Editing Magnetische Multipole

<noinclude>{{Scripthinweis|Elektrodynamik|2|4}}</noinclude>
== (stationär)==

Ausgangspunkt ist
:<math>\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|}</math>
(mit der Coulomb- Eichung <math>\nabla \cdot \bar{A}(\bar{r})=0</math>)


mit den Randbedingungen
:<math>\bar{A}(\bar{r})\to 0</math> für r→ unendlich

Taylorentwicklung nach
:<math>\frac{1}{\left| \bar{r}-\bar{r}\acute{\ } \right|}</math>
von analog zum elektrischen Fall:

Die Stromverteilung <math>\bar{j}(\bar{r}\acute{\ })</math> sei stationär für <math>r>>r\acute{\ }</math>

:<math>\frac{1}{\left| \bar{r}-\bar{r}\acute{\ } \right|}=\frac{1}{r}+\frac{1}{{{r}^{3}}}\left( \bar{r}\cdot \bar{r}\acute{\ } \right)+...</math>

:<math>\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)+...</math>

===Monopol- Term===

'''Mit'''

:<math>{{\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)</math>

Im stationären Fall folgt aus der {{FB|Kontinuitätsgleichung}}:

:<math>{{\nabla }_{r\acute{\ }}}\cdot \bar{j}(\bar{r}\acute{\ })=0</math>

:<math>{{\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}}</math>

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

:<math>\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</math>

<u>Somit verschwindet der Monopolterm in der Theorie.</u>

=== Dipol- Term ===


mit <math>\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]</math> und mit

:<math>\begin{align}
& {{\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{align}</math>

Folgt:

:<math>\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</math>

Da

:<math>\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</math>
weil der Strom verschwindet!
Somit gibt der Term

:<math>\left[ \left( \bar{r}\bar{r}\acute{\ } \right)\bar{j}+\left( \bar{r}\bar{j} \right)\bar{r}\acute{\ } \right]</math>

'''keinen Beitrag zum'''

:<math>\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)</math>

Also:

:<math>\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}</math>

Als {{FB|Dipolpotenzial}}!!

:<math>\begin{align}
& \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{align}</math>

das magnetische Dipolmoment!

Analog zu

:<math>\begin{align}
& \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{align}</math>

dem elektrischen Dipolmoment

Die magnetische Induktion des Dipolmomentes ergibt sich als:

:<math>\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]</math>

Wegen:

:<math>\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)</math> mit <math>\begin{align}
& \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{align}</math>

Analog ergab sich als elektrisches Dipolfeld:

:<math>\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]</math>
{{Beispiel|1=
Beispiel: Ebene Leiterschleife L:



:<math>\begin{align}
& 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{align}</math>

Mit I = Strom durch den Leiter

:<math>\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}</math>

Dabei ist

:<math>\bar{n}</math>
die Normale auf der von L eingeschlossenen Fläche F

Also: Ein Ringstrom bedingt ein {{FB|magnetisches Dipolmoment}} <math>\bar{m}</math> }}

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


=== Bewegte Ladungen ===

N Teilchen mit den Massen m<sub>i</sub> und den Ladungen q<sub>i</sub> bewegen sich.

Dabei sei die spezifische Ladung <math>\frac{{{q}_{i}}}{{{m}_{i}}}=\frac{q}{m}</math> konstant:

:<math>\begin{align}
& \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{align}</math>

Das {{FB|magnetische Dipolmoment}} beträgt:

:<math>\begin{align}
& \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{align}</math>

Mit dem {{FB|Bahndrehimpuls}} <math>\bar{L}</math>:

:<math>\bar{m}=\frac{q}{2m}\bar{L}</math>
gilt aber auch für starre Körper!
* Allgemeines Gesetz!

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

:<math>\begin{align}
& \bar{m}=g\frac{e}{2m}\bar{S} \\
& g\approx 2 \\
\end{align}</math>

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

==== Kraft auf eine Stromverteilung ====

:<math>\bar{j}(\bar{r}\acute{\ })={{\rho }_{i}}(\bar{r}\acute{\ })\bar{v}(\bar{r}\acute{\ })</math>

im Feld einer externen {{FB|magnetischen Induktion}} <math>\bar{B}(\bar{r}\acute{\ })</math>:

Spürt die {{FB|Lorentzkraft}}

:<math>\bar{F}=\int_{{}}^{{}}{{}}{{d}^{3}}r\acute{\ }\bar{j}(\bar{r}\acute{\ })\times \bar{B}(\bar{r}\acute{\ })</math>

Talyorentwicklung liefert:

:<math>\begin{align}
& \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{align}</math>

im stationären Fall gilt wieder:

:<math>\left[ \int_{{}}^{{}}{{}}{{d}^{3}}r\acute{\ }\bar{j}(\bar{r}\acute{\ }) \right]=0</math> (keine Monopole)
Also:

:<math>\begin{align}
& \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{align}</math>

Man fordert:

:<math>\left[ {{\nabla }_{r}}\times \bar{B}(\bar{r}) \right]=0</math>

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

:<math>\begin{align}
& \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{align}</math>

(Vergl. S. 34)