Eichtransformation der Lagrangefunktion

Mechanikvorlesung von Prof. Dr. E. Schöll, PhD

Der Artikel Eichtransformation der Lagrangefunktion basiert auf der Vorlesungsmitschrift von Franz- Josef Schmitt des 2.Kapitels (Abschnitt 3) der Mechanikvorlesung von Prof. Dr. E. Schöll, PhD.

Eichtransformation der Lagrangefunktion	Das Hamiltonsche Prinzip, Das d'Alembertsche Prinzip
	Variationsprinzipien Das hamiltonsche Wirkungsprinzip Eichtransformation der Lagrangefunktion Forminvarianz der Lagrangegleichungen	Das d'Alembertsche Prinzip Das Hamiltonsche Prinzip Symmetrien und Erhaltungsgrößen Der Hamiltonsche kanonische Formalismus Die Hamilton-Jacobi-Theorie Mechanik des starren Körpers Dynamische Systeme und deterministisches Chaos

Uneindeutigkeit der Lagrangefunktion

Die Lagarangefunktion wird duch die Lagrangegleichung nicht eindeutig festgelegt.

Betrachten wir beispielsweise ein geladenes Teilchen im elektrischen Feld:

$\left({{q}_{1}},{{q}_{2}},{{q}_{3}}\right)=\left({{x}_{1}},{{x}_{2}},{{x}_{3}}\right)$

e sei die Ladung

Bewegungsgleichung:

$m{\frac {{{d}^{2}}^{}}{d{{t}^{2}}}}\left({{q}_{1}},{{q}_{2}},{{q}_{3}}\right)=m{\ddot {\bar {q}}}=e{\bar {E}}({\bar {q}},t)+e{\dot {\bar {q}}}\times {\bar {B}}({\bar {q}},t)$

Die Lorentzkraft ist typischerweise nicht konservativ

Die Darstellung des elektrischen und magnetischen Feldes erfolgt über die Potenziale:

${\begin{aligned}&{\bar {E}}({\bar {q}},t)=-\nabla \Phi ({\bar {q}},t)-{\frac {\partial }{\partial t}}{\bar {A}}({\bar {q}},t)\\&{\bar {B}}({\bar {q}},t)=\nabla \times {\bar {A}}({\bar {q}},t)\\\end{aligned}}$

Dabei ist $\Phi$ Skalar und A ein Vektorpotenzial (MKSA- System)

Ziel: Suche eine Lagrangefunktion $L(q,{\dot {q}},t)=T-V$ in der Art, dass ${\frac {\partial L}{\partial {{q}_{k}}}}-{\frac {d}{dt}}{\frac {\partial L}{\partial {{\dot {q}}_{k}}}}=0$

Die Bewegungsgleichung $m{\frac {{{d}^{2}}^{}}{d{{t}^{2}}}}\left({{q}_{1}},{{q}_{2}},{{q}_{3}}\right)=m{\ddot {\bar {q}}}=e{\bar {E}}({\bar {q}},t)+e{\dot {\bar {q}}}\times {\bar {B}}({\bar {q}},t)$ ergeben.

Ansatz:

$L(q,{\dot {q}},t)={\frac {m}{2}}{{\dot {\bar {q}}}^{2}}+e\left({\dot {\bar {q}}}{\bar {A}}({\bar {q}},t)-\Phi ({\bar {q}},t)\right)$

Probe:

${\begin{aligned}&{\frac {\partial L}{\partial {{\dot {q}}_{k}}}}=m{{\dot {q}}_{k}}+e{{A}_{k}}\\&{\frac {d}{dt}}{\frac {\partial L}{\partial {{\dot {q}}_{k}}}}=m{{\ddot {q}}_{k}}+e{\frac {d}{dt}}{{A}_{k}}({\bar {q}}(t),t)\\&{\frac {d}{dt}}{\frac {\partial L}{\partial {{\dot {q}}_{k}}}}=m{{\ddot {q}}_{k}}+e\left({\frac {\partial }{\partial t}}{{A}_{k}}+\sum \limits _{l}{{\frac {\partial {{A}_{k}}}{\partial {{q}_{l}}}}{{\dot {q}}_{l}}}\right)\\&{\frac {d}{dt}}{\frac {\partial L}{\partial {{\dot {q}}_{k}}}}=m{{\ddot {q}}_{k}}+e\left({\frac {\partial }{\partial t}}{{A}_{k}}+\left({\dot {\bar {q}}}\cdot \nabla \right){{A}_{k}}\right)\\\end{aligned}}$

Weiter:

