= Thermodynamic and Rayleigh-Jeans antenna temperatures
{title2=$T_A=\frac{h\nu/k_B}{e^{h\nu/(k_BT)}-1}$}
For a matched receiver accepting one spatial mode and one <polarization of waves>, the thermal contribution to the available spectral <power> of a <blackbody> is
$$
\frac{dP}{d\nu}=\frac{h\nu}{e^{h\nu/(k_BT)}-1}.
$$
This expression excludes a zero-point term; it is the thermally emitted contribution. Defining the <Rayleigh-Jeans law> equivalent <antenna temperature> by $dP/d\nu=k_BT_A$ gives the displayed relation. With $x=h\nu/(k_BT)$,
$$
\frac{T_A}{T}=\frac{x}{e^x-1}=1-\frac x2+\frac{x^2}{12}+O(x^4).
$$
Thus $T_A\simeq T$ requires $x\ll1$. It is an approximation with a frequency-dependent error, rather than a redefinition of thermodynamic <temperature>.
For the stipulated $T=3\,\mathrm K$ and $\nu=4.080\,\mathrm{GHz}$, $h\nu/k_B\simeq0.19581\,\mathrm K$ and $T_A\simeq2.90316\,\mathrm K$. 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.
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