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Griffith's Introduction to Electrodynamics when introducing Faraday's Law, considers two scenarios:

  1. Moving loop on a magnetic field. Current flows due to the motional EMF.
  2. The same scenario but on the frame of reference on of the loop.

I understand that even if $\left.\frac{\partial \mathbf{B}}{\partial t}\right\vert_{\text{wire}} = 0$ having $\left.\frac{\partial B}{\partial t}\right\vert_{\text{other region}} = \left.\nabla\times \mathbf{E}\right\vert_{\text{other region}} \neq 0$ can induce an $\left.\mathbf{E}\right\vert_{\text{wire}} \neq 0$ over the wire.

But, consider you have an infinite uniform stationary field $\mathbf{B}$. Then on an "stationary" frame of reference a charge moving with velocity $v$ would experience a force. On any other inertial frame of reference it could experience any other force (no force, in particular, if the frame you chose is the one of the charge). What's the argument there? $\frac{\partial \mathbf{B}}{\partial t} = 0$ all over the space.

I've read the answer might reside in special relativity. But electromagnetic theory was concived before special relativity, and this seems a discrepancy way too big to left over without an explanation. I mean, I'm not concerned about speeds comparable to speed of light, it's just galilean invariance.

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    $\begingroup$ Electromagnetism truly resides within special relativity. It has is no Gallilean invariance. However, you can use the special relativistic equations and take $v/c << 1$ to see what happens. The answer is that in a different frame of reference, there won't just be a constant $B$ field, but also a constant $E$ field, which explains the force felt by the particle. $E$ and $B$ transform into each other in a certain way under Lorentz transformations. $\endgroup$ – user1379857 Feb 16 at 0:33
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    $\begingroup$ Also, while electromagnetism was conceived of before relativity, it is ultimately inconsistent with Gallliean invariance. For instance, Maxwell's laws predict waves which move at $c$.... but in whose reference frame? It was Einstein who was asking questions like, "what would happen if I were to ride alongside a beam of light" and stuff like that. Before Einstein, people thought Maxwell's laws were only true in one special frame, the frame of the "aether." Einstein realized that the laws were actually true in all frames, i.e. light moves at $c$ for all observers. $\endgroup$ – user1379857 Feb 16 at 0:36
  • $\begingroup$ Ok. I think your second comment is pointing the issue... There is a hidden premise in my 101 electromagnetism course where we assumed we were working on a special frame of reference ("aether" frame). They (Maxwell & Co.) belived "aether" 's frame followed earth? $\endgroup$ – MarcoCiafa Feb 16 at 0:58
  • $\begingroup$ I don't think Maxwell specifically was a big 'aether theorist' (I could be wrong) but I do know that people didn't think the earth was particularly special with respect to the aether. Because the earth orbits the sun, its velocity with respect to the aether would be different at different points in the year. The Michelson Morley experiment was supposed to detect this difference throughout the year, but didn't find anything. It became one of the foundational experiments in favor of special relativity. $\endgroup$ – user1379857 Feb 16 at 2:32
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    $\begingroup$ This is how "Lorentz contraction" was actually discovered, before special relativity was known. They thought that all objects moving with a velocity with respect to the aether had their lengths contracted by $\sqrt{1 - v^2 / c^2}$. en.wikipedia.org/wiki/Length_contraction#History $\endgroup$ – user1379857 Feb 16 at 19:45
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Ultimately the explanation, like you said, requires us to invoke special relativity. However, this explanation might be opaque at first, so I'll add some physics-y interpretation after sketching the SR calculation.

The electric and magnetic fields form components of a 2-dimensional tensor, $F_{\mu\nu}=\partial_\mu A_\nu - \partial_\nu A_\mu$, where $A_\mu$ is a four-vector$^\star$ whose components are given by the normal scalar and vector potentials: $A_0=\phi$ and $A_i$ are the components of the vector potential $\vec{A}$, in units where $c=1$.

