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

[mbond]

About the electromagnetic 4-vector $A^\mu=(\Phi/c,\mathbf{A})$. If it is indeed a four-vector, then its squared length $A_\mu A^\mu=\Phi^2/c^2-\mathbf{A}^2$ should be a Lorentz invariant. What is the physical significance of $\Phi^2/c^2-\mathbf{A}^2$?
Thanks for any help.

 

[PeterDonis]

A 4-vector doesn't have to have a direct physical interpretation to be a 4-vector. As long as it satisfies the mathematical properties of a 4-vector, it is one.

 

[PeroK]

About the electromagnetic 4-vector $A^\mu=(\Phi/c,\mathbf{A})$. If it is indeed a four-vector, then its squared length $A_\mu A^\mu=\Phi^2/c^2-\mathbf{A}^2$ should be a Lorentz invariant. What is the physical significance of $\Phi^2/c^2-\mathbf{A}^2$?
Thanks for any help.

It is a four-vector. The absolute value of a potential is generally not important. The gravitational potential, for example, can be set to zero at an arbitrary point. The electromagnetic potential can be varied by a suitable gauge transformation and still represent the same EM field. So, again, the magnitude of the potential is not physically significant.

 

[Orodruin]

More generally, the 4-potential is only physical up to a gauge transformation $A_\mu \to A_\mu + \partial_\mu \phi$, where $\phi$ is an arbitrary scalar field.

 

[Sagittarius A-Star]

As others mentioned, physical effects in the classical EM theory depend only on related gauge invariant quantities. In the quantum-mechanical Aharonov-Bohm effect, this is also valid for measurable quantities, like the phase difference between the wave functions of both coherent electron beams. But the phase of the wave function of a single beam itself, wich is not measurable, depends in theories, that allow only local interactions, on the electric and magnetic potentials directly.

Source (see free PDF):
https://journals.aps.org/pr/abstract/10.1103/PhysRev.115.485
via:
https://en.wikipedia.org/wiki/Aharonov–Bohm_effect

additional source:
https://web.archive.org/web/2017092...ontent.cgi?article=1118&context=physicsfacpub

 

[pervect]

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

The Aharonov–Bohm effect, sometimes called the Ehrenberg–Siday–Aharonov–Bohm effect, is a quantum-mechanical phenomenon in which an electrically charged particle is affected by an electromagnetic potential, despite being confined to a region in which both the magnetic field and electric field E are zero.

 

[samalkhaiat]

About the electromagnetic 4-vector $A^\mu=(\Phi/c,\mathbf{A})$. If it is indeed a four-vector, then its squared length $A_\mu A^\mu=\Phi^2/c^2-\mathbf{A}^2$ should be a Lorentz invariant. What is the physical significance of $\Phi^2/c^2-\mathbf{A}^2$?
Thanks for any help.

The potential $A_{\mu}$ is NOT a 4-vector. The fact that we can always choose $A_{0}$ to be zero in ALL Lorentz frame show that $A_{\mu}$ cannot be a 4-vector. Indeed, there exists no 4-vector representation $(1/2,1/2)$ for massless particle of helicity $\pm 1$. Under the action of the Lorentz group, the potential transforms according to $$A_{\mu}(x) \to \Lambda^{\nu}{}_{\mu}A_{\nu}(\Lambda x) + \partial_{\mu} f (x , \Lambda).$$ This is why some people say that $A_{\mu}$ is a 4-vector up to a $U(1)$ gauge transformation. The same situation occurs in general relativity where $\Gamma^{\mu}_{\nu \rho}(x)$ is not a tensor. Both $A_{\mu}$ and $\Gamma^{\mu}_{\nu \rho}(x)$ are CONNECTIONS in their respective group.
We cannot associate physical meaning to $A_{\mu}A^{\mu}$ simply because it is NOT Lorentz invariant.

 

[Orodruin]

NOT Lorentz invariant.

For a particular fixed gauge it is Lorentz invariant (and here I am considering gauges that fix $A_0=0$ and similar to be different if fixing that in different frames). What it is not is gauge invariant.

 

[samalkhaiat]

What it is not is gauge invariant.

Gauge invariance must hold in all Lorentz frames, otherwise no physical meaning can be associated to a given quantity.

Also, the gauge-invariant Maxwell equations are Lorentz covariant provided that the connection transforms as $$\bar{A}^{\mu}(\bar{x}) = \Lambda^{\mu}{}_{\nu}A^{\nu}(x) + \partial^{\mu}\lambda (x, \Lambda).$$ In the radiation gauge $A^{0} = 0, \ \partial_{j}A^{j} = 0$, you can show that the gauge function is given by $$\lambda (x , \omega) = - \omega^{0}{}_{j} \int d^{3}y \frac{1}{4 \pi |\vec{x} - \vec{y}|} \partial_{t}A^{j}(t , \vec{y}).$$

 

[martinbn]

The potential $A_{\mu}$ is NOT a 4-vector. The fact that we can always choose $A_{0}$ to be zero in ALL Lorentz frame show that $A_{\mu}$ cannot be a 4-vector.

