Magnetostatische Feldgleichungen

Elektrodynamikvorlesung von Prof. Dr. E. Schöll, PhD

Der Artikel Magnetostatische Feldgleichungen basiert auf der Vorlesungsmitschrift von Franz- Josef Schmitt des 2.Kapitels (Abschnitt 3) der Elektrodynamikvorlesung von Prof. Dr. E. Schöll, PhD.

Magnetostatische Feldgleichungen	Stationäre Ströme und Magnetfeld	Elektrodynamik Schöll
	Kontinuitätsgleichung Magnetische Induktion Magnetostatische Feldgleichungen Magnetische Multipole	Elektrostatik Stationäre Ströme und Magnetfeld Die Maxwell-Gleichungen Elektromagnetische Wellen Materie in elektrischen und magnetischen Feldern Relativistische Formulierung der Elektrodynamik

Sie gelten auch in quasistaischer Näherung: Die zeitliche Änderung muss viel kleiner sein als die räumliche!!

Mit dem Vektorpotenzial

${\bar {A}}({\bar {r}})={\frac {{\mu }_{0}}{4\pi }}\int _{{R}^{3}}^{}{}{{d}^{3}}r{\acute {\ }}{\frac {{\bar {j}}({\bar {r}}{\acute {\ }})}{\left|{\bar {r}}-{\bar {r}}{\acute {\ }}\right|}}$

Welches nicht eindeutig ist, sondern beliebig gemäß

${\bar {A}}({\bar {r}})\to {\bar {A}}+\nabla \Psi$

umgeeicht werden kann.($\Psi ({\bar {r}})$ beliebig möglich, da $\nabla \times \nabla \Psi =0$)

Mit diesem Vektorpotenzial also kann man schreiben:

${\bar {B}}=rot{\bar {A}}({\bar {r}})=\nabla \times {\frac {{\mu }_{0}}{4\pi }}\int _{{R}^{3}}^{}{}{{d}^{3}}r{\acute {\ }}{\frac {{\bar {j}}({\bar {r}}{\acute {\ }})}{\left|{\bar {r}}-{\bar {r}}{\acute {\ }}\right|}}$

Beweis:

${\begin{aligned}&rot{\bar {A}}({\bar {r}})=\nabla \times {\frac {{\mu }_{0}}{4\pi }}\int _{{R}^{3}}^{}{}{{d}^{3}}r{\acute {\ }}{\frac {{\bar {j}}({\bar {r}}{\acute {\ }})}{\left|{\bar {r}}-{\bar {r}}{\acute {\ }}\right|}}={\frac {{\mu }_{0}}{4\pi }}\int _{{R}^{3}}^{}{}{{d}^{3}}r{\acute {\ }}{{\nabla }_{r}}{\frac {1}{\left|{\bar {r}}-{\bar {r}}{\acute {\ }}\right|}}\times {\bar {j}}({\bar {r}}{\acute {\ }})\\&{{\nabla }_{r}}{\frac {1}{\left|{\bar {r}}-{\bar {r}}{\acute {\ }}\right|}}=-{\frac {{\bar {r}}-{\bar {r}}{\acute {\ }}}{{\left|{\bar {r}}-{\bar {r}}{\acute {\ }}\right|}^{3}}}\\&\Rightarrow rot{\bar {A}}({\bar {r}})={\frac {{\mu }_{0}}{4\pi }}\int _{{R}^{3}}^{}{}{{d}^{3}}r{\acute {\ }}{\bar {j}}({\bar {r}}{\acute {\ }})\times {\frac {{\bar {r}}-{\bar {r}}{\acute {\ }}}{{\left|{\bar {r}}-{\bar {r}}{\acute {\ }}\right|}^{3}}}={\bar {B}}({\bar {r}})\\\end{aligned}}$

Folgende Aussagen sind äquivalent: Es existiert ein Vektorpotenzial mit

${\begin{aligned}&{\bar {B}}=rot{\bar {A}}({\bar {r}})\\&\Leftrightarrow \\\end{aligned}}$

$div{\bar {B}}=0$

Beweis:

$div(rot{\bar {A}}({\bar {r}}))=0$

es gibt keine Quellen der magnetischen Induktion (es existieren keine "magnetischen Ladungen".

