Elektrodynamikvorlesung von Prof. Dr. E. Schöll, PhD
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Der Artikel Eichinvarianz basiert auf der Vorlesungsmitschrift von Franz- Josef Schmitt des 3.Kapitels (Abschnitt 6) der Elektrodynamikvorlesung von Prof. Dr. E. Schöll, PhD.
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{{#set:Urheber=Prof. Dr. E. Schöll, PhD|Inhaltstyp=Script|Kapitel=3|Abschnitt=6}}
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Die Felder
![{\displaystyle {\bar {E}},{\bar {B}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a4656d3ca4d5059ea86ae94578f3b718c495d22e)
werden durch die Potenziale
![{\displaystyle \Phi \left({\bar {r}},t\right),{\bar {A}}\left({\bar {r}},t\right)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/470ad1613e599872b9973144b07d95c7a5c49fa3)
dargestellt.:
![{\displaystyle {\begin{aligned}&{\bar {E}}=-\nabla \Phi \left({\bar {r}},t\right)-{\frac {\partial }{\partial t}}{\bar {A}}\left({\bar {r}},t\right)\\&{\bar {B}}=\nabla \times {\bar {A}}\left({\bar {r}},t\right)\\\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/af805834337963f084d21f9b71804a0c6d5132f3)
Dabei drängt sich die Frage auf, welche die allgemeinste Transformation
![{\displaystyle {\begin{aligned}&\Phi \left({\bar {r}},t\right)\to \Phi {\acute {\ }}\left({\bar {r}},t\right)\\&{\bar {A}}\left({\bar {r}},t\right)\to {\bar {A}}{\acute {\ }}\left({\bar {r}},t\right)\\\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/68655ef921ce8b27d52ad5812d349eb95779d2e1)
ist, welche die Felder E und B unverändert läßt.
Also:
![{\displaystyle {\begin{aligned}&{\bar {E}}=-\nabla \Phi \left({\bar {r}},t\right)-{\frac {\partial }{\partial t}}{\bar {A}}\left({\bar {r}},t\right)=-\nabla \Phi {\acute {\ }}\left({\bar {r}},t\right)-{\frac {\partial }{\partial t}}{\bar {A}}{\acute {\ }}\left({\bar {r}},t\right)\\&{\bar {B}}=\nabla \times {\bar {A}}\left({\bar {r}},t\right)=\nabla \times {\bar {A}}{\acute {\ }}\left({\bar {r}},t\right)\\&\Rightarrow {\bar {A}}{\acute {\ }}\left({\bar {r}},t\right)={\bar {A}}\left({\bar {r}},t\right)+\nabla G\left({\bar {r}},t\right)\\&\Rightarrow -\nabla \Phi \left({\bar {r}},t\right)-{\frac {\partial }{\partial t}}{\bar {A}}\left({\bar {r}},t\right)=-\nabla \Phi {\acute {\ }}\left({\bar {r}},t\right)-{\frac {\partial }{\partial t}}\left({\bar {A}}\left({\bar {r}},t\right)+\nabla G\left({\bar {r}},t\right)\right)\\&\Rightarrow \nabla \left(\Phi {\acute {\ }}\left({\bar {r}},t\right)-\Phi \left({\bar {r}},t\right)+{\frac {\partial }{\partial t}}G\left({\bar {r}},t\right)\right)=0\\&\Rightarrow \left(\Phi {\acute {\ }}\left({\bar {r}},t\right)-\Phi \left({\bar {r}},t\right)+{\frac {\partial }{\partial t}}G\left({\bar {r}},t\right)\right)=g(t)(r-unabh{\ddot {a}}ngig)\\\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/57dc62babac67d78b4cb1fc970c468e71c2d21ae)
Mit
![{\displaystyle {\begin{aligned}&F\left({\bar {r}},t\right):=G\left({\bar {r}},t\right)-\int _{to}^{t}{dt{\acute {\ }}g(t{\acute {\ }})}\\&\Rightarrow {\bar {A}}{\acute {\ }}\left({\bar {r}},t\right)={\bar {A}}\left({\bar {r}},t\right)+\nabla F\left({\bar {r}},t\right)\\&\Phi {\acute {\ }}\left({\bar {r}},t\right)=\Phi \left({\bar {r}},t\right)-{\frac {\partial }{\partial t}}F\left({\bar {r}},t\right)\\\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/fb33574263850145a6456e0beb3c6148dbff88a5)
mit eine völlig beliebigen Eichfunktion
.
