Was There Really a Big Bang? Expansion, Background Radiation, and the Ratio of the Elements

0. Key points

• What Big Bang theory actually claims is that the universe was once far hotter and denser than it is now. It does not say that a point of nothingness exploded. Confusing the two makes the theory look far more suspect than it is.
• The first pillar is expansion. The Hubble–Lemaître law, according to which more distant galaxies recede faster, follows inevitably from uniform expansion. The expansion has no center anywhere.
• The second pillar is the cosmic microwave background (CMB): an almost perfect blackbody spectrum at $2.7\ \mathrm{K}$ arriving from every direction in the sky, which no scenario other than “it used to be hot” produces naturally.
• The third pillar is the abundance ratio of the light elements. A pencil-and-paper calculation of the helium mass fraction produced during the first three minutes gives $0.25$, in agreement with the observed $0.245 \pm 0.003$.
• The three can be measured independently of one another, yet they return the same numbers for the amount of baryons and the age of the universe. That is why coincidence is not an adequate explanation.
• Not everything fits, however. The amount of lithium-7 is only about one third of the prediction, and the discrepancy remains unresolved.

1. Motivation: reconstructing last night’s fire from the ashes

In the previous article, Why is the night sky dark?, we saw that the utterly commonplace fact that the night sky is dark in fact suggests that the universe had a beginning (Olbers' paradox（Theorem 3.4）［Why Is the Night Sky Dark? Olbers' Paradox and the Finite Age of the Universe］ and the night sky of a static universe of finite age（Proposition 5.1）［Why Is the Night Sky Dark? Olbers' Paradox and the Finite Age of the Universe］). What sort of beginning was it?

The standard answer today is remarkably unassuming. The universe was once much hotter and much denser than it is now. That is the content of Big Bang theory. It does not say that the universe arose from nothing, nor that a single point exploded. On the contrary, about what happened at $t = 0$ the theory is cleanly silent.

Being unassuming, the claim is easy to test. Suppose someone insists that a campfire burned here last night. You would hold your hand over the ashes to feel the warmth, count the leftover charcoal, and inspect the scorching on the surrounding grass. If all three are consistent, and if even the amount matches — “a fire like that should leave about this much charcoal” — you would believe it. The three pillars of the Big Bang have exactly this structure.

1.1. Who proposed it, and when

• In 1922 Alexander Friedmann found, from the equations of general relativity, solutions describing an expanding or contracting universe.
• In 1927 Georges Lemaître (a physicist who was also a Catholic priest) derived the same solutions independently, predicted that the recession velocity of galaxies should be proportional to their distance, and, running time backwards, proposed a dense initial state he called the “primeval atom”.
• In 1929 Edwin Hubble demonstrated that proportionality observationally. In 2018 the International Astronomical Union recommended that the law be called the Hubble–Lemaître law (the Hubble–Lemaître law（Proposition 2.3）［The Edge and the Age of the Universe］).
• In 1948 George Gamow and his collaborators reasoned that if the early universe had been hot, the elements could have been made there, and as a by-product predicted that the embers should fill the whole sky as radio waves at around $5\ \mathrm{K}$.
• Fred Hoyle and others, meanwhile, advocated the steady-state theory, in which the universe expands but matter is continuously created so that it always looks the same. Hoyle called the rival hypothesis the “Big Bang” on BBC radio in 1949, and that is where the name comes from. The remark is often described as a jibe, but Hoyle himself denied that in later years.
• In 1965 Arno Penzias and Robert Wilson accidentally picked up those embers, and the contest was settled.

Note

The phrase “Big Bang theory” refers, in the narrow sense, to the claim that the early universe was hot and dense, and in the broad sense to the entire modern standard cosmological model including inflation and dark matter. What we test in this article is the narrow sense. For the broad sense see Dark matter and dark energy.

2. Preliminaries: three tools and one timeline

Before the main argument, let us define the three tools we shall use repeatedly.

Definition 2.1（Redshift）

Let $\lambda_{\mathrm{emit}}$ be the wavelength of light emitted by an object and $\lambda_{\mathrm{obs}}$ the wavelength received on Earth. The quantity

$$z = \frac{\lambda_{\mathrm{obs}} - \lambda_{\mathrm{emit}}}{\lambda_{\mathrm{emit}}}$$

is called the redshift of the object. A value $z > 0$ means that the wavelength has been stretched, shifting it toward the red.

We know $\lambda_{\mathrm{emit}}$ because the wavelengths of the emission and absorption lines produced by atoms have been measured precisely in the laboratory. The $\mathrm{H}\alpha$ line of the Balmer series of hydrogen, for instance, sits at $656.3\ \mathrm{nm}$ at rest. If in the spectrum of a distant galaxy the whole pattern of lines is displaced by the same ratio, that displacement is a redshift.

Definition 2.2（Scale factor）

Suppose that the distance $D(t)$ between two galaxies not gravitationally bound to each other can be written, for every pair of galaxies, as

$$D(t) = a(t)\, D_0$$

Then $a(t)$ is called the scale factor of the universe. Here $D_0$ is a constant determined by the pair, and we normalize the present value by $a(t_0) = 1$.

