Stationäre Ströme und Magnetfeld

2vorlesung von Prof. Dr. E. Schöll, PhD

Der Artikel Stationäre Ströme und Magnetfeld basiert auf der Vorlesungsmitschrift von Franz- Josef Schmitt des Elektrodynamik.Kapitels der 2vorlesung von Prof. Dr. E. Schöll, PhD.

Kontinuitätsgleichung

Bewegte Ladungen entsprechen elektrischem Strom I

Experimentelle Erfahrung: Die Ladung bleibt erhalten:

$Q(t)=\int _{V}^{}{}{{d}^{3}}r\rho ({\bar {r}},t)$

Damit folgt ein globaler Erhaltungssatz:

${\frac {d}{dt}}Q(t)={\frac {d}{dt}}\int _{V}^{}{}{{d}^{3}}r\rho ({\bar {r}},t)=-\oint \limits _{\partial V}{}\delta I$

$\delta I={\frac {\rho dV}{dt}}={\frac {\rho \left|v\right|dt\left|df\right|\cos \alpha }{dt}}=\rho {\bar {v}}d{\bar {f}}$

Also gerade die Ladung, die durch $d{\bar {f}}$ pro zeit aus V herausströmt Als eine lokale Größe findet man die elektrische Stromdichte:

${\bar {j}}({\bar {r}},t):=\rho ({\bar {r}},t){\bar {v}}({\bar {r}},t)$

$\Rightarrow {\frac {d}{dt}}\int _{V}^{}{}{{d}^{3}}r\rho ({\bar {r}},t)=-\oint \limits _{\partial V}{}d{\bar {f}}{\bar {j}}({\bar {r}},t)=-\int _{V}^{}{{{d}^{3}}r}\nabla \cdot {\bar {j}}({\bar {r}},t)$ ( Gauß !) für alle Volumina V ( einfach zusammenhängend)

Somit folgt die Kontinuitätsgleichung als LOKALER Erhaltungssatz:

$\Rightarrow {\frac {\partial }{\partial t}}\rho ({\bar {r}},t)+\nabla \cdot {\bar {j}}({\bar {r}},t)=0$

Speziell bei stationären Ladungsverteilungen gilt die Divergenzfreiheit des Stroms:

$\nabla \cdot {\bar {j}}({\bar {r}},t)=0$

Aber : natürlich muss deswegen nicht ${\bar {j}}({\bar {r}},t)=0$ gelten. Der Strom muss räumlich lediglich stationär sein !

Magnetische Induktion

Experimentelle Erfahrung:

Es existieren Wechselwirkungen zwischen den Ladungen: Eine Kraft wirkt auf Ladungen q, die sich mit v bewegen:

${\bar {F}}=q{\bar {v}}\times {\bar {B}}({\bar {r}})$

Die sogenannte Lorentz- Kraft !

${\bar {B}}({\bar {r}})$ ist die magnetische Induktion am Ort ${\bar {r}}$ , die erzeugt wird von den anderen Ladungen mit einer zugeordneten Stromdichte ${\bar {j}}({\bar {r}}{\acute {\ }})$ .

Die Erzeugung dieser Magnetischen Induktion erfolgt gemäß des Ampereschen Gesetzes:

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

Dies läuft völlig analog zur Coulomb- Wechselwirkung in der Elektrostatik:

${\begin{aligned}&{\bar {F}}=q{\bar {E}}({\bar {r}})\\&{\bar {E}}({\bar {r}})={\frac {1}{4\pi {{\varepsilon }_{0}}}}\int _{}^{}{{{d}^{3}}r{\acute {\ }}}\rho ({\bar {r}}{\acute {\ }}){\frac {{\bar {r}}-{\bar {r}}{\acute {\ }}}{\left|{\bar {r}}-{\bar {r}}{\acute {\ }}\right|}}\\\end{aligned}}$

Die Einheiten im SI- System lauten:

$\left[B\right]={\frac {1Ns}{Cm}}={\frac {1kg{{m}^{2}}}{C{{s}^{2}}}}\cdot {\frac {s}{{m}^{2}}}=1V{\frac {s}{{m}^{2}}}=1T$

