 Solution

ID: past-exam-of-the-thermal-and-statistical-physics-course-of-the-university-of-cambridge/2021/section-b/5/b/solution

For

N

Z_{N} = \frac{1}{N ! h ^{3 N}} \int_{V^{N}} \prod_{i = 1}^{N} d^{3} r_{i} \int \prod_{i = 1}^{N} d^{3} p_{i} exp [- β \sum_{i} \frac{p _{i}^{2}}{2 m} - \frac{β}{2} \sum_{i \neq = j} ϕ (r_{ij})] .

The factor

1/ N!

removes overcounting under particle permutations, and

h^{3 N}

makes the classical phase-space measure dimensionless. The pair potential depends only on positions, so all momentum integrations separate. The three-dimensional Gaussian integral for each particle gives

\frac{1}{h ^{3}} \int_{R^{3}} e^{- β p^{2} / (2 m)} d^{3} p = \frac{1}{h ^{3}} (\frac{2 πm}{β})^{3/2} = λ^{- 3}, λ = \frac{2 π ℏ ^{2} β}{m} .

Ξ = \sum_{N = 0}^{\infty} e^{β μ N} Z_{N}

μ

controls the mean number of particles. Therefore

Ξ = N = 0 \sum \infty \frac{1}{N !} (\frac{e ^{β μ}}{λ ^{3}})^{N} \int_{V^{N}} i = 1 \prod N d^{3} r_{i} exp - \frac{β}{2} i \neq = j \sum ϕ (r_{ij}) .

For

N = 0

, the empty products and configurational integral equal one. The factor

1/2

in the interaction prevents double counting of unordered particle pairs.

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