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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 $\mathbf A$?

Engr. Ravi
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    Surely you are just replacing magnetic flux with $\oint {\bf A} \cdot d{\bf l}$ ? I think there is also a problem with hypothesising an infinite solenoid and changing the current in it. As for locality, well when you change the current in a wire, the B-field changes around it. Does that violate the principle of locality too? – ProfRob Dec 21 '19 at 18:39
  • @RobJeffries, that was my first thought too (a worry about changing the current through an infinitely long solenoid which, it seems to me, must have infinite inductance). Thinking out loud here, would a better approach be to first assume a finite length solenoid here and then (carefully) take the limit as the length becomes arbitrarily large. – Alfred Centauri Dec 21 '19 at 20:32
  • … since the magnetic field, ${\bf B}=0$ at all points in the vicinity of the loop … Why would you think that ${\bf B}=0$ outside the solenoid if the current is non stationary? Also, why would you think that Stokes' theorem is inapplicable here, and why do you think that Faraday's law written in terms of potentials is somehow more general? – A.V.S. Dec 23 '19 at 18:11
  • @A.V.S., - "It has been noted [5] that the example of a linearly rising current which has persisted forever in an infinite solenoid is a special case in that Maxwell’s equations are satisfied by an electric field as calculated above and a magnetic field that is zero outside the solenoid." - The Fields Outside a Long Solenoid with a Time-Dependent Current – Alfred Centauri Dec 23 '19 at 18:46
  • This may be required reading: The transient magnetic field outside an infinite solenoid – Alfred Centauri Dec 23 '19 at 18:50
  • @AlfredCentauri: If the solenoid is infinitely long and the current is eternally linearly rising, then the field outside is really static ${\bf E} \sim {\bf e}_\phi / \rho $ so it is no more evidence for the reality of vector potential than any other static electric field. If the solenoid has finite (if large) length then magnetic field outside is nonzero and could be approximated (near the coil) by monopole-antimonopole pair (with growing charges) placed at the ends of solenoid, again vector potential is not required. – A.V.S. Dec 23 '19 at 21:13
  • @AlfredCentauri Thanks for the links. Unfortunately, a satisfactory conclusion is not reached. The 1st paper says that the analysis given by the 2nd (return wire modeled as another solenoid wound over the original in the opposite direction) is "not convincing", and concludes (paraphrased) "[during the transient phase] The small remaining time-dependent magnetic field is sufficient to induce locally an electric field of the instantaneous quasistatic value. There is no need to invoke action at a distance, as might be required in a view that emphasizes only the quasistatic limit." – Engr. Ravi Dec 24 '19 at 09:45
  • Both Feynman's and your argument is flawed, because they make an unjustified logical jump: from vanishing magnetic field to the idea that the local thing causing the effect on charged particle has to be the magnetic vector potential or some other non-vanishing quantity related to magnetic field. In fact, the thing causing the local effect can also be the induced electric field of the accelerated current in the solenoid, and this is how the induced EMF in the loop in your example is explained in the standard way. – Ján Lalinský Dec 03 '23 at 05:33
  • The Maxwell equation related to the Faraday law does not fail to be valid, it only fails to be useful (in differential form) to find the induced electric field outside the solenoid. But it is useful when written in its integral form, and one can find the induced electric field outside the solenoid from it, because the integral form takes into account the magnetic flux inside the solenoid. – Ján Lalinský Dec 03 '23 at 05:35
  • Faraday's law does not fail in the context of classical E&M. It is exact (in that regime) and fully explains what's going on in this scenario. – tparker Dec 03 '23 at 19:41

2 Answers2

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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 not zero everywhere. In fact, it’s a Dirac delta function: $$ B(r)=\Phi\int\delta(r-se_z)e_zds $$ with $\Phi$ the magnetic flux crossing the solenoid. You can therefore calculate the magnetic flux using Stokes theorem etc.

In fact, the situation is identical to an infinite line of 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.

A quantum example is necessary to argue for the reality of $A$. 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 can only see it over finite fluxes. This is why it is more convenient to think in terms of $A$, and argue for its more fundamental nature.

Hope this helps.

LPZ
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  • The vector potential is a model, a mathematical conception that only exists in human minds. It is not fundamental. Experiments are the foundations of physics. Experiments demonstrate that this model accurately captures certain real phenomena, hooray for physics! But never, ever, make the mistake of confusing models with reality, nor forget that no mere model may be considered fundamental. – John Doty Nov 02 '23 at 13:54
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    The magnetic field isn't a Dirac delta function for a solenoid. The current density is. – tparker Dec 03 '23 at 19:38
  • @tparker Not quite. Recall that for a solenoid of finite radius and surface current, the magnetic field is hat function. Letting the radius go to zero and current density to infinity, you’ll get a Dirac delta. The current density is therefore a derivative of Dirac delta, it’s many magnetic dipoles stacked along their common axis. If the current density is a Dirac delta, you rather have a thin wire. – LPZ Dec 03 '23 at 19:45
  • Uh... why are you letting the radius go to zero and the current density to infinity? Why not just consider a finite solenoid? – tparker Dec 03 '23 at 20:22
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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.

tparker
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