Magnetische Multipole

Elektrodynamikvorlesung von Prof. Dr. E. Schöll, PhD

Der Artikel Magnetische Multipole basiert auf der Vorlesungsmitschrift von Franz- Josef Schmitt des 2.Kapitels (Abschnitt 4) der Elektrodynamikvorlesung von Prof. Dr. E. Schöll, PhD.

Magnetische Multipole	{{#ask: Kapitel::2 Abschnitt::0 Urheber::Prof. Dr. E. Schöll, PhD}}	{{#ask: Abschnitt::0 Kapitel::0 Urheber::Prof. Dr. E. Schöll, PhD}}
	{{#ask: Kapitel::2 Abschnitt::!0 Urheber::Prof. Dr. E. Schöll, PhD	format=ol	order=ASC	sort=Abschnitt }}	{{#ask: Abschnitt::0 Urheber::Prof. Dr. E. Schöll, PhD Kapitel::!0	format=ol	order=ASC	sort=Kapitel }}

{{#set:Urheber=Prof. Dr. E. Schöll, PhD|Inhaltstyp=Script|Kapitel=2|Abschnitt=4}} __SHOWFACTBOX__

(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{{#set:Fachbegriff=Kontinuitätsgleichung|Index=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{{#set:Fachbegriff=Dipolpotenzial|Index=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{{#set:Fachbegriff=magnetisches Dipolmoment|Index=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 m_i und den Ladungen q_i 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{{#set:Fachbegriff=magnetische Dipolmoment|Index=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{{#set:Fachbegriff=Bahndrehimpuls|Index=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{{#set:Fachbegriff=magnetischen Induktion|Index=magnetischen Induktion}} ${\bar {B}}({\bar {r}}{\acute {\ }})$:

Spürt die Lorentzkraft{{#set:Fachbegriff=Lorentzkraft|Index=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)