Past exam of the mathematics course of the University of Cambridge / 2026 / ii / Paper 2 / 37E / b

Table of contents
- i b
  - Solution i
- ii b
  - Solution ii
- iii b
  - Solution iii
- iv b
  - Solution iv

i

Solution

Two polarizations and periodic wavevectors give

g (ω) = A ω / (π c_{s}^{2})

.

ii

Solution

Requiring

\int_{0}^{ω_{D}} g (ω) d ω = 2 N

gives

ω_{D} = 2 c_{s} π N / A

.

iii

Solution

For

k_{B} T ≪ ℏ ω_{D}

, extend the integral to infinity:

lo g Z = - \int_{0}^{\infty} g (ω) lo g (1 - e^{- β ℏ ω}) d ω = \frac{A k _{B}^{2} T ^{2}}{π c _{s}^{2} ℏ ^{2}} \sum_{n \geq 1} \frac{1}{n ^{3}} .

Thus $X = ζ (3) / π$ .

iv

Solution

Since

F = - k_{B} T lo g Z

,

S = - \partial F / \partial T = 3 ζ (3) A k_{B}^{3} T^{2} / (π c_{s}^{2} ℏ^{2})

.

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