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In all the text, they do a hand waving exercise where they say the re-ionizing temperature of Hydrogen in the plasma is 3,000K, the current temperature of space is 2.75K, so you divide 3000/2.75 and get 1090. But it turns out the real calculation is considerably more difficult. I've seen two versions:

  1. The redshift is calculated as the point at which the Optical Depth is 1. Meaning that it's very likely the photon has collided just once on it's trip from the Surface of Last Scattering to the observer.

enter image description here

  1. The redshift is calculated as the point where the Visibility Function reaches its maximum.

enter image description here

Which is it?

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  • $\begingroup$ Does this answer your question? Understanding the recombination and decoupling periods in the early universe $\endgroup$
    – ProfRob
    Apr 30, 2020 at 22:49
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    $\begingroup$ @RobJeffries - No, it doesn't. It states (almost) the same question, but doesn't answer it. The option where OpticalDepth = 1 is one way to calculate it, but it doesn't answer the big question. 1090 is widely quoted as the redshift of the CMB. We know that recombination happened over thousands of years over a range of temperatures. When they quote 1090, how (specifically) did they pick a value from that range? $\endgroup$
    – Quark Soup
    Apr 30, 2020 at 22:58
  • $\begingroup$ As that question says, 1090 is the redshift where the optical depth to Thomson scattering reaches 1. The last scattering surface. $\endgroup$
    – ProfRob
    Apr 30, 2020 at 23:04
  • $\begingroup$ It says that method gives a lower redshift, it doesn't say which method was used to determine the redshift of CMB was 1090. That's what my question is: I've seen both methods quoted in various papers, is there a definitive version? $\endgroup$
    – Quark Soup
    Apr 30, 2020 at 23:06
  • $\begingroup$ Where you get the $T_{0}$=2.75 K? I think that you calculte the redshift of recombination with $T_{0}=2.725$ K i.e the themperature of the CMB. $\endgroup$
    – Nothing
    Jun 1, 2020 at 0:55

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$z=1090$ (or $z\simeq 1100$) is often quoted and is approximately where the integrated optical depth of a photon to Thomson scattering reaches 1. The exact value depends precisely how much physics you are able or willing to put into the calculation.

An optical depth of 1 corresponds to the mean free path of a photon (i.e. when a photon travels through a scattering medium, the average path length it travels before scattering corresponds to an optical depth of 1), so this represents how far back a typical CMB photon has travelled from.

You can use other definitions if you like. For example the visibility function represents the probability distribution that any particular photon in the CMB originated at redshift $z$. So you could choose the peak of this probability function and it is obviously closely related to how the optical depth ramps up with increasing $z$. But it is not exactly the same (although you would be hard pressed to see any difference in the x-axis values in the plots shown in the question), in the same sense that the peak of a probability distribution is not the mean value unless the distribution is perfectly symmetric. This is shown more clearly in the plot below (upper panel). This shows the asymmetry in the visibility function versus redshift and indicates that the peak of the probability distribution is at a slightly higher $z$ than the mean.

Visibility function

There is no one definition. The CMB formed over tens of thousands of years and at a range of redshifts. See Understanding the recombination and decoupling periods in the early universe and For how long CMB was being emitted?

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  • $\begingroup$ I understand that the CMB formed over tens of thousands of years. The redshift of 1090 represents a very specific point in this range and that is the essence of this question: how was this one value selected from all the possible values in the range. I've also seen the quote The visibility function is defined as the probability density that a photon is last scattered at redshift z, so I'm trying to work out which one is right. $\endgroup$
    – Quark Soup
    Apr 30, 2020 at 23:18
  • $\begingroup$ Also, An optical depth of 1 corresponds to the mean free path of a photon how does that work? The mean free path of a photon is essential the particle horizon at z. How how does units of 1 correspond to something measured in Gpc? $\endgroup$
    – Quark Soup
    Apr 30, 2020 at 23:21
  • $\begingroup$ $P_{\rm scat} \propto e^{-\tau}$. So $\bar{\tau} = \int^{\infty}_{0} \tau e^{-\tau}\, d\tau/\int^{\infty}_0 e^{-\tau} d\tau = 1$. The average photon travels from an optical depth of 1. That is A definition for where the CMB is formed. $\endgroup$
    – ProfRob
    Apr 30, 2020 at 23:47
  • $\begingroup$ I'm having trouble interpreting the output of the Optical Depth function. The mean free path to CMB is roughly 13.8 Gpc. Are we saying that $\tau(13.8 Gpc)=<number of collisions>$? And so, $\tau(13.8 Gpc)=1$ means that, at 13.8 Gpc, we're likely to have at least one, but no more, collisions, and that defines the redshift of the surface of last scattering? $\endgroup$
    – Quark Soup
    May 1, 2020 at 0:38
  • $\begingroup$ @GluonSoup If the integrated optical depth to scattering back to a point in time corresponding to z, is equal to 1, it means that the average photon emitted at that time will (just) reach us without scattering. $\endgroup$
    – ProfRob
    May 1, 2020 at 7:08

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