<p>For the most part, the CMB photons travel directly to our telescopes from the surface of last scattering.  Some corrections need to be made to determine the blackbody nature of the spectrum, but they are not corrections for <em>absorption</em> of the photons.</p>
<p>The two main corrections are shown clearly by this COBE image sequence:</p>
<p><a href="https://i.sstatic.net/cWgI7.gif" rel="noreferrer"><img src="https://i.sstatic.net/cWgI7.gif" alt="three panel map of COBE CMB measurements" /></a></p>
<p>(from <a href="https://lambda.gsfc.nasa.gov/product/cobe/dmr_image.cfm" rel="noreferrer">here</a>).</p>
<p>First, the motion of us as observers, relative to the rest frame of the CMB, distorts its single-temperature blackbody shape, making it appear hotter in one direction and cooler in the opposite direction.</p>
<p>Once this is removed (by a sky-position-dependent frequency shift), the foreground <em>emission</em> of the Milky Way becomes apparent (middle panel), and that must be modeled and subtracted. The dust in our Galaxy is on average hotter than the CMB, so its contribution can be measured well by observing over a range of wavelengths, and of course it is most heavily concentrated toward the Galactic plane.</p>
<p>After that is subtracted, the remaining emission (bottom panel) is that of a blackbody to very high precision, with of course the remaining structure that gives us so much interesting information about the early universe.</p>
<p>Here is part of the key concluding paragraph from the <a href="https://ui.adsabs.harvard.edu/abs/1994ApJ...420..439M/abstract" rel="noreferrer">Mather et al. 1992 paper</a> on measuring the spectrum:</p>
<blockquote>
<p>The FIRAS spectrum of the cosmic microwave background radiation  agrees with a blackbody spectrum to high accuracy.  The CMBR spectrum is the result of fitting a model including a dipole and a dust map derived from 240 micron data, excluding a region around the Galactic center. The final temperature is 2.726 +/- 0.010 K (95% CL systematic), where the error is dominated by our estimate of the thermometry errors.</p>
</blockquote>
<p>So they are shifting the emission and subtracting the foreground emission, but not applying any overall correction for <em>absorption</em> of the emission.</p>

CC BY-SA 4.0
2021-04-19T13:40:45.560Z
Thank you for this concise yet thorough answer! `+1`
uhoh
CC BY-SA 4.0
2022-07-08T20:06:53.990Z
Do they also subtract things from Andromeda and other observable light sources, or is the Milky Way enough somehow?
amara
CC BY-SA 4.0
2022-07-09T01:12:12.603Z
In the COBE data above, the spatial resolution is low enough that small things like Andromeda don’t matter much.  But in more recent CMB maps, like from Planck, then yes, they have to subtract foreground emission from other objects like Andromeda too. In practice they don’t model individual objects separately, but use the overall emission at a frequency sensitive to dust emission to generate a mask that can be subtracted from the data.
Eric Jensen
CC BY-SA 4.0
2021-04-19T13:35:24.593Z
2021-04-19T13:35:24.593Z
Eric Jensen
<p>Let me add a minor addendum to Eric's excellent answer.</p>
<p>The primary way in which CMB photons interact with matter is via scattering off of electrons in plasmas. After recombination (redshift <span class="math-container">$\sim 1100$</span>, about 370,000 years after the Big Bang), the baryonic matter of the universe was overwhelmingly <em>not</em> ionized, so there weren't any free electrons to scatter CMB photons. But starting around a redshift or 10 or so (a few hundred million years later), something began <a href="https://en.wikipedia.org/wiki/Reionization" rel="noreferrer">&quot;reionizing&quot;</a> the H and He atoms, to the point that most of the universe was ionized again by a redshift of around 6 (about a billion years after the Big Bang). (This &quot;something&quot; is generally understood to be UV radiation from the first generations of stars, aided by UV radiation from active galactic nuclei.)</p>
<p>CMB photons traveling through the reionized universe can interact with the electrons in the reionized plasma -- mostly earlier on, since the expanding universe dilutes the plasma and makes it harder for photons to encounter an electron. For electrons in plasmas with temperatures of <span class="math-container">$\sim 100,000$</span> K or less (which is most of the universe!), this takes the form of <a href="https://en.wikipedia.org/wiki/Thomson_scattering" rel="noreferrer">Thomson scattering</a>, which has the net effect of altering the paths of the photons and increasing their polarization; the former effect does introduce a slight blurring effect. It does not, however, change the overall distribution of photon energies, and so the CMB blackbody spectrum is unaffected. As the <a href="https://physics.stackexchange.com/questions/276890/how-to-calculate-how-many-photons-are-in-the-universe">Physics stackexchange answer by pela</a> notes, this effect is computed as part of analyzing the CMB, and the current estimates are a total optical depth <span class="math-container">$\tau$</span> (from recombination to us) of <span class="math-container">$\sim 0.06$</span>, which means that only <span class="math-container">$\sim 6$</span>% of the CMB photons end up scattered this way.</p>
<p>In massive galaxy clusters, however, the intergalactic plasma is denser and <em>hotter</em>, with temperatures reaching <span class="math-container">$\sim 10$</span> million K or higher. Electrons with temperatures like this are moving at significant fractions of the speed of light, and scatter photons in a process called <a href="https://en.wikipedia.org/wiki/Compton_scattering" rel="noreferrer">inverse Compton scattering</a>. This includes the deflection and polarization seen in Thomson scattering; but it also, on average, <em>boosts</em> the photon energies. This has the effect of shifting the observed CMB temperature to slightly higher values. This &quot;Sunyaev-Zeldovich (SZ) effect&quot; has actually been measured, and is used for some cosmological calculations. However, since only a very small fraction of the universe's volume is in the form of galaxy clusters, this has a very small overall effect; the SZ optical depth through a massive cluster is only <span class="math-container">$\sim 0.01$</span>, so even there only about 1% of the photons are affected.</p>

