In electromagnetism, we can re-write the electric field in terms of the electric scalar potential, and the magnetic vector potential. That is:
E=−∇ϕ−∂A∂t, where A is such that B=∇×A.
I have an intuitive understanding of ϕ as the electric potential, as I am familiar with the formula F=−∇V, where V is the potential energy. Therefore since E=F/q, it is easy to see how ϕ can be interpreted as the electric potential, in the electrostatic case.
I also know that F=dpdt, where p is momentum, and thus this leads me to believe that A should be somehow connected to momentum, maybe like a "potential momentum". Is there such an intuitive way to understand what A is physically?
Answer
1) OP wrote (v1):
[...] and thus this leads me to believe that A should be somehow connected to momentum, [...].
Yes, in fact the magnetic vector potential A (times the electric charge) is the difference between the canonical and the kinetic momentum, cf. e.g. this Phys.SE answer.
2) Another argument is that the scalar electric potential ϕ times the charge
qϕ
does not constitute a Lorentz invariant potential energy. If one recalls the Lorentz transformations for the ϕ and A potentials, and one goes to a boosted coordinate frame, it is not difficult to deduce the correct Lorentz invariant generalization
U = q(ϕ−v⋅A)
that replaces qϕ. The caveat of eq. (2) is that U is a velocity-dependent potential, so that the force is not merely (minus) a gradient, but rather on the form of (minus) a Euler-Lagrange derivative
F = ddt∂U∂v−∂U∂r.
One may show that eq. (3) reproduces the Lorentz force
F = q(E+v×B),
see e.g. Ref. 1.
References:
- Herbert Goldstein, Classical Mechanics, Chapter 1.
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