 Thermodynamic and Rayleigh-Jeans antenna temperatures

For a matched receiver accepting one spatial mode and one polarization of waves, the thermal contribution to the available spectral power of a blackbody isThis expression excludes a zero-point term; it is the thermally emitted contribution. Defining the Rayleigh-Jeans law equivalent antenna temperature by $dP/d\nu =k_B{T}_A$ gives the displayed relation. With $x=h\nu /(k_{B}T)$,Thus $T_A\simeq T$ requires $x\ll 1$. It is an approximation with a frequency-dependent error, rather than a redefinition of thermodynamic temperature.

For the stipulated $T=3\mathrm{K}$ and $\nu =4.080\mathrm{GHz}$, $h\nu /k_B\simeq 0.19581\mathrm{K}$ and $T_A\simeq 2.90316\mathrm{K}$. Replacing $T_A$ by $T$ here would differ by about $3.23\%$. At much higher frequencies, the approximation fails more strongly. A real instrument averages over its measured frequency response and antenna beam pattern, so using a single central frequency adds a narrow-band approximation.