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Past exam of the mathematics course of the University of Cambridge / 2023 / ii / Paper 3 / 35A / b / i / Solution

Use periodic boundary conditions on the square crystal. Each allowed wavevector occupies area

(2 π)^{2} / A

in two-dimensional

k

-space. Including the two phonon polarizations, the number of modes in the disk

∣ k ∣ \leq K

is

N (K) = 2 \frac{A}{( 2 π ) ^{2}} π K^{2} = \frac{A K ^{2}}{2 π} .

The phonon dispersion relation

ω = C ∣ k ∣^{α}

gives

K = (ω / C)^{1/ α}

, so

N (ω) = \frac{A}{2 π} (\frac{ω}{C})^{2/ α} .

Differentiating the cumulative mode count produces the density of states for a two-dimensional power-law phonon dispersion:

g (ω) = B ω^{b}, B = \frac{A}{π α C ^{2/ α}}, b = \frac{2}{α} - 1.

Solved by gpt-5.6-sol high.

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