From: Toby Bartels Subject: Re: diffeomorphism invariance in QFT (better version) Date: Fri, 7 Dec 2001 00:02:18 +0000 (UTC) Newsgroups: sci.physics.research Summary: Aharonov-Bohm effect depends on nontrivial topology of spacetime Jim Heckman wrote: >How can A fields which differ by something other than a change >of gauge give rise to the same F fields? Mathematically, this can happen only if spacetime has nontrivial topology, specifically if it has nontrivial 1st de Rham cohomology. Of course, spacetime near Earth has trivial topology. In practice, the observed Bohm Aharanov effect is one where the F fields are experimentally indistinguishable in a certain region (which region has nontrivial topology) but vastly different elsewhere. For example, F may be large inside a cylindrical solenoid but arbitrarily close to zero outside that solenoid. Nevertheless, physical effects that occur only in the region of nearly indistinguishable F, far from the region of different F, can nevertheless depend strongly on the difference in [A] (where [A] is the equivalence class of A under gauge transformations), so long as the effect takes advantage of the nontrivial topology. In the example, you transport an electron around the solenoid, and it picks up a phase when it wouldn't if [A] were truly 0. -- Toby toby@math.ucr.edu ============================================================================== From: vanesch@ill.fr (Patrick Van Esch) Subject: Re: diffeomorphism invariance in QFT (better version) Date: Fri, 7 Dec 2001 03:12:27 GMT Newsgroups: sci.physics.research "Jim Heckman" wrote in message news:... > > How can A fields which differ by something other than a change > of gauge give rise to the same F fields? In a simply connected space, they can't. The whole trick of the B-A effect is that one only considers the F-fields in a swap of space with a hole in it. In that case, all the different A-fields that have only different F fields in the hole, but the same F-fields outside the hole, are to be considered. That has nothing to do with gauge changes. They are completely different A-fields. So, to answer your question, no they don't give rise to the same F-field *everywhere*. cheers, Patrick. ============================================================================== From: helling@ariel.physik.hu-berlin.de (Robert C. Helling) Subject: Re: diffeomorphism invariance in QFT (better version) Date: Fri, 7 Dec 2001 03:12:54 GMT Newsgroups: sci.physics.research On Thu, 6 Dec 2001 03:05:07 +0000 (UTC), Jim Heckman wrote: >How can A fields which differ by something other than a change >of gauge give rise to the same F fields? Nobody? So let me try to give an answer: This can only be in spaces with non-trivial topology. The classic example is R^3 \ {(x,y,z) |x=y=0}, that is three-space minus some axis. In that case A~(x,y,z,t)=(0,0,0) and A(x,y,z,t) = ( -y/(x^2+y^2), x/(x^2+y^2), 0) are _not_ related by a gauge transformation. Why not? Well, if there were such a transformation, there would be a function Lambda, such that the gradient of Lambda equals A. We could try to obtain it by integrating A along some path, viz. Lambda(x,y,z) = int_p^(x,y,z) A.ds where p is some point you keep fixed and . is the scalar product. But this choice only works if Lambda(x,y,z) is independent of the choice of path from p to that point. This is the case if the integral around closed paths vanishes. But lets consider the path gamma(s) = (r cos(s), r sin(s),0). Then A(gamma(s)) = (-sin(s)/r, cos(s)/r,0) and the integral is int_0^2pi ds A(gamma(s)). gamma'(s)= int_o^2pi ds (sin^2(s)+cos^2(s))= 2 pi Thus Lambda is not well definded and there cannot be such a gauge transformation. Why is that? Note that I took out some part of R^3. You would get the same A if there were an infinitely thin soleonid with some finite magnetic flux at x=y=0. The point is that integrals int A.ds around closed paths are gauge invariant (as you can easily check by applying a gauge transformation delta A=grad Lambda). If you can find a surface that has the path as a boundary you could use Cauchy's theorem to translate this integral into a surface integral measuring the magnetix flux thru that surface (involving only the fieldstrengths E and B). But in our case above there is no such surface since there's a hole in the space. Therefore the integral can be non-zero even if E and B vanish everywhere in space. The Aharonov-Bohm effekt tells you that in quantum mechanics such A fields can have a measurable effect even though the electrons are only allowed to be in places of vanishing field strengths. Robert -- .oOo.oOo.oOo.oOo.oOo.oOo.oOo.oOo.oOo.oOo.oOo.oOo.oOo.oOo.oOo.oOo.oOo.oOo.oOo.oO Robert C. Helling Institut fuer Physik Humboldt-Universitaet zu Berlin print "Just another Fon +49 30 2093 7964 stupid .sig\n"; http://www.aei-potsdam.mpg.de/~helling