4

To identify the Noether 4-current with the electric 4-current, one would in principle have to show that the Noether 4-current indeed appears as the source term in Maxwell's equations. The Maxwell equations with sources (Gauss's + Ampere's laws) are derived by adding the Maxwell Lagrangian $-\frac{1}{4}F_{\mu\nu}F^{\mu\nu}$ to a minimally coupled, gauge-...


2

There are at least 2 lessons to be learned from OP's set-up: Noether's theorem is not necessarily about strict symmetry of the action. It is enough if the action has an (infinitesimal) quasi-symmetry, i.e. symmetry up to boundary terms. There's no free lunch. To prove energy conservation, one must use a non-trivial assumption: In this case, that the ...


1

So the other answers have given a solid response but it is a little high-level, I wanted to give a more verbose explanation of what has happened. In your procedure you have an action integral $S = \int_T\mathrm dt~L(q(t),\dot q(t), t)$ over some time domain $T$. You now want to vary the time coordinate. This means that the domain $T$ is changing, so that ...


1

Let's first clarify which formulation of Noether's theorem we'll be using: The Lagrangian will be a function $$ L = L(x,v,t) $$ and the action a functional $$ S[q] = \int_{t_1}^{t_2} L(q(t),\dot q(t),t) \,dt $$ Proposition. If the transformation $$ t\to t'(t) = t + \epsilon T(t) $$ $$ x\to x'(x,t) = x + \epsilon X(t)$$ $$ q'(t') = q(t(t')) + \epsilon X(t(t'...


1

Edit: The Noether Procedure instructs you to take $\varepsilon = \varepsilon(t)$ to be time dependent. $$ \delta q = \varepsilon(t) \dot q \hspace{1 cm} \delta \dot q = \dot \varepsilon(t) \dot q + \varepsilon(t) \ddot q $$ \begin{align*} \delta L &= (\varepsilon \dot q) \frac{\partial L}{\partial q} + (\dot \varepsilon \dot q + \varepsilon \ddot q) \...


Only top voted, non community-wiki answers of a minimum length are eligible