<p>The Intergalactic medium at the relevant redshift is made of neutral hydrogen. What we can measure is the brightness temperature (the temperature that the IGM would have if it emitted as a blackbody) relative to the CMB. This quantity depends crucially on the following expression:</p>

<p>$\frac{T_S - T_{CMB}}{T_S}$</p>

<p>where $T_S$ is called spin temperature.</p>

<p>The spin temperature is just a measure of the ratio between the number of hydrogen atoms in the first excited hyperfine state (spin parallel) and ground state (spin antiparallel).
It turns out that the spin temperature can be modified in three ways: one of them is Lyman resonant scattering (the other two are collisional coupling and scattering of CMB photons). 
Using atomic physics it's not difficult to see that $T_S$ is a weighted mean of the temperature of the CMB and gas temperature.</p>

<p>UV photons can change the spin temperature because a hydrogen atom in the lowest state, n=1 (if you are familiar with chemistry, a s state) with antiparallel spins, can absorb an UV photon and jump in a p (n=2) state, and then fall again in a s state but with parallel spins.
This mechanism is known as Wouthuysen–Field coupling.</p>

<p>Before the emission takes place, the gas is at the same temperature of the CMB and both are equal to spin temperature, so the relative brightness temperature is zero.
When Lyman alpha emission begins the spin temperature decreases resulting in the observed absorption peak. 
At some point the peak stops because the first stars emit X-rays and the gas becomes hotter than the background radiation.</p>

CC BY-SA 3.0
2018-04-04T04:26:22.850Z
Thanks for your answer. This focuses mainly on altering the spin temperature, but the last two sentences of my question explain that I'm asking how this "...would cause the "blip" in the (now red-shifted) 21 cm part of the Cosmic Microwave Background..." The linked question includes the line shape, and it's a dip, a decrease in intensity. How does the modified $T_S$ produce this feature?
uhoh
CC BY-SA 3.0
2018-04-03T22:01:02.133Z
2018-04-03T22:01:02.133Z
Giuseppe Rossi
<p>Guiseppe Rossi's answer is excellent; I just want to add <em>why</em> the mentioned UV radiation modifies the spectrum of the background radiation.</p>

<p>(one word took the other, and it became a rather long comment.)</p>

<h3>The hyperfine level</h3>

<p>Neutral hydrogen in its ground state can be in two different configurations; either the proton and the electron may have <em>parallel</em> spins ($\uparrow\uparrow$), or they may have <em>antiparallel</em> spins ($\uparrow\downarrow$). When the spins are parallel, the atom has a slightly higher energy than when they're antiparallel. The atoms "wants" to make a spin flip to the lower energy configuration$^\dagger$, and will eventually do so, but since the line is forbidden, the lifetime of the parallel state is of the order $10^7\,\mathrm{yr}$.</p>

<p>The relative population of the states is given by the Boltzmann distribution
$$
\begin{array}{rcl}
\frac{n_1}{n_0} &amp; = &amp; \frac{g_1}{g_0} \, e^{-\Delta E \, / \, k_\mathrm{B} T_S} \\
                &amp; = &amp; 3 \, e^{-0.068\,\mathrm{K} \, / \, T_S}\label{a}\tag{1},
\end{array}
$$
where subscripts 1 and 0 denote the $\uparrow\uparrow$ and $\uparrow\downarrow$ states, respectively, $n$ is the density, $g$ is the statistical weights (with $g_0,g_1 = 1,3$), $\Delta E = 5.9\times10^{-6}\,\mathrm{eV}$ is the energy difference of the states, $k_\mathrm{B}$ is the Boltzmann constant, and $T_S$ is the <a href="https://en.wikipedia.org/wiki/Excitation_temperature" rel="nofollow noreferrer">spin temperature</a>, which I think is better thought of as "a number that describes the relative populations" than an actual temperature.</p>

