Aharonov-Bohm topological explanation

[TrickyDicky]

Are do you simply agree on the locality of the action principle and on the non-locality of the effect?

Sure, did you understand the irrelevance of the locality of the action principle in this context? It is nicely explained in the quote in the post.

Have you understood the difference between nonlocality (which plays an important role here) and "action at a distance" which is irrelevant and confusing in the AB context?

I thought we had already clarified this strictly semantic issue, in your particular view "action at a distance" is restricted to quantum nonlocality, this is not the general view but it's ok, I agreed that quantum nonlocality is not relevant to the AB effect.

 

[tom.stoer]

Sure, ...

fine

I thought we had already clarified this strictly semantic issue, in your particular view "action at a distance" is restricted to quantum nonlocality, this is not the general view but it's ok, I agreed that quantum nonlocality is not relevant to the AB effect.

fine

... did you understand the irrelevance of the locality of the action principle in this context?

It's not irrelevant (that means I don't fully agree with the post). One can formulate (quantum) electrodynamics and quantum mechanics using a gauge fixed formalism; this is not the well-known formulation used in textbooks, but it reduces the gauge theory to physical d.o.f. w/o gauge dependence; in that sense non-locality is introduced in the Hamiltonian, but it's still not subject to any action at a distance phenomenon.

 

[andrien]

${B}_z^\prime = \gamma {B}_z - (\gamma-1){B}_z = B_z$
this holds true even in non-relativistic case.because γ=1 simply in that case,and you get the same result.But still it seems to me that the idea is rather proving the existence of vector potential as there is B,if locality is preserved.

 

[tom.stoer]

${B}_z^\prime = \gamma {B}_z - (\gamma-1){B}_z = B_z$
this holds true even in non-relativistic case.because γ=1 simply in that case,and you get the same result.

Of course; all what I wanted to explain is that you cannot simply transform away the B-field

But still it seems to me that the idea is rather proving the existence of vector potential as there is B,if locality is preserved.

Don't know what you mean by that

 

[andrien]

Don't know what you mean by that

I am just saying that in all accepted physical theories there is no violation of principle of locality.If that is the case with AB effect also,then in the usual experiment on AB effect it is vector potential due to which a complex phase is obtained,so it is better to treat vector potential also as some physical quantity as is magnetic field.

 

[tom.stoer]

I am just saying that in all accepted physical theories there is no violation of principle of locality.

Hm, yes, of course with some caveats
(in QM we know that there is some kind of non-locality as proven by EPR- and Bell-like experiments)

then in the usual experiment on AB effect it is vector potential due to which a complex phase is obtained,...

yes, of course

so it is better to treat vector potential also as some physical quantity as is magnetic field.

yes; a careful analysis in QED shows that the vector potential and the electric field can be treated as the fundamental d.o.f., whereas the B-field is a derived quantity.

 

[andrien]

Hm, yes, of course with some caveats
(in QM we know that there is some kind of non-locality as proven by EPR- and Bell-like experiments)

but bell's theorem does predict something which is not compatible with quantum mechanics(I am not interested in quantum information type thing)

yes; a careful analysis in QED shows that the vector potential and the electric field can be treated as the fundamental d.o.f., whereas the B-field is a derived quantity.

Do you mean A⁰ by electric field(I hope so),in minimal coupling it is only $\phi$ and A(vector potential) appear.

 

[tom.stoer]

Do you mean A⁰ by electric field(I hope so),in minimal coupling it is only $\phi$ and A(vector potential) appear.

No.

In minimal coupling you have $D_\mu = \partial_\mu -ieA_\mu$

with four components. But these are not all physical (b/c for a massless gauge field there are two transversal polarizations). So you have to gauge-fix in order to identify the true dynamical / physical d.o.f. In order to that you must keep in mind what the field strength tensor $F_{\mu\nu}$ tells you (you don't need it of for the Schrödinger equation to work b/c there you may assume that A satisfies the Maxwell equations and is not subject to any coupling to the matter field); but for gauge fixing you have to look at the field strength tensor as well.

