Formula Dispersion Relation for a Crystal with a Diatomic Basis Angular frequency Angular wavenumber Spring constant Mass Lattice constant
$$\begin{align}
\omega_{\pm}^2 ~&=~ D \, \left( \frac{1}{m_1} + \frac{1}{m_2} \right) \\\\
~&\pm~ D \, \sqrt{\left(\frac{1}{m_1} + \frac{1}{m_2}\right)^2 ~-~ \frac{4}{m_1 \, m_2}\,\sin^2\left(\frac{k\,a}{2}\right) }
\end{align}$$
\omega_{\pm}^2 ~&=~ D \, \left( \frac{1}{m_1} + \frac{1}{m_2} \right) \\\\
~&\pm~ D \, \sqrt{\left(\frac{1}{m_1} + \frac{1}{m_2}\right)^2 ~-~ \frac{4}{m_1 \, m_2}\,\sin^2\left(\frac{k\,a}{2}\right) }
\end{align}$$
Angular frequency
$$ \omega_{\pm} $$ Unit $$ \frac{\mathrm{rad}}{\mathrm s} $$ This dispersion relation \(\omega_{\pm}(k)\) describes the relation between the frequency (energy) and the wavenumber (wavelength) of a diatomic chains of a crystal. The oscillation is purely longitudinal (or transversal) and only the interaction between the neighboring chains is considered here.
In a crystal with a diatomic basis there are two vibrational frequencies: \(\omega_{+}(k)\) optical branch and \(\omega_{-}(k)\) acoustic branch.
The angular frequency is related to the frequency \(f\) via \(\omega = 2\pi \, f \).
Angular wavenumber
$$ k $$ Unit $$ \frac{\mathrm{rad}}{\mathrm{m}} $$ Wavenumber is related to wavelength \(\lambda\) via \(k = 2\pi / \lambda \).
Spring constant
$$ D $$ Unit $$ \frac{\mathrm{kg}}{\mathrm{s}^2} $$ Spring constant (or coupling constant) comes from the Hooke spring law and describes how strongly a diatomic lattice plane is coupled to its neighboring lattice planes.
Mass
$$ m_1, m_2 $$ Unit $$ \mathrm{kg} $$ The two masses of a diatomic basis.
Lattice constant
$$ a $$ Unit $$ \mathrm{m} $$ Lattice constant is the distance between two adjacent lattice planes when they are in equilibrium (i.e. not deflected).