most recent 30 from http://mathoverflow.net 2013-05-23T00:21:22Z http://mathoverflow.net/feeds/question/102304 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/102304/is-there-a-mathematical-explanation-for-the-aharonov-casher-effect Dmitri Pavlov 2012-07-15T18:38:34Z 2012-07-17T04:43:11Z

<p>Recall that the <a href="http://en.wikipedia.org/wiki/Aharonov-Bohm_effect" rel="nofollow">Aharonov-Bohm effect</a> can be interpreted mathematically as follows.</p> <p>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.</p> <p>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.</p> <p>In other words, a line bundle with connection can be flat without being trivial, in particular, it can have nontrivial holonomy.</p> <p>Recently I learned that there is a dual (in a certain physical sense) effect to the Aharonov-Bohm effect, namely, the <a href="http://en.wikipedia.org/wiki/Aharonov-Casher_effect" rel="nofollow">Aharonov-Casher effect</a>, in which a charged particle is replaced by a neutral particle with a magnetic moment.</p> <p><strong>Can we interpret the Aharonov-Casher effect mathematically in a similar way to the Aharonov-Bohm effect?</strong></p>

http://mathoverflow.net/questions/102304/is-there-a-mathematical-explanation-for-the-aharonov-casher-effect/102318#102318 Chris Gerig 2012-07-15T22:55:21Z 2012-07-17T04:43:11Z

<p>This is the same mathematical effect (from reading their paper on it): </p> <p>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 <em>dual</em> aspect is.</p> <p><em>Clarification</em>: 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, "<em>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).</em>" Equation 1 here is the standard Lagrangian for particle motion, and involves the vector potential (that <em>is</em> how you get a potential term in the Lagrangian).</p> <p><em>Aside</em>: 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!</p>