User:Teethonreed/sandbox

Derivation of Isotropic Antenna Aperture
Imagine two cavities in thermal equilibrium at temperature T. An antenna in one cavity is connected to a matched resistor inside the second cavity. Using the Rayleigh-Jeans approximation


 * $$B_{\nu} = \frac{2 k T}{\lambda^2},$$

the power received by the antenna over a narrow frequency band is


 * $$P_{\nu}=A_{e}S_{\mathrm{matched}}=A_{e}\frac{S}{2}=\frac{1}{2}\left(\int_{4\pi}A_{e}B_{\nu}d\Omega\right)d\nu.$$

Since the cavities are in thermal equilibrium, this equals the Nyquist spectral power of the resistor


 * $$P_{\nu}=k T d\nu$$

in that same frequency band, thus


 * $$\frac{1}{2}\left(\int_{4\pi}A_{e}B_{\nu}d\Omega\right)d\nu=k T d\nu.$$


 * $$\frac{1}{2}\left(\int_{4\pi}A_{e}\frac{2 k T}{\lambda^2} d\Omega\right)d\nu=k T d\nu.$$

Being isotropic, Ae is constant in every direction, thus


 * $$A_{e}\frac{k T}{\lambda^2} \int_{4\pi} d\Omega = k T.$$


 * $$A_{e} = \frac{\lambda^2}{4\pi}.$$