The cosmic microwave background

Builds on: Sections 13.2, 13.4 and 4.6
Reading time: 40 min
Exercises: 3
Olympiad problems: 1

In 1964 Arno Penzias and Robert Wilson, testing a horn-shaped radio antenna in New Jersey, found a faint hiss of microwaves that came equally from every direction, by day and by night, and that no cleaning of the antenna could remove. It corresponded to a temperature of about 3.5 K. They had found the afterglow of the Big Bang. In the 1990s the COBE satellite measured its spectrum and found it to be the most perfect blackbody ever seen, at 2.725 K, with differences from place to place on the sky of only about one part in a hundred thousand. In this section we see where this radiation comes from and what it tells us.

13.7.1Light from when the Universe became transparent

Before you readMake a guess first

Hydrogen atoms lose their electrons to photons of more than 13.6 eV, which corresponds to a temperature of about 158 000 K. At about what temperature did the early Universe become cool enough for hydrogen atoms to survive?

In the young Universe, atoms could not hold on to their electrons. Free electrons scatter light very easily, so a photon could not travel far before bouncing off one, and the Universe was opaque, like the inside of a star. As the Universe expanded and cooled to about 3000 K, electrons and protons joined into neutral hydrogen atoms, an event called recombination. Neutral atoms scatter light very weakly, so the Universe became transparent, and the photons of that moment have travelled freely ever since. This happened at $z \approx 1100$ , about 380 000 years after the Big Bang (363 000 years in our simple model).

The radiation has been stretched by $1 + z = 1101$ on its way ((13.4)). It left as the glow of gas at 3000 K, mostly in near-infrared and red light, and it arrives as microwaves at 2.725 K, with its peak near 1.06 mm by Wien's law ((4.11)), or near 160 GHz in frequency. This is the cosmic microwave background. Every cubic metre of space holds about 411 million of its photons, against roughly 0.27 atoms.

Figure 13.6The background radiation through time

Back in time to

Temperature
Peak wavelength
Number of photons

The figure shows the blackbody spectrum of the background radiation against wavelength on a logarithmic scale, at the chosen redshift (heavy) and today (dotted), each scaled to the same height. P marks the peak, given by Wien's law. Going back in time, the whole curve slides to shorter wavelengths without changing its shape, which is why the radiation still has a perfect blackbody spectrum today. The table gives the temperature, the peak wavelength and the number of photons in each cubic centimetre, which grows as

(1 + z)^{3}

. The figure can be set to , , or .

Quick checkThe colour of the Universe at recombination

At what wavelength did the spectrum of the background radiation peak at recombination, when its temperature was 3000 K?

13.7.2The dipole and the ripples

The background is not quite the same in every direction. One half of the sky is slightly warmer than the other, by about 3.36 mK. This dipole comes from our own motion. The Sun moves at about 370 km/s relative to the background, and the Doppler shift makes the radiation ahead of us slightly hotter and that behind slightly cooler, by the fraction $v / c$ .

ResultThe temperature of the background(13.14)

\begin{aligned} T & = T_{0} (1 + z), T_{0} = 2.725 K \\ \frac{Δ T}{T_{0}} & = \frac{v}{c} \cos θ \end{aligned}

$T$	the temperature of the background radiation at redshift $z$
$Δ T$	the change of the measured temperature caused by the observer's motion
$v$	the observer's speed relative to the background, much less than $c$
$θ$	the angle between the direction looked at and the direction of motion

For the Sun,

v / c = 1.23 \times 10^{- 3}

, which gives the 3.36 mK dipole. Once the dipole and the glow of our own Galaxy are removed, what remains are ripples of a few tens of microkelvin, about one part in

10^{5}

The ripples are the most important part. They show where the matter of the young Universe was slightly denser or thinner, the seeds from which galaxies later grew by gravity. Their typical size on the sky, about 1°, matches what flat space predicts, which is how we know that the total density is so close to the critical density (Section 13.3).

