advertisement

Final solutions Problem 1 (a) We can write Si Sj = −m2 + m(Si + Sj ) + δSi δSj The last term is due to fluctuations. We then have the mean field Hamiltonian X 1 Si HM F = N zJm2 − (zJm + B) 2 i (b) The partition function is N Z = e−βF = e − 21 βN zJm2 X eβ(zJm+B)S S=−1,0,+1 from which we get F = 1 zJm + B N zJm2 − N kB T ln 1 + 2 cosh . 2 kB T In terms of dimensionless quantities we get m+h 1 2 f = m − θ ln 1 + 2 cosh . 2 θ Problem 2 The first condition gives ∂P ∂v =− T RT 3a + 4 =0 2 (v − b) v which is equivalent to RT = 3a (v − b)2 . v4 The second condition gives 2 2RT 12a ∂ P = − 5 =0 ∂v 2 T (v − b)3 v 1 or, equivalently (v − b)3 . v5 Equating the two equations for RT we get vc = 2b. Substituting this result in anyone of the equations for RT we obtain RT = 6a Tc = 3a . 16b2 R Substituting vc and Tc in the equation of state we get a . 16b3 Pc = Finally RTc 3 = Pc v c 2 which is a very low number. Problem 3 As explained in class, we can assume that the chemical potential is zero in the region of Bose-Einstein condensation. The number of particles in the excited state is thus Z ∞ N̄ g()d, Nexc = 0 where N̄ = and g()d = 1 e/kB T −1 , V 4πp2 dp V 4π 2/s 1 1/s−1 = 3 d 3 h h A s A so that Z Nexc = const × V 0 ∞ 3/s−1 d, e/kB T − 1 If we set x = /kB T , we get Nexc = const0 × V (kB T )3/s . Therefore, since TB is determined by the condition Nexc = N we find that TB ∝ N V s/3 T TB . The fraction Ngs =1− N 2 3/s . Finally, we find that Z ∞ N̄ g()d ∝ T 3/s+1 , U= 0 from which we find CV ∝ T 3/s , and Z S= 0 T CV dT ∝ T 3/s . T 3