Statistical Mechanics

Ensembles and the Meaning of Temperature

Three ensembles and what each one holds fixed. Temperature enters as a property of the counting.

Updated May 2026

An ensemble is a bookkeeping device. We cannot follow one system through its history, so instead we imagine infinitely many copies of it prepared identically and ask what fraction sit in each microstate. Which ensemble you pick depends entirely on what you are holding fixed.

The three ensembles

Microcanonical

Fixed E, V, N

The system is isolated. Every accessible microstate is equally likely, and the entropy is defined by counting them:

S = k_{B} ln Ω (E, V, N)

This is the equation on Boltzmann's headstone. It is taken as a definition here rather than derived, and the rest of the subject is built on top of it.

Canonical

Fixed T, V, N

The system exchanges energy with a reservoir. Microstate probabilities are Boltzmann-weighted, and everything follows from the partition function $Z$ through the Helmholtz free energy $F = - k_{B} T ln Z$ .

Grand canonical

Fixed T, V, μ

The system exchanges both energy and particles. Useful whenever particle number is not conserved: photons, phonons, or an open region of a larger system.

Temperature as a derivative of the count

Put two isolated systems in thermal contact and let them share a fixed total energy $E = E_{1} + E_{2}$ . The combined number of microstates is $Ω_{1} (E_{1}) Ω_{2} (E_{2})$ , and the equilibrium division of energy is whichever one maximises it. Setting the derivative to zero:

\frac{\partial ln Ω _{1}}{\partial E _{1}} = \frac{\partial ln Ω _{2}}{\partial E _{2}}

Theorem

Temperature

Both sides of that equality are the same quantity, and we name it:

\frac{1}{T} \equiv \frac{\partial S}{\partial E}_{V, N}

Temperature is not a substance and not an average energy. It is the rate at which a system's microstate count grows as you feed it energy.

This reframing pays off immediately. Heat flows from hot to cold not because of any force, but because the total microstate count increases when it does. There is nothing pushing.

Negative temperature is not cold

If $Ω$ has a maximum, which happens whenever the energy spectrum is bounded above, as in a spin system, then past that maximum $\partial S / \partial E$ turns negative, and so does $T$ .

A system at negative temperature is not colder than absolute zero. It is hotter than infinitely hot: put it in contact with anything at positive temperature and energy flows out of it. The pathology is in the parametrisation, not the physics: $β = 1/ k_{B} T$ runs smoothly from $+ \infty$ through zero to $- \infty$ without incident.

Equipartition, and where it fails

For a Hamiltonian quadratic in some coordinate, the canonical ensemble gives that coordinate a mean energy of $\frac{1}{2} k_{B} T$ regardless of the coefficient. Hence $\frac{3}{2} k_{B} T$ for a monatomic gas, and $3 k_{B}$ per atom for the heat capacity of a solid.

That last result, the Dulong–Petit law, is right at room temperature and badly wrong as $T \to 0$ , where measured heat capacities fall to zero. The failure is not subtle and it is not fixable classically: equipartition assumes energy is continuous. Once the level spacing exceeds $k_{B} T$ , a mode simply cannot be excited and it stops contributing. Explaining that curve is one of the problems that produced quantum mechanics.