Now $F_{\mu\nu}$ is antisymmetric, therefore $\partial_0A_0$ does not exist and $A_0$ is no dynamical d.o.f. but acts as a Lagrange multiplier. So you can gauge fix $A_0 = 0$ eliminating the unphysical d.o.f. but keep its Euler-Lagrange equation (Gauss law). This additional symmetry allows you to fix the residual gauge symmetry using time-indep. gauge trfs respecting the gauge condition $A_0 = A_0^\prime = 0$. Doing that you end up with a "instantaneous interaction" term (which is not relevant here) plus 2*2 dynamical; transversal d.o.f., i.e. $A_\perp,\,E_\perp$. The B-field is calculated as $\nabla \times A_\perp$

In our case of the AB effect it makes sense to chose the axial gauge $A_z = 0$ (or the Coulomb gauge) as second gauge fixing condition (b/c this corresponds to the well-known AB field configuration). Then the physical d.o.f. are $A_\perp = A_i,\,E_\perp = E_i;\;i=1,2$. The electric field is irrelevant in the case of the AB effect in QM, it is only relevant to identify the physical d.o.f., or in the case of full QED

 

[andrien]

what you are saying is definitely correct.I have seen something similar here
http://www.damtp.cam.ac.uk/user/tong/qft/six.pdf
coulomb gauge eliminates the longitudinal polarization.It is non-covariant gauge but still the results one get by applying it in quantum theory of radiation are covariant!But it seems that there is only $A_\perp$ which is relevant and this is the only one from which both electric and magnetic field can be calculated,since $A_0=0$

 

[tom.stoer]

nice reference; one has to be careful whether all gauge fixings are correct in R³ \ R with B=0 b/c one can show that gauge fixing and solving for A° can be implemented via unitary trf's.; but these may be non-trivial on R³ \ R;

anyway, the Hamiltonian to start with is on page 140; we are not interested in the A-field dynamics and "freeze out" the E; in addition we set e² ~ 0 which eliminates the matter self-coupling due to charge density ρ; with B=0 on R³ \ R we get

$$H = -\int_{\mathbb{R}^3} d^3x\,\psi^\dagger\,\gamma^0\,(i\gamma^i\,D_i +m)\,\psi$$

from which we can derive the Dirac equation via

$$\partial_o\,\psi = \frac{\delta H}{\delta (i\psi^\dagger)}$$
$$(-i\gamma^i\,D_i - m)\,\psi = 0$$

with

$$D_i = \partial_i +ie\,A_i$$

Now one could in principle use the Dirac equation to re-derive the AB effect; but I don't expect any fundamental difference

 

[andrien]

Now one could in principle use the Dirac equation to re-derive the AB effect; but I don't expect any fundamental difference

are you talking about the reality of A.from the hamiltonian for free dirac field one can readily get dirac eqn.Just minimal coupling is required for getting interaction with EM field.How can dirac eqn be used to derive AB effect.(schrodinger can also be used then,right!.it does not make difference.)

 

[tom.stoer]

as already said: I don't see any fundamental difference; all I wanted to explain is that the standard approach for the AB effect (non-rel. Schrödinger equation) can be derived from full QED (QED → Dirac eq. → Pauli eq. → non-rel. Schrödinger eq.)

 

[andrien]

yes,sure.Now please can you tell something about the conclusion of this thread in terms of topological consideration.thank you.

 

[tom.stoer]

One conclusion is that an impenetrable solenoid with a given flux can be replaced by a space R³ with a line R removed, i.e. R³ \ R on which a non-vanishing A-field exists. This A-field is "pure gauge" locally, i.e. A ~ U* dU with dA = 0, but not globally b/c the line R prevents us from constructing a global gauge function U' to gauge away this A → A' = 0 globally.

This can be extended to more than one flux line R, R', R'', ..., i.e. several lines removed: R³ \ R \R' \ R'' ...