Mit diesen Einheiten ist dann ${{\mu }_{0}}=1,26\cdot {{10}^{-6}}{\frac {Vs}{Am}}$ festgelegt, wie die Dielektrizitätskonstante jedoch frei wählbar !! Die magnetische Induktion beschreibt keine neue, von der Coulomb- Wechselwirkung unabhängige WW: Man betrachte dazu lediglich die Transformation auf das lokale Ruhesystem einer bewegten Ladung:

Im Gauß System:

${\bar {F}}={\frac {q}{c}}{\bar {v}}\times {\bar {B}}({\bar {r}})$

${\begin{aligned}&{\bar {B}}({\bar {r}})={\frac {1}{c}}\int _{}^{}{}{{d}^{3}}r{\acute {\ }}{\bar {j}}({\bar {r}}{\acute {\ }})\times {\frac {{\bar {r}}-{\bar {r}}{\acute {\ }}}{{\left|{\bar {r}}-{\bar {r}}{\acute {\ }}\right|}^{3}}}\\&\\\end{aligned}}$

Die Kraft zwischen 2 stromdurchflossenen Leitern:

Betrachten wir zwei infinit. dünne Leiter L, L´, die mit konstanten Strömen I und I´ durchflossen werden:

Der Strom durch L´:

${\begin{aligned}&{\bar {j}}({\bar {r}}{\acute {\ }}){{d}^{3}}r{\acute {\ }}=\rho {{d}^{3}}r{\acute {\ }}{\bar {v}}{\acute {\ }}={\frac {d}{dt}}\rho {{d}^{3}}r{\acute {\ }}d{\bar {r}}{\acute {\ }}\\&{\frac {d}{dt}}\rho {{d}^{3}}r{\acute {\ }}=I{\acute {\ }}\\&\Rightarrow {\bar {j}}({\bar {r}}{\acute {\ }}){{d}^{3}}r{\acute {\ }}=I{\acute {\ }}d{\bar {r}}{\acute {\ }}\\\end{aligned}}$

Somit folgt das Biot- Savartsche Gesetz für unendlich lange Leiter L´:

Die magnetische Induktion ist gerade:

${\begin{aligned}&{\bar {B}}({\bar {r}})={\frac {{\mu }_{0}}{4\pi }}I{\acute {\ }}\int _{L{\acute {\ }}}^{}{}d{\bar {r}}{\acute {\ }}\times {\frac {{\bar {r}}-{\bar {r}}{\acute {\ }}}{{\left|{\bar {r}}-{\bar {r}}{\acute {\ }}\right|}^{3}}}\\&\\\end{aligned}}$

Die Kraft auf eine Ladung im Volumenelement d³r von L ist damit gerade:

$d{\bar {F}}=\rho {\bar {v}}\times {\bar {B}}({\bar {r}}){{d}^{3}}r={\bar {j}}\times {\bar {B}}{{d}^{3}}r=Id{\bar {r}}\times {\bar {B}}$

Also:

${\bar {F}}={\frac {{\mu }_{0}}{4\pi }}II{\acute {\ }}\int _{L}^{}{}d{\bar {r}}\times \int _{L{\acute {\ }}}^{}{}d{\bar {r}}{\acute {\ }}\times {\frac {{\bar {r}}-{\bar {r}}{\acute {\ }}}{{\left|{\bar {r}}-{\bar {r}}{\acute {\ }}\right|}^{3}}}$

Dies ist dann die gesamte Kraft von L´ auf L

mit

${\begin{aligned}&d{\bar {r}}\times \left(d{\bar {r}}{\acute {\ }}\times \left({\bar {r}}-{\bar {r}}\right)\right)=\left(d{\bar {r}}\left({\bar {r}}-{\bar {r}}\right)\right)d{\bar {r}}{\acute {\ }}-\left(d{\bar {r}}d{\bar {r}}{\acute {\ }}\right)\left({\bar {r}}-{\bar {r}}\right)\\&und\\&\int _{L}^{}{}d{\bar {r}}{\frac {{\bar {r}}-{\bar {r}}{\acute {\ }}}{{\left|{\bar {r}}-{\bar {r}}{\acute {\ }}\right|}^{3}}}=-\left.{\frac {1}{\left|{\bar {r}}-{\bar {r}}{\acute {\ }}\right|}}\right|_{L-ANfang}^{L-Ende}=0\\\end{aligned}}$