The proviso “not gravitationally bound” is not decoration. The interior of a galaxy cluster, the solar system, an atom, and your own height do not expand, because gravity and electromagnetism overcome the expansion. What expands is only the distance between regions far enough apart that no binding is effective.

Definition 2.3（Blackbody radiation）

The electromagnetic radiation emitted by a body of temperature $T$ in thermal equilibrium is called blackbody radiation. Its photon number density per unit wavelength is given by Planck’s formula

$$n_\lambda(\lambda)\, d\lambda = \frac{8\pi}{\lambda^{4}}\, \frac{d\lambda}{e^{hc/(\lambda k T)} - 1}$$

whose shape is fixed by the single parameter $T$. Here $h$ is Planck’s constant, $k$ is Boltzmann’s constant and $c$ is the speed of light.

The crucial property of blackbody radiation is that its shape is determined by the temperature alone. Conversely, if an observed spectrum has the blackbody shape, we know that the place that produced it was in thermal equilibrium. This is a strong constraint, and it will do real work later.

Let us also fix a convention for temperature. In particle physics it is customary to express temperature as an energy $kT$; the value $kT = 1\ \mathrm{MeV}$ corresponds to $T = 1.16 \times 10^{10}\ \mathrm{K}$, that is, about ten billion degrees. Wherever we use $\mathrm{MeV}$ we shall append the conversion to kelvin.

Finally, here is the timeline we shall refer to repeatedly. Every number in it is computed somewhere in this article.

flowchart LR
A["about 1 s<br/>about 9 billion K<br/>neutron-to-proton ratio freezes out"] --> B["about 3 min<br/>about 0.8 billion K<br/>helium forms"] --> C["about 380,000 yr<br/>about 3000 K<br/>the universe becomes transparent"] --> D["a few hundred million yr<br/>first stars and galaxies"] --> E["about 13.8 billion yr<br/>2.7 K<br/>now"]

The life history of the universe as drawn by Big Bang theory. Each of the three pillars looks at traces from a different epoch.

3. The first pillar: the universe is expanding

3.1. What was observed

What Hubble established in 1929 amounts to two facts. First, the spectra of almost all galaxies are redshifted. Second, the more distant the galaxy, the larger the redshift. For $z \ll 1$ the redshift may be read with the same formula as the Doppler effect, so putting the recession velocity at $v \approx cz$ we obtain the proportionality

$$v = H_0 d$$

The constant $H_0$ is called the Hubble constant, and its currently measured value is roughly $70\ \mathrm{km/s/Mpc}$. One $\mathrm{Mpc}$ (megaparsec) is about $3.26$ million light years, and the statement reads: “for every additional $\mathrm{Mpc}$ of distance the recession velocity increases by $70\ \mathrm{km/s}$”.

This is where many people stumble. If every galaxy is receding from us, is the Earth the center of the universe? The answer is no, and remarkably it can be shown not by observation but by arithmetic.

3.2. Uniform expansion necessarily gives Hubble’s law

Proposition 3.1（Inevitability of Hubble's law）

Assume the following three things.

1. (Homogeneity) An observer riding on any galaxy measures the same function $v(d)$ for the speed at which a galaxy at distance $d$ recedes.
2. (Composition of velocities) The speeds are small compared with the speed of light, so that velocities simply add.
3. (Continuity) $v(d)$ is a continuous function of $d$.

Then there is a constant $H$ such that $v(d) = H d$ for every $d > 0$.

Proof（Proposition 3.1）

Consider three galaxies $\mathrm{O}$, $\mathrm{A}$, $\mathrm{B}$ lying on a straight line, with the distance from $\mathrm{O}$ to $\mathrm{A}$ and the distance from $\mathrm{A}$ to $\mathrm{B}$ both equal to $d$.

As seen from $\mathrm{O}$, the galaxy $\mathrm{A}$ recedes at speed $v(d)$. By assumption 1, as seen from $\mathrm{A}$ the galaxy $\mathrm{B}$ also recedes at speed $v(d)$ (since $\mathrm{AB}$ is likewise at distance $d$). By assumption 2 the speed of $\mathrm{B}$ as seen from $\mathrm{O}$ is the sum of the two, and since the distance from $\mathrm{O}$ to $\mathrm{B}$ is $2d$ we get

$$v(2d) = v(d) + v(d) = 2\,v(d)$$

Repeating the same argument for $n$ equally spaced galaxies gives $v(nd) = n\,v(d)$ for every natural number $n$.

Next consider the length $d/n$ obtained by dividing $d$ into $n$ equal parts. Applying the identity just proved to $d/n$ gives $v(d) = v\!\left(n \cdot \frac{d}{n}\right) = n\, v\!\left(\frac{d}{n}\right)$, so $v(d/n) = v(d)/n$. Hence for every positive rational $q = m/n$

$$v(qd) = v\!\left(m \cdot \frac{d}{n}\right) = m\, v\!\left(\frac{d}{n}\right) = \frac{m}{n}\, v(d) = q\, v(d)$$

The positive rationals are dense in the positive reals, so by assumption 3 (continuity) we have $v(td) = t\, v(d)$ for every real $t > 0$.

