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I see this beautiful graph of the CMB in Wikipediaenter image description here Apparently the measured data-points match the theoretical curve for black body radiation very exactly and the discrepancies and error-bars are simply too small to observe on the graph.

I am curious if the amplitude of the theoretical curve is taken as a free fitting parameter in constructing the graph? Does it actually look like we are surrounded by a hollow black-body shell at 2.7K, or is it just the spectrum that matches and the actual photon flux is much less or much greater? i.e. is the expansion factor that stretches the spectrum to longer wavelengths the same as the expansion factor that gives us the current level of irradiance? Are there any photons missing? Would a black body isolated in deep space equilibrate to 2.7K?

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    $\begingroup$ Maybe I don't understand what you're getting at, but that graph gives the actual measured intensity of the CMB at each of the marked frequencies (well, inverse wavelengths), in units of megajanskys per steradian. $\endgroup$ – PM 2Ring Mar 27 at 4:03
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    $\begingroup$ @PM 2Ring What determines the amplitude of the blue curve? Is it a free fitting parameter or is it really the flux amplitude you'd expect inside a black-body sphere at 2.7k? $\endgroup$ – Roger Wood Mar 27 at 5:54
  • $\begingroup$ Just to be clear, you are asking if this intensity curve is a 1-parameter curve (temperature only) or a 2-parameter curve (temperature and, "max amplitude")? $\endgroup$ – Cort Ammon Mar 27 at 17:39
  • $\begingroup$ @Cort Ammon Yes, that is exactly the question. I'll accept the answer below. The answer is 1-parameter, not 2-parameters. $\endgroup$ – Roger Wood Mar 27 at 20:32
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The cosmic microwave background is a highly isotropic and homogeneous blackbody radiation field in which we are embedded. The specific intensity of a blackbody radiation field isn't a free parameter and just depends on the temperature and frequency. $$B_\nu = \frac{2h\nu^3}{c^2}\left(\exp\left[h\nu/k_BT\right] -1\right)^{-1}\ {\rm Wm}^{-2}{\rm sr}^{-1}{\rm Hz}^{-1}$$ Any systematic uncertainty in the flux calibration of the observations would lead to uncertainty in the temperature estimate and/or the conclusion that a blackbody wasn't a good fit.

This plot, taken from Samtleben et al. (2008) shows how some (pre-Planck) data (with error bars) is matched to blackbody curves with different temperatures. As far as I know, temperature is the only free parameter here and the curves are just the Planck function at different temperatures. Using Wien's law we find the peak should be at frequency of 160 GHz for $T=2.725$ K and then should have a specific intensity of $3.3\times 10^{-17}$ Wm$^{-2}$sr$^{-1}$Hz$^{-1}$, as illustrated.

In summary, the specific intensity of the CMB is completely specified by its temperature (bar a small dipole anisotropy, which depends on our motion with respect to the co-moving reference frame, and the even smaller anisotropies that depend on cosmological parameters). An isolated body placed in space would equilibriate towards 2.7K. Photons at all frequencies have those frequencies reduced by a factor of $(1+z)^{-1}$ and the Planck function retains its form, but characterised by a temperature that is also shifted by $(1+z)^{-1}$.

CMB spectrum

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  • $\begingroup$ Thanks, it seems clear - there are no free parameters used other than the temperature. I like the term "black body cavity radiation" mentioned in Andrew's answer. It really is truly remarkable result. Both from the perspective of measurement accuracy and simply from the fact that the universe works that way. $\endgroup$ – Roger Wood Mar 27 at 20:58
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This is to agree with ProfRob and add a few more details.

Black body radiation, also called cavity radiation, has the special feature that many of its features depend on only one thing: the temperature. For example:

  1. energy per unit volume $u = 3 a T^4$
  2. entropy per unit volume $s = 4 a T^3$
  3. pressure $p = a T^4$
  4. power per unit area incident on a surface $I = \frac{1}{4}uc = \sigma T^4$

where $$ a = \frac{4 \sigma}{3 c} $$ and $\sigma$ is the Stefan Boltzmann constant.

This means that if this kind of radiation is falling on a detector, then the energy flux, also called intensity, is not a freely variable parameter: once the temperature is given, so is the intensity $I$.

