Aharonov-Bohm topological explanation

[TrickyDicky]

no, I don't see that; there is an A-field in R³ \ R; write it down and there it is; where's the problem?

The problem is that the topology of the experiment prevents it, you can write it down but in a experiment you must justify its existence. In this experiment its existence is usually justified by the switching on of a magnetic field.

why do you think that there is a B-field which can be derived from an A-field that according to your reasoning does not exist?

No, I don't think that. I'm just going by the experiment and in it there is a B field switched on, the condition to derive the existence of an A-field from that B-field is that the space be simply connected, if that condition is not fulfilled in the experiment(and that is precisely what is used as explanation of the effect) the space where the electrons are has not only B=0 but A=0.

 

[TrickyDicky]

there is an A-field in R³ \ R

No but of course in R³ if for instance you have a toroidal inductor of circular cross section, and fix the Coulomb gauge you have nonzero A where B=0.

 

[tom.stoer]

please tell me: why should a non-simply connected manifold (or anything else) forbid the existence of an A-field? there is absolutely no reason for that; A-fields = 1-forms exist on weirdest manifolds

 

[Jano L.]

My entire point was that the Stokes theorem requires simply-connectednes to hold.

This is true, but the Stokes theorem is just mathematical transformation. Vector potential could still be used to describe the effect.


OK, I think I understand your question. You assume hypothetical world which has an infinite cylindrical hole inside which the field is B is nonzero, and worry whether the explanation of the B-A shift really works for such setup (because perhaps there is no vector potential). I believe the answer is no, but for a different reason.



First, I think it is not necessary for the standard quantum-theoretical explanation of the shift. One should get it for ordinary simply-connected space, otherwise there should not have been so much fuss about it.

But let's adopt the above assumption to see where the argument for the shift fails.


1) The toroidal space T we consider has zero magnetic field everywhere. The vector potential can be introduced, but in order to do that, we have to require some conditions. The obvious one is

<br /> \nabla\times\mathbf A = 0,<br />

but this is not sufficient, for there are more solutions to this equation. One of them is \mathbf A = 0 everywhere in T, which gives zero loop integral and no reason for the shift. So, in order to get some, we have to arrive at a different potential, like that which circles around the hole.

2) What is the reason for any definite choice different from A = 0? Besides the above equation, we have to impose another condition, and the only thing left are the boundary conditions in the infinity and at the inner radius.

Now, in ordinary physical space one uses Coulomb vector potential, which is nonzero and circling on the surface of the coil. If one takes this condition to the inner boundary of our space T, one gets the same vector potential and we have a reason to expect some shift.

Now assume the current in the coil is reversed; then the direction of the potential on the surface is opposite, and if we copy this again into our boundary of T, the predicted shift changes sign.

So, there is vector potential mathematically, but we have to copy the boundary conditions from the real situation where the coil is part of the system. For this reason, I think that the idea that the effect has something to do with breakdown of simple-connectedness is wrong.

 

[TrickyDicky]

please tell me: why should a non-simply connected manifold (or anything else) forbid the existence of an A-field? there is absolutely no reason for that; A-fields = 1-forms exist on weirdest manifolds

It doesn't forbid its existence per se, but in a physical experiment you have to justify where it comes from, and usually it is related to a magnetic field, or at least in this case , as usually explained, is. Do you mean that in the AB effect experiment the A-field has nothing to do with the magnetic field?

 

[TrickyDicky]

OK, I think I understand your question. You assume hypothetical world which has an infinite cylindrical hole inside which the field is B is nonzero, and worry whether the explanation of the B-A shift really works for such setup (because perhaps there is no vector potential).

Right.

I believe the answer is no, but for a different reason.

First, I think it is not necessary for the standard quantum-theoretical explanation of the shift. One should get it for ordinary simply-connected space, otherwise there should not have been so much fuss about it.

If this was the case of course my objections dissolve.

But let's adopt the above assumption to see where the argument for the shift fails.
...

So, there is vector potential mathematically, but we have to copy the boundary conditions from the real situation where the coil is part of the system. For this reason, I think that the idea that the effect has something to do with breakdown of simple-connectedness is wrong.

