Is the magnetic vector potential "real" in classical electromagnetism?

Question

From how I've learned it in school the magnetic vector potential is used as a mathematical tool to simplify problems with current-carrying wires in classical electromagnetism, but is never treated as bearing any physical meaning. After doing some research on it myself, I found Feynman's lecture on the subject, which outlines how this vector potential actually takes center-stage in QED, not only featuring in its primary equations for concepts like phase change but also as capable of producing physical changes such the Aharonov-Bohm effect that can't be explained by the magnetic field alone. Are there any comparable physical effects the vector potential has in classical electromagnetism?

Answer 1

The vector potential is gauge-dependent and unobservable in both classical and quantum mechanics. Only gauge-invariant quantities — including the electric and magnetic fields — are observable. Even in the Aharonov-Bohm effect, the vector potential is not directly observable; what we measure is the gauge-invariant quantity \begin{align} \Phi = \oint \vec{A}\cdot \mathrm{d}\vec{\ell} \end{align}

Answer 2

One way of thinking about this issue of what quantities are 'real' is to compare it to the more familiar situation with position.

There are quantities in physics with the property that the quantity itself is coordinate-system dependent, but differences in the quantity between one instance and another have an absolute physical meaning. (See John Baez talking about 'torsors' for more.)

The difference between two positions is the 'length', and has an absolute meaning. But the positions themselves are coordinate-dependent and have no absolute meaning. 'Position' relative to what? To talk about positions, you have to anchor them to something real, and speak relatively. We sometimes talk about fixed positions on Earth, but the Earth is spinning and orbiting the sun, which is orbiting around the galaxy. Will the real 'position' among these please stand up?

Coordinate systems are made up. They are entirely imaginary constructs, in our heads. Only the relative distances between positions and angles between orientations are physically meaningful. And yet using 'position' and 'orientation' as if they are real and absolute is an immensely useful way of thinking about things, that we can't easily do without.

Choosing a gauge is like picking a coordinate system. It's a somewhat arbitrary choice, subject to some mild consistency conditions. But whichever one we pick, differences between the potential values calculated in the same gauge are always the same.

As Baez points out, there are lots of these quantities in physics where only differences are physically meaningful and the values themselves are coordinate-system dependent. 'Time' is another one. The length of a time interval is meaningful, but the question "What is the date and time?" needs an arbitrary choice of calendar, time zone, etc. to answer. 'Gauge' is an unfamiliar name for a familiar concept.

Answer 3

Let's take the definition of a "real field" from the chapter in Feynman's lectures you linked to:

What we mean here by a “real” field is this: a real field is a mathematical function we use for avoiding the idea of action at a distance. If we have a charged particle at the position P , it is affected by other charges located at some distance from P . One way to describe the interaction is to say that the other charges make some “condition”—whatever it may be—in the environment at P . If we know that condition, which we describe by giving the electric and magnetic fields, then we can determine completely the behavior of the particle—with no further reference to how those conditions came about.

In classical physics, we can describe the motion of particles using Newton's laws. The electromagetic force on a particle is given by the Lorentz force law $$ \vec F = q(\vec E + \vec v \times \vec B) $$ So any classical electromagnetic effect can be described in terms of $B$, and therefore we never need $\vec A$ to describe a "real" field.

What happens in quantum mechanics is that Newton's laws no longer hold, so the above argument breaks down. We still have gauge invariance, so in order for $\vec A$ to appear as a "real" field requires something subtle to happen. The Aahronov-Bohm effect uses quantum mechanics in a very clever way to get the gauge invariant line integral $\oint \vec{A} \cdot d \vec \ell$ to appear as the phase difference in the amplitudes for two paths a particle could take to get around a coil of wire. This phase difference can then be measured by preparing a state where the particle is in a superposition of taking both paths. In other words, the Aahronov-Bohm effect crucially relies on quantum properties (e.g., superposition) that do not have a classical analogue.

Answer 4

The vector potential is no more or no less real than the $\mathbf B$ induction field itself. Think of a tightly wound long thin coil or a toroid around a high permeability core, one whose outside induction field is essentially zero. The coil is driven by an ac voltage. If you place a wire loop $\mathcal L$ around and outside the coil, then despite the wire loop being in an essentially zero B-field a voltage will be induced that is proportional to the flux rate inside the coil. This effect was demonstrated by Oliver Lodge in the late 1880's, and it is called the Maxwell-Lodge effect. In the experiment Lodge used a toroid to make sure that the $\mathbf B$-field outside the toroid where the wire loop $\mathcal L$ be located was as close to zero as possible. Here is Lodge p476-477 describing his observation:

So I led a short wire round the outside of the ring and brought its ends into the box where the needle was, one on each side of the needle. A distinct and proper deflexion was now observed. A wire was then taken 10 times around the ring, and connected to a common quadrant electrometer. The deflexion was easy to see. It could also be just seen with only one turn of wire.

