# How can I think off the dispersion relation of a monoatomic chain or a crystal?

In a monoatomic chain, the dispersion relation is:

$$\omega = 2 \sqrt{\frac{K}{M}} \left|sin(k\frac{a}{2}) \right|$$However, does that mean that the phonons have a higher frequency (or energy $$\hbar \omega$$) at specific points (k) in the chain? Would k then be a direction in which the phonons have a higher frequency in a crystal? If so, how can one think of it?

I did see: Dispersion relation of a monoatomic chain but it did not help me.

• A value of k does not specify a point in the chain but a specific wavelength. The vector k does specify a direction in the crystal. Do you understand the linear "dispersion" relation $\omega = ck$ which is the relationship between ferquency and wavelength? Dispersion simply means that the relationship is not linear.
– nasu
Jul 22, 2022 at 14:02
• Are you talking about electrons, phonons or some other kind of excitations? In either case, where does the absolute value sign comes from? Jul 22, 2022 at 14:30
– P M
Jul 22, 2022 at 19:43
• @nasu Thanks for your answer. But the Brillouin zone is in k space. So it is an inverse length. Thus it has to do something with the position. I know this $\omega = c \cdot k$ relationship. I can interpret it for photons but not for phonons.
– P M
Jul 22, 2022 at 19:46
• @RogerVadim The absolute comes from $\sqrt{sin(0.5ka)^2}$ derivation: openphysicslums.files.wordpress.com/2012/08/…
– P M
Jul 22, 2022 at 20:48

• @Mose, I think I got it. $k = \frac{2 \pi}{\lambda}$ . k is just another representation of the wavelength. However, in the 3 D crystal, I have $\vec{k}$ which has a direction like [100] and length corresponding to $k = \frac{2 \pi}{\lambda}$.
• Yes, that is correct. Furthermore, the direction of k typically coincides with the propagation direction of the waves momentum. Plane waves of the form $\exp( \mathbf{k} \cdot \mathbf{x})$ are a nice example. Boldface here denotes vectors as usual. The momentum interpretation follows from equations like $\mathbf{p} = \hbar \mathbf{k}$. $\mathbf{k}$ is sometimes called the quasimomentum. Jul 25, 2022 at 21:13