<p>Say that you have an image with <span class="math-container">$N \times N$</span> pixels that covers a region of sky of <span class="math-container">$\theta \times \theta$</span> steradians. For each pixel <span class="math-container">$i$</span> you have a value of flux density <span class="math-container">$J_i$</span>, expressed in Jy/sr. The image contains a galaxy and you want to calculate its the total flux.</p>
<p>If this were a continuum problem, you would need to do an integral</p>
<p><span class="math-container">$$F = \int_{\Omega_{gal}} J(\theta,\phi) d\Omega$$</span></p>
<p>Since the problem is discrete, you have to do a sum instead</p>
<p><span class="math-container">$$F = \sum_{i=0}^M J_i \Delta \Omega$$</span></p>
<p>Where the pixels <span class="math-container">$i=0,..,M$</span> are the ones that contain the galaxy.
To carry out the sum you only need to notice that <span class="math-container">$\Delta \Omega$</span>, which is the solid angle covered by one single pixel, can be expressed as <span class="math-container">$\Delta \Omega = {\theta^2 \over N^2}$</span>. And you are done:</p>
<p><span class="math-container">$$F={\theta^2 \over N^2} \sum_{i=0}^M J_i$$</span></p>

CC BY-SA 4.0
2021-09-03T09:30:26.853Z
Thanks you very much for this explanation ! This is exactly what I was looking for !
PsycOx
CC BY-SA 4.0
2021-10-03T11:54:55.097Z
@PsycOx you also have the options of up-voting and clicking "accept" on the answer post if you so choose.
uhoh
CC BY-SA 4.0
2021-09-03T09:28:29.073Z
2021-09-03T09:28:29.073Z
Prallax
<p>I have values of total infrared fluxes (SED) in literature expressed in Jy (Jansky) for galaxies. I want to assure that my infrared radial profiles of the same galaxies in MJy/sr (and other in Jy/pixel) give me the same integrated value as in the literature. How to get rid of the angular counterpart in steradian ? The data used to create radial profiles are HERSCHEL SPIRE 250um &amp; 500um.</p>

CC BY-SA 4.0
2021-09-03T08:40:09.863Z
Don't you just need to integrate over the solid angle/pixels?
Prallax
CC BY-SA 4.0
2021-09-03T09:01:21.777Z
Yes I think so, but the problem is to know how to do it when it is over the solid angle in steradian ?
PsycOx
<observational-astronomy><galaxy><data-analysis><units><infrared>
How to calculate an infrared SED from the specific intensities?
PsycOx
# How to calculate an infrared SED from the specific intensities?

(Asked by: username_0 on 2021-09-02)

I have values of total infrared fluxes (SED) in literature expressed in Jy (Jansky) for galaxies. I want to assure that my infrared radial profiles of the same galaxies in MJy/sr (and other in Jy/pixel) give me the same integrated value as in the literature. How to get rid of the angular counterpart in steradian ? The data used to create radial profiles are HERSCHEL SPIRE 250um & 500um.

**Tags:** observational-astronomy, galaxy, data-analysis, units, infrared

> **Comments:**
> **username_1**: Don't you just need to integrate over the solid angle/pixels?  
> **username_0**: Yes I think so, but the problem is to know how to do it when it is over the solid angle in steradian ?  

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## 1 answers

### Answer by username_1 on 2021-09-03. Score: 2 | Accepted answer

Say that you have an image with $N \times N$ pixels that covers a region of sky of $\theta \times \theta$ steradians. For each pixel $i$ you have a value of flux density $J\_i$, expressed in Jy/sr. The image contains a galaxy and you want to calculate its the total flux.

If this were a continuum problem, you would need to do an integral

$$F = \int\_{\Omega\_{gal}} J(\theta,\phi) d\Omega$$

Since the problem is discrete, you have to do a sum instead

$$F = \sum\_{i=0}^M J\_i \Delta \Omega$$

Where the pixels $i=0,..,M$ are the ones that contain the galaxy.
To carry out the sum you only need to notice that $\Delta \Omega$, which is the solid angle covered by one single pixel, can be expressed as $\Delta \Omega = {\theta^2 \over N^2}$. And you are done:

$$F={\theta^2 \over N^2} \sum\_{i=0}^M J\_i$$

> **Comments:**
> **username_0**: Thanks you very much for this explanation ! This is exactly what I was looking for !  
> **username_2**: @username_0 you also have the options of up-voting and clicking "accept" on the answer post if you so choose.  

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