EM: Vector potential vs. Field tensor: Which is fundamental?

[magicfountain]

In my lecture we were discussing the Lagrangian construction of Electromagnetism.
We built it from the vector potential $A^\mu$.
We introduced the field tensor $F^{\mu \nu}$.
We could write the Langrangian in a very short fashion as $-\frac{1}{4}F_{\mu \nu}F^{\mu \nu}$
In the end we derived the equations of motion for the vector potential:
$\Box A^\mu = -J^\mu$
But again we could write that as:
$\partial_\mu F^{\mu \nu} = J^\mu$
The professor then told us that the vector potential is more 'fundamental', even though we could write Lagrangian and EOM using the field tensor.
Why is that? I've thought about it, but couldn't come up with a good reason. One thought would be, that the EOM for the vector potential are a wave equation, but is that good explanation? Who's telling us, that wave equations are more fundamental?
It would be great if anybody could help me out understanding this.

 

[fzero]

The Lagrangian that leads to the equation of motion with sources includes a term $A^\mu J_\mu$, so you can't write it in terms of the field strength alone.

 

[magicfountain]

Hey, thanks for the quick reply. That's a pretty satisfying answer. Much appreciated! So it's pure coincidence that the EOM of F don't have A in them, I guess.

 

[fzero]

Hey, thanks for the quick reply. That's a pretty satisfying answer. Much appreciated! So it's pure coincidence that the EOM of F don't have A in them, I guess.

The underlying reason for that is because $F$ and $J$ are gauge-invariant. It turns out that the action will be gauge-invariant if the current satisfies the continuity equation $\partial_\mu J^\mu=0$. This is similar to the Lorentz-gauge condition $\partial^\mu A_\mu =0$ that you should have needed to obtain the wave equation for $A_\mu$.

 

[Khashishi]

Which is more fundamental is a matter of opinion. Also, it's up to opinion whether the Euler-Lagrange formulation or force-based equations of motion is more fundamental.

One advantage of fields over potentials is that potentials have an arbitrary choice of gauge. The gauge is not fundamental and just an artifact of our mathematical formulation. But maybe the equivalence class of potential modulo choice of gauge is the more fundamental quantity.

 

[stedwards]

I notice that every aspect of electromagnetic theory can be expressed in terms of derivatives of the potential $(\phi, A)$--the wave equation, charge continuity, F, $(\rho, J)$, and the absence of non-zero magnetic charge. Personally, I think of $(\phi, A)$ as the fundamental, or the primitive field. But I can't objectively justify this. They're all gaugable in some manner, including the vector potential.

 

[vanhees71]

From a really fundamental point of view (quantum theory) the equivalence classes of the four-potential defined such that all four-potentials that deviate only by a gauge transformation are identified are the fundamental mathematical entities describing the electromagnetic field. There are observable facts, which underline this point of view. The most simple example is the Aharonov-Bohm effect.

 

[stedwards]

From a really fundamental point of view (quantum theory) the equivalence classes of the four-potential defined such that all four-potentials that deviate only by a gauge transformation are identified are the fundamental mathematical entities describing the electromagnetic field. There are observable facts, which underline this point of view. The most simple example is the Aharonov-Bohm effect.

Yes, but the charge enters separately, rather than as an aspect of the potential.

 

[vanhees71]

What do you mean by that? The AB effect is explained by the gauge-invariant phase factor
$\exp(\mathrm{i} q \int_C \mathrm{d} \vec{r} \cdot \vec{A}),$
where the line integral is along a closed path encircling the magnetic filament. That this is a gauge invariant quantity becomes clear via Stokes's integral theorem,
$\int_{C} \mathrm{d} \vec{r} \cdot \vec{A}=\int_{F} \mathrm{d}^2 \vec{F} \cdot (\vec{\nabla} \times \vec{A}) = \int_{F} \mathrm{d}^2 \vec{F} \cdot \vec{B} = \Phi_B,$
i.e., it's the magnetic flux through an area with boundary curve $C$, which is a gauge invariant quantity. Of course the charge of the particle $q$ is also gauge invariant.

