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How to picture the magnetic vector potental A

  1. Sep 23, 2013 #1
    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. jcsd
  3. Sep 23, 2013 #2
    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 nonzero--its net circulation around a point gives the B field at that point, according to the right-hand rule. It is important to remember though that you can always write down different A's to produce the same B field--this 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 B--instead they directly interact with A. A good example is the Aharanov-Bohm effect: http://en.wikipedia.org/wiki/Aharanov-Bohm_effect
  4. Sep 23, 2013 #3


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    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##.
  5. Sep 24, 2013 #4
    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 current-element 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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