${\frac {\partial L}{\partial {{q}_{k}}}}=e\left[{\frac {\partial }{\partial {{q}_{k}}}}\left({\dot {\bar {q}}}\cdot {\bar {A}}\right)-{\frac {\partial }{\partial {{q}_{k}}}}\Phi \right]$

Somit:

${\begin{aligned}&0={\frac {\partial L}{\partial {{q}_{k}}}}-{\frac {d}{dt}}{\frac {\partial L}{\partial {{\dot {q}}_{k}}}}=m{{\ddot {q}}_{k}}+e\left({\frac {\partial }{\partial t}}{{A}_{k}}+\left({\dot {\bar {q}}}\cdot \nabla \right){{A}_{k}}\right)-e\left[{\frac {\partial }{\partial {{q}_{k}}}}\left({\dot {\bar {q}}}\cdot {\bar {A}}\right)-{\frac {\partial }{\partial {{q}_{k}}}}\Phi \right]\\&=m{{\ddot {q}}_{k}}+e\left({\frac {\partial }{\partial t}}{{A}_{k}}+{\frac {\partial }{\partial {{q}_{k}}}}\Phi \right)+e\left[-{\frac {\partial }{\partial {{q}_{k}}}}\left({\dot {\bar {q}}}\cdot {\bar {A}}\right)+\left({\dot {\bar {q}}}\cdot \nabla \right){{A}_{k}}\right]\\&=m{{\ddot {q}}_{k}}-e{{E}_{k}}-{{\left[e{\dot {\bar {q}}}\times \left(\nabla \times {\bar {A}}\right)\right]}_{k}}\\&=m{{\ddot {q}}_{k}}-e{{E}_{k}}-{{\left[e{\dot {\bar {q}}}\times {\bar {B}}\right]}_{k}}\\\end{aligned}}$

Somit erfüllt unser Ansatz die Bewegungsgleichungen

Eichtransformationen

Die Potenziale lassen sich umeichen mit Hilfe der Eichfunktion $\chi$

${\begin{aligned}&{\bar {A}}({\bar {q}},t)\to {\bar {A}}{\acute {\ }}({\bar {q}},t)={\bar {A}}({\bar {q}},t)+\nabla \chi ({\bar {q}},t)\\&\Phi ({\bar {q}},t)\to \Phi {\acute {\ }}({\bar {q}},t)=\Phi ({\bar {q}},t)-{\frac {\partial }{\partial t}}\chi ({\bar {q}},t)\\\end{aligned}}$

Durch Eisnetzen sieht man schnell, dass sich die Felder nicht ändern:

${\begin{aligned}&{\bar {E}}{\acute {\ }}({\bar {q}},t)=-\nabla \Phi {\acute {\ }}({\bar {q}},t)-{\frac {\partial }{\partial t}}{\bar {A}}{\acute {\ }}({\bar {q}},t)=-\nabla \left(\Phi ({\bar {q}},t)-{\frac {\partial }{\partial t}}\chi ({\bar {q}},t)\right)-{\frac {\partial }{\partial t}}\left({\bar {A}}({\bar {q}},t)+\nabla \chi ({\bar {q}},t)\right)={\bar {E}}({\bar {q}},t)\\&{\bar {B}}{\acute {\ }}({\bar {q}},t)=\nabla \times {\bar {A}}{\acute {\ }}({\bar {q}},t)=\nabla \times \left({\bar {A}}({\bar {q}},t)+\nabla \chi ({\bar {q}},t)\right)={\bar {B}}({\bar {q}},t)\\\end{aligned}}$

Betrachten wir die Lagrangefunktion, so ergibt sich:

${\begin{aligned}&L{\acute {\ }}(q,{\dot {q}},t)={\frac {m}{2}}{{\dot {\bar {q}}}^{2}}+e\left({\dot {\bar {q}}}{\bar {A}}{\acute {\ }}({\bar {q}},t)-\Phi {\acute {\ }}({\bar {q}},t)\right)\\&L{\acute {\ }}(q,{\dot {q}},t)={\frac {m}{2}}{{\dot {\bar {q}}}^{2}}+e\left({\dot {\bar {q}}}{\bar {A}}({\bar {q}},t)+{\dot {\bar {q}}}\cdot \nabla \chi -\Phi ({\bar {q}},t)+{\dot {\chi }}\right)\\&L{\acute {\ }}(q,{\dot {q}},t)=L+e\left({\dot {\chi }}+{\dot {\bar {q}}}\cdot \nabla \chi \right){\acute {\ }}=L+{\frac {d}{dt}}\left(e\chi ({\bar {q}},t)\right)\\\end{aligned}}$

Einsetzen zeigt: L´ führt zu denselben Lagrangegleichungen wie L.

