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The four-potential is not an observable because it is not invariant under a change of gauge. And no predictions of any physical theory are dependent on the choice of gauge, so the four-potential is not observable. What is gauge invariant and observable is the integral of the four potential around a loop, and that is what is observed in the AB effect. ...


3

The reason is confinement. Yang Mills theories with $SU(2)$ and $SU(3)$ gauge groups exhibit confinement, while for example $U(1)$ electrodynamics does not. Whether a theory is confining or not can be found out by studying the properties of Wilson loops.


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It's first important to note that in classical electromagnetism, the $\mathbf{A}$ and $\phi$ fields are not physical in the same way that the $\mathbf{E}$ and $\mathbf{B}$ are. We can't measure them and they aren't uniquely defined. Gravitational potential energy is a good analogy. Suppose an object is on a table a height $h$ above the ground. We could say ...


2

1) Yes, the charge is truly and exactly conserved. What it is confusing you, I think, is that the current for a scalar field explicitly depends on 4-potential $A$, whereas that for a spin-1/2 does not. This is obviously related to the number of derivatives in the Lagrangian kinetic term and, likewise, to the number of derivatives in the current. It can help ...


2

We use gauge theories because they - as an experimental fact - describe the world correctly. Asking why that which we use to describe the world describes the world is not a meaningful physics question. Since hitherto gauge theories have been amazingly successful in describing fundamental interactions, we would of course look for a gauge description of a new ...


2

We must distinguish between the gauge group $G$, stereotypically the Lie group $\mathrm{SU}(N)$, and the group of gauge transformations $\mathcal{G}$, which are all $G$-valued smooth functions of spacetime. There is no issue if you only write down quantities that transform in proper representations of the group of gauge transformations $\mathcal{G}$. The ...


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I only had a brief look at the paper, but to me it looks like that even after choosing $A^{0}$, in such a way that $\nabla_{l}l=0$ and $l^{\mu}=(0,1,0,0)$ then you still have some freedom in what they call $\theta^0$ which will affect $m^{\mu}$ and $\overline{m}^{\mu}$ but not the first two. I will agree that there is no more freedom in $A^{0}$, only for ...


1

Disclaimer: I do particle physics / cosmology, so this is definitely outside my field, apply grains of salt to this answer appropriately. I think Reference [29] (Lin et al, arxiv reference: 1008.4864) honestly does a better job of explaining what is going on (which makes sense, the impression I get is that 1008.4864 is a foundational paper in this ...


1

A gauge field transforms in the adjoint of the gauge group, but not in the adjoint (or any other) representation of the group of gauge transformations. In detail: Let $G$ be the gauge group, and $\mathcal{G} = \{g : \mathcal{M} \to \mathcal{G} \vert g \text{ smooth}\}$ the group of all gauge transformations. A gauge field $A$ is a connection form on a ...


1

Try and think about the gravitational potential on Earth $V(x)$. If you add a constant in space to the potential $V(x)+c$, its gradient $F=-\nabla V(x)$ remains unaltered. So the description of physical observables (the forces) in terms of potential is somewhat overabundant. When you solve a problem of a falling object on earth, you usually put the potential ...


1

Let't me give it a shot. One possible explanation is to imagine we soften the integer condition for $m$'s in $S_{2}$. Then, the cosine term in $S_{1}$ is what we want to add because when $K$ is large, it enforce the integer condition and go back to $S_{2}$.



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