I am confused here. Surely $A^\mu$ is a vector and $A_\mu$ is a one-form?

 

[Orodruin]

Gauge invariance must hold in all Lorentz frames, otherwise no physical meaning can be associated to a given quantity.

Irrelevant to the point. Given a gauge fixing $A_\mu$ are the components of a 4-vector. When you fix the gauge by letting $A_0 =0$, this states that the zero component of $A$ is zero in a particular frame. This is obviously not the same gauge fixing condition as applying the same kind of condition in a different frame.

I am confused here. Surely $A^\mu$ is a vector and $A_\mu$ is a one-form?

In relativity this distinction is many times glossed over as the metric provides a natural mapping between the two.

 

[martinbn]

In relativity this distinction is many times glossed over as the metric provides a natural mapping between the two.

I am OK with that. My confusion comes from the statement that it is not a vector.

 

[PeterDonis]

My confusion comes from the statement that it is not a vector.

I don't think the OP meant "is a 1-form" by "is not a vector".

 

[martinbn]

I don't think the OP meant "is a 1-form" by "is not a vector".

I mean the statement in post #7. May be I misunderstood it, but it seemed to me to say that $A$ is not a vector.

 

[PeterDonis]

I mean the statement in post #7. May be I misunderstood it, but it seemed to me to say that $A$ is not a vector.

That subthread is more complicated, but as far as I can tell, it also did not mean "is a 1-form" by "is not a vector".

"Vector" can be a general term meaning "is an element of a vector space", rather than the more specific "has one upper index". 1-forms are elements of vector spaces just as much as (upper index) vectors are. I think post #7 is using "vector" in that more general sense.

 

[martinbn]

That subthread is more complicated, but as far as I can tell, it also did not mean "is a 1-form" by "is not a vector".

"Vector" can be a general term meaning "is an element of a vector space", rather than the more specific "has one upper index". 1-forms are elements of vector spaces just as much as (upper index) vectors are. I think post #7 is using "vector" in that more general sense.

It is not about vectors vs 1-forms. In that post it says that it is not a tensor of any kind. Just like the Christoffel symbols $\Gamma$. They are both conections.

 

[PeterDonis]

In that post it says that it is not a tensor of any kind. Just like the Christoffel symbols $\Gamma$. They are both conections.

As I understand the subthread in question, that's because @samalkhaiat is requiring that the vector space of which $A_\mu$ would have to be an element would have to be the same regardless of gauge--which condition cannot be satisfied for $A_\mu$ alone, you have to include the extra term $\partial_\mu \lambda$ that appears as soon as you change gauge.

Whereas @Orodruin is allowing the vector space of which $A_\mu$ is an element to change when you change gauges--so once you've picked a gauge, $A_\mu$ by itself, in that gauge, is a vector. But in a different gauge, the $A_\mu$ in that gauge would be an element of a different vector space; it's not "the same vector".

 

[Orodruin]

Whereas @Orodruin is allowing the vector space of which Aμ is an element to change when you change gauges--so once you've picked a gauge, Aμ by itself, in that gauge, is a vector. But in a different gauge, the Aμ in that gauge would be an element of a different vector space; it's not "the same vector".

My point was that $A^\mu A_\mu$, given that the gauge is fully fixed, is a Lorentz scalar and therefore Lorentz invariant contrary to the statement in #7 that it is not. The reason it cannot be assigned physical relevance is that it is not gauge invariant.

 

[samalkhaiat]

I am confused here. Surely $A^\mu$ is a vector and $A_\mu$ is a one-form?

The potential with index up is not a contravariant Lorentz vector because it does not transform like $A^{\mu} \to \Lambda^{\mu}{}_{\nu}A^{\nu}$. And the potential with index down (i.e., the components of the connection 1-form) is not a covariant Lorentz vector because it does not transform like $A_{\mu} \to (\Lambda^{-1})^{\nu}{}_{\mu}A_{\nu}$.

 

[Orodruin]

The potential with index up is not a contravariant Lorentz vector because it does not transform like $A^{\mu} \to \Lambda^{\mu}{}_{\nu}A^{\nu}$. And the potential with index down (i.e., the components of the connection 1-form) is not a covariant Lorentz vector because it does not transform like $A_{\mu} \to (\Lambda^{-1})^{\nu}{}_{\mu}A_{\nu}$.

This is nonsense. Given a gauge fixing, it transforms exactly like that under Lorentz transformations.