CC BY-SA 4.0
2021-04-21T00:30:28.683Z
This answer is gold, thank you very much! I'd read "...which leads to a blurring of distant objects which is not observed" in the [link](http://www.astro.ucla.edu/~wright/tiredlit.htm) (at)connorgarcia's comment under the question, but now realize they just meant the amount of blurring from Compton scattering necessary for the "tired light" explanation to work is not observed. But I see now there is believed to be a small amount of blurring and polarization change (but not a temperature shift) due to Thomas scattering and these effects are taken into account when interpreting CMB measurements.
uhoh
CC BY-SA 4.0
2021-04-21T00:34:57.847Z
I see that the [Sunyaev–Zeldovich effect](https://en.wikipedia.org/wiki/Sunyaev%E2%80%93Zeldovich_effect#Observations) due to massive clusters has indeed been observed several times.
uhoh
CC BY-SA 4.0
2021-04-20T19:55:08.347Z
2021-04-20T19:55:08.347Z
Peter Erwin
<h3>Background</h3>
<p>The question <a href="https://astronomy.stackexchange.com/q/43454/7982">Why do some electromagnetic waves continue travelling while others disappear?</a> is interesting, and in addition to the answer there I started to write:</p>
<blockquote>
<p>This is a supplementary answer to @ConnorGarcia's answer using the example in the question. It's not very accurate but it does address part of the question:</p>
<blockquote>
<p>Cosmic background radiation emitted when the Universe was very young still exists. <strong>But my wifi signal seems to disappear a short distance from my apartment.</strong> Why?</p>
</blockquote>
<p>Imagine walking around in the middle of a city with a receiver that can pull in 900 MHz to 5 GHz. It will tell you that you are in a &quot;soup&quot; of radio photons; no matter where you go <em>inside</em> the city you'll...</p>
</blockquote>
<p>and then realized I didn't know what I was talking about.</p>
<h3>Question</h3>
<p>So instead I'd like to ask about the following:</p>
<p>I think that CMB photons are believed to be &quot;pristine photons&quot; from soon after the big bang rather than absorbed and re-emitted and thermalized, and I seem to remember recently reading an answer here or in Physics SE saying that a measured CMB spectra (in any given direction) is fit with a <a href="https://en.wikipedia.org/wiki/Planck%27s_law" rel="noreferrer">Planck distribution</a> with unity emissivity; there was no &quot;fudge&quot; or scaling factor in front for the absolute <a href="https://en.wikipedia.org/wiki/Radiance#Spectral_radiance" rel="noreferrer">spectral radiance</a> measured.</p>
<p><em>But now I can't find that answer anywhere!</em></p>
<p>Together those suggest to me that <em>macroscopically</em> the universe is essentially transparent to CMB, and the fraction intercepted by stars and dust between us and it is so tiny that it doesn't require a correction factor.</p>
<p><strong>Question:</strong>  Have I got this right?</p>

