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In electromagnetism, we can re-write the electric field in terms of the electric scalar potential, and the magnetic vector potential. That is:

$E = -\nabla\phi - \frac{\partial A}{\partial t}$, where $A$ is such that $B = \nabla \times A$.

I have an intuitive understanding of $\phi$ as the electric potential, as I am familiar with the formula $F = -\nabla V$, where $V$ is the potential energy. Therefore since $E = F/q$, it is easy to see how $\phi$ can be interpreted as the electric potential, in the electrostatic case.

I also know that $F = \frac{dp}{dt}$, 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?

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Kenshin
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1 Answers1

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1) OP wrote (v1):

[...] and thus this leads me to believe that ${\bf A}$ should be somehow connected to momentum, [...].

Yes, in fact the magnetic vector potential ${\bf A}$ (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 $\phi$ times the charge

$$\tag{1} q\phi$$

does not constitute a Lorentz invariant potential energy. If one recalls the Lorentz transformations for the $\phi$ and ${\bf A}$ potentials, and one goes to a boosted coordinate frame, it is not difficult to deduce the correct Lorentz invariant generalization

$$\tag{2} U ~=~ q(\phi - {\bf v}\cdot {\bf A})$$

that replaces $q\phi$. 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

$$\tag{3}{\bf F}~=~\frac{d}{dt} \frac{\partial U}{\partial {\bf v}} - \frac{\partial U}{\partial {\bf r}}. $$

One may show that eq. (3) reproduces the Lorentz force

$$\tag{4}{\bf F}~=~q({\bf E}+{\bf v}\times {\bf B}), $$

see e.g. Ref. 1.

References:

  1. Herbert Goldstein, Classical Mechanics, Chapter 1.
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