Editing Theorem von Noether
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Latest revision | Your text | ||
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<math>L({{q}_{1}},...,{{\dot{q}}_{1}},...,t)</math> | |||
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Die Lagrangefunktion | Die Lagrangefunktion | ||
<math>L({{q}_{1}},...,{{\dot{q}}_{1}},...,t)</math> | |||
eines autonomen Systems sei unter der Transformation | eines autonomen Systems sei unter der Transformation | ||
<math>\bar{q}\to {{h}^{s}}(\bar{q})</math> | |||
invariant. Dabei ist s ein eindimensionaler Parameter und | invariant. Dabei ist s ein eindimensionaler Parameter und | ||
<math>{{h}^{s=0}}(\bar{q})=\bar{q}</math> | |||
die Identität. | die Identität. | ||
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<math>I(\bar{q},\dot{\bar{q}})=\sum\limits_{i=1}^{f}{\frac{\partial L}{\partial {{{\dot{q}}}_{i}}}{{\left( \frac{d}{ds}{{h}^{s}}({{q}_{i}}) \right)}_{s=0}}}</math> | |||
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Sei | Sei | ||
<math>\bar{q}=\bar{q}(t)</math> | |||
eine Lösung der Lagrangegleichung. Dann ist auch | eine Lösung der Lagrangegleichung. Dann ist auch | ||
<math>\bar{q}(s,t):={{h}^{s}}(\bar{q},t)</math> | |||
Lösung, das heißt: | Lösung, das heißt: | ||
<math>\frac{d}{dt}\frac{\partial L(\bar{q}(s,t),\dot{\bar{q}}(s,t))}{\partial {{{\dot{q}}}_{i}}}=\frac{\partial L(\bar{q}(s,t),\dot{\bar{q}}(s,t))}{\partial {{q}_{i}}}</math> | |||
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<math>\begin{align} | |||
& \frac{d}{ds}L(\bar{q}(s,t),\dot{\bar{q}}(s,t))=\sum\limits_{i=1}^{f}{\left( \frac{\partial L}{\partial {{q}_{i}}}\left( \frac{d{{q}_{i}}}{ds} \right)+\frac{\partial L}{\partial {{{\dot{q}}}_{i}}}{{\left( \frac{d{{{\dot{q}}}_{i}}}{ds} \right)}_{{}}} \right)=}0 \\ | & \frac{d}{ds}L(\bar{q}(s,t),\dot{\bar{q}}(s,t))=\sum\limits_{i=1}^{f}{\left( \frac{\partial L}{\partial {{q}_{i}}}\left( \frac{d{{q}_{i}}}{ds} \right)+\frac{\partial L}{\partial {{{\dot{q}}}_{i}}}{{\left( \frac{d{{{\dot{q}}}_{i}}}{ds} \right)}_{{}}} \right)=}0 \\ | ||
& \Rightarrow \frac{d}{dt}I(\bar{q},\dot{\bar{q}})=\sum\limits_{i=1}^{f}{\frac{d}{dt}\left( \frac{\partial L}{\partial {{{\dot{q}}}_{i}}}{{\left( \frac{d}{ds}{{h}^{s}}({{q}_{i}}) \right)}_{s=0}} \right)=}\sum\limits_{i=1}^{f}{\left( \frac{d}{dt}\frac{\partial L}{\partial {{{\dot{q}}}_{i}}}\left( \frac{d{{q}_{i}}}{ds} \right)+\frac{\partial L}{\partial {{{\dot{q}}}_{i}}}\frac{d}{dt}{{\left( \frac{d{{q}_{i}}}{ds} \right)}_{{}}} \right)} \\ | & \Rightarrow \frac{d}{dt}I(\bar{q},\dot{\bar{q}})=\sum\limits_{i=1}^{f}{\frac{d}{dt}\left( \frac{\partial L}{\partial {{{\dot{q}}}_{i}}}{{\left( \frac{d}{ds}{{h}^{s}}({{q}_{i}}) \right)}_{s=0}} \right)=}\sum\limits_{i=1}^{f}{\left( \frac{d}{dt}\frac{\partial L}{\partial {{{\dot{q}}}_{i}}}\left( \frac{d{{q}_{i}}}{ds} \right)+\frac{\partial L}{\partial {{{\dot{q}}}_{i}}}\frac{d}{dt}{{\left( \frac{d{{q}_{i}}}{ds} \right)}_{{}}} \right)} \\ | ||
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<math>\frac{d}{ds}L(\bar{q}(s,t),\dot{\bar{q}}(s,t))=\sum\limits_{i=1}^{f}{\left( \frac{\partial L}{\partial {{q}_{i}}}\left( \frac{d{{q}_{i}}}{ds} \right)+\frac{\partial L}{\partial {{{\dot{q}}}_{i}}}{{\left( \frac{d{{{\dot{q}}}_{i}}}{ds} \right)}_{{}}} \right)=}0</math> | |||
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<math>\frac{d}{dt}I(\bar{q},\dot{\bar{q}})=\frac{d}{ds}L=0</math> |