{\omega ,{\rm {\bf k}}} \right) \\
=\delta _{ik} +\frac{4\pi i}{\omega }(\sigma _{ik}^{(0)}\left( {\omega } \right)+\sigma_{ikl}k_l)$$ where $\delta _{ik}$ is the Kronecker symbol and $\sigma _{ik}\left( {\omega ,{\rm {\bf k}}} \right)$ is the effective conductivity tensor in the reciprocal space, at a given frequency [@Melrose]. $\sigma_{ikl}$ is the third-rank tensor associated with the linear dependence of the effective conductivity tensor on the wave vector, and $\sigma _{ik}^{(0)}$ is the effective conductivity tensor at zero wave vector. Combining Eqs. (\[eq19\]) with Eq. (\[eq15\]) yields: $$\label{eq20}
\gamma _{ikl} =\frac{4\pi}{\omega }\sigma_{ikl}\\
=\frac{4\pi}{\omega }(\sigma_{ikl}^S(\omega)+\sigma _{ikl}^A(\omega))$$ where $$\label{eq16b}
\sigma_{ijk}^A =\frac{1}{2}(\sigma_{ijk}-\sigma_{jik})$$ and $$\label{eq16c}
\sigma _{ijk}^S =\frac{1}{2}(\sigma_{ijk}+\sigma_{jik})$$ Moreover, using the results of Ref. [@Malashevich] and working at nonabsorbing frequencies (i.e., frequencies, such as GHz in ferroelectrics, for which the corresponding energy is below the band gap of the material), one can write

$$\label{eq16}
\sigma _{ijk}^A =ic\left( {e_{jkl} \beta _{il} -e_{ikl} \beta _{jl} }
\right)+\omega \xi _{ijk}$$

with $$\label{eq17}
\beta _{ij} =i\mathrm{Im}(\chi _{ij}^{em}) \mathop = -i\mathrm{Im}(\chi _{ji}^{me})$$ and $$\label{eq18}
\xi _{ijk} =\frac{1}{2}\left[ {\frac{dQ_{kj} }{dE_i }-\frac{dQ_{ki} }{dE_j
}} \right]$$ where $\mathrm{Im}$ stands for the imaginary part and $Q$ is the quadrupole moment of the system [@Raab]. $\chi^{me}$ is the response of the magnetization, [**M**]{}, to an electric field [**E**]{}, while 