advanced29 min studyLesson 31 in Physics

Entropy and the microcanonical ensemble

Count accessible states, define entropy and temperature, and distinguish typical behavior from a deterministic law for every small system.

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What you’ll understand

  • Construct a microcanonical state count with stated macroscopic constraints.
  • Derive the equilibrium condition for weakly coupled systems exchanging energy.
  • Explain why thermodynamic extensivity and sharply defined temperature require appropriate limits.

A cup of gas has far more microscopic coordinates than any experiment can track. Statistical physics starts by specifying what is known, then counting the compatible microscopic possibilities. The resulting entropy describes a macroscopic constraint, rather than measuring how visually untidy one particular configuration looks. We begin with a finite two-state model because its probabilities are exactly countable. The familiar thermodynamic laws emerge when the count becomes large and sharply concentrated, under assumptions that must remain visible.

Define the ensemble before counting

An isolated system has fixed particle number, volume, and energy. In quantum mechanics, count energy eigenstates inside a narrow shell around the specified energy, including degeneracies. The shell should be narrow on a macroscopic energy scale but contain many levels when a thermodynamic description is intended. We use Boltzmann's shell entropy, rather than the alternative Gibbs entropy counting all states below the energy; their finite-system derivatives can differ. The microcanonical postulate assigns equal probabilities to accessible shell states. It is an equilibrium ensemble assumption, not a claim that every isolated initial state immediately explores every eigenstate.

For NN distinguishable, noninteracting two-state units, let an excitation cost energy ϵ>0\epsilon>0. Exactly qq excitations give energy qϵq\epsilon and multiplicity (Nq)\binom Nq. The labels could refer to localized spins, so no additional division by N!N! is appropriate. Entropy is the logarithm of this multiplicity times Boltzmann's constant. Taking a logarithm turns products of independent state counts into sums, which explains why it is compatible with combining weakly interacting systems.

S(E,V,N)=kBln⁡Ω(E,V,N),ΩN(q)=(Nq),E=qϵS(E,V,N)=k_B\ln\Omega(E,V,N),\qquad \Omega_N(q)=\binom Nq,\qquad E=q\epsilon
Ω counts states in the specified energy shell. The binomial expression is exact for distinguishable two-state units with fixed excitation count.↔ Scroll wide equations horizontally.

Energy exchange selects an entropy maximum

Put systems A and B in weak thermal contact while keeping their combined energy fixed. Negligible interaction energy lets us write the count for a chosen allocation as ΩA(EA)ΩB(Etot−EA)\Omega_A(E_A)\Omega_B(E_{\mathrm{tot}}-E_A). The most probable allocation maximizes the sum of entropies. In a smooth large-system approximation, differentiating gives equality of their entropy slopes. Defining inverse temperature through that slope converts a counting statement into the familiar condition that bodies in thermal equilibrium have equal temperatures.

A maximum also requires suitable curvature: equality of slopes alone can describe an unstable stationary allocation. For ordinary short-range systems away from unusual constraints, concave entropy gives stable sharing. A bounded spectrum such as the two-state model can have decreasing multiplicity above half excitation, producing a negative entropy derivative and a negative absolute temperature in this definition. That population-inverted regime is possible only with an appropriate upper energy bound; it is not colder than a positive-temperature reservoir.

1T=(∂S∂E)V,N,P(EA)∝exp⁡ ⁣[SA(EA)+SB(Etot−EA)kB]\frac1T=\left(\frac{\partial S}{\partial E}\right)_{V,N},\qquad P(E_A)\propto\exp\!\left[\frac{S_A(E_A)+S_B(E_{\mathrm{tot}}-E_A)}{k_B}\right]
Temperature is defined from a smooth entropy derivative. For a tiny discrete system, the full probability table is more fundamental than a derivative or a unique finite-difference convention.↔ Scroll wide equations horizontally.