CC BY-SA 4.0
2021-04-19T03:56:35.730Z
There is a good discussion of the Plank distribution of the CMB used to try to debunk the "tired light" explanation for the redshift of distant galaxies here: http://www.astro.ucla.edu/~wright/tiredlit.htm .  I think you are correct that VERY little of the CMB (as observed from Earth) has been absorbed and re-emitted. In some sense, perhaps large portions of the CMB never got a chance to hit a lot of the matter since the CMB has been subject to universal expansion along with the rest of us. I'd post an answer, but I tend to bungle universe expansion.
Connor Garcia
CC BY-SA 4.0
2021-04-19T04:01:53.170Z
I tend to *be boggled by it* so the first part of my question documents my own bungle-detection system in action.
uhoh
CC BY-SA 4.0
2021-04-19T05:07:22.510Z
I started writing an answer, but I decided to can it due to lack of solid references. ;) But anyway, the CMB photons were in thermal equilibrium with matter before they were released from the surface of last scattering, that's why it was (and is) a blackbody spectrum. The interstellar gas and dust is (mostly) in thermal equilibrium with the CMB, so that stuff has (almost) no impact on the CMB spectrum.
PM 2Ring
CC BY-SA 4.0
2021-04-19T08:41:19.677Z
@PM2Ring I don’t see how you can say that “the interstellar gas and dust is (mostly) in thermal equilibrium with the CMB”. The ISM has temperatures ranging from a few to *tens of millions* of kelvins. Except for rare things like the Sunyaev-Zel’dovich effect, the ISM for the most part just doesn’t interact with CMB photons.
Peter Erwin
CC BY-SA 4.0
2021-04-19T08:46:11.457Z
@Peter I was only talking about diffuse isolated cold gas and dust that's at the same temperature as the CMB.
PM 2Ring
CC BY-SA 4.0
2021-04-19T09:12:22.480Z
@PM2Ring I'm out of my element here but I still think that "The interstellar gas and dust is (mostly) in thermal equilibrium with the CMB" in *present tense* form is interesting because to be in thermal equilibrium means at least some of the photons need to be absorbed and re-emitted which would blur the spatial structure of the CMB if the re-emission were also Planck-shaped. However, if the re-emission were in specific bands of frequencies then the photons would not contribute directly to the broad measured spectrum and there would be a net reduction in the spectral radiance(?).
uhoh
CC BY-SA 4.0
2021-04-19T09:28:40.810Z
I am out of domain too but in line with PM 2Ring. Just the matter that screens the background is too dilute to be seen. Actually those tiny dots are what we see observing with the various em windows, ie the beautiful astronomical objects we can detect.
Alchimista
CC BY-SA 4.0
2021-04-19T15:38:35.610Z
Pela's answer here might have been what you were looking for: https://physics.stackexchange.com/questions/276890/how-to-calculate-how-many-photons-are-in-the-universe
Connor Garcia
<observational-astronomy><cosmic-microwave-background><sunyaev-zeldovich-effect>
Is the universe macroscopically transparent to CMB? Is the fraction intercepted by stars and dust so tiny that it doesn't have a correction factor?
uhoh
# Is the universe macroscopically transparent to CMB? Is the fraction intercepted by stars and dust so tiny that it doesn't have a correction factor?

(Asked by: username_0 on 2021-04-19)

### Background

The question [Why do some electromagnetic waves continue travelling while others disappear?](https://astronomy.stackexchange.com/q/43454/7982) is interesting, and in addition to the answer there I started to write:

> This is a supplementary answer to @ConnorGarcia's answer using the example in the question. It's not very accurate but it does address part of the question:
>
> > Cosmic background radiation emitted when the Universe was very young still exists. **But my wifi signal seems to disappear a short distance from my apartment.** Why?
>
> Imagine walking around in the middle of a city with a receiver that can pull in 900 MHz to 5 GHz. It will tell you that you are in a "soup" of radio photons; no matter where you go *inside* the city you'll...

and then realized I didn't know what I was talking about.

