# How to convert from Black body intensity to MJy/sr?

How does one convert from $${\rm ergs\cdot cm^{-2}\cdot s^{-1} \cdot sr^{-1}\cdot Hz^{-1}}$$ to $${\rm MJy \cdot sr^{-1}}$$

and from $${\rm ergs\cdot cm^{-2}\cdot s^{-1} \cdot sr^{-1}\cdot cm^{-1}}$$ to $${\rm MJy \cdot sr^{-1}}$$

• The first one is not difficult. Per Wikipedia you would multiply by $10^{+23}$ if it were Jy/sr, but since it's MJy/sr you multiply by only $10^{+17}$.
– uhoh
May 10, 2020 at 1:22
• The second one has likely been answered before in this site, have a look around. However if you'd like to try it yourself, start with $$f = \frac{c}{\lambda}$$ take the derivative to get $$\frac{df}{d \lambda} = -c\frac{1}{\lambda^2}$$ and rearrange to get $$df = -c\frac{1}{\lambda^2} d \lambda$$ and use $c \approx 2.9979 \times 10^{10} \text{cm s}^{-1}$ Use that to convert your number in $\text{cm}^{-1}$ to $\text{Hz}^{-1}$ then use the info in the first comment to finish the conversion.
– uhoh
May 10, 2020 at 1:33
• I can't guarantee these (haven't had my coffee yet), so posting as comments only, not as an answer.
– uhoh
May 10, 2020 at 1:37
• @uhoh I compiled an answer, although you already answered it in principle in the comments (which I cite). The complication was actually the typesetting. Mar 10, 2021 at 22:07
• @B--rian no, it's a compliment! When I do these kinds of things I inevitably make little errors, and if I do the same thing five times I get three different answers. It's easy to write the equations, but to execute and get the correct result with certainty and then to post it as an answer is brave :-)
– uhoh
Mar 11, 2021 at 6:52

On the one hand, $${\rm 1\, ergs\cdot cm^{-2}\cdot s^{-1} \cdot sr^{-1}\cdot Hz^{-1}} = 10^{-3} {\rm J}/({\rm m^2} \cdot {\rm sr})$$ has basic dimensions $${\rm mass} \cdot {\rm time}^{-2} \cdot {\rm angle}^{-2}$$ and is a radio brightness.
On the other hand, there is $${\rm 1 MJy}\cdot{\rm sr}^{-1} = 10^{-20} {\rm J}/({\rm m^2} \cdot {\rm sr})$$ which has indeed the same basic units. A possible difficulties could be the conversion from Jansky to SI units: $$1\, {\rm Jy} = 10^{-26} {\rm W} \cdot {\rm m}^{-2} \cdot {\rm Hz}^{-1}$$ and (less likely) the fact that $${\rm M}=10^6$$. This said, there is a factor of $$10^{-17}$$ between the two values.
We start with $${\rm 1\, ergs\cdot cm^{-2}\cdot s^{-1} \cdot sr^{-1}\cdot cm^{-1}} = 0.1 {\rm W}/({\rm sr \cdot m^2})/{\rm m}$$ which is easily identified by Wolframalpha as unit for a volume scattering emission coefficient. This is not the same as the expression in part 1. However, since you state that it is about Black body intensity, you can follow the argumentation of uhoh in the comments to convert from a wavenumber (given in $${\rm cm}^{-1}$$) to a frequency (in $${\rm Hz}$$).
However if you'd like to try it yourself, start with $$f= c \lambda$$, take the derivative to get $$\frac{{\rm d}f}{{\rm d}\lambda} = -c \frac{1}{\lambda^2}$$ and rearrange to get $${{\rm d}\lambda} = -c \frac{1}{\lambda^2} {\rm d}\lambda$$ derivative to get $${\rm d}f = -c \frac{1}{\lambda^2} {\rm d}\lambda$$ and use $$c \approx 2.9979\cdot10^{10} {\rm cm}\cdot{\rm s^{−1}}$$. Use that to convert your number in $${\rm cm}^{−1}$$ to $${\rm Hz}^{−1}$$ [...]