Well, it is widely used.

 

[tom.stoer]

It doesn't forbid its existence per se, but in a physical experiment you have to justify where it comes from, and usually it is related to a magnetic field, or at least in this case , as usually explained, is. Do you mean that in the AB effect experiment the A-field has nothing to do with the magnetic field?

as I said, you do not need the B-field; it's a derived quantity, and all physics can be formulated entirely based on the A-field; of course physically the source of the A-field is a current i.e. the solenoid and one has to check whether this works (this is where Maxwell's equations enter the stage; everything else can be formulated w/o any reference to these equations) of course the solenolid solution is an idealization, but besides that there should be no physical problem

 

[TrickyDicky]

For this reason, I think that the idea that the effect has something to do with breakdown of simple-connectedness is wrong.

The not simply connected region is needed so that dA=0 doesn't lead to A being a constant which gives trivial physics and no effect(no change). So I guess it is a necessary condition.

 

[tom.stoer]

not really; dA=0 does not automatically imply A=const; dA=0 means that A is pure-gauge, but here it cannot be gauged away due to the non-trivial topology of the vector bundle defined by R³ * A

but I agree, there is some non-trivial topology involved; whether it's due to R³ \ R or due to the vector bundle or physically due to the solenoid is a matter of taste; but you need something like that

 

[TrickyDicky]

as I said, you do not need the B-field; it's a derived quantity, and all physics can be formulated entirely based on the A-field;

And as I said this statement has nothing to do with what I'm explaining in this thread.

of course physically the source of the A-field is a current i.e. the solenoid and one has to check whether this works (this is where Maxwell's equations enter the stage; everything else can be formulated w/o any reference to these equations)

Sure you can calculate the vector potential from the source current geometry, but a magnetic field is always implied since the existence of A comes from ∇.B=0→∇XA=B , but this derivation needs the additional requirement that the space is simply connected. The opposite way:∇XA=B→∇.B=0 needs no topological assumptions, solenoidal vector fields are divergnce free. Do you agree with this?

 

[TrickyDicky]

not really; dA=0 does not automatically imply A=const; dA=0 means that A is pure-gauge, but here it cannot be gauged away due to the non-trivial topology of the vector bundle defined by R³ * A

but I agree, there is some non-trivial topology involved; whether it's due to R³ \ R or due to the vector bundle or physically due to the solenoid is a matter of taste; but you need something like that

Right, so what's with the "not really"? ;)

 

[TrickyDicky]

Here's one of the sources I'm using for the AB effect:"Topology and Geometry for Physicists (Dover Books (Paperback) by Charles Nash, Siddhartha Sen", pages 301-302. They are freely available in the amazon.com reader page.

One thing I note in their explanation is that they use differential forms rather than vector fields and say(calling F the magnetic field): F=0→dA=F which assumes the form is exact but that requires trivial topology and however the AB effect is based on the nontrivial topology.

Besides they are assuming the magnetic field to be conservative, when I was always told the magnetic field is solenoidal but not conservative.

 

[tom.stoer]

the 'not really' means that it's the gauge field itself which induces the non-trivial topology, not necessarily the underlying manifold; but as I said this is a matter of taste

one can discuss this based on differential forms, but of course the results do depend on the formulation; I would recommend Nakahara which seems to better accessable than Nash & Sen

 

[Jano L.]

I have a hard time accepting that simple-connectedness has something to do with this. One gets non-zero loop integral of A easily in normal space. For example, imagine a finite cylinder with surface currents, whose field approaches that of a coil. Let the magnetic field inside is B and the cross section S. The symmetry of the current distribution is such that the vector potential in Coulomb gauge

<br /> \oint \mathbf A \cdot d\mathbf l = BS \neq 0 <br />

over any reasonable curve drawn around the cylinder. One gets the same result even with continuous 3D distribution of currents in the thick wires. It does not matter whether the coil is finite or infinite or whether it creates some mathematical division of space into two parts.