### Question

So instead I'd like to ask about the following:

I think that CMB photons are believed to be "pristine photons" from soon after the big bang rather than absorbed and re-emitted and thermalized, and I seem to remember recently reading an answer here or in Physics SE saying that a measured CMB spectra (in any given direction) is fit with a [Planck distribution](https://en.wikipedia.org/wiki/Planck%27s_law) with unity emissivity; there was no "fudge" or scaling factor in front for the absolute [spectral radiance](https://en.wikipedia.org/wiki/Radiance#Spectral_radiance) measured.

*But now I can't find that answer anywhere!*

Together those suggest to me that *macroscopically* the universe is essentially transparent to CMB, and the fraction intercepted by stars and dust between us and it is so tiny that it doesn't require a correction factor.

**Question:** Have I got this right?

**Tags:** observational-astronomy, cosmic-microwave-background, sunyaev-zeldovich-effect
> **Comments:**
> **username_1**: There is a good discussion of the Plank distribution of the CMB used to try to debunk the "tired light" explanation for the redshift of distant galaxies here: http://www.astro.ucla.edu/~wright/tiredlit.htm .  I think you are correct that VERY little of the CMB (as observed from Earth) has been absorbed and re-emitted. In some sense, perhaps large portions of the CMB never got a chance to hit a lot of the matter since the CMB has been subject to universal expansion along with the rest of us. I'd post an answer, but I tend to bungle universe expansion.  
> **username_0**: I tend to *be boggled by it* so the first part of my question documents my own bungle-detection system in action.  
> **username_2**: I started writing an answer, but I decided to can it due to lack of solid references. ;) But anyway, the CMB photons were in thermal equilibrium with matter before they were released from the surface of last scattering, that's why it was (and is) a blackbody spectrum. The interstellar gas and dust is (mostly) in thermal equilibrium with the CMB, so that stuff has (almost) no impact on the CMB spectrum.  
> **username_3**: @PM2Ring I don’t see how you can say that “the interstellar gas and dust is (mostly) in thermal equilibrium with the CMB”. The ISM has temperatures ranging from a few to *tens of millions* of kelvins. Except for rare things like the Sunyaev-Zel’dovich effect, the ISM for the most part just doesn’t interact with CMB photons.  
> **username_2**: @Peter I was only talking about diffuse isolated cold gas and dust that's at the same temperature as the CMB.  
> **username_0**: @PM2Ring I'm out of my element here but I still think that "The interstellar gas and dust is (mostly) in thermal equilibrium with the CMB" in *present tense* form is interesting because to be in thermal equilibrium means at least some of the photons need to be absorbed and re-emitted which would blur the spatial structure of the CMB if the re-emission were also Planck-shaped. However, if the re-emission were in specific bands of frequencies then the photons would not contribute directly to the broad measured spectrum and there would be a net reduction in the spectral radiance(?).  
> **username_4**: I am out of domain too but in line with PM 2Ring. Just the matter that screens the background is too dilute to be seen. Actually those tiny dots are what we see observing with the various em windows, ie the beautiful astronomical objects we can detect.  
> **username_1**: Pela's answer here might have been what you were looking for: https://physics.stackexchange.com/questions/276890/how-to-calculate-how-many-photons-are-in-the-universe  

---

## 2 answers

### Answer by username_3 on 2021-04-20. Score: 6 | Accepted answer

Let me add a minor addendum to Eric's excellent answer.

The primary way in which CMB photons interact with matter is via scattering off of electrons in plasmas. After recombination (redshift $\sim 1100$, about 370,000 years after the Big Bang), the baryonic matter of the universe was overwhelmingly *not* ionized, so there weren't any free electrons to scatter CMB photons. But starting around a redshift or 10 or so (a few hundred million years later), something began ["reionizing"](https://en.wikipedia.org/wiki/Reionization) the H and He atoms, to the point that most of the universe was ionized again by a redshift of around 6 (about a billion years after the Big Bang). (This "something" is generally understood to be UV radiation from the first generations of stars, aided by UV radiation from active galactic nuclei.)

CMB photons traveling through the reionized universe can interact with the electrons in the reionized plasma -- mostly earlier on, since the expanding universe dilutes the plasma and makes it harder for photons to encounter an electron. For electrons in plasmas with temperatures of $\sim 100,000$ K or less (which is most of the universe!), this takes the form of [Thomson scattering](https://en.wikipedia.org/wiki/Thomson_scattering), which has the net effect of altering the paths of the photons and increasing their polarization; the former effect does introduce a slight blurring effect. It does not, however, change the overall distribution of photon energies, and so the CMB blackbody spectrum is unaffected. As the [Physics stackexchange answer by pela](https://physics.stackexchange.com/questions/276890/how-to-calculate-how-many-photons-are-in-the-universe) notes, this effect is computed as part of analyzing the CMB, and the current estimates are a total optical depth $\tau$ (from recombination to us) of $\sim 0.06$, which means that only $\sim 6$% of the CMB photons end up scattered this way.