Now fix a reference distance $d_0$ and set $H = v(d_0)/d_0$. Any $d > 0$ can be written as $d = (d/d_0)\, d_0$, whence

$$v(d) = \frac{d}{d_0}\, v(d_0) = H d$$

which is the assertion.

∎

Corollary 3.2（The expansion has no center）

Under the assumptions of Proposition 3.1, an observer riding on any galaxy observes that they themselves are at rest and that every other galaxy is receding from them. Consequently no observation can identify a center of the expansion.

Proof（Corollary 3.2）

Write the velocity field as $\boldsymbol{v}(\boldsymbol{r}) = H \boldsymbol{r}$ (by Proposition 3.1 the speed is proportional to the distance, and the direction is radial because the motion is an expansion). Let $\boldsymbol{r}_A$ be the position of a galaxy $\mathrm{A}$. By the composition of velocities of assumption 2, the velocity measured by an observer riding on $\mathrm{A}$ is obtained by subtracting $\boldsymbol{v}(\boldsymbol{r}_A)$ from the whole field. Writing positions as seen from $\mathrm{A}$ as $\boldsymbol{r}' = \boldsymbol{r} - \boldsymbol{r}_A$, we get

$$\boldsymbol{v}'(\boldsymbol{r}') = H\boldsymbol{r} - H\boldsymbol{r}_A = H(\boldsymbol{r} - \boldsymbol{r}_A) = H \boldsymbol{r}'$$

which is a law of exactly the same form as the original one. Since $\mathrm{A}$ was arbitrary, the same conclusion holds for every galaxy. In other words, “appearing to be at the center” happens equally to every observer.

∎

Let us draw the situation.

Uniform expansion has no center. The middle and bottom rows show exactly the same configuration; they differ only in which galaxy was held fixed on the page. In either picture the more distant galaxies appear to have moved farther.

Tip

Picture a loaf of raisin bread rising. Whichever raisin you put your face next to, the others move away, and the farther ones move away faster. The dough has no “center”. The only place the analogy breaks down is that a loaf of bread has a surface and an outside.

Example 3.3（The distance to the Coma cluster）

The recession velocity of the Coma cluster is measured to be about $6900\ \mathrm{km/s}$. Using $H_0 = 70\ \mathrm{km/s/Mpc}$,

$$d = \frac{v}{H_0} = \frac{6900\ \mathrm{km/s}}{70\ \mathrm{km/s/Mpc}} = 98.6\ \mathrm{Mpc}$$

Converting to light years, $98.6 \times 3.26 \times 10^{6} = 3.2 \times 10^{8}$ light years, about $320$ million. What we see now is light that left before the dinosaurs had appeared on Earth.

This estimate borrows the value of $H_0$. That constant is itself fixed by measuring nearby objects by other means, a procedure treated in How do we know the distances to the stars? (the chain from parallax to Cepheids to supernovae is summarized in how the rungs are joined and how errors propagate（Proposition 6.1）［How We Know the Distance to a Star］).

3.3. Running time backwards

If the universe is expanding, then going back into the past the density rises. How far back can we go? The crudest estimate assumes that the speed of expansion has always been constant.

Proposition 3.4（Hubble time）

If every galaxy has receded throughout the past at the constant speed $v = H_0 d$, then every galaxy was at the same place a time $1/H_0$ ago. For $H_0 = 70\ \mathrm{km/s/Mpc}$ one has $1/H_0 \approx 1.40 \times 10^{10}$ years.

Proof（Proposition 3.4）

If a galaxy now at distance $d$ has receded at the constant speed $v = H_0 d$, then the distance was $0$ at a time $t = d/v = d/(H_0 d) = 1/H_0$ before the present. This value does not contain $d$. In other words the time is the same for every galaxy, and at that time everything coincided.

Now the numbers. Since $1\ \mathrm{Mpc} = 3.086 \times 10^{19}\ \mathrm{km}$,

$$H_0 = \frac{70\ \mathrm{km/s}}{3.086 \times 10^{19}\ \mathrm{km}} = 2.27 \times 10^{-18}\ \mathrm{s^{-1}}$$$$\frac{1}{H_0} = 4.41 \times 10^{17}\ \mathrm{s}$$

One year is $3.156 \times 10^{7}\ \mathrm{s}$, so

$$\frac{1}{H_0} = \frac{4.41 \times 10^{17}}{3.156 \times 10^{7}}\ \text{years} = 1.40 \times 10^{10}\ \text{years}$$

that is, about $14$ billion years.

∎

The actual expansion is not constant. The gravity of matter decelerates it and dark energy accelerates it. The current estimate including both is $13.8$ billion years, only $2\ \%$ away from the crude estimate above, because the effects of the decelerating era and the accelerating era nearly cancel. The calculation obtaining this $13.8$ billion years from an integral is given in the age of a flat universe of matter and dark energy（Theorem 3.6）［The Edge and the Age of the Universe］. For details see The edge and the age of the universe.

