Past exam of the physics A course of the University of Cambridge / 2016 / Paper 2 / Section A / 4

Table of contents
- Solution 4

Solution

Use periodic boundary conditions temporarily to count the allowed wavevectors. Each state occupies

(2 π)^{3} / V

in reciprocal space. For one acoustic polarization, the spherical shell between

k

and

k + d k

contains

d N = \frac{V}{( 2 π ) ^{3}} 4 π k^{2} d k .

(1)

Both positive and negative wavevector directions are already included in that sphere; no further factor of two is needed. With the linear dispersion relation

k = ω / v_{s}

g_{1} (ω) = \frac{d N}{d ω} = \frac{V ω ^{2}}{2 π ^{2} v _{s}^{3}} .

(2)

A three-dimensional monatomic crystal has three acoustic branches. Under the stated common-speed approximation,

g (ω) = \frac{3 V ω ^{2}}{2 π ^{2} v _{s}^{3}} .

(3)

This is a phonon density of states per unit angular frequency, not per unit ordinary frequency. If longitudinal and transverse speeds differ, the isotropic expression becomes