The Debye Model

Calculating the number of particles for a linear dispersion relation we get from Eq (3.32) a limiting frequency

$ω_{D}^{3} = 6 π^{2} v^{3} \frac{N}{V} .$

(3.35)

The corresponding density of states is

$D (ω) = \frac{V ω^{2}}{2 π^{2} v^{3}} .$

(3.36)

Taking into account the three orientation in space we get for the inner energy

$U = 3 \int d ω \frac{V ω^{2}}{2 π^{2} v^{3}} \frac{ℏ ω}{exp (\frac{ℏ ω}{k T}) - 1} .$

(3.37)

leading to:

$C_{V} = \frac{d U}{d T} = 9 N k {(\frac{T}{Θ})}^{3} \int_{0}^{x_{D}} d x \frac{x^{4} e^{x}}{{(e^{x} - 1)}^{2}} .$

(3.38)

with

$x_{D} : = \frac{ℏ ω_{D}}{k T} : = \frac{Θ}{T} .$

(3.39)

and $Θ$ : Debye temperature.
The limiting cases are:
I: $T ≪ Θ$ , i.e. $x_{D} \to \infty$

$C_{V} = 9 N k {(\frac{T}{Θ})}^{3} \frac{π^{4}}{15}$

(3.40)

II: $T ≫ Θ$ , i.e. $x_{D} \to 0$

$\frac{x^{4} e^{x}}{{(e^{x} - 1)}^{2}} \approx x^{2}$

(3.41)

and consequently

$C_{V} = 9 N k {(\frac{T}{Θ})}^{3} \int_{0}^{x_{D}} x^{2} d x = 3 N k$

(3.42)

This is the expected classical result (the Hamiltonian is a bilinear function of the coordinates).