Editing Forminvarianz der Lagrangegleichungen

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:<math>\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</math>
<math>\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</math>




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:<math>F:\left\{ {{q}_{k}} \right\}\to \left\{ {{Q}_{k}} \right\}</math>
<math>F:\left\{ {{q}_{k}} \right\}\to \left\{ {{Q}_{k}} \right\}</math>




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Sei
Sei
:<math>F:\left\{ {{q}_{k}} \right\}\to \left\{ {{Q}_{k}} \right\}</math>
<math>F:\left\{ {{q}_{k}} \right\}\to \left\{ {{Q}_{k}} \right\}</math>
ein C²- Diffeomorphismus,
ein C²- Diffeomorphismus,


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:<math>F,{{F}^{-1}}</math>
<math>F,{{F}^{-1}}</math>
beide zweimal stetig differenzierbar, dann ist
beide zweimal stetig differenzierbar, dann ist




:<math>\left\{ {{Q}_{k}}(t) \right\}</math>
<math>\left\{ {{Q}_{k}}(t) \right\}</math>
Lösung der Lagrangegleichung zur transformierten Lagrangefunktion:
Lösung der Lagrangegleichung zur transformierten Lagrangefunktion:




:<math>\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)</math> mit <math>\begin{align}
<math>\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)</math> mit <math>\begin{align}
   & {{f}_{k}}({{Q}_{i}},t)={{q}_{k}} \\
   & {{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}} \\
  & \sum\limits_{i}{{}}\frac{\partial {{f}_{k}}}{\partial {{Q}_{i}}}{{{\dot{Q}}}_{i}}+\frac{\partial {{f}_{k}}}{\partial t}={{{\dot{q}}}_{k}} \\
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:<math>\left\{ {{q}_{k}}(t) \right\}</math>
<math>\left\{ {{q}_{k}}(t) \right\}</math>
sind Lösung der Lagrangegleichungen zu
sind Lösung der Lagrangegleichungen zu
:<math>L({{q}_{k}},{{\dot{q}}_{k}},t)</math>
<math>L({{q}_{k}},{{\dot{q}}_{k}},t)</math>




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:<math>\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)}</math> wegen <math>\begin{align}
<math>\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)}</math> wegen <math>\begin{align}
   & {{f}_{k}}({{Q}_{i}},t)={{q}_{k}} \\
   & {{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}} \\
  & \sum\limits_{i}{{}}\frac{\partial {{f}_{k}}}{\partial {{Q}_{i}}}{{{\dot{Q}}}_{i}}+\frac{\partial {{f}_{k}}}{\partial t}={{{\dot{q}}}_{k}} \\
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:<math>\begin{align}
<math>\begin{align}
   & \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\}} \\
   & \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\}} \\
  & =\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\}} \\
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:<math>\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)}</math>
<math>\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)}</math>




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:<math>\begin{align}
<math>\begin{align}
   & \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\}} \\
   & \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\}} \\
  & =\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\}} \\
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:<math>\frac{\partial {{q}_{l}}}{\partial {{Q}_{k}}}</math>
<math>\frac{\partial {{q}_{l}}}{\partial {{Q}_{k}}}</math>
die Transformationsmatrix, die nichtsingulär sein muss, also
die Transformationsmatrix, die nichtsingulär sein muss, also
:<math>\det \frac{\partial {{q}_{l}}}{\partial {{Q}_{k}}}\ne 0</math>
<math>\det \frac{\partial {{q}_{l}}}{\partial {{Q}_{k}}}\ne 0</math>




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Sei
Sei
:<math>F:\left\{ {{q}_{k}} \right\}\to \left\{ {{Q}_{k}} \right\}</math>
<math>F:\left\{ {{q}_{k}} \right\}\to \left\{ {{Q}_{k}} \right\}</math>
ein C²- Diffeomorphismus,
ein C²- Diffeomorphismus,


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:<math>F,{{F}^{-1}}</math>
<math>F,{{F}^{-1}}</math>
beide zweimal stetig differenzierbar.
beide zweimal stetig differenzierbar.


Nur dann ist
Nur dann ist
:<math>\left\{ {{Q}_{k}}(t) \right\}</math>
<math>\left\{ {{Q}_{k}}(t) \right\}</math>
Lösung der Lagrangegleichung zur transformierten Lagrangefunktion.
Lösung der Lagrangegleichung zur transformierten Lagrangefunktion.


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:<math>\begin{align}
<math>\begin{align}
   & {{Q}_{i}}={{F}_{i}}({{q}_{1}},...{{q}_{f}},t) \\
   & {{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}}}\ne 0 \\
  & {{q}_{k}}={{f}_{k}}({{Q}_{1}},...,{{Q}_{f}},t)\quad mit\quad \det \frac{\partial {{f}_{k}}}{\partial {{Q}_{i}}}\ne 0 \\
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:<math>\frac{d}{dt}\frac{\partial \tilde{L}}{\partial {{{\dot{Q}}}_{k}}}-\frac{\partial \tilde{L}}{\partial {{Q}_{k}}}</math>
<math>\frac{d}{dt}\frac{\partial \tilde{L}}{\partial {{{\dot{Q}}}_{k}}}-\frac{\partial \tilde{L}}{\partial {{Q}_{k}}}</math>
ist kovariant unter diffeomorphen Transformationen der generalisierten Koordinaten
ist kovariant unter diffeomorphen Transformationen der generalisierten Koordinaten


Also gibt es auch unendlich viele äquivalente Sätze generalisierter Koordinaten.
Also gibt es auch unendlich viele äquivalente Sätze generalisierter Koordinaten.
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