13.7.3The horizon problem

Light that left at recombination had travelled for only about 380 000 years. In that time the furthest any signal could have travelled is the horizon distance at recombination, which today covers only about 1.2° of the sky. Two patches of the background more than a few degrees apart had never been in contact, yet they have the same temperature to one part in $10^{5}$ . This is the horizon problem. The usual answer is inflation, a very brief burst of extremely fast expansion in the first tiny fraction of a second, which took a region that had come to one temperature and stretched it far beyond the horizon.

Example 13.10Photons against atoms

The background now holds about 411 photons in each cubic centimetre. Ordinary matter, with $Ω_{b} = 0.049$ and $H_{0} = 70$ km/s/Mpc, has a density of $4.5 \times 10^{- 28}$ kg/m³. How many photons are there for each atom?

Solution

Taking each atom as a proton of mass $1.673 \times 10^{- 27}$ kg, there are $4.5 \times 10^{- 28} / 1.673 \times 10^{- 27} = 0.27$ atoms in each cubic metre. The photons number $4.11 \times 10^{8}$ in each cubic metre.

AnswerThere are about 1.5 billion photons for every atom. Both numbers fall as

(1 + z)^{- 3}

, so the ratio has stayed the same since the first few minutes, and it is why recombination waited until the Universe had cooled to 3000 K. The value used in Section 13.8,

η = 6.1 \times 10^{- 10}

baryons for each photon, gives about 1.6 billion, since it rests on a slightly smaller density of ordinary matter than the one taken here.

Common errorsSection 13.7

The background radiation was not given out at 2.725 K. It left at about 3000 K and has been stretched by the expansion, and its blackbody shape survives because every wavelength is stretched by the same factor.
The dipole is caused by our motion, not by the Universe being hotter on one side. Removing it leaves the much smaller ripples.
Recombination did not happen at the temperature that matches 13.6 eV. The huge number of photons for each atom kept hydrogen ionised until about 3000 K.
The background comes from one moment in the history of the Universe, and in every direction we see the same surface of last scattering, about 13.6 Gpc away in comoving distance in our model.

ExercisesSection 13.73 questions

13.7.1
What was the temperature of the background radiation at $z = 3$ ?
K
13.7.2
Today there are about 411 background photons in each cubic centimetre. How many were there at recombination, at $z = 1100$ ?
× 10¹¹ per cm³
13.7.3
An observer moves at 600 km/s relative to the background. How large is the dipole, from the average temperature to the hottest point?
mK

Olympiad problemsPast IOAA problems, restated1 problem

2008The horizon at recombination100 points

Adapted from IOAA 2008 (Bandung), theory problem 3. We have restated it in our own words and kept the original data. The solution was last checked on 21 October 2026.

The background radiation has almost the same temperature in every direction. Assume that light given out at recombination, at $T_{r} \approx 3000$ K and $t_{r} \approx 300 000$ years, reaches us only now, at $T_{0} \approx 3$ K and $t_{0} \approx 1.5 \times 10^{10}$ years. Let the scale factor $S$ be 1 today, with $T \propto 1 / S$ . The Universe was ruled by radiation, with $S \propto t^{1 / 2}$ , from the end of inflation at $t = 10^{- 32}$ s until recombination, and by matter, with $S \propto t^{2 / 3}$ , after that.

(a) Estimate the horizon distance at recombination, as an angle on the sky today. Here the horizon is the largest separation of two points in the background that could have exchanged light by the time it was given out. [40 points]

(b) Two points of the background are now seen 5° apart. Could they have exchanged light? Answer yes or no, with a reason. [20 points]

I solved it without the solution

Answer(a) About 1.15°. (b) No, since 5° is more than the horizon angle. (c) About 3.4 m. The problem's model is rough. It takes the horizon as $c t$ and the distance as $c t_{0}$ , and its two ways of scaling back, by temperature (1000) and by time (1360), do not quite agree. Our model gives a horizon of about 1.2°, close to the problem's answer.