Aber: Magnetische Monopole wurden 1936 von Dirac postuliert, um die Quantelung der Ladung zu erklären. (aus der quantenmechanischen Quantisierung des Drehimpulses!) Dies wurde durch die vereinheitlichte Feldtheori4e wieder aufgenommen! Es wurden extrem schwere magnetische Monopole postuliert, die beim Urknall in den ersten ${{10}^{-35}}s$ erzeugt worden sein sollen.

Sehr umstritten ist ein angeblicher experimenteller Nachweis von 1982 (Spektrum der Wissenschaft, Juni 1982, S. 78 ff.) Der Zusammenhang zwischen ${\bar {B}}({\bar {r}})$ und ${\bar {j}}({\bar {r}})$:

${\begin{aligned}&\nabla \times {\bar {B}}({\bar {r}})=\nabla \times \left(\nabla \times {\bar {A}}({\bar {r}})\right)=\nabla \left(\nabla \cdot {\bar {A}}({\bar {r}})\right)-\Delta {\bar {A}}({\bar {r}})\\&\nabla \cdot {\bar {A}}({\bar {r}})=\nabla \cdot {\frac {{\mu }_{0}}{4\pi }}\int _{{R}^{3}}^{}{}{{d}^{3}}r{\acute {\ }}{\frac {{\bar {j}}({\bar {r}}{\acute {\ }})}{\left|{\bar {r}}-{\bar {r}}{\acute {\ }}\right|}}={\frac {{\mu }_{0}}{4\pi }}\int _{{R}^{3}}^{}{}{{d}^{3}}r{\acute {\ }}{{\nabla }_{r}}\cdot \left({\frac {{\bar {j}}({\bar {r}}{\acute {\ }})}{\left|{\bar {r}}-{\bar {r}}{\acute {\ }}\right|}}\right)={\frac {{\mu }_{0}}{4\pi }}\int _{{R}^{3}}^{}{}{{d}^{3}}r{\acute {\ }}{\bar {j}}({\bar {r}}{\acute {\ }}){{\nabla }_{r}}\cdot {\frac {1}{\left|{\bar {r}}-{\bar {r}}{\acute {\ }}\right|}}\\&{{\nabla }_{r}}\cdot {\frac {1}{\left|{\bar {r}}-{\bar {r}}{\acute {\ }}\right|}}=-{{\nabla }_{r{\acute {\ }}}}\cdot {\frac {1}{\left|{\bar {r}}-{\bar {r}}{\acute {\ }}\right|}}\\&\Rightarrow \nabla \cdot {\bar {A}}({\bar {r}})=-{\frac {{\mu }_{0}}{4\pi }}\int _{{R}^{3}}^{}{}{{d}^{3}}r{\acute {\ }}\left[{{\nabla }_{r{\acute {\ }}}}\cdot \left({\frac {{\bar {j}}({\bar {r}}{\acute {\ }})}{\left|{\bar {r}}-{\bar {r}}{\acute {\ }}\right|}}\right)+{\frac {1}{\left|{\bar {r}}-{\bar {r}}{\acute {\ }}\right|}}{{\nabla }_{r{\acute {\ }}}}\cdot {\bar {j}}({\bar {r}}{\acute {\ }})\right]\\&{{\nabla }_{r{\acute {\ }}}}\cdot {\bar {j}}({\bar {r}}{\acute {\ }})=-{\frac {\partial }{\partial t}}\rho =0\\&\Rightarrow \nabla \cdot {\bar {A}}({\bar {r}})=-{\frac {{\mu }_{0}}{4\pi }}\int _{{R}^{3}}^{}{}{{d}^{3}}r{\acute {\ }}{{\nabla }_{r{\acute {\ }}}}\cdot \left({\frac {{\bar {j}}({\bar {r}}{\acute {\ }})}{\left|{\bar {r}}-{\bar {r}}{\acute {\ }}\right|}}\right)\\\end{aligned}}$

Wobei die verwendete Kontinuitätsgleichung natürlich nur für statische Ladungsverteilungen gilt!

