Aharonov-Bohm topological explanation

[The_Duck]

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 path-independent.

2. In a non-simply-connected space where curl(A) = 0 everywhere, line integrals of the A field may be path-dependent.

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 non-simply-connected 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 non-simply-connected space. Therefore whatever topology has to say about line integrals in the non-simply-connected 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 non-simply-connected 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 non-simply-connected 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.

 

[TrickyDicky]

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 path-independent.

2. In a non-simply-connected space where curl(A) = 0 everywhere, line integrals of the A field may be path-dependent.

Agreed.

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:

Thanks for trying to understand my point. Your interpretation is not exactly what I was getting at but not so far either. I can see the difference between our actual space and the space of the experiment, and see the logic behind the path dependence of the R^3 \ (solenoid) space of the experiment.

3. In a simply connected space where curl(A) is NOT everywhere zero, line integrals of the A field may be path dependent.

But the electrons are always in the non-simply connected space, no?

I think this addresses another point you seem to be making, namely, "People talk about this non-simply-connected space, but simultaneously use Stokes' theorem, which does not hold in such a space, to discuss the magnetic flux."

This is closer to my point, yes.

Only in the simply connected space can you apply Stokes' theorem. Only in the non-simply-connected 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.

I'm just not sure you get the same line integrals and therefore the same physics if the Stokes theorem doesn't apply to the space of the experiment.
I don't know, I guess I'll just study it better, I might as well just leave it here.
Thanks everyone.

 

[tom.stoer]

thanks, I think this should contribute to clarification; my arguments seemed to be - at least partially - distracting.

Regarding the_duck's the last paragraph, a) simply connected space and application of Stokes' theorem and b) non-simply-connected space i.e. nontrivial topology and winding numbers: that was the reason for me to cite Wikipedia. I wanted to make clear that they are talking about two entirely different things, namely a non-vanishing loop integral over A (which is the general case for the AB effect) and the solenoid case which is a rather special experimental setup.

 

[TrickyDicky]

This might help understand my point too:
http://unapologetic.wordpress.com/2012/02/18/a-short-rant-about-electromagnetism-texts/

 

[tom.stoer]

I am not sure if I get your point; are you saying that the existence of A satisfying B = rot A is ruled out b/c the space may not be contractable? Or what exactly is your conclusion derived from http://unapologetic.wordpress.com/2012/02/18/a-short-rant-about-electromagnetism-texts/ ?

 

[TrickyDicky]

I am not sure if I get your point; are you saying that the existence of A satisfying B = rot A is ruled out b/c the space may not be contractable?

Yes, not so much as rule out, but certainly it makes one wonder what's going on.

 

[tom.stoer]

Yes, not so much as rule out, but certainly it makes one wonder what's going on.

The problem is that "existence" has not yet been well defined.

What topology tells us is that the existence of a global A-field and a global B-field is indeed problematic; that's the reason we have to use R³ \ R instead of R³. A- and B-field are not well-behaved on R³.

This can be seen by using the A-field outside the solenoid

$$A = \frac{\Phi}{2\pi r^2}(-y,x,0) = \frac{\Phi}{2\pi}\nabla\theta$$

and taking the limit

$$r \to 0; \Phi = \text{const.}$$

Physically this seems to be unproblematic b/c the electrons only "feel" the flux as a physical observable via the phase

$$e^{i\Phi}$$

which remains constant. But the A-field as defined above and the gauge function

$$U = e^{-i\Phi\theta / 2\pi}$$

from which it can be derived as

$$A^\prime = 0 \to A = U^\dagger\,(A + i\nabla)\,U$$

cannot be defined globally; r=0 has to be excluded from our configuration space, i.e. from the base manifold, which results in R³ \ R.

That means that we can define

$$F = dA = 0$$

locally but not globally.

$$dA = 0$$

means that A is locally flat i.e. has vanishing curvature 2-form F (i.e. vanishing el. mag. field strength, i.e. E=0, B=0) locally.

