0
$\begingroup$

I read that a noether current occurs when the lagrangian assume vector values. Well, what are noether current and noether charge in comparison to elementary classical mechanics notions of Noether's theorem?

$\endgroup$
2
  • $\begingroup$ "in comparison to elementary classical mechanics notions of noether theorem" - can you define "elementary classical mechanics notions of noether theorem", so that we know what to compare to? $\endgroup$ Nov 28, 2018 at 19:53
  • 1
    $\begingroup$ Also, where did you read that "a noether current occurs when the lagrangian assume vector values"? $\endgroup$ Nov 28, 2018 at 20:06

1 Answer 1

1
$\begingroup$

If $Q$ is a configuration manifold and $G$ is a Lie group acting on $Q$ (usually $G = \Bbb R$ with a $1$-parameter group of diffeomorphisms of $Q$) leaving a Lagrangian $L:TQ\to \Bbb R$ invariant, for every $X\in \mathfrak{g}$ we have the Noether charge generated by $X$, the map $\mathscr{J}^X\colon TQ\to \Bbb R$ given by $$\mathscr{J}^X(x,v) =\mathbb{F}L(x,v)X^\#_x,$$where $X^\#\in\mathfrak{X}(Q)$ is the action field of $X$ and $\mathbb{F}L$ is the fiber derivative of $L$. The content of Noether's theorem is that $\mathscr{J}^X$ is constant along critical points of the action functional associated to $L$.

If $m>1$ we look at multivariable Lagrangians $L:TQ^{\oplus m}\to \Bbb R$, which have partial fiber derivatives. For $X\in \mathfrak{g}$, the Noether current generated by $X$ is the map $\mathscr{J}^X:TQ^{\oplus m}\to \Bbb R^m$ whose $j$-th component is the $j$-th partial fiber derivative of $L$ evaluated in the action field of $X$, as above. Here, Noether's theorem says that if $L$ is $G$-invariant and we have a critical point $x:\Omega\subseteq \Bbb R^m\to Q$ of the action functional associated to $L$, then the composition of $\mathscr{J}^X$ with the $1$-jet of $x$, which is a bona fide vector field on $\Omega$, has zero divergence.

For better or worse, I have some notes about this which include proofs of these results, as stated above.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.