Mechanikvorlesung von Prof. Dr. E. Schöll, PhD
{{#set:Urheber=Prof. Dr. E. Schöll, PhD|Inhaltstyp=Script|Kapitel=2|Abschnitt=4}}
Kategorie:Mechanik
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Eine schwächere Form der Invarianz (als die Eichinvarianz) ist die Forminvarianz.
Dabei gilt als Forminvarianz:
![{\displaystyle {\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}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5fee726ce028d2b675373346a3dbb53437e1be94)
Für welche Trnsformationen der generalisierten Koordinaten
![{\displaystyle F:\left\{{{q}_{k}}\right\}\to \left\{{{Q}_{k}}\right\}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c2e664b51ad998aefd55940cdd0d663768ce132f)
sind nun die Lagrangegleichungen forminvariant ?
Satz:
Sei
![{\displaystyle F:\left\{{{q}_{k}}\right\}\to \left\{{{Q}_{k}}\right\}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c2e664b51ad998aefd55940cdd0d663768ce132f)
ein C²- Diffeomorphismus,
also eine umkehrbare und eindeutige Abbildung und sind
![{\displaystyle F,{{F}^{-1}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7106b20b68a165b58d50dcef08f52073cf9b375a)
beide zweimal stetig differenzierbar, dann ist
![{\displaystyle \left\{{{Q}_{k}}(t)\right\}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/adda495c6ae16072a0e07f464d0e2aabacf658fd)
Lösung der Lagrangegleichung zur transformierten Lagrangefunktion:
mit ![{\displaystyle {\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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/987ff7b2e795d9a0a4377def42e340ae2f81b8fe)
Diese Aussage ist äquivalent zur Aussage:
![{\displaystyle \left\{{{q}_{k}}(t)\right\}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/98adc7bea6f9a0a405f1c600e5fa516c429abda4)
sind Lösung der Lagrangegleichungen zu
![{\displaystyle L({{q}_{k}},{{\dot {q}}_{k}},t)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/22b452d20e4023176d29c642ebb41344811762f8)
Beweis:
wegen ![{\displaystyle {\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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/987ff7b2e795d9a0a4377def42e340ae2f81b8fe)
Nun:
![{\displaystyle {\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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/fa97a323059479f244461974db907e3beed1408b)
und auf der anderen Seite:
![{\displaystyle {\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)}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e71266aae05276dcc80faa1378e86e493c8be3ea)
Somit:
![{\displaystyle {\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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/59f482e920841839b9bdd4a587c0f261dc03c369)
Dabei bildet
![{\displaystyle {\frac {\partial {{q}_{l}}}{\partial {{Q}_{k}}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/961c97f27492d923fc8a08df26108812d016c87b)
die Transformationsmatrix, die nichtsingulär sein muss, also
![{\displaystyle \det {\frac {\partial {{q}_{l}}}{\partial {{Q}_{k}}}}\neq 0}](https://wikimedia.org/api/rest_v1/media/math/render/svg/eb90073026788a3a8d00e29463684395a9f83500)
Daher die Bedingung, dass
Sei
![{\displaystyle F:\left\{{{q}_{k}}\right\}\to \left\{{{Q}_{k}}\right\}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c2e664b51ad998aefd55940cdd0d663768ce132f)
ein C²- Diffeomorphismus,
also eine umkehrbare und eindeutige Abbildung und
![{\displaystyle F,{{F}^{-1}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7106b20b68a165b58d50dcef08f52073cf9b375a)
beide zweimal stetig differenzierbar.
Nur dann ist
![{\displaystyle \left\{{{Q}_{k}}(t)\right\}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/adda495c6ae16072a0e07f464d0e2aabacf658fd)
Lösung der Lagrangegleichung zur transformierten Lagrangefunktion.
Denn diese Aussage ist äquivalent zu
![{\displaystyle {\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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/af56cfdb0048f457fde898a67a67ff52304bc0f3)
Man sagt, die Variationsableitung
![{\displaystyle {\frac {d}{dt}}{\frac {\partial {\tilde {L}}}{\partial {{\dot {Q}}_{k}}}}-{\frac {\partial {\tilde {L}}}{\partial {{Q}_{k}}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d3289140419c84825fdff9b2abeb0c3907566cf3)
ist kovariant unter diffeomorphen Transformationen der generalisierten Koordinaten
Also gibt es auch unendlich viele äquivalente Sätze generalisierter Koordinaten.