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Microcanonical Ensemble

The microcanonical ensemble describes an isolated system with fixed total energy EEParticle number NNAnd volume VV.

14.1 Density of States

The number of microstates with energy between EE and E+δEE + \delta E is:

Ω(E,V,N)=E<H<E+δEd3Nqd3NpN!h3N\Omega(E, V, N) = \int_{E < \mathcal{H} < E + \delta E} \frac{d^{3N}q\, d^{3N}p}{N!h^{3N}}

The entropy (Boltzmann entropy):

S(E,V,N)=kBlnΩ(E,V,N)S(E, V, N) = k_B \ln \Omega(E, V, N)

The temperature is defined via:

1T=SE\frac{1}{T} = \frac{\partial S}{\partial E}

14.2 The Ideal Gas in the Microcanonical Ensemble

For NN non-interacting particles in volume VV with total energy EE:

Ω=VNN!(2πmE)3N/2EΓ(3N/2)h3NδEE\Omega = \frac{V^N}{N!}\frac{(2\pi m E)^{3N/2}}{E\, \Gamma(3N/2)\, h^{3N}} \cdot \frac{\delta E}{E}

Using Stirling”s approximation and the large-argument expansion of the Gamma function:

S=NkB[ln ⁣(VN)+32ln ⁣(4πmE3Nh2)+52]S = Nk_B\left[\ln\!\left(\frac{V}{N}\right) + \frac{3}{2}\ln\!\left(\frac{4\pi m E}{3Nh^2}\right) + \frac{5}{2}\right]

This is the Sackur—Tetrode equation, identical to the canonical ensemble result (as expected by ensemble equivalence).

From 1/T=S/E1/T = \partial S/\partial E:

E=32NkBTE = \frac{3}{2}Nk_B T

Reproducing the equipartition theorem.

14.3 Classical Virial Theorem

For a system with Hamiltonian H=ipi2/(2mi)+U(r1,,rN)\mathcal{H} = \sum_i p_i^2/(2m_i) + U(\mathbf{r}_1, \ldots, \mathbf{r}_N):

ipiHpi=3NkBT\left\langle \sum_i \mathbf{p}_i \cdot \frac{\partial \mathcal{H}}{\partial \mathbf{p}_i} \right\rangle = 3Nk_B T

iriHri=3NkBT\left\langle \sum_i \mathbf{r}_i \cdot \frac{\partial \mathcal{H}}{\partial \mathbf{r}_i} \right\rangle = -3Nk_B T

For a power-law potential UrnU \propto r^nThis gives:

K=n2U\langle K \rangle = \frac{n}{2}\langle U \rangle

(For the harmonic oscillator, n=2n = 2: K=U\langle K \rangle = \langle U \rangle.)

Worked Example 14.1: Density of States for $N$ Harmonic Oscillators

For NN independent harmonic oscillators with frequency ω\omegaTotal energy EE:

Ω(E)=EN1(N1)!(ω)N\Omega(E) = \frac{E^{N-1}}{(N-1)!\,(\hbar\omega)^N}

Proof: The number of ways to distribute E/(ω)E/(\hbar\omega) energy quanta among NN oscillators is the stars-and-bars problem:

Ω=(n+N1N1)=(n+N1)!n!(N1)!\Omega = \binom{n + N - 1}{N - 1} = \frac{(n+N-1)!}{n!(N-1)!}

Where n=E/(ω)n = E/(\hbar\omega). For large nn using Stirling’s approximation:

S=kB[(n+N)ln(n+N)nlnnNlnN]S = k_B\left[(n+N)\ln(n+N) - n\ln n - N\ln N\right]

1T=SE=kBω[ln(n+N)lnn]=kBωln ⁣(1+Nn)\frac{1}{T} = \frac{\partial S}{\partial E} = \frac{k_B}{\hbar\omega}\left[\ln(n+N) - \ln n\right] = \frac{k_B}{\hbar\omega}\ln\!\left(1 + \frac{N}{n}\right)

At high TT (nNn \gg N): ENkBTE \approx Nk_B T (equipartition, each oscillator has energy kBTk_B T).