How does magnetic moment transform under Lorentz transformation? If I have a particle with magnetic moment $\vec{\mu}$ in its rest system $S$, what will its new magnetic moment $\vec{\mu}'$ be if it is moving at speed $\vec{v}$ in a reference system $S'$?
 A: The magnetic moment is given by
$$ \mu = \frac{1}{2} \int r \times J \, \mathrm{d}^3 x $$
In order to know how this transforms, we have to re-express it in terms of Lorentz-covariant quantities. For starters we can write $(r \times J)_i = \epsilon_{ijk} r_j J_k $ so
$$ \mu_i = \frac{1}{2} \int \epsilon_{ijk} r_j J_k \, \mathrm{d}^3 x $$
Therefore
\begin{align}
\epsilon_{ilm} \mu_i &= \frac{1}{2} \int \epsilon_{ilm} \epsilon_{ijk} r_j J_k \, \mathrm{d}^3 x \\
&= \int (\delta_{lj} \delta_{mk} - \delta_{lk} \delta_{mj}) r_j J_k \, \mathrm{d}^3 x \\
&= \int r_l J_m - r_m J_l \, \mathrm{d}^3 x
\end{align}
where we have used a common identity for the contraction of two $\epsilon$ symbols.
We could replace $r$ with $x$ (the four-dimensional space-time position) and $J$ with the four-current, but the right-hand side would still not be a Lorentz tensor because the volume element $\mathrm{d}^3 x$ isn't Lorentz invariant. So we must write $\mu_i$ as the volume integral of the magnetization, whereupon
$$ \int \epsilon_{ilm} M_i \, \mathrm{d}^3x = \int r_l J_m - r_m J_l \, \mathrm{d}^3x $$
Finally, we can drop the integration to get a relation between two tensors of order 2 representing volume density. The antisymmetric tensor $\epsilon_{ilm} M_i$ in dimension 3 is replaced by an antisymmetric Lorentz tensor $M^{lm}$ in dimension 4 having $\epsilon_{ilm} M_i$ as the spatial submatrix:
$$ M^{lm} = x^l J^m - x^m J^l $$
By construction, there are 6 nonzero space/space components, of which only three are independent (due to antisymmetry) and those three independent components correspond with the components of the magnetization. It turns out (exercise for the reader) that the mixed (space/time) components of $M$ are the electric polarization. The tensor $M$ is called the polarization-magnetization tensor.
It follows that in order to determine the magnetic dipole moment in another frame, you need to know the electric dipole moment in this frame. Just knowing the magnetic dipole moment alone, you don't have enough information to determine the magnetic dipole moment in another frame.
If we assume that the electric dipole moment in the original frame is zero, then, after a Lorentz boost is applied, we find that the magnetization in the frame where the dipole is moving with velocity $v \hat{z}$ is given by $(\gamma \mu_x, \gamma \mu_y, \mu_z)$. The dipole is also length contracted in the z direction by a factor of $\gamma$, so its volume is reduced by a factor of $\gamma$. It follows that the new dipole moment is $(\mu_x, \mu_y, \frac{1}{\gamma} \mu_z)$.
This transformation law is, not surprisingly, the same as that of a pure electric dipole: the boost results in a reduction by a factor of $\gamma$ in the extent of charge separation along the direction of the boost only (leaving the charges themselves and the perpendicular directions unaffected) so the dipole moment in that direction is reduced by a factor of $\gamma$.
A: The magnetic moment is part of a skew symmetric 2-tensor $M_{\mu\nu}$. In the rest frame only the space components $M_{ij}= \epsilon_{ijk} \mu_k$ are non zero. In a boosted frame  non-zero electric dipole terms $M_{i0}$ are induced by the Lorentz transformation of the magnetic field.