Die Eichtransformation

$L(q,{\dot {q}},t)\to L{\acute {\ }}(q,{\dot {q}},t)=L+{\frac {d}{dt}}\left(M({\bar {q}},t)\right)$

Mit einer beliebigen Eichfunktion M ( skalar) läßt die Lagrangegleichungen invariant.

Allgemein gilt:

Sei $M({\bar {q}},t)=M({{q}_{1}},...,{{q}_{f}},t)\in {{C}^{3}}$ beliebig

und ${\begin{aligned}&L{\acute {\ }}(q,{\dot {q}},t)=L+\left({\dot {M}}+{\dot {\bar {q}}}\cdot \nabla M\right)=L+{\frac {d}{dt}}\left(M({\bar {q}},t)\right)\\&L{\acute {\ }}(q,{\dot {q}},t)=L+\sum \limits _{k=1}^{f}{{\frac {\partial M}{\partial {{q}_{k}}}}{{\dot {q}}_{k}}+{\frac {\partial M}{\partial t}}}\\\end{aligned}}$

dann erfüllen die

$\left\{{{q}_{k}}(t)\right\}$

das hamiltonsche Prinzip

Also:

$\delta \int {L{\acute {\ }}dt=0\Leftrightarrow \delta \int {Ldt=0}}$

Das bedeutet, die Euler- Lagrangegleichungen sind invariant unter Transformationen der Art

$L(q,{\dot {q}},t)\to L{\acute {\ }}(q,{\dot {q}},t)=L+{\frac {d}{dt}}\left(M({\bar {q}},t)\right)$

mit $M({\bar {q}},t)=M({{q}_{1}},...,{{q}_{f}},t)\in {{C}^{3}}$ beliebig.

Beweis:

${\begin{aligned}&{\frac {\partial L{\acute {\ }}}{\partial {{q}_{k}}}}-{\frac {d}{dt}}{\frac {\partial L{\acute {\ }}}{\partial {{\dot {q}}_{k}}}}={\frac {\partial L}{\partial {{q}_{k}}}}+{\frac {\partial }{\partial {{q}_{k}}}}\left(\sum \limits _{l=1}^{f}{{\frac {\partial M}{\partial {{q}_{l}}}}{{\dot {q}}_{l}}+{\frac {\partial M}{\partial t}}}\right)-{\frac {d}{dt}}{\frac {\partial L}{\partial {{\dot {q}}_{k}}}}-{\frac {d}{dt}}{\frac {\partial }{\partial {{\dot {q}}_{k}}}}\left(\sum \limits _{l=1}^{f}{{\frac {\partial M}{\partial {{q}_{l}}}}{{\dot {q}}_{l}}+{\frac {\partial M}{\partial t}}}\right)\\&={\frac {\partial L}{\partial {{q}_{k}}}}-{\frac {d}{dt}}{\frac {\partial L}{\partial {{\dot {q}}_{k}}}}+{\frac {\partial }{\partial {{q}_{k}}}}{\frac {dM}{dt}}-{\frac {d}{dt}}{\frac {\partial M}{\partial {{q}_{k}}}}={\frac {\partial L}{\partial {{q}_{k}}}}-{\frac {d}{dt}}{\frac {\partial L}{\partial {{\dot {q}}_{k}}}}\\\end{aligned}}$

mit

${\frac {\partial }{\partial {{\dot {q}}_{k}}}}\left(\sum \limits _{l=1}^{f}{{\frac {\partial M}{\partial {{q}_{l}}}}{{\dot {q}}_{l}}+{\frac {\partial M}{\partial t}}}\right)={\frac {\partial M}{\partial {{q}_{k}}}}$

Einzige Nebenbedingung:

$M({\bar {q}},t)=M({{q}_{1}},...,{{q}_{f}},t)\in {{C}^{3}}$ darf nicht explizit von ${{\dot {q}}_{k}}$ abhängen.

