Statistical Mechanics - Physics at Oregon State University

242 APPENDIX A. SOLUTIONS TO SELECTED PROBLEMS.
Finally, we have CV =
∂U
∂T = −kBβ 2
∆ ≫ ɛ ⇒ CV ≈ kB(βɛ) 2
∂U
∂β and hence in these cases we find:
e βɛ
(e βɛ + 1) 2
∆ = ɛ ⇒ CV = kB(βɛ) 2 eβɛ (eβɛ − 1) 2
∆ ≪ ɛ ⇒ CV ≈ kB(for kBT ≫ ∆)
A.3 Solutions for chapter 3
Problem 3.
A system contains an ideal gas of atoms with spin 1
2 in a magnetic field B(r).
The concentration of the spin up (down) particles is n↑(r) ( n↓(r) ). The
temperature is T.
(A) Evaluate the total chemical potential for the spin up and down particles.
The magnetic moment of an atom with spin S is m = γµB S. Assuming
that the quantization axis is along the B-field, and that the energy per atom
is Umag = −m · B, we get:
µ↑(r) = kBT log( n↑(r) 1
) −
nQ(T ) 2 γµBB(r)
µ↓(r) = kBT log( n↓(r) 1
) +
nQ(T ) 2 γµBB(r)
(B) These two chemical potentials have to be the same and independent of r.
Explain why.
If either chemical potential were position dependent, there would be no
equilibrium and a flow of particles would result. Hence both the spin-up
and spin-down chemical potentials are independent of position. If these
two potentials were not the same, the spins of the atoms would flip, and
again that would be a non equilibrium situation.
(C) Calculate the magnetic moment of this gas as a function of position.
gives
µ = kBT log( n↑(r) 1
) −
nQ(T ) 2 γµBB(r)