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## The temperature of the Universe at recombination (decoupling)

Last week someone (“Cosm”) posted an interesting comment/question on my post “What is the redshift of the Cosmic Microwave Background (CMB)?” . The question asked how we can say that the temperature of the Universe at recombination is about 3000K when the energy of electrons with such a temperature would be

$E = \frac{ 3 }{ 2 } k T = 6.2 \times 10^{-20} \text{ J } = 0.39 \text{ eV }$

and the ionisation potential of hydrogen is 13.6 eV. This would imply, naively, that recombination would occur at a higher temperature, roughly $(13.6/0.39) \times 3000 \approx 105,000 \text{ Kelvin}$. But, it does not occur at a temperature of just over 100,000 K, but at about 3000 K. Why is this?

“Cosm” asked me this question on my blogpost about how do we know the redshift of the CMB

I decided that it was a sufficiently interesting point that I would go through the detail of why recombination (or decoupling as I prefer to call it) happened when the Universe was about 3,000K; even though electrons cease to be able to thermally ionise hydrogen at a much higher temperature of about 100,000K, so one might think that decoupling would happen earlier. The clue is that it has to do with something called equilibrium theory.

A cartoon of recombination (decoupling). As the temperature of the Universe fell, the free electrons combined with naked protons to produce neutral hydrogen, and the Universe became transparent. This is when the radiation (which was already there) was able to travel through the Universe, and it is this radiation which we see as the Cosmic Microwave Background (CMB).

## The temperature of decoupling based on equilibrium theory

One can get a reasonable estimate of the temperature (and redshift) when recombination (decoupling) took place by using equilibrium theory. This was developed in the 1920s and is based on the Saha ionisation equation, named after Indian astrophysicist Meghnad Saha.

We are going to assume that the process

$p + e^{-} \leftrightarrow H + \gamma$

is the dominant reaction for creating neutral hydrogen. That is, electrons and bare protons combining. When a free electron combines with a proton to create neutral hydrogen it will also create a photon, and that is what the $\gamma$ represents (it is not a gamma-ray photon, it is more likely to be a UV or visible-light photon). But, it is a two way process, as obviously electrons can also be ionised to go back to free electrons and bare protons.

We are going to denote the number density of free electrons (the number per unit volume) as $n_{e}$, the number density of free protons (ionised hydrogen) as $n_{p}$, and the number density of neutral hydrogen as $n_{H}$.

We can calculate the relative abundance of the free electrons to protons and neutral hydrogen via the Saha equation

$\frac{ n_{p} n_{e} }{ n_{H} } = \left( \frac{ m_{e} k_{B} T }{ 2 \pi \hbar^{2} } \right)^{3/2} \; exp \left( - \frac{ E_{1} }{ k_{B} T } \right) \text{ (1) }$

Where $m_{e}$ is the mass of an electron, $k_{B}$ is Boltzmann’s constant, $T$ is the temperature, $\hbar$ is the reduced Planck constant (which is defined as $\hbar=h/2\pi$ where $h$ is Planck’s constant), and $E_{1}$ is the ionisation potential of hydrogen (13.6 eV). We know from charge neutrality that $n_{e} = n_{p}$. We are going to define $x_{e}$ as the fraction of free electrons, so by definition

$x_{e} = \frac{ n_{e} }{ (n_{p} + n_{H} ) } \text{ (2) }$

Because $n_{e} = n_{p}$ we can re-write Equation (1) as

$\frac{ n_{e}^{2} }{ n_{H} } = \left( \frac{ m_{e} k_{B} T }{ 2 \pi \hbar^{2} } \right)^{3/2} \; exp \left( - \frac{ E_{1} }{ k_{B} T } \right) \text{ (3) }$

and from (2) we can write

$x_{e}^{2} = \frac{ n_{e}^{2} }{ ( n_{p} + n_{H} )^{2} } \text{ (4) }$

We can also write
$1 - x_{e} = 1 - \frac{ n_{e} }{ (n_{p} + n_{H} ) } = \frac{ n_{p} + n_{H} - n_{e} }{ (n_{p} + n_{H} ) } = \frac{ n_{H} }{ (n_{p} + n_{H} ) } \text{ (5) }$

Dividing Eq. (4) by Eq. (5) we have

$\frac{x_{e}^{2} }{ ( 1 - x_{e} ) } = \frac{ n_{e}^{2} }{ ( n_{p} + n_{H} )^{2} } \; \cdot \; \frac{ (n_{p} + n_{H} ) }{ n_{H} } = \frac{ n_{e}^{2} }{ (n_{p} + n_{H} ) } \; \cdot \; \frac{ 1 }{ n_{H} } \text{ (6) }$

But

$\frac{ n_{e}^{2} }{ n_{H} } = \left( \frac{ m_{e} k_{B} T }{ 2 \pi \hbar^{2} } \right)^{3/2} \; exp \left( - \frac{ E_{1} }{ k_{B} T } \right)$

so we can re-write Eq. (6) as

$\frac{x_{e}^{2} }{ ( 1 - x_{e} ) } = \frac{ 1 }{ (n_{p} + n_{H} ) } \left( \frac{ m_{e} k_{B} T }{ 2 \pi \hbar^{2} } \right)^{3/2} exp \left( - \frac{ E_{1} }{ k_{B} T } \right) \text{ (7) }$

