# Coordinate free description of Noether current

by Ivo Terek   Last Updated July 10, 2019 05:20 AM - source

If $$Q$$ is a differentiable manifold and $$L:TQ\to \Bbb R$$ is a Lagrangian invariant under a $$1$$-parameter group $$(\varphi_s)_{s\in \Bbb R}$$ of diffeomorphisms of $$Q$$, then the Noether charge $$\mathscr{J}:TQ\to \Bbb R$$ defined by $$\mathscr{J}(x,v) = \mathbb{F}L(x,v)\left(\frac{\rm d}{{\rm d}s}\bigg|_{s=0}\varphi_s(x)\right),$$where $$\mathbb{F}L:TQ\to T^*Q$$ is the fiber derivative of $$L$$, is constant along curves $$x:[a,b]\to Q$$ which are critical points of the action functional of $$L$$. This is Noether's Theorem stated in the cleanest way I know.

I am interested in the version of the theorem for Lagrangians whose domain is $$TQ^{\oplus m}$$ for some $$m\geq 1$$, i.e., with more than one tangent vector as input. Coordinates $$(q^1,\ldots, q^n)$$ in $$Q$$ induce coordinates in each copy of $$TQ$$ inside $$TQ^{\oplus m}$$, and so we get coordinates $$(q^1,\ldots , q^n, v^1_{(1)},\ldots ,v^n_{(1)},\ldots, v^1_{(m)},\ldots, v^n_{(m)}).$$Fix $$\Omega\subseteq \Bbb R^m$$ a compact subset with non-empty interior and regular boundary (and coordinates $$(u^1,\ldots u^m)$$), so that the domain of the action functional of $$L:TQ^{\oplus m}\to \Bbb R$$ consists of "$$m$$-surfaces" $$x:\Omega\to Q$$. For a Lagrangian like this invariant over $$(\varphi_s)_{s \in \Bbb R}$$, I got that that $$\sum_{\ell=1}^m\frac{\partial}{\partial u^\ell}\left( \sum_{k=1}^n \frac{\partial L}{\partial v^k_{(\ell)}}(x(u),\nabla x(u))\frac{\partial q^k}{\partial s}(0,u)\right) = 0.$$ This is clearly the divergence of something. I don't know how to describe in an intrinsic way what is this something, in terms of (partial?) fiber derivatives or whatever. I would like to possibly describe this as some map $$TQ^{\oplus m}\to ?$$ that is constant along critical $$m$$-surfaces.

Physics texts are completely unintelligible for me, and the few mathematics texts that could possibly say something useful about this discuss a level of generality that goes way over my head. Help?

Tags :