# Spectrum of CMB vs. duration of last scattering

The epoch of last scattering took over 100,000 years. The Visibility Function has maximum at about 370,000 years after the Big Bang when the temperature was about 5,600 K. 50,000 years earlier the temperature was around 6,000 K and 50,000 years after the maximum it was approximately 5,200 K (I used this tool to do the calculations).

The temperature difference at the beginning and the end of the last scattering epoch is about 800 K. At 6,000 K the spectrum peaks at 483 nm while at 5,200 K the peak is 557 nm. Some of the CMB photons we see today come from the earlier stage of last scattering, some of them from the later stage, so the resulting spectrum would be a convolution of spectra from various times. However the spectrum is not distorted: it looks perfectly like a blackbody spectrum.

Does it mean the CMB photons were cooling at the same rate as the plasma? So all photons which decoupled earlier cooled down (due to expansion of the space) just right to perfectly aligh their wavelengths with the photons that decoupled later on?

• I think I can answer this, but would like a clarification first. "...the spectrum is not fuzzy: it seems to be quite sharp." - which spectrum are you talking about? People plot all sorts of different things as "the spectrum of the CMB". They're all related, but which one you have in mind will be useful to know. May 1, 2014 at 18:01
• @Kyle I am referring to the intensity as a function of wavelength. The spectrum that looks like a 2.7 Kelvin blackbody spectrum: astro.rug.nl/~hidding/ao/cobe_spectrum.png
– mpv
May 1, 2014 at 21:19
• Maybe you could consider that, radiation emitted at higher temperature also suffer from more redshift, though I haven't done the calculation myself.
– DFJ
May 2, 2014 at 21:41
• @DFJ Yes, I did consider that - it is foreshadowed in the last paragraph of my question. However I'm not 100% sure if this is the cause of the phenomenon and I cannot find any paper dealing with this in particular.
– mpv
May 3, 2014 at 6:20
• Er ... "the spectrum is not fuzzy: it seems to be quite sharp" You wouldn't expect it to be "fuzzy" you would expect it to be a convolution: $I(\lambda) = \int \mathrm{d}t \,f(t)I(\lambda,t)$ with $\int \mathrm{d}t\,f(t) = 1$ which will still be a well defined spectrum. May 27, 2014 at 15:05