Boltzmann Distribution · aiwiki.page

The Boltzmann distribution is a probability distribution that describes how a physical system occupies its possible microscopic states at thermal equilibrium. Each state receives a statistical weight that decreases exponentially with its energy at positive temperature. A foundational result of statistical mechanics, it connects microscopic states with macroscopic thermal behavior and supplies the equilibrium probabilities for the canonical ensemble. (damtp.cam.ac.uk)

Mathematical definition

For discrete states indexed by $i$ , with energies $E_i$ , the probability of state $i$ is

p_i=\frac{e^{-\beta E_i}}{Z}, \qquad \beta=\frac{1}{k_{\mathrm B}T}, \qquad Z=\sum_j e^{-\beta E_j}.

Here $k_{\mathrm B}$ is the Boltzmann constant, $T$ is absolute temperature, and $Z$ is the partition function. The numerator, called the Boltzmann factor, is an exponential function of energy. Division by $Z$ ensures that the probabilities sum to one; the distribution requires this normalization to be finite. (damtp.cam.ac.uk)

For two individual states,

\frac{p_i}{p_j} =\exp\!\left[-\frac{E_i-E_j}{k_{\mathrm B}T}\right].

Thus only energy differences determine their relative probabilities. Adding the same constant to every energy leaves the distribution unchanged. The quantity $k_{\mathrm B}T$ provides the relevant thermal energy scale: an energy separation much larger than this scale strongly suppresses the higher state. These consequences follow directly from the canonical probability formula. (damtp.cam.ac.uk)

Physical origin and assumptions

The standard derivation considers a system exchanging energy with a much larger thermal reservoir while its volume and particle number remain fixed. The combined system and reservoir are isolated, with total energy $E_{\mathrm{tot}}$ . Assuming equally probable accessible microstates of the combined object, the probability of system state $i$ is proportional to the reservoir’s number of states at energy $E_{\mathrm{tot}}-E_i$ . (interactivetextbooks.tudelft.nl)

Writing this multiplicity in terms of reservoir entropy gives

p_i\propto \exp\!\left[\frac{S_R(E_{\mathrm{tot}}-E_i)} {k_{\mathrm B}}\right].

For a sufficiently large reservoir, its temperature changes negligibly as energy passes between the two parts. Expanding its entropy to first order,

S_R(E_{\mathrm{tot}}-E_i) \simeq S_R(E_{\mathrm{tot}})-\frac{E_i}{T},

produces the Boltzmann factor after normalization. The derivation therefore depends on equilibrium and on a reservoir whose temperature is effectively fixed. It describes equilibrium occupancy, rather than the rate at which equilibrium is reached. (interactivetextbooks.tudelft.nl)

States, energy levels, and degeneracy

An individual microstate must be distinguished from an energy level containing several states. If level $a$ has energy $\varepsilon_a$ and degeneracy $g_a$ , its total probability is

P_a= \frac{g_a e^{-\beta\varepsilon_a}} {\sum_b g_b e^{-\beta\varepsilon_b}}.

The degeneracy counts independent states sharing that energy. A highly degenerate level can consequently have a larger total population than a less degenerate lower level, even though each of its individual states has a smaller probability. This distinction matters when interpreting thermal populations in spectroscopy. (physchem.no)

As an illustrative consequence, two nondegenerate levels with energies $0$ and $\Delta>0$ have upper-level probability

P_{\mathrm{upper}}=\frac{1}{1+e^{\Delta/(k_{\mathrm B}T)}}.

It approaches zero as $T\to0^+$ and one-half as $T\to\infty$ . These limits describe this finite two-level model, not a general uniform distribution over an unlimited number of states. (physchem.no)

Thermodynamic quantities and fluctuations

The partition function connects state probabilities with thermodynamics. The expected value of energy, identified with the system’s internal energy, is

U=\sum_i p_iE_i =-\frac{\partial\ln Z}{\partial\beta}.

Its Helmholtz free energy is

F=-k_{\mathrm B}T\ln Z.

Thus $Z$ is more than a normalization constant: its dependence on temperature and other system parameters determines equilibrium thermodynamic properties. Computing it can nevertheless be difficult when the number of configurations is large or interactions are complicated. (interactivetextbooks.tudelft.nl)

Energy fluctuates because the system exchanges heat with its reservoir. For temperature-independent state energies at fixed volume and particle number,

\operatorname{Var}(E) =\frac{\partial^2\ln Z}{\partial\beta^2} =k_{\mathrm B}T^2 C_V,

where $C_V$ is the constant-volume heat capacity. This relation links microscopic energy variance to a measurable macroscopic response. (interactivetextbooks.tudelft.nl)

Classical systems

In classical mechanics, states form a continuous phase space of positions and momenta. The corresponding probability density is proportional to $e^{-\beta H}$ , where $H$ is the system’s Hamiltonian, and sums over states become integrals with an appropriate phase-space measure. (damtp.cam.ac.uk)

For a nonrelativistic ideal gas, the kinetic energy $mv^2/2$ produces Gaussian velocity components. Accounting for the number of velocity vectors having a given speed yields the three-dimensional Maxwell–Boltzmann speed distribution,

f(v)=4\pi \left(\frac{m}{2\pi k_{\mathrm B}T}\right)^{3/2} v^2e^{-mv^2/(2k_{\mathrm B}T)}, \qquad v\ge0.

The factor $v^2$ shows why the speed distribution is not simply the Boltzmann factor evaluated at kinetic energy. (damtp.cam.ac.uk)

Relation to quantum statistics

The canonical Boltzmann distribution also applies in quantum mechanics: its states can be the energy eigenstates of an entire many-particle system. It is therefore not restricted to classical particles. (damtp.cam.ac.uk)

However, probabilities for complete system states are different from mean occupations of single-particle states. For ideal quantum gases, those occupations follow Bose–Einstein statistics or Fermi–Dirac statistics. Both reduce to the classical exponential form when occupation numbers are small:

\bar n_i\simeq e^{-\beta(\varepsilon_i-\mu)},

where $\mu$ is the chemical potential. Quantum occupation laws therefore do not invalidate canonical Boltzmann weights; they arise from applying equilibrium statistical mechanics with the appropriate many-particle state counting. (damtp.cam.ac.uk)