Then there is a non-local flux observable ∫A which corresponds to a topological quantity, the cohomology of the one-form A of the fibre bundle of the U(1) gauge group over the base manifold R³ \ R. This ∫A depends on A and on the the path C via an integer winding number w[C] of the path C w.r.t. the line R.

This can be expressed in terms of the gauge function U = exp(iχ) from which A can be constructed as U* dU, therefore already this U carries the full information on the topological structure.

Note that in principle it's not ∫A which is observed directly but only the fractional part of the flux mod 2π/q (where q is the value of the electric charge of the particle). This is due to the fact that we study the compact gauge group U(1) with U ~ exp(iχ), not only the gauge function χ. This is related to the problem of the logarithm χ of U on R³ \ R. There are continuous single-valued gauge functions U where χ jumps by 2πw along a path C for winding number w[C], the simplest one is χ ~ θ where θ is the cylindrical angle coordinate w.r.t. the z-axis.

So this results in a quite complex structure of the gauge group defined over R³ \ R; one may see this by applying the following gauge fixings:
1) gauging away A° = 0
2) gauging away A³ = 0 (we assume that the line R corresponds to the z-axis) with a time-indep. gauge function respecting A° = 0
(up to now this is standard)
3) now there is a residual gauge symmetry with t- and z-indep. gauge transformations U respecting A° = 0 and A³ = 0 such that the integral part of the flux is still unobservable, i.e. related to a global residual gauge symmetry

So the Aharonov-Bohm effect is related to a physical observable

$$\phi_C = \oint_C A \;\text{mod}\; 2\pi$$

which is the "fractional part of the flux".

This is defined in terms of a pure gauge vector potential

$$A = U^\dagger i\nabla U$$

which is sensitive to the topology of the base manifold, i.e. we have

[tex]\phi_C = \phi_C[/tex]

So neither the base manifold nor the gauge group alone are sufficient top describe the Aharonov-Bohm effect; what really matters is the topology of the fibre bundle F which locally (i.e. for each point of the base manifold M) looks like a direct product F ~ M * U(1); but a non-trivial structure of M, i.e. R³ \ R induces a non-trivial structure of F, and we finally arrive at the conclusion that the physical observable is sensitive to

$$\phi_C = \phi_C[F]$$

 

[tom.stoer]

A final remark: this consideration can be generalized to arbitrary base manifolds (not necessarily 3-dim., complicated singularity structure, ...), arbitrary gauge groups and arbitary compact integration sub-manifolds (complicated knots, higher dimensional submanifolds, ...).

Doing that one is no longer doing physics but rather differential geometry and algebraic topology, i.e. one is studying topological structures of manifolds using fibre bundles constructed over these manifolds (or one is studying knot invariants expressed in terms of gauge field configurations in R³).

 

[TrickyDicky]

One conclusion is that an impenetrable solenoid with a given flux can be replaced by a space R³ with a line R removed, i.e. R³ \ R on which a non-vanishing A-field exists. This A-field is "pure gauge" locally, i.e. A ~ U* dU with dA = 0, but not globally b/c the line R prevents us from constructing a global gauge function U' to gauge away this A → A' = 0 globally.

This can be extended to more than one flux line R, R', R'', ..., i.e. several lines removed: R³ \ R \R' \ R'' ...

Then there is a non-local flux observable ∫A which corresponds to a topological quantity, the cohomology of the one-form A of the fibre bundle of the U(1) gauge group over the base manifold R³ \ R. This ∫A depends on A and on the the path C via an integer winding number w[C] of the path C w.r.t. the line R.

This can be expressed in terms of the gauge function U = exp(iχ) from which A can be constructed as U* dU, therefore already this U carries the full information on the topological structure.

Note that in principle it's not ∫A which is observed directly but only the fractional part of the flux mod 2π/q (where q is the value of the electric charge of the particle). This is due to the fact that we study the compact gauge group U(1) with U ~ exp(iχ), not only the gauge function χ. This is related to the problem of the logarithm χ of U on R³ \ R. There are continuous single-valued gauge functions U where χ jumps by 2πw along a path C for winding number w[C], the simplest one is χ ~ θ where θ is the cylindrical angle coordinate w.r.t. the z-axis.