( Der Leiter ist entweder geschlossen oder die Enden liegen im Unendlichen) folgt:

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

für parallele Ströme:

$Id{\bar {r}}I{\acute {\ }}d{\bar {r}}{\acute {\ }}>0$ folgt Anziehung für antiparallele Ströme:

$Id{\bar {r}}I{\acute {\ }}d{\bar {r}}{\acute {\ }}<0$ dagegen Abstoßung

Man sieht außerdem das dritte Newtonsche Gesetz:

${\begin{aligned}&{\bar {r}}\leftrightarrow {\bar {r}}{\acute {\ }}\\&d{\bar {r}}\leftrightarrow d{\bar {r}}{\acute {\ }}\\&I\leftrightarrow I{\acute {\ }}\\\end{aligned}}$

Somit:

${\bar {F}}\leftrightarrow -{\bar {F}}$ ( actio gleich reactio)

Die magnetostatischen Feldgleichungen

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)

Magnetische Multipole

( stationär)

Ausgangspunkt ist ${\bar {A}}({\bar {r}})={\frac {{\mu }_{0}}{4\pi }}\int _{{R}^{3}}^{}{}{{d}^{3}}r{\acute {\ }}{\frac {{\bar {j}}({\bar {r}}{\acute {\ }})}{\left|{\bar {r}}-{\bar {r}}{\acute {\ }}\right|}}$ (mit der Coulomb- Eichung $\nabla \cdot {\bar {A}}({\bar {r}})=0$ )

mit den Randbedingungen ${\bar {A}}({\bar {r}})\to 0$ für r-> unendlich

Taylorentwicklung nach ${\frac {1}{\left|{\bar {r}}-{\bar {r}}{\acute {\ }}\right|}}$ von analog zum elektrischen Fall: Die Stromverteilung ${\bar {j}}({\bar {r}}{\acute {\ }})$ sei stationär für $r>>r{\acute {\ }}$

${\frac {1}{\left|{\bar {r}}-{\bar {r}}{\acute {\ }}\right|}}={\frac {1}{r}}+{\frac {1}{{r}^{3}}}\left({\bar {r}}\cdot {\bar {r}}{\acute {\ }}\right)+...$

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

Monopol- Term

Mit

${{\nabla }_{r{\acute {\ }}}}\cdot \left[{{x}_{k}}{\acute {\ }}{\bar {j}}({\bar {r}}{\acute {\ }})\right]={{x}_{k}}{\acute {\ }}\left({{\nabla }_{r{\acute {\ }}}}\cdot {\bar {j}}({\bar {r}}{\acute {\ }})\right)+{\bar {j}}({\bar {r}}{\acute {\ }})\cdot \left({{\nabla }_{r{\acute {\ }}}}{{x}_{k}}{\acute {\ }}\right)$

Im stationären Fall folgt aus der Kontinuitätsgleichung:

${{\nabla }_{r{\acute {\ }}}}\cdot {\bar {j}}({\bar {r}}{\acute {\ }})=0$

${{\nabla }_{r{\acute {\ }}}}\cdot \left[{{x}_{k}}{\acute {\ }}{\bar {j}}({\bar {r}}{\acute {\ }})\right]={\bar {j}}({\bar {r}}{\acute {\ }})\cdot \left({{\nabla }_{r{\acute {\ }}}}{{x}_{k}}{\acute {\ }}\right)={{j}_{l}}{{\delta }_{kl}}={{j}_{k}}$

Mit ${{\nabla }_{r{\acute {\ }}}}\cdot \left[{{x}_{k}}{\acute {\ }}{\bar {j}}({\bar {r}}{\acute {\ }})\right]={{j}_{k}}$ folgt dann:

$\int _{}^{}{{{d}^{3}}r{\acute {\ }}}{{j}_{k}}({\bar {r}}{\acute {\ }})=\int _{}^{}{{{d}^{3}}r{\acute {\ }}}{{\nabla }_{r{\acute {\ }}}}\cdot \left[{{x}_{k}}{\acute {\ }}{\bar {j}}({\bar {r}}{\acute {\ }})\right]=\oint \limits _{S\infty }{d{\bar {f}}}\left[{{x}_{k}}{\acute {\ }}{\bar {j}}({\bar {r}}{\acute {\ }})\right]=0$

