I am not sure which form of Maxwell's equations is fundamental, integral form or differential form. Imagine an ideal infinitely long solenoid. When a current is changing in time, can we detect classical effects outside a solenoid, for example generating a circular current around solenoid by Faraday's law. If the differential form is fundamental, we won't get any current, but the integral form is fundamental we will get a current.
[Physics] Which form of Maxwell’s equations is fundamental, in integral form or differential form
electromagnetismmaxwell-equations
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An electromagnetic wave with a well-defined frequency and direction, i.e. $\vec k$, only has two possible truly physical i.e. transverse polarizations, i.e. the linearly polarized waves in the $x$ and $y$ direction (or the two circularly polarized ones). That implies that a truly physical counting of polarizations gives you 2, more generally $D-2$ in $D$ spacetime dimensions.
Starting from the $A_\mu$ potential fields, one component is unphysical because it's pure gauge, $A_\mu\sim\partial_\mu\lambda$, and one of them is forbidden due to Gauss' constraint $\rm div\,\vec D=0$ etc. that already constrains the allowed initial state of the electromagnetic field. Both of these killed polarizations are ultimately linked to the $U(1)$ gauge symmetry.
If one is allowed to count off-shell and unphysical fields, there may be many more components than two. But it's always possible to deduce that there are two physical polarizations at the end. For example, when we view $\vec B,\vec E$ as basic fields, there are six components, a lot. But these fields only enter Maxwell's equations through first derivatives, and not second as expected for "normal" bosonic fields, so these fields are simultaneously the canonical momenta for themselves. This brings us to three polarizations but one of them is killed by the constraints, the Maxwell's equations that don't contain time derivatives.
The Hertz vector is just the most famous "non-standard" example how to write the electromagnetic field as a combination of derivatives of some other fields. One must understand that the room for mathematical redefinitions etc. is unlimited and it is a matter of pure maths. All these descriptions may describe the same physics. At the end, the only "truly invariant because measurable" number of "fields" that all these approaches must agree about is the number of linearly independent physical polarizations of a wave/photon with a given $\vec k$.
If you can analyze any mathematical formulation of electromagnetism or another field theory and derive that there are $D-2$ physical polarizations (this usually boils down to the difference of the number of a priori fields minus the number of independent constraints and the number of parameters defining identifications i.e. gauge symmetries – but the independence is sometimes hard to see and requires you to make many steps of the counting), then you have proved everything that is "really forced to be true". Various formalisms may offer you other ways to count the number of off-shell fields (with different answers) and they may be useful (because they satisfy certain conditions or enter some laws) but to discuss them, one has to know what the laws where they enter actually are.
A truly physical approach is only one that counts the physical polarizations. The gauge symmetry is just a redundancy, a mathematical trick to get the right theory with 2 physical polarizations out of a greater number of fields with certain extra constraints or identifications. The precise number of constraints or identifications may depend on the chosen mathematical formalism and it is not a physically meaningful question – it is a question of a subjectively preferred mathematical formalism because the physics is equivalent for all of them.
I think a source of difficulty lies in the fact that the solenoid is infinitely long, so strictly speaking there shouldn't be any field lines outside the solenoid. But imagine for a minute the field that the loop sets up due to the induced current. The solenoid's windings will intersect these field lines and so there will be a force on the solenoid. By Newton's third law, one would expect an equal and opposite force on the loop.
Best Answer
I'm not sure how you came to that conclusion, but it's not true. Both the differential and integral forms of Maxwell's equations are saying exactly the same thing. Either can be derived from the other, and both of them predict the exact same physical consequences in any situation.
Most physicists would say the differential form is more fundamental, but that's just an artifact of how we think about modern physics, in terms of fields which interact at specific points. It's really a philosophical issue, not a physical one, because it doesn't matter for the purpose of doing calculations which form you consider to be more fundamental.
In the specific situation you're asking about, with the solenoid, you will get a current in the loop around the solenoid. It may be easier to see that by using the integral form of Faraday's law, but the differential form makes the exact same prediction.
Let me demonstrate this explicitly. Suppose you have an ideal solenoid of radius $r_0$, with $n$ turns per unit length, oriented along the $z$ axis. Its magnetic field is given by
$$\vec{B} = \begin{cases}\mu_0 n I\hat{z} & r < r_0 \\ 0 & r > r_0\end{cases}$$
As you've noticed, this implies that $\nabla\times\vec{E} = 0$ outside the solenoid. Now, you might think that implies the integral $\oint\vec{E}\cdot\mathrm{d}\ell$ around a loop outside the solenoid, which gives the EMF, must be zero. But that's not actually the case. The relationship between $\nabla\times\vec{E}$ and $\oint\vec{E}\cdot\mathrm{d}\ell$ comes from Stokes' theorem, and it says
$$\oint_{\mathcal{C}}\vec{E}\cdot\mathrm{d}\ell = \iint_{\mathcal{S}}(\nabla\times\vec{E})\cdot\mathrm{d}^2\vec{A}$$
So the line integral around the loop is determined by the curl of $\vec{E}$ everywhere inside the loop, including inside the solenoid where
$$\nabla\times\vec{E} = -\mu_0 n \frac{\partial I}{\partial t}\hat{z}\quad(r < r_0)$$
Performing the integral gives you
$$\mathcal{E} = \oint_{\mathcal{C}}\vec{E}\cdot\mathrm{d}\ell = \iint_{\mathcal{S}}(\nabla\times\vec{E})\cdot\mathrm{d}^2\vec{A} = -\int_0^{2\pi}\int_{0}^{r_0}\mu_0 n \frac{\partial I}{\partial t}\hat{z}\cdot r\mathrm{d}r\,\mathrm{d}\theta\,\hat{z} = -\mu_0 \pi r_0^2 n\frac{\partial I}{\partial t}$$
so you can see that any time-varying current in the solenoid will create an EMF and induce a current.