add first theo lecture without graphics
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\documentclass{lecture}
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\begin{document}
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\chapter{Phänomenologie der Maxwell-Gleichungen}
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\section{Elektrisches Feld}
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\begin{satz}[Colomb Gesetz]
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Kraft zwischen zwei Ladungen
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\[
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F = k \frac{q_1 q_2}{| \vec{r}_1 - \vec{r}_2 |^2}
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.\]
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Im SI-System ist $k = \frac{1}{4 \pi \epsilon_0}$ mit
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$\epsilon_0 = 8.854 \cdot 10^{-12} \frac{\text{As}}{\text{Vm}} = \frac{1}{4 \pi \cdot 9 \cdot 10^{9}}
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\frac{\text{As}}{\text{Vm}}$ mit A für Ampère und As $=$ C für Coulomb.
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Im Gauß-System ist $q = \frac{q^{*}}{\sqrt{4 \pi \epsilon_0} }$, so dass $k = 1$.
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\end{satz}
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\begin{figure}[h]
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\centering
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\begin{tikzpicture}
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\draw[fill] (-1, 0) circle (0.1cm) node[below left] {$q_1$};
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\draw[fill] (1, 0) circle (0.1cm) node[below left] {$q_2$};
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\end{tikzpicture}
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\caption{Zwei Ladungen}
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\end{figure}
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\section{Elektrische Feldstärke}
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\begin{satz}[Coulomb Kraft]
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\[
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\vec{F} = q \vec{E}
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.\]
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\end{satz}
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\begin{satz}[Lorentz Kraft]
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\[
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\vec{F} = q \vec{v} \times \vec{B}
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.\]
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\end{satz}
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\section{Maxwell-Gleichungen}
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Die Maxwell Gleichungen bilden die axiomatische Verbindung zwischen Feldern
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$\vec{E}, \vec{B}, \vec{D}, \vec{H}$ und den Ladungen $\rho, \vec{\jmath}$
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in Inertialsystemen (oder lokal in frei fallenden Systemen).
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\begin{enumerate}[(1)]
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\item
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\begin{salign*}
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\text{div } \vec{E} &= \nabla \cdot \vec{E} = \partial_i E^{i} = 4 \pi \rho
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\intertext{Integralform}
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\int_{V}^{} \text{div } \vec{E} \d[3]{x} &\stackrel{\text{Gauß}}{=}
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\int_{\partial V}^{} \d{\vec{S}} \cdot \vec{E} = \underbrace{\Psi}_{\text{elektrischer Fluss}}
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= \int_{V}^{} 4 \pi\rho \d[3]{r} = \underbrace{4 \pi q}_{\text{Ladung}}
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\intertext{Im Fall von sphärischer Symmetrie gilt}
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\int_{\partial V}^{} \d{\vec{S}} \cdot \vec{E} &= 4 \pi r^2 | \vec{E}(r) | = 4 \pi q
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\implies |E| \sim \frac{q}{r^2}
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.\end{salign*}
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\item
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\begin{salign*}
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\text{div } \vec{B} &= \nabla \times \vec{B} = \partial_i B^{i} = 0
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\intertext{Integralform}
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\int_{V}^{} \d[3]{r} \text{div } \vec{B} &= \int_{\partial V}^{} \d{\vec{S}} \cdot \vec{B}
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= \underbrace{\Phi}_{\text{magnetischer Fluss}} = 0
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.\end{salign*}
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Es gibt keine magnetische Ladung, es gilt also immer $\rho_m = 0$. Elektromagnetische Dualität.
