\[
w=-\frac{1}{V}\frac{\partial}{\partial \beta} ln \sum_i \exp(-\beta E_i).
\]
- \item Evaluate the expression of the energy density.
- {\bf Hint:}
- The energy levels of a harmonic crystal of N ions
- can be regarded as 3N independent oscillators,
- whose frequencies are those of the 3N classical normal modes.
- The contribution to the total energy of a particular normal mode
- with angular frequency $\omega_s({\bf k})$
- ($s$: branch, ${\bf k}$: wave vector) is given by
- $(n_{{\bf k}s} + \frac{1}{2})\hbar\omega_s({\bf k})$ with the
- excitation number $n_{{\bf k}s}$ being restricted to integers greater
- or equal zero.
- The total energy is given by the sum over the energies of the individual
- normal modes.
- Use the totals formula of the geometric series to expcitly calculate
- the sum of the exponential functions.
- \item Separate the above result into a term vanishing as $T$ goes to zero and
- a second term giving the energy of the zero-point vibrations of the
- normal modes.
- \item Write down an expression for the specific heat.
- Consider a large crystal and thus replace the sum over the discrete
- wave vectors with an integral.
- \item Debye replaced all branches of the vibrational spectrum with three
- branches, each of them obeying the dispersion relation
- $w=ck$.
- Additionally the integral is cut-off at a radius $k_{\text{D}}$
- to have a total amount of N allowed wave vectors.
- Determine $k_{\text{D}}$.
- Evaluate the simplified integral and introduce the
- Debye frequency $\omega_{\text{D}}=k_{\text{D}}c$
- and the Debye temperature $\Theta_{\text{D}}$ which is given by
- $k_{\text{B}}\Theta_{\text{D}}=\hbar\omega_{\text{D}}$.
- Write down the resulting expression for the specific heat.
+ \item \begin{itemize}
+ \item Total energy contribution of a particular normal mode:
+ $(n_{{\bf k}s} + \frac{1}{2})\hbar\omega_s({\bf k})$
+ with $n_{{\bf k}s}=0,1,2,\ldots$
+ \item A state of the crystal is specified by the excitation numbers
+ of the 3N normal modes.
+ \item The total energy is the sum of the energies of the individual
+ normal modes:\\
+ $E=\sum_{{\bf k}s}(n_{{\bf k}s}+
+ \frac{1}{2})\hbar\omega_s({\bf k})$
+ \end{itemize}
+ \begin{eqnarray}
+ \Rightarrow
+ w&=&-\frac{1}{V}\frac{\partial}{\partial \beta} ln\left(
+ \prod_{{\bf k}s}(\exp(-\beta\hbar\omega_s({\bf k})/2)+
+ \exp(-3\beta\hbar\omega_s({\bf k})/2)+
+ \exp(-5\beta\hbar\omega_s({\bf k})/2)+
+ \ldots)
+ \right)\nonumber\\
+ &=&-\frac{1}{V}\frac{\partial}{\partial \beta} ln \prod_{{\bf k}s}
+ \frac{\exp(-\beta\hbar\omega_s({\bf k})/2)}
+ {1-\exp(-\beta\hbar\omega_s({\bf k}))}\nonumber\\
+ &=&-\frac{1}{V}\frac{\partial}{\partial \beta} \sum_{{\bf k}s} ln
+ \frac{\exp(-\beta\hbar\omega_s({\bf k})/2)}
+ {1-\exp(-\beta\hbar\omega_s({\bf k}))}\nonumber\\
+ &=&-\frac{1}{V}\sum_{{\bf k}s}
+ \frac{1-\exp(-\beta\hbar\omega_s({\bf k}))}
+ {\exp(-\beta\hbar\omega_s({\bf k})/2)}\nonumber\\
+ &&\times
+ \frac{(1-e^{-\beta\hbar\omega_s({\bf k})})
+ e^{-\beta\hbar\omega_s({\bf k})/2}(-\hbar\omega_s({\bf k})/2)+
+ e^{-\beta\hbar\omega_s({\bf k})/2}
+ e^{-\beta\hbar\omega_s({\bf k})}(-\hbar\omega_s({\bf k}))}
+ {(1-e^{-\beta\hbar\omega_s({\bf k})})^2}\nonumber\\
