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- Thread starter Rocky Raccoon
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The Aharonov–Bohm effect proves that the potentials are physical quantities.

However, it is a quantum effect, and I think that in the pure classical domain, only the E and B fields are necessary to describe the behaviour of any physical system.

In quantum mechanics, although a particle might feel no field at all in the experimental setup, it can "feel the potential".

See the site below, or many other sites, explaining this Aharonov–Bohm effect in detail.

I have always been impressed that nearly 200 years ago, physicists have been able to identify a peculiar structure in classical mechanics (Lagrange, Hamilton) that is deeply related to quantum mechanics. So true that the Lagragian formulation is the 'shortest path' to quantum mechanics and the Aharonov-Bohm effect more specifically. Amazing!

See for example:

http://www.applet-magic.com/bohm.htm

http://www.weylmann.com/aharonov.pdf

However, it is a quantum effect, and I think that in the pure classical domain, only the E and B fields are necessary to describe the behaviour of any physical system.

In quantum mechanics, although a particle might feel no field at all in the experimental setup, it can "feel the potential".

See the site below, or many other sites, explaining this Aharonov–Bohm effect in detail.

I have always been impressed that nearly 200 years ago, physicists have been able to identify a peculiar structure in classical mechanics (Lagrange, Hamilton) that is deeply related to quantum mechanics. So true that the Lagragian formulation is the 'shortest path' to quantum mechanics and the Aharonov-Bohm effect more specifically. Amazing!

See for example:

http://www.applet-magic.com/bohm.htm

http://www.weylmann.com/aharonov.pdf

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- #4

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The Aharonov–Bohm effect proves that the potentials are physical quantities.

This is misleading for a couple reasons. In general, under a gauge transform, A and phi can be changed and leave the E and B fields unchanged. Yet the Aharonov-Bohm effect is gauge invariant. This implies that we cannot deduce the value of A from the AB experiment, and actually, from the form of the quantum Hamiltonian, from any experiment.

We can glean no more information than is already contained in the fields. In particular, the AB effect can be written entirely in terms of the magnetic field (specifically, the magnetic flux modulo some fundamental flux constant).

The only reason to favor the interpretation of the potential as "real" and not the fields is the field interpretation implies non-locality. Since the AB effect takes place in a static B field, we have no right to complain about nonlocality because the information has had time to propagate from the B field to everywhere in space. Even if we were to complain about nonlocality and therefore adopt A and phi as real, in E&M, A and phi are not necessarily local (c.f. Coulomb gauge).

Perhaps it is the case that A and phi are somehow more real than E and B, but the AB experiment certainly doesn't prove it.

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Rephrasing your question in geometrical terms, I say: Why is a connection more fundamental than curvature ? Because curvature comes from connection and not connection comes from curvature. That's the story in geometry.

You can write down the E-m lagrangian in terms of E&B (D and H) and build the Hamiltonian formalism on it without any problem, too.

The E-m potential captures the gauge freedom of the theory. We know that dF=0. For a cute manifold as flat Minkowski space-time, the de Rham cohomology is trivial and so we end up with F=dA. But that automatically leads to gauge invariance. Different potentials could give the same field. So even starting with F will eventually force you to use A and encounter gauge invariance.

The potential becomes fundamental at quantum level, where gauge invariance really intervenes. Actually, in QFT one can show that the e-m field's consistent couplings with matter occur only through the potential and not through its curvature.

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You can write down the E-m lagrangian in terms of E&B (D and H) and build the Hamiltonian formalism on it without any problem, too.

Can you provide a reference for this?

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How would you couple these fields to matter? Phenomenologically, we require things to couple through a four current, and so we need a four vector to which we can couple the current. Otherwise, we can't make a lorentz invariant action.

- #9

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Every theory that is based on a least action principle that I know of assumes that the Lagrangian depends on at most, first derivatives of the field. When we perform a variation, we get equations that involve second derivatives of the field. Maxwell's equations involve only first derivatives with respect to the electric and magnetic fields. Keeping in mind that the electromagnetic fields are expressed as first derivatives of the electromagnetic potentials, it means that maxwell's equations are second order equations with respect to the potentials. Then, one can formulate a Lagrangian which gives an action, which, upon varying with respect to the the potentials, not the fields, gives the proper Maxwell's equations.

Then, if you write the Lorentz force acting on a point charge in terms of the potentials:

[tex]

\mathbf{F} = q \left[-\frac{1}{c} \frac{\partial \mathbf{A}}{\partial t} - \nabla \Phi + \frac{1}{c} (\mathbf{v} \times (\nabla \times \mathbf{A}))\right]

[/tex]

Then, using the double crossed product formula:

[tex]

(\mathbf{v} \times (\nabla \times \mathbf{A})) = \nabla(\mathbf{v} \cdot \stackrel{\downarrow}{\mathbf{A}}) - (\mathbf{v} \cdot \nabla) \mathbf{A} = \nabla(\mathbf{v} \cdot \mathbf{A}) - (\mathbf{v} \cdot \nabla) \mathbf{A}

[/tex]

and the fact that the total time derivative:

[tex]

\frac{d \mathbf{A}}{d t} = \frac{\partial \mathbf{A}}{\partial t} + (\mathbf{v} \cdot \nabla) \mathbf{A}

[/tex]

we can write the Lorentz force as:

[tex]

\mathbf{F} = \frac{d}{d t}\left(-\frac{q}{c} \mathbf{A}\right) - \nabla\left[q \left(\Phi - \frac{1}{c} (\mathbf{v} \cdot \mathbf{A})\right)\right]

[/tex]

Then, if we introduce a

[tex]

U = q \left[\Phi - \frac{1}{c}(\mathbf{v} \cdot \mathbf{A})\right]

[/tex]

we can write the Lorentz force as:

[tex]

\mathbf{F} = \frac{d}{d t} \frac{\partial U}{\partial \mathbf{v}} - \nabla U

[/tex]

So, the force is expressible through a potential of the form as the Lagrange's equations. This potential enters in the Lagrange function for the particle in an external electromagnetic field. As you can see, this potential energy is expressible in terms of the electromagnetic potentials.

Finally, according to QFT, the electromagnetic potentials form a 4-vector of a gauge field of the local U(1) symmetry.

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