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Noether's theorem states that there exists a conservation law for every continuous (in fact, differentiable) symmetry. Reflection is a discrete symmetry, so the theorem is not applicable here. But, in quantum mechanics, you have the parity operator $P$, that reflects the coordinates $$P\psi(\vec{r}) = \psi(-\vec{r})$$ Since $P^2 = I$, the operator $P$ has ...


1

You're correct. To find the equations of motion, we have: \begin{align*}c_i&=\frac{\partial L(v^2)}{\partial v_i}\\ &=L'(v^2) 2 v_i \end{align*} so that $L'(v^2) v_i$ is constant for all of time. Firstly, you could imagine a world in which all paths ${\bf x}(t)$ are valid mechanical paths. Then the Galilean transform of a valid mechanical path ...


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It follows from $L$ being a function $\propto\dot{x}^2$. With this at hand, you are left with two choices: $\left(\nabla_{\dot{x}}L\right)'\sim\left(\dot{x}\right)'=0$ implies $\dot{x}=\rm const.$ $L=0$ implies $\dot{x}=0=\rm const.$ Either way, you get that the velocity is constant in time (for this particular, free-particle case).


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It only is a symmetry when $F$ is independent of $\dot q$ (disregarding more complicated cases like Qmechanic mentions), exactly for the reason that you state: only then the equations of motion remain the same. Your possible solutions in order: Do we have to introduce boundary conditions so that q˙ is not varied at the endpoints as well? This seems ...


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OP is asking about (1) Euler-Lagrange (EL) equations and (2) Noether's (first) theorem. 1) Let us start with EL eqs. OP is pondering what happens to EL eqs. if we change the Lagrangian $L$ with a total time derivative $$\tag{1} \tilde{L}~:=~L+\frac{dF}{dt}$$ in various settings. Often we assume that Lagrangians do not depend on $\ddot{q},\dddot{q}, ...


2

Comments to the question (v2): Let there be given a Lagrangian $$\tag{1} L(q,v,a,t), \qquad v^i~:=~\dot{q}^i,\qquad a^i~:=~\dot{v}^i,\qquad \jmath^i~:=~\dot{a}^i, $$ that depends on up to second time derivative. Let $$\tag{2} \delta q^i~=~\varepsilon Y^i(q,v,a,t) ,$$ be a (global, vertical) quasi-symmetry of the Lagrangian, i.e. there exists a ...



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