So this results in a quite complex structure of the gauge group defined over R³ \ R; one may see this by applying the following gauge fixings:
1) gauging away A° = 0
2) gauging away A³ = 0 (we assume that the line R corresponds to the z-axis) with a time-indep. gauge function respecting A° = 0
(up to now this is standard)
3) now there is a residual gauge symmetry with t- and z-indep. gauge transformations U respecting A° = 0 and A³ = 0 such that the integral part of the flux is still unobservable, i.e. related to a global residual gauge symmetry

So the Aharonov-Bohm effect is related to a physical observable

$$\phi_C = \oint_C A \;\text{mod}\; 2\pi$$

which is the "fractional part of the flux".

This is defined in terms of a pure gauge vector potential

$$A = U^\dagger i\nabla U$$

which is sensitive to the topology of the base manifold, i.e. we have

[tex]\phi_C = \phi_C[/tex]

So neither the base manifold nor the gauge group alone are sufficient top describe the Aharonov-Bohm effect; what really matters is the topology of the fibre bundle F which locally (i.e. for each point of the base manifold M) looks like a direct product F ~ M * U(1); but a non-trivial structure of M, i.e. R³ \ R induces a non-trivial structure of F, and we finally arrive at the conclusion that the physical observable is sensitive to

$$\phi_C = \phi_C[F]$$

This is all fine, but when we restrict ourselves to 3D with time as parameter NRQM/classical ED(prior to relativistic formulation) space the following are relevant points:
-In R^3/R the classical Kelvin-Stokes theorem is not valid , so no observable ( fractional part of) flux in this topology.
-Given this we are again stuck with an A field that has gauge freedom, and therefore can't be a physical observable.
-In a purely NRQM or classical ED interpretation(without using the relativistic covariant formulation), with only three dimensional space and time as parameter, local action is not respected and therefore there is "action at a distance".
However:
-In the context of QED which is the one your explanation is valid in, local action is respected and in that sense there is no "action at a distance", and yet we still have nonlocality since there is no separability, nor observability which are required together with local action to consider the effect local.

The conclusion is that the A-field interpretation of the effect respects local action principle but that is because the Lagrangian density that integrates over spacetime is used, since we are calling A-field to the 4-potential while I was actually restricting to the 3-vector potential of classical Maxwellian ED without covariant formulation, R^3 and the Lagrangian with the integral over time.

Finally, as commented in post #120, the QFT-QED interpretation itself is not without severe ontological problems, for instance:"First there is the general problem faced by any interpretation of a quantum field theory. In this case this involves understanding the relation between the quantized electron field and the electrons which are its quanta on the one hand, and the relation between the quantized and classical electromagnetic fields on the other. Without some account of the ontology of quantum fields one can give no description of either electromagnetism outside the solenoid or the passage of the electrons through the apparatus, still less a separable description. And one is therefore in no position to show that interactions between these two processes conform to Local Action", but that belongs more to the philosophy of science.

 

[TrickyDicky]

So neither the base manifold nor the gauge group alone are sufficient top describe the Aharonov-Bohm effect; what really matters is the topology of the fibre bundle F which locally (i.e. for each point of the base manifold M) looks like a direct product F ~ M * U(1); but a non-trivial structure of M, i.e. R³ \ R induces a non-trivial structure of F, and we finally arrive at the conclusion that the physical observable is sensitive to

$$\phi_C = \phi_C[F]$$

Right, we have here not just the manifold non trivial topology but the gauge group non trivial topology. So we need for this interpretation additional structure to the base manifold, that of the principle bundle with its non trivial gauge potential, and this additional mathematical structure is not governed by the field equations which leads to indeterminism of the AB effect.

 

[tom.stoer]

DickyTricky, I just summarized some well-known topological consideration andrien asked for. Everything is standard textbook, well estanblished, known for yearsm and there is no reason for any further discussion.

Anyway ...

