From 6893272d5705b9dff8c01c3bfbf6410631caf49e Mon Sep 17 00:00:00 2001
From: hackbard <hackbard@hackdaworld.org>
Date: Wed, 4 Jun 2008 02:04:02 +0200
Subject: [PATCH] tarting of exc 2

---
 solid_state_physics/tutorial/2_03s.tex | 24 ++++++++++++++++++++++++
 1 file changed, 24 insertions(+)

diff --git a/solid_state_physics/tutorial/2_03s.tex b/solid_state_physics/tutorial/2_03s.tex
index 882df74..fb0e83b 100644
--- a/solid_state_physics/tutorial/2_03s.tex
+++ b/solid_state_physics/tutorial/2_03s.tex
@@ -150,6 +150,30 @@ w=\frac{1}{V}\frac{\sum_i E_i \exp(-\beta E_i)}{\sum_i \exp(-\beta E_i)}.
        \[
    w=-\frac{1}{V}\frac{\partial}{\partial \beta} ln \sum_i \exp(-\beta E_i).
        \]
+ \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
+       \end{eqnarray}
+
  \item Evaluate the expression of the energy density.
        {\bf Hint:}
        The energy levels of a harmonic crystal of N ions
-- 
2.20.1