Most of the quantities on the right hand side are physical constants, and the rest are known functions of redshift $z$. For example, the temperature history of the CMB at any redshift $z$ is just given by

$T(z) = 2.728 (1 + z) \text{ (8) }$

where $T=2.728$ is the CMB’s current temperature. The total number density of hydrogen (neutral and ionised), $(n_{p} + n_{H})$ is also a function of redshift, and is given by

$(n_{p} + n_{H}) (z) = 1.6(1+z)^{3} \text{ per m}^{3} \text{ (9) }$

where $1.6 \text{ per m}^{3}$ is the currently observed density. So, we will re-write Eq. (7) as

$\frac{x_{e}^{2} }{ ( 1 - x_{e} ) } = \frac{ 1 }{ 1.6(1+z)^{3} } \left( \frac{ m_{e} k_{B} }{ 2 \pi \hbar^{2} } \right)^{3/2} T^{3/2} \; exp \left( - \frac{ E_{1} }{ k_{B} T } \right) \text{ (10) }$

Equation (10) cannot be solved analytically, only numerically.

We will assume that we need a fraction of free electrons of $x_{e} = 50\%$ for decoupling to have occurred (in other words, $50\%$ of the electrons have bound to protons to form neutral hydrogen). We will try different temperatures to see what fraction $x_{e}$ they give. Note: when we assume a particular temperature, this will fix the redshift, because of Eq. (8).

Let us first try a temperature of $T=3000 \text{ Kelvin}$. From Eq. (8) this gives $z \approx 1100$. Plugging these values of $T$ and $z$ into Eq. (10) gives

$\frac{x_{e}^{2} }{ ( 1 - x_{e} ) } = 2.8 \times 10^{-6} \rightarrow x_{e}^{2} + 2.8 \times 10^{-6} x_{e} - 2.8 \times 10^{-6} = 0$

Solving this quadratic equation gives $\boxed{ x_{e} = 1.67 \times 10^{-3} \text{ if } T=3000 \text{ K } }$

This is obviously much less than $50\%$, by the time the Universe has cooled to 3000K there are very few free electrons, decoupling is essentially complete.

What if we use $T=4000 \text{ K}$? Doing the same thing we find $z \approx 1500$ which then gives

$\frac{x_{e}^{2} }{ ( 1 - x_{e} ) } = 0.85 \rightarrow x_{e}^{2} + 0.85 x_{e} - 0.85 = 0$

Solving this quadratic gives $\boxed{ x_{e} = 0.6 \text{ if } T=4000 \text{ K } }$. This means that $60\%$ of the electrons are free when the temperature is 4000K, meaning $40\%$ of the hydrogen atoms have become neutral.

Thirdly, let us try 3800K. This gives $z \approx 1400$. Plugging this into Eq. (9) gives

$\frac{x_{e}^{2} }{ ( 1 - x_{e} ) } = 0.1225 \rightarrow x_{e}^{2} + 0.1225 x_{e} - 0.1225 = 0$

Solving this gives $\boxed{x_{e} = 0.29 \text{ if } T=3800 \text{ K}}$

At a temperature of T=3800K just under $30\%$ of the electrons are free, so some $70\%$ of the hydrogen atoms are neutral.

So, we can surmise from this that the Universe had decoupled enough to be about $50\%$ transparent to radiation by the time the temperature was a little below $\boxed{ T=4000 \text{ Kelvin} }$. This is why the temperature of decoupling (recombination) is usually given as lying between a temperature of $T= 4000 \text{K and } 3000\text{K}$.

## Why is this just an approximation?

As I mentioned above, this is just an approximation, but not a bad one. It’s an approximation because it is a simplification of what is actually going on. In 1968, Jim Peebles (who had done the work in the 1965 Dicke etal. paper) and, independently, Yakov Zel’dovich (he of the Sunyaev-Zel’dovich effect) in the USSR, worked out a more complete theory where the hydrogen has three energy levels, rather than what we have done here where we assume the free electrons go straight into the ground-state (i.e. only two energy levels). Their third level was the n=2 state, the energy level just above the ground state. This is an important energy level for hydrogen, as it is transitions down to the n=2 level which give rise to visible-light photons. Using this more complicated 3-level model gives a Universe which is 90% neutral at $z \approx 1100$, which would correspond to a temperature of T=3000K, which is why this temperature is most often quoted. You can read more about their 3-level model here.

## Summary

Using equilibrium theory, which is an oversimplification, gives the following fractions of neutral hydrogen for three different temperatures

• At T=3000K the Universe would have been more than 99% neutral
• At T=4000K the Universe would have been about 40% neutral
• At T=3800K the Universe would have been about 70% neutral

Using a more correct 3-level model developed by Peebles and, independently, by Zel’dovich, gives that the Universe would have been about 90% neutral by the time the temperature had dropped to T=3000K. It is this temperature which is usually quoted when we talk about the temperature of the Universe when recombination (decoupling) occurred.

### 5 Responses

1. Excellent article and information! Thank you!

2. Thanks, this is explanation is nice and clear.

• Good. That was my hope! 😉

3. on 05/12/2018 at 20:48 | Reply Robin Jeffries

Your present day hydrogen number density is too high by an order of magnitude. The baryon density is 4.6% of the critical density. Did you mistakenly use the matter density instead? I make it 0.19 H atoms per cubic metre.

4. Yeah, you’re off by a factor of 10 on the present day density of H. 0.14 to 0.13 atoms per cubic m is a more accurate value.