In other words A is pure gauge locally i.e. A ~ A' = 0 but not globally.

What Stokes' theorem (and cohomology) tells us is that

$$\oint_C A$$

does only depend on the homotopy class of the loop C w.r.t. r=0, i.e. it depends only on the winding number

$$w[C] = n = 0,\pm 1,\pm 2, ...$$

of the loop C around the r=0.

Using Stokes' theorem naively

$$\int_{\partial M} \omega = \int_M d\omega\;\;\to\;\; \int_{C} A = \int_D \nabla\times A = \int_D B = 0$$

one would derive a vanishing magnetic flux. But this is wrong due to the fact that the extension of A from the loop C to the disc D is not allowed due to the above mentioned singularity in r=0. Therefore the naive application of Stokes theorem is wrong, but this is due to the r.h.s. (= the surface integral) not due to the l.h.s.

Physically one can save the theorem by explaining the B-field via the solenoid, i.e. via replacing the A-field and the B-field by the solenoid solution on a disc D. Mathematically one can save the story by using the locally flat A-field outside and vanishing B-field! Instead one has to cut out the disc D with shrinking radius such that one arrives at R³ \ R where the A-field is well-defined.

That means that the physical solenoid can be replaced by the non-trivial base manifold R³ \ R.

Now the question remains what is physically meaningful. At first glance one could argue in favor of the solenoid with non-vanishing B-field, but even this may be obscure: iff (if and only if) we know that the A-field has been prepared using a solenoid everything is fine. But we (and the electrons) are not able to penetrate the solenoid, so there is no physical mechanism to distinguish between the "physical solenoid" and the "R³ \ R".

What I am saying is that if somebody prepares an impenetrable solenoid with locally flat but non-zero A-field outside w/o telling you any details, you are not able to distinguish between the following two scenarios:
a) the impenetrable solenoid contains a non-vanishing B-field
b) the impenetrable solenoid wraps a topological singularity

 

[Jano L.]

Tom, I do not understand how you can write solenoidal field as a gradient in that way. Is the angle $\theta$ in your writings a multiple-valued function of position? Otherwise we get sudden jump of A at $\theta = 0$, which makes the loop integral zero...

 

[tom.stoer]

I am not sure whether I understand you problem. It's true that θ is in [0,2π[ and that therefore θ itself is not a "global coordinate". But that doesn't matter b/c all physical quantities are well-behaved:

The first definition of A is fine.
The loop integral becomes a multipliation by the length i.e. by 2π for constant and is OK.
The gauge transformation has been introduced via U which is periodic in θ and therefore well-defined.
Therefore the gradient acting on U is well-defined, too; acting on θ may be problematic, but this is due to θ, not due to the gradient

θ itself is not a globally well-defined coordinate, and in principle one would have to use a "cut R²" with several charts using different angle coordinates; but as usual in physics I am a bit sloppy and avoid all this stuff.

 

[Jano L.]

OK, now I get it, your prescription gives correct solenoidal A without delta function in A. I mistakenly thought that we would get delta function in A due to chain rule and jump in theta. But now I realize the chain rule does not apply for $e^{i\theta}$, so it is OK.

Tricky Dicky: I read the passage in the book by Nash&Sen and also sec. 2.2 in D. J. Thouless, Topological quantum numbers in nonrelativistic physics, (1998) which Wikipedia cites and was surprised that they do not say more than that topology is somehow relevant to BA shift. They do not elaborate on that.

Furthermore, the papers on BA effect I have seen in PR do not seem to claim that considerations of topology are necessary.

So can you give some reference to work where the authors really show there is some connection ?