-In R^3/R the classical Kelvin-Stokes theorem is not valid , so no observable ( fractional part of) flux in this topology.

Please try to understand my reasoning (or if you like read some standard textbooks) to learn why the fractional part of the flux is an observable; and forget about the Stokes theorem; you need it to convert the integral from a line to a surface integral; but I don't need this theorem in this context b/c I do not use a surface integral; the flux is encoded in the loop integral over A !

-Given this we are again stuck with an A field that has gauge freedom, and therefore can't be a physical observable.

I never said that A is an observable; the loop integral mod 2π is an observable !

-In a purely NRQM or classical ED interpretation(without using the relativistic covariant formulation), with only three dimensional space and time as parameter, local action is not respected and therefore there is "action at a distance".

There is no physical result which depends on the formulation (using 4-vectors or not); ED is always covariant, even if you don't use 4-vectors; the action is local, and there is no action at a distance.

-In the context of QED which is the one your explanation is valid in ...

My explanation (all explanations on the AB effect) is valid in non-rel. QM; I never used QED, I never quantized the el.-mag. field.

... since we are calling A-field to the 4-potential while I was actually restricting to the 3-vector potential of classical Maxwellian ED without covariant formulation, R^3 and the Lagrangian with the integral over time.

As I already said; the explanation and the effect does in no way depend on the formulation; the formulation of electrodynamics and its gauge symmetry is more transparent using 4-vectors, but physics is not affected, of course (classical electrodynamics didn't change after introduction of 4-vectors; the Maxwell equations today predict the same physics as they did 150 years ago)

"First there is the general problem faced by any interpretation of a quantum field theory. In this case this involves understanding the relation between the quantized electron field and the electrons which are its quanta on the one hand, and the relation between the quantized and classical electromagnetic fields on the other. Without some account of the ontology of quantum fields one can give no description of either electromagnetism outside the solenoid or the passage of the electrons through the apparatus, still less a separable description. And one is therefore in no position to show that interactions between these two processes conform to Local Action", but that belongs more to the philosophy of science.

Sorry, but as long we do not agree on basic problems in quantum mechanics it seems too early to try to discuss ontological problems in quantum field theory.

 

[andrien]

I just summarized some well-known topological consideration andrien asked for. Everything is standard textbook, well established, known for years and there is no reason for any further discussion.

Book?which book.I have never seen any book describing AB effect in terms of topology of fiber bundle.

 

[tom.stoer]

I think Nakahara has a nive chapter on the AB effect; he discusses fibre bundles, perhaps w/o explicitly mentioning the AB effect, but the relation should not be too difficult. Nash and Sen have no chapter on the AB effect afaik, but they provide the necessary topological tools as well. I never studied Frankel's book but from the contents I see that he discusses the AB effect in a chapter on vector bundles.

Edit:

Perhaps this paper is interesting: http://128.84.158.119/abs/0909.0370v2

Edit 2:

DickyTricky, that was a comment on your way of questioning the explanations; they are not mine, they are standard in mathematical physics; and they are even used by mathematicians themselves. Of course my explanations can be unclear to you, and other texts may be better formulated (pedagogically). But it is really frustrating that you are questioning arguments I have never put forward; two examples: you always argue against the Stokes theorem which is never used in the topological context; you always argue against A being an observable which I never claimed. In order to make progress you should focus on facts, not on guesses

 

[TrickyDicky]

DickyTricky, that was a comment on your way of questioning the explanations; they are not mine, they are standard in mathematical physics; and they are even used by mathematicians themselves. Of course my explanations can be unclear to you, and other texts may be better formulated (pedagogically). But it is really frustrating that you are questioning arguments I have never put forward; two examples: you always argue against the Stokes theorem which is never used in the topological context; you always argue against A being an observable which I never claimed. In order to make progress you should focus on facts, not on guesses

Sorry if I give that impression, part of it has to do with my struggling with english (not to mention with math and physics ;-)).
So yes I know you are not claiming A is an observable, and no, my last post was not arguing against anything in your post nor saying the standard explanation is plain wrong or anything like that.
Wrt the Stokes theorem which to save further confusion I should clarify I'm not referring to the general theorem but to the classical Stokes-Kelvin one used in vector calculus in the Euclidean 3D setting, this one requires the topological condition of simply connected space, I'm just making that distinction.
By the way you claim you are not mentioning QFT but I associate loop integrals with Feynman diagrams.
You are right that a particular formulation should not be of relevance to the physics, but I'm actually referring to the difference between considering a 3 dim. space vs a 4 dim. spacetime, I hope you admit this makes a difference, you just have to take a look to the last 100 years in physics.