+ &=&-\frac{1}{V}\sum_{{\bf k}s}\hbar\omega_s({\bf k})
+ \frac{e^{-\beta\hbar\omega_s({\bf k})}-
+ \frac{1}{2}(1-e^{-\beta\hbar\omega_s({\bf k})})}
+ {1-e^{-\beta\hbar\omega_s({\bf k})}}\nonumber\\
+ &=&-\frac{1}{V}\sum_{{\bf k}s}\hbar\omega_s({\bf k})
+ \frac{\frac{1}{2}e^{-\beta\hbar\omega_s({\bf k})}-\frac{1}{2}}
+ {1-e^{-\beta\hbar\omega_s({\bf k})}}
+ =\frac{1}{V}\sum_{{\bf k}s}\hbar\omega_s({\bf k})\frac{1}{2}
+ \frac{1+e^{\beta\hbar\omega_s({\bf k})}}
+ {e^{\beta\hbar\omega_s({\bf k})}-1}\nonumber\\
+ &=&\frac{1}{V}\sum_{{\bf k}s}\hbar\omega_s({\bf k})\frac{1}{2}
+ \frac{1+e^{\beta\hbar\omega_s({\bf k})}}
+ {e^{\beta\hbar\omega_s({\bf k})}-1}
+ =\frac{1}{V}\sum_{{\bf k}s}\hbar\omega_s({\bf k})\frac{1}{2}
+ \frac{2+e^{\beta\hbar\omega_s({\bf k})}-1}
+ {e^{\beta\hbar\omega_s({\bf k})}-1}\nonumber\\
+ &=&\frac{1}{V}\sum_{{\bf k}s}\hbar\omega_s({\bf k})
+ (\frac{1}{e^{\beta\hbar\omega_s({\bf k})}-1}
+ +\frac{e^{\beta\hbar\omega_s({\bf k})}-1}
+ {2(e^{\beta\hbar\omega_s({\bf k})}-1)})
+ =\frac{1}{V}\sum_{{\bf k}s}\hbar\omega_s({\bf k})
+ (\underbrace{\frac{1}{e^{\beta\hbar\omega_s({\bf k})}-1}}_{n_s({\bf k})}
+ +\frac{1}{2})\nonumber
+ \end{eqnarray}
+ $n_s({\bf k})$: Mean excitation number of the normal mode ${\bf k}s$ at
+ temperature $T$.
+
+ \item \[
+ w=w_{\text{eq}}+
+ \frac{1}{V}\sum_{{\bf k}s}\frac{1}{2}\hbar\omega_s({\bf k})+
+ \frac{1}{V}\sum_{{\bf k}s}
+ \frac{\hbar\omega_s({\bf k})}{e^{\beta\hbar\omega_s({\bf k})}-1}
+ \]
+ \item \[
+ c_{\text{V}}=\frac{1}{V}\sum_{{\bf k}s}\frac{\partial}{\partial T}
+ \frac{\hbar\omega_s({\bf k})}{e^{\beta\hbar\omega_s({\bf k})}-1}
+ \]
+ Large crystal:
+ \[
+ \lim_{v\rightarrow\infty}c_{\text{V}}=\frac{1}{V}\sum_{{\bf k}s}
+ \frac{\partial}{\partial T}
+ \frac{\hbar\omega_s({\bf k})}{e^{\beta\hbar\omega_s({\bf k})}-1}
+ =\frac{\partial}{\partial T}
+ \sum_s\int\frac{d{\bf k}}{(2\pi)^3}
+ \frac{\hbar\omega_s({\bf k})}{e^{\beta\hbar\omega_s({\bf k})}-1}
+ \]
+ \item \begin{itemize}
+ \item {\color{red}3} branches with Debye dispersion relation
+ $w={\color{green}ck}$
+ \item Volume of $k$-space per wave vector: $(2\pi)^3/V$\\
+ $\Rightarrow (2\pi)^3N/V=4\pi k_{\text{D}}^3/3
+ \Rightarrow n=\frac{N}{V}=\frac{k_{\text{D}}^3}{6\pi^2}$
+ and $dn={\color{blue}\frac{k^2}{2\pi^2}dk}$
+ \item Debye frequency: $\omega_{\text{D}}=k_{\text{D}}c$
+ \item Debye temperature:
+ $k_{\text{B}}\Theta_{\text{D}}=\hbar\omega_{\text{D}}$,
+ $\Theta_{\text{D}}=\hbar ck_{\text{D}}/k_{\text{B}}$
+ \end{itemize}
+ Integral:
+ \[
+ c_{\text{V}}=\frac{\partial}{\partial T}\, {\color{red}3}\int_0^{k_D}
+ {\color{blue}\frac{k^2}{2\pi^2}dk} \frac{\hbar {\color{green}ck}}
+ {e^{\beta\hbar {\color{green}ck}}-1}=
+ \frac{\partial}{\partial T}\frac{3\hbar c}{2\pi^2}\int_0^{k_D}
+ \frac{k^3}{e^{\beta\hbar ck}-1}dk=
+ \frac{3\hbar c}{2\pi^2}\int_0^{k_D}
+ \frac{k^3e^{\beta\hbar ck}\beta\hbar ck\frac{1}{T}}
+ {(e^{\beta\hbar ck}-1)^2}dk
+ \]
+ Change of variables: $\beta\hbar ck=x$
+ \[
+ \Rightarrow
+ k=\frac{x}{\beta\hbar c} \quad \textrm{, } \quad
+ dk=\frac{1}{\beta\hbar c} dx
+ \]
+ \[
+ c_{\text{V}}=
+ \]
\end{enumerate}
\end{document}
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