The reason to bring this up is that it doesn't make sense to say you only have a magnetic field in two different frames. In one of the two frames you mentioned, you must have an electric field. This is a consequence of the transformation laws \begin{eqnarray} \vec{E}'_{||} &=& \vec{E}_{||} \\ \vec{B}'_{||} &=& \vec{B}_{||} \\ \vec{E}'_{\perp} &=& \gamma (\vec{E}_{\perp} + \vec{v} \times \vec{B}_\perp )\\ \vec{B}'_{||} &=& \gamma (\vec{B}_{\perp} - \vec{v} \times \vec{E}_\perp ) \end{eqnarray} where primed quantities are evaluated in the boosted frame and unprimed in the original frame, $\gamma=\sqrt{1-v^2}$ is the usual Lorentz factor, $\vec{B}_{||}$ refers to the components of $\vec{B}$ parallel to the velocity of the boost, $\vec{B}_\perp$ refers to the component of the magnetic field perpendicular to the velocity, I've made the same definitions for the electric field, and as a reminder I've set $c=1$.

So now we apply these formulas. Let's suppose that in the original frame, the ring is moving in the $x$ direction, and the field is purely magnetic and the magnetic field is pointed in the $z$ direction. Then the Lorentz force $\vec{F}\propto \vec{v}\times \vec{B}$ will point in the $y$ direction. Then in the rest frame of the ring, we find that there must be an electric field with a $y$ component given by \begin{equation} E_y' = - \gamma v_x B_z \end{equation} This electric field provides the force in the $y$ direction you were worried about.

OK but this doesn't exactly answer the question of where does this electric force come from? The answer is that you have made an idealization by considering an infinitely large magnetic field, and in doing so you've lost track of what is generating the magnetic field in the first place. There must evidently be some currents "at infinity", but no net charge because there is no electric field. You need to account for length contraction when you apply the boost, which will tend to create a net charge density, following an analysis present in (for example) the book by Purcell.

Having said all of this, what will happen in this case is that negative charges will tend to "pile up" at the maximum and "deplete" at the minimum of the ring along the $y$ direction. This charge separation will provide a counter electric force that will oppose and eventually stop any current. So there is not a current when the ring reaches equilibrium. This may be a reason why this case is not typically studied explicitly.

Einstein did consider a related problem as motivation for special relativity. He not consider the problem of a constant magnetic field over all space, but rather a finite conductor generating the magnetic field which could generate a steady current via induction. The fact that in different frames there were different explanations for why there was a current (motional emf vs the Lorentz force) bothered him, and he states his motivation for developing special relativity in his classic 1905 paper is to give a deeper explanation that works in any inertial reference frame.

$^\star$ well, if you are careful about gauge invariance, $A_\mu$ isn't really a 4 vector, but let's ignore this.

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  • $\begingroup$ Thanks very much for your answer. I will certantly read your it, but since I'm not literate on SR (I know you formulated in a way it's not necesary, but it seems unfeasible to fill that patch without it) I want to do it thoughtfully. $\endgroup$ – MarcoCiafa Feb 16 at 1:12
  • $\begingroup$ @MarcoCiafa Please feel free to ask questions if I can help to bridge the gap. I think the main points are: (a) in the "pure magnetic field" frame with $B$ pointing in the $z$ direction and the ring moving in the $x$ direction, the Lorentz force will generate a force in the $y$ direction, (b) special relativity will generate an electric field $E'_y=\gamma v_x B_z$ in the "ring frame" that provides the force in the $y$ direction, (c) this field comes from length contraction applied to the currents that generated the original magnetic field, (d) however, there isn't a steady current in the ring. $\endgroup$ – Andrew Feb 16 at 1:17
  • $\begingroup$ Sorry I took so long in replying. I'll check for the transformations laws for electric and magnetic field in my CM notes and that Purcell's apendix you mentioned. Thanks. $\endgroup$ – MarcoCiafa Mar 9 at 3:28

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