Alle physikalischen Aussagen müssen invariant sein! Aber nicht nur
![{\displaystyle {\bar {E}},{\bar {B}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a4656d3ca4d5059ea86ae94578f3b718c495d22e)
sondern auch
![{\displaystyle \Phi \left({\bar {r}},t\right),{\bar {A}}\left({\bar {r}},t\right)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/470ad1613e599872b9973144b07d95c7a5c49fa3)
sind physikalisch relevant.
So muss auch
![{\displaystyle \oint \limits _{\partial F}{d{\bar {s}}}{\bar {A}}\left({\bar {r}},t\right)=\int _{F}^{}{d{\bar {f}}{\bar {B}}\left({\bar {r}},t\right)=\Phi \left({\bar {r}},t\right)}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/04e238ab3099b05a74e0f0ad663f9f7fe05c8704)
erfüllt sein.
Dies ist gewährleistet, wenn die Maxwellgleichungen erfüllt sind.
Durch
![{\displaystyle {\begin{aligned}&{\bar {E}}=-\nabla \Phi \left({\bar {r}},t\right)-{\frac {\partial }{\partial t}}{\bar {A}}\left({\bar {r}},t\right)\\&{\bar {B}}=\nabla \times {\bar {A}}\left({\bar {r}},t\right)\\\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/af805834337963f084d21f9b71804a0c6d5132f3)
sind die homogenen Maxwellgleichungen bereits erfüllt:
![{\displaystyle {\begin{aligned}&\nabla \times {\bar {E}}=-\nabla \times \nabla \Phi \left({\bar {r}},t\right)-{\frac {\partial }{\partial t}}\nabla \times {\bar {A}}\left({\bar {r}},t\right)=-{\frac {\partial }{\partial t}}{\bar {B}}\\&\nabla \cdot {\bar {B}}=\nabla \cdot \left(\nabla \times {\bar {A}}\left({\bar {r}},t\right)\right)=0\\\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/3c8d06f2ada9bdf04f9037e727b558f5171a0c13)
Auch die Umkehrung gilt:
![{\displaystyle {\begin{aligned}&\nabla \cdot {\bar {B}}=0\\&\Rightarrow \exists {\bar {A}}\left({\bar {r}},t\right)\Rightarrow \nabla \times {\bar {A}}\left({\bar {r}},t\right)={\bar {B}}\\&\nabla \times {\bar {E}}=-{\frac {\partial }{\partial t}}{\bar {B}}=-\nabla \times {\frac {\partial }{\partial t}}{\bar {A}}\left({\bar {r}},t\right)\Rightarrow \nabla \times \left({\bar {E}}+{\frac {\partial }{\partial t}}{\bar {A}}\left({\bar {r}},t\right)\right)=0\\&\Rightarrow \exists \Phi \left({\bar {r}},t\right)\Rightarrow {\bar {E}}+{\frac {\partial }{\partial t}}{\bar {A}}\left({\bar {r}},t\right)=-\nabla \Phi \left({\bar {r}},t\right)\\\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d58c7d94f13cdb2553716c3add3191ce7b9b0825)
Wähle nun eine Eichung derart, dass die inhomogenen Maxwellgleichungen besonders einfach werden
Ziel: Entkopplung der DGLs für
![{\displaystyle {\bar {A}}\left({\bar {r}},t\right),\Phi \left({\bar {r}},t\right)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a50ed8090e8a89a1a10cef3febd0b8cd38fe8fc5)
- Lorentz- Eichung:
![{\displaystyle \nabla \cdot {\bar {A}}\left({\bar {r}},t\right)+{{\varepsilon }_{0}}{{\mu }_{0}}{\frac {\partial }{\partial t}}\Phi \left({\bar {r}},t\right)=0}](https://wikimedia.org/api/rest_v1/media/math/render/svg/47c0d0c5f2c42b98a642d7559c11f9d57782c5be)
Genau dadurch werden die Feldgleichungen entkoppelt:
1)
![{\displaystyle {\begin{aligned}&-\nabla \cdot {\bar {E}}=\nabla \cdot \left(\nabla \Phi \left({\bar {r}},t\right)+{\frac {\partial }{\partial t}}{\bar {A}}\left({\bar {r}},t\right)\right)=-{\frac {\rho }{{\varepsilon }_{0}}}\\&\Delta \Phi \left({\bar {r}},t\right)+{\frac {\partial }{\partial t}}\nabla \cdot {\bar {A}}\left({\bar {r}},t\right)=-{\frac {\rho }{{\varepsilon }_{0}}}\\\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/140aa9a7e035be1f04ff3776cac9e14910308eb1)
