From 4b4fe331cae31ee4043bc2aec02b8bcdca28395d Mon Sep 17 00:00:00 2001
From: hackbard <hackbard@sage.physik.uni-augsburg.de>
Date: Thu, 12 Jun 2008 16:01:10 +0200
Subject: [PATCH] seminar

---
 solid_state_physics/tutorial/2_04s.tex | 55 +++++++++++++++++---------
 1 file changed, 37 insertions(+), 18 deletions(-)

diff --git a/solid_state_physics/tutorial/2_04s.tex b/solid_state_physics/tutorial/2_04s.tex
index 287c275..bb9024c 100644
--- a/solid_state_physics/tutorial/2_04s.tex
+++ b/solid_state_physics/tutorial/2_04s.tex
@@ -89,6 +89,16 @@
        \left.\frac{\partial G}{\partial T}\right|_p=-S
        \]
  \item Maxwell relations:\\
+       Internal energy: $dE=TdS-pdV$
+       \[
+       \frac{\partial}{\partial S}
+       \left(\left.\frac{\partial E}{\partial V}\right|_S\right)_V=
+       \frac{\partial}{\partial V}
+       \left(\left.\frac{\partial E}{\partial S}\right|_V\right)_S
+       \Rightarrow
+       \left.-\frac{\partial p}{\partial S}\right|_V=
+       \left.\frac{\partial T}{\partial V}\right|_S
+       \]
        Enthalpy: $dH=TdS+Vdp$
        \[
        \frac{\partial}{\partial S}
@@ -109,31 +119,40 @@
        \left.-\frac{\partial S}{\partial V}\right|_T=
        \left.-\frac{\partial p}{\partial T}\right|_V
        \]
- \item For a thermodynamic potential $\Phi(X,Y)$ the following identity
-       expressing the permutability of derivatives holds:
+       Gibbs free energy: $dG=Vdp-SdT$
        \[
-       \frac{\partial^2 \Phi}{\partial X \partial Y} =
-       \frac{\partial^2 \Phi}{\partial Y \partial X}
+       \frac{\partial}{\partial p}
+       \left(\left.\frac{\partial G}{\partial T}\right|_p\right)_T=
+       \frac{\partial}{\partial T}
+       \left(\left.\frac{\partial G}{\partial p}\right|_T\right)_p
+       \Rightarrow
+       \left.-\frac{\partial S}{\partial p}\right|_T=
+       \left.\frac{\partial V}{\partial T}\right|_p
        \]
-       Derive the Maxwell relations by taking the mixed derivatives of the
-       potentials in (b) with respect to the variables they depend on.
-       Exchange the sequence of derivation and use the identities gained in (b).
 \end{enumerate}
 
 \section{Thermal expansion of solids}
 
-It is well known that solids change their length $L$ and volume $V$ respectively
-if there is a change in temperature $T$ or in pressure $p$ of the system.
-The following exercise shows that
-thermal expansion cannot be described by rigorously harmonic crystals.
-
 \begin{enumerate}
- \item The coefficient of thermal expansion of a solid is given by
-       $\alpha_L=\frac{1}{L}\left.\frac{\partial L}{\partial T}\right|_p$.
-       Show that the coefficient of thermal expansion of the volume
-       $\alpha_V=\frac{1}{V}\left.\frac{\partial V}{\partial T}\right|_p$
-       equals $3\alpha_L$ for isotropic materials.
- \item Find an expression for the pressure as a function of the free energy
+ \item Coefficients of thermal expansion:\\
+       Consider a cube with side lengthes $L_1,L_2,L_3$.
+       Isotropic material: $\frac{1}{L_1}\frac{\partial L_1}{\partial T}=
+                            \frac{1}{L_2}\frac{\partial L_2}{\partial T}=
+			    \frac{1}{L_3}\frac{\partial L_3}{\partial T}=
+			    \alpha_L$.
+       \begin{eqnarray}
+       \alpha_V&=&\frac{1}{V}\frac{\partial V}{\partial T}=
+       \frac{1}{L_1L_2L_3}\frac{\partial}{\partial T}(L_1L_2L_3)=
+       \frac{1}{L_1L_2L_3}\left(L_2L_3\frac{\partial L_1}{\partial T}+
+                                L_1L_3\frac{\partial L_2}{\partial T}+
+                                L_1L_2\frac{\partial L_3}{\partial T}\right)
+				\nonumber\\
+       &=&\frac{1}{L_1}\frac{\partial L_1}{\partial T}+
+          \frac{1}{L_2}\frac{\partial L_2}{\partial T}+
+          \frac{1}{L_3}\frac{\partial L_3}{\partial T}=3\alpha_L\nonumber
+       \end{eqnarray}
+ \item 
+       Find an expression for the pressure as a function of the free energy
        $F=E-TS$.
        Rewrite this equation to express the pressure entirely in terms of
        the internal energy $E$.
-- 
2.20.1