Calculating the number of particles for a linear dispersion relation we get from Eq (3.32) a limiting frequency
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| \begin{equation*} \omega_D^3=6\pi^2v^3\frac{N}{V} \qquad . \end{equation*} | (3.35) |
The corresponding density of states is
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| \begin{equation*} D(\omega) =\frac{V\omega^2}{2\pi^2v^3} \qquad . \end{equation*} | (3.36) |
Taking into account the three orientation in space we get for the inner energy
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| \begin{equation*} U = 3 \int d\omega \frac{V\omega^2}{2\pi^2v^3} \frac{\hbar \omega}{\exp\left(\frac{\hbar \omega}{k T}\right)-1} \qquad . \end{equation*} | (3.37) |
leading to:
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| \begin{equation*} C_V = \frac{dU}{dT}=9 N k \left(\frac{T}{\Theta}\right)^3 \int_0^{x_D} dx \frac{x^4 e^x}{\left(e^x-1\right)^2} \qquad . \end{equation*} | (3.38) |
with
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| \begin{equation*} x_D:=\frac{\hbar \omega_D}{k T}:=\frac{\Theta}{T} \qquad . \end{equation*} | (3.39) |
and \(\Theta\): Debye temperature.
The limiting cases
are:
I: \(T \ll \Theta\), i.e. \(x_D \rightarrow \infty\)
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| \begin{equation*} C_V = 9 Nk \left(\frac{T}{\Theta}\right)^3 \frac{\pi^4}{15} \end{equation*} | (3.40) |
II: \(T \gg \Theta\), i.e. \(x_D \rightarrow 0\)
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| \begin{equation*} \frac{x^4 e^x}{\left(e^x-1\right)^2} \approx x^2 \end{equation*} | (3.41) |
and consequently
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| \begin{equation*} C_V = 9 N k \left(\frac{T}{\Theta}\right)^3 \int_0^{x_D} x^2 dx = 3 N k \end{equation*} | (3.42) |
This is the expected classical result (the Hamiltonian is a bilinear function of the coordinates).
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© J. Carstensen (Stat. Meth.)