 

[tom.stoer]

It depends on what you call the AB effect ;-)

According to Wikipedia The Aharonov–Bohm effect ... is a quantum mechanical phenomenon in which an electrically charged particle is affected ... despite being confined to a region in which both the magnetic field B and electric field E are zero. The underlying mechanism is the coupling of the electromagnetic potential with the complex phase of a charged particle's wavefunction ... Following that line of reasoning one may conclude that the effect is due to locally but not globally flat gauge field which indicates a non-trivial vector bundle structure, so in this general setup it's a topological effect.

http://www.encyclopediaofmath.org/index.php/Bohm-Aharonov_effect
http://iopscience.iop.org/0143-0807/9/3/007

But physically one could simply study the most commonly described case, sometimes called the Aharonov–Bohm solenoid effect, ... which does by no means require any topological reasoning.

As you may have observed I think that topology of gauge field and vector bundles is more fundamental than one specific experimental setup (just as quantization of action is a more fundamental principle than the black body radiation). I hope the above mentioned reference clarifies some of the topological concepts.

 

[TrickyDicky]

It depends on what you call the AB effect ;-)

According to Wikipedia The Aharonov–Bohm effect ... is a quantum mechanical phenomenon in which an electrically charged particle is affected ... despite being confined to a region in which both the magnetic field B and electric field E are zero. The underlying mechanism is the coupling of the electromagnetic potential with the complex phase of a charged particle's wavefunction ... Following that line of reasoning one may conclude that the effect is due to locally but not globally flat gauge field which indicates a non-trivial vector bundle structure, so in this general setup it's a topological effect.

http://www.encyclopediaofmath.org/index.php/Bohm-Aharonov_effect
http://iopscience.iop.org/0143-0807/9/3/007

I don't have access to your second reference, so let me follow line by line the encyclopedia of math one:
"A phenomenon in which the topological non-triviality of a gauge field is measurable physically [a1]. Moreover, this topological non-triviality, which can be expressed as a number , say, is a global topological invariant and so is not expressible by a local formula; this latter point being in contrast to a simpler topological invariant such as the dimension of the underlying space, which is deducible locally."
I think we all agree with this paragraph, the main reason of the need for globality being IMO that homotopy relations always contain global information as it is well known.
In the next line is where I find my conundrum rather than with the effect itself:
"To demonstrate this effect physically, one arranges that a non-simply-connected region of space has zero electromagnetic field . This electromagnetic field is related to the gauge field by the usual relation F=dA"
The usual relation F=dA for a non-simply connected region is not granted, that is what I mean by my doubts about the existence of the A-field in this particular not simply connected set up(see my link "A short rant..." from the Unapologetic mathematician to check this).
Here in the 5th paragraph:
"If we flip over to the language of differential forms, we know that the curl operator on a vector field corresponds to the operator $\alpha\mapsto *d\alpha$ on 1-forms, while the gradient operator corresponds to $f\mapsto df$. We indeed know that *ddf=0 automatically — the curl of a gradient vanishes — but knowing that $d\alpha=0$ is not enough to conclude that $\alpha=df$ for some f. In fact, this question is exactly what de Rham cohomology is all about!"
In our case F is a 2-form and A is a one-form. See also the wikipedia page "Closed and exact differential forms" to see that a closed form like F=0 needs a simply connected space to imply F=dA.

But physically one could simply study the most commonly described case, sometimes called the Aharonov–Bohm solenoid effect, ... which does by no means require any topological reasoning.

I don't know why you say the solenoid effect, that is an example of the AB effect, "by no means require any topological reasoning". I don't think your distinction between a topological general effect and a non-topological special magnetic case is founded. See this: http://rugth30.phys.rug.nl/quantummechanics/ab.htm

 

[tom.stoer]

"To demonstrate this effect physically, one arranges that a non-simply-connected region of space has zero electromagnetic field . This electromagnetic field is related to the gauge field by the usual relation F=dA"
The usual relation F=dA for a non-simply connected region is not granted, that is what I mean by my doubts about the existence of the A-field in this particular not simply connected set up

The relation F=dA is granted locally but not globally, so F and A may required patching, cutting out singularities etc. In the case of the A-field as described above one has to remove r=0. On this R³ / R the relation F=dA=0 is valid (so "not globally" means "not on R³ but on R³ / R").

