I will try to explain the answer in a really symple way. Proabably as you know, the dispersion relation that you have written above is obtained doing some approximations (i.e. the first neighbour interaction, and the approximation that the interaction between atoms of the chain is harmonic). At this point you can imagine your 1-D mono-atomic crystal as a set of harmonic oscillators, so, if you perturbe your system, you will get an "harmonic response". The reason why in this case you can talk about speed of sound, is that you can map the nuclear part of the Born-Oppeneimer equation into a 1-D D'Almbert equation in which this speed of sound $v_{s}$ appears (i will not demonstrate it, but trust me on the fact that you can do this). As you know this equation reads $$ (\frac{1}{v_{s}^{2}} \frac{\partial^{2}}{\partial t^{2}}- \frac{\partial^{2}}{\partial x^{2}}) f(x,t) = 0 $$ Now, let's consider the classical elastic theory for continuos medium, in which the atomic nature is negligible and let's construct a parallelism with the atomic chain theory. If you perturbe a continuos medium you will obtain a wave travelling into the medium with a dispersion relation: $$\omega= v_{s} k$$ where the speed of sound (i.e. 795
the speed of propagation of the wave due to the perturbation induced) is classicaly defined as: $$v_{s} = \sqrt{\frac{C}{\rho}}$$ where $C$ is the Young-modulus of the medium and $\rho$ is the density of the medium. At this point, let's link this classical elastic theory with the one associated to the mono-atomic chain. In other words, let's consider that your medium is made by mono-atomic chains in such a way that you have a set of atomic layers of mass $M$, section $S$ and each layer is distant $a$ from another one ($a$ is the reticular constant). For the sake of simplicity imagine that in every layers there are $N$ atoms and, finally, imagine that this medium has got a total lenght at rest $l= Na$, where N is the number of atoms that compose the linear chain. If you ideally compress it by a quantity $dl$ the response of the system is harmonic (it is, beacuase of the approximation that you have done in your linear-chain theory) and so, you will have that the response-force can be written as: $$dF = (C S)\frac{dl}{l} $$ Now setting $dl \approx a$ and remembering that the force is harmonic $dF = k(dl) \approx k a$, one has: $$ka = (C S)\frac{a}{N a}$$ notice that $S a$ is clearly the volume $V$ occupied by a layer of atoms, then multiply both members of the equation by $a$. In this way: $$k a^2 = \frac{CV}{N}$$ and using $\rho =\frac{M}{V} \rightarrow V= \frac{M}{\rho}$ one gets: $$\frac{k a^2}{M} = \frac{C}{\rho N}$$ Let's identify $m= \frac{M}{N}$ as the mass of a single atom, and so, in conclusion we get to: $$\boxed{ \frac{K a^2}{m} = \frac{C}{\rho} = v_{s}^{2} } (1)$$ In this way you have identified the speed of sound by the parameters of your mono-atomic chain theory. Now, if you see at the dispersion relation obtained in the thoery of the linear chain, you can simple derive the same exact expression that is at the left side of the equation (1) for the group velocity $\frac{d \omega(k)}{d k}$ computed for small $k$, exactly as you did in your question (apart from a dimensional factor that is in front of the dispersion relation that you have omitted). This means that you can identify the oscillation modes of the linear chain at big wave-lenght ( or $k \approx 0$) like a sound wave that is propagating into a continuos medium! 453