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	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

(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:

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

{{{1}}}

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 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)