2 The Hot Universe

Equilibrium states are characterised by a number of macroscopic quantities. These will be dealt with in detail in the Statistical Physics course, but here we summarise some key facts.

The most important characteristic of an equilibrium state is temperature. This is related to the average energy of the state in a way that we will make precise below. The reason that temperature plays such an important role is due to the following property: suppose that we have two different systems, each individually in equilibrium, one at temperature $T_1$ and the other at temperature $T_2$. We then bring the two systems together and allow them to exchange energy. If $T_1=T_2$, then the two systems remain unaffected by this, and the combined system is in equilibrium. In contrast, if $T_1\ne T_2$, then then a net energy will flow from the hotter system to the colder system, and the combined system will eventually settle down to a new equilibrium state at some intermediate temperature. Two systems which have the same temperature are said to be in thermal equilibrium.

Other kinds of equilibria are also possible. One that we will meet later in this section arises when two systems are able to exchange particles. Often we will be interested in this when one type of particle can transmute into another. In this case, we characterise the system by another quantity known as the chemical potential. (The name comes from chemical reactions although, in this course, will be more interested in processes in atomic or particle physics.) The chemical potential has the property that if two systems have the same value then, when brought together, there will not be a net transfer of particles from one system to the other. In this case, the systems are said to be in chemical equilibrium.

A microscopic understanding of thermal equilibrium was first provided by Boltzmann. It turns out that the result is somewhat easier to state in the language of quantum mechanics, although it also applies to the classical world. Consider a system with discrete energy eigenstates $|n⟩$, each with energy $E_n$. In thermal equilibrium at temperature $T$, the probability that the system sits in the state $|n⟩$ is given by the Boltzmann distribution,

$p(n)={\displaystyle \frac{e^{-E_n/k_{B}T}}{Z}}$

(2.89)

Here $k_B$ is the Boltzmann constant, defined to be

$k_B\approx 1.381\times {10}^{-23}\mathrm{J}{\mathrm{K}}^{-1}$

This fundamental constant provides a translation between temperatures and energies. Meanwhile $Z$ is simply a normalisation constant designed to ensure that

${\displaystyle \sum _n}p(n)=1\mathit{\hspace{1em}\hspace{1em} }\Rightarrow \mathit{\hspace{1em}\hspace{1em} }Z={\displaystyle \sum _n}{e}^{-E_n/k_{B}T}$

This normalisation factor $Z$ has its own name: it is called the partition function and it plays a starring role in most treatments of statistical mechanics. For our purposes, it will suffice to keep $Z$ firmly in the background.

It is possible to derive the Boltzmann distribution from more elementary principles. (Such a derivation can be found in the lectures on Statistical Physics.) Here, we will simply take the distribution (2.89) to be the definition of both thermal equilibrium and the temperature.

The Boltzmann distribution gives us some simple intuition for the meaning of thermal equilibrium. We see that the any state with $E_n\ll k_{B}T$ has a more or less equal chance of being occupied, while any state with $E_n\gg k_{B}T$ has a vanishingly small chance of being occupied. In this way $k_{B}T$ sets the characteristic energy scale of the system.

We’ll see many variations of the Boltzmann distribution in what follows. It gets tedious to keep writing $1/k_{B}T$. For this reason we define

$\beta ={\displaystyle \frac{1}{k_{B}T}}$

We will be careless in what follows and also refer to $\beta$ as “temperature”: obviously it is actually (proportional to) the inverse temperature. The Boltzmann distribution then reads

$p(n)={\displaystyle \frac{e^{-\beta E_n}}{Z}}$

Above, we mentioned the key property of temperature: it determines whether two systems sit in thermal equilibrium. We should check that this is indeed obeyed by the Boltzmann distribution. Suppose that we have two systems, $A$ and $B$, both at the same temperature $\beta$, but with different microscopic constituents, meaning that their energy levels differ. If we bring the two systems together, we expect that the combined system also sits in a Boltzmann distribution at temperature $\beta$. Happily, this is indeed the case. To see this note that we have independent probability distributions for $A$ and $B$, so the combined probability distribution is given by

$p(n,m)={\displaystyle \frac{e^{-\beta E_n^A}}{Z_A}}{\displaystyle \frac{e^{-\beta E_m^B}}{Z_B}}={\displaystyle \frac{e^{-\beta (E_n^A+E_m^B)}}{Z_A{Z}_B}}$

But this is again of the Boltzmann form. The denominator $Z_A{Z}_B$ can be written as

$Z_A{Z}_B=\left({\displaystyle \sum _n}{e}^{-\beta E_n^A}\right)\left({\displaystyle \sum _m}{e}^{-\beta E_m^B}\right)={\displaystyle \sum _{n,m}}{e}^{-\beta E_n^A}{e}^{-\beta E_m^B}={\displaystyle \sum _{n,m}}{e}^{-\beta (E_n^A+E_m^B)}$

where we recognise this final expression as $Z_{A+B}$, the partition function of the combined system $A+B$. This had to be the case to ensure that the joint probability distribution $p(n,m)$ is correctly normalised.

It’s worth re-iterating what we have learned. You might think that if we combined two systems, separately in equilibrium, then there would be no energy transfer from one to the other if the average energies coincide, i.e. $⟨E^A⟩=⟨E^B⟩$, with

$⟨E⟩={\displaystyle \frac{1}{Z}}{\displaystyle \sum _n}{E}_n{e}^{-\beta E_n}$

However, this is not the right criterion. As we have seen above, the average energies of the two systems can be very different. It is the temperatures that must coincide.

We consider a gas of particles. We’ll assume that each particle is independent of the others, and focus on the state of a just single particle, specified by the momentum $𝐩$ or, equivalently, the velocity $𝐯=𝐩/m$. If the momentum is continuous (or finely spaced) we should talk about the probability that the velocity lies in some some volume $d^{3}v$ centred around $𝐯$. We denote the probability distribution as $f(𝐯)d^{3}v$. The Boltzmann distribution (2.89) tells us that this is

$f(𝐯)d^{3}v={\displaystyle \frac{e^{-\beta m{v}^2/2}}{Z}}{d}^{3}v$

(2.91)

where $Z$ is a normalisation factor that we will determine shortly.

Our real interest lies in the speed $v=|𝐯|$. The corresponding speed distribution $f(v)dv=f(𝐯)d^{3}v$ is

$f(v)dv={\displaystyle \frac{4\pi v^2}{Z}}{e}^{-\beta m{v}^2/2}dv$

(2.92)

Note that we have an extra factor of $4\pi v^2$ when considering the probability distribution over speeds $v$, as opposed to velocities $𝐯$. This reflects the fact that there’s “more ways” to have a high velocity than a low velocity: the factor of $4\pi v^2$ is the area of the sphere swept out by a velocity vector $𝐯$.