 

[tom.stoer]

For a finite cylinder we run into difficulties:

The flux is

\Phi[C] = \oint_C A

Now let's use a pure-gauge A-field

U = e^{-i\chi(x)}
A = U^\dagger \;id\;U = d\chi

where U has winding number n

U = e^{-in\phi};\;\phi\in [0,2\pi];\;n=0,\pm 1,\pm 2,\ldots

The integral is related to this winding number which is a topological invariant; b/c of periodicity of U the integer n cannot be changed continuously (*)

\Phi[C] \sim n

Now assume that the loop C is deformed smoothly and moved continuously through space. Therefore we know that
a) n changes continuously -- but is ruled out b/c of (*)
b) n changes discontinuously -- but is ruled out b/c we allow only for smooth deformations of C
c) n does not change at all
b/c (a) and (b) are rules out we know that no matter how we move C, n must not change.

But with a finite cylinder the loop C could be moved freely which may contradict integer n ≠ 0. Therefore moving C freely is ruled out by topology of the pure gauge A-field.

 

[Jano L.]

Can you please explain what you did above? I am sorry, but I do not see relevance of your calculation. We can calculate the loop integral through any closed contour by using standard parametrization. Deforming the contour into point will result in continuous change of flux from initial value into 0. The contour relevant for calculation is such that the flux is non-zero; apart from that, there is no reason to insist it is integer multiple of something else...

 

[The_Duck]

It doesn't forbid its existence per se, but in a physical experiment you have to justify where it comes from, and usually it is related to a magnetic field, or at least in this case , as usually explained, is. Do you mean that in the AB effect experiment the A-field has nothing to do with the magnetic field?

I think in principle nontrivial A fields need not be associated with B fields (of course, in the case of a solenoid we always get both). Suppose we *actually* lived in R^3 \ (an infinite cylinder). Then we might well find that when we passed charged particles around the defect, we got an Aharonov-Bohm interference effect. This would demonstrate the existence of an A field around the defect. Perhaps someone will correct me on this, but I think this gives a perfectly consistent theory of electrodynamics. Physicists in this universe should include a parameter in their models which they might poetically call "the magnetic flux through the defect," even though no B field can be detected. They would have to determine this parameter by experiment.

 

[TrickyDicky]

I think in principle nontrivial A fields need not be associated with B fields (of course, in the case of a solenoid we always get both). Suppose we *actually* lived in R^3 \ (an infinite cylinder). Then we might well find that when we passed charged particles around the defect, we got an Aharonov-Bohm interference effect. This would demonstrate the existence of an A field around the defect. Perhaps someone will correct me on this, but I think this gives a perfectly consistent theory of electrodynamics. Physicists in this universe should include a parameter in their models which they might poetically call "the magnetic flux through the defect," even though no B field can be detected. They would have to determine this parameter by experiment.

Yes, to me the existence of an A field is not actually in doubt. But I'm finding it paradoxical under the usual assumptions with respect to Maxwell equations, closed line integrals and Stokes theorem dependence on trivial topology and the effect itself dependence on nontrivial one, so far I haven't found a way out.

 

[tom.stoer]

A question: what do you mean by "non-existence of the B-field"; do you mean that B=0? or do you mean that the B-field does not exist mathematically?

 

[TrickyDicky]

A question: what do you mean by "non-existence of the B-field"; do you mean that B=0? or do you mean that the B-field does not exist mathematically?

No, I'm of course referring to the region where B=0, outside the solenoid, otherwise it wouldn't make sense to talk about the A-field either.
But I don't recall saying those words.

 

[strangerep]

My entire point was that the Stokes theorem requires simply-connectednes to hold.

And mine is (partly) that regarding the AB setup as a plane with a point (or small disc) removed (hence not simply connected) is an unphysical idealization. The space of the (physical) experiment does not have a hole. It has a region where curl A is nonzero. The topological arguments are a distracting waste of time.

[...] I'm finding it paradoxical under the usual assumptions with respect to Maxwell equations, closed line integrals and Stokes theorem dependence on trivial topology and the effect itself dependence on nontrivial one, so far I haven't found a way out.

You could try solving the Maxwell equations expressed in terms of just A and a source current j. I.e., for an AB-like current configuration, find A. Then integrate your A solution around a closed loop...

 

[tom.stoer]

Can you please explain what you did above? I am sorry, but I do not see relevance of your calculation.

