# Symmetry associated with current conservation

arlesterc
As I understand it for every symmetry there is associated a conserved quantity - so for time symmetry there is energy conservation. I understand as well that charge conservation is associated with a 'mathematical' local symmetry - something turning in a mathematical space at a point so to speak. What I am not clear about is what symmetry is associated with conserved currents? I understand Kirchoff type conservation but I am not understanding the use of 'current' in the context of symmetry. For instance this sentence from another post - " For continuous systems, the conserved quantities become conserved "currents". " (https://www.physicsforums.com/threa...robability-current-for-wave-functions.188784/) - I don't understand this. Does this mean that for continuous systems it is not energy that is conserved but 'energy currents' - if so what exactly is that. I have also seen the phrase Noether current but have not been able to grasp what that means. I have seen a lot of J's along the way that are meant to stand for current I believe but have not been able to fit them into my understanding. I understand current conservation in Kirchoff systems - and it ends there. Ultimate goal here is understanding what symmetry is associated with 'current' conservation, the concepts needed to get to that goal. Thanks in advance for any assistance with this.

Mentor
Really they shouldn’t say conserved current. It is the Noether’s charge that is conserved. The current is the flow of that charge from one location to another. The Noether current and the Noether charge are related through a continuity equation which essentially defines a local conservation law.

So for electromagnetism the symmetry is the gauge symmetry of the potential, and the resulting Noether charge is electric charge and the resulting Noether current is electric current.

• • vanhees71 and hutchphd
aclaret
exercise ;) ;)

effect variation ##\delta A_{\mu} = \partial_{\mu} \eta## of ##A_{\mu}## where ##\eta: R^4 \rightarrow R## arbitrary function, and more, do calculate resulting variation in the lagrangian ##\mathscr{L} = -\frac{1}{4} F_{\mu \nu} F^{\mu \nu} - A_{\mu} j^{\mu}##. now proof that ##j^{\mu}## is a conserved... :)

with that lemma, proposition follow by recalling that electric charge obtained by volume integral of the ##j^0##

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arlesterc
Dale -

Thanks - that's very helpful. You write:
" The Noether current and the Noether charge are related through a continuity equation which essentially defines a local conservation law. "

What is the continuity equation?

Where does the J I see so often come in? Aclaret's post as is is a few levels above my mental pay grade I am afraid. Maybe if I had a worked example it might help - actual numbers, actual case.

Mentor
What is the continuity equation?
It is an equation of the form $$\frac{\partial \rho}{\partial t}+\nabla \cdot \vec j = 0$$ where ##\rho## is the charge density and ##\vec j## is the current density. Basically, it says that if charge accumulates somewhere then it cannot simply magically appear there but must flow in as current that doesn’t flow out.

aclaret
not as hard, as you are making it out to be... let me show you. here I will skimp on a bit of rigour, but hope give general idea :)

Under that variation ##\delta A_{\mu} = \partial_{\mu} \eta## that i mention (fancy-schmancy term: gauge transformation), it clear that resulting variation in lagrangian is ##\delta \mathscr{L} = - j^{\mu} \partial_{\mu} \eta##.

For theorem of noether to apply, transformed lagrangian must be same up to some divergence; i.e. ##\mathscr{L} \mapsto \mathscr{L} + \partial_{\mu} v^{\mu}## for some ##v##. That in mind, we then must assert that

##\delta \mathscr{L} = - j^{\mu} \partial_{\mu} \eta \overset{!}{=} -\partial_{\mu} (j^{\mu} \eta) \implies \eta \partial_{\mu} j^{\mu} = 0##

this hold for arbitrary ##\eta##, thus ##\partial_{\mu} j^{\mu} = 0##. this express conservation of charge :)

$$\hat{Q}=\int_{\mathbb{R}^3} \mathrm{d}^3 x :\hat{\bar{\psi}}(t,\vec{x}) \gamma^0 \hat{\psi}(t,\vec{x}):.$$