How do I transform flux density into temperature? I have flux density data from Planck mission  (in Jy)  - measure in a certain frequency - and I would like to transform it to temperature data (Kelvin). I'm not sure if I should use Stefan Boltzmann's Law or not because the units don't match.
 A: We start from the Planck's law, describing the spectral density of electromagnetic radiation:
\begin{equation}
B_{\nu}(T) = \dfrac{2 h \nu^3}{c^2} \dfrac{1}{\mathrm{exp} \left( \dfrac{h \nu}{k T} \right) -1}
\end{equation}
Where $k$ is the Boltzmann constant, $\nu$ the frequency of the signal and $c$ the speed of light.
A Taylor expansion of Planck's law with $h \nu \ll kT$ (which is a valid assumption in radioastronomy) yields the Rayleigh-Jeans approximation:
\begin{equation}
B_{RJ}(\nu, T) = \dfrac{2 \nu^2}{c^2} k T
\end{equation}
With this relation a given intensity $I_{\nu}$ [Jy] can be linked to a blackbody temperature $T$ [K]. Indeed, this is the temperature of a blackbody that would emit an amount  of light given by $I_{\nu} = B_{\nu}(T)$:
\begin{equation}
T = \dfrac{\lambda ^ 2}{2 k} I_{\nu}
\end{equation}
Also, it is possible to convert the integrated quantity $\int T dv$ to $\int I_{\nu} \mathrm{d} \nu$ by considering the following identity (Doppler effect):
\begin{equation}
\dfrac{dv}{c} = \dfrac{\mathrm{d} \nu}{\nu}
\end{equation}
In this case we have:
\begin{equation}
\int T dv = \dfrac{\lambda ^ 3}{2 k}  \int I_{\nu} \mathrm{d} \nu
\end{equation}
Where $I_{\nu}$ is integrated over the frequency, and $T$ over (Doppler) velocities.
