Past exam of the condensed matter physics course of the University of Oxford / 2013 / 3

Let $a$ be the equilibrium spacing of this monatomic chain and $u_j$ the displacement from $ja$. The right nearest-neighbour bond exerts $K(u_{j+1}-u_j)$ and the left bond $K(u_{j-1}-u_j)$. The two next-nearest-neighbour bonds similarly connect $j$ to $j\pm 2$. ThusEvery uniform displacement gives zero force, as translation symmetry requires. There is one displacement coordinate per primitive cell. Adding further-neighbour springs does not introduce a second atom or turn this into a diatomic-chain problem.

Substitute the traveling normal mode $u_j=u\exp [i(kja-\omega t)]$. Combining the opposite-neighbour phases givesso the positive-frequency branch isFor positive spring constants this is a stable acoustic branch. At long wavelengths,The next-nearest-neighbour bond spans twice the distance, hence contributes the factor four. Relative to $G=0$, positive $G$ increases the sound speed by $\sqrt{1+4G/K}$. The chain still has one acoustic branch and no optical branch.

The group speed near $k=0$ and the maximum frequency need not change in the same way. The zone-edge value is always ${\omega }_0(\pi /a)=2\sqrt{K/m}$, because the next-nearest-neighbour sites then move together. For $G>K/4$ the frequency maximum moves inside the zone, with $\cos (ka)=-K/(4G)$; the small-$k$ sound speed nevertheless keeps increasing.

For a viscously damped monatomic-chain phonon, the extra force gives $m{\ddot{u}}_j+\Gamma {\dot{u}}_j$ on the left of the preceding equation. Keep a real wave number $k$ and permit complex frequency in the convention $u_j\propto e^{i(kja-\omega t)}$. The dispersion equation isWriting $\gamma =\Gamma /(2m)$, its two roots areThe negative imaginary part corresponds to decay with the chosen time convention. Here $\Gamma$ has units of mass per time, while $\gamma$ is a decay rate. A driven experiment at specified real frequency would instead solve for a generally complex wave number; it is a different way of specifying the response, not a contradictory sign choice.

For passive friction $\Gamma >0$, the time dependence of each spatial normal coordinate follows its damping regime:

For any fixed positive damping, sufficiently small nonzero $k$ is overdamped because ${\omega }_0\simeq c_s|k|$. The slow branch then has $\omega \simeq -i(m{c}_s^2/\Gamma )k^2$, a diffusive long-wavelength relaxation. At exactly $k=0$, uniform translation has no restoring force, so $u_0(t)=A+B{e}^{-\Gamma t/m}$ and the constant displacement need not decay. Thus saying every amplitude simply decays as $e^{-\Gamma t/(2m)}$ is valid only for the underdamped envelope, not for all wave numbers.