Discussion
In a two-dimensional square lattice, frequency \(\omega\) of phonons in the long wavelength limit changes linearly with the wave vector \(k\). Then the density of states of phonons is proportional to
Answer: A
The density of states is proportional to \(\omega\).
In two dimensions the number of modes with wave vector less than \(k\) is proportional to the area of a circle in \(k\)-space, \(N \propto \pi k^{2}\). With the linear (acoustic) dispersion \(\omega = v k\), \(N \propto \omega^{2}\), and
\[ g(\omega) = \frac{dN}{d\omega} \propto \omega \]
\(\omega^{2}\) is the Debye result for three dimensions. \(\sqrt{\omega}\) is the 3D density of states for free particles with quadratic dispersion, and \(1/\sqrt{\omega}\) belongs to one dimension with quadratic dispersion.
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