radiated by the dipole $\bm{p}$. According to Larmor, the cycle-averaged scattered power is [@jacksonbook] $P=\frac{4\pi k^4}{3 Z}|\bm{p}|^2.$ Hence one obtains the well known cross sections $$\sigma_{\mathrm{ext}}=4\pi k \mathrm{Im}\alpha_{EE}
\quad\mbox{and}\quad
\sigma_{\mathrm{scatt}}=\frac{8\pi}{3}k^4|\alpha_{EE}|^2.\label{Eq:scalarcrossect}$$ Equating extinction to scattering for nonabsorbing particles to impose energy conservation, gives rise to the optical theorem for the polarizability $$\mathrm{Im}\alpha=\frac{2}{3}k^3|\alpha_{EE}|^2
\label{Eq:scalarOpticalTheorem}$$ This equation for instance shows the well-known fact that a real (electrostatic) $\alpha_0$, such as Rayleigh’s polarizability $\alpha=3V(\epsilon -1)/(\epsilon +2)$ for a small sphere of dielectric constant $\epsilon$, never satisfies the optical theorem [@vdHulst]. An electrostatic $\alpha_0$ can be made to satisfy the optical theorem by adding radiation damping [@Pedro98; @Abajo07] to obtain the dynamic polarizability $$\frac{1}{\alpha}=\frac{1}{\alpha_{0}}-i\frac{2}{3}k^3.
\label{Eq:scalarRadiationDamping}$$ It is easy to verify that the albedo of a scatterer with polarizability given by Eq. (\[Eq:scalarRadiationDamping\]) is $$a=\frac{\sigma_\mathrm{scat}}{\sigma_\mathrm{ext}}=\frac{1}{1+\frac{2}{3}k^3\mathrm{Im}\alpha_0},$$ confirming that radiation damping indeed transforms any lossless electrostatic polarizability ($\mathrm{Im} \alpha_0=0$) into a scatterer that satisfies the optical theorem. Also material loss included in $\alpha_0$ via $\epsilon$ evidently leads to a lossy scatterer $a<1$, as expected. Many alternative derivations of Eq. (\[Eq:scalarRadiationDamping\]) have appeared, for instance by making a size parameter expansion of dipolar Mie coefficients [@wokaun].

Inspired by the case of a simple electric dipole, we now generalize the 