Remark 3.5（The Hubble constant is still disputed）

The value of $H_0$ disagrees depending on how it is measured. Determined indirectly from the fluctuations of the CMB it is $67.4 \pm 0.5\ \mathrm{km/s/Mpc}$; determined directly from nearby Cepheid variables and supernovae it is $73 \pm 1\ \mathrm{km/s/Mpc}$, a difference hard to attribute to chance given the error bars. This is called the Hubble tension. Where the discrepancy comes from on the side of the direct measurement is treated in the great controversy caused by 0.17 magnitudes（Example 6.2）［How We Know the Distance to a Star］. Note, however, that the dispute is about how fast the expansion is, not about whether there is one; the first pillar itself is not shaken.

Remark 3.6（Redshift is the stretching of space）

Strictly speaking, the redshift of a distant galaxy is not a Doppler effect. Using general relativity one derives the relation

$$1 + z = \frac{1}{a(t_{\mathrm{emit}})}$$

according to which the wavelength is stretched by exactly the amount by which space stretched while the light was in flight (we normalize the present by $a(t_0) = 1$). Think of a wave drawn on a rubber band being stretched along with the band. For $z \ll 1$ this agrees with the Doppler formula $v \approx cz$, so nearby it may be read as a velocity; but a galaxy at $z = 3$ is not moving at three times the speed of light. The derivation is in the chapter on cosmic expansion in Ryden’s textbook.

4. The second pillar: the cosmic microwave background

4.1. Predicted first, found afterwards

In 1948 Gamow and his collaborators pointed out that if the early universe was hot, the thermal radiation of that era should still be present in cooled form, and Alpher and Herman produced the figure of about $5\ \mathrm{K}$. Almost nobody set out seriously to look for it. Radio waves at $5\ \mathrm{K}$, so the conventional wisdom went, would be buried in noise and invisible.

In 1964 Penzias and Wilson at Bell Labs were measuring radio waves at a wavelength of $7.35\ \mathrm{cm}$ with a horn antenna built for satellite communications. There was an excess noise that refused to go away, corresponding to an antenna temperature of $3.5 \pm 1.0\ \mathrm{K}$. Suspecting the apparatus, they evicted a pair of pigeons nesting inside the antenna and cleaned out the “white dielectric material” the birds had left behind. The noise remained. It had the same strength in every direction of the sky and through every season.

At about the same time Dicke and his group at Princeton were building an instrument to search for precisely Gamow’s embers. A single telephone call connected the two efforts, and in 1965 two papers appeared side by side in the same issue. Penzias and Wilson reported only their observations, plainly; Dicke and his collaborators interpreted them as the embers of the universe. The discovery earned the 1978 Nobel Prize in Physics.

4.2. What being a blackbody means

What makes this radiation decisive evidence is the shape of its spectrum. In 1990 the FIRAS instrument on the COBE satellite showed that the CMB spectrum agrees with a blackbody at $2.7255\ \mathrm{K}$ to better than $10^{-4}$ of the peak intensity. It is the most perfect blackbody spectrum humanity has ever measured.

Example 4.1（Computing the peak wavelength）

The peak wavelength of blackbody radiation is fixed by Wien’s displacement law, $\lambda_{\max} T = 2.898 \times 10^{-3}\ \mathrm{m \cdot K}$. Substituting $T = 2.7255\ \mathrm{K}$,

$$\lambda_{\max} = \frac{2.898 \times 10^{-3}\ \mathrm{m \cdot K}}{2.7255\ \mathrm{K}} = 1.06 \times 10^{-3}\ \mathrm{m}$$

about $1.1\ \mathrm{mm}$, in the microwave to millimetre band. That is where the name “cosmic microwave background” comes from. Incidentally, radiation at $2.7\ \mathrm{K}$ amounts to about $411$ photons per $\mathrm{cm^3}$. Even in the volume of space at the tip of your thumb there are some $400$ photons that have been flying since the earliest days of the universe.

A blackbody spectrum is the fingerprint of thermal equilibrium. When dust absorbs starlight and re-emits it, dust at many temperatures is mixed together, so one never gets so clean a single-temperature blackbody. Nobody has yet managed to reach this precision with any scenario other than the universe as a whole having once been thoroughly in thermal equilibrium.

4.3. Cooling preserves the blackbody form

To say that radiation which was once hot is now at $2.7\ \mathrm{K}$, we must show that blackbody radiation diluted by expansion remains blackbody radiation. This can be checked by calculation.

Proposition 4.2（Expansion maps blackbody radiation to blackbody radiation）

Suppose that photons are neither created nor destroyed and that the scale factor of the universe grows from $a(t_1)$ to $a(t_2)$ by a factor $s = a(t_2)/a(t_1) > 1$. Then what was blackbody radiation of temperature $T$ at time $t_1$ is blackbody radiation of temperature $T/s$ at time $t_2$. That is, the temperature is inversely proportional to the scale factor, and combining this with Remark 3.6 gives

$$T(z) = T_0\,(1+z)$$

where $T_0$ is the present temperature.

Proof（Proposition 4.2）

Only two facts are needed.

First, by Remark 3.6, the wavelength of every photon is stretched along with space, $\lambda \to s\lambda$.

Second, photons are neither created nor destroyed, so their number is conserved, while the volume grows by a factor $s^3$. Hence the photon number density is multiplied by $s^{-3}$.