We require that

${\displaystyle {\int }_0^{\mathrm{\infty }}}𝑑vf(v)=1\mathit{\hspace{1em}\hspace{1em} }\Rightarrow \mathit{\hspace{1em}\hspace{1em} }Z={\left({\displaystyle \frac{2\pi k_{B}T}{m}}\right)}^{3/2}$

Finally, we find the probability that the particle has speed between $v$ and $v+dv$ to be

$f(v)dv=4\pi v^2{\left({\displaystyle \frac{m}{2\pi k_{B}T}}\right)}^{3/2}{e}^{-m{v}^2/2k_{B}T}dv$

(2.93)

This is known as the Maxwell-Boltzmann distribution. It tells us the distribution of the speeds of gas molecules in this room.

The expression (2.95) holds for both relativistic and non-relativistic systems, a fact that we will make use of later. For now, we care only for the non-relativistic case with $𝐩=m𝐯$. Here we have

$P={\displaystyle \frac{4\pi n}{3}}{\left({\displaystyle \frac{m}{2\pi k_{B}T}}\right)}^{3/2}{\displaystyle \int 𝑑vm{v}^4{e}^{-m{v}^2/2k_{B}T}}$

The integral is straightforward: it is given by

${\displaystyle {\int }_0^{\mathrm{\infty }}}𝑑x{x}^4{e}^{-a{x}^2}={\displaystyle \frac{3}{8}}\sqrt{{\displaystyle \frac{\pi }{a^5}}}$

Using this, we find a familiar friend

$P=n{k}_{B}T$

This is the equation of state for an ideal gas.

We can also calculate the average kinetic energy. If the gas contains $N$ particles, the total energy is

$⟨E⟩={\displaystyle \frac{N}{2}}m⟨v^2⟩=N{\displaystyle {\int }_0^{\mathrm{\infty }}}𝑑v{\displaystyle \frac{1}{2}}m{v}^{2}f(v)={\displaystyle \frac{3}{2}}N{k}_{B}T$

(2.96)

This confirms the result (1.37) that we met when we first introduced non-relativistic fluids.

The state of a single photon is specified by its momentum $𝐩=\mathrm{\hslash }𝐤$, with $𝐤$ the wavevector. The energy of the photon is given by

$E=pc=\mathrm{\hslash }\omega$

where $\omega =ck$ is the (angular) frequency of the photon.

Blackbody radiation comes with a new conceptual ingredient, because the number of photons is not a conserved quantity. This means that when considering the possible states of the gas, we should include states with an arbitrary number of photons. We do this by stating how many photons $N(𝐩)$ sit in the state $𝐩$.

In thermal equilibrium, we will not have a definite number of photons $N(𝐩)$, but rather some probability distribution over the number of photons, Focussing on a fixed state $𝐩=\mathrm{\hslash }𝐤$, the average number of particles is dictated by the Boltzmann distribution

$⟨N(𝐩)⟩={\displaystyle \frac{1}{Z}}{\displaystyle \sum _{n=0}^{\mathrm{\infty }}}n{e}^{-\beta n\mathrm{\hslash }\omega }\mathit{\hspace{1em}\hspace{1em}}\mathrm{with}Z={\displaystyle \sum _{n=0}^{\mathrm{\infty }}}{e}^{-\beta n\mathrm{\hslash }\omega }$

We can easily do both of these sums. Defining $x=e^{-\beta \mathrm{\hslash }\omega }$, the partition function is given by

$Z={\displaystyle \sum _{n=0}^{\mathrm{\infty }}}{x}^n={\displaystyle \frac{1}{1-x}}$

Meanwhile the numerator of $⟨N(𝐩)⟩$ takes the form

${\displaystyle \sum _{n=0}^{\mathrm{\infty }}}n{x}^n=x{\displaystyle \sum _{n=0}^{\mathrm{\infty }}}n{x}^{n-1}=x{\displaystyle \frac{dZ}{dx}}={\displaystyle \frac{x}{{(1-x)}^2}}$

We learn that the average number of particles with momentum $𝐩$ is

$⟨N(𝐩)⟩={\displaystyle \frac{1}{e^{\beta \mathrm{\hslash }\omega }-1}}$

(2.97)

For $k_{B}T\ll \mathrm{\hslash }\omega$, the number of photons is exponentially small. In contrast, when $k_{B}T\gg \mathrm{\hslash }\omega$, the number of photons grows linearly as $⟨N(𝐩)⟩\approx k_{B}T/\mathrm{\hslash }\omega$.

It’s easier to count objects that are discrete rather than continuous. For this reason, we’ll put our system in a square box with sides of length $L$. At the end of the calculation, we can happily send $L\to \mathrm{\infty }$. In such a box, the wavevector is quantised: it takes values

$k_i={\displaystyle \frac{2\pi q_i}{L}}\mathit{\hspace{1em}\hspace{1em}}{q}_i\in 𝐙$

This is true for both a classical wave or a quantum particle; in both cases, an integer number of wavelengths must fit in the box.

Different states are labelled by the integers $q_i$. When counting, or summing over such states, we should therefore sum over the $q_i$. However, for very large boxes, so that $L$ is much bigger than any other length scale in the game, we can approximate this sum by an integral,

${\displaystyle \sum _q}\approx {\displaystyle \frac{L^3}{{(2\pi )}^3}}{\displaystyle \int d^{3}k}={\displaystyle \frac{4\pi V}{{(2\pi )}^3}}{\displaystyle {\int }_0^{\mathrm{\infty }}}𝑑k{k}^2$

(2.98)

where $V=L^3$ is the volume of the box. The formula above counts all states. But the final form has a simple interpretation: the number of states with the magnitude of the wavevector between $k$ and $k+dk$ is $4\pi V{k}^2/{(2\pi )}^3$. Note that the $4\pi k^2$ term is reminiscent of the $4\pi v^2$ term that appeared in the Maxwell-Boltzmann distribution; both have the same origin.

We would like to compute the number of states with frequency between $\omega$ and $\omega +d\omega$. For this, we simply use

$\omega =ck\mathit{\hspace{1em}\hspace{1em} }\Rightarrow \mathit{\hspace{1em}\hspace{1em} }{\displaystyle \frac{4\pi V}{{(2\pi )}^3}}{\displaystyle \int 𝑑k{k}^2}={\displaystyle \frac{4\pi V}{{(2\pi c)}^3}}{\displaystyle \int 𝑑\omega {\omega }^2}$

This tells us that the number of states with frequency between $\omega$ and $\omega +d\omega$ is $4\pi V{\omega }^2/{(2\pi c)}^3$.

There is one final fact that we need. Photons come with two polarisation states. This means that the total number of states is twice the number above. We can now combine this with our earlier result (2.97). In thermal equilibrium, the average number of photons with frequency between $\omega$ and $\omega +d\omega$ is

$⟨N(\omega )⟩d\omega =2\times {\displaystyle \frac{4\pi V}{{(2\pi c)}^3}}{\displaystyle \frac{{\omega }^2}{e^{\beta \mathrm{\hslash }\omega }-1}}d\omega$

We usually write this in terms of the number density $n=N/V$. Moreover, we will be a little lazy and drop the expectation value $⟨n⟩$ signs. The distribution of photons in a thermal bath is then written as

$n(\omega )={\displaystyle \frac{1}{{\pi }^2{c}^3}}{\displaystyle \frac{{\omega }^2}{e^{\beta \mathrm{\hslash }\omega }-1}}$

(2.99)

This is the Planck blackbody distribution. For a fixed temperature, $\beta =1/k_{B}T$, the distribution tells us how many photons of a given frequency – and hence, of a given colour – are present. The distribution peaks in visible light for temperatures around $6000$ K, which is the temperature of the surface of the Sun. (Presumably the Sun evolved to be at exactly the right temperature so that our eyes can see it. Or something.)

