The CMB is a near-perfect black-body spectrum, and assuming this has been true since the de-coupling, we should have been able to see the glow. In fact, at a certain point, it should have been almost exactly like staring into the sun in all directions. Is this correct?

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    $\begingroup$ Sure. See physics.stackexchange.com/a/133943/123208 but also see David Hammen's comment on a related question. $\endgroup$
    – PM 2Ring
    Feb 15, 2022 at 23:31
  • $\begingroup$ I'm not able to parse the question title. Is there some word missing in "... would have been visible to the naked eye" ? Perhaps "it"? That would make the title read "... would it have been visible to the naked eye". $\endgroup$ Feb 16, 2022 at 1:40
  • $\begingroup$ @DavidHammen You caught some typos. Thanks. $\endgroup$
    – zucculent
    Feb 16, 2022 at 2:13
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    $\begingroup$ A minor point, but because the CMB is a black body spectrum, it does still have a visible component right now. It's just that visible component is soo diffuse that it's too dim for the human eye to pick up on. $\endgroup$
    – zephyr
    Feb 16, 2022 at 3:39
  • $\begingroup$ @zephyr For large frequencies, the intensity decays exponentially with the photon energy. And the visible spectrum would need many hundred times larger photon energy than the average CMB photon. I doubt than even a single free eye-visible CMB photon exists today in the Universe. $\endgroup$
    – peterh
    Feb 17, 2022 at 9:33

2 Answers 2


It wouldn't be like staring into the Sun so much as being inside the Sun (its photosphere, anyway).

What we really see is not distant objects, but light that enters our eyeballs. The CMB radiation that we detect now is here, not billions of light years away. If you were magically transported back to a time when the C“M”B temperature was 3000 K, you would be immersed in a photon gas of that temperature, and I'm not sure you would survive long enough to consciously perceive anything.

The reason the Sun doesn't melt us is that it occupies only a small part of our field of view (about 0.0005% of the 4π sr total), and the rest glows at a much lower temperature.

If you flew a heat-shielded spaceship from here to the Sun's photosphere, the Sun would seem hotter and hotter as it occupied more of the sky, though its areal brightness and associated temperature wouldn't change. Ultimately it would fill the entire sky, not just half of it, because the photosphere has a significant thickness. What you'd see and feel at that point, for as long as you survived, is similar to what you would see and feel in the early universe at around the photon decoupling time, except that the early universe was much more homogeneous and somewhat cooler.


The cosmic microwave background (CMB) was emitted as a blackbody at a temperature of about 3000 K. The CMB temperature then falls as the universe expands and is currently at a temperature of 2.73 K. Since brightness scales as $T^4$ (and the brightness of a blackbody spectrum only depends on temperature$^{1}$), then the brightness has fallen by a factor of $\sim 10^{12}$ over that time.

In contrast, the solar photosphere has a temperature of 5780 K. Thus the CMB has never been as hot or bright as the photosphere of the Sun.

To be sure, the CMB would have been visible and very bright when it was first emitted at 3000 K. It would be a factor of 10 fainter (over all wavelengths) than the Sun, but with the peak wavelength shifted into the near infrared. In fact using this calculator for the blackbody spectrum, the brightness at 500 nm (centre of the visible band) would be about a factor of 100 fainter than the Sun. However, this occurs only a few hundred thousand years after the big bang and there were no stars and planets then.

The first stars probably formed at about a redshift of 20, corresponding to around 100 million years after the big bang and also about 100 million years after the CMB formed.To work out what the temperature was at any given redshift, we can use $$ T(z) = 2.73 (1+z)\ ,$$ where $z$ denotes the redshift.

At $z=20$ the CMB had already cooled to 57 K. This is far, far to cool to emit any significant amount of visible radiation, and would be about $10^8$ times less bright than the Sun overall, with most of that radiation at far infrared wavelengths. The calculator suggests it would be hundreds of orders of magnitude fainter at 500 nm!

$^{1}$ Note added: Brightness and luminosity aren't synonyms. Brightness is a flux per unit area at the detector (your eye in this case). If a source of radiation covers a solid angle on the sky that is large enough to be resolved, then the brightness (the flux per unit area at any point on the image of the source in the back of your eye) is independent of the size of the thing emitting the radiation or how far away it is. In this question, this applies both to the Sun, which is half a degree across, and the isotropic CMB. For example, how bright the Sun is would not change if you got closer to it - clearly your eye receives more flux, but the Sun's image is spread over a larger area at the back of the eye (or any other detector) and these two effects exactly cancel. It is also true for the CMB which arrives at your eye almost isotropically from the whole sky.

[NB: Stars appear to be of varying brightness with distance/size because their images are unresolved point sources on our eye however far or big they are, unlike the Sun.]

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    $\begingroup$ But to the OP's implied question - was the irradiance (power density) at any time sufficiently high for the equivalent of our Mark-I eyeball to detect or be blinded by? $\endgroup$ Feb 16, 2022 at 13:56
  • $\begingroup$ @CarlWitthoft I think, if GR or Lambda CDM does not affect things too strongly, then we can approximate the scenario with the brightness inside an infinite, 3000K hot black body radiating gas. In this case, thermodynamics says, the intensity was enough to heat the observers to 3000K, and it was partially in the visible spectrum. But I think it depends a lot on the density and the expansion rate of the 300000 year old Universe. $\endgroup$
    – peterh
    Feb 16, 2022 at 19:05
  • $\begingroup$ Luminosity (brightness) depends on temperature and the area over which it is emitted, as well as the distance from the source (hence red giants being brighter than hotter white dwarfs). I think it would be more precise to say flux depends only on temperature. $\endgroup$
    – Justin T
    Feb 17, 2022 at 3:16
  • $\begingroup$ @JustinTackett brightness and luminosity are not synonyms. Brightness is a flux per unit area (in this case, at the back of your eye) and is $\sigma T^4$ for a blackbody. If the blackbody is resolved - I E. not a point source, (true here for the surface of the Sun, or an isotropic CMB) then the flux per unit area at the back of your eye does not depend on the area of the source or how far away it is. $\endgroup$
    – ProfRob
    Feb 17, 2022 at 6:26
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    $\begingroup$ @ProfRob Yeah you’re right on both accounts; I was thinking comparing absolute magnitudes where the observed flux distance terms cancel out and you can use luminosity to compare relative brightnesses. I also forgot that for resolvable bodies like galaxies (or in this case, all of space), surface brightness doesn’t change by distance. My bad, sorry about that! $\endgroup$
    – Justin T
    Feb 17, 2022 at 7:21

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