I'll come back to it later

 

[tom.stoer]

Let's summarize:
- the B-field is neither required nor sufficient to explain the Aharonov-Bohm effect
- the B-field can be regarded as a derived entity, whereas the A-field is fundamental
- this holds not only in QM but already in classical electrodynamics (Maxwell's equiations can be formulated w/o using the B-field)
- it is wrong (although it is often claimed) that the A-field is unphysical or unobservable; the loop integral represents a (gauge invariant) classical observable

... regarding the AB setup as a plane with a point removed is an unphysical idealization. The space of the (physical) experiment does not have a hole. ... The topological arguments are a distracting waste of time.

Not really; I agree that physically the situation must be described using the solenoid rather than the punctured plane; but for the calculation of the phase shift there is no difference; the electron can't distinguish between the "physical" and the "topological" scenario b/c it can't penetrate the solenoid. Therefore the topological consideration is relevant.

 

[PhilDSP]

Don't the mechanics of "spin" enter into consideration here also? A rotating particle containing an extended charge or sub-charge will generate a magnetic dipole protruding from both ends of the rotation axis. But if the rotation is complex, possibly consisting of several rotations about several axes, then magnetic monopoles or quasi-monopoles would exist in the microscopic realm. The fields would no longer be simply-connected.

There seems to be a history, starting from Heaviside, to allow for the possibility of the existence of magnetic monopoles in an initial, primitive invocation of the Maxwell equations for particular problems where those monopoles are removed prior to the finalization of a particular solution to the EM problem. Doing so could be interpreted as invoking SU(n) topology in the beginning and resolving it into U(1), couldn't it?

 

[TrickyDicky]

And mine is (partly) that regarding the AB setup as a plane with a point (or small disc) removed (hence not simply connected) is an unphysical idealization. The space of the (physical) experiment does not have a hole. It has a region where curl A is nonzero. The topological arguments are a distracting waste of time.

Hmm, your opininion contradicts all the texts I've consulted.

You could try solving the Maxwell equations expressed in terms of just A and a source current j. I.e., for an AB-like current configuration, find A. Then integrate your A solution around a closed loop...

Integrating around the closed loop independently of path requires the trivial topology, so only in case you were right and topology was irrelevant would you have a point.

 

[TrickyDicky]

Let's summarize:
- the B-field is neither required nor sufficient to explain the Aharonov-Bohm effect
...

What do you mean is not required, there is no effect without a magnetic flux inside the solenoid, is there?

 

[tom.stoer]

I think it doesn't make sense to repeat everything umpteen times; the B-field is a derived quantity; the A-field is fundamental and sufficient; the A-field is pure gauge locally, so no B-field is assoiciated with it; the B-field is zero outside; the electron can't "look" into the solenoid and therefore does not see any B-field; the magnetic flux is not the same as the magnetic field, but it's a loop-integral over the A-field, the flux around the loop does not require a B-field inside b/c the electron cannot distinguish between the B-field in the solenoid and a pure gauge w/o any B-field ...

As long as you insist on using the B-field you will never be able to understand the AB effect, local gauge theory with global gauge effects and other consequences.

 

[tom.stoer]

Integrating around the closed loop independently of path requires the trivial topology, so only in case you were right and topology was irrelevant would you have a point.

No!

Integrating around closed loops tells you something regarding topology, winding number etc. The integral is NOT path-independent in general but only path-indep. 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.

 

[TrickyDicky]

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?

 

[TrickyDicky]

I think it doesn't make sense to repeat everything umpteen times; the B-field is a derived quantity; the A-field is fundamental and sufficient; the A-field is pure gauge locally, so no B-field is assoiciated with it; the B-field is zero outside; the electron can't "look" into the solenoid and therefore does not see any B-field; the magnetic flux is not the same as the magnetic field, but it's a loop-integral over the A-field, the flux around the loop does not require a B-field inside b/c the electron cannot distinguish between the B-field in the solenoid and a pure gauge w/o any B-field ...

As long as you insist on using the B-field you will never be able to understand the AB effect, local gauge theory with global gauge effects and other consequences.

I agree with all the points you list, that shows to me you are not really getting my point.