<h3>Departure from equilibrium</h3>

<p>In thermal equilibrium, the spin temperature is equal to the "real", kinetic temperature. Just after <a href="https://en.wikipedia.org/wiki/Decoupling_(cosmology)" rel="nofollow noreferrer">decoupling</a> of the radiation from matter at a redshift of $z\simeq1100$, the gas and the photons share the same energy, and since $T\gg 1$, we have that $n_1/n_0 \simeq 3$. But when the first stars begin to shine, they produce massive amounts of hard UV radiation which ionizes their surrounding medium. The ionized gas quickly recombines (in the beginning, at least), with $\sim2/3$ of the recombinations resulting in the emission of a Lyman $\alpha$ photon, i.e. a photon with an energy corresponding to the energy difference between the first excited state (one of the three $2P$ states) and the ground state (the $1S$ state) of the hydrogen atom (10.2 eV).</p>

<p>The Ly$\alpha$ photons scatter multiple times on the neutral hydrogen. Each scattering excites an atom from $1S\rightarrow 2P$, which subsequently de-excites and emits an Ly$\alpha$ photon in another direction. But since the energy difference between the $2P$ and the $1S$ state is a million times larger than between the hyperfine states, there is equal chance of ending in the $\uparrow\uparrow$ and the $\uparrow\downarrow$ state.</p>

<p>That is, $n_1/n_0$ is no longer $\simeq 3$, but is driven toward $\sim 1$. This is the <a href="https://en.wikipedia.org/wiki/Wouthuysen%E2%80%93Field_coupling" rel="nofollow noreferrer">Wouthuysen–Field effect</a> that Guiseppe Rossi mentions; from eq. \ref{a}, you see that this corresponds to a much smaller spin temperature, and thus the factor that Guiseppe Rossi mentions becomes negative. The full equation describing the brightness (or, equivalently, the flux received) as a function of redshift can be written (e.g. <a href="https://arxiv.org/abs/astro-ph/0311514" rel="nofollow noreferrer">Zaldarriaga et al. 2004</a>)
$$
T(z) = 23\,\mathrm{mK} \,
       \frac{T_S - T_\mathrm{CMB}}{T_S} \,
       (1+\delta)
       x_\mathrm{HI}(z)
       \frac{\Omega_\mathrm{b}h^2}{0.02}
       \left( \frac{0.15}{\Omega_\mathrm{m}h^2} \, \frac{1+z}{10} \right)^{1/2}\label{b}\tag{2}
$$
and when the $(T_S - T_\mathrm{CMB})\,/\,T_S$ factor is negative, you will get an absorption line.</p>

<p>(In eq. \ref{b},
$\delta$,
$x_\mathrm{HI}(z)$,
$\Omega_\mathrm{b}$,
$\Omega_\mathrm{m}$, and
$h$,
are
the local overdensity,
the neutral fraction of hydrogen,
the baryon and matter density parameter, and
the dimensionless (reduced) Hubble constant,
respectively, but this is of less importance.)</p>

<p>Since the observed absorption line (<a href="http://adsabs.harvard.edu/abs/2018Natur.555...67B" rel="nofollow noreferrer">Bowman et al. 2018</a>) starts to drop around an observed frequency of $\nu_\mathrm{obs} = 65\text{–}70\,\mathrm{MHz}$, and since the rest frequency of the hyperfine line is $\nu_\mathrm{rest} = 1420\,\mathrm{MHz}$, this means that the first stars appeared around a redshift of $z = \nu_\mathrm{rest}/\nu_\mathrm{obs} - 1 \simeq 20$, corresponding to an age of the Universe of $\sim 180\,\mathrm{Myr}$ (i.e. million years — the largest absorption is reached at $z\simeq 17$, or $t\simeq 200\,\mathrm{Myr}$).</p>

<p>Now the big question is, according to eq. \ref{b} the dip should be of the order a few tens of mK, but is in fact roughly 0.5 K, i.e. an order of magnitude larger. One possible mechanism that could produce this effect is coupling of the gas with dark matter, something which is not usually considered possible but could happen if the dark matter particle has a very small charge (<a href="https://arxiv.org/abs/1803.03091" rel="nofollow noreferrer">Barkana et al. 2018</a>).</p>

<h3>Time evolution of the 21 cm signal</h3>

<p>The figure below (from a great review by <a href="https://arxiv.org/abs/1109.6012" rel="nofollow noreferrer">Pritchard &amp; Loeb 2012</a>) shows how the 21 cm signal evolves with time. The dip discussed in this answer is the orange and red part.</p>

