Infinite wavelength for the 1D harmonic chain of oscillators corresponds to a uniform translation, but why does it still have a finite phase velocity?

Question

The dispersion for the 1d chain of harmonic oscillators is

$\nu = \sqrt{\frac{\alpha}{m}} \frac{|\sin(\pi k d)|}{\pi}$

Where I'm explicitly not using angular frequencies ($\nu = \frac{1}{T}, k = \frac{1}{\lambda}$), $d$ is the lattice spacing, $\alpha$ the spring constant. The phase velocity would be

$V_{p} = \frac{\nu}{k} = d\sqrt{\frac{\alpha}{m}} \text{sinc}(\pi kad)$

If $k \rightarrow 0$ then $\nu \rightarrow 0$ so all solutions become constants (they lose their oscillatory parts) but the phase velocity becomes $V_{p} = d\sqrt{\frac{\alpha}{m}}$.

Why is there still a finite phase velocity? If I take the continuum limit of $d\rightarrow 0$ I get that the medium becomes a dispersionless string but this seems quite different than an arbitrary translation.

Answer 1

I think that we should treat the case $k=0$ and $k \to 0$ separately.

First, if you take $k=0$ (which means that you are rigidly translating every atom in the lattice by the same amount), then $\nu = 0$, which is telling you that there won't be any dynamics in the system (atoms don't oscillate because they are maintained at the equilibrium distance by the perturbation). I guess that in this case the definition of phase velocity makes no sense at all.

Second, if you take $k \to 0$, then you are perturbing the system with a periodic perturbation on a length scale $1/k \gg d$, and in this case you can expect a dynamical response. In fact you find $\nu \sim d \sqrt{\frac{\alpha}{m}} k$, which is an oscilation with phase velocity equivalent to the group veocity $V_p = V_g = d\sqrt{\frac{\alpha}{m}}$.

Hope this helps :) let me know