Somit verschwindet der Monopolterm in der Theorie

Dipol- Term

mit

$\left[{\bar {r}}{\acute {\ }}\times {\bar {j}}({\bar {r}}{\acute {\ }})\right]\times {\bar {r}}=\left({\bar {r}}{\bar {r}}{\acute {\ }}\right){\bar {j}}-\left({\bar {r}}{\bar {j}}\right){\bar {r}}{\acute {\ }}=2\left({\bar {r}}{\bar {r}}{\acute {\ }}\right){\bar {j}}-\left[\left({\bar {r}}{\bar {r}}{\acute {\ }}\right){\bar {j}}+\left({\bar {r}}{\bar {j}}\right){\bar {r}}{\acute {\ }}\right]$

und mit

${\begin{aligned}&{{\nabla }_{r{\acute {\ }}}}\left[{{x}_{k}}{\acute {\ }}\left({\bar {r}}{\bar {r}}{\acute {\ }}\right){\bar {j}}\right]=\left[\left({\bar {r}}{\bar {r}}{\acute {\ }}\right){{j}_{k}}+{{x}_{k}}{\acute {\ }}\left({\bar {r}}{\bar {j}}\right)+{{x}_{k{\acute {\ }}}}\left({\bar {r}}{\bar {r}}{\acute {\ }}\right){{\nabla }_{r{\acute {\ }}}}\cdot {\bar {j}}\right]\\&{{\nabla }_{r{\acute {\ }}}}\cdot {\bar {j}}=0\\&\Rightarrow {{\nabla }_{r{\acute {\ }}}}\left[{{x}_{k}}{\acute {\ }}\left({\bar {r}}{\bar {r}}{\acute {\ }}\right){\bar {j}}\right]=\left[\left({\bar {r}}{\bar {r}}{\acute {\ }}\right){{j}_{k}}+{{x}_{k}}{\acute {\ }}\left({\bar {r}}{\bar {j}}\right)\right]\\\end{aligned}}$

Folgt:

$\int _{{R}^{3}}^{}{}{{d}^{3}}r{\acute {\ }}{{\nabla }_{r{\acute {\ }}}}\left[{{x}_{k}}{\acute {\ }}\left({\bar {r}}{\bar {r}}{\acute {\ }}\right){\bar {j}}\right]=\int _{{R}^{3}}^{}{}{{d}^{3}}r{\acute {\ }}\left[\left({\bar {r}}{\bar {r}}{\acute {\ }}\right){{j}_{k}}+{{x}_{k}}{\acute {\ }}\left({\bar {r}}{\bar {j}}\right)\right]=0$

Da

$\int _{{R}^{3}}^{}{}{{d}^{3}}r{\acute {\ }}{{\nabla }_{r{\acute {\ }}}}\left[{{x}_{k}}{\acute {\ }}\left({\bar {r}}{\bar {r}}{\acute {\ }}\right){\bar {j}}\right]=\oint \limits _{S\infty }{d{\bar {f}}}\left[{{x}_{k}}{\acute {\ }}\left({\bar {r}}{\bar {r}}{\acute {\ }}\right){\bar {j}}\right]=0$ weil der Strom verschwindet ! Somit gibt der Term

$\left[\left({\bar {r}}{\bar {r}}{\acute {\ }}\right){\bar {j}}+\left({\bar {r}}{\bar {j}}\right){\bar {r}}{\acute {\ }}\right]$

keinen Beitrag zum

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

Also:

${\bar {A}}({\bar {r}})={\frac {{\mu }_{0}}{4\pi {{r}^{3}}}}{\frac {1}{2}}\int _{{R}^{3}}^{}{}{{d}^{3}}r{\acute {\ }}\left({\bar {r}}{\acute {\ }}\times {\bar {j}}({\bar {r}}{\acute {\ }})\right)\times {\bar {r}}$

Als DIPOLPOTENZIAL !!