<h2><a href="https://i.sstatic.net/TVRZs.jpg" rel="nofollow noreferrer"><img src="https://i.sstatic.net/TVRZs.jpg" alt="21cm"></a></h2>

<p>$^\dagger$<sub>An analogy would be two magnets aligned parallel to each other with north in the same direction, preferring to flip around, but note Ken G's comment below; the transition doesn't necessarily involve a spin flip, and the analogy is not to be taken literally, since parallel magnets are alike, whereas parallel electrons/protons have opposite charges.</sub></p>

CC BY-SA 3.0
2018-04-04T09:23:40.947Z
This is extremely helpful, thank you for taking the time to write this up and explain so clearly. I'll take some time to read it and the references as well.
uhoh
CC BY-SA 3.0
2018-04-04T12:23:32.447Z
@uhoh I added another good reference to a review.
pela
CC BY-SA 3.0
2018-04-04T14:07:04.807Z
Minor correction to that wonderful answer: The 21 cm transition does not necessarily involve a spin flip, indeed the ground state involves a state where the spins of neither the proton nor the electron are specified (they are only specified to be opposite each other in an antisymmetric combination).  The upper level can have parallel definite spins, or indefinite spins that are known only to be antiparallel in a symmetric combination.  There is a loss of angular momentum in the transition, and given all this complexity, you can see why people claim it's a spin flip, but it's not really true.
Ken G
CC BY-SA 3.0
2018-04-04T14:31:36.160Z
Another interesting point:  the electron and proton have opposite charges, so you'd think that if their spins pointed the same way, that would be like magnets aligned in opposite directions, which is the *lowest* energy, not the highest.  The reason it is the highest energy when the spins are aligned and definite is actually that the strongest interaction between electron and proton is when the electron is inside the proton, so it's like anti-aligned magnets with one inside the other-- that's the highest energy configuration.
Ken G
CC BY-SA 3.0
2018-04-04T14:33:06.580Z
And when the spins are anti-aligned and indefinite, that's also the highest energy in the symmetric configuration ,because the symmetric (two swapping particle states) configuration is the one where the electron is going to be outside the proton, so that's the only situation that acts like two normal magnets.
Ken G
CC BY-SA 3.0
2018-04-04T15:22:42.943Z
Thanks for those comments, @KenG. I'll edit my answer accordingly.
pela
CC BY-SA 3.0
2018-04-04T17:24:54.357Z
It was such a splendid answer, may as well go all out!
Ken G
CC BY-SA 3.0
2018-04-04T09:06:04.147Z
2018-04-05T13:44:27.093Z
2018-04-05T13:44:27.093Z
pela
<p>In <a href="https://astronomy.stackexchange.com/q/25337/7982">this question</a> I discuss the recent (open access) paper in Nature <a href="https://www.nature.com/articles/nature25792" rel="noreferrer">An absorption profile centred at 78 megahertz in the sky-averaged spectrum</a> at length. The abstract begins:</p>

<blockquote>
  <p>After stars formed in the early Universe, their ultraviolet light is expected, eventually, to have penetrated the primordial hydrogen gas and altered the excitation state of its 21-centimetre hyperfine line. This alteration would cause the gas to absorb photons from the cosmic microwave background, producing a spectral distortion that should be observable today at radio frequencies of less than 200 megahertz(1).</p>
</blockquote>

<p>where reference 1 is Jonathan Pritchard and Abraham Loeb (2012) <a href="http://iopscience.iop.org/article/10.1088/0034-4885/75/8/086901/meta" rel="noreferrer">21 cm cosmology in the 21st century</a>. From there I've found Leonid Chuzhoy and Zheng Zheng (2007) <a href="http://iopscience.iop.org/article/10.1086/522491/meta" rel="noreferrer">Radiative Transfer Effect on Ultraviolet Pumping of the 21 cm Line in the High-Redshift Universe</a>.</p>

<p>I understand some basics about the hyperfine transition in hydrogen, and that the "spin temperature" of a gas in space can differ from the temperature of other partitions if it's being pumped, but these papers are more than a bit hard to read.</p>

<p>Is there a simple way to explain the basics behind how the exposure of hydrogen to the UV light produced in early stars would cause the "blip" in the (now red-shifted) 21 cm part of the <a href="https://en.wikipedia.org/wiki/Cosmic_microwave_background" rel="noreferrer">Cosmic Microwave Background</a> radiation spectrum we see today?</p>

