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:<math>L({{q}_{1}},...,{{\dot{q}}_{1}},...,t)</math>
<math>L({{q}_{1}},...,{{\dot{q}}_{1}},...,t)</math>




'''Theorem (E.Noether, 1882-1935)'''
'''Theorem ( E.Noether, 1882-1935)'''


Die Lagrangefunktion
Die Lagrangefunktion
:<math>L({{q}_{1}},...,{{\dot{q}}_{1}},...,t)</math>
<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>
<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>
<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>
<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>
<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>
<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>
<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}
<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)} \\
\end{align}</math> Mit <math>\begin{align}
\end{align}</math>
 
 
Mit
 
 
<math>\begin{align}
   & \frac{d}{dt}\frac{\partial L}{\partial {{{\dot{q}}}_{i}}}=\frac{\partial L}{\partial {{q}_{i}}} \\
   & \frac{d}{dt}\frac{\partial L}{\partial {{{\dot{q}}}_{i}}}=\frac{\partial L}{\partial {{q}_{i}}} \\
  & \frac{d}{dt}\left( \frac{d{{q}_{i}}}{ds} \right)=\left( \frac{d{{{\dot{q}}}_{i}}}{ds} \right) \\
  & \frac{d}{dt}\left( \frac{d{{q}_{i}}}{ds} \right)=\left( \frac{d{{{\dot{q}}}_{i}}}{ds} \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>
<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>
<math>\frac{d}{dt}I(\bar{q},\dot{\bar{q}})=\frac{d}{ds}L=0</math>
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