Now, photons whose wavelength lay in the range $[\lambda,\ \lambda + d\lambda]$ at time $t_1$ lie in the range $[s\lambda,\ s\lambda + s\,d\lambda]$ at time $t_2$. By the second fact their number density is

$$s^{-3}\, n_\lambda(\lambda)\, d\lambda = s^{-3}\, \frac{8\pi}{\lambda^{4}}\, \frac{d\lambda}{e^{hc/(\lambda k T)} - 1}$$

Let us check whether this coincides with a Planck distribution of temperature $T/s$. Substituting into the formula of Definition 2.3 the temperature $T/s$, the wavelength $\lambda' = s\lambda$ and the width $d\lambda' = s\, d\lambda$, we get

$$\frac{8\pi}{(s\lambda)^{4}}\, \frac{s\, d\lambda}{e^{hc/\left(s\lambda \cdot k T/s\right)} - 1} = \frac{8\pi}{s^{4}\lambda^{4}}\, \frac{s\, d\lambda}{e^{hc/(\lambda k T)} - 1} = s^{-3}\, \frac{8\pi}{\lambda^{4}}\, \frac{d\lambda}{e^{hc/(\lambda k T)} - 1}$$

The essential point is the content of the exponential: because the wavelength is multiplied by $s$ and the temperature by $1/s$, the factors of $s$ cancel and $hc/(\lambda k T)$ survives unchanged. The two expressions agree exactly.

Hence the photon distribution at time $t_2$ is precisely a Planck distribution of temperature $T/s$. This proves $T \propto 1/a$, and substituting $a = 1/(1+z)$ gives $T(z) = T_0 (1+z)$.

∎

This proposition turns “a hot past” into a quantitative claim: measure a temperature and you learn the scale factor of that epoch.

Example 4.3（The moment the universe became transparent）

Above a few thousand kelvin a hydrogen atom cannot hold on to its electron and the gas becomes a plasma. Free electrons scatter light efficiently, so the universe is opaque, like a thick fog. When the temperature falls enough for electrons to be captured by protons, the scattering abruptly stops mattering and light can travel in straight lines. This is called the recombination, or the moment the universe becomes transparent, and it happens at $T \approx 3000\ \mathrm{K}$.

Using Proposition 4.2, the redshift of that moment is

$$1 + z = \frac{T}{T_0} = \frac{3000\ \mathrm{K}}{2.7255\ \mathrm{K}} = 1101$$

that is, $z \approx 1100$. Lengths in the universe then were $1/1100$ of their present values and volumes $1/1100^3 \approx 1/(1.3 \times 10^{9})$, less than a billionth. In terms of time this is about $380{,}000$ years after the beginning.

The CMB we see is light arriving from this “wall at the moment the fog lifted”. On a sphere centred on the Earth, that wall lies in every direction. The Earth is not special: by the same reasoning as Corollary 3.2, the inhabitants of any galaxy see the same wall centred on themselves.

4.4. The predictions keep coming

If the CMB really is the residue of the whole universe cooling down, then $T(z) = T_0(1+z)$ from Proposition 4.2 should hold in the past as well. This can actually be tested.

Example 4.4（Measuring the temperature of the early universe directly）

When light from a distant quasar passes through an intervening gas cloud, the molecules and atoms in the gas absorb particular wavelengths. Molecules have closely spaced energy levels, and how the population is distributed among them is determined by the temperature of the surrounding radiation. In other words, the ratio of the strengths of the absorption lines is a thermometer for the CMB at the epoch of that cloud.

Measuring this in a cloud at redshift $z \approx 2.7$ gives $T \approx 10\ \mathrm{K}$. The theoretical prediction is

$$T = 2.7255 \times (1 + 2.7) = 10.1\ \mathrm{K}$$

in agreement with observation. Had the CMB been “just something that has always sat there at $2.7\ \mathrm{K}$”, the temperature would have been independent of $z$. The prediction succeeded and the rival hypothesis failed.

Another prediction concerns fluctuations in the temperature. If matter had been perfectly uniform at recombination, no galaxies would have formed in the $13.8$ billion years since. Seeds are needed for gravity to gather matter around. Big Bang theory predicted that the CMB should show slight temperature differences, and in 1992 COBE found them. Their size is $\Delta T / T \sim 10^{-5}$, one part in $100{,}000$. This discovery earned the 2006 Nobel Prize in Physics.

Caution

The CMB also shows a comparatively large temperature difference of $3.4\ \mathrm{mK}$: a dipole component, with one side of the sky hot and the opposite side cold. This is not a cosmological fluctuation but a Doppler effect due to the solar system moving at about $370\ \mathrm{km/s}$ relative to the CMB. What remains after subtracting it is the genuine fluctuation at the $10^{-5}$ level.

Tip

Before the switch to digital broadcasting, the snow on an empty television channel contained a share of the CMB, estimated at around $1\ \%$ of the total. For a long time we dismissed the light of the beginning of the universe as bad reception.

5. The third pillar: the ratio of hydrogen to helium

5.1. Why is the universe so full of helium?

Counting ordinary matter in the universe by mass, roughly $75\ \%$ is hydrogen, $25\ \%$ is helium, and less than a few percent is everything else. Carbon, oxygen and iron are, on a cosmic scale, mere rounding.

