Just a quick question relating to the thermal Blackbody temperature of a celestial object. In the c.g.s regime of astrophysics, is it more 'sophisticated' to quote the thermal Blackbody temperature in units of Kelvin ($\mathrm{K}$) or in units of $\mathrm{keV}$? That is to say to use $k_{B}T_{\mathrm{bb}}$ and convert to an energy?
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I'm not entirely sure what you mean, but the (Planck's) formula for blackbody radiation is given by
$$S_{\lambda} = \frac{8 \pi h c}{\lambda^5} \frac{1}{e^{hc/\lambda kT} - 1}$$
where $h$ is in $\mathrm{J \cdot s}$, $c$ in $\mathrm{m/s}$, $\lambda$ in $\mathrm{m}$, $k$ in $\mathrm{J/K}$ and $T$ in $\mathrm{K}$. So, the temperature is just in kelvin, not in energy. This formula, with these units, gives $S$ in $\mathrm{W/m^2/m}$, which describes the amount of energy per temperature and wavelength. There is no need to convert a given temperature to energy.
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$\begingroup$ Did you mean to write $\mathrm{W/ m^2/m}$? $\endgroup$ Oct 25, 2016 at 21:25