Mechanikvorlesung von Prof. Dr. E. Schöll, PhD
{{#set:Urheber=Prof. Dr. E. Schöll, PhD|Inhaltstyp=Script|Kapitel=2|Abschnitt=3}}
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Uneindeutigkeit der Lagrangefunktion[edit | edit source]
Die Lagarangefunktion wird duch die Lagrangegleichung nicht eindeutig festgelegt.
Betrachten wir beispielsweise ein geladenes Teilchen im elektrischen Feld:
![{\displaystyle \left({{q}_{1}},{{q}_{2}},{{q}_{3}}\right)=\left({{x}_{1}},{{x}_{2}},{{x}_{3}}\right)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4aaa8abf62472860773b90bc48957a4876d12010)
e sei die Ladung
Bewegungsgleichung:
![{\displaystyle m{\frac {{{d}^{2}}^{}}{d{{t}^{2}}}}\left({{q}_{1}},{{q}_{2}},{{q}_{3}}\right)=m{\ddot {\bar {q}}}=e{\bar {E}}({\bar {q}},t)+e{\dot {\bar {q}}}\times {\bar {B}}({\bar {q}},t)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/364f36b8312fa1ef9a2daf181ebf0efdea26b6ec)
Die Lorentzkraft{{#set:Fachbegriff=Lorentzkraft|Index=Lorentzkraft}} ist typischerweise nicht konservativ
Die Darstellung des elektrischen und magnetischen Feldes erfolgt über die Potenziale:
![{\displaystyle {\begin{aligned}&{\bar {E}}({\bar {q}},t)=-\nabla \Phi ({\bar {q}},t)-{\frac {\partial }{\partial t}}{\bar {A}}({\bar {q}},t)\\&{\bar {B}}({\bar {q}},t)=\nabla \times {\bar {A}}({\bar {q}},t)\\\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5cf868e18074e039e6d4193610d9e239b726306e)
Dabei ist
Skalar und A ein Vektorpotenzial (MKSA- System)
Ziel: Suche eine Lagrangefunktion
in der Art, dass
Die Bewegungsgleichung
![{\displaystyle m{\frac {{{d}^{2}}^{}}{d{{t}^{2}}}}\left({{q}_{1}},{{q}_{2}},{{q}_{3}}\right)=m{\ddot {\bar {q}}}=e{\bar {E}}({\bar {q}},t)+e{\dot {\bar {q}}}\times {\bar {B}}({\bar {q}},t)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/364f36b8312fa1ef9a2daf181ebf0efdea26b6ec)
ergeben.
Ansatz:
![{\displaystyle L(q,{\dot {q}},t)={\frac {m}{2}}{{\dot {\bar {q}}}^{2}}+e\left({\dot {\bar {q}}}{\bar {A}}({\bar {q}},t)-\Phi ({\bar {q}},t)\right)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d94304c7daddcd47fb51142363382b5cd8eac887)
Probe:
![{\displaystyle {\begin{aligned}&{\frac {\partial L}{\partial {{\dot {q}}_{k}}}}=m{{\dot {q}}_{k}}+e{{A}_{k}}\\&{\frac {d}{dt}}{\frac {\partial L}{\partial {{\dot {q}}_{k}}}}=m{{\ddot {q}}_{k}}+e{\frac {d}{dt}}{{A}_{k}}({\bar {q}}(t),t)\\&{\frac {d}{dt}}{\frac {\partial L}{\partial {{\dot {q}}_{k}}}}=m{{\ddot {q}}_{k}}+e\left({\frac {\partial }{\partial t}}{{A}_{k}}+\sum \limits _{l}{{\frac {\partial {{A}_{k}}}{\partial {{q}_{l}}}}{{\dot {q}}_{l}}}\right)\\&{\frac {d}{dt}}{\frac {\partial L}{\partial {{\dot {q}}_{k}}}}=m{{\ddot {q}}_{k}}+e\left({\frac {\partial }{\partial t}}{{A}_{k}}+\left({\dot {\bar {q}}}\cdot \nabla \right){{A}_{k}}\right)\\\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0db443a86125f6ea0a282a7744a9d32f85f601e4)