Beispiel: eindimensionaler Oszi $L=T-V={\frac {m}{2}}{{\dot {q}}^{2}}-{\frac {m{{\omega }^{2}}}{2}}{{q}^{2}}$ Beispielhafte Eichfunktion: $M(q):={\frac {m{{\omega }^{2}}}{2}}{{q}^{2}}\Rightarrow {\frac {dM}{dt}}=m{{\omega }^{2}}q{\dot {q}}$ $L{\acute {\ }}={\frac {m}{2}}{{\dot {q}}^{2}}-{\frac {m{{\omega }^{2}}}{2}}\left({{q}^{2}}-2q{\dot {q}}\right)$ Die Lagrangegleichungen lauten: ${\begin{aligned}&{\frac {d}{dt}}{\frac {\partial L{\acute {\ }}}{\partial {{\dot {q}}_{}}}}=m{\ddot {q}}+m{{\omega }^{2}}{\dot {q}}\\&{\frac {\partial L{\acute {\ }}}{\partial {{q}_{k}}}}=-m{{\omega }^{2}}q+m{{\omega }^{2}}{\dot {q}}\\\end{aligned}}$ Es folgt als Bewegungsgleichung ${\ddot {q}}+{{\omega }^{2}}q=0$

Forminvarianz der Lagrangegleichung

Eine schwächere Form der Invarianz ( als die Eichinvarianz) ist die Forminvarianz.

Dabei gilt als Forminvarianz:

${\frac {\partial L}{\partial {{q}_{k}}}}-{\frac {d}{dt}}{\frac {\partial L}{\partial {{\dot {q}}_{k}}}}=0\Rightarrow {\frac {\partial L}{\partial {{Q}_{k}}}}-{\frac {d}{dt}}{\frac {\partial L}{\partial {{\dot {Q}}_{k}}}}=0$

Für welche Trnsformationen der generalisierten Koordinaten

$F:\left\{{{q}_{k}}\right\}\to \left\{{{Q}_{k}}\right\}$

sind nun die Lagrangegleichungen forminvariant ?

Satz:

Sei $F:\left\{{{q}_{k}}\right\}\to \left\{{{Q}_{k}}\right\}$ ein C²- Diffeomorphismus,

also eine umkehrbare und eindeutige Abbildung und sind

$F,{{F}^{-1}}$ beide zweimal stetig differenzierbar, dann ist

$\left\{{{Q}_{k}}(t)\right\}$ Lösung der Lagrangegleichung zur transformierten Lagrangefunktion:

${\tilde {L}}({{Q}_{k}},{{\dot {Q}}_{k}},t):=L({{f}_{k}}({{Q}_{i}},t),\sum \limits _{i}{}{\frac {\partial {{f}_{k}}}{\partial {{Q}_{i}}}}{{\dot {Q}}_{i}}+{\frac {\partial {{f}_{k}}}{\partial t}},t)$

mit

${\begin{aligned}&{{f}_{k}}({{Q}_{i}},t)={{q}_{k}}\\&\sum \limits _{i}{}{\frac {\partial {{f}_{k}}}{\partial {{Q}_{i}}}}{{\dot {Q}}_{i}}+{\frac {\partial {{f}_{k}}}{\partial t}}={{\dot {q}}_{k}}\\\end{aligned}}$

Diese Aussage ist äquivalent zur Aussage:

$\left\{{{q}_{k}}(t)\right\}$ sind Lösung der Lagrangegleichungen zu $L({{q}_{k}},{{\dot {q}}_{k}},t)$

Beweis:

${\frac {d}{dt}}{\frac {\partial {\tilde {L}}}{\partial {{\dot {Q}}_{k}}}}=\sum \limits _{l=1}^{f}{{\frac {d}{dt}}{\frac {\partial L}{\partial {{\dot {q}}_{l}}}}{\frac {\partial {{\dot {q}}_{l}}}{\partial {{\dot {Q}}_{k}}}}=}\sum \limits _{l=1}^{f}{{\frac {d}{dt}}\left({\frac {\partial L}{\partial {{\dot {q}}_{l}}}}{\frac {\partial {{q}_{l}}}{\partial {{Q}_{k}}}}\right)}$ wegen

${\begin{aligned}&{{f}_{k}}({{Q}_{i}},t)={{q}_{k}}\\&\sum \limits _{i}{}{\frac {\partial {{f}_{k}}}{\partial {{Q}_{i}}}}{{\dot {Q}}_{i}}+{\frac {\partial {{f}_{k}}}{\partial t}}={{\dot {q}}_{k}}\\\end{aligned}}$

Nun:

${\begin{aligned}&{\frac {d}{dt}}{\frac {\partial {\tilde {L}}}{\partial {{\dot {Q}}_{k}}}}=\sum \limits _{l=1}^{f}{\left\{\left[{\frac {d}{dt}}\left({\frac {\partial L}{\partial {{\dot {q}}_{l}}}}\right)\right]{\frac {\partial {{q}_{l}}}{\partial {{Q}_{k}}}}+{\frac {\partial L}{\partial {{\dot {q}}_{l}}}}{\frac {d}{dt}}\left({\frac {\partial {{q}_{l}}}{\partial {{Q}_{k}}}}\right)\right\}}\\&=\sum \limits _{l=1}^{f}{\left\{\left[{\frac {d}{dt}}\left({\frac {\partial L}{\partial {{\dot {q}}_{l}}}}\right)\right]{\frac {\partial {{q}_{l}}}{\partial {{Q}_{k}}}}+{\frac {\partial L}{\partial {{\dot {q}}_{l}}}}\left({\frac {\partial {{\dot {q}}_{l}}}{\partial {{Q}_{k}}}}\right)\right\}}\\\end{aligned}}$

und auf der anderen Seite:

