Forminvarianz der Lagrangegleichungen - Revision history

*>SchuBot: Interpunktion, replaced: ( → (

2010-09-12T22:28:19Z

Interpunktion, replaced: ( → (

← Older revision		Revision as of 00:28, 13 September 2010
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	<noinclude>{{Scripthinweis\|Mechanik\|2\|4}}</noinclude>		<noinclude>{{Scripthinweis\|Mechanik\|2\|4}}</noinclude>
	Eine schwächere Form der Invarianz ( als die Eichinvarianz) ist die Forminvarianz.		Eine schwächere Form der Invarianz (als die Eichinvarianz) ist die Forminvarianz.

	Dabei gilt als Forminvarianz:		Dabei gilt als Forminvarianz:

*>SchuBot: Einrückungen Mathematik

2010-09-12T15:26:38Z

Einrückungen Mathematik

← Older revision		Revision as of 17:26, 12 September 2010
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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>


Line 19:		Line 19:

	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,

Line 25:		Line 25:


	<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>


Line 91:		Line 91:

	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,

Line 97:		Line 97:


	<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.

Line 107:		Line 107:


	<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.

*>SchuBot: Einrückungen Mathematik

2010-09-12T15:06:01Z

Einrückungen Mathematik

← Older revision		Revision as of 17:06, 12 September 2010
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	<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>		<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}


	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>\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>		<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}
	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}} \\

Schubotz: Die Seite wurde neu angelegt: „{{Scripthinweis|Mechanik|2|4}} Eine schwächere Form der Invarianz ( als die Eichinvarianz) ist die Forminvarianz. Dabei gilt als Forminva…“

2010-08-28T17:18:35Z

Die Seite wurde neu angelegt: „<noinclude>{{Scripthinweis|Mechanik|2|4}}</noinclude> Eine schwächere Form der Invarianz ( als die Eichinvarianz) ist die Forminvarianz. Dabei gilt als Forminva…“

New page

<noinclude>{{Scripthinweis|Mechanik|2|4}}</noinclude>
Eine schwächere Form der Invarianz ( als die Eichinvarianz) ist die Forminvarianz.

Dabei gilt als Forminvarianz:

<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>

Für welche Trnsformationen der generalisierten Koordinaten

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

sind nun die Lagrangegleichungen forminvariant ?

Satz:

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

also eine umkehrbare und eindeutige Abbildung und sind

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

<math>\left\{ {{Q}_{k}}(t) \right\}</math>
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}
& {{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{align}</math>

Diese Aussage ist äquivalent zur Aussage:

<math>\left\{ {{q}_{k}}(t) \right\}</math>
sind Lösung der Lagrangegleichungen zu
<math>L({{q}_{k}},{{\dot{q}}_{k}},t)</math>

'''Beweis:'''

<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}} \\
& \sum\limits_{i}{{}}\frac{\partial {{f}_{k}}}{\partial {{Q}_{i}}}{{{\dot{Q}}}_{i}}+\frac{\partial {{f}_{k}}}{\partial t}={{{\dot{q}}}_{k}} \\
\end{align}</math>

Nun:

<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\}} \\
& =\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{align}</math>

und auf der anderen Seite:

<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>

Somit:

<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\}} \\
& =\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{align}</math>

Dabei bildet

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

Daher die Bedingung, dass

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

also eine umkehrbare und eindeutige Abbildung und

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

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

Denn diese Aussage ist äquivalent zu

<math>\begin{align}
& {{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 \\
\end{align}</math>

Man sagt, die Variationsableitung

<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

Also gibt es auch unendlich viele äquivalente Sätze generalisierter Koordinaten.