
#55
Dec412, 03:07 AM

P: 2,889





#56
Dec412, 03:13 AM

P: 2,889





#57
Dec412, 03:27 AM

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P: 5,307

I think it doesn't make sense to repeat everything umpteen times; the Bfield is a derived quantity; the Afield is fundamental and sufficient; the Afield is pure gauge locally, so no Bfield is assoiciated with it; the Bfield is zero outside; the electron can't "look" into the solenoid and therefore does not see any Bfield; the magnetic flux is not the same as the magnetic field, but it's a loopintegral over the Afield, the flux around the loop does not require a Bfield inside b/c the electron cannot distinguish between the Bfield in the solenoid and a pure gauge w/o any Bfield ...
As long as you insist on using the Bfield you will never be able to understand the AB effect, local gauge theory with global gauge effects and other consequences. 



#58
Dec412, 03:32 AM

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P: 5,307

Integrating around closed loops tells you something regarding topology, winding number etc. The integral is NOT pathindependent in general but only pathindep. for homotopic paths. That's one key lesson: the magnetic flux IS a topological quantity related to the first homotopy group of the base manifold. 



#59
Dec412, 03:32 AM

P: 2,889

So do we agree at least that the solenoid cross section acts as a hole of the space of the paths of the electrons?
If so, how does the independent of path closed line integral of A work? It is usually stated there must be no holes for the Green and Stokes theorems to hold. Is this wrong? 



#60
Dec412, 03:37 AM

P: 2,889





#61
Dec412, 03:41 AM

P: 2,889





#62
Dec412, 04:39 AM

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P: 5,307

TrickyDicky, I don't get your point b/c there seems to be no point at all!
I do not start with the negative! I simply say that the integral is not pathindependent in general (this is contary your statement). The pathdependence related to the first homotopy group is is one key factor in understanding the topology of the AB effect. So I do not simply repeat what you are saying "in other words", there is a fundamental difference And I do not answer specific questions b/c I don't see them. So what are your questions? (and what is missing in our explanations?) 



#63
Dec412, 06:17 AM

P: 2,889





#64
Dec412, 06:31 AM

P: 561





#65
Dec412, 06:40 AM

P: 2,889

This is how the magnetic AB effect is described in wikipedia just for reference:
"The most commonly described case, sometimes called the Aharonov–Bohm solenoid effect, takes place when the wave function of a charged particle passing around a long solenoid experiences a phase shift as a result of the enclosed magnetic field, despite the magnetic field being negligible in the region through which the particle passes and the particle's wavefunction being negligible inside the solenoid. This phase shift has been observed experimentally. The magnetic Aharonov–Bohm effect can be seen as a result of the requirement that quantum physics be invariant with respect to the gauge choice for the electromagnetic potential, of which the magnetic vector potential A forms part. Electromagnetic theory implies that a particle with electric charge q travelling along some path P in a region with zero magnetic field B, but nonzero A (by [itex] \mathbf{B} = 0 = \nabla \times \mathbf{A}[/itex]), acquires a phase shift [itex] \varphi[/itex], given in SI units by [itex] \varphi = \frac{q}{\hbar} \int_P \mathbf{A} \cdot d\mathbf{x},[/itex] Therefore particles, with the same start and end points, but travelling along two different routes will acquire a phase difference [itex] \Delta \varphi[/itex] determined by the magnetic flux [itex]\Phi_B[/itex] through the area between the paths (via Stokes' theorem and [itex] \nabla \times \mathbf{A} = \mathbf{B})[/itex], and given by: [itex] \Delta\varphi = \frac{q\Phi_B}{\hbar}. [/itex] In quantum mechanics the same particle can travel between two points by a variety of paths. Therefore this phase difference can be observed by placing a solenoid between the slits of a doubleslit experiment (or equivalent). An ideal solenoid (i.e. infinitely long and with a perfectly uniform current distribution) encloses a magnetic field B, but does not produce any magnetic field outside of its cylinder, and thus the charged particle (e.g. an electron) passing outside experiences no magnetic field B. However, there is a (curlfree) vector potential A outside the solenoid with an enclosed flux, and so the relative phase of particles passing through one slit or the other is altered by whether the solenoid current is turned on or off. This corresponds to an observable shift of the interference fringes on the observation plane." End quote I'm only centering on this part of the quote: "but nonzero A (by [itex] \mathbf{B} = 0 = \nabla \times \mathbf{A}[/itex])", to make the addition that is not mentioned in the wikipedia that what is between parenthesis only hold when the space is simply connected, and that requirement is precisely the one demanded by the explanation of the effect that not be fulfilled. Is this apparent contradiction so hard to see? 



