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3

"but how can you have a "basis" that changes at every point?" This is really the root of your problem. Mathematically (and physically) speaking, such a basis works fine. You just have a (hopefully temporary) conceptual problem. Maybe try thinking of it this way: How does $\hat{x}$ know to point "to the right" in cartesian coordinates, at an arbitrary ...


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It is a Taylor expansion. You might be a little freaked out because they are treating $L$ as a function of $v^2$, but that doesn't matter, consider $$ L(x + \delta) \sim L(x) + \frac{\partial L}{\partial x} \delta + O(\delta^2) $$ but take $x=v^2$ and $\delta = 2 \boldsymbol{v} \cdot \boldsymbol{\epsilon} + \epsilon^2 $, $$ \begin{align*} L(v'^2) &= ...


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Symmetry of potential alone is not enough to guarantee a symmetric orbit. The initial and boundary conditions must also be symmetric. Also, for non-linear systems, numerical treatment has to be sufficiently accurate.


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In case anyone is interested I'll post here what I learnt. If anyone has any corrections or comments please let me know. The thing that I was missing was that the metric makes the kinetic energy well defined, i.e. independent of coordinates. Working in $\mathbb{R}^2$ with standard coordinates $(x^1,x^2)$ we have the induced coordinates $(x^1,x^2,v^1,v^2)$ ...


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The force that each of the pistons can drive with is given by $$F=\frac14 \pi d^2 P$$ How much lift that produces depends on the exact geometry of the table and in general will be a function of height. In essence if the table moves $x_1$ when the pistons move $x_2$ the force will be $$F_{table}=2F_{piston}\frac{x_2}{x_1}$$


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It seems like a very arbitrary question. It would be much easier if you could just elaborate on one given "specific" and well defined scenario, that you wish to solve, then one can help. From which then you would analogously try to solve other similar scenarios. From what you've asked thus far, your starting point should be Conservation of momentum. Since ...


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The answer is quite simple. You can't see it because you've forgotten that you make an unphysical simplification when considering "tied to an immovable wall" type situations. What's unphysical about the situation is quite simple : there's no such thing as an immovable wall. Let's say the rocket is tied to the earth. To say where the lost chemical potential ...



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