# How to picture the magnetic vector potental A

1. Sep 23, 2013

### si22

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. Sep 23, 2013

### Jolb

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

3. Sep 23, 2013

### WannabeNewton

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. Sep 24, 2013

### jjustinn

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.