Noether theorem - fixed volume symmetry breaking

Question

I have a question related to Noether theorem.

The Theorem is the following: (Noether’s Theorem). Suppose $F$ is a one-parameter family of symmetries of the Lagrangian system, $L : T Q \to \mathbb{R}$. Then, \begin{align*} p^i \delta q_i - l \end{align*} is conserved, that is, it's time derivative is zero for any path $q \in \Gamma$ satisfying the Euler-Lagrange equations.

A standard application is to show spatial translation symmetry. For a free particle in $\mathbb{R}^n$ , we have $Q = \mathbb{R}^n$ and $L = \frac{1}{2} m \dot{x}^2$, with mass $m$ and velocity $\dot{x}$. Under infinitesimal space translation \begin{align*} \begin{cases} x^\prime = x + a & \to \delta x = a \\ \dot{x}^\prime = \dot{x} & \to \delta \dot{x} = 0 \end{cases} \end{align*} the variation of the Lagrangian becomes \begin{align*} \delta L = \frac{\partial L}{\partial x} \delta x + \frac{\partial L}{\partial \dot{x}} \delta \dot{x} = 0 \end{align*} and the final result (I skip here some steps, since it is a standard calculation, given in all of literature about the topic) is that the momentum is constant.

My question is now, what happens if the volume of the system is fixed, because I read that this would directly break symmetry of spatial translation. I guess that $Q$ is then just part of $\mathbb{R}^n$, but does it mean that Noether is not valid at all, or in other words only valid for an infinte large space?

Answer 1

First of all, a fixed volume is bad for dealing with derivatives that are needed in Noether theorem (how do you differentiate something on the boundary of the volume?). So, let's impose a smoother but essentially identical restriction: the space is still $\mathbb R^n$, but there is a term in potential energy that starts rapidly (but smoothly) going to infinity outside of some volume (this is what I often do in simulations, by the way). The system looses translational symmetry: upon translation, the volume & the potential get translated as well, preventing the Lagrangian from being translation-invariant.

You can see directly why conservation of momentum fails here: a particle hitting the boundary of allowed volume instantaneously changes its momentum, thus changing the overall momentum as well. Instantaneous change of momentum means a force was applied, and this is exactly what causes pressure on the boundary of the volume.

Answer 2

If the position variable $q$ is confined to a compact interval $[a,b]$, it's effectively as if there is an infinite well potential $V(q)$, which clearly breaks translational symmetry, and thereby in turn, formally breaks the conclusion of Noether's theorem (NT).

However pragmatically, if we are only interested in describing the system within some open interval $]c,d[$ away from the endpoints $a$ and $b$, then we can for all practical purposes remove the infinite well potential $V(q)$, and work as if $q$ lives on the whole real line $\mathbb{R}$, so that translational symmetry (and the conclusion of NT, namely momentum conservation) are restored.

Of course, momentum is not conserved at the endpoints $a$ and $b$.