Editing Inhomogene Maxwellgleichungen im Vakuum

<noinclude>{{Scripthinweis|Elektrodynamik|6|3}}</noinclude>
 (Erregungsgleichungen)

:<math>\begin{align}
& {{\varepsilon }_{0}}\nabla \cdot \bar{E}=\rho  \\
& \Leftrightarrow {{\partial }_{1}}{{E}^{1}}+{{\partial }_{2}}{{E}^{2}}+{{\partial }_{3}}{{E}^{3}}=\frac{1}{{{\varepsilon }_{0}}c}c\rho  \\
& \Leftrightarrow {{\partial }_{1}}{{F}^{10}}+{{\partial }_{2}}{{F}^{20}}+{{\partial }_{3}}{{F}^{30}}=\frac{1}{{{\varepsilon }_{0}}c}{{j}^{0}} \\
& \Leftrightarrow {{\partial }_{\nu }}{{F}^{\nu 0}}=\frac{1}{{{\varepsilon }_{0}}c}{{j}^{0}} \\
& wegen{{\partial }_{0}}{{F}^{00}}=0 \\
& auch{{\partial }_{i}}{{F}^{i0}}=\frac{1}{{{\varepsilon }_{0}}c}{{j}^{0}} \\
\end{align}</math>

#
# <math>\nabla \times \bar{B}-\frac{1}{{{c}^{2}}}\frac{\partial }{\partial t}\bar{E}={{\mu }_{0}}\left( \nabla \times \bar{H}-{{\varepsilon }_{0}}\frac{\partial }{\partial t}\bar{E} \right)={{\mu }_{0}}\bar{j}</math>
#

# Komponente

:<math>\begin{align}
& {{\partial }_{2}}{{B}^{3}}-{{\partial }_{3}}{{B}^{2}}={{\mu }_{0}}{{j}^{1}}+{{\varepsilon }_{0}}{{\mu }_{0}}\frac{\partial }{\partial t}{{E}^{1}} \\
& {{\mu }_{0}}c=\frac{1}{{{\varepsilon }_{0}}c} \\
& \Leftrightarrow {{\partial }_{2}}{{F}^{21}}-.{{\partial }_{3}}{{F}^{13}}=\frac{1}{{{\varepsilon }_{0}}c}{{j}^{1}}+.{{\partial }_{0}}{{F}^{10}} \\
& {{\partial }_{2}}{{F}^{21}}+{{\partial }_{3}}{{F}^{31}}+{{\partial }_{0}}{{F}^{01}}=\frac{1}{{{\varepsilon }_{0}}c}{{j}^{1}} \\
& \Leftrightarrow {{\partial }_{\nu }}{{F}^{\nu 1}}=\frac{1}{{{\varepsilon }_{0}}c}{{j}^{1}} \\
& wegen{{\partial }_{1}}{{F}^{11}}=0 \\
\end{align}</math>

Dies kann analog für die zweite und dritte Komponente durchgeixt werden. Aus der Nullten Komponente hatten wir die Nullte des Stroms (Erregungsgleichung des elektrischen Feldes), so dass insgesamt folgt:

:<math>\begin{align}
& {{\partial }_{\nu }}{{F}^{\mu \nu }}=-\frac{1}{{{\varepsilon }_{0}}c}{{j}^{\mu }} \\
& {{\partial }_{\nu }}{{F}^{\nu \mu }}=\frac{1}{{{\varepsilon }_{0}}c}{{j}^{\mu }} \\
\end{align}</math>

Die Viererdivergenz des elektrischen Feldstärketensors!

