Statistical Mechanics - Physics at Oregon State University

254 APPENDIX A. SOLUTIONS TO SELECTED PROBLEMS.
n(1 + nλ 3 5 −
T 2 2 ) ≈ 2λ −3
T λ
λ ≈ n 1
2 λ3T (1 + nλ 3 5 −
T 2 2 )
Inserting in the pressure, and only keeping terms to second order:
p ≈ 2kBT λ −3
5
T λ(1 − λ2− 2 )
p ≈ 2kBT λ −3
T n1
2 λ3T (1 + nλ 3 5 −
T 2 2 )(1 − n 1
2 λ3 5 −
T 2 2 )
p ≈ kBT n(1 + nλ 3 5 −
T 2 2 − n 1
2 λ3 5 −
T 2 2 )
B2(T ) = λ 3 7 −
T 2 2
This is positive, increasing the pressure, due to the fact that fermions do not
like to be in the same place.
Solution Problem 6.
The chemical potential follows from the equation for N:
N =
fF D(ɛo; T, µ)
which is in this case
N = 2V
(2π) 3
o
d 3 kfF D( 2 c 2 k 2 + m 2 c 4 ; T, µ)
and there are no convergence problems, like in the notes. This gives
N = 2V
(2π) 3
d 3 1
k
eβ(√2c2 k2 +m2c4−µ) + 1
with β = 1
kBT
and
The equation for N gives
. At T = 0 this reduces to
N = 2V
(2π) 3
ɛF =
k