Was mit Hilfe der Lorentzeichung wird zu
![{\displaystyle \Delta \Phi \left({\bar {r}},t\right)-{{\varepsilon }_{0}}{{\mu }_{0}}{\frac {{\partial }^{2}}{\partial {{t}^{2}}}}\Phi \left({\bar {r}},t\right)=-{\frac {\rho }{{\varepsilon }_{0}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5ec897488624c41561fc0abdd74a6864b49c75c6)
Für A:
2)
![{\displaystyle {\begin{aligned}&{\frac {1}{{\mu }_{0}}}\nabla \times {\bar {B}}-{{\varepsilon }_{0}}{\frac {\partial }{\partial t}}{\bar {E}}={\bar {j}}\\&\Rightarrow \nabla \times \left(\nabla \times {\bar {A}}\left({\bar {r}},t\right)\right)+{{\varepsilon }_{0}}{{\mu }_{0}}{\frac {\partial }{\partial t}}\left(\nabla \Phi \left({\bar {r}},t\right)+{\frac {\partial }{\partial t}}{\bar {A}}\left({\bar {r}},t\right)\right)={{\mu }_{0}}{\bar {j}}\\&\nabla \times \left(\nabla \times {\bar {A}}\left({\bar {r}},t\right)\right)=+\nabla \left(\nabla \cdot {\bar {A}}\left({\bar {r}},t\right)\right)-\Delta {\bar {A}}\left({\bar {r}},t\right)\\&\Rightarrow \Delta {\bar {A}}\left({\bar {r}},t\right)-{{\varepsilon }_{0}}{{\mu }_{0}}{\frac {{\partial }^{2}}{\partial {{t}^{2}}}}{\bar {A}}\left({\bar {r}},t\right)-\nabla \left(\nabla \cdot {\bar {A}}\left({\bar {r}},t\right)+{{\varepsilon }_{0}}{{\mu }_{0}}{\frac {\partial }{\partial t}}\Phi \left({\bar {r}},t\right)\right)=-{{\mu }_{0}}{\bar {j}}\\\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f3d55863ad80310d8a9f87bb6e96222c036809a4)
Was mit der Lorentz- Eichung
![{\displaystyle \nabla \cdot {\bar {A}}\left({\bar {r}},t\right)+{{\varepsilon }_{0}}{{\mu }_{0}}{\frac {\partial }{\partial t}}\Phi \left({\bar {r}},t\right)=0}](https://wikimedia.org/api/rest_v1/media/math/render/svg/47c0d0c5f2c42b98a642d7559c11f9d57782c5be)
wird zu
![{\displaystyle \Delta {\bar {A}}\left({\bar {r}},t\right)-{{\varepsilon }_{0}}{{\mu }_{0}}{\frac {{\partial }^{2}}{\partial {{t}^{2}}}}{\bar {A}}\left({\bar {r}},t\right)=-{{\mu }_{0}}{\bar {j}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5962b8a304a370d6d236ee3ac47d9cb26dc05469)
Dies kann in Viererschreibweise mit dem dÁlembertschen Operator # mit
![{\displaystyle \#:=\Delta -{\frac {1}{{c}^{2}}}{\frac {{\partial }^{2}}{\partial {{t}^{2}}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/836b61241d65085e5fe49dea837ec20a141b531c)
zusammengefasst werden:
![{\displaystyle {\begin{aligned}&\#\Phi \left({\bar {r}},t\right)=-{\frac {\rho }{{\varepsilon }_{0}}}\\&\#{\bar {A}}\left({\bar {r}},t\right)=-{{\mu }_{0}}{\bar {j}}\\\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/da0211640b94d753a07108ca1c3be8906bb49a98)
Dies sind die inhomogenen Wellengleichungen für die Potenziale (entkoppelt mittels Lorentz- Eichung)
Es ergibt sich im SI- System:
![{\displaystyle {\frac {1}{\sqrt {{{\varepsilon }_{0}}{{\mu }_{0}}}}}:=c=2,994\cdot {{10}^{8}}{\frac {m}{s}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a1c101fbeb51d0a2f203e2b8d174b731894998d2)
als Lichtgeschwindigkeit
Dies ist einfach die ermittelte Ausbreitungsgeschwindigkeit der elektromagnetischen Wellen im Vakuum!
