10. Appendix

Solution to Problem 3.17 617
forces holding the atoms together, and a second one stems from the vibration
of the atoms. Since phonons are bosons (like photons), the calculation of
the free energy term due to phonons is the same as that for photons and can
be found in many standard textbooks on statistical mechanics (see, for example,
C. Kittel and H. Kroemer: Thermal Physics, second edition. Freeman, San
Francisco, 1980. p. 112). The contribution Fvib to the total free energy from
the phonons can be calculated from the partition function Zvib of phonons:
Fvib kBT ln Zvib where kB is the Boltzmann’s constant and
Zvib
1 (n
e 2)
i,n
ˆi
kBT e
i
ˆi 1
2kBT 1 e ˆ
i
kBT 1
(3.17.3)
ˆi
i 2 sinh
2kBT
with the summation i over all the phonon modes.
From this partition function we obtain:
Fvib kBT
ˆi
ln 2 sinh
(3.17.4)
2kBT
i
The energy º is assumed now to be a function of v only. Using (3.17.4) for
Fvib we obtain an expression for P:
dº
P kBT
dv
ˆi dˆi
coth
(3.17.5)
2kBT 2kBT dv
i
To simplify this expression we assume that the mode Grüneisen parameters
for all the phonon modes are roughly the same and can be approximated by
an average Grüneisen parameter 〈Á〉 ∼d(ln ˆi)/d(ln v) for all phonon modes.
Within this approximation the expression for P simplifies to:
dº 〈Á〉
ˆi ˆi
P ≈
coth
(3.17.6)
dv v
2kBT 2
i
i
The average energy (like the average energy for photons) of a phonon mode
with frequency ˆ is given by:
U(ˆ) {n(1/2)}(ˆ) where n is the phonon occupancy or Bose-Einstein
distribution function (see p. 126), given by: n [exp(ˆ/kBT)1] 1 . Using the
result that {n (1/2)} (1/2) coth[(ˆ/2kBT)] we obtain:
ˆi ˆi
coth
U (3.17.7)
2kBT 2
where U represents the internal energy of the crystal due to vibrational modes
only. Hence we arrive at:
P (dº/dv) 〈Á〉U/v (3.17.8)
We can regard this relation between P and v as an equation-of-state of the
crystal.
Using this equation we can obtain the relation between the coefficient of
thermal expansion ‚ and 〈Á〉 under a quasi-harmonic approximation. In this