# What is the current value for the temperature at which Recombination took place?

I'm looking for the most recent, most accepted value for the temperature at which the photon first decoupled from the atoms during the Epoch of Recombination? That is, what was the temperature of the CMB at the moment it was made? (and a reference, please, if you have it)

EDIT: Is this calculation independent of cosmological model? I've seen versions that reference $$\Omega_b$$ and $$h_{100}$$ as well as versions of the Saha equation that only reference $$z$$.

There is no single value that can be given, only a range. The process is that hydrogen ions (protons) (and also helium ions actually), gradually capture electrons as the universe cools. This does not all happen instantly at some temperature.

Restricting ourselves to a consideration of hydrogen recombination, the ionisation fraction is given by $$x_e$$, then using the Saha equation, one can obtain the following relationship (e.g. lecture notes here)

$$\frac{x_e^2}{1-x_e} = \frac{5.8\times 10^{15}}{\Omega_b h^2 T_4^{3/2}} \exp(-15.8/T_4),$$ where $$\Omega_b$$ is the baryon density parameter, $$h$$ is the the Hubble parameter divided by 100 km/s Mpc$$^{-1}$$ and $$T_4$$ is the temperature of the universe in units of $$10^4$$ K.

Using $$\Omega_b h^2 = 0.023$$, then the ionisation fraction changes as follows $$x_e =0.5$$ when $$T_4 = 0.374$$,; $$x_e = 0.1$$ when $$T_4 = 0.342$$ and $$x_e=0.01$$ when $$T_4 = 0.310$$

So, depending on what you want to call "recombined" then the temperature at which 50%, 90% and 99% of the hydrogen has recombined would be 3740 K, 3420 K or 3100 K respectively.

As PM2Ring points out, the numbers above are approximate. Real calculations need to account for the multi-level nature of the atoms, the presence of helium and a host of other smaller effects. A brief scan of the available literature (e.g. Seager et al. 1999; Wong 2008; Chluba & Thomas 2011 and indeed further on in the lecture notes I reference above) suggest $$x_e =0.1$$ at redshifts of about 1000, corresponding to a temperature of about 2730K. The small corrections that have been added in later papers seem to affect the (small) ionisation fraction at later times and lower redshifts.

Note, that I think this is a different question to what is the temperature at which the CMB originates. The redshift of last scattering is a smooth "visibility function" that peaks at around $$z=1100$$, corresponding to temperatures of 3000K (e.g. The initial conditions of the CMB spectrum ).

• That's a pretty wide range. Can you tell me which of these is generally used to calculate the distance to the Surface of Last Scattering? – user32023 Nov 6 '18 at 12:59
• N. Steinle provide a link to a useful article on the subject. Equation 10 in the article (cited below) gives a model-independent calculation of the temperature. How dependent are these values that you quote on the cosmological model used? – user32023 Nov 6 '18 at 13:45
• @DonaldAirey They depend on $\Omega_b h^2$ as shown. No other dependences that I can think of since the ionisation state of hydrogen at a given temperature has nothing to do with cosmological models other than through the baryon density which is given by the aforementioned parameter. What you calculate as "the" surface of last scattering depends on how it is defined. There is no "the" surface. – Rob Jeffries Nov 6 '18 at 15:23
• @DonaldAirey Equation 10 in the cited source is almost the same as the one I quote (bar the sign error, which I have now corrected). It also uses $\Omega_b h^2$ because it relies (see equation 9) on the current baryon density of the universe. – Rob Jeffries Nov 6 '18 at 18:18
• @DonaldAirey The simple model underlying the Saha equation means that Rob's results for various levels of recombination are only rough estimates. Wikipedia describes a more accurate 3-level atom model, and it also discusses modern refinements to that model which are expected to be accurate to 0.1%. The variations due to these different models of recombination have more impact than the variations due to cosmological model (those cosmological parameters impact the timing, though). – PM 2Ring Nov 6 '18 at 18:45

OK, it appears that the most correct answer is 1090 from this document: Plank 2013 Results. The team employed the Peebles and Zeldovich model of 3-state transitions and crunch it through three algorithms, HyRec, CosmoRec and RecFast to get values that agree to within 0.05%. The CMB power spectrum is sensitive to to $$x_e$$ because it changes the sound horizon at recombination and affects the thickness of the last scattering surface.

Having said that, I'm upvoting Rob Jeffries answer because it has the best amount of background for a mortal trying to understand the mechanics of CMB.

• The 1090 number in that paper has a very specific meaning. It is the redshift at which the optical depth to Thomson scattering is 1 between now and that redshift. It is not the answer to the question you asked. It is the peak of the so-called "visibility function". Photons in the microwave background arrive at Earth from a range of redshifts. – Rob Jeffries Nov 7 '18 at 12:22
• I agree that the question appears to be confusing and that's related to my naive understanding of the subject matter. The CMB is a picture of the universe at some point in time. I want to know what the temperature of the universe was when that picture was created. If it's a range of values, then how do you accurately calculate the Sound Horizon and the distance to the Surface of Last Scattering? – user32023 Nov 7 '18 at 13:13
• The CMB is not a picture of the universe at some point in time. I have explained that the universe does not become transparent to the CMB at a single temperature. Neither can the estimate of what that range of temperatures is be entirely divorced from a model of what the baryonic density is as a function of redshift, since ionisation depends on temperature and density. The whole point of CMB studies is that CMB properties do depend on cosmological models and parameters. – Rob Jeffries Nov 7 '18 at 21:36
• @RobJeffries - Redshift is model independent. If I know the temperature now (2.75K) I can determine the temperature at z=1090 by the formula $T = T_0(1+z)$. Why can't I do the same thing with the density of hydrogen if I know the present value: $(n_p+n_H)(z)=1.6(1+z)^3$? That would remove the Cosmological model from at least the Sound Horizion and Distance to LS calculations. – user32023 Nov 7 '18 at 21:49
• As I explained, you don't "know" the current value. It is $\Omega_b$ times the critical density. – Rob Jeffries Nov 7 '18 at 22:04

The temperature of the last scattering surface is precisely determined here, which for WMAP data calibrated to FIRAS data is,

$$T = 2.7260 (0.0013) K$$

and from the literature is,

$$T = 2.72548 (0.00057) K$$

Table 2 in the paper above shows several experimental CMB temperature results.

And as far as I understand, the Planck mission is more focused on aspects of the temperature fluctuations, rather than the temperature itself, so I think the source cited here is about as recent as we have, but I could be wrong.

• That's the temperature today. I want to know what the temperature was when the CMB radiation was originally made. – user32023 Nov 6 '18 at 0:57
• We don't measure that. We calculate it due to cosmological redshift: universeadventure.org/big_bang/cmb-origins.htm – N. Steinle Nov 6 '18 at 1:00
• I didn't ask for the measurement, I'm looking for the value. The value is a property of the temperature and the relative abundance of hydrogen nuclei and electrons. I believe the cosmological model also plays a part, but I wanted to address that separately. I've not seen any literature that claims it's a function of redshift (though, if you have the temperature at which the ionization takes place, you can calculate the redshift). – user32023 Nov 6 '18 at 1:03
• I think this article should clarify things: thecuriousastronomer.wordpress.com/2016/06/13/… – N. Steinle Nov 6 '18 at 2:53
• I was asking about the temperature at Recombination, not the current temperature. Rob Jefferies got it right, though that last article you reference was very helpful as well. – user32023 Nov 7 '18 at 1:25