# Why acoustic phonon dispersion cross $\omega=0$ at $k=0$?

When I am learning about phonons, it is taught that acoustic phonons necessarily have $$\omega=0$$ at $$k=0$$. while optical phonons have a finite $$\omega$$ at $$k=0$$. But I am confused about two things:

1. For the acoustic mode, $$k=0$$ means all lattice points oscillate in phase. But still, they are oscillating so each lattice point must have some finite frequency (otherwise they would hold still or travel in a single direction indefinitely). How comes this frequency is zero?
2. The energy of the phonon is given by $$\hbar\omega$$. This implies at $$k=0$$, the acoustic phonon has no energy, how? Isn't all atoms are still moving at the same pace with the same phase? On the other hand, for the optical mode, it's simply two sublattices moving opposite to each other. How comes an opposite motion of two sublattices add energy to the system?

I think there must be some mistake in my understanding of the concepts. Please point them out, thank you.

• At $k=0$ in the acoustic branch, the atoms aren't moving at all. At $k=0$ in the optical branch, the two sublattices are moving relative to each other, so there is both non-zero kinetic energy and potential energy (due to the stretching/compressing of springs between the atoms in different sublattices). Feb 22 at 20:23

1. For the acoustic mode, $$k=0$$ means all lattice points oscillate in phase. But still, they are oscillating so each lattice point must have some finite frequency (otherwise they would hold still or travel in a single direction indefinitely). How comes this frequency is zero?
It's zero because the dispersion relation forces it to be zero. The frequency $$\omega$$ is directly proportional to the wavenumber $$k$$. So, as $$k$$ goes to zero so does $$\omega$$.
1. The energy of the phonon is given by $$\hbar\omega$$. This implies at $$k=0$$, the acoustic phonon has no energy, how?
The energy of the acoustic phonon at $$k=0$$ is zero, because $$k=0$$ is an infinite wavelength and infinite period "oscillation." I put "oscillation" in quotes because at $$k=0$$ the mode is just a overall translation of the crystal in space. The overall translation in space of the cryptal can't increase the total energy, so the energy of the associated excitation has to be zero.