# Aharonov-Bohm topological explanation

by TrickyDicky
Tags: aharonovbohm, explanation, topological
 P: 1,020 but the cyclic integral of A can be made zero in a suitably chosen frame in which there is no magnetic field,because it is only the curl which is important.In that case it is ∅ which appears because of the simple action principle of feynman in which $e^(iS)$,where $S=∫Ldt$,and in it appears -i∫∅dt,which is a time integral.(physics must be same for two inertial observers)
P: 5,464
 Quote by andrien but the cyclic integral of A can be made zero in a suitably chosen frame in which there is no magnetic field,because it is only the curl which is important.
To which integral are you referring?

The scenario w/o B-field and R³/R base manifold we already discussed in detail; the Stokes' theorem fails due to topology, so one must not rewrite the line integral to a surface integral over a (vanishing) curl;

The physical scenario with B-field and trivial topology plus Lorentz transformation sending B to zero may be interesting; I think the Schrödinger equation fails b/c it is not Lorentz covariant so one must take the Dirac equation plus relativistic effects into account.
P: 1,020
 The physical scenario with B-field and trivial topology plus Lorentz transformation sending B to zero may be interesting; I think the Schrödinger equation fails b/c it is not Lorentz covariant so one must take the Dirac equation plus relativistic effects into account.
How it fails?How will dirac eqn helps here.I don't know how relativistic effects are really of concern here.
P: 5,464
 Quote by andrien How it fails?How will dirac eqn helps here.I don't know how relativistic effects are really of concern here.
It's simple: you start with a non-zero B-field inside the solenoid in the lab frame. Now you apply the Lorentz transformation:
- this changes the 4-current density as source of the B-field
- it changes the B-field (for a specific trf. it sends the B-field to zero)
- it creates an E-field
- in case you study the A-field only you have to calculate how the trf. affects the A-field.

And of course the trf. affects the wave function!

But how? this is undefined b/c for the wave function in non-rel. QM you cannot apply the Lorentz trf.; you have to use a fully relativistic description to analyze the effect of a Lorenzt trf. on the wave function and the interference pattern. Otherwise the calculation incomplete.

EDIT:

Looking at the R³/R case w/o B-field and with the A-field as defined above one observes that an Lorentz transformation in z-direction does not alter the A-field (with A°=0, and spatial part perpendicular to z). In this case it's rather trivial that the interference pattern related to a line integral in the xy-plane doesn't change when the whole setup is boosted in z-direction.
 P: 1,020 but even in non-relativistic case,it is possible to eliminate B field. schrodinger wavefunction DOES GET affected because of vector potential as ψ=ψ0exp(ie∫A.ds),and in absence of magnetic field the line integral does vanish and the contribution will come from ∅.
P: 5,464
 Quote by andrien but even in non-relativistic case,it is possible to eliminate B field.
How?

 Quote by andrien schrodinger wavefunction DOES GET affected because of vector potential as ψ=ψ0exp(ie∫A.ds)
Yes, we all know this.

 Quote by andrien and in absence of magnetic field the line integral does vanish
No, why? The line integral is over A, not over B, so it does not vanish.

 Quote by andrien and the contribution will come from ∅.
Sorry, what is ∅? (a problem with my browser?)
 P: 1,020 if we assume validity of stokes theorem,(no topology),it does vanish.∅ is simply scalar potential(A0).i am just thinking with something like a charge in motion,and we go to it's rest frame and we can eliminate B.it does not depend on non-relativistic motion or relativistic one.Is this not possible to do?
 Sci Advisor P: 5,464 I tried to find some references explaining the topological and gauge theoretic aspects of the Aharonov-Bohm effect: here's a nice article paying attention to single-valued exponentials with multi-valued gauge transforms and cohomologies: http://bolvan.ph.utexas.edu/~vadim/Classes/11f/abm.pdf
 P: 1,020 is relativity really important here?The reference seems just a copy of sakurai,to which I have already seen.
P: 5,464
 Quote by andrien is relativity really important here?
Yes, it becomes relevant once you want to use it to transform away the B-field (that's your idea, not Sakurai's, so you have to deal with it)

 Quote by andrien i am just thinking with something like a charge in motion,and we go to it's rest frame and we can eliminate B.it does not depend on non-relativistic motion or relativistic one.
There is no single moving charge in the case of the solenoid.

Let's start with the A-field outside the solenoid. It's pure gauge, so it could be gauged away locally (not globally!!) and it defines vanishing B-field (and E-field of course). An arbitrary Lorentz trf. sends a vanishing el.-mag. field to a vanishing el.-mag. field b/c the Lorentz trf. is linear in E and B. Therefore the Lorentz trf. does not create E- or B-fields "from nothing". The transformed A-field is of course non-zero but still pure gauge (again this is related to the fact that both LOrentz trf. and gauge trf. are linear).

Now let's look at the A-field inside the solenoid and use Lorentz transformations for electromagnetic fields: http://en.wikipedia.org/wiki/Classic...ial_relativity Starting with constant B-field in z-direction and vanishing E-field the formulas simplify to

$${E}^\prime = \gamma {v} \times {B}$$
$${B}^\prime = \gamma {B} - (\gamma-1)({B}{e}_v){e}_v$$

By inspection you see that a pure z-Boost cannot eliminate

$${B}_z^\prime = \gamma {B}_z - (\gamma-1){B}_z = B_z$$

In addition any boost with xy-components produces an E-field, therefore the AB-effect would be (partially) due to a non-vanishing E-field in the boosted frame!

Forget about transforming away B; it does not help.
P: 3,060
 Quote by tom.stoer But at the same time the underlying theory i.e. the classical electromagnetism and the interaction of the A-field with matter fields in the Schrödinger or the relativistic Dirac Lagrangian are interactions defined via a local action principle. So whereas $$\oint_C A$$ is a non-local entity (posts #16, #45, #77), the underlying Lagrangian $$\mathcal{L} \sim \bar{\psi}\gamma^\mu D_\mu \psi$$ for a Dirac field using the covariant derivative D with a classical A-field is local.

See #119.
P: 5,464
 Quote by TrickyDicky See #119.
so you agree?
 P: 3,060 No, sorry I meant the quote in #120
P: 5,464
 Quote by TrickyDicky No, sorry I meant the quote in #120
Do you want me to read the entire post again and to guess what you want me to say by referring to it? or do you want to stress some important ideas and consequences?

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

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?
P: 3,060
 Quote by tom.stoer 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.
P: 5,464
 Quote by TrickyDicky Sure, ...
fine

 Quote by TrickyDicky 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

 Quote by TrickyDicky ... 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 text books, 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.
 P: 1,020 ${B}_z^\prime = \gamma {B}_z - (\gamma-1){B}_z = B_z$ 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.
 Quote by 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.