyle E=\sum \epsilon _{i}n_{i}\,\!}$

and total number of particles

${\displaystyle N=\sum n_{i}.\,\!}$

The N particles are arranged into microstates given by

${\displaystyle W=(g_{1}^{n_{1}}g_{2}^{n_{2}}g_{3}^{n_{3}}...){\frac {N!}{n_{1}!n_{2}!n_{3}!...}}.\,\!}$

Here, W is the probability of a given distribution. To find the distribution of particles in our system, we will need to find the most probable microstate, or the distribution that maximizes W. To accomplish this, we will equivalently find a maximum in lnW.

Solving for lnW from our previous expression, we get

${\displaystyle lnW=n_{1}lng_{1}+n_{2}lng_{2}+...+lnN!-n_{1}lnn_{1}-n_{2}lnn_{2}-...+n_{1}+n_{2}+...\,\!}$
${\displaystyle lnW=n_{1}(1+ln{\frac {g_{1}}{n_{1}}})+n_{2}(1+ln{\frac {g_{2}}{n_{2}}})+...+lnN!\,\!}$

where we implemented Stirling’s approximation for large N, ${\displaystyle lnN!\approx NlnN-N}$, to simplify our expression. Using Lagrange’s method of undetermined multipliers, we seek a maximum in

${\displaystyle lnW-\alpha \sum n_{i}-\beta \sum \epsilon _{i}n_{i},\,\!}$

where ${\displaystyle \alpha }$ and ${\displaystyle \beta }$ are constant values. A maximum in this expression is equivalent to a maximum in lnW or in W because the second and third terms consist of only constant values, where we have defined our total energy and our total number of particles to be constant (recall that we are considering an isolated ensemble). Thus, we solve for the maximum value of this expression with respect to ${\displaystyle n_{i}}$ to find the most probable microstate.

${\displaystyle {\frac {d}{dn_{i}}}(lnW-\alpha \sum n_{i}-\beta \sum \epsilon _{i}n_{i})={\frac {dlnW}{dn_{i}}}-\alpha -\beta \epsilon _{i}=0\,\!}$

Recalling our equation for lnW from earlier,

${\displaystyle {\frac {dlnW}{dn_{i}}}={\frac {d}{dn_{i}}}(n_{i}+n_{i}lng_{i}-n_{i}lnn_{i})\,\!}$
${\displaystyle {\frac {dlnW}{dn_{i}}}=1+lng_{i}-lnn_{i}-n_{i}{\frac {1}{n_{i}}}\,\!}$
${\displaystyle {\frac {dlnW}{dn_{i}}}=ln{\Big (}{\frac {g_{i}}{n_{i}}}{\Big )}\,\!}$

Plugging this in to our maximum probability expression, we then get

${\displaystyle ln{\Big (}{\frac {g_{i}}{n_{i}}}{\Big )}-\alpha -\beta \epsilon _{i}=0\,\!}$
${\displaystyle {\frac {g_{i}}{n_{i}}}e^{-\alpha }e^{-\beta \epsilon _{i}}=1\,\!}$
${\displaystyle n_{i}=g_{i}e^{-\alpha }e^{-\beta \epsilon _{i}}\,\!}$

Finally, taking a ratio between two energy states, we get

${\displaystyle {\frac {n_{i}}{n_{j}}}={\frac {g_{i}}{g_{j}}}e^{-\beta (\epsilon _{i}-\epsilon _{j})},\,\!}$

which is a generalized form of the Boltzmann equation. Here, ${\displaystyle \beta ={\frac {1}{kT}}}$. The ${\displaystyle \alpha }$ term in our derivation, though it dropped out of our Boltzmann distribution, is given by ${\displaystyle \alpha ={\frac {\mu }{kT}}}$, where ${\displaystyle \mu }$ is the chemical potential.

## 3 Degeneracies of the hydrogen atom

The energy degeneracies corresponding to each E_n for the hydrogen atom are given by:

${\displaystyle g_{n}=\sum _{l=0}^{n-1}(2l+1)=n^{2}\,\!}$

Taking the two electron spin states and the two proton spin states of hydrogen into account, our degeneracies then ultimately become

${\displaystyle g_{n}=4n^{2}.\,\!}$

Using this, we can then show the ratios of excited states of hydrogen to the ground state as a function of temperature:

Plot of the ratios of hydrogen excited states to the ground state as a function of temperature