${\begin{aligned}&{\bar {A}}({\bar {r}}):={\frac {{\mu }_{0}}{4\pi {{r}^{3}}}}{\bar {m}}\times {\bar {r}}\\&{\bar {m}}={\frac {1}{2}}\int _{{R}^{3}}^{}{}{{d}^{3}}r{\acute {\ }}\left({\bar {r}}{\acute {\ }}\times {\bar {j}}({\bar {r}}{\acute {\ }})\right)\\\end{aligned}}$

das magnetische Dipolmoment !

Analog zu

${\begin{aligned}&\Phi ({\bar {r}}):={\frac {1}{4\pi {{\varepsilon }_{0}}{{r}^{3}}}}{\bar {p}}\cdot {\bar {r}}\\&{\bar {p}}:=\int _{{R}^{3}}^{}{}{{d}^{3}}r{\acute {\ }}{\bar {r}}{\acute {\ }}\rho ({\bar {r}}{\acute {\ }})\\\end{aligned}}$

dem elektrischen Dipolmoment

Die magnetische Induktion des Dipolmomentes ergibt sich als:

${\bar {B}}({\bar {r}}):=\nabla \times {\frac {{\mu }_{0}}{4\pi {{r}^{3}}}}{\bar {m}}\times {\bar {r}}={\frac {{\mu }_{0}}{4\pi {{r}^{5}}}}\left[3\left({\bar {m}}\cdot {\bar {r}}\right){\bar {r}}-{{r}^{2}}{\bar {m}}\right]$

Wegen:

$\nabla \times \left({\bar {a}}\times {\bar {b}}\right)=\left({\bar {b}}\cdot \nabla \right){\bar {a}}-\left({\bar {a}}\cdot \nabla \right){\bar {b}}+{\bar {a}}\left(\nabla \cdot {\bar {b}}\right)-{\bar {b}}\left(\nabla \cdot {\bar {a}}\right)$

mit

${\begin{aligned}&{\bar {a}}={\frac {\bar {m}}{{r}^{3}}}\\&{\bar {b}}={\bar {r}}\\&\Rightarrow div{\bar {a}}=-3{\frac {{\bar {m}}\cdot {\bar {r}}}{{r}^{5}}}\\&div{\bar {b}}=3\\&\left({\bar {b}}\cdot \nabla \right){\bar {a}}=-3{\frac {{\bar {m}}\cdot {{r}^{2}}}{{r}^{5}}}\\&\left({\bar {a}}\cdot \nabla \right){\bar {b}}={\frac {\bar {m}}{{r}^{3}}}\\\end{aligned}}$

Analog ergab sich als elektrisches Dipolfeld:

${\bar {E}}({\bar {r}}):={\frac {1}{4\pi {{\varepsilon }_{0}}{{r}^{5}}}}\left[3\left({\bar {p}}\cdot {\bar {r}}\right)-{{r}^{2}}{\bar {p}}\right]$

Beispiel: Ebene Leiterschleife L:

${\begin{aligned}&d{\bar {f}}{\acute {\ }}={\frac {1}{2}}{\bar {r}}{\acute {\ }}\times d{\bar {s}}{\acute {\ }}\\&{{d}^{3}}{\bar {r}}{\acute {\ }}j({\bar {r}}{\acute {\ }})=d{\bar {s}}{\acute {\ }}I\\\end{aligned}}$

Mit I = Strom durch den Leiter

$\Rightarrow {\bar {m}}={\frac {1}{2}}\oint \limits _{L}{}{{d}^{3}}r{\acute {\ }}\left({\bar {r}}{\acute {\ }}\times {\bar {j}}({\bar {r}}{\acute {\ }})\right)={\frac {I}{2}}\oint \limits _{L}{}{\bar {r}}{\acute {\ }}\times d{\bar {s}}{\acute {\ }}=I\int _{F}^{}{}d{\bar {f}}{\acute {\ }}=IF{\bar {n}}$

Dabei ist

${\bar {n}}$ die Normale auf der von L eingeschlossenen Fläche F

Also: Ein Ringstrom bedingt ein magnetisches Dipolmoment ${\bar {m}}$

analog: 2 Punktladungen bedingen ein elektrisches Dipolmoment ${\bar {p}}=q{\bar {a}}$ , welches von der positiven zur negativen Ladung zeigt.

Bewegte Ladungen N Teilchen mit den Massen mi und den Ladungen qi bewegen sich.