Coulomb- Eichung
(sogenannte Strahlungseichung):
![{\displaystyle \nabla \cdot {\bar {A}}\left({\bar {r}},t\right)=0}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1f3c1b8f5b82dabf4be211a430ed051f848f53bd)
Vergleiche Kapitel 2.3 (Magnetostatik):
Für
![{\displaystyle {\begin{aligned}&{\dot {\bar {D}}}=0\\&\Rightarrow \nabla \times {\bar {B}}=\nabla \left(\nabla \cdot {\bar {A}}\right)-\Delta {\bar {A}}={{\mu }_{0}}{\bar {j}}\\\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2247b1769723ce605da7ef25591983f61da539d1)
(Poissongleichung der Magnetostatik)
Zerlegung in longitudinale und transversale Anteile :
Allgemein kann man
![{\displaystyle {\bar {E}}=-\nabla \Phi \left({\bar {r}},t\right)-{\frac {\partial }{\partial t}}{\bar {A}}\left({\bar {r}},t\right)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/72f6fc0c0e54949dd88028c45642930a2d8c19b2)
in ein wirbelfreies Longitudinalfeld:
![{\displaystyle {{\bar {E}}_{l}}:=-\nabla \Phi \left({\bar {r}},t\right)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c6f843555a606cb752e9580794834ff990b1583e)
und ein quellenfreies Transversalfeld
![{\displaystyle {{\bar {E}}_{t}}=-{\frac {\partial }{\partial t}}{\bar {A}}\left({\bar {r}},t\right)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9a2f80385cd132b6af35b16d80f890373473f52f)
zerlegen.
Tatsächlich gilt:
![{\displaystyle \nabla \times {{\bar {E}}_{l}}:=-\nabla \times \left(\nabla \Phi \left({\bar {r}},t\right)\right)=0}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4eabc14c7801cfe7b080f1a12a6c2e103ec14f9b)
![{\displaystyle \nabla \cdot {{\bar {E}}_{t}}=-{\frac {\partial }{\partial t}}\nabla \cdot {\bar {A}}\left({\bar {r}},t\right)=0}](https://wikimedia.org/api/rest_v1/media/math/render/svg/96ff0d8d3f7ad31811a9b45cccbff46d12f20570)
Da
![{\displaystyle {\bar {B}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/989ca009963f2cdcdf47ad8e9610946bfba335c9)
quellenfrei ist, ist B auch immer transversal:
![{\displaystyle \nabla \cdot {\bar {B}}:=\nabla \cdot \left(\nabla \times {\bar {A}}\right)=0}](https://wikimedia.org/api/rest_v1/media/math/render/svg/870a738e092bc014c8c09b021ef85a75ca73735a)
Also:
![{\displaystyle \Phi \left({\bar {r}},t\right)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0c7a47f037ae16229ec700eb90b53060d5dbbc27)
ergibt die longitudinalen Felder und
![{\displaystyle {\bar {A}}\left({\bar {r}},t\right)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a20aecfcc454446b978df9c655623b73da954bba)
die transversalen Felder.
Merke: Felder, die Rotation eines Vektorfeldes sind (Vektorpotenzials) sind grundsätzlich transversaler Natur. (Divergenz verschwindet). Divergenzfelder (als Gradienten eines Skalars) sind immer longitudinal! (Rotation verschwindet).
