Group velocity at Brillouin zone boundary

1. Apr 1, 2012

infk

I am working on an assignment here;

A linear chain with a two-atom primitive basis, both atoms of the same mass but different nearest neighbor separation and thus different force constants.

I have made a rigorous calculation in order to find the dispersion relation ω(k), with extensive help from Kittel's book. But I wonder if there is a way to predict ω(k) qualitatively(?)
Anyway, I plugged in some values for the constants in my formula for ω(k) and drew the graps in mathematica. I (obviously) get different curves when I choose the plus sign and the minus sign in the quadratic formula. How should I interpret that? Are those the acoustical and optical branches?

I am asked to say something about the relative motion of the atoms at the brillouin zone boundary, within a cell AND from cell to cell, and relate that to the group velocity. Again, what can be qualitatively deducted here?
I know that the group velocity is the derivative of ω(k) so if my curves have a horizontal tangent at k =$\pm$ $\pi$/a the group velocity is zero there. I have read the wikipedia entry on group velocity but I can't figure out how that applies to the relative motion of the atoms. Are the cells seen as wave packages here?

2. Apr 1, 2012

Polyrhythmic

There is a very nice website http://physics-animations.com/Physics/English/phon_txt.htm explaining your example generally (it has different masses instead of different force constants, but you can just rename the constant in the first equation on the webpage).

The animation in the bottom left corner of the website illustrates the standard (qualitative) behaviour of such systems.

Yes.
According to the picture on the webpage, there are two branches (optical and acoustical -- actually, d acoustical and d*(n-1) optical branches, where d is the dimension and n the number of particles), the acoustical is the bottom one with the (for small k) linear dispersion relation, the optical the top one with the (for small k) constant disp.rel. The naming is explained on the webpage.

In the solutions of the acoustic branch, all atoms move in the same direction at a given time. In the optical branch, the atoms of one sort moves to the right while the atoms of the other sort move to the left (like I said, different force constants and masses are the same thing here).

The group velocity is d\omega(k)/dk. This means that, if you form a wave packet that has a central frequency k_0, its maximum will move with the velocity d\omega(k_0)/dk. As the derivative vanishes at the borders, the maximum will not move.
Something I didn't know is also explained on the webpage: At the B.Z. borders, one of the atom sorts doesn't move at all.

No. A wave package is formed by superposing (adding up) waves around a central frequency k_0, similar to a Fourier transform. This wave package has a group velocity.