CC BY-SA 3.0
2018-03-01T15:12:36.123Z
Sounds like multi-photon absorption, where the second absorption can only take place when the electron is in a short-lived excited state.  Is that what you're thinking of?
Carl Witthoft
CC BY-SA 3.0
2018-03-01T15:36:34.030Z
@CarlWitthoft I'm guessing it has something to do with the 21 cm line being associated with the electron in the n=1 level. If excited by UV, the hyperfine splitting is going to be very different and there won't be a transition at 21 cm anymore. But how that affects the blackbody spectrum is complicated.
uhoh
<cosmology><radio-astronomy>
How did UV from the earliest stars 'alter the state of the 21 cm line' such that it shows up in CMB today?
uhoh
# How did UV from the earliest stars 'alter the state of the 21 cm line' such that it shows up in CMB today?

(Asked by: username_0 on 2018-03-01)

In [this question](https://astronomy.stackexchange.com/q/25337/7982) I discuss the recent (open access) paper in Nature [An absorption profile centred at 78 megahertz in the sky-averaged spectrum](https://www.nature.com/articles/nature25792) at length. The abstract begins:

> After stars formed in the early Universe, their ultraviolet light is expected, eventually, to have penetrated the primordial hydrogen gas and altered the excitation state of its 21-centimetre hyperfine line. This alteration would cause the gas to absorb photons from the cosmic microwave background, producing a spectral distortion that should be observable today at radio frequencies of less than 200 megahertz(1).

where reference 1 is Jonathan Pritchard and Abraham Loeb (2012) [21 cm cosmology in the 21st century](http://iopscience.iop.org/article/10.1088/0034-4885/75/8/086901/meta). From there I've found Leonid Chuzhoy and Zheng Zheng (2007) [Radiative Transfer Effect on Ultraviolet Pumping of the 21 cm Line in the High-Redshift Universe](http://iopscience.iop.org/article/10.1086/522491/meta).

I understand some basics about the hyperfine transition in hydrogen, and that the "spin temperature" of a gas in space can differ from the temperature of other partitions if it's being pumped, but these papers are more than a bit hard to read.

Is there a simple way to explain the basics behind how the exposure of hydrogen to the UV light produced in early stars would cause the "blip" in the (now red-shifted) 21 cm part of the [Cosmic Microwave Background](https://en.wikipedia.org/wiki/Cosmic_microwave_background) radiation spectrum we see today?

**Tags:** cosmology, radio-astronomy
> **Comments:**
> **username_1**: Sounds like multi-photon absorption, where the second absorption can only take place when the electron is in a short-lived excited state.  Is that what you're thinking of?  
> **username_0**: @CarlWitthoft I'm guessing it has something to do with the 21 cm line being associated with the electron in the n=1 level. If excited by UV, the hyperfine splitting is going to be very different and there won't be a transition at 21 cm anymore. But how that affects the blackbody spectrum is complicated.  

---

## 2 answers

### Answer by username_3 on 2018-04-04. Score: 5 | Accepted answer

Guiseppe Rossi's answer is excellent; I just want to add *why* the mentioned UV radiation modifies the spectrum of the background radiation.

(one word took the other, and it became a rather long comment.)

### The hyperfine level

Neutral hydrogen in its ground state can be in two different configurations; either the proton and the electron may have *parallel* spins ($\uparrow\uparrow$), or they may have *antiparallel* spins ($\uparrow\downarrow$). When the spins are parallel, the atom has a slightly higher energy than when they're antiparallel. The atoms "wants" to make a spin flip to the lower energy configuration$^\dagger$, and will eventually do so, but since the line is forbidden, the lifetime of the parallel state is of the order $10^7\,\mathrm{yr}$.

