Elektrodynamikvorlesung von Prof. Dr. E. Schöll, PhD
{{#set:Urheber=Prof. Dr. E. Schöll, PhD|Inhaltstyp=Script|Kapitel=2|Abschnitt=3}}
Kategorie:Elektrodynamik
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Sie gelten auch in quasistaischer Näherung{{#set:Fachbegriff=quasistaischer Näherung|Index=quasistaischer Näherung}}: Die zeitliche Änderung muss viel kleiner sein als die räumliche!!
Mit dem Vektorpotenzial{{#set:Fachbegriff=Vektorpotenzial|Index=Vektorpotenzial}}
![{\displaystyle {\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|}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a964632c7ef0c64192bb0f9ae029431513a8ef72)
Welches nicht eindeutig ist, sondern beliebig gemäß
![{\displaystyle {\bar {A}}({\bar {r}})\to {\bar {A}}+\nabla \Psi }](https://wikimedia.org/api/rest_v1/media/math/render/svg/fd619b69c1fffef002a27b5cb5fd053143b2896b)
umgeeicht werden kann.(
beliebig möglich, da
)
Mit diesem Vektorpotenzial also kann man schreiben:
![{\displaystyle {\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|}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b4161feb2c66df87d9680d5ed866b0731dcda7d7)
Beweis:
![{\displaystyle {\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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9f0ad24a06806a0f7668409349e1046e179c4165)
Folgende Aussagen sind äquivalent:
Es existiert ein Vektorpotenzial mit
![{\displaystyle {\begin{aligned}&{\bar {B}}=rot{\bar {A}}({\bar {r}})\\&\Leftrightarrow \\\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8f104ba6d789390d10b7142a301cdfb2c486d3ee)
![{\displaystyle div{\bar {B}}=0}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f4abe61512c270648aec2cbeb26ee0c1cced2df2)
Beweis:
![{\displaystyle div(rot{\bar {A}}({\bar {r}}))=0}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ffa175f1d1abc7a8bddd8c6fd3faa1b831e1899c)
es gibt keine Quellen der magnetischen Induktion (es existieren keine "magnetischen Ladungen".
Aber: Magnetische Monopole{{#set:Fachbegriff=Magnetische Monopole|Index=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
erzeugt worden sein sollen.
Sehr umstritten ist ein angeblicher experimenteller Nachweis von 1982 (Spektrum der Wissenschaft, Juni 1982, S. 78 ff.)
Der Zusammenhang zwischen
und
:
![{\displaystyle {\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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/71ccd82038a449ba956cd7dec5a46a6d0e3280c6)
Wobei die verwendete Kontinuitätsgleichung{{#set:Fachbegriff=Kontinuitätsgleichung|Index=Kontinuitätsgleichung}} natürlich nur für statische Ladungsverteilungen gilt!
Im Allgemeinen Fall gilt dagegen:
![{\displaystyle {\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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/77e8f0e3911583826f74718cae0a16ce8ea2fab3)
Mit dem Gaußschen Satz{{#set:Fachbegriff=Gaußschen Satz|Index=Gaußschen Satz}}.
Wenn das Potenzial jedoch ins unendliche hinreichend rasch abfällt, so gilt:
![{\displaystyle \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}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f11bd22c576d9fd6510c38658bd8b301c7b1678d)
Also:
![{\displaystyle \nabla \cdot {\bar {A}}({\bar {r}})=-{{\mu }_{0}}{{\varepsilon }_{0}}{\frac {\partial }{\partial t}}\Phi ({\bar {r}},t)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/fb377291f04163edc3ce3d49cfbd4b75e6e78a86)
Also:
![{\displaystyle \nabla \left(\nabla \cdot {\bar {A}}({\bar {r}})\right)={{\mu }_{0}}{{\varepsilon }_{0}}{\frac {\partial }{\partial t}}{\bar {E}}({\bar {r}},t)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/02631e0af192657d8f2c0ff171b3510b5eeeca88)
Auf der anderen Seite ergibt sich ganz einfach
wegen ![{\displaystyle {{\Delta }_{r}}\cdot \left({\frac {1}{\left|{\bar {r}}-{\bar {r}}{\acute {\ }}\right|}}\right)=4\pi \delta \left({\bar {r}}-{\bar {r}}{\acute {\ }}\right)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/10215b001369b18a8732e18d45ab3f135342ab05)
Also:
![{\displaystyle \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)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/301277f25d939311a28431ab3b350672b39b2a8d)
Für stationäre Ströme, die gerade bei stationären Ladungsverteilungen vorliegen, folgt:
![{\displaystyle {\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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/20c373f2121f2671c2b30ffc14b5512070acd295)
Dies ist die differenzielle Form des Ampereschen Gesetzes{{#set:Fachbegriff=Ampereschen Gesetzes|Index=Ampereschen Gesetzes}}.
Die Ströme sind die Wirbel der magnetischen Induktion!!
Integration über eine Fläche F mit Rand
liefert die Intgralform:
![{\displaystyle {\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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f5eb653a073b764b7ec838b3b0ea6e1511acca85)
Mit dem Satz von Stokes{{#set:Fachbegriff=Satz von Stokes|Index=Satz von Stokes}}
Das sogenannte Durchflutungsgesetz{{#set:Fachbegriff=Durchflutungsgesetz|Index=Durchflutungsgesetz}}!
(quellenfreiheit)
![{\displaystyle {\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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/94fb0d0296b3a47111dc3a239e2d787348420e5c)
Gilt jedoch nur im Falle der Coulomb-Eichung{{#set:Fachbegriff=Coulomb-Eichung|Index=Coulomb-Eichung}}:
![{\displaystyle \nabla \cdot {\bar {A}}=0}](https://wikimedia.org/api/rest_v1/media/math/render/svg/42bb1465257ddadb3a67b5add05d312cb22e8907)
Dies geschieht durch die Umeichung
![{\displaystyle {\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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7602a8934f829f402dc40aca7be4ea081e7e73eb)
(Wirbelfreiheit)
![{\displaystyle {\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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/96a451a39e99d60cfcc7c12047935e0a13a475c3)
differenzielle Form / integrale Form
(Poissongleichung{{#set:Fachbegriff=Poissongleichung|Index=Poissongleichung}})