Of course, Faraday's induction law is about flux rate, $\frac{d\Phi}{dt}$, that is not zero, and it is not about the field intensity directly, so Maxwell's equations are valid but there is no simple or intuitive way to explain this effect unless one accepts the reality of the vector potential field that is not zero necessarily even when associated induction field is zero.

This is in contrast to other apparent paradoxes, such as involving the homopolar dynamo, which can be explained by the Lorentz force acting on the free charge carriers, for example.

Furthermore, $\frac{d\Phi}{dt} = \oint_{\mathcal L} \mathbf A \cdot d\ell$, and in Lodge's experiment the magnetic vector potential is not zero for either a tightly wound toroid or a tightly wound thin core anywhere. So in this sense, it seems, that the reality of the vector potential manifests itself explicitly and directly as being the primary magnetic field.

Answer 5

The vector potential is a mathematical abstraction employed in physical theory, just like other abstractions such as voltage, momentum, etc. Mathematical abstractions are not "real", but they are often useful in models of physical phenomena. How useful a particular abstraction is for model construction depends on the phenomenon, the model, and the physicist.

Added:

There's a second question at the end: "Are there any comparable physical effects the vector potential has in classical electromagnetism?" An example is radiation from antennas, where in models vector potential often appears either directly or in the form of Hertz vectors.

Answer 6

The Hamiltonian, i.e. the energy, of a nonrelativistic particle in an external electromagnetic field is $$H=\frac{\left(\mathbf{P}-q\mathbf{A}\right)^{2}}{2m}+q\phi.$$ In other words, the energy of the system is written in terms of the potentials, and so it stands to argue that they have as much reality as the energy. Additionally, since the motion is derived from the Hamiltonian together with the initial conditions you could argue that all physics is determined by the potentials.

Yet, once you derive the equations of motion from the Hamiltonian, only the combinations $\nabla\times\mathbf{A}$ and $-\nabla \phi - \partial_t \mathbf{A}$ will appear. These are of course the magnetic and electric fields, respectively. So one might think that maybe the fields are still more fundamental because they appear in the equations of motions. But then, the energy doesn't appear in the equations of motion and yet it is a constant of motion unlike the momentum $\mathbf{P}$ which appears in both the Hamiltonian and the equations of motion and which we undoubtedly consider a meaningful physical quantity.

My point is that what you consider physically meaningful is a question of your point of view. To take an extreme point of view it is possible to study the motion of charges without considering any fields at all by careful integrating over all particles trajectories to find all the pairs which are on each others' light cones at some point in their histories and then deriving their effects on each other from that. From that point of view the fields are just a tool we use to make that complex history somehow manageable and eliminate the need to keep complete track of all trajectories, but the actual content of the physics would only be the ensemble of the trajectories of all charged particles. This point of view was discussed e.g. in this question.

Answer 7

...such the Aharonov-Bohm effect that can't be explained by the magnetic field alone.

That's the orthodox position and Feynman explained it well. However, notice that even in the QT analysis of the BA system and in the BA experiments, the measured things depend on $\oint \mathbf A\cdot d\mathbf s$, which is magnetic flux, not on value of gauge-dependent vector potential at any single point or some other gauge-dependent function of $\mathbf A$. That is, the BA experiment does not reveal which gauge is the real one.

Correlation of phase shift with magnetic flux in the solenoid is interesting physics, but it should not be entirely convincing us that "vector potential value at point $\mathbf x$ is just as real as that of magnetic induction $\mathbf B$ at point $\mathbf x$".

Are there any comparable physical effects the vector potential has in classical electromagnetism?

If by "effects of the vector potential" you mean correlation with $\oint \mathbf A\cdot d\mathbf s$, then yes - the example hyportnex gave. The induced EMF in a loop of wire going around the solenoid equals rate of change of $\oint \mathbf A\cdot d\mathbf s$. This does not show "realness" of $\mathbf A$, only "realness" of $\oint \mathbf A\cdot d\mathbf s$. The real physical field that causes the induced EMF in the wire is induced electric field. Yes, magnetic field outside the solenoid is close to zero, but not the electric field.

Vector potential is too ambiguous a concept to base EM theory on. It is not possible to rewrite deterministic Maxwell's equations in terms of vector potential and retain the property that initial value problem has a unique solution. The resulting equations based on Maxwell's equations have a gauge freedom and thus do not have unique solution. The initial condition is not enough, there is infinity of possible solutions, diverging from the initial condition.

To restore determinism in the equations, one has to impose further equation (one or more), sometimes called the gauge fixing condition, like $\nabla \cdot \mathbf A = 0$, or others. These additional conditions are not physical laws, there is infinity of them and choosing one is arbitrary; changing this choice does not change anything in physically significant quantities.