Consider longitudinal vibrations of particles on a line connected by springs. Setting all constants to one, the Lagrangian is $$L = \frac12 \sum_i \dot{\phi}_i^2 - (\phi_i - \phi_{i-1})^2.$$ Here, $\phi_i$ is the displacement of particle $i$ from its equilibrium position, and the canonical momenta are $\pi_i = \dot{\phi_i}$. The system has translational symmetry, $$\phi_i \to \phi_i + a$$ and the resulting conserved quantity is the total canonical momentum $\sum \pi_i$, which makes sense.
Now suppose we take the continuum limit, giving Lagrangian density $$\mathcal{L} = \frac12 \dot{\phi}^2 - \frac12 (\partial_x \phi)^2.$$ This system is supposed to be basically the same, but there are now two symmetries, $$\phi(x) \to \phi(x+a) \quad \text{and} \quad \phi(x) \to \phi(x) + a.$$ Both of these seem to be some sort of translational symmetry. The conserved quantity resulting from the first symmetry is what we usually call 'the momentum', and it is $$P = -\int \pi \partial_x \phi \, dx.$$ The conserved quantity resulting from the second symmetry is the total canonical momentum $$\Pi = \int \pi\, dx.$$ So we started with one conserved momentum, took the continuum limit, and now we have two! They're totally different quantities, not even the same order in the fields. What is going on?
- What is the physical difference between these two symmetries? Since this is a longitudinal wave, they both seem to be the very same translational symmetry.
- Exactly what part of the continuum limit 'doubles the symmetry'? Why didn't we see this in the original system?
- What is the physical interpretation of $\Pi$?
Some more calculations about $P$ and $\Pi$ follow. Let's impose the reasonable boundary conditions that $\phi(x) \to 0$ for $x \to \infty$, in some frame. There are two reasonable-looking boost transformations. The first one is $$\phi(x) \to \phi(x) + vt, \quad \pi(x) \to \pi(x) + v.$$ The quantity $\Pi$ changes by $vL$, where $L$ is the length of the line. But the quantity $P$ changes to $$P \to P - v \int \frac{d\phi}{dx} dx = P - v(vt - vt) = P$$ where I applied the boundary conditions. Then $P$ is boost invariant.
The second reasonable-looking boost transformation is $$\phi(x) \to \phi(x) + v t \frac{d\phi}{dx}$$ which is equivalent to $x \to x + vt$. By similar calculations, $\Pi$ is invariant under this boost, but $P$ is not.
Physically, I think the second boost is "boosting to a new coordinate frame", while the first boost is "boosting the medium with respect to the wave". That means that $P$ should be interpreted "the total momentum" and $\Pi$ should be interpreted as "the momentum of the wave in the medium". But this all sounds like nonsense to me because there is no 'wave' and 'medium'. There are just masses on springs. What's the difference?