$\chi^{em}$ is the response of the electrical polarization, [**P**]{}, to a magnetic field [**B**]{}, that is: $$\label{eq22}
\chi _{ij}^{me} =\frac{dM_i }{dE_j } \\ \nonumber ~~~{\rm and} ~~~
\chi _{ji}^{em} =\frac{dP_j }{dB_i}$$

Inserting Eq. (\[eq16\]) into Eq. (\[eq20\]) provides : $$\label{eq21}
\gamma _{ijk} =\frac{4\pi }{\omega }\left[ {c\left( {e_{jkl} \mathrm{Im}\,\chi
_{li}^{me} -e_{ikl} \mathrm{Im}\,\chi _{lj}^{me} } \right)+\omega \xi _{ijk} } \right] + \gamma_{ijk}^S$$ where $\gamma_{ijk}^S=(4\pi/\omega)\sigma_{ijk}^S$ is the contribution of the symmetric part of the conductivity to the $\gamma$ tensor. As a result, $\gamma_{ijk}^S$ is non-zero only when the system is magnetized or possesses a spontaneous magnetic order [@Landau].

Let us now focus on the magnetization, which can be written as [@Raab]: $$\label{eqm1}
{\bf M} =\frac{1}{2cV} \int {\left ( {\bf r} \times {\bf {\cal J}(r)}\right ) d^3r }$$ where $c$ is the speed of light, $V$ is the volume of the system, [ **r**]{} is the position vector, and ${\bf {\cal J}(r)}$ is the current density. We consider here the following contributions to this density: $$\label{eq6}
{\bf {\cal J}(r)}=\dot {{\cal P}}({\bf r)}+c~ {\bm \nabla}\times {\bf {\cal M}_0(r)}$$ where the dot symbol refers to the partial derivative with respect to time. ${\bf {\cal P}(r)}$ is the polarization [*field*]{}, that is, the quantity for which the 