The problem is the $25\ \%$ of helium. One is tempted to say that hydrogen fused into helium inside stars, but the numbers do not work. There are two reasons.

First, the quantity is insufficient. Even summing over an entire galaxy the hydrogen burned by stars during their lifetimes, the helium produced is of order a few percent, nowhere near $25\ \%$.

Second — and this is decisive — when stars make helium they also make heavier elements such as carbon and oxygen. If stars had made it, objects poorer in heavy elements should also be poorer in helium. Indeed, galaxies with very little oxygen (that is, almost uncontaminated by stellar processing) do have a smaller helium mass fraction, but it bottoms out at a definite value. Extrapolating to zero oxygen gives $0.245 \pm 0.003$. Before a single star existed, the universe already contained this much helium.

Definition 5.1（Helium mass fraction）

The fraction of the total mass of baryons (ordinary matter made of protons and neutrons) accounted for by $^4\mathrm{He}$ is written $Y$ and called the helium mass fraction. The value of cosmological origin, obtained after subtracting production by stars, is called the primordial helium mass fraction $Y_p$.

5.2. What happened during the first three minutes

Big Bang theory answers as follows. In the era when the universe was hot, protons and neutrons were in thermal equilibrium, converting into one another. As the temperature fell the reactions could no longer keep up and the ratio froze. Afterwards essentially all the surviving neutrons were locked into helium-4. Thus $Y$ is determined by the number ratio of neutrons to protons alone.

Lemma 5.2（The neutron-to-proton number ratio in thermal equilibrium）

Suppose that neutrons and protons are in thermal equilibrium at temperature $T$ through the weak interaction (reactions such as $n + \nu_e \leftrightarrow p + e^-$), that $kT \ll m_p c^2$, and that the chemical potentials of the electrons and neutrinos are negligible. Then the ratio of their number densities is

$$\frac{n_n}{n_p} = \left(\frac{m_n}{m_p}\right)^{3/2} \exp\!\left(-\frac{(m_n - m_p)c^2}{kT}\right) \approx \exp\!\left(-\frac{1.29\ \mathrm{MeV}}{kT}\right)$$

Proof（Lemma 5.2）

For a non-relativistic particle with $kT \ll mc^2$ in thermal equilibrium at temperature $T$, the Maxwell–Boltzmann distribution gives the number density

$$n = g\left(\frac{2\pi m k T}{h^{2}}\right)^{3/2} e^{-mc^{2}/kT}$$

where $g$ is the number of internal degrees of freedom. The temperatures at issue here are $kT \sim 1\ \mathrm{MeV}$, while the rest energy of a nucleon is $m c^2 \approx 939\ \mathrm{MeV}$, so $kT/mc^2 \sim 10^{-3}$ and the non-relativistic condition is amply satisfied.

Neutrons and protons both have spin $1/2$ and hence $g = 2$, so on taking the ratio the factors $g$ and $(2\pi kT/h^2)^{3/2}$ cancel and there remains

$$\frac{n_n}{n_p} = \left(\frac{m_n}{m_p}\right)^{3/2} \exp\!\left(-\frac{(m_n - m_p)c^{2}}{kT}\right)$$

Now the numbers. Since $m_n c^2 = 939.565\ \mathrm{MeV}$ and $m_p c^2 = 938.272\ \mathrm{MeV}$ we have $m_n/m_p = 1.0014$ and $(1.0014)^{3/2} = 1.002$. This is a correction of less than $1\ \%$, which we drop. The mass difference is

$$(m_n - m_p)c^{2} = 939.565 - 938.272 = 1.293 \approx 1.29\ \mathrm{MeV}$$

which gives the stated formula.

The argument of $\exp$ is negative, so $n_n < n_p$: neutrons, being slightly heavier, are slightly harder to make. That is all there is to it, and yet this difference of $1.29\ \mathrm{MeV}$ determines the elemental composition of the universe.

∎

Example 5.3（Freeze-out at one second and the losses over three minutes）

The formula of Lemma 5.2 is valid only while equilibrium is maintained. The rate of the weak reactions that maintain it falls steeply as the fifth power of the temperature, whereas the expansion rate of the universe scales only as the square, so as the temperature drops the reactions must at some point fail to keep up. The dividing line is at $kT \approx 0.8\ \mathrm{MeV}$ (about $9$ billion K, roughly one second after the beginning), after which the ratio freezes out. Its value there is

$$\frac{n_n}{n_p} = \exp\!\left(-\frac{1.29}{0.8}\right) = e^{-1.61} = 0.20 = \frac{1}{5}$$

Helium synthesis, however, does not begin at once. To make helium one must first make deuterium $\mathrm{D} = {}^2\mathrm{H}$, but the binding energy of deuterium is only $2.22\ \mathrm{MeV}$, and the universe contains about $1.6$ billion photons per nucleon — an overwhelming excess. High-energy photons in the tail of the distribution destroy deuterium as fast as it forms, so deuterium cannot survive until the temperature has fallen to about

$$kT \approx \frac{2.22\ \mathrm{MeV}}{\ln(16 \times 10^{8})} \approx \frac{2.22}{21} \approx 0.1\ \mathrm{MeV}$$

A careful calculation gives $kT \approx 0.07\ \mathrm{MeV}$ (about $0.8$ billion K), which is about three minutes after the beginning. This is called the deuterium bottleneck.

