I have a very basic question on the Aharonov-Bohm effect.
Why is the curve integral ∮ΓA⋅dr non-zero ? Γ is the "difference" of both paths P1 and P2. If the magnetic field is limited to the interior of the solenoid curlA=0 along the integral path Γ, so I can conclude that I can write A=∇f. A closed curve integral of a gradient function is zero.
I guess it is related with a possible singularity of A in the very center of the solenoid.
Nevertheless if I travel around a point-like source of a gravitational field and compute the integral ∮ΓF⋅dr where F=−∇V(r) the closed curve integral over an conservative force field is certainly zero, whereas V(r) even has a singularity and F consequently too. I would be very grateful for an explanation.
Answer
From curlA=0
To show that (1) does not imply (2) for 3-space with a line removed, we can assume that the line is the z-axis and, consider A={∇arctan2(y,x)(x,y)≠(0,−1)(0,−1,0)(x,y)=(0,−1)
It less trivial (or rather, one first needs to construct some sledgehammers, and then it's a 2-line proof) to show that for 3-space with a point (1) does indeed imply (2).
Update: If you want to understand why a point can be removed but not a line, one can think of it as needing to make a hole that is big enough. Suppose we have two curves γ1 and γ2 and ∫γiA, where curlA=0. If γ1 can be deformed to γ2 one can apply Stokes's theorem to argue that the integrals are equal. Suppose that the space is such that any curve can be deformed into any other curve. Then a standard prescription exists to find f: pick any point x0 and let f(x)=∫γA
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