assume that all losses are introduced via $\gamma$. To determine the sign of $\eta_E$, $\eta_H$ and $\eta_C$, we expect that for very slow driving the charge (current) on the capacitor directly follows the driving $E$ ($H$)-field, implying $\eta_E>0$ and $\eta_H>0$. The sign of $\eta_C$ follows similar reasoning, After charge build-up, charge associated with a $\bm{p}_x=\mathrm{Re}(\alpha(\omega) \eta_E e^{-i \omega t} E_x)$ relaxes as counter-clockwise current, giving rise to a negative $\bm{m}_z=\mathrm{Re}(\alpha(\omega) i \eta_C e^{-i \omega t} E_x)$, implying that $\mathrm{sign}\,\eta_C=\mathrm{sign}\,\eta_E$.

Limit on magneto-electric coupling\[magnetoelectricsection\]
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Having constructed an electrostatic polarizability in accordance with RLC circuit models proposed in earlier reports, we apply radiation damping according to Eq. (\[Eq:TensorRadiationDamping\]) to obtain a scatterer that has a correct energy balance: $$\boldsymbol{\alpha}_{{SRR}}^{-1}=(\boldsymbol{\alpha}_{SRR}^{\mathrm{static}})^{-1}-\frac{2}{3}k^3\mathbb{I}.\label{Eq:SRRradiationdamp}$$ So far we have not explicitly discussed absorption loss, except through the inclusion of the material damping constant $\gamma$ in the quasi-static polarizability. Starting from a quasi-static polarizability with quasi-static eigenpolarizabilities $A^{\mathrm{static}}_i$, the albedo for each eigenillumination $\bm{v}_i$ can be expressed as $$a_i=\frac{1}{1+\frac{2}{3}k^3\mathrm{Im}A^{\mathrm{static}}_i}.\label{Eq:albedo}$$ It follows that for any lossy scatterer the imaginary part of each eigenvalue $A^\mathrm{static}_i$ of the electrostatic polarizability tensor must be positive to ensure $0\leq a\leq1$. In the case of a tensorial $\boldsymbol{\alpha}$ with loss included as in Eq. 