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Unformatted text preview: Solutions to problem set 5 (141A Sp07) 1 1. (ISSP 5.4) (a) The density of modes is given by D ( ω ) d ω = 2 2 π kdk (2 π/ L ) 2 = A π v 2 s ω d ω where A is the area of the system, v s is the sound velocity, and the factor of 2 comes from the two di ff erent branches (along a high symmetry direction, the two di ff erent polarizations, one longitudinal and one transverse). Therefore U = Z ω D d ω D ( ω ) ~ ω e ~ ω/ k B T 1 = A ~ π v 2 s Z ω D d ω ω 2 e ~ ω/ k B T 1 = A ~ π v 2 s k B T ~ ! 3 Z T D / T dx x 2 e x 1 As T D / T → ∞ , U = ( const ) T 3 and so C V ∼ T 2 . (b) Let z be the direction in which the binding is weak, and let the dispersion be ω 2 = v 2 1 ( k 2 x + k 2 y ) + v 2 2 k 2 z with v 1 v 2 . For extremely low temperatures, ie k B T ~ v 1 2 π L only modes with k x = k y = 0 will be occupied and so the system is e ff ectively one dimensional. The transition temperature is (for a best case scenario I choose a high v 1 , such as one might find for graphene) T * ’ ~ k B v 1 2 π a 1 N 1 / 3 ’ (1000 K )10 8 ’ 10 5 K This is a very low temperature indeed. For higher temperatures, the surface of constant angular frequency in kspace is a prolate ellipsoid, with an zaxis intersection at ω/ v 2 , and x and yaxes intersections at...
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 Spring '08
 SOUZA
 Thermodynamics, Energy, Solid State Physics, Heat

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