# Can Coupled Mode Theory be used to calculate Antenna Aperture?

A typical metric when discussing electromagnetic wave scattering problems is the scattering cross section, typically defined as the ratio of scattered power to field power density: $$\sigma_{scat} = P_{scat} / I_0$$

In this paper discussing plane wave scattering coupled mode theory, they define the scattering cross section as: $$\sigma_{scat}=\frac{ (\frac{1}{\tau_{rad}})^2}{(\omega_0-\omega)^2+(\frac{1}{\tau_{rad}} + \frac{1}{\tau_{abs}})^2} (2l+1)\frac{\lambda^2}{2\pi}$$

Where $$\lambda$$ is the wavelength, $$\tau_{rad}$$ is the radiation time constant, and $$\tau_{abs}$$ is the absorption time constant. If we assume a lossless dipole at resonance, $$l=1$$ and $$1/\tau_{abs} \rightarrow \infty$$ and $$\omega = \omega_0$$. This simplifies the cross section to: $$\sigma_{scat}(\omega_0)=\frac{3\lambda^2}{2}$$ Separately, in antenna engineering, a common metric used is the antenna aperture. This is defined as the ratio of power delivered to load and field power density: $$A_e = P_L / I_0$$

Antenna aperture can be calculated from the gain of an antenna by $$A_e = \frac{G \lambda^2}{4 \pi}$$ For a very small dipole, $$G = 3/2$$ and the antenna aperture is therefore $$A_e = 3\lambda^2/8\pi$$.

This is a factor of 4 smaller than the scattering cross section. I'd assume the cross section for totally elastic scattering would be twice the cross section for the antenna aperture, because only 50% of the power would be delivered to the load under ideal load matching. This still leaves a factor of 2 unaccounted for. I'm wondering if there's any mistake in my logic, or a justification for why the scattering cross section for a small dipole should be 4 times larger than its antenna aperture.