A time translation invariant system without energy conservation
Noether's theorem relating continuous Lie symmetries to conserved quantities famously implies that given a time translation symmetry
\begin{equation} \phi(t, x^i) \to \phi'(t) = \phi(t + T, x^i) \end{equation}with infinitesimal action
\begin{equation} \delta \phi = \varepsilon \dot{\phi} \end{equation}The Lagrangian density changes by
\begin{eqnarray} \delta \mathcal{L} &=& \frac{\partial \mathcal{L}}{\partial \phi} \varepsilon \dot{\phi} + \frac{\partial \mathcal{L}}{\partial \dot{\phi}} \varepsilon \ddot{\phi}\\ &=& \varepsilon \frac{d}{dt} \mathcal{L} \end{eqnarray}...
The hidden assumption here is that the field has the proper values and variation on the boundary. If we forgo this condition, energy is not conserved. There are two ways we could break this condition.
The first one is to have an extra boundary to the manifold. This will usually correspond to the condition of a naked singularity in our spacetime. The simplest possible example is to consider the static spacetime made from the slice $\mathbb{R}^3 \setminus \{ 0 \}$, Minkowski space minus the origin.
A good way to convince yourself is to first consider Minkowski space with the charge distribution
\begin{equation} j^\mu(x) = (\theta(t) \delta(x^i), 0) \end{equation}This is not a very good charge distribution (it has a discontinuous total charge), but this won't matter much. Using the Lorenz gauge, the Maxwell equation is then
\begin{eqnarray} \Box A^0 &=& \theta(t) \delta(x^i)\\ \Box A^i &=& 0 \end{eqnarray}The exact set of solutions doesn't matter too much here, we are only after a particular one. The simplest will be to assume $A_i = 0$, and using the retarded D'Alembert green function,
\begin{eqnarray} A^0 &=& \int d^4x \theta(t) \delta(x^i) \frac{\delta((t - t_0) + |x^i - x^i_0|)}{4\pi |x^i - x^i_0|}\\ &=& \int dt \theta(t) \frac{\delta(t - (t_0 - |x^i_0|)}{4\pi |x^i_0|}\\ &=& \theta(t_0 - r_0) \frac{1}{4\pi r} \end{eqnarray}So that, for $r > t$, the potential is the Coulomb potential and simply $0$ otherwise. The electric field $E^i = - \nabla^i A^0$ is then
\begin{eqnarray} E^i &=& -\nabla^i (\theta(t - r) \frac{1}{4\pi r})\\ &=& -\frac{1}{4\pi r} \nabla^i \theta(t - r) - \theta(t - r) \nabla^i \frac{1}{4\pi r}\\ &=& -\frac{x^i}{4\pi r^2} (\delta(t - r) - \theta(t - r) r^{-1}) \end{eqnarray}The energy of our electric field is then
\begin{eqnarray} H &=& \int E^i E_i d^3x\\ &=& -\int \frac{1}{4\pi r} (\delta(t - r) + \theta(t - r) r^{-1}) d^3x\\ &=& 4\pi \int \frac{r}{4\pi} (\delta(t - r) + \theta(t - r) r^{-1}) dr\\ &=& 4\pi (\frac{t}{4\pi} + \ldots) \end{eqnarray}Posted on 2019-09-18 10:17:41