During this waiting period, free neutrons beta-decay into protons with a mean lifetime $\tau = 880\ \mathrm{s}$. Starting from $0.2$ neutrons per proton at freeze-out, let us compute the situation at $t = 250\ \mathrm{s}$.

$$n_n = 0.2 \times e^{-250/880} = 0.2 \times 0.753 = 0.151$$

The decayed neutrons turn into protons, so

$$n_p = 1 + (0.2 - 0.151) = 1.049$$

and therefore

$$\frac{n_n}{n_p} = \frac{0.151}{1.049} = 0.143 = \frac{1}{7}$$

A ratio that was $1/5$ has been whittled down to $1/7$ during the wait.

Proposition 5.4（The helium mass fraction comes out to 1/4）

Suppose that at the onset of light-element synthesis the number ratio of neutrons to protons is $n_n : n_p = 1 : 7$, that all the neutrons are locked into $^4\mathrm{He}$, and that the amount of nuclei heavier than $^4\mathrm{He}$ produced is negligible. Then the mass fraction of the $^4\mathrm{He}$ produced is

$$Y = \frac{2\,(n_n/n_p)}{1 + n_n/n_p} = \frac{1}{4}$$

Proof（Proposition 5.4）

Let us first look at concrete numbers. Take $16$ nucleons; at a ratio of $1:7$ these are $2$ neutrons and $14$ protons. A $^4\mathrm{He}$ nucleus consists of $2$ protons and $2$ neutrons. By the assumption that all neutrons are incorporated, the $2$ neutrons pair up with $2$ protons to make one $^4\mathrm{He}$. The remaining $12$ protons are hydrogen nuclei as they stand.

Counting masses in approximate units of one nucleon mass, the total is $16$ and the $^4\mathrm{He}$ is $4$. Hence

$$Y = \frac{4}{16} = 0.25$$

The same computation works for a general ratio. Given $N_n$ neutrons and $N_p$ protons (with $N_n \le N_p$), the number of $^4\mathrm{He}$ nuclei produced is $N_n/2$, whose mass in units of the nucleon mass is $4 \times (N_n/2) = 2N_n$. The total mass is $N_n + N_p$, so

$$Y = \frac{2N_n}{N_n + N_p} = \frac{2\,(N_n/N_p)}{1 + N_n/N_p}$$

Substituting $N_n/N_p = 1/7$ from Example 5.3 gives

$$Y = \frac{2 \times \frac{1}{7}}{1 + \frac{1}{7}} = \frac{\frac{2}{7}}{\frac{8}{7}} = \frac{2}{8} = \frac{1}{4}$$

as claimed.

∎

Against the $0.25$ obtained with nothing but pencil and paper, the observed value is $Y_p = 0.245 \pm 0.003$. For a rough estimate the agreement is almost too good. A careful numerical calculation gives $0.247$.

5.3. Deuterium as a baryometer

There is stronger evidence still. As we saw in Example 5.3, the time at which synthesis begins is set by the ratio $\eta$ of nucleons to photons (the baryon-to-photon ratio). If there are many baryons the nuclear reactions run to completion and no deuterium is left; if there are few, the reactions end half-finished and deuterium is left over. In other words, the amount of surviving deuterium is a ruler for the amount of baryons in the universe.

Moreover, deuterium cannot be made inside stars. Stars only burn deuterium and destroy it. If we find deuterium anywhere in the universe, it is a survivor of the Big Bang.

Example 5.5（Two utterly different methods giving the same answer）

Measuring the deuterium-to-hydrogen ratio in primordial gas clouds that absorb the light of distant quasars gives $\mathrm{D}/\mathrm{H} = 2.5 \times 10^{-5}$. Feeding this into the calculation of light-element synthesis yields, for the baryon density parameter,

$$\Omega_{\mathrm{b}} h^{2} = 0.022$$

On the other hand, analysing the angular pattern of the temperature fluctuations of the CMB gives, entirely independently,

$$\Omega_{\mathrm{b}} h^{2} = 0.0224 \pm 0.0002$$

One is nuclear reactions three minutes after the beginning of the universe; the other is acoustic oscillations $380{,}000$ years later. The physics is different and so are the telescopes. And they agree to three decimal places.

Let us convert this value into something one can picture. Here $\Omega_{\mathrm{b}}$ is the density parameter, the density divided by the critical density (critical density and density parameters（Definition 5.5）［Dark Matter and Dark Energy］), and $h$ is $H_0$ measured in units of $100\ \mathrm{km/s/Mpc}$. Taking $h = 0.67$ gives $\Omega_{\mathrm{b}} = 0.022/0.67^2 = 0.049$: ordinary matter is only about $5\ \%$ of the energy of the universe (the identity of the remaining $95\ \%$ is the subject of Dark matter and dark energy). In terms of number density, the cosmic average is $0.25$ hydrogen atoms per $\mathrm{m^3}$.

