
#1
Sep2313, 08:24 AM

P: 4

whats a good way to picture the vector potental A in terms of B & like what exactly is A & how does it even exist outside a torus where B & etc =0
for example its easy to see the electric potential uses the electric field E like E*ds & its quite obvious, wheras how does A not even contain the B field also why is A sometimes said to not even exist or is just a paper shortcut when it actualy seems to work or exist in some way. thanks 



#2
Sep2313, 11:42 AM

P: 419

Since the curl of the vector potential A is equal to the magnetic field B, a good way to think of it is that A circulates around any point where B is nonzeroits net circulation around a point gives the B field at that point, according to the righthand rule. It is important to remember though that you can always write down different A's to produce the same B fieldthis is called choosing a gauge. For example, a uniform B field in the z direction could be represented by any of the following:
A = By i A = Bx j A = By/2 i + Bx/2 j where i is the unit vector in the x direction, and j is the unit vector in the y direction, and B is the magnitude of B. If you plot these, you will see that they all look quite different, but they all circulate around in a similar fashion. In classical E&M, the B field is the measurable quantity, so A is said to just be a mathematical convenience. However, in quantum physics, particles can be affected by magnetism even if they never pass through a region of nonzero Binstead they directly interact with A. A good example is the AharanovBohm effect: http://en.wikipedia.org/wiki/AharanovBohm_effect 



#3
Sep2313, 11:45 AM

C. Spirit
Sci Advisor
Thanks
P: 4,938

What do you mean the vector potential ##A## isn't given in terms of the magnetic field ##B##? ##\nabla \times A = B## so you can picture it in terms of the usual geometric interpretation of the curl (think of the vorticity of velocity fields of fluids). The reason classically that ##A## is said to simply be a purely mathematical field (and not a physical field) is because it is not a gauge invariant quantity. I can take ##A \rightarrow A + \nabla \varphi## and I will still get the same physical magnetic field ##B## i.e. ##\nabla \times (A + \nabla \varphi) =\nabla \times A##.




#4
Sep2413, 12:05 AM

P: 148

how to picture the magnetic vector potental A
I've found it helpful to look at the vector potential in the Lorenz gauge  where each component of the vector potential acts like an independent scalar potential for the corresponding current component...so you can imagine each infinitesimal currentelement in the <x, y, z> direction as a source for a corresponding 1/r A field whose vector points in the same <x, y, z> direction. What you lose, though, is the ability to see the direction of the Lorentz force by just comparing the directions of two vectors at a single point.



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