I don't know why you say the solenoid effect, that is an example of the AB effect, "by no means require any topological reasoning". I don't think your distinction between a topological general effect and a non-topological special magnetic case is founded.

In the solenoid case one has a finite disc with non-vanishing B-field; one has a regular solution inside the solenoid; one can use Stokes' theorem w/o any problem. In the case with a point-like singularity at r=0 there is a vanishing B-field; there is an A-field with dA=0 in R³ / R, so the explanation via a B-field is invalid b/c it's zero. The solenoid case can be realized experimentally; this is not possible for the R³ / R case. Therefore this distinction is relevant.

 

[TrickyDicky]

The relation F=dA is granted locally but not globally, so F and A may required patching, cutting out singularities etc. In the case of the A-field as described above one has to remove r=0.On this R³ / R the relation F=dA=0 is valid (so "not globally" means "not on R³ but on R³ / R").

But you need it to be granted globally, at least at the global region where electrons are influenced by the A-field so their wave function is shifted.The homotopy from R^3 to R^3/R always carries global information.
In any case I don't think is granted locally either for the R^3/R case and you haven't explained why.

In the solenoid case one has a finite disc with non-vanishing B-field; one has a regular solution inside the solenoid; one can use Stokes' theorem w/o any problem. In the case with a point-like singularity at r=0 there is a vanishing B-field; there is an A-field with dA=0 in R³ / R, so the explanation via a B-field is invalid b/c it's zero. The solenoid case can be realized experimentally; this is not possible for the R³ / R case. Therefore this distinction is relevant.

The solenoid case should work just like the idealized case, you guarantee the simply- connectedness just by making the electrons unable to traverse the solenoid. So you are removing r=0 just the same.

 

[tom.stoer]

But you need it to be granted globally, at least at the global region where electrons are influenced by the A-field so their wave function is shifted.The homotopy from R^3 to R^3/R always carries global information.
In any case I don't think is granted locally either for the R^3/R case and you haven't explained why.

Come on! it is granted locally! you can see A (it has been written down ;-) and you can check dA=0 by explicit calculation of the curl; and it is granted globally on R³/R b/c the check works for all (x,y,z) when r=0 is excluded

 

[TrickyDicky]

See also this for the validity of the solenoid experiment: http://physics.stackexchange.com/questions/34779/on-aharonovbohm-effect

 

[andrien]

In case,if one transform to a frame which moves uniformly with respect to any original one it is possible to eliminate magnetic field,in that frame of course the so far gauge invariant integral is zero.In that case the effect comes from A₀, by a phase like-q/h^-(∫∅dt),negative in sign with original one.Now here no line integral,no stokes theorem,or I should say no topology?

 

[TrickyDicky]

Come on! it is granted locally! you can see A (it has been written down ;-) and you can check dA=0 by explicit calculation of the curl; and it is granted globally on R³/R b/c the check works for all (x,y,z) when r=0 is excluded

You should realize that you are taking for granted A everytime, if we are debating the existence of A, you cannot write dA=0 to beging with bc then you are already assuming it. You are affirming the consequent to quote the UM. ;)

Wrt to the trick of cutting out the singularity r=0, well it is a bit silly since then we get back to R^3, and it defeats any further consideration about the problem of a non simply connected space for the existence of an A-field.

 

[TrickyDicky]

I guess my doubt is more related to topology and vector calculus than to the quantum mechanical effect itself since it involves a step previous to the effect proper, so I might take it to the Topology and geometry subforum.

 

[tom.stoer]

In case,if one transform to a frame which moves uniformly with respect to any original one it is possible to eliminate magnetic field,in that frame of course the so far gauge invariant integral is zero.In that case the effect comes from A₀, by a phase like-q/h^-(∫∅dt),negative in sign with original one.Now here no line integral,no stokes theorem,or I should say no topology?

Can you show us the calculation? From what I understand you start in the A°=0 gauge in the rest frame and then transform to an inertial frame which moves parallel to the solenoid.

