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# p09fl17 - ξ n 1 B lattice with a basis L16C C Two more...

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Lecture 17: Oscillator chains (12 Oct 09) A. linear chain – complete L16B 1. Chain of longitudinal oscillators L = T - V = m 2 X j ˙ η 2 j - K 2 X j ( η j +1 - η j ) 2 m ¨ η j = - K [2 η j - η j +1 - η j - 1 ] 2. Construct solutions: η j = y 0 exp( i [ jq‘ - ω ( q ) t ]) ( q ) 2 = 2 K [1 - cos( q‘ )] select q by boundary conditions at end of chain, e.g., periodic boundary condition q n = n 2 π/N‘ and fixed ends q n = nπ/N‘ and mapping to ± symmetry - π/‘ < q < π/‘ . 3. Then form the total energy using (periodic b. condition case) complex ξ n = Ξ exp( n ) η j = N X n =1 [ ξ n exp( i [ jq n - ω n t ]) + c.c. ] / 2 N X j =1 ˙ η 2 j = N X n ω 2 n | ξ n | 2 / 2 X j K ( η j - η j - 1 ) 2 = X n 2 K [1 - cos( q n )] | ξ n | 2 / 2 E = 1 2 X n ( q n ) 2 | ξ n | 2 4. The energy is the sum of the energies in the normal modes. Count of variables to specify the solution: 2 N , here two components of the

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Unformatted text preview: ξ n . 1 B. lattice with a basis: L16C C. Two more examples 1. FW #4.11: stretched spring. Rest length = a , average length along ˆ x is = ‘ with 2 directions for displacement: R = q ( x j-x j-1 ) 2 + ( y j-y j-1 ) 2 = q ( ‘ + η j-η j-1 ) 2 + ( y j-y j-1 ) 2 R-a ’ ( ‘-a ) + η j-η j-1 + (1 / 2 ‘ )( y j-y j-1 ) 2 ( K/ 2)( R-a ) 2 ’ ( K/ 2)[( ‘-a ) 2 + 2( ‘-a )( η j-η j-1 ) + ( η j-η j-1 ) 2 + (1-[ a/‘ ])( y j-y j-1 ) 2 ] The sign of the “force constant” for the y-motion depends on the sign of ‘-a . 2. FW #4.13. 2...
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