Suppose the detector is not perfectly efficient. In this case the signal strength will depend on the efficiency. One can model this by supposing there is an absorbing layer between the incident radiation and a perfect detector. In this case one of two things can happen. If the absorbing layer is itself passive then eventually it will reach the temperature of the radiation and then it has no net effect so the signal amplitude goes to full strength. Or, if the absorbing layer carries some energy away (e.g. by producing an electric current) then the detector gets a weaker signal. This inefficiency issue can be studied beforehand for any given detector, and thus the detector is calibrated for amplitude. Once calibrated, it can give a precise reading for incident amplitude with the known inefficiency accounted for.

It is a non-trivial fact about General Relativity that, in an expanding space, all these properties of thermal radiation are preserved, and thus, amazing as it may seem, the amplitude of the CMB is not a function of distance from the last scattering surface except through the way cosmic expansion affects the temperature and all the other properties together.

Added note

This added note is to address the fact that one can have a body such as a star or an electric filament light bulb that emits radiation with a black body spectrum, and that radiation diminishes in intensity with distance from the source. Such radiation can legitimately be called 'thermal' but it should not be called 'cavity radiation' and it is a moot point whether or not the terminology should allow it to be called 'black body radiation' because it is not homogeneous, so it does not have all the properties of cavity radiation described above. The CMB is (to very good approximation) cavity radiation and at each spatial location there exists a local reference frame in which it is isotropic.

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    $\begingroup$ Thanks. I like your clarification. I think the term "black-body cavity radiation" is very helpful and spells out that the intensity and 'color' are inseparable. $\endgroup$ – Roger Wood Mar 27 at 20:52
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From where did the CMBR originate? It came into existence when protons and electrons recombined to form atoms. The protons had a velocity distribution similar to molecules in a gas:

enter image description here

This already resembles the picture you have shown. A gas emits black-body radiation which has the same form as the picture above:

enter image description here

This same form is imprinted to the energy distribution of the photons of the CMB-gas. Which is what has been measured. In a way, you can say that the CMBR is the black-body radiation of the protons and electrons recombining in the early universe. So there are no photons missing. It's the real spectrum. An object put in empty space will eventually obtain a temperature of 2,7 Kelvin.

Is the intensity correctly predicted? Well, the form is correctly predicted. The graph you have shown is the shows the measured values of the intensities for different photon frequencies. this intensity depends on the number of protons and electrons that once recombined. A body placed in outer space would attain 2.7 Kelvin faster if the intensity were higher than measured.

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  • $\begingroup$ re. "A body placed in outer space would attain 2.7 Kelvin faster if the intensity were higher than measured". I think that statement is not correct and that a black body would equilbrate at a temperature higher than 2.7K. But I don't know how to create an intensity higher than black body cavity radiation. $\endgroup$ – Roger Wood Mar 27 at 20:49
  • $\begingroup$ @RogerWood I think you are right. Every black body shines with the same intensity. $\endgroup$ – Deschele Schilder Mar 27 at 20:55
  • $\begingroup$ @ Deschele Schilder I am not correct. It's obviously very easy to take a bunch of powerful microwave generators at closely-spaced frequencies and simulate the spectrum of black-body radiation at any desired level. $\endgroup$ – Roger Wood Mar 27 at 21:45
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    $\begingroup$ This answer is very misleading. The intensity of the CMB has absolutely nothing to do with the "number of protons and electrons that once recombined". No radiation energy was created or destroyed during recombination-- the presence of this radiation prior to recombination is a critical aspect of Big Bang cosmology. As discussed in other answers, the spectrum is set entirely by the temperature of recombination, and this was essentially set by the binding energy of hydrogen. Note that the baryon-to-photon ratio is a parameter of cosmology independent of the CMB temperature. $\endgroup$ – jawheele Mar 29 at 3:23
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    $\begingroup$ All recombination did was neutralize the primordial soup, drastically reducing the electromagnetic interactions between radiation and matter, so that the radiation went from constantly scattering to free-streaming until the present day. This is why recombination is known as the "time of last scattering". So the radiation wasn't created at this time; it just stopped being obstructed. Tagging OP (@RogerWood) in case this helps his concern. $\endgroup$ – jawheele Mar 29 at 3:26

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