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Recall that the Aharonov-Bohm effect can be interpreted mathematically as follows.

Consider an electromagnetic field A on some smooth manifold M, i.e., A is an element in the first differential cohomology of M, or in other words, an isomorphism class of complex hermitian line bundles with a metric connection over M.

The field strength of A is defined to be its curvature. It can happen that the field strength is zero, i.e., a charged particle traveling through M has no forces acting on it, yet we can still observe a nontrivial change in phase if the particle travels around a loop with a nontrivial holonomy. This change of phase can then be observed in a physical experiment.

In other words, a line bundle with connection can be flat without being trivial, in particular, it can have nontrivial holonomy.

Recently I learned that there is a dual (in a certain physical sense) effect to the Aharonov-Bohm effect, namely, the Aharonov-Casher effect, in which a charged particle is replaced by a neutral particle with a magnetic moment.

Can we interpret the Aharonov-Casher effect mathematically in a similar way to the Aharonov-Bohm effect?

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  • $\begingroup$ I don't quite understand the difference between the effects. As you are writing, A is an electromagentic field, and the particle is charged under it. Whether field or particle are purely electrical or magnetical depends on the choice of a reference frame, and I am tempted to say that it can thus be considered irrelevant. $\endgroup$
    – Konrad Waldorf
    Commented Jul 16, 2012 at 6:34
  • $\begingroup$ @Waldorf, no matter what frame you are in the particle is neutral. $\endgroup$
    – Chris Gerig
    Commented Jul 16, 2012 at 7:03
  • $\begingroup$ @Chris Gerig: Note that the particle in the AC effect is supposed to have a magnetic moment. So it is not neutral with respect to the electromagentic field A. $\endgroup$
    – Konrad Waldorf
    Commented Jul 16, 2012 at 7:16
  • $\begingroup$ Sorry I thought you meant relativistic frames where E-fields or B-fields disappear. But yes, the main difference here is that either the particle carries a charge or it carries a moment, and both can be considered to interact with A, as I mentioned. $\endgroup$
    – Chris Gerig
    Commented Jul 16, 2012 at 7:29

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This is the same mathematical effect (from reading their paper on it):

The action-functional is effectively (ignoring kinetic term) $\bar{v}\cdot \bar{E}\times\bar{\mu}$. But this is equivalent to the precession of the magnetic moment in a magnetic field, $\bar{E}\times\bar{\mu}=e\bar{A}$ (a paper of Kan and Koh, 1992, actually explains this in great detail). From here the Lagrangian for the AC-effect is effectively the Lagrangian for the AB-effect, and is due to the vector potential... this is what the dual aspect is.

Clarification: In fact, their paper came across this effect by simply manipulating the view of the AB-effect in the case of a solenoid (the standard example). A solenoid can be represented as a bunch of magnetic moments lined up, and this is what they do to get the AC-effect. They explicitly attribute this to the vector potential, quote, "Is it possible to generate a situation in which a neutral particle exhibits the A-B effect? We will show that this is indeed possible and is actually a necessary consequence of the physics described by Eq. (1)." Equation 1 here is the standard Lagrangian for particle motion, and involves the vector potential (that is how you get a potential term in the Lagrangian).

Aside: This is related to how us physics students learn [in Electrodynamics] that the only physical quantities are the E-field and B-field, and the vector potential $A$ and scalar potential $\phi$ are simply mathematical constructs to help computations... this is indeed true, as the electromagnetic field is described by virtual photons. Yet the AB-effect (and hence AC-effect) shows how through the vector potential we realize a topological condition on our fields!

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    $\begingroup$ I wonder why physicists would make such a big deal out of it (and write several papers about it) if it is merely a change in the coordinate system. In particular, I don't understand why a mere change in the coordinate system would require an independent experimental confirmation, as claimed in this article, for example: atomwave.org/rmparticle/ao%20refs/… $\endgroup$
    – Dmitri Pavlov
    Commented Jul 16, 2012 at 13:52
  • $\begingroup$ No, the math is the same, the physics is different. The topological phase comes about due to the vector potential A mathematically, but this achieved in two different ways... One via an electric charge, one via a magnetic moment. $\endgroup$
    – Chris Gerig
    Commented Jul 16, 2012 at 21:53
  • $\begingroup$ But it's still merely a choice of the reference frame, isn't it? Won't an observer moving at relativistic speeds observe AB effect as AC and vice versa? $\endgroup$
    – Dmitri Pavlov
    Commented Jul 16, 2012 at 23:53
  • $\begingroup$ No, because the particle will never gain a charge... $\endgroup$
    – Chris Gerig
    Commented Jul 17, 2012 at 0:59
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    $\begingroup$ We were just clarifying that regardless of the E- or B-field, the magnetic moment will interact (or the charge will interact, depending on which particle we're considering). "Reference frame" in that sense wasn't meaning move at a speed (which is what I originally thought when he made the comment, and then said 'sorry' after). $\endgroup$
    – Chris Gerig
    Commented Jul 17, 2012 at 16:51

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