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-rw-r--r--vorlesungen/punktgruppen/slides.pdfbin36038 -> 50470 bytes
-rw-r--r--vorlesungen/punktgruppen/slides.tex115
2 files changed, 112 insertions, 3 deletions
diff --git a/vorlesungen/punktgruppen/slides.pdf b/vorlesungen/punktgruppen/slides.pdf
index bd7afc9..0851fd8 100644
--- a/vorlesungen/punktgruppen/slides.pdf
+++ b/vorlesungen/punktgruppen/slides.pdf
Binary files differ
diff --git a/vorlesungen/punktgruppen/slides.tex b/vorlesungen/punktgruppen/slides.tex
index 25761c0..e800a87 100644
--- a/vorlesungen/punktgruppen/slides.tex
+++ b/vorlesungen/punktgruppen/slides.tex
@@ -8,6 +8,7 @@
\usepackage{tikz}
\usetikzlibrary{positioning}
\usetikzlibrary{arrows.meta}
+\usetikzlibrary{calc}
% Theme
\beamertemplatenavigationsymbolsempty
@@ -95,7 +96,7 @@
\draw[white, very thick, -Circle] (pos.east) to ++ (1,0) node (p) {};
\draw[white, very thick, -Circle] (neg.east) to ++ (1,0) node (n) {};
- \draw[white, thick, ->] (p) to[out = -80, in = 80] node[midway, right] {\(U = 0\)} (n);
+ \draw[white, thick, ->] (p) to[out = -70, in = 70] node[midway, right] {\(U = 0\)} (n);
\end{scope}
\begin{scope}[
node distance = 0cm,
@@ -124,7 +125,7 @@
\draw[white, very thick, -Circle] (pos.east) to ++ (1,0) node (p) {};
\draw[white, very thick, -Circle] (neg.east) to ++ (1,0) node (n) {};
- \draw[white, thick, ->] (p) to[out = -80, in = 80] node[midway, right] {\(U \neq 0\)} (n);
+ \draw[white, thick, ->] (p) to[out = -70, in = 70] node[midway, right] {\(U \neq 0\)} (n);
\end{scope}
\end{tikzpicture}
\end{center}
@@ -142,6 +143,114 @@
\section{Matrizen}
\section{Kristalle}
+\begin{frame}[fragile]{}
+ \begin{columns}[T]
+ \begin{column}{.5\textwidth}
+ Kristallgitter:
+ \(n_i \in \mathbb{Z}\),
+ \(\vec{a}_i \in \mathbb{R}^3\)
+ \[
+ \vec{r} = n_1 \vec{a}_1 + n_2 \vec{a}_2 + n_3 \vec{a}_3
+ \]
+ \begin{center}
+ \begin{tikzpicture}[
+ dot/.style = {
+ draw, circle, thick, white, fill = gray!40!background,
+ minimum size = 2mm,
+ inner sep = 0pt,
+ outer sep = 1mm,
+ },
+ ]
+
+ \begin{scope}
+ \clip (-1,-1) rectangle (4,3);
+ \foreach \y in {-5,-4,...,5} {
+ \foreach \x in {-5,-4,...,5} {
+ \node[dot, xshift=3mm*\y] (N\x\y) at (\x, \y) {};
+ }
+ }
+ \end{scope}
+
+ \draw[white, thick] (-1, -1) rectangle (4,3);
+
+ \draw[red!80!background, thick, ->] (N00) to node[midway, below] {\(\vec{a}_1\)} (N10);
+ \draw[cyan!80!background, thick, ->] (N00) to node[midway, left] {\(\vec{a}_2\)} (N01);
+
+ \end{tikzpicture}
+ \end{center}
+ Invariant (symmetrisch) unten Translation
+ \[
+ Q_i(\vec{r}) = \vec{r} + \vec{a}_i
+ \]
+ \end{column}
+ \begin{column}{.5\textwidth}
+ Wie kombiniert sich \(Q_i\) mit der anderen Symmetrien?
+ \begin{center}
+ \begin{tikzpicture}[
+ dot/.style = {
+ draw, circle, thick, white, fill = gray!40!background,
+ minimum size = 2mm,
+ inner sep = 0pt,
+ outer sep = 1mm,
+ },
+ ]
+
+ \node[dot] (A1) at (0,0) {};
+ \node[below left] at (A1) {\(A\)};
+
+ \node[dot] (A2) at (2.5,0) {};
+ \node[below right] at (A2) {\(A'\)};
+
+ \draw[red!80!background, thick, ->]
+ (A1) to node[midway, below] {\(\vec{Q}\)} (A2);
+
+ \node[dot] (B1) at (120:2.5) {};
+ \node[above left] at (B1) {\(B\)};
+
+ \draw[green!70!background, thick, ->]
+ (A1) ++(.5,0) arc (0:120:.5) node[midway, above, xshift=1mm] {\(C_n\)};
+ \draw[red!80!background, dashed, thick, ->] (A1) to (B1);
+
+
+ \node[dot] (B2) at ($(A2)+(60:2.5)$) {};
+ \node[above right] at (B2) {\(B'\)};
+
+ \draw[green!70!background, thick, dashed, ->] (A2) ++(-.5,0) arc (180:60:.5);
+ \draw[red!80!background, dashed, thick, ->] (A2) to (B2);
+
+ \draw[yellow!80!background, thick, ->] (B1) to node[above, midway] {\(\vec{Q}'\)} (B2);
+
+ \draw[gray, dashed, thick] (A1) to (A1 |- B1) node (X) {};
+ \draw[gray, dashed, thick] (A2) to (A2 |- B2);
+
+ \node[above left, xshift=-2mm] at (X) {\(x\)};
+ \end{tikzpicture}
+ \end{center}
+ Sei \(q = |\vec{Q}|\), \(\alpha = 2\pi/n\) und \(n \in \mathbb{N}\)
+ \begin{align*}
+ q' = n q &= q + 2x \\
+ nq &= q + 2q\sin(\alpha - \pi/2) \\
+ n &= 1 - 2\cos\alpha
+ \end{align*}
+ \end{column}
+ \end{columns}
+\end{frame}
+
+\frame{
+ \begin{columns}[T]
+ \begin{column}{.5\textwidth}
+ Somit muss
+ \[
+ \alpha = \cos^{-1}\left(\frac{m-1}{2}\right)
+ \]
+ \begin{gather*}
+ \alpha \in \left\{ 0, 60^\circ, 90^\circ, 120^\circ, 180^\circ \right\}
+ \end{gather*}
+ \end{column}
+ \begin{column}{.5\textwidth}
+ \end{column}
+ \end{columns}
+}
\section{Anwendungen}
\begin{frame}[fragile]{}
@@ -154,7 +263,7 @@
]
\matrix [nodes = {box, align = center}, column sep = 1cm, row sep = 1.5cm] {
- & \node (A) {32 Punktgruppe}; \\
+ & \node (A) {32 Punktgruppen}; \\
\node (B) {11 Mit\\ Inversionszentrum}; & \node (C) {21 Ohne\\ Inversionszentrum}; \\
& \node[fill=red!20!background] (D) {20 Piezoelektrisch}; & \node (E) {1 Nicht\\ piezoelektrisch}; \\
};