Classical solenoid example to demonstrate the physical reality of the magnetic vector potential $\mathbf A$

Question

To argue for the physical reality of the magnetic vector potential, $\mathbf A$, Feynman refers to the quantum mechanical Aharonov–Bohm solenoid effect, Vol II, Ch:15–4 $\mathbf B$ versus $\mathbf A$.

But, isn't the following classical solenoid induction example enough to make the point that $\mathbf A$ is physically real and not just a mathematical abstraction?

Consider an [infinitely] long solenoid with a conducting loop around it, somewhere in the middle, far away from the ends. Whatever the current in the solenoid coils, the magnetic field outside the solenoid core is $\mathbf 0$.

Yet, if the current thorough the solenoid is changed at a constant rate, as per experimental observation, there is a constant induced current in the loop (say shown by an indicator LED connected in series).

However, since the magnetic field, $\mathbf B = \mathbf 0$ at all points in the vicinity of the loop, even as the current changes, the only way to explain the induced current, without violating the Principle of locality, is to invoke the magnetic vector potential which is not zero (outside the solenoid core) and does change as the current in the solenoid changes.

Furthermore, this is another case where the usual presentation of Faraday's Law as: $$\nabla \times \mathbf{E} = -\frac{ \partial \mathbf{B} }{ \partial t }$$ fails (applying Stokes' Law, its corollary, the flux rule is also inapplicable in this case, though, it gives the correct answer), while the more general formulation of Faraday's Law as (Wikipedia: Magnetic vector potential): $$\mathbf{E} = -\nabla\phi - \frac{ \partial \mathbf{A} }{ \partial t } \,,\quad \mathbf{B} = \nabla \times \mathbf{A}$$ holds.

Am I missing something, as to why Feynman did not use this simpler classical solenoid example to justify the physical reality of $\mathbf A$?

Answer 1

Classically, everything is explainable in terms of $B$. This follows from gauge invariance.

For your example, even if $B$ is zero outside the solenoid, it is still not zero inside the solenoid. Even in the limiting case of a vanishingly narrow solenoid, the flux is concentrated as a Dirac delta. You can therefore calculate the magnetic flux using Stokes theorem etc.

In fact, the situation is identical to an infinite line of current, magnetic flux being the analogue of electric current. You just replace the current density $j$ with $B$ and $B$ with $A$. There is no reason why $B$ should be more physical than $j$ in this case.

The mere fact that the equations are written as PDE illustrate that the theory is local. Just as there is no violation of locality when a static current distribution generates a long range magnetic field, there is not one in your case either. Plus remember that causality is preserved in Maxwell's theory. If you were to suddenly modify the current of the solenoid, you would need to wait for the perturbation to radiate at the speed of light to see its effect in the conducting loop.

To really understand the convenience of $A$, a quantum example is necessary. This is because the gauge invariance is modified. Check out a previous answer of mine for more details. You cannot detect this change of gauge invariance locally just with the magnetic field. You need to consider the flux through finite surfaces. This is why it is easier to think in terms of $A$. More generally, an interacting quantum theory is difficult to write with $B$, but standard with $A$.

Hope this helps.

Answer 2

It is perfectly possible to have an induced electric field in a region where the magnetic field (and all sources) vanish. Maxwell's equations don't say that electric fields induce magnetic fields or vice versa; they say that changes in electric fields induce magnetic fields and vice versa. You can certainly have a constant electric field in a region with no magnetic fields. This is obviously true from electrostatics, but remains true even if the net charge density vanishes identically, as in your setup.

A standard exercise shows that in the quasistatic limit, the electric field outside a time-varying solenoid is nonzero but constant even through the magnetic field vanishes. The electric field is what induces the current. This quasistatic limit is already accurate enough to answer your question.

The exact solution to Maxwell's equations has a nonzero magnetic field outside of the solenoid, which decays to zero over time. (That's assuming that the time-varying current started finitely far back in the past. In the formal case that $I(t) \propto t$ for all times, there is no magnetic field outside the solenoid ever - but this is highly artificial setup, because it implies an unboundedly large current in the far past.) The electric field also varies but settles down to a nonzero constant configuration a long time after the current starts changing.

Answer 3

This question still requires attention in my view.

The underlying issue is that, as the OP notes, the differential form of the Maxwell equations, in case the Faraday law, does not predict $\vec E$ outside a solenoid with slowly varying current, but only state that its rotation vanishes. The integral form of the Maxwell equations does predict $\vec E$. The question therefore exposes a physical difference between the two. The difference is that the integral form can be understood to follow from the differential form by integration by parts. In the present case this means that the integral form is non-local.

The physicist thus faces the dilemma between incompletely predictive local or completely predictive non-local equations. The use of potentials removes the dilemma. The equations in terms of the potentials $$\partial_\mu \left( \partial^\mu A^\nu - \partial^\nu A^\mu \right) = -j^\nu/\varepsilon_0$$ are local and make the correct prediction.

For those of us who believe in gauge invariance - likely everybody but myself - the question arises which gauge to use. My recommendation is to always use the Lorenz gauge as only then the potentials are causal.

It is however much easier to reject gauge invariance and to adopt the wave theory of electromagnetism$$\partial_\mu\partial^\mu A^\nu = -j^\nu/\varepsilon_0 \,.$$ See my paper at https://arxiv.org/abs/physics/0106078 and a video lecture at https://youtu.be/eCruJN3w-CY?si=YtIjo13kTWmgUAzr.