Im Allgemeinen Fall gilt dagegen:

${\begin{aligned}&\Rightarrow \nabla \cdot {\bar {A}}({\bar {r}})=-{\frac {{\mu }_{0}}{4\pi }}\int _{{R}^{3}}^{}{}{{d}^{3}}r{\acute {\ }}{{\nabla }_{r{\acute {\ }}}}\cdot \left({\frac {{\bar {j}}({\bar {r}}{\acute {\ }})}{\left|{\bar {r}}-{\bar {r}}{\acute {\ }}\right|}}\right)-{\frac {\partial }{\partial t}}{\frac {{\mu }_{0}}{4\pi }}\int _{{R}^{3}}^{}{}{{d}^{3}}r{\acute {\ }}{\frac {\rho ({\bar {r}}{\acute {\ }},t)}{\left|{\bar {r}}-{\bar {r}}{\acute {\ }}\right|}}\\&{\frac {{\mu }_{0}}{4\pi }}\int _{{R}^{3}}^{}{}{{d}^{3}}r{\acute {\ }}{\frac {\rho ({\bar {r}}{\acute {\ }},t)}{\left|{\bar {r}}-{\bar {r}}{\acute {\ }}\right|}}={{\mu }_{0}}{{\varepsilon }_{0}}\Phi ({\bar {r}},t)\\&\Rightarrow \nabla \cdot {\bar {A}}({\bar {r}})=-{\frac {{\mu }_{0}}{4\pi }}\oint \limits _{S\infty }{}{{d}^{3}}{\bar {f}}{\acute {\ }}\left({\frac {{\bar {j}}({\bar {r}}{\acute {\ }})}{\left|{\bar {r}}-{\bar {r}}{\acute {\ }}\right|}}\right)-{{\mu }_{0}}{{\varepsilon }_{0}}{\frac {\partial }{\partial t}}\Phi ({\bar {r}},t)\\\end{aligned}}$

Mit dem Gaußschen Satz. Wenn das Potenzial jedoch ins unendliche hinreichend rasch abfällt, so gilt:

$\oint \limits _{S\infty }{}{{d}^{3}}{\bar {f}}{\acute {\ }}\left({\frac {{\bar {j}}({\bar {r}}{\acute {\ }})}{\left|{\bar {r}}-{\bar {r}}{\acute {\ }}\right|}}\right)=0$

Also:

$\nabla \cdot {\bar {A}}({\bar {r}})=-{{\mu }_{0}}{{\varepsilon }_{0}}{\frac {\partial }{\partial t}}\Phi ({\bar {r}},t)$

Also:

$\nabla \left(\nabla \cdot {\bar {A}}({\bar {r}})\right)={{\mu }_{0}}{{\varepsilon }_{0}}{\frac {\partial }{\partial t}}{\bar {E}}({\bar {r}},t)$

Auf der anderen Seite ergibt sich ganz einfach

${\begin{aligned}&\Delta {\bar {A}}({\bar {r}})={\frac {{\mu }_{0}}{4\pi }}\int _{{R}^{3}}^{}{}{{d}^{3}}r{\acute {\ }}{{\Delta }_{r}}\cdot \left({\frac {{\bar {j}}({\bar {r}}{\acute {\ }})}{\left|{\bar {r}}-{\bar {r}}{\acute {\ }}\right|}}\right)={\frac {{\mu }_{0}}{4\pi }}\int _{{R}^{3}}^{}{}{{d}^{3}}r{\acute {\ }}{\bar {j}}({\bar {r}}{\acute {\ }}){{\Delta }_{r}}\cdot \left({\frac {1}{\left|{\bar {r}}-{\bar {r}}{\acute {\ }}\right|}}\right)\\&={\frac {{\mu }_{0}}{4\pi }}\int _{{R}^{3}}^{}{}{{d}^{3}}r{\acute {\ }}{\bar {j}}({\bar {r}}{\acute {\ }})\delta \left({\bar {r}}-{\bar {r}}{\acute {\ }}\right)=-{{\mu }_{0}}{\bar {j}}({\bar {r}})\\\end{aligned}}$ wegen ${{\Delta }_{r}}\cdot \left({\frac {1}{\left|{\bar {r}}-{\bar {r}}{\acute {\ }}\right|}}\right)=4\pi \delta \left({\bar {r}}-{\bar {r}}{\acute {\ }}\right)$

Also:

