# Why do we like gauge potentials so much?

Today I read articles and texts about Dirac monopoles and I have been wondering about the insistence on gauge potentials. Why do they seem (or why are they) so important to create a theory about magnetic monopoles?

And more generally, why do we like gauge potentials so much?

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## 2 Answers

1) Postponing for a moment the issue of magnetic monopoles, one conventional answer is, that the gauge potential $A_{\mu}$ (as opposed to, e.g., the electric and magnetic $\vec{E}$ and $\vec{B}$ fields) constitute the true fundamental variables and (the photon field) of QED.

At the classically level, by saying that $A_{\mu}$ are fundamental variables, we mean that the Maxwell action $S[A]$ depends on $A_{\mu}$, and that the corresponding Euler-Lagrange equations are the remaining Maxwell equations, namely Gauss' and (modified) Ampere's laws. (The obsolete Maxwell equations are Bianchi identities, which are rendered vacuous by the existence of the gauge potential $A_{\mu}$.)

Quantum mechanically, it seems appropriate to mention the Aharonov Bohm effect, which shows that the $A_{\mu}$ gauge field has physical consequences not already encoded in the gauge-invariant $\vec{E}$ and $\vec{B}$ fields.

2) Up until now we haven't discussed magnetic monopoles and Dirac strings.

At a magnetic monopole, the $A_{\mu}$ gauge potential is not well-defined. This is the main topic of, e.g., this, this, and this questions.

Although magnetic monopoles so far have not been experimentally detected, there are theoretical reasons to believe they do exist, see e.g. this, this, and this questions.

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Aharonov-Bohm effect can be used as an argument if (and only if) elementary particles are considered as point-like objects. This argument is based on the assumption that particle's field is LOCALIZED in the area where $A_{\mu}$ is non-zero, while $E$ and $B$ are zero. – Murod Abdukhakimov Mar 24 '12 at 9:16
@Qmechanic: Tell me if I am right: a non-gauge potential would give the same E and B as a gauge potential, but the action would not be then as simple to calculate; in that sense a gauge potential, though non unique, is more fundamental. Correct? – Isaac Mar 24 '12 at 10:25

It is the gauge potentials, not the fields, that determine the quantum motion of particles. In either the Schrodinger equation or the path integral, the gauge field appears, not the E and B, and for nonabelian theories, this is impossible to fix because you don't have a integral Stokes law relation.

The interaction with charged particles is that a particle moving along a path gets a phase equal to the integral of A along the path. If you want to replace the A with B, you need to use the fact that the integral of A along a closed loop is the magnetic flux enclosed by the loop, and this is a nonlocal condition. So you can't write local equations of motion for a quantum particle using E and B.

The integral relation for B states that if you make a circle, and you want the phase that the charged particle will get if it moves on this circle, you draw some surface whose boundary is the circle, and the magnetic flux through this surface is the phase.

The dirac condition is simply the statement that if you have a monopole and draw a circle around the monopole, the flux through the northern hemisphere is equal to the flux through the southern hemisphere, up to a multiple of 2pi, which is an undetectable phase change. This tells you that the magnetic charge times the electric charge must be an integer multiple of $2\pi$.

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