Finally, we can actually do the integral (2.100). In fact, there’s a couple of quantities of interest. The energy density is

$\rho$

$=$

${\displaystyle \frac{\mathrm{\hslash }}{{\pi }^2{c}^3}}{\displaystyle {\int }_0^{\mathrm{\infty }}}𝑑\omega {\displaystyle \frac{{\omega }^3}{e^{\beta \mathrm{\hslash }\omega }-1}}={\displaystyle \frac{{(k_{B}T)}^4}{{\pi }^2{\mathrm{\hslash }}^3{c}^3}}{\displaystyle {\int }_0^{\mathrm{\infty }}}𝑑y{\displaystyle \frac{y^3}{e^y-1}}$

Meanwhile, the total number density is

$n={\displaystyle {\int }_0^{\mathrm{\infty }}}𝑑\omega n(\omega )={\displaystyle \frac{1}{{\pi }^2{c}^3}}{\displaystyle {\int }_0^{\mathrm{\infty }}}𝑑\omega {\displaystyle \frac{{\omega }^2}{e^{\beta \mathrm{\hslash }\omega }-1}}={\displaystyle \frac{{(k_{B}T)}^3}{{\pi }^2{\mathrm{\hslash }}^3{c}^3}}{\displaystyle {\int }_0^{\mathrm{\infty }}}𝑑y{\displaystyle \frac{y^2}{e^y-1}}$

Both of these integrals take a similar form. Here we just quote the general result without proof:

$I_n={\displaystyle {\int }_0^{\mathrm{\infty }}}𝑑y{\displaystyle \frac{y^n}{e^y-1}}=\mathrm{\Gamma }(n+1)\zeta (n+1)$

(2.101)

The Gamma function is the analytic continuation of the factorial function to the real numbers; when evaluated on the integers it gives $\mathrm{\Gamma }(n+1)=n!$. Meanwhile, the Riemann zeta function is defined, for $\mathrm{Re}(s)>1$, as $\zeta (s)={\sum }_{q=1}{q}^{-s}$. It turns out that $\zeta (4)={\pi }^4/90$, giving us $I_3={\pi }^4/15$. In contrast, there is no such simple expression for $\zeta (3)\approx 1.20$. It is sometimes referred to as Apéry’s constant. A derivation of (2.101) can be found in Section 3.5.3 of the lectures on Statistical Physics.

We learn that the energy density is

$\rho ={\displaystyle \frac{{\pi }^2}{15{\mathrm{\hslash }}^3{c}^3}}{(k_{B}T)}^4$

(2.102)

Meanwhile, the total number density is

$n={\displaystyle \frac{2\zeta (3)}{{\pi }^2{\mathrm{\hslash }}^3{c}^3}}{(k_{B}T)}^3$

(2.103)

Notice, in particular, that the number density of photons varies with the temperature. This will be important in what follows.

From the temperature (2.104), we can determine the energy density and number density in photons. From (2.102), the energy density is given by

${\rho }_{\gamma }\approx 4.3\times {10}^{-14}\mathrm{kg}{\mathrm{m}}^{-1}{s}^{-2}$

We can compare this to the critical energy density (1.69), ${\rho }_{\mathrm{crit},0}=8.5\times {10}^{-10}\mathrm{kg}{\mathrm{m}}^{-1}{s}^{-2}$ to find

${\mathrm{\Omega }}_{\gamma }={\displaystyle \frac{{\rho }_{\gamma }}{{\rho }_{\mathrm{crit},0}}}\approx 5\times {10}^{-5}$

This is the value (1.68) that we quoted previously. There are, of course, further photons in starlight, but they are dwarfed in both energy and number by the CMB.

From (2.103), the number density of CMB photons is

$n_{\gamma }=4\times {10}^8{\mathrm{m}}^{-3}=400{\mathrm{cm}}^{-3}$

We can compare this to the number of baryons (i.e. protons and neutrons). The density of baryons is (1.72) ${\mathrm{\Omega }}_B\approx 0.05$, so the total mass in baryons is

${\rho }_B\approx {\mathrm{\Omega }}_B{\rho }_{\mathrm{crit},0}\approx 4\times {10}^{-11}\mathrm{kg}{\mathrm{m}}^{-1}{s}^{-2}$

The mass of the proton and neutron are roughly the same, at $m_p\approx 1.7\times {10}^{-27}\mathrm{kg}$. This places the number density of baryons as

$n_B={\displaystyle \frac{{\rho }_B}{m_p{c}^2}}\approx 0.3{\mathrm{m}}^{-3}$

We see that there are many more photons in the universe than baryons: the ratio is

$\eta \equiv {\displaystyle \frac{n_B}{n_{\gamma }}}\approx {10}^{-9}$

(2.105)

This is one of the fundamental numbers in cosmology. As we will see, this ratio has been pretty much constant since the first second or so after the Big Bang and plays a crucial role in both nucleosynthesis (the formation of heavier nuclei) and in recombination (the formation of atoms). We do not, currently, have a good theoretical understanding of where this number fundamentally comes from: it is something that we can only derive from observation.

Recall that equilibrium is a property that arises when particles are constantly interacting. Yet the CMB photons have barely spoken to anyone for the past 13 billion years. The occasional photon may bump into a planet, or an infra-red detector fitted to a satellite, but most just wend their merry way through the universe, uninterrupted.

How then did the CMB photons come to form a perfect equilibrium spectrum? The answer is that this dates from a time when the photons were interacting frequently with matter. Fluids like this, that have long since fallen out of thermal equilibrium, but nonetheless retain their thermal character, are called relics.

There are a couple of questions that we would like to address. The first is: when were the photons last interacting and, hence, last genuinely in equilibrium? This is called the time of last scattering, $t_{\mathrm{last}}$ and we will compute it in Section 2.3 below. The second question is: what happened to the distribution of photons subsequently?

We start by answering the second of these questions. Once the photons no longer interact, they are essentially free particles. As the universe expands, each photon is redshifted as explained in Section 1.1.3. This means that the wavelength is stretched and, correspondingly, the frequency is decreased as the universe expands.

