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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.

## What is the redshift of the Cosmic Microwave Background (CMB)?

Last week, as I mentioned in this blog here, I had an article on the Cosmic Microwave Background’s accidental discovery in 1965 published in The Conversation. Here is a link to the article. As of writing this, there have been two questions/comments. One was from what I, quite frankly, refer to as a religious nutter, although that may be a bit harsh! But, the second comment/question by a Mark Robson was very interesting, so I thought I would blog the answer here.

This article on the Cosmic Microwave Background was published in The Conversation last Thursday (23rd July 2015)

Mark asked how we know the redshift of the CMB if it has no emission or absorption lines, which is the usual way to determine redshifts of e.g. stars and galaxies. I decided that the answer deserves its own blogpost – so here it is.

## How does the CMB come about

As I explain in more detail in my book on the CMB, the origin of the CMB is from the time that the Universe had cooled enough so that hydrogen atoms could form from the sea of protons and electrons that existed in the early Universe. Prior to when the CMB was “created”, the temperature was too high for hydrogen atoms to exist; electrons were prevented from combining with protons to form atoms because the energy of the photons in the Universe’s radiation (given by $E=h \nu$ where $\nu$ is the frequency) and of the thermal energy of the electrons was high enough to ionise any hydrogen atoms that did form. But, as the Universe expanded it cooled.

In fact, the relationship between the Universe’s size and its temperature is very simple; if $a(t)$ represents the size of the Universe at time $t$, then the temperature $T$ at time $t$ is just given by

$T(t) \propto \frac{ 1 }{ a(t) }$

This means that, as the Universe expands, the temperature just decreases in inverse proportion to its size. Double the size of the Universe, and the temperature will halve.

When the Universe had cooled to about 3,000K it was cool enough for the electrons to finally combine with the protons and form neutral hydrogen. At this temperature the photons were not energetic enough to ionise any hydrogen atoms, and the electrons had lost enough thermal energy that they too could not ionise electrons bound to protons. Finally, for the first time in the Universe’s history, neutral hydrogen atoms could form.

For reasons that I have never properly understood, astronomers and cosmologists tend to call this event recombination, although really it was combination, without the ‘re’ as it was happening for the first time. A term I prefer more is decoupling, it is when matter and radiation in the Universe decoupled, and the radiation was free to travel through the Universe. Before decoupling, the photons could not travel very far before they scattered off free electrons; after decoupling they were free to travel and this is the radiation we see as the CMB.

## The current temperature of the CMB

It was shown by Richard Tolman in 1934 in a book entitled Relativity, Thermodynamics, and Cosmology that a blackbody will retain its blackbody spectrum as the Universe expands; so the blackbody produced at the time of decoupling will have retained its blackbody spectrum through to the current epoch. But, because the Universe has expanded, the peak of the spectrum will have been stretched by the expansion of space (so it is not correct to think of the CMB spectrum as having cooled down, rather than space has expanded and stretched its peak emission to a lower temperature). The peak of a blackbody spectrum is related to its temperature in a very precise way, it is given by Wien’s displacement law, which I blogged about here.

In 1990 the FIRAS instrument on the NASA satellite COBE (COsmic Background Explorer) measured the spectrum of the CMB to high precision, and found it to be currently at a temperature of $2.725 \text{ Kelvin}$ (as an aside, the spectrum measured by FIRAS was the most perfect blackbody spectrum ever observed in nature).

The spectrum of the CMB as measured by the FIRAS instrument on COBE in 1990. It is the most perfect blackbody spectrum in nature ever observed. The error bars are four hundred times larger than normal, just so one can see them!

It is thus easy to calculate the current redshift of the CMB, it is given by

$z \text{ (redshift)} = \frac{3000}{2.725} = 1100$

and “voilà”, that is the redshift of the CMB.  Simples 😉