Typical is not certain

For macroscopic weakly correlated systems, the entropy maximum commonly overwhelms other allocations, giving reproducible bulk behavior despite microscopic motion. A six-spin example has no such overwhelming concentration: visibly different allocations retain substantial probability. The second law is therefore not a claim that every microscopic fluctuation raises entropy. Reversible microscopic equations and overwhelmingly probable macroscopic relaxation answer different questions, provided the initial preparation and coarse-graining have been specified.

Extensivity means that entropy grows proportionally to system size when energy, volume, and particle number are scaled together at fixed densities. It is a thermodynamic-limit property, not an exact identity for every finite count. Stirling corrections, boundaries, and constraints produce subleading terms. Long-range nonadditive interactions can require a different analysis. Always separate the exact finite combinatorics from a large-system approximation, and do not infer the time needed to equilibrate merely by counting equilibrium states.

Four localized units with two excitations

Four distinguishable two-state units have exactly two excitations. Find their multiplicity and entropy, then compare with the multiplicity for one excitation.

  1. Choosing the two excited labels gives Ω4(2)=4!/(2!2!)=6\Omega_4(2)=4!/(2!2!)=6. The energy is exactly 2ϵ2\epsilon.

  2. The entropy is S2=kBln⁡6≈1.79176kBS_2=k_B\ln6\approx1.79176k_B. This is the entropy of the equal-weight ensemble over the six accessible configurations.

  3. One excitation gives Ω4(1)=4\Omega_4(1)=4 and S1=kBln⁡4≈1.38629kBS_1=k_B\ln4\approx1.38629k_B. The entropy difference is kBln⁡(6/4)≈0.405465kBk_B\ln(6/4)\approx0.405465k_B, although a four-unit finite difference is not a unique bulk temperature.

The result

The half-excited macrostate has more compatible configurations. Individual configurations still have definite labels; the ensemble describes uncertainty under the stated energy constraint.

Explore the model

Interactive lab

Interactive lab · probability

Multiplicity versus a fixed-energy constraint

Set success probability p to one half and vary the number of independent trials n. The exact binomial bars reveal the combinatorial enhancement of middle excitation counts in an unconstrained independent-spin ensemble.

Multiplicity versus a fixed-energy constraintThese are exact binomial probabilities for independent trials with one fixed success probability. Playback tours the distribution; it does not resample outcomes. Current readings: Mean · np: 5; Variance · np(1−p): 2.5; Total probability: 1. All parameters are adjustable with the labeled controls below.03581000.070.150.220.3Successes kP(X = k)k = 0 · P = 9.77e-4
Exact probabilityHighlighted outcomeMean np
Mean · np
5
Variance · np(1−p)
2.5
Total probability
1
t = 0 s

These are exact binomial probabilities for independent trials with one fixed success probability. Playback tours the distribution; it does not resample outcomes. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forThe lab shows a binomial distribution over different total energies, not a microcanonical ensemble at one fixed energy. Conditioning on one excitation count selects the corresponding fixed-energy shell; its multiplicity is the binomial coefficient.

Check your understanding

What is fixed when defining the microcanonical ensemble of the isolated two-state model?

Try it yourself

Two groups contain three distinguishable two-state units each and share exactly two excitations. What are the probabilities that group A has zero, one, or two excitations?

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

For each allocation, multiply the multiplicity of A by that of B, then normalize over all allocations.

Compare with the solution

The counts are (30)(32)=3\binom30\binom32=3, (31)(31)=9\binom31\binom31=9, and (32)(30)=3\binom32\binom30=3. Their sum is 15=(62)15=\binom62, so the probabilities are 1/51/5, 3/53/5, and 1/51/5. Equal energy sharing is most likely but has only 60 percent probability. The finite system illustrates why an entropy maximum need not make fluctuations negligible.

Take it with you

  • A microcanonical ensemble requires explicit energy-shell and accessibility assumptions.
  • Temperature equality follows from maximizing the combined state count under energy exchange.
  • Finite-size fluctuations and nonadditive interactions limit naive thermodynamic conclusions.

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