Elektrodynamikvorlesung von Prof. Dr. E. Schöll, PhD
{{#set:Urheber=Prof. Dr. E. Schöll, PhD|Inhaltstyp=Script|Kapitel=6|Abschnitt=3}}
Kategorie:Elektrodynamik
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Ziel: 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:
![{\displaystyle {\begin{aligned}&\delta W=0\\&W=\int _{1}^{2}{}ds\\\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6d9580f8cc01f086697f4cf6d35e984f71b0df80)
letzteres: Wirkungsintegral
Wichtig:
![{\displaystyle {{\left.\delta {{x}^{i}}\right|}_{1,2}}=0}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2d0b2e6832a0a3f5cbe801228e3572ea38123031)
Newtonsche Mechanik ist Grenzfall:
![{\displaystyle W=-{{m}_{0}}c\int _{1}^{2}{}ds}](https://wikimedia.org/api/rest_v1/media/math/render/svg/65cf3e659b3fb99403c53f19cb23eda1e251111b)
Wechselwirkung eines Massepunktes mit einem 4- Vektor- Feld
![{\displaystyle {\begin{aligned}&\left({{\phi }^{i}}\right)({{x}^{j}})\\&\Rightarrow \\\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8bb7b7ce449fbc68e9e744d57d519e83412db0f4)
![{\displaystyle W=\int _{1}^{2}{}\left\{-{{m}_{0}}cds-{{\phi }^{i}}d{{x}_{i}}\right\}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7407024e3e3c91d30dfb3dd11aff3205b8d5c464)
mit den Lorentz- Invarianten
und ![{\displaystyle {{\phi }^{i}}d{{x}_{i}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a10dfffe9fd568bd5be4544f7d1bdf8c141c3c61)
Variation:
![{\displaystyle \delta W=\int _{1}^{2}{}\left\{-{{m}_{0}}c\delta \left(ds\right)-\delta \left({{\phi }^{\mu }}d{{x}_{\mu }}\right)\right\}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/67241a1ce8a26f2f7793e50a3fb763af572e51b3)
Nun:
![{\displaystyle {\begin{aligned}&\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{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6cdb53d192ff8a6ca295cdd818fecbc56b652648)
Außerdem:
![{\displaystyle \delta \left({{\phi }^{\mu }}d{{x}_{\mu }}\right)=\delta {{\phi }^{\mu }}d{{x}_{\mu }}+{{\phi }^{\mu }}d\left(\delta {{x}_{\mu }}\right)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/714b689eeabaf820142fc6e408f0d9ad09ae22b2)
Somit:
![{\displaystyle \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\}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e84095401f7e9525ffc8da1edcb79754d252bdfb)
Weiter mit partieller Integration:
![{\displaystyle {\begin{aligned}&\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{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/773edc2d7707153bd4528a22fbeec90de28394c9)
Weiter:
![{\displaystyle \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)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9916c0d44cd66cdcda4fd99af6c5049d19a4720d)
Mit
![{\displaystyle {\begin{aligned}&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{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/05e193ed09c552395d3f581fa4b96e9ba2702185)
Einsetzen in
![{\displaystyle \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\}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e84095401f7e9525ffc8da1edcb79754d252bdfb)
liefert:
![{\displaystyle \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 }}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/cad83cd8738cb5f8894f7b23a0163fc3f4865015)
Wegen
![{\displaystyle {\begin{aligned}&\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{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1c3071dd1348dfc947673cc3fc602bcc33bf5b9b)
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:
![{\displaystyle {\begin{aligned}&{{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{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/dd700789ad4868b688eb47e714c487b5a80a6e33)
Man bestimmt die Ortskomponenten
![{\displaystyle \alpha =1,2,3}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e70c36a23e94c1aa3e6b4778e3db3fc598ad458d)
über
![{\displaystyle {\begin{aligned}&{\frac {d}{dt}}{\bar {p}}=q\left({\bar {E}}+{\bar {v}}\times {\bar {B}}\right)\\&\\\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1edf521243260b24912ee616137e36c033c0ab0e)
überein, denn mit
![{\displaystyle {\begin{aligned}&{{u}^{0}}=\gamma \\&{{u}^{\alpha }}={\frac {\gamma }{c}}{{v}^{\alpha }}=-{{u}_{\alpha }}\\\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d1899d017cf01a7c235e607205db36695c0d8b48)
folgt dann:
mit ![{\displaystyle ds={\frac {c}{\gamma }}dt}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f5fecbfd8502df3e5722a93e4d0e786ff5580491)
![{\displaystyle {\frac {d}{ds}}{{p}^{1}}={\frac {q}{c}}{{F}^{1\mu }}{{u}_{\mu }}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6685e3f5363c5fd14c58fa09c30b1957d020d411)
Die zeitartige Komponente
![{\displaystyle \mu =0}](https://wikimedia.org/api/rest_v1/media/math/render/svg/3753282c0ad2ea1e7d63f39425efd13c37da3169)
gibt wegen
![{\displaystyle {{p}^{0}}={\frac {E}{c}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/536d78441c3a0a3350ceacbb304167339d268454)
![{\displaystyle {\begin{aligned}&{\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{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f939428937ca016b500befd3a67aaa54b96288cd)
Dies ist die Leistungsbilanz: Die Änderung der inneren Energie ist gleich der reingesteckten Arbeit