$\lambda (t)={\lambda }_{\mathrm{last}}{\displaystyle \frac{a(t)}{a(t_{\mathrm{last}})}}\mathit{\hspace{1em}\hspace{1em} }\Rightarrow \mathit{\hspace{1em}\hspace{1em} }\omega (t)={\omega }_{\mathrm{last}}{\displaystyle \frac{a(t_{\mathrm{last}})}{a(t)}}$

(2.106)

At the same time, the number of photons is diluted by a factor of ${\left(a(t_{\mathrm{last}})/a(t)\right)}^3$ as the universe expands. Putting these two effects together, an initial blackbody distribution (2.99) will, if left alone, evolve as

$n({\omega }_{\mathrm{last}};T_{\mathrm{last}},t)d{\omega }_{\mathrm{last}}={\displaystyle \frac{1}{{\pi }^2{c}^3}}{\left({\displaystyle \frac{a(t_{\mathrm{last}})}{a(t)}}\right)}^3{\displaystyle \frac{{\omega }_{\mathrm{last}}^2}{e^{\beta \mathrm{\hslash }{\omega }_{\mathrm{last}}}-1}}d{\omega }_{\mathrm{last}}$

The $1/a^3$ dilution factor is absorbed into the frequency in the ${\omega }^2$ and $d\omega$ terms. But not in the exponent. However, the resulting distribution can be put back into blackbody form if we think of the temperature as time dependent

$n(\omega ;T,t)d\omega ={\displaystyle \frac{1}{{\pi }^2{c}^3}}{\displaystyle \frac{\omega {(t)}^2}{e^{\beta (t)\mathrm{\hslash }\omega (t)}-1}}d\omega (t)$

where the $\beta (t)=1/k_{B}T(t)$, with the time varying temperature

$T(t)=T_{\mathrm{last}}{\displaystyle \frac{a(t_{\mathrm{last}})}{a(t)}}$

(2.107)

We see that, left alone, a blackbody distribution will keep the same overall form, but with the temperature scaling as $T\sim 1/a$.

This means that, if we can figure out the temperature $T_{\mathrm{last}}$ when the photons were last in equilibrium, then we can immediately determine the redshift at which this occurred $1+z_{\mathrm{last}}=a{(t_{\mathrm{last}})}^{-1}$. We’ll compute both of these in Section 2.3.

To their surprise, they found a background noise which did not depend on the direction in which they pointed their antenna. Nor did it depend on the time of day or the time of a year. Taken seriously, this suggested that the noise was a message from the wider universe.

There was, however, an alternative, more mundane explanation. Maybe the noise was coming from the antenna itself, some undiscovered systematic effect that they had failed be take into account. Indeed, they soon found a putative source of the noise: a pair of pigeons had taken roost and deposited what Penzias called “a white dielectric material” over much of the antenna. They removed this material (and shot the pigeons), but the noise remained. What Penzias and Wilson had on their hands was not pigeon shit, but one of the great discoveries of the twentieth century: the afterglow of the Big Bang itself, with a temperature that they measured to lie between 2.5 K and 4.5 K. In 1965 they published their result with the attention-grabbing title: “A Measurement of Excess Antenna Temperature at 4080 Mc/s”.

Penzias and Wilson were not unaware of the significance of their finding. In the year since they first found the noise, they had done what good scientists should always do: they talked to their friends. They were soon put in touch with the group in nearby Princeton where Jim Peebles, a theoretical cosmologist, had recently predicted a background radiation with a temperature of a few degrees, based on the idea of nucleosynthesis in the very early universe (an idea we will describe in Section 2.5.3). Meanwhile, three experimental colleagues, Dicke, Roll and Wilkinson had cobbled together a small antenna in the hope of searching for this radiation. These four scientists wrote a companion paper, outlining the importance of the discovery. In 1978, Penzias and Wilson were awarded the Nobel prize. It took another 39 years before Peebles gained the same recognition.

In fact there had been earlier predictions of the CMB. In the 1940s, Gammow together with Alpher and Herman suggested that the early universe began only with neutrons and, through somewhat dodgy calculations, concluded that there should be a background radiation at 5 K. Later other scientists, including Zel’dovich in the Soviet Union, and Hoyle and Taylor in England, used nucleosynthesis to predict the existence of the CMB at a few degrees. Yet none of these results were taken sufficiently seriously to search for the signal before Penzias and Wilson made their serendipitous discovery.

Detecting the CMB was just the beginning of the story. The radiation is not, it turns out, perfectly uniform but contains small anisotropies. These contain precious information about the make-up of the universe when it was much younger. A number of theorists, including Harrison, Zel’dovich, and Peebles and Yu, predicted that these anisotropies could be observed at a level of ${10}^{-4}$ to ${10}^{-5}$. These were finally detected by the NASA COBE satellite in the early 1990s. Since then a number of ground based telescopes, including BOOMERanG and MAXIMA, and a two full sky maps from the satellites WMAP and Planck, have mapped out the CMB in exquisite detail. We will describe these anisotropies in Section 3.

Although our ultimate goal is to describe atomic reactions, we can first introduce the chemical potential in a more mundane setting. Suppose that we have a fixed number of atoms $N$ in a box of size $V$. If we focus attention on some large, fixed sub-volume $V^{\prime }\subset V$, then we would expect the gas in $V^{\prime }$ to share the same macroscopic properties, such as temperature and pressure, as the whole gas in $V$. But particles can happily fly in and out of $V^{\prime }$ and the total number in this region is not fixed. Instead, there is some probability distribution which has the property that the average number density coincides with $N/V$.

In this situation, it’s clear that we should consider states of all possible particle number in $V^{\prime }$. There is a possibility, albeit a very small one, that $V^{\prime }$ contains no particles at all. There is also a small possibility that it contains all the particles.

If we work in the language of quantum mechanics, each state $|n⟩$ in the system can be assigned both an energy $E_n$ and a particle number $N_n$. Correspondingly, equilibrium states are characterised by two macroscopic properties: the temperature $T$ and the chemical potential $\mu$. These are defined through the generalised Boltzmann distribution

$p(n)={\displaystyle \frac{e^{-\beta (E_n-\mu N_n)}}{𝒵}}$

(2.109)

where $𝒵={\sum }_n{e}^{-\beta (E_n-\mu N_n)}$ is again the appropriate normalisation factor. In the language of statistical mechanics, this is referred to as the grand canonical ensemble.

Clearly, the distribution has the same exponential form as the Boltzmann distribution. This is important. We learned in Section 2.1.1 that two isolated systems which sit at the same temperature will remain in thermal equilibrium when brought together, meaning that there will be no transfer of energy from one system to the other. Exactly the same argument tells us that if two isolated systems have the same chemical potential then, when brought together, there will be no net flux of particles from one system to the other. In this case, we say that the systems are in chemical equilibrium.

Notice that the requirement for equilibrium is not that the number densities of the systems are equal: it is the chemical potentials that must be equal. This is entirely analogous to the statement that it is temperature, rather than energy density, that determines whether systems are in thermal equilibrium.

We’ll see examples of how to wield the chemical below, but before we do it’s worth mentioning a few issues.

• •

  In general, we can introduce a different chemical potential for every conserved quantity in the system. This is because conserved quantities commute with the Hamiltonian, and so it makes sense to label microscopic states by both the energy and a further quantum number. One familiar example is electric charge $Q$. Here, the corresponding chemical potential is voltage.

  This leads to an almost-contradictory pair of statements. First, we can only introduce a chemical potential for any conserved quantity. Second, the purpose of the chemical potential is to allow this conserved quantity to fluctuate! If you’re confused about this, then think back to the volume $V^{\prime }\subset V$, or to the meaning of voltage in electromagnetism, both of which give examples where these statements hold.

