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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?

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

Roger V.
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P M
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  • 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 '22 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? – Roger V. Jul 22 '22 at 14:30
  • @RogerVadim Phonons – P M Jul 22 '22 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 '22 at 19:46
  • @RogerVadim The absolute comes from $ \sqrt{sin(0.5ka)^2} $ derivation: https://openphysicslums.files.wordpress.com/2012/08/latticevibrations.pdf – P M Jul 22 '22 at 20:48
  • Exactly because is an inverse length it does not represent a position in the real space, so not in the crystal. See answer by @Moose. – nasu Jul 23 '22 at 00:16
  • If you understand that relationship, then you understand dispersion. It is a dispersion relationship and means the same for any type of wave. Just that for phonons is not linear. Even for photons in a medium is not linear – nasu Jul 23 '22 at 00:19

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I think nasu has already answered your question, but let me elaborate. k refers to the phonon wavenumber which is the inverse of wavelength. This is consistent with your observation that k is an inverse length. Nevertheless, k is defined in the reciprocal space not the real space. Small and large k corresponds to long and short wavelengths respectively. Short (long) wavelengths has many (few) oscillations giving a high (low) energy/frequency.

MOOSE
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  • @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} $. – P M Jul 23 '22 at 11:02
  • 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. – MOOSE Jul 25 '22 at 21:13