Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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?

share|improve this question
    
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. –  Konrad Waldorf Jul 16 '12 at 6:34
    
@Waldorf, no matter what frame you are in the particle is neutral. –  Chris Gerig Jul 16 '12 at 7:03
    
@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. –  Konrad Waldorf Jul 16 '12 at 7:16
    
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. –  Chris Gerig Jul 16 '12 at 7:29
add comment

1 Answer 1

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!

share|improve this answer
2  
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/… –  Dmitri Pavlov Jul 16 '12 at 13:52
    
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. –  Chris Gerig Jul 16 '12 at 21:53
    
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? –  Dmitri Pavlov Jul 16 '12 at 23:53
    
No, because the particle will never gain a charge... –  Chris Gerig Jul 17 '12 at 0:59
    
@Chris: How can this be reconciled with Konrad's claim (to which you apparently agreed) that being electrically charged or having a magnetic moment depends on the reference frame? –  Dmitri Pavlov Jul 17 '12 at 9:49
show 1 more comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.