### Translational symmetry

Suppose, as in figure 2.4, we translate every particle in a system by the same amount in the same direction and that nothing, including the action, changes under this transformation.

There are actually three symmetries - one for each direction in space - but let's just choose the $x$ direction.

Then, the change in each of the coordinates is just $$\delta x_i = \epsilon$$ (and $\delta y_i = \delta z_i = 0$).

In terms of the Noether charge, we are just setting $$f_i(x) = 1, \quad \forall i$$

Thus, as we have already seen, the conserved quantity is given by $$\sum_{i}p_i \; f_i(x) = \sum_{i} m_i \; \dot{x_i}$$ which is the linear momentum in the $x$ direction.

Notice that we know nothing specific about the forces acting on the system - all we knew was that the action did not change under translation. In fact, we can make an even weaker statement - we can still deduce conservation of linear momentum if all we knew was that the potential associated with the system doesn't change under translation.

### Rotational symmetry

We consider a rotation in the $x, y$ plane, which can be written as $$\begin{bmatrix}x'\\y'\end{bmatrix}=\begin{bmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \quad \cos \theta \end{bmatrix} \begin{bmatrix}x\\y\end{bmatrix}$$

Here, the variation in the system, $\theta = \epsilon$, is small, so we can use the approximations $\cos \epsilon \approx 1, \sin \epsilon \approx \epsilon$ and we get $$\begin{matrix}x \to x - \epsilon y \\ y \to y + \epsilon x \end{matrix}$$ or $$\begin{align*}\delta x &= - \epsilon y \\ \delta y &= +\epsilon x \end{align*}$$

Thus, we can substitute $$\begin{align*}f_x(x, y) &= -y \\ f_y(x, y) &= +x \end{align*}$$ into the expression for the Noether charge, $$p_x f_x+ p_y f_y$$ to find the conserved quantity, $$L_z = x p_y - y p_x$$ which is angular momentum about the $z$-axis.

Again, we knew nothing about the forces. All we knew was that the system was symmetric - the action didn't change - under rotations.