# Boltzmann factor

In physics, the Boltzmann factor is a weighting factor that determines the relative probability of a particle to be in a state $i$ in a multi-state system in thermodynamic equilibrium at temperature $T$. The Boltzmann factor is normally used when discussing systems described by the canonical ensemble - for the grand canonical ensemble, it is more appropriate to use the Gibbs' factor which takes into account particle transfer between the system and the environment.

The probability that a system is in a state $E_i$ is given by:

$P(E_i) = \frac{1}{Z}\exp{(-\beta E_i \,)}$

Where $\beta$ is given by

$\beta = \frac{1}{k_BT}$

$Z_{}$ is the partition function, $k_B$ is Boltzmann constant, $T$ is temperature and $E_i$ is the energy of state $i$.

The Boltzmann factor is the term:

$\exp{(-\beta E_i \,)} = \exp{\left (\frac{-E_i}{k_BT} \right )}$

## Derivation

Consider a single atom system with energy states $E_1, E_2, ...$. The system is in contact with a heat reservoir and thus conservation of energy gives total energy

$E = E_s + E_r = const$

with $E_s$ being total system energy and $E_r$ being total reservoir energy. In equilibrium, the number of states in $R$ and $S$ is the multiplicity $\Omega$. Similar to total energy, we can write:

$\Omega_E \approx \Omega_R(E_R) \Omega_S(E_S)$

The probability that an atom is in state $E_j$ is related to the number of states in the reservoir - from Equipartition theorem. Consider taking the ratio of two probabilities:

$\frac{P(E_2)}{P(E_1)} = \frac{\Omega_R(E-E_2)}{\Omega_R(E-E_1)}$

We can relate the number of states to the entropy via

$S_R(E_j) = k_B \ln[\Omega_R(E-E_j)]$

giving:

$\frac{P(E_2)}{P(E_1)} = \frac{\exp{[\frac{S_R(E_2)}{k_B}]}}{\exp{[\frac{S_R(E_1)}{k_B}]}} = \exp\left[ \frac{S_R(E_2) - S_R(E_1)}{k_B}\right]$

The Fundamental thermodynamic relation tells us that for the reservoir (rearranging and neglecting the chemical potential term):

$dS_R = \frac{1}{T}[dU_R + PdV_R]$

Where $S_R$ is the entropy, $U_R$ is the internal energy, $P$ is pressure and $V$ is volume.

For a gas, it is reasonable to assume that $PdV_R \ll dU_R$ so this term also drops out giving:

$\Delta S_R = \frac{1}{T} \Delta U_R$
$\Delta S_R = \frac{1}{T}[U_R(E_2)-U_R(E_1)]$

By energy conservation $E = E_R + E_i$ and $U_R(E_j) = E-E_j$ giving

$\Delta S_R = -\frac{1}{T} (E_1-E_2)$

Substituting into the probability ratio gives:

$\frac{P(E_2)}{P(E_1)} = \exp\left(\frac{-[E_2-E_1]}{k_B T}\right) = \frac{\exp{(-\beta E_2)}}{\exp{(-\beta E_1)}}$

Where we have defined an arbitrary symbol $\beta$, which is the reciprocal of Boltzmann's constant times temperature. Through separation of variables, we can write

$\frac{P(E_2)}{\exp{(-\beta E_2)}} = \frac{P(E_1)}{\exp{(-\beta E_1)}} = const = \frac{1}{Z}$

and hence

$P(E_i) = \frac{1}{Z}\exp{(-\beta E_i)}$

## Notes

The Boltzmann factor is not a probability by itself, because it is not normalized. The normalization factor is one divided by the partition function, the sum of the Boltzmann factors for all states of the system. This gives the Boltzmann distribution.

From the Boltzmann factor it is possible to derive the Maxwell-Boltzmann statistics, Bose-Einstein statistics, and Fermi-Dirac statistics that govern classical particles as well as quantum mechanical bosons, and fermions, respectively.