 

[tom.stoer]

You should realize that you are taking for granted A everytime, if we are debating the existence of A, you cannot write dA=0 to beging with bc then you are already assuming it. You are affirming the consequent to quote the UM. ;)

TrickyDicky, it becomes a little bit annoying!

I have written down (explicitly) an A-field valid for r>0. So this A-field certainly does exist on R³/R. And of course you can (and should) calculate rot A which again is perfectly valid for r>0.

What else do you need to be convinced that A does exist ?

 

[andrien]

Can you show us the calculation? From what I understand you start in the A°=0 gauge in the rest frame and then transform to an inertial frame which moves parallel to the solenoid.

http://books.google.co.in/books?id=f9viHUx3H5gC&pg=SA15-PA8&dq=vector+potential+and+quantum+mechanics+feynman+lectures+vol.+2&hl=en#v=onepage&q=vector%20potential%20and%20quantum%20mechanics%20feynman%20lectures%20vol.%202&f=false
15.9 onwards,However it is not written here that one can eliminate magnetic field,but it is possible to do so.In very simple cases,whenever there is a motion of charge producing magnetic field,one can transform to an inertial frame in which this field is zero and only electric field remains.Still there is a possibility of vector potential,but since the gauge invariant integral only depends on curl of A,it is zero because of absence of magnetic field.The law given in the reference is directly a consequence of lagrangian of charged particle coupled to electromagnetic field,as you are already aware of it.

 

[TrickyDicky]

TrickyDicky, it becomes a little bit annoying!

I have written down (explicitly) an A-field valid for r>0. So this A-field certainly does exist on R³/R. And of course you can (and should) calculate rot A which again is perfectly valid for r>0.

What else do you need to be convinced that A does exist ?

Ok, Tom, it is obvious to me you still think I am not aware that it is obviously the A-field that produces the effect, the OP is about something different as others understood, but as I said it is in part my fault for choosing the wrong subforum.

 

[tom.stoer]

Sorry, I don't get it. It's not the wrong subforum but perhaps the wrong way asking questions

 

[TrickyDicky]

I believe this is the typical case where "you can't have your cake and eat it too".
We have here a potential A-field that we've agreed that is not globally defined in the space of the AB effect, we have to cut out the origin where the solenoid is, leaving a singularity. The only problem being that by removing the origin we eliminate the justification for having the A-field at all, that is the EM field that is switched on in the experiment to obtain the shift in wave function pattern.

 

[tom.stoer]

We have here a potential A-field that we've agreed that is not globally defined in the space of the AB effect, we have to cut out the origin

Yes.

The only problem being that by removing the origin we eliminate the justification for having the A-field at all

No, sorry, it seems you understood nothing.

 

[TrickyDicky]

No, sorry, it seems you understood nothing.

You mean the A-field in the AB experiment is unrelated to the EM field?

 

[tom.stoer]

I mean that the EM-field is zero!

 

[TrickyDicky]

I mean that the EM-field is zero!

I 'm not referring to that, you still don't have a clue what I'm asking do you? Either that or you are pulling my leg. Just in case I'll try with a kindergarten explanation.
What has to happen in order to observe the phase shift in the interference pattern of the electrons? Hint: It's got to do with switching something on.
So we have a situation for the AB experimental setup in which something is switched off, where the EM-field is zero for the electrons and there's no A-field, and a situation in which something is switched on where the EM field is still zero for the electrons but the A-field is nonzero and produces a phase shift. See? We can relate that nonzero A-field with the switching on of something, I'll let you call that something however you want, but I think it's called an EM-field. The take away point is that the A-field is related with switching it on, fine so far?
What I want to underline is that I'm not doubting the effect, I'm only concerned about the usual explanation of it because it seems to mix two incompatible scenarios in a contradictory way, the R^3 scenario and the R^3/R scenario.
The situation before the switching on is compatible with both spaces, but the situation after is compatible with the R^3/R space only, however an effect that is only causally compatible with the activation of something that doesn't exist for the R^3/R space is produced. If that is ok for you guys , then fine. I won't bother about it anymore

 

[TrickyDicky]

I wonder why in the quantum physics forum nobody has mentioned the non-locality of the effect. Probably that's all my quibble amounts to.

