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

• magicfountain
In summary: 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 \
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.

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.

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.

magicfountain said:
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##.

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.

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.

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.

nsaspook
vanhees71 said:
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.

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.

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.

vanhees71 said:
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

vanhees71 said:
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.

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..

Last edited:
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!

vanhees71 said:
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.

## 1. What is the difference between vector potential and field tensor?

Vector potential and field tensor are both mathematical quantities used to describe electromagnetic fields. The main difference between them is that the vector potential is a three-dimensional vector while the field tensor is a four-dimensional tensor. This means that the vector potential has three components (x, y, z) while the field tensor has 16 components (4x4 matrix).

## 2. Which one is more fundamental in describing electromagnetic fields?

Both the vector potential and field tensor are equally fundamental in describing electromagnetic fields. They are different mathematical representations of the same physical phenomenon. The choice of which one to use depends on the specific problem being studied and personal preference.

## 3. Can the vector potential and field tensor be used interchangeably?

No, the vector potential and field tensor cannot be used interchangeably. They are mathematically different quantities and represent different aspects of the electromagnetic field. While the vector potential can be derived from the field tensor, the reverse is not possible.

## 4. How are the vector potential and field tensor related?

The vector potential and field tensor are related through the Maxwell's equations, which describe the behavior of electromagnetic fields. Specifically, the vector potential is related to the electric field through the curl operator, while the field tensor is related to both the electric and magnetic fields through the covariant derivative.

## 5. Which one is easier to work with in calculations?

The choice of which one is easier to work with in calculations depends on the specific problem at hand and personal preference. Some problems may be easier to solve using the vector potential, while others may be easier with the field tensor. It is important to have a good understanding of both quantities and their respective mathematical representations to be able to choose the appropriate one for a given problem.

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