spatial average is the macroscopic polarization. Similarly, ${\bf {\cal M}_0(r)}$ is the magnetization field, that is, the quantity for which the spatial average is the part of the macroscopic magnetization that does not originate from the time derivative of the polarization field [@gauge]. By plugging this latter equality into Eq. (\[eqm1\]), we have: $$\label{eqn6}
{\bf M} =\frac{1}{2cV} \int{ \left ( {\bf r} \times  \dot {{\cal P}}({\bf r)}\right ) d^3r}+
\frac{1}{2V} \int{ \left ( {\bf r} \times  {\bm \nabla}\times {\bf {\cal M}_0(r)} \right ) d^3r}
=\frac{1}{2cV} \int{ \left ( {\bf r} \times \dot {{\cal P}}({\bf r)}\right ) d^3r} + {\bf M}_0$$ The analytical expression of this latter equation bears some similarities with the definition of the electrical toroidal moment, [**G**]{}, that is [@Dubovik] $$\label{eqn1}
{\bf G}=\frac{1}{2V}\int {\left ( {\bf r} \times {\bf {\cal P}(r)}  \right ) d^3r}~~~,$$

More precisely, taking the time derivative of [**G**]{} gives: $$\label{eqntds}
\dot{\bf G} \simeq \frac{1}{2V} \int{ \left ( {\bf r} \times \dot {{\cal P}}({\bf r)}\right ) d^3r}$$ when omitting the time dependency of the volume (the numerical simulations presented below indeed show that one can safely neglect this dependency when computing the time derivative of the electric toroidal moment).

As a result, combining Eq. (\[eqntds\]) 