Zerlegung der Stromdichte:
mit ![{\displaystyle \nabla \times {{\bar {j}}_{l}}=0}](https://wikimedia.org/api/rest_v1/media/math/render/svg/92b079b50826bbb9fd73b996386d6e7695697b48)
![{\displaystyle \nabla \cdot {{\bar {j}}_{t}}=0}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0dec95ec88dd79d6234dfdf3fe9bb1de5f2ac45b)
Mit
![{\displaystyle {\begin{aligned}&{\frac {\partial }{\partial t}}\rho +\nabla \cdot {{\bar {j}}_{l}}+\nabla \cdot {{\bar {j}}_{t}}=0\\&\rho ={{\varepsilon }_{0}}\nabla \cdot {{\bar {E}}_{l}}\\&\nabla \cdot {{\bar {j}}_{t}}=0\\&\Rightarrow \nabla \cdot \left({{\bar {j}}_{l}}+{{\varepsilon }_{0}}{\frac {\partial }{\partial t}}{{\bar {E}}_{l}}\right)=0\\\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ff7e4f664141e6201bd73c52e2c45c7cb90236c1)
Außerdem gilt nach der Definition von longitudinal:
![{\displaystyle \nabla \times \left({{\bar {j}}_{l}}+{{\varepsilon }_{0}}{\frac {\partial }{\partial t}}{{\bar {E}}_{l}}\right)=0}](https://wikimedia.org/api/rest_v1/media/math/render/svg/253659fa446701fe9b4bb03e72568d413e1ac805)
Also:
![{\displaystyle \left({{\bar {j}}_{l}}+{{\varepsilon }_{0}}{\frac {\partial }{\partial t}}{{\bar {E}}_{l}}\right)=const}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5454f33ba837e57693bc2455f922015e82fce6e5)
Da beide Felder aber für r→ 0 verschwinden folgt:
![{\displaystyle \left({{\bar {j}}_{l}}+{{\varepsilon }_{0}}{\frac {\partial }{\partial t}}{{\bar {E}}_{l}}\right)=0}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f05ac9b3509e106377df75ebf8e220c2f598553e)
Also:
![{\displaystyle {{\bar {j}}_{l}}={{\varepsilon }_{0}}\nabla {\frac {\partial \Phi }{\partial t}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0542fe59a710ba5fd5ed6adc37574980a78c4c31)
Also:
Die Feldgleichungen
und ![{\displaystyle {\begin{aligned}&\Delta {\bar {A}}\left({\bar {r}},t\right)-{{\varepsilon }_{0}}{{\mu }_{0}}{\frac {{\partial }^{2}}{\partial {{t}^{2}}}}{\bar {A}}\left({\bar {r}},t\right)-\nabla \left(\nabla \cdot {\bar {A}}\left({\bar {r}},t\right)+{{\varepsilon }_{0}}{{\mu }_{0}}{\frac {\partial }{\partial t}}\Phi \left({\bar {r}},t\right)\right)=-{{\mu }_{0}}{\bar {j}}\\&\nabla \cdot {\bar {A}}\left({\bar {r}},t\right)=0\\&\nabla {{\varepsilon }_{0}}{\frac {\partial }{\partial t}}\Phi \left({\bar {r}},t\right)={{\bar {j}}_{l}}\\\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e3051a5d61f15f597980d256b59821ce9674d18e)
erhalten dann die Form:
und ![{\displaystyle {\begin{aligned}&\#{\bar {A}}\left({\bar {r}},t\right)=-{{\mu }_{0}}{{\bar {j}}_{t}}\\&\\\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0fac72247839bd0cdbf49b0d76eec22f777fef78)
In der Coulomb- Eichung!
Also.
![{\displaystyle \Delta \Phi =-{\frac {\rho }{{\varepsilon }_{0}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7d8c5a9f4c2b25f5ea284bb0793947b798854715)
- longitudinale Felder entsprechend der Elektrostatik
![{\displaystyle \#{\bar {A}}\left({\bar {r}},t\right)=-{{\mu }_{0}}{{\bar {j}}_{t}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c47d68a4f5c48f499bff42bff95933d595d4257b)
als transversale Felder entsprechend elektromagnetischen Wellen.
Das bedeutet : Die Coulombeichung ist zweckmäßig bei Strahlungsproblemen!
Sie liefert eine Poissongleichung für
![{\displaystyle \Phi }](https://wikimedia.org/api/rest_v1/media/math/render/svg/aed80a2011a3912b028ba32a52dfa57165455f24)
und eine Wellengleichung für
.