Momentum of stationary electron in a curl free vector potential
The essence of this question is simplicity itself: There is an electron in a curl-free $\vec{A}$ field. The electron is stationary so its m$\vec{v}$ momentum is 0. However, it has "momentum" from the term q$\vec{A}$ of the equation $\vec{p}$ = m$\vec{v}$ + q$\vec{A}$. Note that the usual notion that $\vec{A}$ is arbitrary/nonphysical seems to be contradicted by the empirical validation of the existence of the Aharonov-Bohm Effect. Also there is, a time-domain Ahronov-Bohm Effect reported by <PERSON>, <PERSON> et al in "Detection of the Time-dependent Electromagnetic Potential at 1.3 GHz" equation (20). Also consider <PERSON>, N.J. "On the Fields of a Torus and the Observability of the Vector Potential" (7.13) formulating Maxwell's Equations without gauge transformations.
What is the intuitive description of this "momentum"?
Note further that the aforelinked paper by <PERSON> specifically states that for the case in question (a single reference frame involving a torus, electron and observer at rest with respect to each other) Maxwell's Equations may be formulated without gauge transformations. This gauge invariant formulation, while not being <PERSON> invariant, is, nevertheless, convenient for analysis in our single inertial frame. 346
Can anyone now provide the answer to my question without invoking the "arbitrary nature of gauge choice"? Here is the relevant passage from <PERSON>:
Any further complication of this question is made necessary only because some answers attempt to attribute this q$\vec{A}$ "momentum" to the electron's m$\vec{v}$ momentum being somehow converted to or stored as "potential" momentum in the $\vec{A}$ field. That is equivalent to saying it is not possible to position the electron in the $\vec{A}$ field without contributing to its source current. Really? Please do not allow the following thought experiment (intending to demonstrate the positioning of an electron in an $\vec{A}$ field without contributing to its source current) to cloud the simplicity of the original question. If you are asserting that it impossible to position an electron in an $\vec{A}$ field without contributing to the $\vec{A}$ field's source current, please explain how the following process does so:
An electron is approaching along the torus's axis of symmetry. There are exactly 2 directions in which it can make this approach (there being only 2 directions of travel on a line). The electron is slowed to a stop. The way the electron is slowed to a stop is to place a static charge on the torus that goes to zero when the electron stops. One may posit that this, somehow, increases the current sheet $\vec{J}$ but one may not make that argument for both directions of approach. Hence there is at least one scenario in which the electron can come to a rest in the $\vec{A}$ field without contributing to $\vec{J}$ (hence without somehow "storing" its m$\vec{v}$ as q$\vec{A}$ "potential" momentum).
Now, back to the original, simple, question:
One way of intuitively explaining what q$\vec{A}$ means would be to describe the system in two states:
1) The poloidal current sheet, $\vec{J}$, is zero hence $\vec{J}$=$\vec{A}$=q$\vec{A}$=0
2) The poloidal current sheet, $\vec{J}$, is non-zero hence q$\vec{A}$ is non-zero.
How does the system's behavior differ in these two cases since its property given by q$\vec{A}$ differs in these two cases? If this momentum is not conserved then in why may we say this system embodies conservation of momentum? 780