<h3>Part 1</h3>
<p>On the one hand, <span class="math-container">${\rm 1\, ergs\cdot cm^{-2}\cdot s^{-1} \cdot sr^{-1}\cdot Hz^{-1}} = 10^{-3} {\rm J}/({\rm m^2} \cdot {\rm sr})$</span> has basic dimensions <span class="math-container">${\rm mass} \cdot {\rm time}^{-2} \cdot {\rm angle}^{-2}$</span> and is a <strong>radio brightness</strong>.</p>
<p>On the other hand, there is <span class="math-container">${\rm 1 MJy}\cdot{\rm sr}^{-1} = 10^{-20} {\rm J}/({\rm m^2} \cdot {\rm sr})$</span> which has indeed the same basic units. A possible difficulties could be the conversion from <a href="https://en.wikipedia.org/wiki/Jansky" rel="nofollow noreferrer">Jansky</a> to <a href="https://en.wikipedia.org/wiki/International_System_of_Units" rel="nofollow noreferrer">SI units</a>: <span class="math-container">$1\, {\rm Jy} = 10^{-26} {\rm W} \cdot {\rm m}^{-2} \cdot {\rm Hz}^{-1}$</span> and (less likely) the fact that <span class="math-container">${\rm M}=10^6$</span>. This said, there is a factor of <span class="math-container">$10^{-17}$</span> between the two values.</p>
<h3>Part 2</h3>
<p>We start with <span class="math-container">${\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} $</span> which is easily identified by <a href="https://www.wolframalpha.com/input/?i=1+ergs+%2F+cm%5E3+%2F+s+%2F+sr" rel="nofollow noreferrer">Wolframalpha</a> as unit for a <strong>volume scattering emission coefficient</strong>. This is not the same as the expression in part 1. However, since you state that it is about <em>Black body intensity</em>, you can follow the argumentation of <a href="https://astronomy.stackexchange.com/users/7982/uhoh">uhoh</a> in the comments to convert from a wavenumber (given in <span class="math-container">${\rm cm}^{-1}$</span>) to a frequency (in <span class="math-container">${\rm Hz}$</span>).</p>
<blockquote>
<p>However if you'd like to try it yourself, start with <span class="math-container">$f= c \lambda$</span>, take the derivative to get
<span class="math-container">$$\frac{{\rm d}f}{{\rm d}\lambda} = -c \frac{1}{\lambda^2} $$</span>
and rearrange to get
<span class="math-container">$${{\rm d}\lambda} = -c \frac{1}{\lambda^2} {\rm d}\lambda$$</span>
derivative to get
<span class="math-container">$${\rm d}f = -c \frac{1}{\lambda^2} {\rm d}\lambda$$</span>
and use <span class="math-container">$c \approx 2.9979\cdot10^{10} {\rm cm}\cdot{\rm s^{−1}}$</span>. Use that to convert your number in <span class="math-container">${\rm cm}^{−1}$</span> to <span class="math-container">${\rm Hz}^{−1}$</span> [...]</p>
</blockquote>

CC BY-SA 4.0
2021-03-10T21:48:33.147Z
2021-03-10T22:06:23.087Z
2021-03-10T22:06:23.087Z
B--rian
<p>How does one convert from  <span class="math-container">${\rm ergs\cdot cm^{-2}\cdot s^{-1} \cdot sr^{-1}\cdot Hz^{-1}}$</span> to <span class="math-container">${\rm MJy \cdot sr^{-1}}$</span></p>
<p>and from <span class="math-container">${\rm ergs\cdot cm^{-2}\cdot s^{-1} \cdot sr^{-1}\cdot cm^{-1}}$</span> to <span class="math-container">${\rm MJy \cdot sr^{-1}}$</span></p>

CC BY-SA 4.0
2020-05-10T01:22:55.093Z
The first one is not difficult. Per [Wikipedia](https://en.wikipedia.org/wiki/Jansky) you would multiply by $10^{+23}$ if it were Jy/sr, but since it's MJy/sr you multiply by only $10^{+17}$.
uhoh
CC BY-SA 4.0
2020-05-10T01:33:04.730Z
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
CC BY-SA 4.0
2020-05-10T01:37:08.147Z
I can't guarantee these (haven't had my coffee yet), so posting as comments only, not as an answer.
uhoh
CC BY-SA 4.0
2021-03-10T22:07:04.677Z
@uhoh I compiled an answer, although you already answered it in principle in the comments (which I cite). The complication was actually the typesetting.
B--rian
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2021-03-11T02:37:10.730Z
@B--rian you are brave!
uhoh
CC BY-SA 4.0
2021-03-11T06:25:57.793Z
@uhoh I do not quite understand, why brave? Because of citing you?
B--rian
CC BY-SA 4.0
2021-03-11T06:52:09.923Z
@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
<units><black-body>
How to convert from Black body intensity to MJy/sr?
RUNN
# How to convert from Black body intensity to MJy/sr?

(Asked by: username_0 on 2020-05-09)

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 ${\rm MJy \cdot sr^{-1}}$

**Tags:** units, black-body
> **Comments:**
> **username_1**: The first one is not difficult. Per [Wikipedia](https://en.wikipedia.org/wiki/Jansky) you would multiply by $10^{+23}$ if it were Jy/sr, but since it's MJy/sr you multiply by only $10^{+17}$.  
> **username_1**: 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.  
> **username_1**: I can't guarantee these (haven't had my coffee yet), so posting as comments only, not as an answer.  
> **username_2**: @uhoh I compiled an answer, although you already answered it in principle in the comments (which I cite). The complication was actually the typesetting.  
> **username_1**: @B--rian you are brave!  
> **username_2**: @uhoh I do not quite understand, why brave? Because of citing you?  
> **username_1**: @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 :-)  

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## 1 answers

### Answer by username_2 on 2021-03-10. Score: 1

### Part 1

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](https://en.wikipedia.org/wiki/Jansky) to [SI units](https://en.wikipedia.org/wiki/International_System_of_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.

### Part 2

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](https://www.wolframalpha.com/input/?i=1+ergs+%2F+cm%5E3+%2F+s+%2F+sr) 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](https://astronomy.stackexchange.com/users/7982/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 ${\rm Hz}^{−1}$ [...]


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