Weiter:
![{\displaystyle {\frac {\partial L}{\partial {{q}_{k}}}}=e\left[{\frac {\partial }{\partial {{q}_{k}}}}\left({\dot {\bar {q}}}\cdot {\bar {A}}\right)-{\frac {\partial }{\partial {{q}_{k}}}}\Phi \right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/94ce91c26a23a0b95492735ab13776ebedae936a)
Somit:
![{\displaystyle {\begin{aligned}&0={\frac {\partial L}{\partial {{q}_{k}}}}-{\frac {d}{dt}}{\frac {\partial L}{\partial {{\dot {q}}_{k}}}}=m{{\ddot {q}}_{k}}+e\left({\frac {\partial }{\partial t}}{{A}_{k}}+\left({\dot {\bar {q}}}\cdot \nabla \right){{A}_{k}}\right)-e\left[{\frac {\partial }{\partial {{q}_{k}}}}\left({\dot {\bar {q}}}\cdot {\bar {A}}\right)-{\frac {\partial }{\partial {{q}_{k}}}}\Phi \right]\\&=m{{\ddot {q}}_{k}}+e\left({\frac {\partial }{\partial t}}{{A}_{k}}+{\frac {\partial }{\partial {{q}_{k}}}}\Phi \right)+e\left[-{\frac {\partial }{\partial {{q}_{k}}}}\left({\dot {\bar {q}}}\cdot {\bar {A}}\right)+\left({\dot {\bar {q}}}\cdot \nabla \right){{A}_{k}}\right]\\&=m{{\ddot {q}}_{k}}-e{{E}_{k}}-{{\left[e{\dot {\bar {q}}}\times \left(\nabla \times {\bar {A}}\right)\right]}_{k}}\\&=m{{\ddot {q}}_{k}}-e{{E}_{k}}-{{\left[e{\dot {\bar {q}}}\times {\bar {B}}\right]}_{k}}\\\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4e2e6ae3db72dda0fc3e0d92f2a01161676e3bb4)
Somit erfüllt unser Ansatz die Bewegungsgleichungen
Die Potenziale lassen sich umeichen mit Hilfe der Eichfunktion{{#set:Fachbegriff=Eichfunktion|Index=Eichfunktion}}
:
![{\displaystyle {\begin{aligned}&{\bar {A}}({\bar {q}},t)\to {\bar {A}}{\acute {\ }}({\bar {q}},t)={\bar {A}}({\bar {q}},t)+\nabla \chi ({\bar {q}},t)\\&\Phi ({\bar {q}},t)\to \Phi {\acute {\ }}({\bar {q}},t)=\Phi ({\bar {q}},t)-{\frac {\partial }{\partial t}}\chi ({\bar {q}},t)\\\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1f276a4e4ba8abd251ba0e0ba58cac4282ca9a94)
Durch Einsetzen sieht man schnell, dass sich die Felder nicht ändern:
![{\displaystyle {\begin{aligned}&{\bar {E}}{\acute {\ }}({\bar {q}},t)=-\nabla \Phi {\acute {\ }}({\bar {q}},t)-{\frac {\partial }{\partial t}}{\bar {A}}{\acute {\ }}({\bar {q}},t)=-\nabla \left(\Phi ({\bar {q}},t)-{\frac {\partial }{\partial t}}\chi ({\bar {q}},t)\right)-{\frac {\partial }{\partial t}}\left({\bar {A}}({\bar {q}},t)+\nabla \chi ({\bar {q}},t)\right)={\bar {E}}({\bar {q}},t)\\&{\bar {B}}{\acute {\ }}({\bar {q}},t)=\nabla \times {\bar {A}}{\acute {\ }}({\bar {q}},t)=\nabla \times \left({\bar {A}}({\bar {q}},t)+\nabla \chi ({\bar {q}},t)\right)={\bar {B}}({\bar {q}},t)\\\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/3372fd10be324adc81c5fc43af5121ae809333b0)
Betrachten wir die Lagrangefunktion, so ergibt sich:
