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Use periodic boundary conditions on the square crystal. Each allowed wavevector occupies area $(2\pi {)}^2/A$ in two-dimensional $k$-space. Including the two phonon polarizations, the number of modes in the disk $|k|\le K$ isThe phonon dispersion relation $\omega =C|k{|}^{\alpha }$ gives $K=(\omega /C{)}^{1/\alpha }$, soDifferentiating the cumulative mode count produces the density of states for a two-dimensional power-law phonon dispersion: