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PU-203 / L27 · lecture

The Boltzmann factor

18 min read · use the Tier toggle to reveal standard (T2) and advanced (T3) depth

BEFORE YOU START

Prerequisites

You need the microcanonical picture and the definition of entropy from L24 · Ensembles, plus Lagrange multipliers from D-018.

Why this matters

Almost every "how likely is this state?" question in thermal physics has the same answer: states cost energy, and higher-energy states are exponentially rarer.

Why does water vapour thin out with altitude? Why do only a lucky few molecules have enough energy to react? Why does a magnet lose its alignment when you heat it? These look like unrelated puzzles, but they share one engine. A system in contact with a large reservoir at temperature T does not sit in its lowest-energy state — thermal jostling constantly kicks it around. The Boltzmann factor is the single rule that says exactly how often those kicks succeed, and it is the most reused result in the whole unit.

The energy penalty

Picture the state you care about as one small system pressed against a huge reservoir. Every unit of energy the small system borrows is a unit the reservoir loses — and the reservoir has vastly more ways to arrange itself when it keeps that energy. So the probability of the small system occupying a state is not set by that state's energy alone, but by how much reservoir entropy is sacrificed to pay for it. Expand the reservoir's entropy to first order in the borrowed energy and the penalty turns out to be exponential, with temperature setting the exchange rate.

The derivation

◆ DERIVATION EMBED · not re-derived here

D-013 · The Boltzmann distribution from maximum entropy

Maximise the Gibbs entropy over all probability distributions subject to fixed normalisation and fixed mean energy; the multipliers fix the form uniquely.

pi = e−Ei/kBT / Z,    Z = Σi e−Ei/kBT
Open full derivation →

Reading the result

The factor e−Ei/kBT is a raw weight, not yet a probability — dividing by the partition function Z normalises the weights so they sum to one. Two features carry all the physics. First, only energy differences matter: the ratio pi/pj = e−(Ei−Ej)/kBT is blind to where you set the zero of energy. Second, kBT is the natural yardstick for energy: states within about kBT of the ground state are comparably populated, while states far above it are frozen out.

Tier 3 · watch out

The factor weights states, not energy levels. If an energy level E is g-fold degenerate, the probability of being at that energy is g·e−E/kBT/Z — and the degeneracy can beat the exponential, so the most probable energy is generally not the ground state. Forgetting the g is the single most common error in canonical calculations. C

CHECK YOURSELF

Quick questions

Two states differ in energy by exactly kBT. What is the ratio of their populations? Now double T — does the ratio move toward 1 or away from it, and why?

PROBLEM SET

Practise on real systems

Isothermal atmospheres, paramagnets, and reaction rates — the problems that turn this one factor into working intuition.

Derivations cited: D-013 · D-018 · Threads: chance · energy