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ID: past-exam-of-the-physics-course-of-the-year-4-of-the-natural-sciences-course-of-the-university-of-cambridge/2015/paper-1/section-b/b2/a/solution

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Past exam of the physics course of the year 4 of the natural sciences course of the University of Cambridge / 2015 / Paper 1 / Section B / B2 / a / Solution by

Codex  0 2026-10-20

Standing-wave wave numbers are

k_{n} = nπ / L

, with one positive mode index and two polarizations. Since

ϵ = ℏ c k

, the continuum density of states is

g (ϵ) = \frac{2 L}{π ℏ c} .

(1)

The factor two counts polarizations; counting independent positive and negative traveling momenta in addition would double count standing modes. The thermal energy is

U = \int_{0}^{\infty} g (ϵ) \frac{ϵ d ϵ}{e ^{ϵ / (k_{B} T)} - 1} = \frac{2 L ( k _{B} T ) ^{2}}{π ℏ c} \frac{π ^{2}}{6} .

(2)

Thus the one-dimensional photon-gas thermodynamics gives

U = \frac{π L k _{B}^{2} T ^{2}}{3ℏ c} .

(3)

Replacing the mode sum by an integral requires the stated temperature to be large compared with the level-spacing scale. This is thermal energy above the vacuum contribution; a separate boundary-dependent vacuum energy is not included.

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