The Microcanonical Ensemble and Entropy

Introduction

An isolated system has a fixed total energy but can rearrange that energy among many microscopic configurations. The microcanonical ensemble treats compatible states as equally likely in equilibrium. Its central quantity is the number Ω of accessible states in a narrow energy shell. Taking k times the logarithm of this count gives entropy, and the way that entropy changes with energy defines temperature.

Core explanation

For fixed E, V and N, choose a narrow energy interval E to E + δE containing many microscopic states but small on a macroscopic energy scale. Let Ω(E,V,N;δE) count those states. In the equilibrium microcanonical description, each has probability 1/Ω. Boltzmann entropy is S(E,V,N) = k ln Ω, with a consistent choice of energy shell. The logarithm is crucial: for two weakly interacting independent systems, state counts multiply, Ω total ≈ Ω AΩ B, while entropies add, S total ≈ S A + S B.

Suppose two subsystems exchange a small amount of energy while combined total E is fixed. The number of combined states for a division E A and E B = E total − E A is Ω A(E A)Ω B(E B). The most probable division maximises ln Ω A + ln Ω B, equivalently S A + S B. Differentiating with respect to E A gives (∂S A/∂E A) V,N − (∂S B/∂E B) V,N = 0. Define 1/T = (∂S/∂E) V,N. The equilibrium condition is 1/T A = 1/T B, hence equal positive temperatures.

The derivative definition explains heat flow. If A has smaller inverse temperature, it has higher T; moving energy from hotter A to cooler B increases total entropy until their slopes agree. For very small systems, fluctuations around the most probable energy split can be noticeable; for macroscopic systems the entropy maximum is sharply dominant. Volume and particle exchange yield analogous equilibrium conditions involving pressure and chemical potential through entropy derivatives.

One must distinguish a single microscopic state from an energy level with degeneracy. Counting energy levels once can grossly undercount states if many distinct microstates share that energy. The exact numerical value of Ω can depend on energy-bin convention or classical phase-space normalisation, but physically meaningful entropy differences and derivative relations are handled consistently. A sparse system may require care with derivatives and continuum approximations.

Step-by-step reasoning

Define the system boundary and fixed E,V,N. Count all distinct compatible microstates in an appropriately narrow energy shell, including degeneracy. Take S = k ln Ω. For two energy-exchanging subsystems, multiply their state counts at a proposed split, then maximise the log at fixed total energy. Translate equal entropy slopes into equal temperatures.

Visual explanation

Draw two boxes A and B sharing an energy-permeable wall inside a sealed larger box. Above a horizontal E A axis, plot S A(E A) + S B(E total − E A) as a peaked curve. Its highest point marks the most probable split and equal slopes. Beneath show Ω AΩ B, a much sharper peak for large systems.

Real-world analogy

Distributing a fixed number of tokens between two groups allows many arrangements for some splits and few for others. The most commonly realised split is the one with the most combined arrangements. Energy distribution in isolated coupled systems follows the same counting logic, but the molecular states have energies and constraints that simple tokens do not capture.

Real-world example

Two solids initially at different temperatures are placed in thermal contact inside an insulated container. Total energy is conserved, while energy flows between them. Their final equilibrium temperature is the state that maximises total entropy subject to that energy conservation, not necessarily the arithmetic average of their initial temperatures because their heat capacities may differ.

Why?

Equal a priori probability assigns no preference among accessible microstates of an isolated equilibrium system. Macrostates with more compatible microstates are therefore more probable. The logarithm converts multiplicative counts into additive entropy, enabling a macroscopic derivative. Temperature emerges as the inverse slope of entropy with respect to energy.

Common misconception

The microcanonical ensemble does not set temperature directly; it sets energy, and T is derived. Entropy does not equal the raw number of states; it is proportional to its logarithm. Also, the most probable energy split is not necessarily equal energies. It is the split where entropy slopes, or temperatures, match.

Worked example

Imagine two weakly coupled subsystems whose accessible-state counts near a proposed energy division scale as Ω A(E A) ∝ E A² and Ω B(E B) ∝ E B³ for positive energies. At fixed E total = 10 units, maximise ln Ω total = 2 ln E A + 3 ln(10 − E A) plus a constant. The derivative is 2/E A − 3/(10 − E A) = 0, giving 2(10 − E A) = 3E A and E A = 4, E B = 6. Equal energy division would not maximise the count. At the optimum, the inverse-temperature slopes 2k/E A and 3k/E B both equal k/2 in these model units.

Quick check

1. Why does the logarithm make entropy additive for independent subsystems? Answer: Independent state counts multiply, Ω total = Ω AΩ B. Then k ln Ω total = k ln Ω A + k ln Ω B = S A + S B, matching extensive entropy addition.

Exam focus

State fixed E,V,N and distinguish an energy shell from exactly one energy value for a continuum. Count microstates rather than merely distinct energy levels. When deriving thermal equilibrium, apply E B = E total − E A and retain the minus sign in differentiation. Use 1/T = (∂S/∂E) V,N only under the stated variables.

Advanced insight

For systems with bounded energy spectra, S(E) can sometimes decrease with increasing E over part of its range, producing a negative Boltzmann temperature under a particular definition. Such states require special preparation and do not mean “colder than zero”; they are hotter than positive-temperature states in the sense of heat flow. Ordinary unbounded translational gases do not support this regime.

Summary

The microcanonical ensemble fixes E,V,N and weights accessible states equally. Entropy S = k ln Ω turns state counting into an additive thermodynamic quantity. Maximising combined entropy for weakly coupled systems yields equality of 1/T = ∂S/∂E and explains thermal equilibrium from microscopic counting.

Practice questions

1. System A has 100 accessible states and independent B has 1,000 at a selected energy split. How many combined states and what entropy expression result? Answer: Ω total = 100 × 1,000 = 100,000. S total = k ln 100,000 = k ln 100 + k ln 1,000, assuming independent weakly coupled state counts. 2. If an isolated coupled pair has maximum total entropy at E A = 3 J and E B = 7 J, must their energies become equal later? Answer: No. The maximum is already the equilibrium split; equal entropy derivatives and temperatures matter, not equal energy amounts. 3. What thermodynamic variable is obtained from (∂S/∂E) V,N? Answer: The reciprocal temperature 1/T. It measures how rapidly entropy grows when energy is added at fixed volume and particle number.