Unformatted text preview: Chapter 12: Solid state physics 63 12.3 Crystal vibrations 12.3.1 A lattice with one type of atoms In this model for crystal vibrations only nearestneighbour interactions are taken into account. The force on atom s with mass M can then be written as: F s = M d 2 u s dt 2 = C ( u s +1 u s ) + C ( u s 1 u s ) Assuming that all solutions have the same timedependence exp( i ω t ) this results in: M ω 2 u s = C ( u s +1 + u s 1 2 u s ) Further it is postulated that: u s ± 1 = u exp( isKa ) exp( ± iKa ) . This gives: u s = exp( iKsa ) . Substituting the later two equations in the ¡st results in a system of linear equations, which has only a solution if their determinant is 0. This gives: ω 2 = 4 C M sin 2 ( 1 2 Ka ) Only vibrations with a wavelength within the ¡rst Brillouin Zone have a physical signi¡cance. This requires that π < Ka ≤ π . The group velocity of these vibrations is given by: v g = d ω dK = Ca 2 M cos( 1 2 Ka ) ....
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 Spring '10
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 Force, Mass, Photon, Solid State Physics, Brillouin zone, M1 M2 ±C, crystal vibrations, ﬁrst Brillouin Zone

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