Phonons: Understanding k-Vectors & Brillouin Zone

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Homework Statement


http://en.wikipedia.org/wiki/Phonon
Can someone explain how the movie above the caption
k-Vectors exceeding the first Brillouin zone (red) do not carry more information than their counterparts (black) in the first Brillouin zone.
is a justification of the statement in the caption?

Homework Equations


The Attempt at a Solution

 
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you see that both the red and the black sinosodial curve represent the "same" physical state (the particle going up and down). They carry the same information.

Since k are multipiles of reciprocal lattice vectors, we see that each k of this kind carry the same information as the smallest one, i.e we only need the k that is contained in the 1st BZ. The other ones, we can "translate back" with a multiple of reciprocal lattice vectors.
 
I see. The key is that the two waves travel at different speeds and in different directions as a result of the dispersion relation, right?
 
ehrenfest said:
I see. The key is that the two waves travel at different speeds and in different directions as a result of the dispersion relation, right?

Hmm yes, I think so. You get an infinite set of soultion, but you can always relate them to the smallest one (smallest magnitude), by substracting or adding reciprocal lattice vectors.
 
Why does that same argument not imply that electromagnetic waves with wavevectors outside of some zone do not carry extra information? I guess I am confused about how the periodicity of the reciprocal lattice makes that argument true? If you put more of those black dots in there, would they just go up and down also, or is there something special about the black dots that they chose?
 
every black dot will show the same behavior.

The difference between EM and phonons is perhaps that phonon is not a particle, EM is a photon. And also that the crystal have translation symmetry-
 
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