 

[tom.stoer]

So we have a situation for the AB experimental setup in which something is switched off, where the EM-field is zero for the electrons and there's no A-field, and a situation in which something is switched on where the EM field is still zero for the electrons but the A-field is nonzero and produces a phase shift.

OK.

We can relate that nonzero A-field with the switching on of something, I'll let you call that something however you want, but I think it's called an EM-field.

In that case of the experimental setup it's nothing else but the electric current in the solenoid.

The take away point is that the A-field is related with switching it on, fine so far?

Fine.

What I want to underline is that I'm not doubting the effect, I'm only concerned about the usual explanation of it because it seems to mix two incompatible scenarios in a contradictory way, the R^3 scenario and the R^3/R scenario.

Hm, I'm curious, what's next.

The situation before the switching on is compatible with both spaces, ...

No current, no A-field, yes, that's allowed in both cases.

... but the situation after is compatible with the R^3/R space only, ...

I don't understand. In the case of the experimental setup with a solenoid, switching on the current produces an A-field outside the solenoid. In the R³/R case we don't care where the A-field comes from; it's there - end-of-story. The funny thing is that the math is the same, so the R³/R case is an idealization.

There are two ways you can look at the experiment:

1) you constructed an apparatus with the solenoid and you can switch the current on and off. In that case you know what you are doing. You can solve the Maxwell equations for the current; you find the non-vanishing EM-field inside the solenoid and the vanishing EM-field but non-vanishing A outside; you can calculate the phase shift of the wave function and you'll find that it agrees with experiment.

2) you haven't constructed the apparatus and you can't switch anything on and off. All there is is an interference pattern. You observe that this pattern deviates from the usual expectation, so it's not symmetric w.r.t. the symmetry axis of the experimental setup. OK, now you may guess that the apparatus has been constructed as described above (1) and that there's a current inside which produces the A-field. Fine. But it could be as well that nothing is inside, except for a singularity, a one-dim. line removed from R³ - and that due to some unknown reason there is an A-field which is pure-gauge, locally flat, w/o any EM-field, w/o any energy stored in the A-field etc.

w/o looking into the solenoid you can't distinguish between
1) the solenoid with a current and
2) the solenoid wrapping vacuum w/o any current, a one-dim. singularity, and a source-less A-field.
The interference patterns are identical.

Physically (2) seems to be unacceptable, but as I said: w/o looking into the solenoid and inspecting the apparatus in detail there is no way to distinguish between (1) and (2).

This is fine for me: mathematical I can do it either way, and physically I know what the clever guys in the lab have constructed. No problem for me.

EDIT:

I wonder why in the quantum physics forum nobody has mentioned the non-locality of the effect. Probably that's all my quibble amounts to.

I mentioned it a couple of times; the loop integral = the holonomy related to the winding number is nothing else but a non-local observable due to the non-trivial topology of the vector bundle.

In #22 I wrote "The A-field ... is pure-gauge locally, but not globally; that's what's measured by the loop integral"; in #77: "in other words A is pure gauge locally i.e. A ~ A' = 0 but not globally"; in #82 you wrote "... this topological non-triviality, which can be expressed as a number , say, is a global topological invariant and so is not expressible by a local formula"; #83: "... F=dA is granted locally but not globally, so F and A may required patching, cutting out singularities etc. In the case of the A-field as described above one has to remove r=0. On this R³ / R the relation F=dA=0 is valid, so #not globally' means 'not on R³ but on R³ / R'".

 

[TrickyDicky]

In the R³/R case we don't care where the A-field comes from; it's there - end-of-story.

It must be just me then, I care.

There are two ways you can look at the experiment:

1) you constructed an apparatus with the solenoid and you can switch the current on and off. In that case you know what you are doing. You can solve the Maxwell equations for the current; you find the non-vanishing EM-field inside the solenoid and the vanishing EM-field but non-vanishing A outside; you can calculate the phase shift of the wave function and you'll find that it agrees with experiment.

