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Unformatted text preview: UIUC Physics 427 Homework 9 Solutions Philip Powell October 27, 2008 1. Problem 1 (K&K 7.3) (a) Consider a Fermi electron gas in its ground state. The energy of the gas is given by K&K U = Z F D ( ) d (1) where the density of states is given by D ( ) = V 2 π 2 2 m ~ 2 3 / 2 √ (2) As noted on page 217 of K&K we obtain the result U = 3 5 N F F = ~ 2 2 m 3 π 2 N V 2 / 3 (3) or equivalently U = 3 ~ 2 10 m 3 π 2 N V 2 / 3 (4) The pressure is given, in general, by p = ∂F ∂V τ,V (5) Since F = U τσ , at τ = 0 we have F = U so that p = ∂U ∂V τ,V (6) Substituting (4) we obtain p = (3 π 2 ) 2 / 3 5 ~ 2 m N V 5 / 3 (7) (b) Next, we seek to find the entropy of a Fermi electron gas for τ F . Recall that the definition of heat capacity is C = dU dτ (8) The thermodynamic identity is dU = τdσ pdV (9) 1 Combining these two expressions and requiring that dV = 0 gives C = τ ∂σ ∂τ V (10) From K&K page 218 we have C = π 2 2 N τ F + ··· (11) Combining these two expressions gives ∂σ ∂τ V = π 2 2 N 1 F + ··· (12) Integrating, we obtain σ = π 2 2 N τ F + ··· (13) where we have noted that the ground state entropy (i.e. σ (0)) is zero, since there is a unique ground state, so the multiplicity is 1. 2. Problem 2 (K&K 7.4) Consider the chemical potential μ ( T ) in d = 1 and d = 3 dimensions. We wish to determine the general behavior of μ ( T ) for small nonzero temperatures. We start by noting that the total number of particles in a system may be expressed in the form N = Z ∞ D...
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This note was uploaded on 12/06/2009 for the course PHYS 427 taught by Professor Flynn during the Fall '08 term at University of Illinois, Urbana Champaign.
 Fall '08
 Flynn
 Physics, Energy, Work

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