in another thermal cavity CR (the characteristic impedance of the antenna, line and filter are all matched). Both cavities are at the same temperature . The filter Fν only allows through a narrow band of frequencies from to . Both cavities are filled with blackbody radiation in equilibrium with the antenna and resistor. Some of this radiation is received by the antenna. The amount of this power within the band of frequencies passes through the transmission line and filter Fν and is dissipated as heat in the resistor. The rest is reflected by the filter back to the antenna and is reradiated into the cavity. The resistor also produces Johnson–Nyquist noise current due to the random motion of its molecules at the temperature . The amount of this power within the frequency band passes through the filter and is radiated by the antenna. Since the entire system is at the same temperature it is in thermodynamic equilibrium; there can be no net transfer of power between the cavities, otherwise one cavity would heat up and the other would cool down in violation of the second law of thermodynamics. Therefore the power flows in both directions must be equal The radio noise in the cavity is unpolarized, containing an equal mixture of polarization states. However any antenna with a single output is polarized, and can only receive one of two orthogonal polarization states. For example, a linearly polarized antenna cannot receive components of radio waves with electric field perpendicular to the antenna's linear elements; similarly a right circularly polarized antenna cannot receive left circularly polarized waves. Therefore the antenna only receives the component of power density S in the cavity matched to its polarization, which is half of the total power density Suppose is the spectral radiance per hertz in the cavity; the power of black body radiation per unit area (meter2) per unit solid angle (steradian) per unit frequency (hertz) at frequency and temperature in the cavity. If is the antenna's aperture, the amount of power in the frequency range the antenna receives from an increment of solid angle in the direction is To find the total power in the frequency range the antenna receives, this is integrated over all directions (a solid angle of ) Since the antenna is isotropic, it has the same aperture in any direction. So the aperture can be moved outside the integral. Similarly the