• •

  The story above is very similar to our derivation of the Planck blackbody distribution for photons. There too we labeled states by both energy and particle number, but we didn’t introduce a chemical potential. What’s different now? This is actually a rather subtle issue. Ultimately it is related to the fact that we ignore interactions while simultaneously pretending that they are crucial to reach equilibrium. As soon as we take these interactions into account, the number of photons is not conserved so we can’t label states by both energy and photon number. This is what prohibits us from introducing a chemical potential for photons. In contrast, we can introduce a chemical potential in situations where particle number (or some other quantity) is conserved even in the presence of interactions.

We consider non-relativistic particles, with energy

$E_{𝐩}={\displaystyle \frac{p^2}{2m}}$

As with our calculation of photons, we now consider states that have arbitrary numbers of particles. We choose to specify these states by stating how many particles $n_{𝐩}$ have momentum $𝐩$. For each choice of momentum, the number of particles⁷⁷ 7 Actually, there is a subtlety here: I am implicitly assuming that the particles are bosons. We’ll look at this more closely in Section 2.4. can be $n_{𝐩}=0,1,2,\mathrm{\dots }$. The generalised Boltzmann distribution (2.109) then tells us that the average number of particles with momentum $𝐩$ is

$⟨N(𝐩)⟩={\displaystyle \frac{1}{{𝒵}_{𝐩}}}{\displaystyle \sum _{n_{𝐩}=0}^{\mathrm{\infty }}}{n}_{𝐩}{e}^{-\beta (n_{𝐩}{E}_{𝐩}-\mu n_{𝐩})}$

where the normalisation factor (or, in fancy language, the grand canonical partition function) is given by the geometric series

${𝒵}_{𝐩}={\displaystyle \sum _{n=0}^{\mathrm{\infty }}}{e}^{-\beta n_{𝐩}(E_{𝐩}-\mu )}={\displaystyle \frac{e^{\beta (E_{𝐩}-\mu )}}{e^{\beta (E_p-\mu )}-1}}$

This is exactly the same calculation as we saw for photons in Section 2.2.1, but with the additional minor complication of a chemical potential. Note that computing ${𝒵}_{𝐩}$ allows us to immediately determine the expected number of particles since we can write

$⟨N(𝐩)⟩={\displaystyle \frac{1}{\beta }}{\displaystyle \frac{\partial }{\partial \mu }}\log {𝒵}_{𝐩}={\displaystyle \frac{1}{e^{\beta (E_{𝐩}-\mu )}-1}}$

(2.110)

This is known as the Bose-Einstein distribution and will be discussed further in Section 2.4.

To compute the average total number of particles, we simply need to integrate over all momenta $𝐩$. We must include the density of states, but this is identical to the calculation we did for photons, with the result (2.98). The total average number of particles is then

$N={\displaystyle \frac{V}{{(2\pi \mathrm{\hslash })}^3}}{\displaystyle \int d^{3}pN(𝐩)}$

where we’ve been a little lazy and dropped the $⟨\cdot ⟩$ brackets on $N(𝐩)$. We usually write this in terms of the particle density $n=N/V$,

$n={\displaystyle \frac{1}{{(2\pi \mathrm{\hslash })}^3}}{\displaystyle \int d^{3}pN(𝐩)}={\displaystyle \frac{4\pi }{{(2\pi \mathrm{\hslash })}^3}}{\displaystyle {\int }_0^{\mathrm{\infty }}}𝑑p{\displaystyle \frac{p^2}{e^{-\beta \mu }{e}^{\beta p^2/2m}-1}}$

(2.111)

where, in the second equality, we have chosen to integrate using spherical polar coordinates, picking up a factor of $4\pi$ from the angular integrals and a factor of $p^2$ in the Jacobian for our troubles. We have also used the explicit expression $E_{𝐩}=p^2/2m$ for the energy in the distribution.

At this stage, we have an annoying looking integral to do. To proceed, let’s pick a value of the chemical potential $\mu$ such that $e^{-\beta \mu }\gg 1$. (We’ll see what this means physically below.) We can then drop the $-1$ in the denominator and approximate the integral as

$n$

$\approx$

${\displaystyle \frac{4\pi }{{(2\pi \mathrm{\hslash })}^3}}{e}^{\beta \mu }{\displaystyle {\int }_0^{\mathrm{\infty }}}𝑑p{p}^2{e}^{-\beta p^2/2m}={\left({\displaystyle \frac{m{k}_{B}T}{2\pi {\mathrm{\hslash }}^2}}\right)}^{3/2}{e}^{\beta \mu }$

(2.112)

Let’s try to interpret this. Read naively, it seems to tell us that the number density of particles depends on the temperature. But that’s certainly not what happens for the gas in this room, where $\rho$ and $P$ depend on temperature but the number density $n=N/V$ is fixed. We can achieve this by taking the chemical potential $\mu$ to also depend on temperature. Specifically, we wish to describe a gas with fixed $n$ , then we simply invert the equation above to get an expression for the chemical potential

$e^{\beta \mu }={\left({\displaystyle \frac{2\pi {\mathrm{\hslash }}^2}{m{k}_{B}T}}\right)}^{3/2}n$

(2.113)

Before we proceed, we can use this result to understand what the condition $e^{-\beta \mu }\gg 1$, that we used to do the integral, is forcing upon us. Comparing to the expression above, it says that the number density is bounded above by

$n\ll {\left({\displaystyle \frac{m{k}_{B}T}{2\pi {\mathrm{\hslash }}^2}}\right)}^{3/2}$

This is sensible. It’s telling us that the ideal gas can’t be too dense. In particular, the average distance between particles should be much larger than the length scale set by $\lambda =\sqrt{2\pi {\mathrm{\hslash }}^2/m{k}_{B}T}$. This is the average de Broglie wavelength of particles at temperature $T$. If $n$ is increased so that the separation between particles is comparable to $\lambda$ then quantum effects kick in and we have to return to our original integral (2.111) and make a different approximation to do the integral and understand the physics. (This path will lead to the beautiful phenomenon of Bose-Einstein condensation, but it is a subject for a different course.)

We can now calculate the energy density and pressure. Once again, taking the limit $e^{\beta \mu }\gg 1$, the energy density is given by

	$\rho$	$=$	${\displaystyle \frac{1}{{(2\pi \mathrm{\hslash })}^3}}{\displaystyle \int d^{3}p{E}_{𝐩}N(𝐩)}$
		$\approx$	${\displaystyle \frac{4\pi }{{(2\pi \mathrm{\hslash })}^3}}{e}^{\beta \mu }{\displaystyle {\int }_0^{\mathrm{\infty }}}𝑑p{\displaystyle \frac{p^4}{2m}}{e}^{-\beta p^2/2m}={\displaystyle \frac{3}{2}}n{k}_{B}T$

This is a result that we have met before (2.96). Meanwhile, we can use our expression (2.95) to compute the pressure,

	$P$	$=$	${\displaystyle \frac{1}{{(2\pi \mathrm{\hslash })}^3}}{\displaystyle \int d^{3}p\frac{𝐯\cdot 𝐩}{3}N(𝐩)}$
		$=$	${\displaystyle \frac{4\pi }{{(2\pi \mathrm{\hslash })}^3}}{e}^{\beta \widehat{\mu }}{\displaystyle {\int }_0^{\mathrm{\infty }}}𝑑p{\displaystyle \frac{p^4}{3m}}{e}^{-\beta p^2/2m}=n{k}_{B}T$

Again, this recovers the familiar ideal gas equation.

