Are there any known spectral lines shifted by ~1100? If not, then how certain is mainstream that the CMB has a redshift of ~1100? All I see is a blackbody radiation curve void of spectral lines.
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$\begingroup$ Possibly relevant: arxiv.org/abs/astro-ph/0607373 $\endgroup$– called2voyage ♦Apr 22, 2016 at 17:38
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2 Answers
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The redshift of the Cosmic Microwave Background (CMB) is not measured in the same way as the redshift of a galaxy. It is inferred from the measured temperature now compared with the calculated temperature at recombination.
The brief explanation is that as the universe expands and cools, it becomes energetically favourable to form bound atoms (known as recombination). The reduction in free electrons allows the universe to become transparent and photons escape as a blackbody radiation field and eventually form the CMB.
The temperature at recombination, $T$, can be calculated using well understood physics (see https://en.m.wikipedia.org/wiki/Recombination_(cosmology)) and if the current temperature $T_0$ is known, then the redshift is simply given by $$ z = \frac{T}{T_0} -1\ .$$
Since the CMB is an almost (see below) perfect example of blackbody radiation, it is a smooth continuum with no lines, and the spectrum is given by the Planck function.
There should be some very weak recombination lines, caused by capture of electrons into upper atomic energy levels, which then cascade to lower energy levels. But these have not yet been observed, since no instruments have had sufficient sensitivity to get the signal to noise required to find them. The reason they are expected to be so weak is that there are $\sim 10^9$ photons for every proton in the universe, and therefore the recombination lines are a tiny perturbation (1 part in a billion at most) on the blackbody background. See for example https://astronomy.stackexchange.com/a/48635/2531
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$\begingroup$ What about the spectral lines OP asked for? Is it valid to say that there are none? Maybe also because the CMB's origin lies within a time frame where there were no atoms (but only plasma) to produce such spectral absorption lines? $\endgroup$– AlfeJul 30, 2017 at 1:22
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$\begingroup$ A nice summary is also given at thecuriousastronomer.wordpress.com/2015/07/30/… (Also perhaps worth a look there: thecuriousastronomer.wordpress.com/2016/06/13/… ) $\endgroup$– Jacob C.Sep 18, 2018 at 19:26
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$$\lambda_\mathrm{peak} \approx \frac{2.898 \times 10^6\ \mathrm{nm \cdot K}}{T}$$
With $T = 5000\ \mathrm{K}$, that $\lambda_\mathrm{peak}$ is about 580 nm, which is visible light.
Because everything was a hot plasma with unbound electrons there are no elemental spectral lines, which are generated by orbital jumps of electrons.
The stars would not have spectral lines either if they did not shine through a colder region of gases around them. The spectral lines from stars are less ionized elements surrounding the star further away than the central plasma, that emit photons at specific wavelengths according to the electron orbits.
Hot plasma itself has atoms where the electrons are dissociated and have variable distances from the nuclei so plasma emits a broad spectrum a plasma tube... That's actually how Bunsen and his colleague discovered what the sun was made of... by analyzing plasma tubes through prisms and lining up a prism on a telescope at the sun, then burning elements in colder flames in front of the plasma to see how the spectral signature of combusting elements adds to the spectrum of a plasma tube.
Check this thread for the maths: https://physics.stackexchange.com/questions/477758/what-wavelength-was-the-cmb-originally