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\item Faraday-Induktionsgesetz
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\begin{salign*}
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\text{rot } \vec{E} &= \nabla \times \vec{E} = \underbrace{-}_{\text{Lenz-Regel}} \partial_{ct} \vec{B}
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\intertext{Integralform}
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\int_{S}^{} \d{\vec{S}} \cdot \text{rot } \vec{E}
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&\stackrel{\text{Stokes}}{=} \int_{\partial S}^{} \d{\vec{r}} \cdot \vec{E}
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= \underbrace{U}_{\text{Spannung}}
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= - \frac{\d}{\d (ct)} \int_{}^{} \d{\vec{S}} \cdot \vec{B} = - \frac{\d}{\d (ct)} \Phi
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.\end{salign*}
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\item Ampère-Gesetz
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\begin{salign*}
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\text{rot } \vec{B} &= \nabla \times \vec{B} = \partial_{ct} \vec{E} + \frac{4 \pi}{c} \vec{\jmath}
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\intertext{Integralform}
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\int_{S}^{} \d{\vec{S}} \cdot \vec{B} &\stackrel{\text{Stokes}}{=}
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\int_{\partial S}^{} \d{\vec{r}} \cdot \vec{B}
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= \frac{\d }{\d ct} \underbrace{\int_{S}^{} \d{\vec{S}} \cdot \vec{E} }_{\Psi}
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+ \frac{4\pi}{c} \underbrace{\int_{}^{} \d{\vec{S}} \cdot \vec{\jmath}}_{I \text{ Strom}}
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\intertext{Im Fall von zylindrischer Symmetrie gilt:}
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\int_{\partial S}^{} \d{\vec{r}} \cdot \vec{B} &=
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4 \pi r | \vec{B}(r)| = \frac{4 \pi}{c} I \implies | \vec{B} | \sim \frac{1}{r}
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.\end{salign*}
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\end{enumerate}
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\section{Eigenschaften der Maxwell-Gleichungen}
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\begin{itemize}
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\item linear ($\to$ Superposition)
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\item partiell
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\item hyperbolisch
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\item Inertialsystem
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\item Überbestimmtheit?
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\end{itemize}
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\section{Erhaltung aus elektrischer Ladung}
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Betrachte Maxwell (1) und (4):
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\begin{salign*}
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\text{div } \vec{E} &= 4 \pi \rho \mid \partial_{ct}\\
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\text{rot } \vec{B} &= \partial_{ct} \vec{E} + \frac{4 \pi}{c} \vec{\jmath} \mid \text{div }
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\intertext{Damit folgt}
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\text{div } \text{rot } \vec{B} = 0 &= \partial_{ct} \text{div } \vec{E} + \frac{4 \pi}{c} \text{div }
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\vec{\jmath} \\
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&= 4 \pi \partial_{ct} \rho + \frac{4\pi}{c} \text{div } \vec{\jmath}
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\intertext{Also insgesamt die sog. Kontinuitätsgleichung}
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\partial_t \rho + \text{div } \vec{\jmath} &= 0 \tageq\label{eq:cont}
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.\end{salign*}
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In Integralform ergibt sich daraus:
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\begin{salign*}
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\frac{\d}{\d t} \int_{V}^{} \d[3]{r} \rho = \frac{\d }{\d t} q = - \int_{V}^{} \d[3]{r} \text{div }
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\vec{\jmath} &\stackrel{\text{Gauß}}{=} - \int_{\partial V}^{} \d{\vec{S}} \cdot \vec{\jmath}
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.\end{salign*}
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Die Elektrodynamik ist eine Kontinuumstheorie. Ladung ist dabei eine Art ,,Fluid``.
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\section{Elektromagnetische Dualität}
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Betrachte die Maxwell-Gleichungen im Vakuum, d.h. $\rho = 0$, $\vec{\jmath} = 0$. Dann gilt
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\begin{enumerate}[(1)]
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\item $\text{div } \vec{E} = 0$
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\item $\text{div } \vec{B} = 0$
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\item $\text{rot } \vec{E} = - \partial_{ct} \vec{B}$
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\item $\text{rot } \vec{B} = \partial_{ct} \vec{E}$
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\end{enumerate}
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Vertauschung $\vec{E} \longrightarrow \vec{B}$, $\vec{B} \longrightarrow - \vec{E}$: Dualität
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Warum existieren eigentlich keine magnetischen Ladungen?
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\begin{enumerate}[(1)]
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\item $\text{div } \vec{E} = 4 \pi \rho$
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\item $\text{div } \vec{B} = 4 \pi \rho_{m}$
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\item $\text{rot } \vec{E} = - \partial_{ct} \vec{B} + \frac{4 \pi}{c} \vec{\jmath}_m$
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\item $\text{rot } \vec{B} = \partial_{ct} \vec{E} + \frac{4 \pi}{c} \vec{\jmath}$
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\end{enumerate}
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Es folgt also direkt $\partial_t \rho_m + \text{div } \vec{\jmath} = 0$ analog zu \ref{eq:cont}.
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\end{document}
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BIN
Binary file not shown.
@@ -22,4 +22,6 @@ Rui Yang (\href{mailto:rui.yang@stud.uni-heidelberg.de}{rui.yang@stud.uni-heidel
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\tableofcontents
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\tableofcontents
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\input{theo1.tex}
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\end{document}
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\end{document}
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