<p>It is. They're called galaxies and stars and we see them against the cosmic microwave <em>background</em>. However, they aren't (all) cooling because of the energy sources they contain.</p>
<p>But perhaps that isn't what you mean. When the universe become transparent it <em>also</em> means that the amount of radiation emitted by the H atoms afterwards is (a) much less and (b) won't have a blackbody spectrum. The two things go together. Blackbodies are the most effective radiators but they must be completely opaque, by definition.</p>
<p>There is also much less radiation emitted after recombination because there are about <span class="math-container">$10^9$</span> microwave background photons for every proton in the universe.</p>

CC BY-SA 4.0
2021-12-04T19:18:50.297Z
2023-05-11T13:04:02.740Z
2023-05-11T13:04:02.740Z
ProfRob
<p>I am confused about why the light released from the moment when gas first formed was so dominant in comparison to the light released afterwards. Why isn't the CMB in interference with a series of cooling blackbodies as the universe continued to expand and cool?</p>

<cosmology><big-bang-theory><cosmic-microwave-background>
Why isn't the CMB blurred by the blackbodies emitted in the time after atoms first formed?
Daniel Turon
# Why isn't the CMB blurred by the blackbodies emitted in the time after atoms first formed?

(Asked by: username_0 on 2021-12-04)

I am confused about why the light released from the moment when gas first formed was so dominant in comparison to the light released afterwards. Why isn't the CMB in interference with a series of cooling blackbodies as the universe continued to expand and cool?

**Tags:** cosmology, big-bang-theory, cosmic-microwave-background

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## 1 answers

### Answer by username_1 on 2021-12-04. Score: 21

It is. They're called galaxies and stars and we see them against the cosmic microwave *background*. However, they aren't (all) cooling because of the energy sources they contain.

But perhaps that isn't what you mean. When the universe become transparent it *also* means that the amount of radiation emitted by the H atoms afterwards is (a) much less and (b) won't have a blackbody spectrum. The two things go together. Blackbodies are the most effective radiators but they must be completely opaque, by definition.

There is also much less radiation emitted after recombination because there are about $10^9$ microwave background photons for every proton in the universe.


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