2) you haven't constructed the apparatus and you can't switch anything on and off. All there is is an interference pattern. You observe that this pattern deviates from the usual expectation, so it's not symmetric w.r.t. the symmetry axis of the experimental setup. OK, now you may guess that the apparatus has been constructed as described above (1) and that there's a current inside which produces the A-field. Fine. But it could be as well that nothing is inside, except for a singularity, a one-dim. line removed from R³ - and that due to some unknown reason there is an A-field which is pure-gauge, locally flat, w/o any EM-field, w/o any energy stored in the A-field etc.

w/o looking into the solenoid you can't distinguish between
1) the solenoid with a current and
2) the solenoid wrapping vacuum w/o any current, a one-dim. singularity, and a source-less A-field.
The interference patterns are identical.

Physically (2) seems to be unacceptable, but as I said: w/o looking into the solenoid and inspecting the apparatus in detail there is no way to distinguish between (1) and (2).

This is fine for me: mathematical I can do it either way, and physically I know what the clever guys in the lab have constructed. No problem for me.

Ok, we have very different perspectives of what is problematic in physics, you don't mind about a magical source-less A-field and I do.


Anyway in the case 1), when you switch the current on, is the nonzero A-field in R^3/R or in R^3? If the former you have to recur to the magic sourceless A-field that somehow knows when you switch it, if the latter you have the A-field but don't have the vanishing B-field so no experiment. You seem content with the former, but I don't, I guess that's all.

 

[tom.stoer]

Ok, we have very different perspectives of what is problematic in physics, you don't mind about a magical source-less A-field and I do.

Don't worry about the A-field but about the string-like singularity. If the singularity is present and space is R³/R then the existence of the A-field is no longer sorcery but well-understood due to the topology of the base manifold.

Refer to #77.

Anyway in the case 1), when you switch the current on, is the nonzero A-field in R^3/R or in R^3?

R³

if the latter you have the A-field but don't have the vanishing B-field so no experiment.

The B-field still vanishes outside the solenoid and the electrons only feel A - so for the electrons w/o your knowledge about (1) both cases are always indistinguishable.

 

[TrickyDicky]

R³


The B-field still vanishes outside the solenoid and the electrons only feel A - so for the electrons w/o your knowledge about (1) both cases are always indistinguishable.

I disagree, as long as you use a solenoid to create a singularity for the electrons so that they "feel" A but have no access to B-field the space can't be R^3.

Further if you declare R^3 indistinguishable from R^3/R in the AB experiment, the logic of the explanation of the experiment based on the latter topology is invalidated. As I said you can't have your cake and eat it too, ;-)

 

[tom.stoer]

I still think you understand nothing :-(

The solenoid does not "create a singularity"; it's a physical solenoid with a current, nothing special. You don't even use delta-functions or something like that. Math is perfectly OK, physics is standard, don't worry. The electrons are shielded by some mechanism from the current, so they don't penetrate the solenoid and feel only the A-field (no EM-field, and not directly the current). That's perfectly valid as a "physical" explanation w/o strange math, singularities R³/R etc.

And of course I can "declare R^3 indistinguishable from R^3/R in the AB experiment" b/c as I said:
a) the solenoid is impenetrable, therefore the electrons can't distinguish between the scenario 1) with the current and the scenario 2) with "vacuum + singularity"
b) the A-field the electrons feel is identical in case 1) and 2)

The A-field the electrons feel in both cases (it can be calculated exactly !) which creates the phase shift is the same (!) for 1) and 2) so the math to solve the Schrödinger equation, to calculate the phase shift etc. is identical. In that sense the scenarios cannot be distinguished, neither experimentally nor mathematically (the guy preparing the solenoid can tell you the difference).

So a minor correction to what I wrote above is ion order: this is not really a "declaration" but a consequence of the math.

Remark: the cake was fine ;-)