From: hackbard <hackbard@sage.physik.uni-augsburg.de>
Date: Wed, 11 Jun 2008 09:14:12 +0000 (+0200)
Subject: final version
X-Git-Url: https://hackdaworld.org/gitweb/?a=commitdiff_plain;ds=sidebyside;h=3c04d03ca53a5321cc777399970f4a43ddbef905;hp=e3dace08d67225de11d0e1b0a829c8a7efafa545;p=lectures%2Flatex.git

final version
---

diff --git a/solid_state_physics/tutorial/2_04.tex b/solid_state_physics/tutorial/2_04.tex
index 08180fb..abaf2c4 100644
--- a/solid_state_physics/tutorial/2_04.tex
+++ b/solid_state_physics/tutorial/2_04.tex
@@ -39,7 +39,7 @@
  {\Large\bf Tutorial 4}
 \end{center}
 
-\vspace{8pt}
+\vspace{4pt}
 
 \section{Legendre transformation and Maxwell relations}
 
@@ -57,7 +57,7 @@
        in such a way that $g$ is immediately identified to be a function of
        the variables $x_1,\ldots,x_r$ and $u_{r+1},\ldots,u_n$,
        where $u_i$ is called the conjugate variable of $x_i$.
-       The transformation is called Legendre transformation.
+       This transformation is called Legendre transformation.
  \item By taking the derivatives of transformed thermodynamic potentials
        with respect to the variables they depend on,
        relations between intensive and extensive variables can be gained.
@@ -66,13 +66,13 @@
        Write down the total differential using the equalities
        $T=\left.\frac{\partial E}{\partial S}\right|_V$ and
        $-p=\left.\frac{\partial E}{\partial V}\right|_S$.
-       Find more relations by doing the transformation to the potentials
+       Use Legendre transformation to get the potentials
        \begin{itemize}
         \item $H=E+pV$ (Enthalpy)
         \item $F=E-TS$ (Helmholtz free energy)
         \item $G=H-TS=E+pV-TS$ (Gibbs free energy)
        \end{itemize}
-       and taking the appropriate derivatives.
+       and find more relations by taking the appropriate derivatives.
  \item For a thermodynamic potential $\Phi(X,Y)$ the following identity
        expressing the permutability of derivatives holds:
        \[
@@ -88,6 +88,8 @@
 
 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
@@ -95,8 +97,28 @@ if there is a change in temperature $T$ or in pressure $p$ of the system.
        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 
- \item
+ \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 $E$.
+       Evaluate the pressure by using the harmonic form of the internal energy.
+       {\bf Hint:}
+       Step 2 introduced an integral over the temperature $T'$.
+       Change the integration variable $T'$ to $x=\hbar\omega_s({\bf k})/T'$.
+       Use integration by parts with respect to $x$.
+ \item The normal mode frequencies of a rigorously harmonic crystal
+       are unaffected by a change in volume.
+       What does this imply for the pressure
+       (Which variables does the pressure depend on)?
+       Draw conclusions for the coefficient of thermal expansion.
+ \item Find an expression for $C_p-C_V$ in terms of temperature $T$,
+       volume $V$, the coefficient of thermal expansion $\alpha_V$ and
+       the inverse bulk modulus (isothermal compressibility)
+       $\frac{1}{B}=-\frac{1}{V}\left.\frac{\partial V}{\partial p}\right|_T$.\\
+       $C_p=\left.\frac{\partial E}{\partial T}\right|_p$ is the heat capacity
+       for constant pressure and
+       $C_V=\left.\frac{\partial E}{\partial T}\right|_V$ is the heat capacity
+       for constant volume.
 \end{enumerate}
 
 \end{document}