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
$V_{p} = frac{nu}{k} = dsqrt{frac{alpha}{m}} text{sinc}(pi kad)$
If $k rightarrow 0$ then $nu rightarrow 0$ so all solutions become constants (they lose their oscillatory parts) but the phase velocity becomes $V_{p} = dsqrt{frac{alpha}{m}}$.
Why is there still a finite phase velocity? If I take the continuum limit of $drightarrow 0$ I get that the medium becomes a dispersionless string but this seems quite different than an arbitrary translation.

Matteo · Accepted Answer

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 $V_p = V_g = dsqrt{frac{alpha}{m}}$.
Hope this helps :) let me know

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

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