Antenna gain-effective-area relation ($A_{\mathrm{eff}}(\Omega )={\lambda }^{2}G(\Omega )/(4\pi )$)

For a reciprocal antenna with matched polarization and load, the available antenna effective area is proportional to the directional antenna gain. For a lossless antenna, a thermal equilibrium argument fixes the constant: one polarization of Rayleigh-Jeans law radiance gives incident power $(k_{B}T/{\lambda }^2)\int A_{\mathrm{eff}}d\Omega d\nu$, and matched thermal electrical noise gives $k_{B}Td\nu$. Thus $\int A_{\mathrm{eff}}d\Omega ={\lambda }^2$. Since $\int Gd\Omega =4\pi$, the displayed relation follows. Reciprocal losses multiply both area and gain by the same efficiency.

Antennas with effective areas $A_{e,1}$ and $A_{e,2}$ receiving the same polarization-matched plane wave have available powers in that ratio. If their matched-load resistances are equal, terminal voltage magnitudes have ratio $\sqrt{A_{e,1}/A_{e,2}}$. With unequal matched resistances the ratio becomes $\sqrt{A_{e,1}{R}_1/(A_{e,2}{R}_2)}$. Open-circuit voltages additionally depend on each antenna's source impedance. Geometry alone therefore fixes available-power estimates more directly than unspecified terminal voltages.