Is the electromagnetic 4-vector indeed a 4-vector?

[Ibix]

I am still confused. What does the Lorentz structure have to do with it? We have a manifold $M$, in this case $\mathbb R^4$. What is a vector and what is 1-form depends only on the manifold. In this case $A^\mu \partial_\mu$ is a vector and $A_\mu dx^\mu$ is a one form. Lorentz transformations have nothing to do with it. What am I missing?

I think the point is this: choose a frame and a gauge fixing such that, in this frame, $A^0=0$. Now transform $A^\mu$ as a four vector (rightly or wrongly) and you will find that $\Lambda^0{}_\mu A^\mu\neq 0$. The result is still a four potential for the same EM field. I think everyone agrees this.

The argument is about what gauge fixing means, if I understand correctly. @samalkhaiat says that fixing the gauge means defining the zeroth component of the four potential to be zero (other gauge conditions are available). Thus $A^0=A'^0=A''^0=\ldots=0$ and since $\Lambda^0{}_\mu A^\mu\neq 0$, clearly $A^\mu$ does not transform as a four vector. @Orodruin says that fixing the gauge means defining the zeroth component of the four potential to be zero in some chosen frame (again, other gauge conditions are available) and notes that this uniquely defines the four potential in all frames - it just doesn't use the same set-zeroth-component-to-zero gauge. In this view, the four potential transforms as a vector but has a different gauge in different frames.

(If I've misrepresented them I'm sure they'll correct me, but that's my understanding.) I don't have the knowledge to offer a view on whether one view is better than the other.

 

[martinbn]

I think the point is this: choose a frame and a gauge fixing such that, in this frame, $A^0=0$. Now transform $A^\mu$ as a four vector (rightly or wrongly) and you will find that $\Lambda^0{}_\mu A^\mu\neq 0$. The result is still a four potential for the same EM field. I think everyone agrees this.

The argument is about what gauge fixing means, if I understand correctly. @samalkhaiat says that fixing the gauge means defining the zeroth component of the four potential to be zero (other gauge conditions are available). Thus $A^0=A'^0=A''^0=\ldots=0$ and since $\Lambda^0{}_\mu A^\mu\neq 0$, clearly $A^\mu$ does not transform as a four vector. @Orodruin says that fixing the gauge means defining the zeroth component of the four potential to be zero in some chosen frame (again, other gauge conditions are available) and notes that this uniquely defines the four potential in all frames - it just doesn't use the same set-zeroth-component-to-zero gauge. In this view, the four potential transforms as a vector but has a different gauge in different frames.

(If I've misrepresented them I'm sure they'll correct me, but that's my understanding.) I don't have the knowledge to offer a view on whether one view is better than the other.

It's ok, I deleted it because I just realized the discussion is about four-vectors as in relativity. I need more sleep.

 

[jbergman]

I wanted to take a stab at answering this question. Wald addresses this in his book on Electromagnetism in chapter 9, "Electromagnetism as a Guage Theory".

In particular, he states the following,

Up to this point, we have treated the 4-vector potential as a dual vector field on spacetime. However, since $A'_{\mu}=A_{\mu} + \partial_{\mu}\chi$ represents the same physical field, an electromagnetic field really corresponds to an equivalence class of 4-vector potentials, where two are considered equivalent if they differ by a gauge transformation... Therefore, it is very useful to give an alternative characterization of $A_{\mu}$ as a uniquely defined quantity on a higher-dimensional space.

He then defines $A_{\mu}$ as part of a connection field on a trivial U(1) principal bundle over space time, $P = \mathbb{R}^4 \times U(1)$. More precisely, since $P$ is 5-dimensional with the U(1) part included, a connection $\mathcal{A}_{\Lambda}$ (with $\Lambda = 0,1,2,3,4$) is a dual vector field on $P$ the the property that $\mathcal{A}_{\Lambda}(x^{\mu}, s)$ is independent of $s$ and $\mathcal{A}_{4} = 1$, where $s$ is the direction tangent to U(1). Since $\mathcal{A}_{\Lambda}(x^{\mu}, s)$ is independent of $s$, we may identify its $\Lambda = 0,1,2,3$ components with a dual vector field $A_{\nu}(x^{\mu})$ on spacetime. However on $P$ we may perform an arbitrary coordinate transformation that preserves the U(1) action,
$$s' = s - \chi(x^{\mu})$$
Under this transformation the components of $\mathcal{A}_{\Lambda}$ transform as,
$$\begin{align*}
\mathcal{A}'_4 &= \mathcal{A}_4 = 1\\
\mathcal{A}'_{\Lambda} &=\mathcal{A}_{\Lambda} + \partial_{\Lambda}\chi, (\Lambda \in {0,1,2,3})
\end{align*}
$$
And the dual vector on spacetime corresponding to $\mathcal{A}'_{\Lambda}$ is
$$A'_{\mu} = A_{\mu} + \partial_{\mu}\chi$$

He then concludes,

Thus we see that the gauge equivalence class of $A_{\mu}$ on spacetime corresponds to different coordinate representations of of the single object $\mathcal{A}_{\Lambda}$ on $P$. It is natural to view this single object $\mathcal{A}_{\Lambda}$ as providing a fundamental description of the electromagnetic field.

Therefore, my answer after reading Wald is that we really should be thinking of the connection field $\mathcal{A}_{\Lambda}$ on $P$ as the fundamental object since we can define it in a coordinate independent way. And based on that I would say that it is a stretch to think of $A_{\mu}$ as a 4-vector.

 

[Sagittarius A-Star]

Also of potential interest is the Aharonov-Bohm effect, https://en.wikipedia.org/wiki/Aharonov–Bohm_effect.

Wald describes in his book "Advanced Classical Electromagnetism" the physical significance of the vector potential (modulo gauge) for the Aharonov-Bohm effect:

However, as we shall see in chapter 9, the coupling of the electromagnetic field to fundamental charged matter (namely, charged fields) can be described only in terms of the potentials, not the field strengths. Furthermore, there are physically relevant situations where $\vec E$ and $\vec B$ do not contain all of the information about the electromagnetic field.

As an example, consider the region outside an infinite solenoid. Suppose that inside the solenoid, there is a nonvanishing, uniform magnetic field, but outside the solenoid, we have $\vec E = \vec B = 0$. Since the region outside the solenoid is not simply connected, the fact that $\vec E$ and $\vec B$ vanish in that region does not imply that the potentials are gauge equivalent to zero there. Indeed, eq. (1.7) implies, via Stokes’s theorem, that when $\vec B\ne 0$ inside the solenoid, we have $\oint \vec A \cdot d \vec l \ne 0$ for any loop outside the solenoid that encloses it. (Note that $\oint \vec A \cdot d \vec l$ is gauge invariant, i.e., its value does not change under eq. (1.13).)

A quantum mechanical charged particle that stays entirely outside the solenoid will be affected by this vector potential, as it will produce a relative phase shift in the parts of the wave function that go around the solenoid in different directions, producing a physically measurable shift in the resulting interference pattern. This phenomenon, known as the Aharonov-Bohm effect, is sometimes attributed to the weirdness of quantum mechanics.

However, the effect has nothing to do with quantum mechanics - the same effect would occur for a classical charged field. And there is nothing weird about the effect, once one recognizes that the electromagnetic field is represented, at a fundamental level, by the potentials $\phi, \vec A$ (modulo gauge), not the field strengths $\vec E, \vec B$.

Source (see preview, scroll down to chapter 1.1, page 5):
https://www.amazon.com/Advanced-Cla...m-Robert-Wald/dp/0691220395?tag=pfamazon01-20