So far, the chemical potential has not bought us anything new. We have simply recovered old results in a slightly more convoluted framework in which the number of particles can fluctuate. But, as we will now see, this is exactly what we need to deal with atomic reactions.

To simplify life, we will assume that the hydrogen atom forms in its ground state, with a binding energy

$E_{\mathrm{bind}}\approx 13.6\mathrm{eV}$

In fact, this turn out to be a bad assumption! We explain why at the end of this section.

Naively, we would expect hydrogen to ionize when we reach temperatures of $k_{B}T\approx E_{\mathrm{bind}}$. It’s certainly true that for temperature $k_{B}T\gg E_{\mathrm{bind}}$, the electrons can no longer cling on to the protons, and any hydrogen atom is surely ripped apart. However, it will ultimately turn out that hydrogen only forms at temperatures significantly lower than $E_{\mathrm{bind}}$.

We’ll treat each of the massive particles – the electron, proton and hydrogen atom – in a similar way to the non-relativistic gas that we met in Section 2.3.2. There will, however, be two differences. First, we include the rest mass energy of the atoms, so each particle has energy

$E_{𝐩}=m{c}^2+{\displaystyle \frac{p^2}{2m}}$

This will be useful as we can think of the binding energy $E_{\mathrm{bind}}$ as the mass difference

$(m_e+m_p-m_H)c^2=E_{\mathrm{bind}}\approx 13.6\mathrm{eV}$

(2.114)

Secondly, each of our particles comes with a number $g$ of internal states. The electron and proton each have $g_e=g_p=2$ corresponding to the two spin states, referred to as “spin up” and “spin down”. (These are analogous to the two polarisation states of the photon that we included when discussing blackbody radiation.) For hydrogen, we have $g_H=4$; the electron and proton spin can either be aligned, to give a spin 0 particle, or anti-aligned to give 3 different spin 1 states.

With these two amendments, our expression for the number density (2.112) of the different species of particles is given by

$n_i=g_i{\left({\displaystyle \frac{m_i{k}_{B}T}{2\pi {\mathrm{\hslash }}^2}}\right)}^{3/2}{e}^{-\beta (m_i{c}^2-{\mu }_i)}$

(2.115)

Note that the rest mass energy $m{c}^2$ in the energy can be absorbed by a constant shift of the chemical potential.

Now we can use the chemical potential for something new. We require that these particles are in chemical equilibrium. This means that there is no rapid change from $e^{-}+p^{+}$ pairs into hydrogen, or vice versa: the numbers of electrons, protons and hydrogen are balanced. This is ensured if the chemical potentials are related by

${\mu }_e+{\mu }_p={\mu }_H$

(2.116)

This follows from our original discussion of what it means to be in chemical equilibrium. Recall that if two isolated systems have the same chemical potential then, when brought together, there will be no net flux of particles from one system to the other. This mimics the statement about thermal equilibrium, where if two isolated systems have the same temperature then, when brought together, there will be no net flux of energy from one to the other.

There is no chemical potential for photons because they’re not conserved. In particular, in addition to the reaction $e^{-}+p^{+}\leftrightarrow H+\gamma$ there can also be reactions in which the binding results in two photons, $e^{-}+p^{+}\leftrightarrow H+\gamma +\gamma$, which is ultimately why it makes no sense to talk about a chemical potential for photons. (Some authors write this, misleadingly, as ${\mu }_{\gamma }=0$.)

We can use the condition for chemical equilibrium (2.116) to eliminate the chemical potentials in (2.115) to find

${\displaystyle \frac{n_H}{n_e{n}_p}}={\displaystyle \frac{g_H}{g_e{g}_p}}{\left({\displaystyle \frac{m_H}{m_e{m}_p}}{\displaystyle \frac{2\pi {\mathrm{\hslash }}^2}{k_{B}T}}\right)}^{3/2}{e}^{-\beta (m_H-m_e-m_p)c^2}$

(2.117)

In the pre-factor, it makes sense to approximate $m_H\approx m_p$. However, in the exponent, the difference between these masses is crucial; it is the binding energy of hydrogen (2.114). Finally, we use the observed fact that the universe is electrically neutral, so

$n_e=n_p$

We then have

${\displaystyle \frac{n_H}{n_e^2}}={\left({\displaystyle \frac{2\pi {\mathrm{\hslash }}^2}{m_e{k}_{B}T}}\right)}^{3/2}{e}^{\beta E_{\mathrm{bind}}}$

(2.118)

This is the Saha equation.

Our goal is to understand the fraction of electron-proton pairs that have combined into hydrogen. To this end, we define the ionisation fraction

$X_e={\displaystyle \frac{n_e}{n_B}}\approx {\displaystyle \frac{n_e}{n_p+n_H}}$

where, in the second equality, we’re ignoring neutrons and higher elements. (We’ll see in Section 2.5.3 that this is a fairly good approximation.) Since $n_e=n_p$, if $X_e=1$ it means that all the electrons are free. If $X_e=0.1$, it means that only 10% of the electrons are free, the remainder bound inside hydrogen.

Using $n_e=n_p$, we have $1-X_e=n_H/n_B$ and so

${\displaystyle \frac{1-X_e}{X_e^2}}={\displaystyle \frac{n_H}{n_e^2}}{n}_B$

The Saha equation gives us an expression for $n_H/n_e^2$. But to translate this into the fraction $X_e$, we also need to know the number of baryons. This we take from observation. First, we convert the number of baryons into the number of photons, using (2.105),

$\eta ={\displaystyle \frac{n_B}{n_{\gamma }}}\approx {10}^{-9}$

Here we need to use the fact that $\eta \approx {10}^{-9}$ has remained constant since recombination. Next, we use the fact that photons sit at the same temperature as the electrons, protons and hydrogen because they are all in equilibrium. This means that we can then use our earlier expression (2.103) for the number of photons

$n_{\gamma }={\displaystyle \frac{2\zeta (3)}{{\pi }^2{\mathrm{\hslash }}^3{c}^3}}{(k_{B}T)}^3$

Combining these gives our final answer

${\displaystyle \frac{1-X_e}{X_e^2}}=\eta {\displaystyle \frac{2\zeta (3)}{{\pi }^2}}{\left({\displaystyle \frac{2\pi k_{B}T}{m_e{c}^2}}\right)}^{3/2}{e}^{\beta E_{\mathrm{bind}}}$

(2.119)

Suppose that we look at temperature $k_{B}T\sim E_{\mathrm{bind}}$, which is when we might naively have thought recombination takes place. We see that there are two very small numbers in the game: the factor of $\eta \sim {10}^{-9}$ and $k_{B}T/m_e{c}^2$, where the electron mass is $m_e{c}^2\approx 0.5\mathrm{MeV}=5\times {10}^5\mathrm{eV}$. These ensure that at $k_{B}T\sim E_{\mathrm{bind}}$, the ionisation fraction $X_e$ is very close to unity. In other words, nearly all the electrons remain free and unbound. In large part this is of the enormous number of photons, which mean that whenever a proton and electron bind, one can still find sufficient high energy photons in the tail of the blackbody distribution to knock them apart.

