Noether's Theorem and conservation of momentum

Question

So as we all know for a system that has translational symmetry Noether's Theorem states that momentum is conserved, more precisely the theorem states that the quantity: $$\frac{\partial L}{\partial \dot{q}}$$ so the generalized momentum is conserved. Here I have a problem: suppose I want to show, that classical momentum $p=mv$ is conserved in a system with traslational symmetry (also of course potential energy in the Lagrangian does not depend on velocity) I then have: $$\frac{\partial L}{\partial \dot{x}}=\frac{\partial K}{\partial \dot{x}}=\frac{\partial}{\partial \dot{x}}\frac{1}{2}m\dot{x}^2=m\dot{x}.$$ Perfect! But suppose that i want to use a parametrization for my system, so: $$x(t)=\Gamma(q(t))$$ as we usually do in Lagrangian Mechanics, then I have that the conserved quantity is still: $$\frac{\partial L}{\partial \dot{q}}.$$ In fact Noether's Theorem states that generalised momentum is conserved and this is by definition the generalized momentum. Well then I have: $$\frac{\partial L}{\partial \dot{q}}=\frac{\partial}{\partial \dot{q}}\frac{1}{2}m\dot{q}^2|\Gamma ' (q)|^2=m\dot{q}|\Gamma ' (q)|^2=mv|\Gamma ' (q)|.$$ What is this? Furthermore, if I choose $\Gamma$ to represent a line with the following parametrization: $$\Gamma = \begin{bmatrix}kq \\ 0 \\ 0\end{bmatrix}.$$ I get: $$\frac{\partial L}{\partial \dot{q}}=mv|k|$$ so the conserved quantity depends on the parametrization?

Now: I know of course that I made a mistake somewhere; maybe on the content of Noether's Theorem (even if i took the content of said theorem straight from my book of Lagrangian Mechanics) or maybe somewhere else. My questions are:

1. Why I get this result?
2. How can I show that momentum $p=mv$ is conserved for a symmetrically translational system using Noether's Theorem and using any parametrization $\Gamma$ I want?
3. Is it true that generalised momentum is conserved for any symmetrically translational system?
4. When conservation of generalised momentum implies conservation of classical momentum?

This is my problem; hope you can help me out. Please try to give me a complete answer, this problem is bugging me a lot.

Answer 1

1. Let us for simplicity consider a 1D system. If the Lagrangian $L(\dot{x},t)$ has a cyclic variable $x$, then the action has an infinitesimal translation symmetry $$\delta x~=~\epsilon,$$ and it is well-known that the conserved Noether charge $$ Q~=~\frac{\partial L}{\partial \dot{x}}\tag{1} $$ is the conjugate momentum.

2. OP considers next a coordinate transformation $$x~=~f(q,t).$$ Note that $q$ is not necessarily a cyclic variable (because $\dot{x}=\frac{\partial f}{\partial q}\dot{q}+\frac{\partial f}{\partial t}$ may depend on $q$). The new symmetry becomes $$ \delta q~=~\epsilon Y,$$ where $$Y~=~\frac{\partial q}{\partial x}~=~\left(\frac{\partial f}{\partial q}\right)^{-1}$$ is the so-called generator. According to Noether's formula, the conserved Noether charge is "momentum times generator": $$ Q~=~\frac{\partial L}{\partial \dot{q}} Y~=~\frac{\partial L}{\partial \dot{x}},\tag{2}$$ which is the same as before because of the chain rule.