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Aharonov-Bohm: realizable?

by nonequilibrium
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nonequilibrium
#1
Apr1-12, 02:36 PM
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So the Aharonov-Bohm effect relies on a magnetic field that is only non-zero within a given range. However, is it possible, even in principle (i.e. theoretically), to have such a magnetic field?
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mfb
#2
Apr1-12, 03:31 PM
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It is possible to get a very good approximation.
The effect does not require zero field strength anywhere - the interesting thing is just that it works with zero field strength, too.
nonequilibrium
#3
Apr1-12, 03:34 PM
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The effect does not require zero field strength anywhere - the interesting thing is just that it works with zero field strength, too.
But the surprising element of the effect relies on the field being (exactly) zero outside the given region; if the field is not exactly zero (again: not even theoretically [because obviously experimentally it will never be exactly zero]), the significance of the effect would be marginal, and the effect wouldn't be near as famous as it is today (at least I don't see any reason for this).

Vanadium 50
#4
Apr1-12, 03:57 PM
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Aharonov-Bohm: realizable?

There are no frictionless planes, no massless pulleys, no stretchless ropes, and no regions with exactly zero magnetic field. You shouldn't let this get in the way of your understanding, because there are regions where the friction, mass, elasticity and exact zeroness of the field are unimportant.
nonequilibrium
#5
Apr1-12, 04:01 PM
P: 1,412
Vanadium, if you'll look at my above posts, I emphasized as often as I could "even theoretically". Theoretically there are frictionless planes, etc. But are there theoretically (i.e. consistent with Maxwell's laws, or maybe even with QED if need be) magnetic fields which are only non-zero within a certain boundary?

If the answer is "no", then the Aharonov-Bohm effect seems to be nothing special.

EDIT: also, I wouldn't call this matter unimportant in the context of the Aharonov-Bohm effect; rather, it's the whole shabang.
strangerep
#6
Apr1-12, 08:43 PM
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Quote Quote by mr. vodka View Post
[...], I emphasized as often as I could "even theoretically".
Well, "theoretically" you could have an infinitely long ideal solenoid (of infinitesimal diameter)...
nonequilibrium
#7
Apr1-12, 08:52 PM
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I don't like your first example, as it contains something infinitely big and hence e.g. infinite energy, but I think I do like your second example. I'm still not completely sold about it, as it contains a limiting procedure. But I think we can avoid that: do you think that specifying the following charge and current distribution will give a zero field outside of a torus?
Charge = 0
Current = along the circumference of the circles that make up the torus

I would think so... and that would answer my question. Thank you :)
Creator
#8
Apr1-12, 09:10 PM
P: 547
Quote Quote by mr. vodka View Post
So the Aharonov-Bohm effect relies on a magnetic field that is only non-zero within a given range. However, is it possible, even in principle (i.e. theoretically), to have such a magnetic field?

Of course it is possible.

The AB effect is most easily seen in a magnetic field excluded environment in a superconducting ring. You can't get much greater magnetic field exclusion in a region than that.
I don't exactly understand the nature of your concern...
It is the Magnetic vector potential that affects the phase of electrons, and it is the strength of the vector potential that determines the phase change....

And the AB effect is very significant....in that it accounts for the magnetic quantum flux requirement in a superconductive ring.

...
Vanadium 50
#9
Apr1-12, 09:23 PM
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It is not possible - even theoretically - to have a frictionless plane, a massless pulley, etc. Or an identically zero magnetic field over a region large enough to perform this experiment. Every ideal experiment is realized in some way where something is not ideal. That's not a reason not to learn from them.

In the case in question, B cannot be made identically zero. However, what one can do is make B too small to have a measurable impact on the outcome, so everything observable is driven by the strength of A. Surely that should be enough.


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