<p>I'm not entirely sure what you mean, but the (Planck's) formula for blackbody radiation is given by</p>

<p>$$S_{\lambda} = \frac{8 \pi h c}{\lambda^5} \frac{1}{e^{hc/\lambda kT} - 1}$$</p>

<p>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.</p>

CC BY-SA 3.0
2016-10-25T21:25:27.663Z
Did you mean to write $\mathrm{W/ m^2/m}$?
Sir Cumference
CC BY-SA 3.0
2015-10-16T09:19:55.497Z
2016-10-25T18:13:59.963Z
2016-10-25T18:13:59.963Z
Vincent Oomen
<p>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?</p>

<astrophysics><high-energy-astrophysics>
In what units to quote the thermal Blackbody temperature
MichaelJRoberts
# In what units to quote the thermal Blackbody temperature

(Asked by: username_0 on 2015-10-14)

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?

**Tags:** astrophysics, high-energy-astrophysics


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## 1 answers

### Answer by username_1 on 2015-10-16. Score: 2 | Accepted answer

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.

> **Comments:**
> **username_2**: Did you mean to write $\mathrm{W/ m^2/m}$?  

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