 

[tom.stoer]

Thanks, that clarifies a lot!

So yes I know you are not claiming A is an observable, and no, my last post was not arguing against anything in your post nor saying the standard explanation is plain wrong or anything like that.

OK, fine, then we agree

Wrt the Stokes theorem which to save further confusion I should clarify I'm not referring to the general theorem but to the classical Stokes-Kelvin one used in vector calculus in the Euclidean 3D setting, this one requires the topological condition of simply connected space, I'm just making that distinction.

Of course, I understand. I only wanted to make clear that nothing follows from this theorem b/c I don't use it.

By the way you claim you are not mentioning QFT but I associate loop integrals with Feynman diagrams.

OK, just to clarify that: I called the line integral $\oint A = \int_C A$ using a closed loop C a loop integral; perhaps that was misleading; it's nothing else but a standard line integral evaluated for a classical field A; sorry for the confusion!

... but I'm actually referring to the difference between considering a 3 dim. space vs a 4 dim. spacetime, I hope you admit this makes a difference, you just have to take a look to the last 100 years in physics.

No, it doesn't make any physical difference. The formulation is different, the math looks different, but the results are identical; whether you use the lengthy Maxwell equations or whether you simply write dj = 0, d*F = 0 makes no physical difference. You can mix all these formulations w/o running into trouble.

 

[andrien]

I think Nakahara has a nive chapter on the AB effect; he discusses fibre bundles, perhaps w/o explicitly mentioning the AB effect, but the relation should not be too difficult. Nash and Sen have no chapter on the AB effect afaik, but they provide the necessary topological tools as well. I never studied Frankel's book but from the contents I see that he discusses the AB effect in a chapter on vector bundles.

ohh,thanks again.Got all of them.

 

[TrickyDicky]

I only wanted to make clear that nothing follows from this theorem b/c I don't use it.

But how do you relate the magnetic flux with the line closed integral of A, then?
By using the gauge function U, right?

No, it doesn't make any physical difference. The formulation is different, the math looks different, but the results are identical; whether you use the lengthy Maxwell equations or whether you simply write dj = 0, d*F = 0 makes no physical difference. You can mix all these formulations w/o running into trouble.

I'm not talking about the formulation only, but I agree the physics happens to come out the same because the topological properties of infinite dimensional Hilbert spaces in the NRQM case allow it (I'm referring to the unitary group U(1)).

 

[TrickyDicky]

Just to be sure whenever you refer to A, you mean the EM 4-potential or the magnetic 3-vector potential?

 

[TrickyDicky]

Tom, the paper you linked in #160 is interesting even if it adds further complexity, some of it helps clarify some of the things I was arguing and you considered wrong, for instance:
"Many authors consider infinitely thin solenoid and the Euclidean space R3 with removed
axis corresponding to the solenoid as the base M. In this case, the base M is not
simply connected, and the Aharonov–Bohm effect could be called topological. As was
shown earlier, this is not necessary. It is sufficient to assume that the magnetic field differs
from zero only in the bounded region on the plane, and the integration contour does
not cross this region. Moreover, if we assume that the base M is the Euclidean space R3
with removed axis then the Stokes formulae is not applicable, and hence the expression
for the Aharonov–Bohm phase through the magnetic field flux (26) requires additional
assumptions. So the Aharonov–Bohm effect must be considered as geometrical rather
then topological."

 

[tom.stoer]

But how do you relate the magnetic flux with the line closed integral of A, then?
By using the gauge function U, right?