Dabei sei die spezifische Ladung

${\frac {{q}_{i}}{{m}_{i}}}={\frac {q}{m}}$ konstant:

${\begin{aligned}&\rho ({\bar {r}})=\sum \limits _{i}{}{{q}_{i}}\delta \left({\bar {r}}-{{\bar {r}}_{i}}\right)\\&{\bar {j}}({\bar {r}})=\sum \limits _{i}{}{{q}_{i}}{{\bar {v}}_{i}}\delta \left({\bar {r}}-{{\bar {r}}_{i}}\right)\\&{{\bar {v}}_{i}}={\frac {d{{\bar {r}}_{i}}}{dt}}\\\end{aligned}}$

Das magnetische Dipolmoment beträgt:

${\begin{aligned}&{\bar {m}}={\frac {1}{2}}\oint \limits _{L}{}{{d}^{3}}r{\acute {\ }}\left({\bar {r}}{\acute {\ }}\times {\bar {j}}({\bar {r}}{\acute {\ }})\right)={\frac {1}{2}}\sum \limits _{i}{}{{q}_{i}}\int _{}^{}{}{{d}^{3}}r{\acute {\ }}{\bar {r}}{\acute {\ }}\times {{\bar {v}}_{i}}\delta \left({\bar {r}}{\acute {\ }}-{{\bar {r}}_{i}}\right)={\frac {1}{2}}\sum \limits _{i}{}{{q}_{i}}{{\bar {r}}_{i}}\times {{\bar {v}}_{i}}={\frac {1}{2}}\sum \limits _{i}{}{\frac {{q}_{i}}{{m}_{i}}}{{m}_{i}}{{\bar {r}}_{i}}\times {{\bar {v}}_{i}}\\&{\frac {{q}_{i}}{{m}_{i}}}={\frac {q}{m}}\\&\Rightarrow {\bar {m}}={\frac {q}{2m}}{\bar {L}}\\\end{aligned}}$

Mit dem Bahndrehimpuls ${\bar {L}}$

${\bar {m}}={\frac {q}{2m}}{\bar {L}}$ gilt aber auch für starre Körper !

• Allgemeines Gesetz !

Jedoch gilt dies nicht für den Spin eines Elektrons !!!

${\begin{aligned}&{\bar {m}}=g{\frac {e}{2m}}{\bar {S}}\\&g\approx 2\\\end{aligned}}$

Somit ist der Spin nicht vollständig durch die Vorstellung von einer rotierenden Ladungsverteilung zu verstehen !

Kraft auf eine Stromverteilung:

${\bar {j}}({\bar {r}}{\acute {\ }})={{\rho }_{i}}({\bar {r}}{\acute {\ }}){\bar {v}}({\bar {r}}{\acute {\ }})$

im Feld einer externen magnetischen Induktion ${\bar {B}}({\bar {r}}{\acute {\ }})$

Spürt die Lorentzkraft

${\bar {F}}=\int _{}^{}{}{{d}^{3}}r{\acute {\ }}{\bar {j}}({\bar {r}}{\acute {\ }})\times {\bar {B}}({\bar {r}}{\acute {\ }})$

Talyorentwicklung liefert:

${\begin{aligned}&{\bar {B}}({\bar {r}}{\acute {\ }})={\bar {B}}({\bar {r}})+\left[\left({\bar {r}}{\acute {\ }}-{\bar {r}}\right)\nabla \right]{\bar {B}}({\bar {r}})+....\\&\Rightarrow {\bar {F}}=\left[\int _{}^{}{}{{d}^{3}}r{\acute {\ }}{\bar {j}}({\bar {r}}{\acute {\ }})\right]\times {\bar {B}}({\bar {r}}{\acute {\ }})+\int _{}^{}{}{{d}^{3}}r{\acute {\ }}{\bar {j}}({\bar {r}}{\acute {\ }})\times \left[\left({\bar {r}}{\acute {\ }}-{\bar {r}}\right)\nabla \right]{\bar {B}}({\bar {r}})+...\\\end{aligned}}$

im stationären Fall gilt wieder:

$\left[\int _{}^{}{}{{d}^{3}}r{\acute {\ }}{\bar {j}}({\bar {r}}{\acute {\ }})\right]=0$ ( keine Monopole) Also:

${\begin{aligned}&{\bar {F}}=\int _{}^{}{}{{d}^{3}}r{\acute {\ }}{\bar {j}}({\bar {r}}{\acute {\ }})\times \left[\left({\bar {r}}{\acute {\ }}\right){{\nabla }_{r}}\right]{\bar {B}}({\bar {r}})-\int _{}^{}{}{{d}^{3}}r{\acute {\ }}{\bar {j}}({\bar {r}}{\acute {\ }})\times \left[\left({\bar {r}}\right){{\nabla }_{r}}\right]{\bar {B}}({\bar {r}})\\&\int _{}^{}{}{{d}^{3}}r{\acute {\ }}{\bar {j}}({\bar {r}}{\acute {\ }})\times \left[\left({\bar {r}}\right){{\nabla }_{r}}\right]{\bar {B}}({\bar {r}})=0,da\int _{}^{}{}{{d}^{3}}r{\acute {\ }}{\bar {j}}({\bar {r}}{\acute {\ }})=0\\&\Rightarrow {\bar {F}}=\int _{}^{}{}{{d}^{3}}r{\acute {\ }}{\bar {j}}({\bar {r}}{\acute {\ }})\times \left[\left({\bar {r}}{\acute {\ }}\right){{\nabla }_{r}}\right]{\bar {B}}({\bar {r}})\\&\left[\left({\bar {r}}{\acute {\ }}\right){{\nabla }_{r}}\right]{\bar {B}}({\bar {r}})={{\nabla }_{r}}\left[\left({\bar {r}}{\acute {\ }}\right)\cdot {\bar {B}}({\bar {r}})\right]-{\bar {r}}{\acute {\ }}\times \left[{{\nabla }_{r}}\times {\bar {B}}({\bar {r}})\right]\\\end{aligned}}$

Man fordert:

$\left[{{\nabla }_{r}}\times {\bar {B}}({\bar {r}})\right]=0$

( Das externe Feld soll keine Stromwirbel im Bereich von ${\bar {j}}({\bar {r}}{\acute {\ }})$ haben:

${\begin{aligned}&{\bar {F}}=\int _{}^{}{}{{d}^{3}}r{\acute {\ }}{\bar {j}}({\bar {r}}{\acute {\ }})\times {{\nabla }_{r}}\left[\left({\bar {r}}{\acute {\ }}\right)\cdot {\bar {B}}({\bar {r}})\right]\\&{\bar {j}}({\bar {r}}{\acute {\ }})\times {{\nabla }_{r}}\left[\left({\bar {r}}{\acute {\ }}\right)\cdot {\bar {B}}({\bar {r}})\right]=-{{\nabla }_{r}}\times \left[\left(\left({\bar {r}}{\acute {\ }}\right)\cdot {\bar {B}}({\bar {r}})\right){\bar {j}}({\bar {r}}{\acute {\ }})\right]+\left[\left({\bar {r}}{\acute {\ }}\right)\cdot {\bar {B}}({\bar {r}})\right]{{\nabla }_{r}}\times {\bar {j}}({\bar {r}}{\acute {\ }})\\&{{\nabla }_{r}}\times {\bar {j}}({\bar {r}}{\acute {\ }})=0\\&\Rightarrow {\bar {F}}=-\int _{}^{}{}{{d}^{3}}r{\acute {\ }}{{\nabla }_{r}}\times \left[\left(\left({\bar {r}}{\acute {\ }}\right)\cdot {\bar {B}}({\bar {r}})\right){\bar {j}}({\bar {r}}{\acute {\ }})\right]=-{{\nabla }_{r}}\times \left({\bar {m}}\times {\bar {B}}({\bar {r}})\right)\\&{\bar {F}}=-{{\nabla }_{r}}\times \left({\bar {m}}\times {\bar {B}}({\bar {r}})\right)=\left({\bar {m}}\cdot {{\nabla }_{r}}\right){\bar {B}}({\bar {r}})=-{{\nabla }_{r}}\left(-{\bar {m}}\cdot {\bar {B}}({\bar {r}})\right)\\\end{aligned}}$

( Vergl. S. 34)