The fact that the CMB has a temperature of 2.7 degrees Kelvin is well publicized, but I have a harder time finding a measure of the wattage per unit area. The Wikipedia article on the CMB is totally silent on the subject.

Then when I do find a source for W/m^2 for the CMB, they use the blackbody formula.

A flux of radiation has a Kelvin temperature. We see the temperature of the CMB as 2.735 Kelvin. We can convert this to watts/meter2 for blackbody radiation with the Stephan and Boltzmann law. K4*5.5698E-8_W/(m2*K4) = _W/m2

2.735 Kelvin = 3.11E-6_W/m2

When one sees something, it is in terms of W/m2 and the inverse square law.

That source does mention the fact that expansion of space decreases photon density. But I don't actually see a revision of the W/m^2 figure. I can't understand if they mean to say that the combination of temperature and density change balances it out so that it still follows the blackbody radiation law.

There's also the matter of whether the gas from which the CMB came was a true blackbody or had some reflectivity to it.

Surely we have a measured value for this. Does the value come anywhere close to the Stefan–Boltzmann law prediction? Does the expansion of space predict a deviation from that intensity or not? Do non-blackbody properties of the primordial gas predict a deviation from that?

  • $\begingroup$ I can't write up an answer at the moment, but this is discussed at length in Weinberg's Cosmology... I'll get around to it possibly tomorrow if nobody beats me to it. I'm pretty sure it obeys Stefan-Boltzmann but I'll double check before making a definitive statement. EDIT:... of course it does. The CMB follows a Planck distribution, up to some matter absorption effects in the intervening space, and Stefan-Boltzmann is just integrating up the Planck distribution. I'll double check tomorrow, but I'll be surprised if it's otherwise... $\endgroup$
    – Michael
    Commented Aug 30, 2013 at 16:42
  • $\begingroup$ @MichaelBrown But I don't think it should! Expansion of space decreases temperature and radiation density at the same time. There's just no reason with only that information that I should imagine those two to continue to correspond according to sigma T^4 law. If they do correspond in that way, surely there is some "current" argument that elegantly establishes why, possibly from the book you reference. I just want to be clear that I'm fighting intuition. $\endgroup$ Commented Aug 30, 2013 at 16:50
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    $\begingroup$ I recall when the COBE data came out they plotted their finding against the Plank law and it looked like the kind of lab report you'd get from a undergrad cheating badly. The error bars were similar in size to the markers and every single one fell bang on the line. $\endgroup$ Commented Aug 30, 2013 at 18:27
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    $\begingroup$ Historical side note: While we have measured the spectrum of the CMB today, the first measurement by Penzias and Wilson was of intensity at a single frequency ($4080\ \mathrm{Mc/s}$ - megacycles per second). The intensity was converted to a temperature ($3.5\pm1.0\ \mathrm{K}$) assuming a blackbody spectrum. More precisely, radio astronomers speak in terms of antenna temperature $T=c^2I_\nu/2k\nu^2$, which approximates the conversion between $T$ and $I_\nu$ assuming you are in the Rayleigh-Jeans part of a blackbody curve. $\endgroup$
    – user10851
    Commented Aug 30, 2013 at 18:33

2 Answers 2


Righteo, I'm basically copying this out of Weinberg Chapter 2. Apologies for copying, but I'm not sure there is really anything of substance that I could add. (I believe this qualifies as fair use as it is an insubstantial portion of the entire work and is being used for educational purposes. Any errors are my own typos.)

The number density of photons in equilibrium with matter at temperature $T$ at photon frequency between $\nu$ and $\nu+d\nu$ is given by the black-body spectrum:

$$ n_T(\nu)d\nu=\frac{8\pi\nu^2d\nu}{\exp(h\nu/k_\mathcal{B}T)-1},\;(2.1.1)$$ where $h$ is the original Planck's constant (which first made its appearance in a formula equivalent to this one), and $k_\mathcal{B}$ is Boltzmann's constant. (Recall that we are using units with $c=1$.)

As time passed, the matter became cooler and less dense, and eventually the radiation began a free expansion, but its spectrum has kept the same form. We can see this mot easily under an extreme assumption, that there was a time $t_L$ when radiation suddenly went from being in thermal equilibrium with matter to a free expansion. (The subscript $L$ stands for "last scattering.") Under this assumption, a photon that has frequency $\nu$ at some later time $t$ when photons are traveling freely would have had frequency $\nu a(t)/a(t_L)$ at the time the radiation went out of equilibrium with matter, and so the number density at time $t$ of photons with frequency between $\nu$ and $\nu+d\nu$ would be

$$n(\nu,t)d\nu=\left(a(t_L)/a(t)\right)^3 n_{T(t_L)}\left(\nu a(t)/a(t_L)\right)d(\nu a(t)/a(t_L)),\;(2.1.2)$$

with the factor $\left(a(t_L)/a(t)\right)^3$ arising from the dilution of photons due to the cosmic expansion. Using Eq. (2.1.1) in (2.1.2), we see that the redshift factors $a(t)/a(t_L)$ all cancel except in the exponential, so that the number density at time $t$ is given by

$$ n(\nu,t)d\nu=\frac{8\pi\nu^2d\nu}{\exp(h\nu/k_\mathcal{B}T(t))-1}=n_{T(t)}(\nu)d\nu,\;(2.1.3)$$


$$ T(t)=T(t_L)a(t_L)/a(t).\;(2.1.4)$$

Thus the photon density has been given by the black-body form even after the photons went out of equilibrium with matter, but with a redshifted temperature (2.1.4).

He then goes to argue that it doesn't affect the result if the transition takes finite time, then a bit later writes

The energy density in this radiation is given by

$$\int_0^\infty h\nu n(\nu)d\nu=a_\mathcal{B} T^4\;(2.1.6)$$

where $a_\mathcal{B}$ is the radiation constant; in c.g.s. units,

$$ a_\mathcal{B} =\frac{8\pi^5 k_\mathcal{B}^4}{15 h^3 c^3} = 7.56577(5)\times 10^{-15}\ \mathrm{erg\ cm^{-3}\ deg^{-4}}\;(2.1.7)$$

I appreciate Weinberg for keeping factors of $h$ (not $\hbar$!) and $k_\mathcal{B}$, but dropping factors of $c$ (sometimes)! Keeping odd choices straight proves his physics machoness. Eh?

EDIT: Chris White rightfully points out that there exists such a thing as observational data! (*gasp* *shock*) One of the first measurements of this distribution was done by the COBE satellite, reproduced below (image from Wikipedia).

COBE CMB spectrum

EDIT 2: Note the above argument also works for a Fermi-Dirac distribution, but it only works for massless particles. I.e., if the photon were massive the redshifted distribution would no longer be perfectly thermal. This provides a means of constraining the mass of a photon, which is observationally consistent with zero.

So while irrelevant for photons, obtaining a non-thermal distribution this way is an important step in old fashioned theories of GUT scale baryogenesis: we postulate a set of very heavy particles that "decouple" early in the universe (i.e. stop interacting with ordinary standard model particles). These then last a while as their spectrum is redshifted out of equilibrium. Eventually they decay into standard model particles (in a CP and B violating reaction), producing a net excess of matter over antimatter. The non-thermal distribution is key to getting an imbalance. Weinberg as well as Kolb and Turner have nice discussions of this scenario.


Or to put it slightly more succinctly


  • $\begingroup$ Doesn't really explain that the spectrum retains its form with redshifting, but still... neat. $\endgroup$
    – Michael
    Commented Aug 31, 2013 at 2:51

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