#66
Dec412, 06:44 AM

P: 2,889

Anyway here the obvious reason for a breakdown of path independence is the nontrivial topology. 



#67
Dec412, 06:46 AM

Sci Advisor
P: 5,307

In #56 you are asking "What do you mean is not required, there is no effect without a magnetic flux inside the solenoid" I said (quite often in the meantime) that there is a difference between the magnetic FIELD (which is zero outside and which is NOT required) and the magnetic FLUX as calculated via the line integral. Please try to get the following statement: the AB effect does not require any Bfield; it can be expressed purely in terms of the Afield; and the Afield is evaluated purely outside the solenoid where the Bfield is zero. The electrons do not interact with an Bfield but with an Afield! In #59 you are asking "how does the independent of path closed line integral of A work?" I think we agree that there is no path independence in general. OK; so what are you specific questions that have not been addressed? (sorry for insisting on these questions, but I want to avoid answering the wrong questions or addressing issues that are already clear to you) 



#68
Dec412, 06:54 AM

P: 2,889





#69
Dec412, 07:39 AM

Sci Advisor
P: 5,307

I am not sure whether I get this point (which was not addressed to me, I guess) "Integrating around the closed loop independently of path requires the trivial topology".
The phase shift measured is [tex]\Delta\phi \sim \oint_C A[/tex] This does neither require trivial topology (or Stokes theorem to calculate the flux) nor any Bfield. Let's cite Wikipedia: The next sentence refers to the very specific case of solenoids, but this is only one possible experimental setup, not the general case: [tex]\oint_C A \neq 0[/tex] 



#70
Dec412, 08:27 AM

P: 2,889

See #65




#71
Dec412, 01:15 PM

P: 788

There seems to be some confusing communication regarding "path dependence." Probably we can all agree on the following unambiguous statements:
1. In a simply connected space where curl(A) = 0 everywhere, line integrals of the A field are pathindependent. 2. In a nonsimplyconnected space where curl(A) = 0 everywhere, line integrals of the A field may be pathdependent. TrickyDicky, I have the sense that you are arguing as follows: "People claim that the line integral of A depends on the winding number around the solenoid, because the space R^3 \ (solenoid) is not simply connected. But this is nonsense. The actual physical space is simply connected. Therefore the line integral of A cannot be path dependent. In particular, it's nonsense to claim that the line integral can depend on something topological like a winding number when the physical space has no nontrivial topology to speak of." Correct me if that's a misrepresentation; it's just a guess at what you are getting at. In any case, the response is that the two cases above neglected the actual physical case: 3. In a simply connected space where curl(A) is NOT everywhere zero, line integrals of the A field may be path dependent. If the physical space is simply connected, what justification do we have for talking about nonsimplyconnected spaces and reasoning topologically? The point is that if you work in the simply connected space, but don't consider line integrals that intrude into the solenoid, you get the same results for line integrals as if you had worked in the nonsimplyconnected space. Therefore whatever topology has to say about line integrals in the nonsimplyconnected space also applies to the same line integrals in the simply connected space. I think this addresses another point you seem to be making, namely, "People talk about this nonsimplyconnected space, but simultaneously use Stokes' theorem, which does not hold in such a space, to discuss the magnetic flux." Only in the simply connected space can you apply Stokes' theorem. Only in the nonsimplyconnected space do you have nontrivial topology and winding numbers. But since you get the same line integrals and the same physics either way, we feel comfortable talking about Stokes theorem and nontrivial topology at the same time, even though strictly speaking we are imagining different spaces in each case. We're just using different mathematical tools to get the same answer. 



#72
Dec412, 01:45 PM

P: 2,889

I don't know, I guess I'll just study it better, I might as well just leave it here. Thanks everyone. 


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