Magnetic Field and Flow of Vector Potential

I am sorry, when my question is not really concrete, but here we go.

Consider the Hamiltonian function $$H(x, \xi) = \frac{1}{2m}\bigl|\xi - eA(x)\bigr|^2$$ corresponding to a charged particle in a magnetic field, where $A$ is a vector potential (i.e. some vector field), $e$ is the charge and $m$ is the mass.

The classical trajectories in $x(t)$ satisfy the second order equation $$m\ddot{x} = e \,\bigl(\dot{x} \times \mathrm{curl}\,A(x)\bigr),$$ In particular, if we replace $A(x)$ by $\tilde{A}(x) := A(x) + \nabla f(x)$ for some function $f$, we obtain the same classical trajectories, because $\mathrm{curl}\,\tilde{A} = \mathrm{curl}\,A$.

My question is: Let $\Theta_t(x)$ be the flow of the vector field $A$. Does $\Theta_t(x)$ have any physical significance (classical or quantum mechanical) related to this problem? If yes what is it?

The answer seems to be "no" at first glance, because the flow does clearly depend very much on the function $f$ above, which on the other hand does not play a role for the classical trajectory. But is there more to it?

• I think your interpretation is correct. $\mathbf A$ is just a vector potential, so it is defined up to a gradient of some doubly differentiable scalar function. What you can physically observe is the flow of the curl, which are just the line of force of the field. Jan 27, 2015 at 12:56

In my course of electromagnetism i learned that the vector potential can be interpreted as a quantity of motion per unit of charge, or as a potential energy per unit of charge and unit of speed, at least in the case of stationary currents, because in that case the magnetic force is

\begin{eqnarray} \boldsymbol{F}&=&q\boldsymbol{v}\times\boldsymbol{B}=q\boldsymbol{v}\times\left(\nabla\times\boldsymbol{A}\right)=q\left[\nabla\left(\boldsymbol{v}\cdot\boldsymbol{A}\right)-\left(\boldsymbol{v}\cdot\nabla\right)\boldsymbol{A}\right]\\&=&-\nabla\left(-q\boldsymbol{v}\cdot\boldsymbol{A}\right)+\frac{d}{dt}\left(-q\boldsymbol{A}\right)=-\nabla U+\frac{d\boldsymbol{p}}{dt} \end{eqnarray}

and then

\begin{eqnarray} U&\equiv&-q\boldsymbol{v}\cdot\boldsymbol{A}\\ \boldsymbol{p}&\equiv&-q\boldsymbol{A} \end{eqnarray}

and the interpretation follows from these expressions. A more profound discussion about the meaning of the vector potential and it's reality can be found in the Feynman Lectures on Physics vol. 2

Since $\vec A$ can't be observed directly, I don't see why it's integral curves (manifestation of flow) could be observed either. It is this "unphysicality" that makes gauge transformations possible. After all, it would not make sense to be able to gauge transform a physical quantity. It would be completely arbitrary!

As for the equation of motion, recall the definition of the magnetic field: $$\vec B=\operatorname{curl}\vec A$$ The thing that we observe is the flow $\phi_t$ of $\vec B$. Remember the thing you did in high school or middle school with the iron shavings and a magnet? Those lines you saw were the integral curves of $\phi_t$.