The Microcanonical Ensemble and S = k ln Omega
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You should be comfortable with the definition of a microstate as a point in phase space and understand that an isolated system evolves on a constant-energy surface. Familiarity with basic calculus, specifically partial derivatives and logarithmic differentiation, is required.
Why this matters
This lecture provides the fundamental bridge between microscopic mechanics and macroscopic thermodynamics.
The central achievement of statistical mechanics is explaining why the Second Law of Thermodynamics holds. By defining entropy not as a vague concept of disorder, but as a precise count of accessible microstates, we derive the direction of time and the equilibrium conditions from first principles. This section establishes the microcanonical ensemble as the foundational framework for all subsequent statistical treatments, linking the mechanical state of particles to measurable quantities like temperature and pressure.
The Core Idea: Counting States
Imagine an isolated box containing a gas. The system is defined by fixed total energy \( U \), volume \( V \), and particle number \( N \). In classical mechanics, the state of the system is a single point in a \( 6N \)-dimensional phase space. However, because the energy is fixed, the system is constrained to move on a hypersurface of constant energy. The microcanonical ensemble postulates that all accessible microstates on this surface are equally probable. The key quantity is \( \Omega(U, V, N) \), the total number of these accessible microstates. Boltzmann’s insight was that the macroscopic property we call entropy, \( S \), is directly proportional to the logarithm of this count. This transforms thermodynamics from a phenomenological science into a statistical one, where "disorder" is rigorously defined as the multiplicity of microscopic arrangements compatible with the same macroscopic observation.
Deriving the Entropy Function
To determine the functional form of entropy, we consider two macroscopic systems, A and B, that can exchange energy but not particles. The total energy \( U_{\text{tot}} = U_A + U_B \) is fixed. The total number of microstates for the composite system is the product of the individual microstate counts: \( \Omega_{\text{tot}} = \Omega_A(U_A) \Omega_B(U_B) \). Equilibrium is reached when the total number of accessible states is maximized, as this represents the most probable macrostate. We maximise \( \ln \Omega_{\text{tot}} \) with respect to \( U_A \), noting that \( U_B = U_{\text{tot}} - U_A \).
This condition implies that \( \frac{\partial \ln \Omega_A}{\partial U_A} = \frac{\partial \ln \Omega_B}{\partial U_B} \). For this quantity to be a state function that is additive for composite systems (a requirement of thermodynamic entropy), it must be proportional to the thermodynamic entropy. Thus, we define \( S = k \ln \Omega \), where \( k \) is Boltzmann’s constant. This definition ensures that entropy is extensive and that the equilibrium condition corresponds to the equality of a new parameter, temperature.
Statistical Temperature
From the definition \( S = k \ln \Omega \), we can identify the statistical temperature. In thermodynamics, temperature is defined by the relation \( \frac{1}{T} = \left( \frac{\partial S}{\partial U} \right)_{V,N} \). Substituting the statistical definition of entropy, we obtain:
This equation is profound. It defines temperature not as a measure of "hotness," but as the rate of change of the logarithm of the number of microstates with respect to energy. A system with a large \( \Omega \) that increases rapidly with energy has a low temperature, while a system where \( \Omega \) increases slowly has a high temperature. This derivation recovers the thermodynamic temperature purely from the counting of states, validating the microcanonical ensemble as a consistent framework.
Connection to the Second Law
The Second Law states that the entropy of an isolated system never decreases. In the microcanonical framework, this is a statistical inevitability. If a system starts in a macrostate with a small \( \Omega \) (low entropy), it will evolve into macrostates with larger \( \Omega \) simply because there are vastly more of them. The probability of finding the system in a low-entropy state is proportional to \( \Omega_{\text{low}} / \Omega_{\text{total}} \), which is exponentially small for macroscopic systems. Thus, the "arrow of time" emerges from the overwhelming statistical weight of high-entropy states. The equilibrium state is the one with the maximum \( \Omega \), and fluctuations away from this state are negligible for large \( N \).
The derivation
Boltzmann Entropy and Statistical Temperature
For an isolated composite of two macroscopic systems exchanging energy at fixed total U = U1 + U2, the equilibrium energy partition is the one that maximises the total number of accessible microstates Ω = Ω1Ω2. Requiring this maximum forces the entropy to take the form S = k ln Ω (additive, monotonic in Ω) and identifies the statistical temperature through 1/T = (∂S/∂U)V,N, reproducing the thermodynamic entropy of the Clausius relation dS = δQ/T.
A common misconception is that \( S = k \ln \Omega \) implies entropy is a discrete integer. In reality, \( \Omega \) is a continuous measure of phase space volume in classical mechanics, and the logarithm renders it extensive. Furthermore, the definition assumes the system is ergodic, meaning it explores all accessible microstates over time. If the system is non-ergodic, the microcanonical ensemble may not correctly predict equilibrium properties, requiring more sophisticated treatments. C
Worked example
Problem. For a monatomic ideal gas of \(N = 1.00 \times 10^{23}\) atoms, the number of microstates depends on energy as \(\Omega(U) = C\,U^{3N/2}\), with \(C\) independent of \(U\). Find the temperature when \(U = 600\ \mathrm{J}\). Use \(k = 1.38 \times 10^{-23}\ \mathrm{J\,K^{-1}}\).
- Statistical temperature: \(\dfrac1T = k\left(\dfrac{\partial \ln\Omega}{\partial U}\right)_{V,N}\).
- \(\ln\Omega = \ln C + \tfrac{3N}{2}\ln U\), so \(\dfrac{\partial \ln\Omega}{\partial U} = \dfrac{3N}{2U}\).
- Hence \(\dfrac1T = \dfrac{3Nk}{2U}\), that is \(U = \tfrac32 NkT\), the equipartition result.
- Substitute: \(T = \dfrac{2U}{3Nk} = \dfrac{2 \times 600}{3 \times 1.00 \times 10^{23} \times 1.38 \times 10^{-23}} = \dfrac{1200}{4.14}\).
- So \(T = 290\ \mathrm{K}\), about room temperature: 0.166 mol of monatomic gas at room temperature holds about 600 J of kinetic energy.
Units check. \(\mathrm{J}/(\mathrm{J\,K^{-1}}) = \mathrm{K}\); \(N\) is a pure number.
Three quick tests
- Why is the logarithm necessary in \( S = k \ln \Omega \)?
Answer
Entropy must be additive for composite systems, while the number of microstates multiplies. The logarithm converts multiplication into addition. - If \( \frac{\partial \ln \Omega}{\partial U} \) is negative, what is the sign of the temperature?
Answer
Negative. This corresponds to systems with an upper bound on energy, such as spin systems, where adding energy reduces the number of available states. - Does the microcanonical ensemble apply to a system in contact with a heat bath?
Answer
No. A system in contact with a heat bath exchanges energy, so its energy is not fixed. This requires the canonical ensemble.
Problem set for PU-203 →
Each derivation in this unit ends with five graded problems with full solutions.
Lecture notes checked line by line by the editor; every number in the worked example was recomputed.