![{\displaystyle {\begin{aligned}&L{\acute {\ }}(q,{\dot {q}},t)={\frac {m}{2}}{{\dot {\bar {q}}}^{2}}+e\left({\dot {\bar {q}}}{\bar {A}}{\acute {\ }}({\bar {q}},t)-\Phi {\acute {\ }}({\bar {q}},t)\right)\\&L{\acute {\ }}(q,{\dot {q}},t)={\frac {m}{2}}{{\dot {\bar {q}}}^{2}}+e\left({\dot {\bar {q}}}{\bar {A}}({\bar {q}},t)+{\dot {\bar {q}}}\cdot \nabla \chi -\Phi ({\bar {q}},t)+{\dot {\chi }}\right)\\&L{\acute {\ }}(q,{\dot {q}},t)=L+e\left({\dot {\chi }}+{\dot {\bar {q}}}\cdot \nabla \chi \right){\acute {\ }}=L+{\frac {d}{dt}}\left(e\chi ({\bar {q}},t)\right)\\\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/61459314896eee2e15118c4176a01c6fa624b19e)
Einsetzen zeigt: L´ führt zu denselben Lagrangegleichungen wie L.
Die Eichtransformation
![{\displaystyle L(q,{\dot {q}},t)\to L{\acute {\ }}(q,{\dot {q}},t)=L+{\frac {d}{dt}}\left(M({\bar {q}},t)\right)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/cd64f42df2e5b0c2d6babc0630ec5ac54bae1eaf)
mit einer beliebigen Eichfunktion M (skalar) läßt die Lagrangegleichungen invariant.
|
{{#set:Definition=Eichtransformation|Index=Eichtransformation}}
Allgemein gilt:
Sei
beliebig und
![{\displaystyle {\begin{aligned}&L{\acute {\ }}(q,{\dot {q}},t)=L+\left({\dot {M}}+{\dot {\bar {q}}}\cdot \nabla M\right)=L+{\frac {d}{dt}}\left(M({\bar {q}},t)\right)\\&L{\acute {\ }}(q,{\dot {q}},t)=L+\sum \limits _{k=1}^{f}{{\frac {\partial M}{\partial {{q}_{k}}}}{{\dot {q}}_{k}}+{\frac {\partial M}{\partial t}}}\\\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/16e16fa4eab5320f09b3146adbe160ff243ef3dc)
dann erfüllen die
![{\displaystyle \left\{{{q}_{k}}(t)\right\}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/98adc7bea6f9a0a405f1c600e5fa516c429abda4)
das hamiltonsche Prinzip
Also:
![{\displaystyle \delta \int {L{\acute {\ }}dt=0\Leftrightarrow \delta \int {Ldt=0}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/14fc130689bbd2dff9811e3e1a7fae1bd1ab4c17)
Das bedeutet, die Euler- Lagrangegleichungen sind invariant unter Transformationen der Art
mit ![{\displaystyle M({\bar {q}},t)=M({{q}_{1}},...,{{q}_{f}},t)\in {{C}^{3}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/16a3233f563ca70573c712e20bce2ff52723c05a)
beliebig.
Beweis:
mit ![{\displaystyle {\frac {\partial }{\partial {{\dot {q}}_{k}}}}\left(\sum \limits _{l=1}^{f}{{\frac {\partial M}{\partial {{q}_{l}}}}{{\dot {q}}_{l}}+{\frac {\partial M}{\partial t}}}\right)={\frac {\partial M}{\partial {{q}_{k}}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1f956016cedfaae586cba68c255559750c0cec2e)
Einzige Nebenbedingung:
![{\displaystyle M({\bar {q}},t)=M({{q}_{1}},...,{{q}_{f}},t)\in {{C}^{3}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/16a3233f563ca70573c712e20bce2ff52723c05a)
darf nicht explizit von
![{\displaystyle {{\dot {q}}_{k}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1cdb234254124fc59d5c6b02d956fb40323a7c88)
abhängen.