'''Bemerkungen'''

# die homogenen Maxwellgleichungen sind durch den Potenzialansatz

:<math>\left\{ {{F}_{\mu \nu }} \right\}=\left\{ {{\partial }_{\mu }}{{\Phi }_{\nu }}-{{\partial }_{\nu }}{{\Phi }_{\mu }} \right\}=\left( \begin{matrix}
0 & \frac{1}{c}{{E}_{x}} & \frac{1}{c}{{E}_{y}} & \frac{1}{c}{{E}_{z}}  \\
-\frac{1}{c}{{E}_{x}} & 0 & -{{B}_{z}} & {{B}_{y}}  \\
-\frac{1}{c}{{E}_{y}} & {{B}_{z}} & 0 & -{{B}_{x}}  \\
-\frac{1}{c}{{E}_{z}} & -{{B}_{y}} & {{B}_{x}} & 0  \\
\end{matrix} \right)</math>

automatisch erfüllt:

:<math>\begin{align}
& {{\varepsilon }^{\alpha \beta \mu \nu }}{{\partial }_{\beta }}{{F}_{\mu \nu }}={{\varepsilon }^{\alpha \beta \mu \nu }}{{\partial }_{\beta }}{{\partial }_{\mu }}{{\Phi }_{\nu }}-{{\varepsilon }^{\alpha \beta \mu \nu }}{{\partial }_{\beta }}{{\partial }_{\nu }}{{\Phi }_{\mu }} \\
& {{\varepsilon }^{\alpha \beta \mu \nu }}{{\partial }_{\beta }}{{\partial }_{\mu }}{{\Phi }_{\nu }}=0, \\
& da:{{\partial }_{\beta }}{{\partial }_{\mu }}{{\Phi }_{\nu }}\quad symmetrisch \\
& {{\varepsilon }^{\alpha \beta \mu \nu }}\quad antisymmetrisch \\
& {{\varepsilon }^{\alpha \beta \mu \nu }}{{\partial }_{\beta }}{{\partial }_{\nu }}{{\Phi }_{\mu }}=0 \\
\end{align}</math>

Aus den inhomogenen Maxwell- Gleichungen

:<math>{{\partial }_{\beta }}{{F}^{\beta \nu }}={{\partial }_{\beta }}{{\partial }^{\beta }}{{\Phi }^{\nu }}-{{\partial }_{\beta }}{{\partial }^{\nu }}{{\Phi }^{\beta }}=\frac{1}{{{\varepsilon }_{0}}c}{{j}^{\nu }}</math>

folgt mit Lorentz- Eichung

:<math>{{\partial }_{\mu }}{{\Phi }^{\mu }}=0</math>

:<math>\begin{align}
& {{\partial }_{\beta }}{{\partial }^{\nu }}{{\Phi }^{\beta }}={{\partial }^{\nu }}{{\partial }_{\beta }}{{\Phi }^{\beta }}=0 \\
& also: \\
\end{align}</math>

:<math>{{\partial }_{\beta }}{{F}^{\beta \nu }}={{\partial }_{\beta }}{{\partial }^{\beta }}{{\Phi }^{\nu }}=\frac{1}{{{\varepsilon }_{0}}c}{{j}^{\nu }}</math>
als inhomogene Wellengleichung

'''Die Maxwellgleichungen'''

:<math>\begin{align}
& {{\varepsilon }^{\alpha \beta \mu \nu }}{{\partial }_{\beta }}{{F}_{\mu \nu }}={{\varepsilon }^{\alpha \beta \mu \nu }}{{\partial }_{\beta }}{{\partial }_{\mu }}{{\Phi }_{\nu }}-{{\varepsilon }^{\alpha \beta \mu \nu }}{{\partial }_{\beta }}{{\partial }_{\nu }}{{\Phi }_{\mu }}=0 \\
& {{\partial }_{\beta }}{{F}^{\beta \nu }}={{\partial }_{\beta }}{{\partial }^{\beta }}{{\Phi }^{\nu }}=\frac{1}{{{\varepsilon }_{0}}c}{{j}^{\nu }} \\
\end{align}</math>

sind ihrerseits nun Lorentz- kovariant, da sie durch 4 Pseudovektoren ausgedrückt sind.
Merke: Pseudo - 4- Vektor stört nicht, da rechte Seite gleich Null!!

<u>'''Gauß- System:'''</u>

:<math>{{\partial }_{\beta }}{{F}^{\beta \nu }}=\frac{4\pi }{c}{{j}^{\nu }}</math>