Transformations in Noether's theorem and point transformations In lagrangian mechanics, I've come across a proof that point transformations don't change the Euler-Lagrange equations of motion. Suppose our new coordinates are simply $\tilde{q_i}=\tilde{q_i}(q_1,q_2,\ldots,t)$ and our new velocities are of a similar form.
Then we can show that $$\tilde{L}(\tilde{q},\dot{\tilde{q}},t)=L(q,\dot{q},t)$$
Moreover, the form of the Euler Lagrange equations remain the same. For example, transformations from Cartesian to polar coordinates can be an example of such point transformations.
My question is then, what exactly are the symmetry transformations in Noether's theorem ? According to what I've studied about Noether's theorem, we need transformations such that $\delta L=0$, i.e. the lagrangian must be the same after the transformation. Well, that is exactly what is happing in case of the point transformations, right ?
However, there can be an infinite number of point transformations, for example, $q\rightarrow q+n$, where $n$ can be any number. This is just changing the coordinate system by shifting the origin. Hence, there are an infinite possible types of point transformations. If all of these are going to keep the lagrangian invariant, shouldn't there be infinite number of conserved quantities according to Noether's theorem. Clearly I'm missing something here, as this is obviously wrong.
I'm getting confused between point transformations, and the symmetry transformations from Noether's theorem. Can someone help me clear this mess ? For some reason, changing from cartesian to polar ( point transformation ), gives us no conserved quantity, as it should, since the lagrangian remains the same.
 A: First of all, given the transformation
$$
q = f(\tilde{q},t),
$$
the new Lagrangian is, by definition,
$$
\tilde{L}(\tilde{q},\dot{\tilde{q}},t) = L(f(\tilde{q},t),\dot{f}(\tilde{q},\dot{\tilde{q}},t),t)\tag{1}
$$
where
$$
\dot{f}(\tilde{q},\dot{\tilde{q}},t) = \frac{\partial f}{\partial\tilde{q}}\cdot\dot{\tilde{q}}+\frac{\partial f}{\partial t}
$$
i.e. $\tilde{L} = L\circ f$.
The Lagrange equations of $\tilde{L}$ are not the same of Lagrange equations of $L$, but if we set
$$
\tau = \frac{d}{dt}\frac{\partial L}{\partial\dot{q}}-\frac{\partial L}{\partial q} \\
\tilde{\tau} = \frac{d}{dt}\frac{\partial \tilde{L}}{\partial\dot{\tilde{q}}}-\frac{\partial \tilde{L}}{\partial \tilde{q}}
$$
we have
$$
\tau \circ f = \tilde\tau \cdot \frac{\partial f}{\partial \tilde{q}}
$$
so
$$
\tau = 0 \iff \tilde{\tau} = 0
$$
and the solutions are the same, modulo the change of variables.
In the theorem of Noether, it is supposed that there exist a group of transformations
$$
q = f_{\lambda}(\tilde{q},t), \quad \lambda\in\mathbb{R}
$$
such that $L$ is invariant with respect to each transformation of the group (i.e. for each $\lambda$). Invariant means
$$
\tilde{L}(\tilde{q},\dot{\tilde{q}},t) = L(\tilde{q},\dot{\tilde{q}},t)
$$
i.e. the new Lagrangian has the same functional form of the old Lagrangian, and this is not the same as $(1)$.
A: *

*On one hand, Noether's first theorem typically considers an $n$-parameter active transformation that is a quasisymmetry of the action.


*On the other hand, the Lagrangian is invariant and the Euler-Lagrange (EL) equations are covariant under a passive point transformation, cf. e.g. this Phys.SE post. In particular, Noether's theorem does not apply to the latter scenario.
