# Do solid particles vibrate?

Aug 10, 2017

Yes. And in lattices, if the atoms themselves are heavier, they vibrate at lower natural frequencies, and vice versa.

This can be modeled by an infinite array of harmonic oscillators, i.e. balls and springs.

For simplicity, consider a segment of a 1D lattice of $N$ identical atoms vibrating with a given wave number $\kappa$ over time $t$, with the atoms at each end fixed:

The position over time for the $j$th atom is given by:

${x}_{j} \left(t\right) = {A}_{\kappa} {e}^{i \left({\omega}_{\kappa} t + \kappa a j\right)}$

They would vibrate with an angular frequency ${\omega}_{\kappa}$ of:

${\omega}_{\kappa} = 2 \sqrt{\frac{k}{m}} | \sin \left(\frac{\kappa a}{2}\right) |$

and, since $0 \le | \sin u | \le 1$, they would reach a maximum frequency, called the Debye frequency ${\omega}_{D}$, of:

${\omega}_{D} = 2 \sqrt{\frac{k}{m}}$,

where:

• $k$ is the force constant between the two identical atoms in ${\text{kg/s}}^{2}$.
• $\kappa$ is the wave number (the number of waves for a unit distance).
• $m$ is the mass of each atom in $\text{kg}$.
• $a$ is the position coordinate of a given atom.

Above ${\omega}_{D}$, the vibrations in the lattice become independent of the lattice, i.e. the atoms in each unit cell vibrate at a well-defined frequency that is distinct from the collective lattice vibrations.

In other words, the wavelength $\lambda$ is lower than the width of the unit cell, given by the Debye (minimum) wavelength ${\lambda}_{D}$.

A full derivation of the above equations can be found here using differential equation math and Lagrangian mechanics.