Recombination only takes place when the $e^{\beta E_{\mathrm{bind}}}$ factor is sufficient to compensate both the $\eta$ and $k_{B}T/m_e{c}^2$ factors. Clearly recombination isn’t a one-off process; it happens continuously as the temperature varies. As a benchmark, we’ll calculate the temperature when $X_e=0.1$, so 90% of the electrons are sitting happily in their hydrogen homes. From (2.119), we learn that this occurs when $\beta E_{\mathrm{bind}}\approx 45$, or

$k_B{T}_{\mathrm{rec}}\approx 0.3\mathrm{eV}\mathit{\hspace{1em}\hspace{1em} }\Rightarrow \mathit{\hspace{1em}\hspace{1em} }{T}_{\mathrm{rec}}\approx 3600\mathrm{K}$

This corresponds to a redshift of

$z_{\mathrm{rec}}={\displaystyle \frac{T_{\mathrm{rec}}}{T_0}}\approx 1300$

This is significantly later than matter-radiation equality which, as we saw in (1.71), occurs at $z_{\mathrm{eq}}\approx 3400$. This means that, during recombination, the universe is matter dominated, with $a(t)\sim {(t/t_0)}^{2/3}$. We can therefore date the time of recombination to,

$t_{\mathrm{rec}}\approx {\displaystyle \frac{t_0}{{(1+z_{\mathrm{rec}})}^{3/2}}}\approx 300,000\mathrm{years}$

After recombination, the constituents of the universe have been mostly neutral atoms. Roughly speaking this means that the universe is transparent and photons can propagate freely. We will look more closely at this statement a little more closely below.

The hydrogen atom doesn’t just have a ground state: there are a whole tower of excited states. These can form without emitting a high energy photon and, indeed, at these low temperatures the thermal bath of photons is in equilibrium with the tower of excited states of hydrogen. There are then two, rather inefficient processes, which populate the 1s state. The 2s state decays down to 1s by emitting two photons (to preserve angular momentum), neither of which have enough energy to re-ionize other atoms. Alternatively, the 2p state can decay to 1s, emitting a photon whose energy is barely enough to excite another hydrogen atom out of the ground state. If this photon experiences redshift, then it can no longer do the job and we increase the number of atoms in the ground state. More details can be found in the book by Weinberg. These issues do not greatly change the values of $T_{\mathrm{rec}}$ and $z_{\mathrm{rec}}$ that we computed above.

Similar scenarios play out a number of times in the early universe: particles, which once interacted frequently, stop talking to their neighbours and subsequently evolve without care for what’s going on around them. This process is common enough that it is worth exploring in a little detail. As we will see, at heart it hinges on what it means for particle to be in “equilibrium”.

Strictly speaking, an expanding universe is a time dependent background in which the concept of equilibrium does not apply. In most situations, such a comment would be rightly dismissed as the height of pedantry. The expansion of the universe does not, for example, stop me applying the laws of thermodynamics to my morning cup of tea. However, in the very early universe this can become an issue.

For a system to be in equilibrium, the constituent particles must frequently interact, exchanging energy and momentum. For any species of particle (or pair of species) we can define the interaction rate $\mathrm{\Gamma }$. A particle will, on average, interact with another particle in a time $t_{\mathrm{int}}=1/\mathrm{\Gamma }$. It makes sense to talk about equilibrium provided that the universe hasn’t significantly changed in the time $t_{\mathrm{int}}$. The expansion of the universe is governed by the Hubble parameter, so we can sensibly talk about equilibrium provided

$\mathrm{\Gamma }\gg H$

In contrast, if $\mathrm{\Gamma }\ll H$ then by the time particles interact the universe has undergone significant expansion. In this case, thermal equilibrium cannot be maintained.

For many processes, both the interaction rate and temperature scale with $T$, but in different ways. The result is that particles retain equilibrium at early times, but decouple from the thermal bath at late time. This decoupling occurs when $\mathrm{\Gamma }\approx H$ and is known as freeze out.

We now apply these ideas to photons, where freeze out also goes by the name of last scattering. In the early universe, the photons are scattered primarily by the electrons (because they are much lighter than the protons) in a process known as Thomson scattering

$e+\gamma \to e+\gamma$

The scattering is elastic, meaning that the energy, and therefore the frequency, of the photon is unchanged in the process. For Thomson scattering, the interaction rate is given by

$\mathrm{\Gamma }=n_e{\sigma }_{T}c$

where ${\sigma }_T$ is the cross-section, a quantity which characterises the strength of the scattering. We computed the cross-section for Thomson scattering in the lectures on Electromagnetism (see Section 6.3.1 of these lectures) where we showed it was given by

${\sigma }_T={\displaystyle \frac{{\mu }_0^2{e}^4}{6\pi m_e^2{c}^2}}\approx 6\times {10}^{-30}{\mathrm{m}}^2$

Note the dependence on the electron mass $m_e$; the corresponding cross-section for scattering off protons is more than a million times smaller.

Last scattering occurs at the temperature $T_{\mathrm{last}}$ such that $\mathrm{\Gamma }(T_{\mathrm{last}})\approx H(t_{\mathrm{last}})$. We can express the interaction rate by replacing the number density of electrons with the number density of photons,

$\mathrm{\Gamma }(T_{\mathrm{last}})=n_B{X}_e(T_{\mathrm{last}}){\sigma }_{T}c=\eta {\sigma }_T{\displaystyle \frac{2\zeta (3)}{{\pi }^2{\mathrm{\hslash }}^3{c}^2}}{(k_B{T}_{\mathrm{last}})}^3{X}_e(T_{\mathrm{last}})$

(2.120)

Meanwhile, we can trace back the current value of the Hubble constant, through the matter dominated era, to last scattering. Meanwhile, to compute $H(T_{\mathrm{last}})$, we use the formula (1.66)

${\left({\displaystyle \frac{H}{H_0}}\right)}^2={\displaystyle \frac{{\mathrm{\Omega }}_r}{a^4}}+{\displaystyle \frac{{\mathrm{\Omega }}_m}{a^3}}+{\displaystyle \frac{{\mathrm{\Omega }}_k}{a^2}}+{\mathrm{\Omega }}_{\mathrm{\Lambda }}$

Evaluated at recombination, radiation, curvature and the cosmological constant are all irrelevant, and this formula becomes

${\left({\displaystyle \frac{H}{H_0}}\right)}^2\approx {\displaystyle \frac{{\mathrm{\Omega }}_m}{a^3}}$

Using the fact that temperature scales as $T\sim 1/a$, we then have

$H(T_{\mathrm{last}})=H_0\sqrt{{\mathrm{\Omega }}_m}{\left({\displaystyle \frac{T_{\mathrm{last}}}{T_0}}\right)}^{3/2}$