The relative population of the states is given by the Boltzmann distribution
$$
\begin{array}{rcl}
\frac{n\_1}{n\_0} & = & \frac{g\_1}{g\_0} \, e^{-\Delta E \, / \, k\_\mathrm{B} T\_S} \\
& = & 3 \, e^{-0.068\,\mathrm{K} \, / \, T\_S}\label{a}\tag{1},
\end{array}
$$
where subscripts 1 and 0 denote the $\uparrow\uparrow$ and $\uparrow\downarrow$ states, respectively, $n$ is the density, $g$ is the statistical weights (with $g\_0,g\_1 = 1,3$), $\Delta E = 5.9\times10^{-6}\,\mathrm{eV}$ is the energy difference of the states, $k\_\mathrm{B}$ is the Boltzmann constant, and $T\_S$ is the [spin temperature](https://en.wikipedia.org/wiki/Excitation_temperature), which I think is better thought of as "a number that describes the relative populations" than an actual temperature.

### Departure from equilibrium

In thermal equilibrium, the spin temperature is equal to the "real", kinetic temperature. Just after [decoupling](https://en.wikipedia.org/wiki/Decoupling_(cosmology)) of the radiation from matter at a redshift of $z\simeq1100$, the gas and the photons share the same energy, and since $T\gg 1$, we have that $n\_1/n\_0 \simeq 3$. But when the first stars begin to shine, they produce massive amounts of hard UV radiation which ionizes their surrounding medium. The ionized gas quickly recombines (in the beginning, at least), with $\sim2/3$ of the recombinations resulting in the emission of a Lyman $\alpha$ photon, i.e. a photon with an energy corresponding to the energy difference between the first excited state (one of the three $2P$ states) and the ground state (the $1S$ state) of the hydrogen atom (10.2 eV).

The Ly$\alpha$ photons scatter multiple times on the neutral hydrogen. Each scattering excites an atom from $1S\rightarrow 2P$, which subsequently de-excites and emits an Ly$\alpha$ photon in another direction. But since the energy difference between the $2P$ and the $1S$ state is a million times larger than between the hyperfine states, there is equal chance of ending in the $\uparrow\uparrow$ and the $\uparrow\downarrow$ state.

That is, $n\_1/n\_0$ is no longer $\simeq 3$, but is driven toward $\sim 1$. This is the [Wouthuysen–Field effect](https://en.wikipedia.org/wiki/Wouthuysen%E2%80%93Field_coupling) that Guiseppe Rossi mentions; from eq. \ref{a}, you see that this corresponds to a much smaller spin temperature, and thus the factor that Guiseppe Rossi mentions becomes negative. The full equation describing the brightness (or, equivalently, the flux received) as a function of redshift can be written (e.g. [Zaldarriaga et al. 2004](https://arxiv.org/abs/astro-ph/0311514))
$$
T(z) = 23\,\mathrm{mK} \,
\frac{T\_S - T\_\mathrm{CMB}}{T\_S} \,
(1+\delta)
x\_\mathrm{HI}(z)
\frac{\Omega\_\mathrm{b}h^2}{0.02}
\left( \frac{0.15}{\Omega\_\mathrm{m}h^2} \, \frac{1+z}{10} \right)^{1/2}\label{b}\tag{2}
$$
and when the $(T\_S - T\_\mathrm{CMB})\,/\,T\_S$ factor is negative, you will get an absorption line.

(In eq. \ref{b},
$\delta$,
$x\_\mathrm{HI}(z)$,
$\Omega\_\mathrm{b}$,
$\Omega\_\mathrm{m}$, and
$h$,
are
the local overdensity,
the neutral fraction of hydrogen,
the baryon and matter density parameter, and
the dimensionless (reduced) Hubble constant,
respectively, but this is of less importance.)

Since the observed absorption line ([Bowman et al. 2018](http://adsabs.harvard.edu/abs/2018Natur.555...67B)) starts to drop around an observed frequency of $\nu\_\mathrm{obs} = 65\text{–}70\,\mathrm{MHz}$, and since the rest frequency of the hyperfine line is $\nu\_\mathrm{rest} = 1420\,\mathrm{MHz}$, this means that the first stars appeared around a redshift of $z = \nu\_\mathrm{rest}/\nu\_\mathrm{obs} - 1 \simeq 20$, corresponding to an age of the Universe of $\sim 180\,\mathrm{Myr}$ (i.e. million years — the largest absorption is reached at $z\simeq 17$, or $t\simeq 200\,\mathrm{Myr}$).

