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


1 Answer 1


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$^1$

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

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

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

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

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

with covariant derivative

$$ D_{\mu}~=~d_{\mu}-ieA_{\mu}, \tag{6} $$ and with Minkowski sign convention (-,+,+,+). (Here we are too lazy to denote various matter masses $m$ and charges $e$ differently.) The matter equations of motion (eom) are

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

(The $\stackrel{m}{\approx}$ symbol means equality modulo matter eom, i.e. an on-shell equality.)

The infinitesimal global off-shell gauge transformation is

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

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

The Noether current is the electric $4$-current$^2$

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

Noether's first Theorem is a theorem about classical field theory. It yields an on-shell continuity equation$^3$

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

Hence the electric charge

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

is conserved on-shell.


  1. M. Srednicki, QFT.


$^1$ Note that the matter Lagrangian density ${\cal L}_{\rm matter}$ may depend on the gauge field $A_{\mu}$

$^2$ Interestingly, the electric $4$-current $j^{\mu}$ depends on the gauge potential $A_{\mu}$ in case of scalar QED matter.

$^3$ Note that the above proof of the continuity equation (10) via Noether's first theorem (as OP requested) never uses Maxwell's equations.

  • $\begingroup$ 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. $\endgroup$
    – FraSchelle
    Mar 24, 2013 at 17:26
  • 3
    $\begingroup$ It is true that [Maxwell eqs. $d_{\mu}F^{\mu\nu}\stackrel{A_{\lambda}}{\approx}-J^{\nu}$] $\Rightarrow$ [Continuity eq. $d_{\mu}J^{\mu}\stackrel{A_{\lambda}}{\approx}0$] $\Rightarrow$ [Electric charge conservation], where $J^{\mu}:=\frac{\delta S_{\rm matter}}{\delta A_{\mu}}$. In fact, here $J^{\mu}$ could be an unspecified background source that knows nothing about the matter theory. 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]. $\endgroup$
    – Qmechanic
    Apr 24, 2013 at 17:19
  • $\begingroup$ for the footnote 2: (Interestingly, the electric ....) This statement is weird. to my view, the on-shell 4-current $j^\mu$ is gauge-invariant and independent on the gauge potential $A_\mu$. This part of scalar QED matter shows this result, $j^\mu \sim - ie\{\phi^{\dagger} D^{\mu}\phi-(D^{\mu}\phi)^{\dagger}\phi\}$ $\endgroup$
    – Zoe Rowa
    Jun 9, 2021 at 6:05
  • $\begingroup$ Hi @Zoe Rowa. Thanks for the feedback. As I see it $j^{\mu}$ is gauge-invariant but depends on $A$. $\endgroup$
    – Qmechanic
    Jun 9, 2021 at 7:11
  • $\begingroup$ Hi @Qmechanic, thanks for your response. I agree your statement. Using the similar logic, one could say $F_{\mu\nu}$ is gauge-invariant and dependent on $A_\mu$. For the terminology, Gauge-invariant is clear, while "dependence" is not clear enough, the latter needs more words to elucidate its content. $\endgroup$
    – Zoe Rowa
    Jun 9, 2021 at 7:29

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