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Classical Physics
Mechanics
Conservation laws from Lagrange's equation
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[QUOTE="Phylosopher, post: 6017175, member: 599850"] [U]My question is related to the book: Classical Mechanics by Taylor. Section 7.8[/U] So, In the book Taylor is trying to derive the conservation of momentum and energy from Lagrange's equation. I understood everything, but I am struggling with the concept and the following equation: $$\delta\mathcal{L}= \epsilon \frac{\partial \mathcal{L}}{\partial x_{1}}+\epsilon \frac{\partial \mathcal{L}}{\partial x_{2}}+...=0$$ He derived this by saying that in theory, displacing the position of all N particles from ##\vec{r_{i}} \rightarrow \vec{r_{i}}+\vec{\epsilon}## will make ##\delta \mathcal{L}=0## "Unchanged". Which I understand. Also this means that: $$\mathcal{L}(\vec{r_{1}},...) = \mathcal{L}(\vec{r_{1}}+\vec{\epsilon},...)$$ Then from this he derived the equation that I am confused by, by letting ##\vec{\epsilon}## be in the direction of ##x## only. I am not sure how to derive the first equation. It seems to be just the chain rule! But I didn't succeed with it. In short I have two questions: 1- How to derive the first equation. 2- Why should all particles be moved by ##\vec{\epsilon}##. I know that this is mathematically useful to have ##\delta \mathcal{L}=0##. If each is displaced differently then there is an energy from outside the system. But what if all were displaced by an energy from outside the system exactly by the same amount ##\vec{\epsilon}## that he assumes, how can I be sure that ##\delta \mathcal{L}=0##. [/QUOTE]
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Mechanics
Conservation laws from Lagrange's equation
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