Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

How to prove conservation of electric charge using Noether's theorem according to classical (non-quantum) mechanics? I know the proof based on using Klein–Gordon field, but that derivation use quantum mechanics particularly.

share|improve this question
    
Possible duplicate: physics.stackexchange.com/q/2721/2451 –  Qmechanic Jan 4 '13 at 0:57
2  
Of course, it's not a duplicate of linked question. If you read the question in details, you'll find a words "..according to classical (non-quantum) mechanics?..". And the linked question is less specific. –  user8817 Jan 4 '13 at 9:46

1 Answer 1

up vote 5 down vote accepted

By the word classical we will mean $\hbar=0$, and we will use the conventions of Ref. 1.

The Lagrangian density for Maxwell theory with various matter content is

$$\tag{1} {\cal L} ~=~{\cal L}_{\rm Maxwell} + {\cal L}_{\rm matter} , $$

$$\tag{2} {\cal L}_{\rm Maxwell}~=~ -\frac{1}{4}F_{\mu\nu}F^{\mu\nu},$$

$$\tag{3} {\cal L}_{\rm matter}~=~{\cal L}_{\rm matter}^{\rm QED}+{\cal L}_{\rm matter}^{\rm scalar QED} + \ldots, $$

$$\tag{4} {\cal L}_{\rm matter}^{\rm QED} ~:=~ \overline{\Psi}( i\gamma^{\mu} D_{\mu}-m)\Psi , $$

$$\tag{5} {\cal L}_{\rm matter}^{\rm scalar QED}~:=~ -(D_{\mu}\phi)^{\dagger} D^{\mu}\phi -m^2\phi^{\dagger}\phi -\frac{\lambda}{4} (\phi^{\dagger}\phi)^2, $$

with covariant derivative

$$ \tag{6} D_{\mu}~=~d_{\mu}-ieA_{\mu}. $$

(Here we are too lazy to denote various matter masses $m$ and charges $e$ differently.) The matter equations of motion (eom) are

$$ \tag{7}( i\gamma^{\mu} D_{\mu}-m)\Psi ~\approx~0, \qquad D_{\mu}D^{\mu}\phi~\approx~m^2\phi+\frac{\lambda}{2} \phi^{\dagger}\phi^2, \qquad \ldots.$$

(The $\approx$ symbol means equality modulo eom, i.e. an on-shell equality.)

The infinitesimal global off-shell gauge transformation is

$$ \delta A_{\mu} ~=~0, \qquad \delta\Psi~=~-i\epsilon \Psi, \qquad \delta\overline{\Psi}~=~i\epsilon \overline{\Psi}, $$ $$ \tag{8} \delta\phi~=~-i\epsilon \phi,\qquad \delta\phi^{\dagger}~=~i\epsilon \phi^{\dagger}, \qquad \ldots, \qquad\delta {\cal L} ~=~0, $$

where the infinitesimal parameter $\epsilon$ does not depend on $x$.

The Noether current is the electric $4$-current$^1$

$$ \tag{9} j^{\mu}~=~e\overline{\Psi}\gamma^{\mu}\Psi - ie\{\phi^{\dagger} D^{\mu}\phi-(D^{\mu}\phi)^{\dagger}\phi\}+\ldots. $$

Noether's Theorem is a theorem about classical field theory. It yields an on-shell conservation law

$$ \tag{10} d_{\mu}j^{\mu}~\approx~0.$$

Hence the electric charge

$$\tag{11} Q~=~\int\! d^3x~ j^0$$

is conserved on-shell.

References:

  1. M. Srednicki, QFT.

--

$^1$ Interestingly, the electric $4$-current $j^{\mu}$ depends on the gauge potential $A_{\mu}$ in case of scalar QED matter.

share|improve this answer
    
I'm wondering if PhysiXxx would not prefer to call a "classical" proof a proof with real fields... then without gauge symmetry. I believe in "classical" Maxwell equations the conservation of charge $dQ / dt = 0$ is a postulate, leading to $\partial_{\mu} j^{\mu} = 0$ through integration over a finite volume. –  FraSchelle Mar 24 '13 at 17:26
    
It is true that [Maxwell eqs. $d_{\mu}F^{\mu\nu}=-j^{\nu}$] $\Rightarrow$ [Continuity eq. $d_{\mu}j^{\mu}=0$] $\Rightarrow$ [Electric charge conservation]. However, OP asked specifically to use Noether's (first) theorem in the proof. According to Noether's (first) theorem, we have [global gauge symmetry of the action]$\Rightarrow$ [Electric charge conservation]. –  Qmechanic Apr 24 '13 at 17:19

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.