 

[PeterDonis]

$A^\mu A_\mu$, given that the gauge is fully fixed, is a Lorentz scalar and therefore Lorentz invariant

The potential with index up is not a contravariant Lorentz vector because it does not transform like $A^{\mu} \to \Lambda^{\mu}{}_{\nu}A^{\nu}$. And the potential with index down (i.e., the components of the connection 1-form) is not a covariant Lorentz vector because it does not transform like $A_{\mu} \to (\Lambda^{-1})^{\nu}{}_{\mu}A_{\nu}$.

These statements seem to contradict each other. Either one is wrong, or there is a missing piece somewhere. And the discussion of this in general is probably "A" level and hence is probably too much for an "I" level thread. Is there an "I" level resolution?

 

[PeterDonis]

there exists no 4-vector representation $(1/2,1/2)$ for massless particle of helicity $\pm 1$.

Why not?

 

[samalkhaiat]

Irrelevant to the point.

It is the point, $A_{\mu}A^{\mu}$ is not physically meaningful quantity because it is neither gauge nor Lorentz invariant.

Given a gauge fixing $A_\mu$ are the components of a 4-vector. When you fix the gauge by letting $A_0 =0$, this states that the zero component of $A$ is zero in a particular frame. This is obviously not the same gauge fixing condition as applying the same kind of condition in a different frame.

I don’t know what you mean by those words. Can you translate your words to mathematical expressions so that I can understand them.

I suggest you read Weinberg QFT Vol 1 book. The second paragraph on page 251, he concludes: “The fact that $a^{0}$ vanishes in all Lorentz frames shows vividly that $a^{\mu}$ cannot be a four-vector. Instead, …… , so that under a general Lorentz transformation $$U(\Lambda)a_{\mu}(x)U^{-1}(\Lambda) = \Lambda^{\nu}{}_{\mu}a_{\nu}(\Lambda x) + \partial_{\mu}\Omega (x, \Lambda), \ \ \ \ \ (5.9.31)$$ ……” At the bottom of the same page he writes: “… $$f_{\mu\nu} = \partial_{\mu}a_{\nu} - \partial_{\nu}a_{\mu} . \ \ \ \ \ (5.9.34)$$Note that this is a tensor even though $a^{\mu}$ is not a four-vector, because the extra term in Eq.(5.9.31) drops out in Eq.(5.9.34).”

You may also need to see the explanation in Bjorken and Drell, Vol 2, page 74 and problem 2 on page 81, which I posted the solution on these forums many years ago.

Look, the reason that QED is gauge and (formally) Lorentz invariant is down to the fact that we couple the potential $A_{\mu}$ to a conserved 4-vector current $j^{\mu}$.

 

[PeterDonis]

Given a gauge fixing $A_\mu$ are the components of a 4-vector. When you fix the gauge by letting $A_0 =0$, this states that the zero component of $A$ is zero in a particular frame. This is obviously not the same gauge fixing condition as applying the same kind of condition in a different frame.

In other words, if I fix the gauge in a particular frame by setting $A_0 = 0$ in that frame, and then I do a Lorentz transformation to a different frame (without changing gauge), I will not have $A_0 = 0$ in the new frame, correct? So "choosing a gauge", at least in this case, presupposes that I have also chosen a particular frame.

 

[PeterDonis]

"choosing a gauge", at least in this case, presupposes that I have also chosen a particular frame.

I say "at least in this case" because there is another choice of gauge, $\partial_\mu A^\mu = 0$, which would appear to be preserved by a Lorentz transformation, so if I make that gauge choice I don't have to choose any particular frame.

 

[Orodruin]

“The fact that $a^{0}$ vanishes in all Lorentz frames shows vividly that $a^{\mu}$ cannot be a four-vector.

But $A^0$ does not vanish in all frames if you have a fixed gauge condition. If you choose to also impose $A'^0 = 0$ as you make a Lorentz transformation, then you have done a Lorentz transformation and a gauge transformation. You are choosing to impose a gauge depending on your frame rather than imposing your gauge condition in a Lorentz invariant manner, i.e., $U \cdot A = 0$ with $U$ being $(1,0,0,0)$ in some given frame.

Perhaps it is an issue of what each of us consider the "Lorentz transformed" $A$, which boils down to what gauge condition is applied. When you say $A^0 = 0$ I would primarily take that to refer to a single frame only. "The same" gauge in another frame would instead be $U\cdot A = 0$ with $U$ being the 4-velocity of the original frame. The alternative is imposing (somewhat artificially in my meaning) that $A^0$ in all frames. To me that is mixing the gauge and Lorentz transformations, but if that is what you want to do ...

 

[Orodruin]

In other words, if I fix the gauge in a particular frame by setting $A_0 = 0$ in that frame, and then I do a Lorentz transformation to a different frame (without changing gauge), I will not have $A_0 = 0$ in the new frame, correct? So "choosing a gauge", at least in this case, presupposes that I have also chosen a particular frame.

