Is the current in a loop accelerating in a uniform magnetic field the same for moving and stationary observer?

Question

By Faraday's law, a changing magnetic flux through a loop produces electromotive force $\varepsilon=-\frac{\mathrm{d}\Phi}{\mathrm{d}t}$. Suppose there is a rectangular loop in XY plane with Ohmic resistance $R$ and dimensions $a \times b$ moving at speed $v$ relative to an observer, in $x$ direction, in a uniform magnetic field $\vec{B}=b\vec{\hat{z}}$. Magnetic flux through the a loop changes for the stationary observer due to length contraction $$\Phi=B a (b / \gamma),$$ inducing current $I=\varepsilon / R = \frac{B a b}{R \gamma^2} \frac{\mathrm{d}\gamma}{\mathrm{d}t}$ ($t$ is time time in stationary frame). This means that when loop accelerates, it will have smaller area, so the flux changes, and therefore there is current induced in the loop.

The formula is the same from the loop's perspective. If we gauge $\vec{B}$ as $\vec{B}=\nabla \times \vec{A}$ where $\vec{A}=B x \vec{\hat{y}}$, we get that in loop's frame $A' = B (x' - vt') / \gamma \vec{\hat{y}}$ so $B'=B / \gamma$, meaning $\Phi = (B / \gamma) a b$ and we recover the same result.

Therefore current induced in the loop in both frames is the same $I=I'$. But (four-)current should also transform under Lorentz transformations - what did I miss?

My first idea was that I am misunderstanding four current transformation and current is a scalar in this context. But if we now imagine the cross-section of the loop, and charge passing through that cross section in time $\delta t$ ($\delta t'$ in moving frame) as $Q$ ($Q'$). Then $\delta t' = \frac{1}{\gamma} \delta t$ by time dilation and $Q=Q'$ since it is scalar. So in that light it should be $I' = \gamma I$ so clearly I am missing something.

Answer 1

Four-current consists of the charge density $\rho$ and current density $J$ at a given point and transforms as a Lorentz vector.

On the other hand, for unsteady currents, the "global" current $I$ is not even well-defined in a single frame. Along a wire loop, the magnitude of $J$ might be different at different points, and it's possible that charge density might accumulate in some regions along the wire (which occurs when $J$ points toward that point (or away from it) from the wire on both sides of it). How would you define the value of $I$ for such a wire loop? Well, if you wait long enough, the current might stabilize or exhibit stable oscillations (i.e. alternating current) and at that point it's possible to talk about whether there is an "overall" transport of charge around the loop over time (and if so, the rate and direction of such transport).

For example, suppose the loop is initially stationary, and suddenly (i.e., within a negligible amount of time) accelerates and then travels at a fixed velocity $v$ perpendicular to the applied magnetic field. We know that the Lorentz force acting on the charges in the wire will result in a charge separation (i.e. the loop will acquire a dipole moment perpendicular to the motion) which will exert a back emf, and eventually this will build up to a point where it cancels the Lorentz force; a steady state will be reached. At that point, we can say the current is 0.

But in the scenario with the accelerating loop, you never reach a steady state, so it's not clear how you would define "the I of the loop". On the other hand you are free to analyze the four-vector $(\rho, J)$ at each point and it should transform as a Lorentz vector. Be careful though—Maxwell's equations in their usual flat space form do not apply in the frame comoving with the wire, since it's not an inertial frame.