${\frac {\partial {\tilde {L}}}{\partial {{Q}_{k}}}}=\sum \limits _{l=1}^{f}{\left({\frac {\partial L}{\partial {{q}_{l}}}}{\frac {\partial {{q}_{l}}}{\partial {{Q}_{k}}}}+{\frac {\partial L}{\partial {{\dot {q}}_{l}}}}\left({\frac {\partial {{\dot {q}}_{l}}}{\partial {{Q}_{k}}}}\right)\right)}$

Somit:

${\begin{aligned}&{\frac {d}{dt}}{\frac {\partial {\tilde {L}}}{\partial {{\dot {Q}}_{k}}}}-{\frac {\partial {\tilde {L}}}{\partial {{Q}_{k}}}}=\sum \limits _{l=1}^{f}{\left\{\left[{\frac {d}{dt}}\left({\frac {\partial L}{\partial {{\dot {q}}_{l}}}}\right)\right]{\frac {\partial {{q}_{l}}}{\partial {{Q}_{k}}}}+{\frac {\partial L}{\partial {{\dot {q}}_{l}}}}\left({\frac {\partial {{\dot {q}}_{l}}}{\partial {{Q}_{k}}}}\right)-\left({\frac {\partial L}{\partial {{q}_{l}}}}{\frac {\partial {{q}_{l}}}{\partial {{Q}_{k}}}}+{\frac {\partial L}{\partial {{\dot {q}}_{l}}}}\left({\frac {\partial {{\dot {q}}_{l}}}{\partial {{Q}_{k}}}}\right)\right)\right\}}\\&=\sum \limits _{l=1}^{f}{\left\{\left[{\frac {d}{dt}}\left({\frac {\partial L}{\partial {{\dot {q}}_{l}}}}\right)\right]{\frac {\partial {{q}_{l}}}{\partial {{Q}_{k}}}}-\left({\frac {\partial L}{\partial {{q}_{l}}}}{\frac {\partial {{q}_{l}}}{\partial {{Q}_{k}}}}\right)\right\}}=\sum \limits _{l=1}^{f}{{\frac {\partial {{q}_{l}}}{\partial {{Q}_{k}}}}\left\{\left[{\frac {d}{dt}}\left({\frac {\partial L}{\partial {{\dot {q}}_{l}}}}\right)\right]-\left({\frac {\partial L}{\partial {{q}_{l}}}}\right)\right\}}\\\end{aligned}}$

Dabei bildet

${\frac {\partial {{q}_{l}}}{\partial {{Q}_{k}}}}$ die Transformationsmatrix, die nichtsingulär sein muss, also $\det {\frac {\partial {{q}_{l}}}{\partial {{Q}_{k}}}}\neq 0$

Daher die Bedingung, dass

Sei $F:\left\{{{q}_{k}}\right\}\to \left\{{{Q}_{k}}\right\}$ ein C²- Diffeomorphismus,

also eine umkehrbare und eindeutige Abbildung und

$F,{{F}^{-1}}$ beide zweimal stetig differenzierbar.

Nur dann ist $\left\{{{Q}_{k}}(t)\right\}$ Lösung der Lagrangegleichung zur transformierten Lagrangefunktion.

Denn diese Aussage ist äquivalent zu

${\begin{aligned}&{{Q}_{i}}={{F}_{i}}({{q}_{1}},...{{q}_{f}},t)\\&{{q}_{k}}={{f}_{k}}({{Q}_{1}},...,{{Q}_{f}},t)\quad mit\quad \det {\frac {\partial {{f}_{k}}}{\partial {{Q}_{i}}}}\neq 0\\\end{aligned}}$

Man sagt, die Variationsableitung

${\frac {d}{dt}}{\frac {\partial {\tilde {L}}}{\partial {{\dot {Q}}_{k}}}}-{\frac {\partial {\tilde {L}}}{\partial {{Q}_{k}}}}$ ist kovariant unter diffeomorphen Transformationen der generalisierten Koordinaten

Also gibt es auch unendlich viele äquivalente Sätze generalisierter Koordinaten.