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AharonovBohm topological explanation 
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#127
Dec1012, 03:27 AM

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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 [itex]e^(iS)[/itex],where [itex]S=∫Ldt[/itex],and in it appears i∫∅dt,which is a time integral.(physics must be same for two inertial observers)



#128
Dec1012, 03:41 AM

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The scenario w/o Bfield 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 Bfield 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. 


#129
Dec1012, 03:55 AM

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#130
Dec1012, 04:08 AM

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 this changes the 4current density as source of the Bfield  it changes the Bfield (for a specific trf. it sends the Bfield to zero)  it creates an Efield  in case you study the Afield only you have to calculate how the trf. affects the Afield. And of course the trf. affects the wave function! But how? this is undefined b/c for the wave function in nonrel. 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 Bfield and with the Afield as defined above one observes that an Lorentz transformation in zdirection does not alter the Afield (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 xyplane doesn't change when the whole setup is boosted in zdirection. 


#131
Dec1012, 06:10 AM

P: 1,020

but even in nonrelativistic case,it is possible to eliminate B field. schrodinger wavefunction DOES GET affected because of vector potential as
ψ=ψ_{0}exp^{(ie∫A.ds)},and in absence of magnetic field the line integral does vanish and the contribution will come from ∅. 


#132
Dec1012, 06:20 AM

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#133
Dec1012, 06:33 AM

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if we assume validity of stokes theorem,(no topology),it does vanish.∅ is simply scalar potential(A^{0}).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 nonrelativistic motion or relativistic one.Is this not possible to do?



#134
Dec1012, 06:40 AM

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I tried to find some references explaining the topological and gauge theoretic aspects of the AharonovBohm effect: here's a nice article paying attention to singlevalued exponentials with multivalued gauge transforms and cohomologies:
http://bolvan.ph.utexas.edu/~vadim/Classes/11f/abm.pdf 


#135
Dec1012, 06:44 AM

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is relativity really important here?The reference seems just a copy of sakurai,to which I have already seen.



#136
Dec1012, 07:22 AM

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Let's start with the Afield outside the solenoid. It's pure gauge, so it could be gauged away locally (not globally!!) and it defines vanishing Bfield (and Efield 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 Bfields "from nothing". The transformed Afield is of course nonzero 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 Afield inside the solenoid and use Lorentz transformations for electromagnetic fields: http://en.wikipedia.org/wiki/Classic...ial_relativity Starting with constant Bfield in zdirection and vanishing Efield the formulas simplify to [tex]{E}^\prime = \gamma {v} \times {B}[/tex] [tex]{B}^\prime = \gamma {B}  (\gamma1)({B}{e}_v){e}_v[/tex] By inspection you see that a pure zBoost cannot eliminate [tex]{B}_z^\prime = \gamma {B}_z  (\gamma1){B}_z = B_z[/tex] In addition any boost with xycomponents produces an Efield, therefore the ABeffect would be (partially) due to a nonvanishing Efield in the boosted frame! Forget about transforming away B; it does not help. 


#137
Dec1012, 09:16 AM

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See #119. 


#138
Dec1012, 09:32 AM

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#139
Dec1012, 10:24 AM

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No, sorry I meant the quote in #120



#140
Dec1012, 10:48 AM

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Are do you simply agree on the locality of the action principle and on the nonlocality 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? 


#141
Dec1012, 01:42 PM

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#142
Dec1012, 03:29 PM

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#143
Dec1112, 01:03 AM

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[itex]{B}_z^\prime = \gamma {B}_z  (\gamma1){B}_z = B_z[/itex]
this holds true even in nonrelativistic 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. 


#144
Dec1112, 01:09 AM

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