What is the physical constraint that gauge invariance is a required condition for electromagnetic fields? What would happen if the electromagnetic fields were not gauge invariant?
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asked Mar 17, 2014 at 14:20
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2$\begingroup$ If the fields did not have a $U(1)$ gauge symmetry, then the conserved Noether charge associated to the global version of the symmetry would not be conserved. $\endgroup$– JamalSCommented Mar 17, 2014 at 14:23
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$\begingroup$ Is possible to expand on the topic of conserved Noether charge? $\endgroup$– linuxfreebirdCommented Mar 17, 2014 at 14:26
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$\begingroup$ Possible duplicate of physics.stackexchange.com/q/90119 $\endgroup$– Kyle KanosCommented Mar 17, 2014 at 14:42
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1 Answer
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By disposing of the $U(1)$ gauge symmetry, we also dispose of the global version of the $U(1)$ symmetry which gives rise to both a conserved current and conserved charge due to Noether's theorem.
Specifically, every continuous global symmetry, by Noether's theorem, gives rise to a conserved current $j^{\mu}$ which satisfies the continuity equation,
$$\partial_\mu j^{\mu} = \frac{\partial j^{0}}{\partial t} + \nabla \cdot \vec{j} = 0.$$
The conserved Noether charge is defined as,
$$Q = \int_V d^3 x \, \, j^{0},$$
and by demanding $\partial_t Q = 0$, we see this implies,
$$\frac{\partial Q}{\partial t} = -\int_V d^3x \, \, \nabla \cdot \vec{j} = -\int_{\partial V} \vec{j} \cdot ndS = 0,$$
or in other words the flux of $\vec{j}$ is zero, and hence $Q$ is conserved locally. By losing $U(1)$ gauge symmetry, we lose a redundancy in the description of the system, but also a conservation law.
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$\begingroup$ Good answer. Where does the broken gauge invariance come into the math? Do we have to start with the Lagrangian? $\endgroup$ Commented Mar 17, 2014 at 14:44
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2$\begingroup$ If you want to break gauge invariance, you need to add a term to the Lagrangian, e.g. a mass term. The reason we need the Higgs mechanism in the standard model for certain fields is because we can't add a mass term that preserves symmetries for those fields. $\endgroup$– JamalSCommented Mar 17, 2014 at 14:48
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$\begingroup$ Whoa, the higgs field breaks charge conservation? $\endgroup$ Commented Mar 17, 2014 at 14:49
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1$\begingroup$ No, certainly not. We introduce the Higgs field to avoid introducing mass terms which break symmetries, in a nutshell $\endgroup$– JamalSCommented Mar 17, 2014 at 14:52
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$\begingroup$ It should be noted many symmetries are actually spontaneously (not explicitly) broken in Nature, for example $SU(2) \times U(1)_{L} \to U(1)_{EM}$. $\endgroup$– JamalSCommented Mar 17, 2014 at 14:53