$\nabla \times {\bar {B}}({\bar {r}})=\nabla \left(\nabla \cdot {\bar {A}}({\bar {r}})\right)-\Delta {\bar {A}}({\bar {r}})={{\mu }_{0}}{\bar {j}}({\bar {r}})+{{\mu }_{0}}{{\varepsilon }_{0}}{\frac {\partial }{\partial t}}{\bar {E}}({\bar {r}},t)$

Für stationäre Ströme, die gerade bei stationären Ladungsverteilungen vorliegen, folgt:

${\begin{aligned}&\nabla \times {\bar {B}}({\bar {r}})={{\mu }_{0}}{\bar {j}}({\bar {r}})\\&{{\mu }_{0}}{{\varepsilon }_{0}}{\frac {\partial }{\partial t}}{\bar {E}}({\bar {r}},t)=0\\\end{aligned}}$

Dies ist die differenzielle Form des Ampereschen Gesetzes.

Die Ströme sind die Wirbel der magnetischen Induktion!!

Integration über eine Fläche F mit Rand $\partial F$ liefert die Intgralform:

${\begin{aligned}&\int _{}^{}{d{\bar {f}}\cdot }\nabla \times {\bar {B}}({\bar {r}})=\oint \limits _{\partial F}{}d{\bar {s}}{\bar {B}}({\bar {r}})=\int _{}^{}{d{\bar {f}}\cdot }{{\mu }_{0}}{\bar {j}}({\bar {r}})={{\mu }_{0}}I\\&\oint \limits _{\partial F}{}d{\bar {s}}{\bar {B}}({\bar {r}})={{\mu }_{0}}I\\\end{aligned}}$

Mit dem Satz von Stokes Das sogenannte Durchflutungsgesetz!

Zusammenfassung

Magnetostatik

$div{\bar {B}}=0\Leftrightarrow {\bar {B}}=rot{\bar {A}}$ (quellenfreiheit)

${\begin{aligned}&rot{\bar {B}}={{\mu }_{0}}{\bar {j}}({\bar {r}})\Leftrightarrow \oint \limits _{\partial F}{}d{\bar {s}}\cdot {\bar {B}}={{\mu }_{0}}I\\&\Rightarrow \Delta {\bar {A}}=-{{\mu }_{0}}{\bar {j}}({\bar {r}})\\\end{aligned}}$

Gilt jedoch nur im Falle der Coulomb-Eichung:

$\nabla \cdot {\bar {A}}=0$

Dies geschieht durch die Umeichung

${\begin{aligned}&{\bar {A}}{\acute {\ }}({\bar {r}})\to {\bar {A}}+\nabla \Psi \\&\nabla \times {\bar {A}}{\acute {\ }}({\bar {r}})\to \nabla \times {\bar {A}}+\nabla \times \nabla \Psi \\&\nabla \times \nabla \Psi =0\Rightarrow \nabla \times {\bar {A}}{\acute {\ }}({\bar {r}})\to \nabla \times {\bar {A}}\\&\Rightarrow \nabla \times \left(\nabla \times {\bar {A}}{\acute {\ }}({\bar {r}})\right)=\nabla \times {\bar {B}}({\bar {r}})={{\mu }_{0}}{\bar {j}}\\&\nabla \times \left(\nabla \times {\bar {A}}{\acute {\ }}({\bar {r}})\right)=\nabla \left(\nabla \cdot {\bar {A}}{\acute {\ }}({\bar {r}})\right)-\Delta {\bar {A}}{\acute {\ }}({\bar {r}})\\\end{aligned}}$

Elektrostatik

$rot{\bar {E}}=0\Leftrightarrow {\bar {E}}=-\nabla \Phi$ (Wirbelfreiheit)

${\begin{aligned}&{{\varepsilon }_{0}}\nabla \cdot {\bar {E}}=\rho \\&\Leftrightarrow {{\varepsilon }_{0}}\oint \limits _{\partial V}{d{\bar {f}}\cdot }{\bar {E}}=Q\\\end{aligned}}$

differenzielle Form / integrale Form

$\Rightarrow \Delta \Phi =-{\frac {1}{{\varepsilon }_{0}}}\rho \left({\bar {r}}\right)$ (Poissongleichung)