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How one can know the gauge field emerging from the local gauge invariance is actually the EM field?

I understood in a simple scalar field whose Lagrangian is given by

$ \mathcal{L} = \frac{1}{2}(\partial_\mu \phi)^{*}(\partial^\mu \phi) - \frac{1}{2} |\phi|^2 $

needs a vector gauge field in order to satisfy the local gauge invariance, which is simply

the electromagnetic field $A_\mu$.

What I'm trying to understand is that how one can prove that gauge field

is exactly an EM field that satisfy the Maxwell's eqn: $\partial_\mu \partial^{[\space\mu} A^{\nu \space]} = J^\nu$.

For making the Lagrangian be locally gauge invariant, the field $A_\mu$, whatever it is,

I showed it should have a gauge transformation relation

$A_\mu \rightarrow A_\mu ' = A_\mu - \partial_\mu \chi $ as $\phi \rightarrow \phi' = e^{-i\chi}\phi$

I think however, it doesn't guarantee that $A_\mu$ should be the EM field

that satisfies above Maxwell's eqn.

Please let me understand. Thank you.

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1 Answer 1

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OP asks (v1):

How one can know the gauge field emerging from the local gauge invariance is actually the EM field?

Assuming that OP is pondering about gauging theoretical models (rather than concerned with our actual world and phenomenological inputs) then the answer is: One cannot know. For starters, the gauge group $G$ could be different than $U(1)$. And even if the gauge group is $G=U(1)$, then it could be another $U(1)$ than $U(1)_{EM}$.

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  • $\begingroup$ OP seems focused on the "EM field" fulfilling Maxwell's equations. Any $\mathrm{U}(1)$ field will fulfill Maxwell's equations, just perhaps for a current other than the electromagnetic one. $\endgroup$
    – ACuriousMind
    Dec 20, 2014 at 23:20
  • $\begingroup$ Yeah, since OP talked about gauging a (matter) theory, I was mainly concerned with the eom for the matter rather than the eom for the gauge field. For all we know the gauge field action could be, say, the Born-Infeld action rather than the Maxwell action. $\endgroup$
    – Qmechanic
    Dec 21, 2014 at 1:19

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