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phys documents (dragged) 69

phys documents (dragged) 69 - Chapter 12 Solid state...

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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 nearest-neighbour 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 time-dependence 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 fist 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 first Brillouin Zone have a physical significance. 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 ) . and is 0 on the edge of a Brillouin Zone. Here, there is a standing wave.
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