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

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

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

• 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 '20 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 '20 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 '20 at 1:37