Specific heat and thermal expansion
The energy of lattice vibrations is quantized (phonons). Phonons are thermally excited
thermal phonons. Their energy can be calculated based on the harmonic oscillator model.
En = (n + )ℏω with n=0,1,2,3…. To get the total energy, one has to sum up N oscillators.
How many oscillators are excited at a given temperature and what is the mean energy at this
temperature? Calculate the mean value of the occupancies <n>. One obtains the mean
thermal equilibrium occupancy of a certain state n as function of the thermal energy kT:
< n >= which is the Bose-Einstein distribution, valid for all quanta with an
exp( ) − 1
integer spin. Therefore the total energy of excited phonons is given by
E = ∑ (< n > + )ℏω (q ) .(*)
The number of allowed values of frequencies per unit volume of q space, for each branch for
a system with propagating waves within a crystal of size L³=(Na)³ is
∆ω (q ) = ( L / 2π )³ = const. . The sum of spectral number densities Sj(ω) over all the branches
ω max 3p ω max
j within the phonon dispersion ∫ S (ω )dω = ∑ ∫ S (ω )dω = 3 pN
j depends on the number
0 j =1 0
of atoms per cell p and number of unit cells N. In 3D space the spectral density of states
(DOS) Sj(ω) is given by S j (ω ) = ( )³∫∫ where dF is an area element with ω(q) =
2π | grad qω |
const. and gradqω the derivative dω/dq in 3D. In general case, the last formula has no
analytic solution. Therefore two models exit:
A: Einstein model, where the optical branch is approximated by a constant
0 ω ≠ ωE
S j (ω ) = where ωE is the Einstein frequency.
3 pN ω = ω E
B: Debye model, where the acoustic branches are approximated by ω=vs q with vs=const. as
the sound velocity up to a certain qmax. Here S j (ω ) = ω ² (for p=1) and ω < ωD, with ωD
3 Nvs3 3
being the Debye frequency given by : ωD = ( L / 2π )²4π .
Both approximations are helpful to calculate the specific heat contribution of phonons. Here
one needs to calculate the temperature dependent part of the equ.(*) which is:
∂E ∞ ℏω kT
cv = ( )V = const = k ∫ dωS (ω )( )² .
∂T 0 kT (exp( ℏω ) − 1)²
Using Einstein model for S(ω) one gets cv = 3k for the case of high T, i.e. ℏω << kT , but a
wrong result for low temperatures. Alternatively the Debye model provides the correct
T 3 x D x 4e x
relation for the low T case: cV = 3Nk ( 3 ) ∫ where ΘD is the Debye temperature
Θ D 0 (e x − 1)²
(Temperature where all possible phonon modes are exited) and xD=ΘD/T.