The loop integral $\oint A$ is the flux; w/o considering a B-field I do not need anything else; the gauge function U is only oused to construct the A-field.

I'm not talking about the formulation only, but I agree the physics happens to come out the same because the topological properties of infinite dimensional Hilbert spaces in the NRQM case allow it (I'm referring to the unitary group U(1)).

I never talked about the "topological properties of the Hilbert space"; I am talking about topology of the base manifold or about the fiber bundle. There is no need to refer to the Hilbert space structure in order to understand the AB effect

Just to be sure whenever you refer to A, you mean the EM 4-potential or the magnetic 3-vector potential?

In principle yes, but it doesn't matter b/c

$$\int_C dx^\mu\,A_\mu = \int_C dx^i A_i$$

b/c C has no timelike direction, i.e. x° is constant along C, i.e. dx°=0 on C (in addition one can fix the A°=0 gauge)

 

[TrickyDicky]

The loop integral $\oint A$ is the flux; w/o considering a B-field I do not need anything else;

You do realize that in order to make that identification between the integral and the flux you are using the Stokes theorem, don't you?

 

[tom.stoer]

No, I am not using the Stokes theorem; at least not in the standard way.

The idea is the following: usually the magnetic flux through a surface S is related to the line integral along a curve C (which is the boundary of S). Now b/c the A-field is the fundamental entity, from now on I never use the B-field through S but only the A-field through C.

So now I call $\oint A$ the magnetic flux; I never use B, I don't care about it; it's a derived quantity, and I can explain all physical effects w/o ever using B. There are even effects (the Aharanov-Bohm effect ;-) where the explanation via B is cumbersome or not sufficient and where I should use A instead.

So instead of using the surface integral $\int_S B$ to define the magnetic flux, and then relating it to the line integral along C via Stokes' theorem, I would argue the other way around: I use the line integral $\oint_C A$ to define the magnetic flux! I don't need any B-field which is
a) not the fundamental entity
b) not sufficient to explain the AB-effect
c) requires the applicability of Stokes's theorem

So from principle of maximum simplicity it follows that we need not (should not) care about B, S and Stokes theorem; all what matters is A and C.

I don't known whether you know the philosopher Ludwig Wittgenstein and Wittgenstein's ladder from Tractatus Logico-Philosophicus:

"My propositions serve as elucidations in the following way: anyone who understands me eventually recognizes them as nonsensical, when he has used them - as steps - to climb beyond them. He must, so to speak, throw away the ladder after he has climbed up it"

Here the ladder is B, S and Stokes' theorem ;-)

 

[TrickyDicky]

I see, nice post, I've always admired Wittgenstein(BTW thanks for bearing with my questions).
Do you agree that starting with the loop integral of A to get the flux is not allowed unless you consider the vector potential physically significant to begin with, but that is something that classically was not considered possible, and that is only considered the right explanation from a prediction in QM and the empirical proof of the AB effect.

 

[tom.stoer]

Do you agree that starting with the loop integral of A to get the flux is not allowed unless you consider the vector potential physically significant to begin with, ...

yes, one must attribute a fundamental, physical role to the vector potential; otherwise this reasoning would not be forbidden, but physically unreasonable (*)

... but that is something that classically was not considered possible, ...

no, it is possible classically, but it's not necessary in classical electrodynamics; there everything, including interaction with charged particles, can be described in terms of E and B; A is only an auxiliary variable;

(*) one could even say that using A introduces gauge invariance and unphysical redundancy which has to be removed via gauge fixing, so in some sense using A instead of E and B could even be seen as pointless

... and that is only considered the right explanation from a prediction in QM and the empirical proof of the AB effect.

yes, it's QM which indicates that one should use A instead of B; $\oint_C A$ is a physical, gauge invariant observable in classical electrodynamics; but it's QM which tells us that we should better use $\oint_C A$ instead od $\int_S B$; in classical electrodynamics based on (*) we would prefer $\int_S B$

 

[TrickyDicky]

Ok, thanks.