Yes. You can, of course, make an additional gauge transformation such that $A_0 = 0$ in the new frame along with your Lorentz transformation if that is your wish, but to me that is artificially imposing a gauge condition that is explicitly Lorentz violating. Then we should not be surprised that the resulting $A$ does not transform as a 4-vector as the gauge condition itself is where the Lorentz violation appears.

Edit: The Lorentz transformed $A_\mu$ is, of course, also a valid 4-potential. It is just not in a gauge where $A_0 = 0$ in the new frame. It is in a gauge where $U\cdot A = 0$ (in all frames) with $U$ being the 4-velocity of the original frame.

 

[Sagittarius A-Star]

I suggest you read Weinberg QFT Vol 1 book. The second paragraph on page 251, he concludes: “The fact that $a^{0}$ vanishes in all Lorentz frames shows vividly that $a^{\mu}$ cannot be a four-vector.

On page 250 in the chapter "5.9 Massless Particle Fields", Weinberg called the $a_\mu(x)$ a "polarization vector":

Let's temporarily close our eyes to this difficulty, and go ahead anyway, using ... to define a polarization vector for arbitrary momentum, and take the field as $a_\mu(x) =$ ...

On page 251, before the sentence you quoted, he wrote:

As we shall see in Chapter 9, these are conditions satisfied by the vacuum vector potential of electrodynamics in what is called Coulomb or radiation gauge.

At this stage in the text, he did not yet call $a_\mu(x)$ a vector potential. Then he solved the earlier mentioned difficulty by replacing the earlier used equation (5.9.6) by equation (5.9.30).

 

[samalkhaiat]

Why not?

The mathematical details is given on the pages 248 and 249 of Weinberg text. The math is easy if you are familiar with the technics of “little group”. If you are not, then observe the following problem: Consider the propagator of the massive vector field representation $(1/2,1/2)$ (which is a true Lorentz vector) $$\Delta_{\mu\nu}(x,y) \sim \int d^{4}x \ e^{ip(x - y)} \frac{\eta_{\mu\nu} + p_{\mu}p_{\nu}/m^{2}}{p^{2}+ m^{2} - i \epsilon} .$$ Since this propagator has singularities at $m = 0$, we cannot obtain the theory of massless field of helicity $\pm 1$ by simply taking the limit $m \to 0$ of the theory of massive vector field of spin 1. Therefore, the massless field of helicity $\pm 1$ cannot be described by a true Lorentz vector.

 

[samalkhaiat]

On page 250 in the chapter "5.9 Massless Particle Fields", Weinberg called the $a_\mu(x)$ a "polarization vector":
On page 251, before the sentence you quoted, he wrote:
At this stage in the text, he did not yet call $a_\mu(x)$ a vector potential. Then he solved the earlier mentioned difficulty by replacing the earlier used equation (5.9.6) by equation (5.9.30).

Weinberg constructs the field $a_{\mu}(x)$ from the polarization “vector” $e_{\mu}(\vec{p}, \sigma)$ and the annihilation and creation operators. Since the little group technics, he uses, is simpler to apply directly to the polarization vector, he derives the properties of the field $a_{\mu}$ from those of the polarization vector $e_{\mu}(\vec{p}, \sigma)$.
In section (5.9) of the book, Weinberg tries to answer the following equivalent questions: Can massless field of helicity $\pm 1$ be described by a true Lorentz vector? Does the field $a_{\mu}(x)$ transform like $a^{\mu} \to \Lambda^{\mu}{}_{\nu}a^{\nu}$? Or, can the series of massless representations of the Lorentz group contain a true Lorentz vector? His answer was negative: “We have thus come to the conclusion that no four-vector field can be constructed from the annihilation and creation operators for a particle of mass zero and helicity $\pm 1$” He then realized that the best one can do is to take the Lorentz transformation of polarization “vector” to be of the form $$e^{\mu}(\vec{p}_{\Lambda}, \pm 1 ) e^{\pm i \theta (\vec{p}, \Lambda)} = D^{\mu}{}_{\nu}(\Lambda) e^{\mu}(\vec{p}, \pm 1) + p^{\mu} \Omega_{\pm} (\vec{p}, \Lambda). \ \ (5.9.30)$$ When translated to the field $a_{\mu}(x)$, Eq(5.9.30) becomes $$U(\Lambda)a_{\mu}(x)U^{-1}(\Lambda) = \Lambda^{\nu}{}_{\mu}a_{\nu}(\Lambda x) + \partial_{\mu}\Omega (x, \Lambda). \ \ \ (5.9.31)$$ The presence of the second term makes the field $a_{\mu}(x)$ to be a gauge 4-potential instead of a true Lorentz 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