nearly finished

diff --git a/solid_state_physics/tutorial/2_03s.tex b/solid_state_physics/tutorial/2_03s.tex
index fb0e83b..f1b880d 100644 (file)
--- a/solid_state_physics/tutorial/2_03s.tex
+++ b/solid_state_physics/tutorial/2_03s.tex
@@ -171,41 +171,77 @@ w=\frac{1}{V}\frac{\sum_i E_i \exp(-\beta E_i)}{\sum_i \exp(-\beta E_i)}.
        \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
+            {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 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 \[
+       w=w_{\text{eq}}+
+         \frac{1}{V}\sum_{{\bf k}s}\frac{1}{2}\hbar\omega_s({\bf k})+
+         \frac{1}{V}\sum_{{\bf k}s}
+        \frac{\hbar\omega_s({\bf k})}{e^{\beta\hbar\omega_s({\bf k})}-1}
+       \]
+ \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: ($\lim_{V\to\infty}\frac{1}{V}\sum_{{\bf k}}F({\bf k})
+                        =\int\frac{d{\bf k}}{(2\pi)^3}F({\bf k})$)
+       \[
+       \Rightarrow
+       c_{\text{V}}=\frac{\partial}{\partial T}
+       \sum_s\int\frac{d{\bf k}}{(2\pi)^3}
+       \frac{\hbar\omega_s({\bf k})}{e^{\beta\hbar\omega_s({\bf k})}-1}
+       \]
+ \item \begin{itemize}
+        \item Debye dispersion relation: $w=ck$
+       \item Volume of $k$-space per wave vector:\\
+             $(2\pi)^3/V \Rightarrow (2\pi)^3N/V=4\pi k_{\text{D}}^3/3
+              \Rightarrow n=\frac{N}{V}=\frac{k_{\text{D}}^3}{6\pi^2}$
+       \item Debye frequency: $\omega_{\text{D}}=k_{\text{D}}c$
+       \item Debye temperature:
+             $k_{\text{B}}\Theta_{\text{D}}=\hbar\omega_{\text{D}}$
+       \end{itemize}
+       Integral:
+       \[
+       c_{\text{V}}=\ldots
+       \]
 \end{enumerate}
 
 \end{document}