Now the big question is, according to eq. \ref{b} the dip should be of the order a few tens of mK, but is in fact roughly 0.5 K, i.e. an order of magnitude larger. One possible mechanism that could produce this effect is coupling of the gas with dark matter, something which is not usually considered possible but could happen if the dark matter particle has a very small charge ([Barkana et al. 2018](https://arxiv.org/abs/1803.03091)).

### Time evolution of the 21 cm signal

The figure below (from a great review by [Pritchard & Loeb 2012](https://arxiv.org/abs/1109.6012)) shows how the 21 cm signal evolves with time. The dip discussed in this answer is the orange and red part.

## [21cm](https://i.sstatic.net/TVRZs.jpg)

$^\dagger$An analogy would be two magnets aligned parallel to each other with north in the same direction, preferring to flip around, but note Ken G's comment below; the transition doesn't necessarily involve a spin flip, and the analogy is not to be taken literally, since parallel magnets are alike, whereas parallel electrons/protons have opposite charges.

> **Comments:**
> **username_0**: This is extremely helpful, thank you for taking the time to write this up and explain so clearly. I'll take some time to read it and the references as well.  
> **username_3**: @uhoh I added another good reference to a review.  
> **username_4**: Minor correction to that wonderful answer: The 21 cm transition does not necessarily involve a spin flip, indeed the ground state involves a state where the spins of neither the proton nor the electron are specified (they are only specified to be opposite each other in an antisymmetric combination).  The upper level can have parallel definite spins, or indefinite spins that are known only to be antiparallel in a symmetric combination.  There is a loss of angular momentum in the transition, and given all this complexity, you can see why people claim it's a spin flip, but it's not really true.  
> **username_4**: Another interesting point:  the electron and proton have opposite charges, so you'd think that if their spins pointed the same way, that would be like magnets aligned in opposite directions, which is the *lowest* energy, not the highest.  The reason it is the highest energy when the spins are aligned and definite is actually that the strongest interaction between electron and proton is when the electron is inside the proton, so it's like anti-aligned magnets with one inside the other-- that's the highest energy configuration.  
> **username_4**: And when the spins are anti-aligned and indefinite, that's also the highest energy in the symmetric configuration ,because the symmetric (two swapping particle states) configuration is the one where the electron is going to be outside the proton, so that's the only situation that acts like two normal magnets.  
> **username_3**: Thanks for those comments, @KenG. I'll edit my answer accordingly.  
> **username_4**: It was such a splendid answer, may as well go all out!  

---

### Answer by username_2 on 2018-04-03. Score: 7

The Intergalactic medium at the relevant redshift is made of neutral hydrogen. What we can measure is the brightness temperature (the temperature that the IGM would have if it emitted as a blackbody) relative to the CMB. This quantity depends crucially on the following expression:

$\frac{T\_S - T\_{CMB}}{T\_S}$

where $T\_S$ is called spin temperature.

The spin temperature is just a measure of the ratio between the number of hydrogen atoms in the first excited hyperfine state (spin parallel) and ground state (spin antiparallel).
It turns out that the spin temperature can be modified in three ways: one of them is Lyman resonant scattering (the other two are collisional coupling and scattering of CMB photons).
Using atomic physics it's not difficult to see that $T\_S$ is a weighted mean of the temperature of the CMB and gas temperature.

UV photons can change the spin temperature because a hydrogen atom in the lowest state, n=1 (if you are familiar with chemistry, a s state) with antiparallel spins, can absorb an UV photon and jump in a p (n=2) state, and then fall again in a s state but with parallel spins.
This mechanism is known as Wouthuysen–Field coupling.

Before the emission takes place, the gas is at the same temperature of the CMB and both are equal to spin temperature, so the relative brightness temperature is zero.
When Lyman alpha emission begins the spin temperature decreases resulting in the observed absorption peak.
At some point the peak stops because the first stars emit X-rays and the gas becomes hotter than the background radiation.

> **Comments:**
> **username_0**: Thanks for your answer. This focuses mainly on altering the spin temperature, but the last two sentences of my question explain that I'm asking how this "...would cause the "blip" in the (now red-shifted) 21 cm part of the Cosmic Microwave Background..." The linked question includes the line shape, and it's a dip, a decrease in intensity. How does the modified $T_S$ produce this feature?  

---