Equating this with (2.120) gives

$X_e(T_{\mathrm{last}}){(k_B{T}_{\mathrm{last}})}^{3/2}={\displaystyle \frac{{\pi }^2{\mathrm{\hslash }}^3{c}^2}{2\zeta (3)}}{\displaystyle \frac{H_0\sqrt{{\mathrm{\Omega }}_m}}{\eta {\sigma }_T{(k_B{T}_0)}^{3/2}}}$

Using (2.119) to solve for $X_e(T_{\mathrm{last}})$ (which is a little fiddly) we find that photons stop interacting with matter only when

$X_e(T_{\mathrm{last}})\approx 0.01$

We learn that the vast majority of electrons must be housed in neutral hydrogen, with only 1% of the original electrons remaining free, before light can happily travel unimpeded. This corresponds to a temperature

$k_B{T}_{\mathrm{last}}\approx 0.27\mathrm{eV}\mathit{\hspace{1em}\hspace{1em} }\Rightarrow \mathit{\hspace{1em}\hspace{1em} }{T}_{\mathrm{last}}\approx 3100\mathrm{K}$

and, correspondingly, a time somewhat after recombination,

$z_{\mathrm{last}}={\displaystyle \frac{T_{\mathrm{last}}}{T_0}}\approx 1100\mathit{\hspace{1em}\hspace{1em} }\Rightarrow \mathit{\hspace{1em}\hspace{1em} }{t}_{\mathrm{last}}={\displaystyle \frac{t_0}{{(1+z_{\mathrm{last}})}^{3/2}}}\approx 350,000\mathrm{years}$

After this time, the universe becomes transparent. The cosmic microwave background is a snapshot of the universe from this time.

In what follows, we will restrict attention to non-interacting particles. In this case, there is a simple way to construct the full set of states $|n⟩$ starting from the single-particle Hilbert space. The state of a single particle is specified by its momentum $𝐩=\mathrm{\hslash }𝐤$. (There may also be some extra, discrete internal degrees of freedom like polarisation or spin; we’ll account for these later.) We’ll denote this single particle state as $|𝐩⟩$. For a relativistic particle, the energy is

$E_{𝐩}=\sqrt{m^2{c}^4+p^2{c}^2}$

(2.121)

To specify the full multi-particle state $|n⟩$, we need to say how many particles $n_{𝐩}$ occupy the state $|𝐩⟩$. The possible values of $n_{𝐩}$ depend on whether the underlying particle is a boson or fermion:

	$\mathrm{Bosons}:$		$\mathrm{ }\mathit{\hspace{1em}}{n}_{𝐩}=0,1,2,\mathrm{\dots }$
	$\mathrm{Fermions}:$		$\mathrm{ }\mathit{\hspace{1em}}{n}_{𝐩}=0,1$

In our previous discussions of blackbody radiation in Section 2.2.1 and the non-relativistic gas in Section 2.3.2, we did the counting appropriate for bosons. This is fine for blackbody radiation, since photons are bosons, but was an implicit assumption in the case of a non-relativistic gas.

The other alternative is a fermion. For these particles, the Pauli exclusion principle says that a given single-particle state $|𝐩⟩$ is either empty or occupied. But you can’t put more than one fermion there. This is entirely analogous to the way the periodic table is constructed in chemistry, by filling successive shells, except now the states are in momentum space. (A better analogy is the way a band is filled in solid state physics as described in the lectures on Quantum Mechanics.) For bosonic particles, there is no such restriction: you can pile up as many as you like.

Now we can compute some quantities, like the average particle number and average energy. We deal with bosons and fermions in turn

For bosons, the calculation is exactly the same as we saw in Section 2.3.2. For a given momentum $𝐩$, the average number of photons is

$⟨N(𝐩)⟩={\displaystyle \frac{1}{{𝒵}_{𝐩}}}{\displaystyle \sum _{n_{𝐩}=0}^{\mathrm{\infty }}}{n}_{𝐩}{e}^{-\beta (n_{𝐩}{E}_{𝐩}-\mu n_{𝐩})}={\displaystyle \frac{1}{\beta }}{\displaystyle \frac{\partial }{\partial \mu }}\log {𝒵}_{𝐩}$

where the normalisation factor is given by the geometric series

${𝒵}_{𝐩}={\displaystyle \sum _{n=0}^{\mathrm{\infty }}}{e}^{-\beta n_{𝐩}(E_{𝐩}-\mu )}={\displaystyle \frac{e^{\beta (E_{𝐩}-\mu )}}{e^{\beta (E_p-\mu )}-1}}$

As in the previous section, we will be a little lazy and drop the expectation value, so $⟨N(𝐩)⟩\equiv N(𝐩)$. Then we have

$N(𝐩)={\displaystyle \frac{1}{e^{\beta (E_{𝐩}-\mu )}-1}}$

(2.122)

This is known as the Bose-Einstein distribution.

For fermions, the calculation is easier still. We can have only $n_{𝐩}=0$ or 1 particles in a given state $|𝐩⟩$ so the average occupation number is

$N(𝐩)={\displaystyle \frac{1}{{𝒵}_p}}{\displaystyle \sum _{n_{𝐩}=0,1}}{n}_{𝐩}{e}^{-\beta (n_{𝐩}{E}_{𝐩}-\mu n_{𝐩})}\mathit{\hspace{1em}\hspace{1em} }\mathrm{with}\mathit{\hspace{1em}\hspace{1em} }{𝒵}_{𝐩}={\displaystyle \sum _{n_{𝐩}=0,1}}{e}^{-\beta (n_{𝐩}{E}_{𝐩}-\mu n_{𝐩})}$

Again, keeping the $⟨\cdot ⟩$ expectation value signs implicit, we have

$N(𝐩)={\displaystyle \frac{1}{e^{\beta (E_{𝐩}-\mu )}+1}}$

(2.123)

This is the Fermi-Dirac distribution.

For both bosons and fermions, the calculation of the density of states (2.98) proceeds as before, so that if we integrate over all possible momenta, it should be weighted by

${\displaystyle \frac{4\pi V}{{(2\pi \mathrm{\hslash })}^3}}{\displaystyle \int d^{3}p}$

with the pre-factor telling us how quantum states are in a small region $d^{3}p$.

If we include the degeneracy factor $g$, which tells us the number of internal states of the particle, the number density $n=N/V$ is given by

$n={\displaystyle \frac{g}{{(2\pi \mathrm{\hslash })}^3}}{\displaystyle \int d^{3}pN(𝐩)}$

(2.124)

Similarly, the energy density is

$\rho ={\displaystyle \frac{g}{{(2\pi \mathrm{\hslash })}^3}}{\displaystyle \int d^{3}p{E}_{𝐩}N(𝐩)}$

(2.125)

and the pressure (2.95) is

$P={\displaystyle \frac{g}{{(2\pi \mathrm{\hslash })}^3}}{\displaystyle \int d^{3}p\frac{𝐯\cdot 𝐩}{3}N(𝐩)}$

(2.126)

We’ll now apply these in various examples.