 

[my2cts]

What is not gauge invariant is the Noether energy-momentum tensor of the Lagrangian given in the first post.
No worries, we have an ad hoc solution, the Belinfante-Rosenfeld patch.
Another non-GI quantity is intrinsic angular momentum (photon spin).
Again no worries, a the total angular momentum that is GI can be constructed from the patched energy-momentum tensor.
This may be the reason why a leading textbook like Jackson defers the discussion of electromagnetic spin to problem 7.19.

 

[andresB]

What do you mean by that? The AB effect is explained by the gauge-invariant phase factor
$\exp(\mathrm{i} q \int_C \mathrm{d} \vec{r} \cdot \vec{A}),$
where the line integral is along a closed path encircling the magnetic filament. That this is a gauge invariant quantity becomes clear via Stokes's integral theorem,
$\int_{C} \mathrm{d} \vec{r} \cdot \vec{A}=\int_{F} \mathrm{d}^2 \vec{F} \cdot (\vec{\nabla} \times \vec{A}) = \int_{F} \mathrm{d}^2 \vec{F} \cdot \vec{B} = \Phi_B,$
i.e., it's the magnetic flux through an area with boundary curve $C$, which is a gauge invariant quantity. Of course the charge of the particle $q$ is also gauge invariant.

The AB phase doesn't need a closed path

http://arxiv.org/pdf/quant-ph/9801013v2.pdf

 

[andresB]

From a really fundamental point of view (quantum theory) the equivalence classes of the four-potential defined such that all four-potentials that deviate only by a gauge transformation are identified are the fundamental mathematical entities describing the electromagnetic field. There are observable facts, which underline this point of view. The most simple example is the Aharonov-Bohm effect.

I am not as educated as I would like, but saying that is not the same as saying that the important thing are the fields?, for example in the clased path magnetic AB effect, the final phase only depend on the magnetic flux through the enclosed region.

 

[my2cts]

The fundamental quantity is the potential. The fundamental equation is the wave equation for the potential.
It is inhomogeneous since the charge-current appears on the right hand side.
This equation implies a one-to-one correspondence between charge-current and potential.
A direct consequence is that the Lorenz condition corresponds to current conservation and thus always holds if there is charge conservation.
The Lagrangian is -\frac{1}{2} d_\mu A_\nu d*\mu A*\nu.
This approach leads to separately conserved electromagnetic spin and resolves all paradoxes that involve energy-momentum and angular momentum..

 

[vanhees71]

I've to look at

http://arxiv.org/abs/quant-ph/9801013

in detail before I can comment on it.

The reason for posting this rather empty message is that one should quote the abstract pages in arXiv not the pdf, because then there's a chance that Physics Forums becomes recognized at arXiv as a blog referring to the article, which is great for PF, because more professional physicists may notice it!

 

[stedwards]

What do you mean by that? The AB effect is explained by the gauge-invariant phase factor
$\exp(\mathrm{i} q \int_C \mathrm{d} \vec{r} \cdot \vec{A}),$
where the line integral is along a closed path encircling the magnetic filament. That this is a gauge invariant quantity becomes clear via Stokes's integral theorem,
$\int_{C} \mathrm{d} \vec{r} \cdot \vec{A}=\int_{F} \mathrm{d}^2 \vec{F} \cdot (\vec{\nabla} \times \vec{A}) = \int_{F} \mathrm{d}^2 \vec{F} \cdot \vec{B} = \Phi_B,$
i.e., it's the magnetic flux through an area with boundary curve $C$, which is a gauge invariant quantity. Of course the charge of the particle $q$ is also gauge invariant.

The Faraday tensor is the exterior derivative of the 4-vector potential, F=dA. F and dA are considered the different names for the same thing. The current-charge three form, J is the exterior derivative of the hodge dual of F, *F. In the same manner, J can considered to be, not a separate phenomena, but synonymous with d*F.