<p>The raw data <em>does</em> show that. However, the sky is surveyed in multiple frequency bands, and the foreground and background are visible to varying degrees in each. WMAP, for example, has 5 bands, here's two of them (<a href="https://lambda.gsfc.nasa.gov/product/wmap/current/m_images.html" rel="noreferrer">more here</a>).</p>
<p>K-Band Map (23 GHz):
<a href="https://i.sstatic.net/Tc9MA.jpg" rel="noreferrer"><img src="https://i.sstatic.net/Tc9MA.jpg" alt="K-Band Map (23 GHz)" /></a></p>
<p>W-Band Map (94 GHz):
<a href="https://i.sstatic.net/BhfFH.jpg" rel="noreferrer"><img src="https://i.sstatic.net/BhfFH.jpg" alt="W-Band Map (94 GHz)" /></a></p>
<p>These are processed together to give a background-only map:
<a href="https://i.sstatic.net/I0iqR.jpg" rel="noreferrer"><img src="https://i.sstatic.net/I0iqR.jpg" alt="Internal Linear Combination Map" /></a></p>
<p>To give a simplified explanation, these are combined as a weighted average with the bands that show the galactic contribution more prominently given negative weights to subtract that contribution out. For example, say you have three measurements that in one band are:</p>
<pre><code>1 3 1
</code></pre>
<p>And in another band are:</p>
<pre><code>1 2 1
</code></pre>
<p>Give the first a weight of -1 and the second a weight of 2, and the weighted linear combination is:</p>
<pre><code>1 1 1
</code></pre>
<p>...and you've subtracted out the strong &quot;foreground&quot; signal in the middle. Of course, what you have is now a mix of wavelengths, so this isn't without loss.</p>
<p>What they actually did is a bit more complicated, using different coefficients for different parts of the map and some smoothing to eliminate discontinuities. <a href="https://arxiv.org/abs/astro-ph/0302208" rel="noreferrer">This paper</a> goes into much more detail into this and some other approaches.</p>

CC BY-SA 4.0
2022-07-15T23:23:12.530Z
Beautiful answer! It looks like that background is *primarily* from the Milky Way, and I wonder what it is from. I don't think it can be thermal radiation from dust since it would have to be colder than the CMB to fall off faster with frequency. As to "from stars and galaxies" other than the milky way, I think that *per pixel* stars don't really contribute much but galaxies probably do a little bit, since ours does so much. Okay I'll go read the paper now to see if I can find out...
uhoh
CC BY-SA 4.0
2022-07-16T12:45:16.847Z
I'm reminded of the Star Trek scene where they're trying to find someone hiding in the ship and McCoy goes around to every crew member, subtracting their heartbeat from the total sound of all the heartbeats on the ship.
Barmar
CC BY-SA 4.0
2022-07-15T21:51:53.287Z
2022-07-15T21:51:53.287Z
Christopher James Huff
<p>Images of the CMB show no sign of stars or galaxies. If they did, shouldn't the Milky Way be a bright band dividing the CMB into hemispheres? Black body radiation in a given wavelength/frequency range gets more intense as the object in question gets hotter, so we can't say stars are too hot to glow in the microwave range. Are they just too few and far between to have a noticeable effect?</p>

CC BY-SA 4.0
2022-07-15T19:13:45.680Z
different question but potentially helpful: [What is the ratio of cosmic microwave background radiation to normal radiation?](https://astronomy.stackexchange.com/q/4813/7982)
uhoh
CC BY-SA 4.0
2022-07-15T19:40:17.173Z
Very interesting question. I have often asked myself the same thing. In fact, the pictures from wmap and some of those from Planck actually show a middle horizontal region that is covered by the emission by the MW. But in other pictures from Planck this region is not present. How did they get rid of it?
Prallax
CC BY-SA 4.0
2022-07-16T20:51:05.540Z
What research did you do? The Question in itself is interesting and any search engine should have given dozens, prolly hundreds and possibly many thousands of detailed responses. Did yours not?
Robbie Goodwin
<cosmic-microwave-background>
How did astronomers get rid of microwave radiation from stars and galaxies in picures of the CMB?
zucculent
# How did astronomers get rid of microwave radiation from stars and galaxies in picures of the CMB?

(Asked by: username_0 on 2022-07-15)

Images of the CMB show no sign of stars or galaxies. If they did, shouldn't the Milky Way be a bright band dividing the CMB into hemispheres? Black body radiation in a given wavelength/frequency range gets more intense as the object in question gets hotter, so we can't say stars are too hot to glow in the microwave range. Are they just too few and far between to have a noticeable effect?

**Tags:** cosmic-microwave-background

> **Comments:**
> **username_1**: different question but potentially helpful: [What is the ratio of cosmic microwave background radiation to normal radiation?](https://astronomy.stackexchange.com/q/4813/7982)  
> **username_2**: Very interesting question. I have often asked myself the same thing. In fact, the pictures from wmap and some of those from Planck actually show a middle horizontal region that is covered by the emission by the MW. But in other pictures from Planck this region is not present. How did they get rid of it?  
> **username_3**: What research did you do? The Question in itself is interesting and any search engine should have given dozens, prolly hundreds and possibly many thousands of detailed responses. Did yours not?  

---

## 1 answers

### Answer by username_4 on 2022-07-15. Score: 11 | Accepted answer

The raw data does show that. However, the sky is surveyed in multiple frequency bands, and the foreground and background are visible to varying degrees in each. WMAP, for example, has 5 bands, here's two of them (more here).
K-Band Map (23 GHz):

W-Band Map (94 GHz):

These are processed together to give a background-only map:

To give a simplified explanation, these are combined as a weighted average with the bands that show the galactic contribution more prominently given negative weights to subtract that contribution out. For example, say you have three measurements that in one band are:
1 3 1

And in another band are:
1 2 1

Give the first a weight of -1 and the second a weight of 2, and the weighted linear combination is:
1 1 1

...and you've subtracted out the strong "foreground" signal in the middle. Of course, what you have is now a mix of wavelengths, so this isn't without loss.
What they actually did is a bit more complicated, using different coefficients for different parts of the map and some smoothing to eliminate discontinuities. This paper goes into much more detail into this and some other approaches.

> **Comments:**
> **username_1**: Beautiful answer! It looks like that background is *primarily* from the Milky Way, and I wonder what it is from. I don't think it can be thermal radiation from dust since it would have to be colder than the CMB to fall off faster with frequency. As to "from stars and galaxies" other than the milky way, I think that *per pixel* stars don't really contribute much but galaxies probably do a little bit, since ours does so much. Okay I'll go read the paper now to see if I can find out...  
> **username_5**: I'm reminded of the Star Trek scene where they're trying to find someone hiding in the ship and McCoy goes around to every crew member, subtracting their heartbeat from the total sound of all the heartbeats on the ship.  

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