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<noinclude>{{Scripthinweis|Elektrodynamik|6|3}}</noinclude> <u>'''Ziel: '''</u>Formulierung der Elektrodynamik als Lagrange- Feldtheorie Die rel. Dynamik eines Massepunktes kann aus dem Extremalprinzip abgeleitet werden, wenn man Die Punkt 1 und 2 als Anfangs- und Endereignis im 4- Raum sieht und wenn man die Ränder bei Variation festhält: :<math>\begin{align} & \delta W=0 \\ & W=\int_{1}^{2}{{}}ds \\ \end{align}</math> letzteres: Wirkungsintegral Wichtig: :<math>{{\left. \delta {{x}^{i}} \right|}_{1,2}}=0</math> Newtonsche Mechanik ist Grenzfall: :<math>W=-{{m}_{0}}c\int_{1}^{2}{{}}ds</math> Wechselwirkung eines Massepunktes mit einem 4- Vektor- Feld :<math>\begin{align} & \left( {{\phi }^{i}} \right)({{x}^{j}}) \\ & \Rightarrow \\ \end{align}</math> :<math>W=\int_{1}^{2}{{}}\left\{ -{{m}_{0}}cds-{{\phi }^{i}}d{{x}_{i}} \right\}</math> mit den Lorentz- Invarianten :<math>{{m}_{0}}cds</math> und <math>{{\phi }^{i}}d{{x}_{i}}</math> '''Variation:''' :<math>\delta W=\int_{1}^{2}{{}}\left\{ -{{m}_{0}}c\delta \left( ds \right)-\delta \left( {{\phi }^{\mu }}d{{x}_{\mu }} \right) \right\}</math> Nun: :<math>\begin{align} & \delta \left( ds \right)=\delta {{\left( d{{x}^{\mu }}d{{x}_{\mu }} \right)}^{\frac{1}{2}}}=\frac{1}{2}\frac{\left( d\delta {{x}^{\mu }} \right)d{{x}_{\mu }}+d{{x}^{\mu }}\left( d\delta {{x}_{\mu }} \right)}{ds} \\ & \left( d\delta {{x}^{\mu }} \right)d{{x}_{\mu }}=d{{x}^{\mu }}\left( d\delta {{x}_{\mu }} \right) \\ & =\frac{d{{x}^{\mu }}}{ds}\left( d\delta {{x}_{\mu }} \right)={{u}^{\mu }}\left( d\delta {{x}_{\mu }} \right) \\ \end{align}</math> Außerdem: :<math>\delta \left( {{\phi }^{\mu }}d{{x}_{\mu }} \right)=\delta {{\phi }^{\mu }}d{{x}_{\mu }}+{{\phi }^{\mu }}d\left( \delta {{x}_{\mu }} \right)</math> Somit: :<math>\delta W=\int_{1}^{2}{{}}\left\{ -{{m}_{0}}c{{u}^{\mu }}\left( d\delta {{x}_{\mu }} \right)-\delta {{\phi }^{\mu }}d{{x}_{\mu }}-{{\phi }^{\mu }}d\left( \delta {{x}_{\mu }} \right) \right\}</math> Weiter mit partieller Integration: :<math>\begin{align} & \int_{1}^{2}{{}}-{{m}_{0}}c{{u}^{\mu }}d\left( \delta {{x}_{\mu }} \right)=\left[ -{{m}_{0}}c{{u}^{\mu }}\left( \delta {{x}_{\mu }} \right) \right]_{1}^{2}+\int_{1}^{2}{{}}{{m}_{0}}cd{{u}^{\mu }}\left( \delta {{x}_{\mu }} \right) \\ & \left[ -{{m}_{0}}c{{u}^{\mu }}\left( \delta {{x}_{\mu }} \right) \right]_{1}^{2}=0,weil\delta {{x}_{\mu }}_{1}^{2}=0 \\ & \Rightarrow \int_{1}^{2}{{}}-{{m}_{0}}c{{u}^{\mu }}d\left( \delta {{x}_{\mu }} \right)=\int_{1}^{2}{{}}{{m}_{0}}cd{{u}^{\mu }}\left( \delta {{x}_{\mu }} \right)=\int_{1}^{2}{{}}{{m}_{0}}c\frac{d{{u}^{\mu }}}{ds}\left( \delta {{x}_{\mu }} \right)ds \\ \end{align}</math> Weiter: :<math>\int_{1}^{2}{{}}-{{\phi }^{\mu }}d\left( \delta {{x}_{\mu }} \right)=-\left[ {{\phi }^{\mu }}\delta {{x}_{\mu }} \right]_{1}^{2}+\int_{1}^{2}{{}}d{{\phi }^{\mu }}\left( \delta {{x}_{\mu }} \right)</math> Mit :<math>\begin{align} & d{{\phi }^{\mu }}={{\partial }^{\nu }}{{\phi }^{\mu }}d{{x}_{\nu }}={{\partial }^{\nu }}{{\phi }^{\mu }}{{u}_{\nu }}ds \\ & \delta {{\phi }^{\mu }}={{\partial }^{\nu }}{{\phi }^{\mu }}\delta {{x}_{\nu }} \\ & \delta {{\phi }^{\mu }}d{{x}_{\mu }}={{\partial }^{\nu }}{{\phi }^{\mu }}\delta {{x}_{\nu }}d{{x}_{\mu }}=i<->k={{\partial }^{\mu }}{{\phi }^{\nu }}\delta {{x}_{\mu }}d{{x}_{\nu }}={{\partial }^{\mu }}{{\phi }^{\nu }}{{u}_{\nu }}\delta {{x}_{\mu }}ds \\ \end{align}</math> Einsetzen in :<math>\delta W=\int_{1}^{2}{{}}\left\{ -{{m}_{0}}c{{u}^{\mu }}\left( d\delta {{x}_{\mu }} \right)-\delta {{\phi }^{\mu }}d{{x}_{\mu }}-{{\phi }^{\mu }}d\left( \delta {{x}_{\mu }} \right) \right\}</math> liefert: :<math>\delta W=\int_{1}^{2}{{}}\left\{ {{m}_{0}}c\frac{d{{u}^{\mu }}}{ds}-\left( {{\partial }^{\mu }}{{\phi }^{\nu }}-{{\partial }^{\nu }}{{\phi }^{\mu }} \right){{u}_{\nu }} \right\}\delta {{x}_{\mu }}</math> '''Wegen''' :<math>\begin{align} & \delta W=\int_{1}^{2}{{}}\left\{ {{m}_{0}}c\frac{d{{u}^{\mu }}}{ds}-\left( {{\partial }^{\mu }}{{\phi }^{\nu }}-{{\partial }^{\nu }}{{\phi }^{\mu }} \right){{u}_{\nu }} \right\}\delta {{x}_{\mu }}=0 \\ & {{m}_{0}}c\frac{d{{u}^{\mu }}}{ds}=\left( {{\partial }^{\mu }}{{\phi }^{\nu }}-{{\partial }^{\nu }}{{\phi }^{\mu }} \right){{u}_{\nu }}:={{f}^{\mu \nu }}{{u}_{\nu }} \\ & {{f}^{\mu \nu }}=\left( {{\partial }^{\mu }}{{\phi }^{\nu }}-{{\partial }^{\nu }}{{\phi }^{\mu }} \right) \\ \end{align}</math> Dies ist dann die aus dem hamiltonschen Prinzip abgeleitete Bewegungsgleichung eines Massepunktes der Ruhemasse m0 und der Ladung q unter dem Einfluss der Lorentz- Kraft. Man setze: :<math>\begin{align} & {{p}^{\mu }}={{m}_{0}}c{{u}^{\mu }} \\ & {{f}^{\mu \nu }}=\frac{q}{c}{{F}^{\mu \nu }}=\left( {{\partial }^{\mu }}{{\phi }^{\nu }}-{{\partial }^{\nu }}{{\phi }^{\mu }} \right) \\ & {{\phi }^{\mu }}=\frac{q}{c}{{\Phi }^{\mu }} \\ & \frac{d}{ds}{{p}^{\mu }}=\frac{q}{c}{{F}^{\mu \nu }}{{u}_{\nu }}\Leftrightarrow \delta W=\delta \int_{1}^{2}{{}}\left\{ -{{m}_{0}}cds-\frac{q}{c}{{\Phi }^{\mu }}d{{x}_{\mu }} \right\}=0 \\ \end{align}</math> Man bestimmt die Ortskomponenten :<math>\alpha =1,2,3</math> über :<math>\begin{align} & \frac{d}{dt}\bar{p}=q\left( \bar{E}+\bar{v}\times \bar{B} \right) \\ & \\ \end{align}</math> überein, denn mit :<math>\begin{align} & {{u}^{0}}=\gamma \\ & {{u}^{\alpha }}=\frac{\gamma }{c}{{v}^{\alpha }}=-{{u}_{\alpha }} \\ \end{align}</math> folgt dann: :<math>\begin{align} & \frac{d}{dt}{{p}^{1}}=q\left( {{E}^{1}}+{{v}^{2}}{{B}^{3}}-{{v}^{3}}{{B}^{2}} \right) \\ & =q\left( {{F}^{10}}+{{F}^{21}}\frac{1}{c}{{v}^{2}}-{{F}^{13}}\frac{1}{c}{{v}^{3}} \right) \\ & =\frac{q}{\gamma }\left( {{F}^{10}}\gamma +{{F}^{21}}\frac{\gamma }{c}{{v}^{2}}-{{F}^{13}}\frac{\gamma }{c}{{v}^{3}} \right)=\frac{q}{\gamma }{{F}^{1\mu }}{{u}_{\mu }} \\ \end{align}</math> mit <math>ds=\frac{c}{\gamma }dt</math> : :<math>\frac{d}{ds}{{p}^{1}}=\frac{q}{c}{{F}^{1\mu }}{{u}_{\mu }}</math> Die zeitartige Komponente :<math>\mu =0</math> gibt wegen :<math>{{p}^{0}}=\frac{E}{c}</math> : :<math>\begin{align} & \frac{d}{ds}\frac{E}{c}=\frac{\gamma }{{{c}^{2}}}\frac{dE}{dt}=\frac{q}{c}\left( {{F}^{01}}{{u}_{1}}+{{F}^{02}}{{u}_{2}}+{{F}^{03}}{{u}_{3}} \right)= \\ & =\frac{q\gamma }{{{c}^{2}}}\left( -{{E}^{1}}{{v}_{1}}-{{E}^{2}}{{v}_{2}}-{{E}^{3}}{{v}_{3}} \right)=\frac{q\gamma }{{{c}^{2}}}\left( {{E}^{1}}{{v}^{1}}+{{E}^{2}}{{v}^{2}}+{{E}^{3}}{{v}^{3}} \right) \\ & \frac{dE}{dt}=q\bar{E}\cdot \bar{v} \\ \end{align}</math> Dies ist die Leistungsbilanz: Die Änderung der inneren Energie ist gleich der reingesteckten Arbeit
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