# Symmetry and conservation

### Symmetry

Given a physical system, a symmetry is a change in the system that doesn't change the action. By change, we mean a transformation of the coordinates of the system.

For classical mechanics, only continuous transformations can possibly lead to conservation laws. That means we are only interested in infinitesimal variations, which then build into large transformations.

Nb. Discrete transformations can lead to conservation laws in quantum mechanics, however.

We can write down a very general definition of an infinitesimal transformation of the coordinates, $q_i$, as $$q_i \to q_i + \epsilon f_i(q)$$ for small $\epsilon$ and arbitrary $f_i$. Notice that $f_i$ is described a function of $q = (q_1,\dots,q_N)$, all of which are functions of $t$.

Alternatively, we can just write down the variation itself, $$\delta q_i = \epsilon f_i(q)$$

For this transformation to be a symmetry we require that the action doesn't change under the transformation. That is, $$\delta A = 0$$

We want to show that this leads to a conservation law.

### Noether's Theorem

This states that if a transformation is a symmetry of a given system, then there exists a quantity associated with the system, called the Noether charge, that is conserved. Further, if the transformation is given by $$q_i \to q_i + \epsilon f_i(q)$$ for small $\epsilon$ and arbitrary $f_i$ then the Noether charge is given by $$\sum_{i} \pi_i \; f_i(q)$$ where $\pi_i$ is the momentum canonically conjugate to $q_i$.

Figure 4.1 – symmetric transformation

Nb. This is not a context in which we can apply the principle of least action, since here, the endpoints of the two trajectories are not necessarily the same.

To see why the theorem is true, we first suppose that a given trajectory, $q_i$, over some arbitrary time period $[t_1, t_2]$, is already a solution of the Euler-Lagrange equations, $$\frac{\partial L}{\partial q_i} - \frac{\mathrm{d} }{\mathrm{d} t}\left( \frac{\partial L}{\partial \dot{q_i}} \right) = 0$$

Since the transformation, $\delta q_i$, is a symmetry then, by definition, the action doesn't change.

Thus, we can write \begin{align*}0 &= \delta A\\&= \delta \int_{t_1}^{t_2} L(q_i, \dot{q_i}) \mathrm{d}t\\&= \int_{t_1}^{t_2} \sum_{i} \left\{ \frac{\partial L}{\partial q_i} \delta q_i + \frac{\partial L}{\partial \dot{q_i}} \delta \dot{q_i} \right\} \mathrm{d}t\\&= \int_{t_1}^{t_2} \sum_{i} \left\{ \left( \frac{\partial L}{\partial q_i} - \frac{\mathrm{d}}{\mathrm{d} t} \left(\frac{\partial L}{\partial \dot{q_i}} \right) \right) \delta q_i \right\} \mathrm{d}t + \left[ \sum_{i} \frac{\partial L}{\partial \dot{q_i}} \delta q_i \right]_{t_1}^{t_2}\\&= \int_{t_1}^{t_2} \sum_{i} \left\{ 0 \;\cdot\; \delta q_i \right\} \mathrm{d}t + \left[ \sum_{i} \frac{\partial L}{\partial \dot{q_i}} \delta q_i \right]_{t_1}^{t_2}\\&= \left[ \sum_{i} \pi_i \; \delta q_i \right]_{t_1}^{t_2}\end{align*}
using the fact that $q_i$ is a solution of the Euler-Lagrange equations and the conjugate momemta are $$\pi_i = \frac{\partial L}{\partial \dot{q_i}}$$

The integrated term - inside the square brackets - is the conserved quantity we are looking for. Why? Because the start time and the end time $t_1, t_2$ were chosen arbitrarily and so when we expand out the bracket we get $$\left. \sum_{i} \pi_i \;\delta q_i \right|_{t_2} = \left. \sum_{i} \pi_i \; \delta q_i \right|_{t_1} \quad \forall t_1, t_2$$

### Noether charge

Furthermore, we can substitute $$\delta q_i = \epsilon f_i(q)$$ into the above result which means that $$\epsilon \sum_{i} \pi_i \;f_i(q)$$ is conserved. But we can also divide this by $\epsilon$ and still have a conserved quantity $$\sum_{i} \pi_i \;f_i(q)$$

This is the Noether charge associated with the symmetry, to which the arbitrary functions $f_i = f_i(q)$ refer. Once you know the symmetries of a system, you can use Noether's theorem to calculate the associated conserved quantites.

Of course, it's an art in itself finding out what the symmetries are. In practise, if you think a certain transformation is a symmetry, then you can check to see if the associated action changes under small deviations. If it doesn't, the conserved quantity - the Noether charge - can be blindly calculated.