How to find the surface charge induced on a perfect conductor when Lorentz transforming electromagnetic fields? Suppose that I have a perfect electrical conductor ($B=0$ inside conductor) in free space with a known magnetic field $\mathbf{B_s}$ outside of it, and no electric field. If I transform to a frame of reference moving non-relativistically with velocity $\mathbf{v_{s'}}$, I obtain an electric field $\mathbf{E_{s'}} = \mathbf{v_{s'}} \times \mathbf{B_s}$ outside the conductor. The boundary conditions imply that there is a surface charge density $\sigma$ in this frame on the conductor.
In general, \begin{equation}
\sigma = \epsilon \mathbf{E}.\mathbf{\hat{n}}|_{\rm surface} \tag{1}
\end{equation}
where $\mathbf{\hat{n}}$ is the normal unit vector to the surface of the conductor.
My question(s) :

*

*Whether the electric field vector $\mathbf{E}$ in $Eq. (1)$ is just the vector $\mathbf{E_{s'}}$ or the composite (or total) electric field vector $\mathbf{E_{s'}} + \mathbf{E_{c'}}$, where $\mathbf{E_{c'}}$ is the secondary field due to the surface charges?


*In both cases, how to calculate the total electric field outside the conductor and the surface charge, since I do not know $\mathbf{E_{c'}}$ and $\sigma$ apriori?
 A: Your problem is just the standard problem of a conducting sphere which is solved n every EM textbook for a uniform E field.  If the original B field is not uniform, a Legendre polynomial, spherical harmonic expansion is necessary.  In any event the
total field is the one that counts.
A: *

*The surface charge density is from Maxwell's equations,
$\vec \nabla \cdot \vec E = \rho/\epsilon_0$, so the electric
field discontinuity is the total field, which is your $E'_s$.


*Since you have the fields in the rest frame, you Lorentz transform to your new frame in the standard way. The discontinuity in the transformed electric field is coming from the discontinuity in the original $\vec B$ field. That is, you say $\vec B = 0 $ in the conductor. Therefore the normal component of $\vec B$ is zero outside, but the tangential component is not. Since it is zero inside, the tangential component is discontinuous and there is a surface current which you calculate in the usual way from the curl  of $\vec B$ Maxwell equation. If you Lorentz transform this surface current you will get the surface charge density.
A: Since $\vec{B}_\text{above} \neq 0$ above the surface and $\vec{B}_\text{below} = 0$ below, there must be a 2-D current density $\vec{K} = \frac{1}{\mu_0} \hat{n} \times \vec{B}_\text{above}$ on the surface. Note also that we must have $\hat{n} \cdot \vec{B}_\text{above} = 0$ (unless Cabrera was right all along.)
We can write $\vec{K}$ as a 3-D current density via throwing in a delta function:
$$
\vec{J} = \vec{K} \delta( \vec{n} \cdot \vec{r})
$$
These will form the components of a four-current density $J^\mu = (0, \vec{J})$.  By applying the Lorentz transformations to this four-vector, we can find the components of the four-current in another frame.  In general, the $t$-component of the transformed four-current ${J'}^\mu$ will be non-zero, and it will correspond to the charge density $\rho$ in this frame;  integrating the result over the normal to the surface (eliminating the delta function) will then yield the surface charge.