In massive galaxy clusters, however, the intergalactic plasma is denser and *hotter*, with temperatures reaching $\sim 10$ million K or higher. Electrons with temperatures like this are moving at significant fractions of the speed of light, and scatter photons in a process called [inverse Compton scattering](https://en.wikipedia.org/wiki/Compton_scattering). This includes the deflection and polarization seen in Thomson scattering; but it also, on average, *boosts* the photon energies. This has the effect of shifting the observed CMB temperature to slightly higher values. This "Sunyaev-Zeldovich (SZ) effect" has actually been measured, and is used for some cosmological calculations. However, since only a very small fraction of the universe's volume is in the form of galaxy clusters, this has a very small overall effect; the SZ optical depth through a massive cluster is only $\sim 0.01$, so even there only about 1% of the photons are affected.

> **Comments:**
> **username_0**: This answer is gold, thank you very much! I'd read "...which leads to a blurring of distant objects which is not observed" in the [link](http://www.astro.ucla.edu/~wright/tiredlit.htm) (at)connorgarcia's comment under the question, but now realize they just meant the amount of blurring from Compton scattering necessary for the "tired light" explanation to work is not observed. But I see now there is believed to be a small amount of blurring and polarization change (but not a temperature shift) due to Thomas scattering and these effects are taken into account when interpreting CMB measurements.  
> **username_0**: I see that the [Sunyaev–Zeldovich effect](https://en.wikipedia.org/wiki/Sunyaev%E2%80%93Zeldovich_effect#Observations) due to massive clusters has indeed been observed several times.  

---

### Answer by username_5 on 2021-04-19. Score: 19

For the most part, the CMB photons travel directly to our telescopes from the surface of last scattering. Some corrections need to be made to determine the blackbody nature of the spectrum, but they are not corrections for *absorption* of the photons.

The two main corrections are shown clearly by this COBE image sequence:

[![three panel map of COBE CMB measurements](https://i.sstatic.net/cWgI7.gif)](https://i.sstatic.net/cWgI7.gif)

(from [here](https://lambda.gsfc.nasa.gov/product/cobe/dmr_image.cfm)).

First, the motion of us as observers, relative to the rest frame of the CMB, distorts its single-temperature blackbody shape, making it appear hotter in one direction and cooler in the opposite direction.

Once this is removed (by a sky-position-dependent frequency shift), the foreground *emission* of the Milky Way becomes apparent (middle panel), and that must be modeled and subtracted. The dust in our Galaxy is on average hotter than the CMB, so its contribution can be measured well by observing over a range of wavelengths, and of course it is most heavily concentrated toward the Galactic plane.

After that is subtracted, the remaining emission (bottom panel) is that of a blackbody to very high precision, with of course the remaining structure that gives us so much interesting information about the early universe.

Here is part of the key concluding paragraph from the [Mather et al. 1992 paper](https://ui.adsabs.harvard.edu/abs/1994ApJ...420..439M/abstract) on measuring the spectrum:

> The FIRAS spectrum of the cosmic microwave background radiation agrees with a blackbody spectrum to high accuracy. The CMBR spectrum is the result of fitting a model including a dipole and a dust map derived from 240 micron data, excluding a region around the Galactic center. The final temperature is 2.726 +/- 0.010 K (95% CL systematic), where the error is dominated by our estimate of the thermometry errors.

So they are shifting the emission and subtracting the foreground emission, but not applying any overall correction for *absorption* of the emission.

> **Comments:**
> **username_0**: Thank you for this concise yet thorough answer! `+1`  
> **username_6**: Do they also subtract things from Andromeda and other observable light sources, or is the Milky Way enough somehow?  
> **username_5**: In the COBE data above, the spatial resolution is low enough that small things like Andromeda don’t matter much.  But in more recent CMB maps, like from Planck, then yes, they have to subtract foreground emission from other objects like Andromeda too. In practice they don’t model individual objects separately, but use the overall emission at a frequency sensitive to dust emission to generate a mask that can be subtracted from the data.  

---