The summary is as follows. The table fixes the baryon density at the CMB value, computes light-element synthesis, and sets the results beside the observations.

Element	Theoretical prediction	Observed value
mass fraction $Y_p$ of $^4\mathrm{He}$	$0.247$	$0.245 \pm 0.003$
$\mathrm{D}/\mathrm{H}$	$2.5 \times 10^{-5}$	$2.5 \times 10^{-5}$
$^3\mathrm{He}/\mathrm{H}$	$\sim 10^{-5}$	comparable
$^7\mathrm{Li}/\mathrm{H}$	$5 \times 10^{-10}$	$1.6 \times 10^{-10}$

Caution

The last line does not agree. The amount of lithium-7 measured at the surfaces of old stars is only about one third of the prediction. This is called the lithium problem, and it has resisted solution for nearly forty years. Competing explanations include lithium sinking and being destroyed inside stars, errors in the measured nuclear reaction rates, and the involvement of unknown particles. Two of the three pillars perfect and one with a hole in it is the honest state of affairs in the 2020s.

6. What the three pillars support

6.1. The theory is shaped so that it can be killed

The strength of a theory is measured not only by the number of successful predictions but by whether it made predictions that would kill it if they failed. On that count Big Bang theory is admirably exposed. Any one of the following would bring it down.

If this were observed	Consequence
the CMB spectrum clearly departing from a blackbody	fatal; there would have been no epoch of thermal equilibrium
an object whose helium mass fraction, after subtracting stellar production, falls well below $20\ \%$	fatal; the framework of Proposition 5.4 would collapse
a star reliably older than the universe	fatal
the CMB temperature at redshift $z$ departing from $2.7255(1+z)\ \mathrm{K}$	fatal; Proposition 4.2 would be refuted
no change in the populations of galaxies and active objects between the distant and the nearby universe	indistinguishable from the steady-state theory; one strong piece of evidence lost

The third of these actually became a crisis. In the 1990s the globular clusters appeared for a while to be older than $1/H_0$. The contradiction was resolved by improvements in distance measurement and by the discovery of dark energy. That the theory came close to being refuted and survived is itself evidence of its health, I think.

The last line is what defeated the steady-state theory. If one claims that the universe looks the same at all times and places, then the distant (that is, the early) universe must look like the nearby one. Yet counting radio sources and quasars shows a clearly higher density at greater distances. The universe has a history.

6.2. Three common misconceptions

“The Big Bang was an explosion at a particular point in space.” No. As Corollary 3.2 shows, uniform expansion has no center. The hot, dense state occurred everywhere in space at the same time. An explosion did not spread within space; space itself stretched.

“What is outside the universe?” The theory says nothing about an outside. What we can observe is only the region light has reached — the observable universe, that is, the interior of the particle horizon（Definition 4.1）［The Edge and the Age of the Universe］ — and what lies beyond it, and whether the whole is finite or infinite, is unsettled. For details see The edge and the age of the universe.

“Big Bang theory explained the beginning of the universe.” It did not. What the three pillars speak about is the interval from one second to $380{,}000$ years after the beginning, and the present. Earlier than that, and especially in the limit $t = 0$, both general relativity and quantum theory are needed, and the theory that combines them is one nobody has. Some feel this candour to be a weakness; I think the opposite. It draws a clear line between what we know and what we do not.

6.3. Where the three pillars meet

The first pillar (expansion) says that the universe was smaller in the past. The second pillar (the CMB) says that there was a hot state of thermal equilibrium in the past, and gets the relation between temperature and redshift right as well. The third pillar (the elemental ratios) reproduces the amounts of the elements from physics at the specific times of one second and three minutes.

And the three constrain one another. The baryon density from the second pillar agrees with the baryon density from the third (Example 5.5). The age of the universe from the first pillar is consistent with the age of the oldest stars. This phenomenon of arriving at the same number by separate routes is what separates a genuine theory from mere bookkeeping.

That is not to say Big Bang theory is finished. The lithium problem remains, the Hubble tension has deepened, and $95\ \%$ of the universe is of unknown nature. The answer to “was there really a Big Bang?” is: there was certainly an era of high temperature and density, but what came before it, and most of what it contained, we still do not know.

7. Exercises

Exercise 7.1Easy

In the spectrum of a galaxy, the $\mathrm{H}\alpha$ line with rest wavelength $656.3\ \mathrm{nm}$ is observed at $721.9\ \mathrm{nm}$. Taking $H_0 = 70\ \mathrm{km/s/Mpc}$ and $c = 3.0 \times 10^{5}\ \mathrm{km/s}$, find the redshift of the galaxy and its approximate distance from the Earth in light years. Use $1\ \mathrm{Mpc} = 3.26 \times 10^{6}$ light years.

Solution

By Definition 2.1,

$$z = \frac{721.9 - 656.3}{656.3} = \frac{65.6}{656.3} = 0.100$$

Since $z = 0.1$ is small compared with $1$, it may be read as a velocity, as noted in Remark 3.6.

$$v \approx cz = 3.0 \times 10^{5} \times 0.100 = 3.0 \times 10^{4}\ \mathrm{km/s}$$

From the law $v = H_0 d$ of Proposition 3.1,

$$d = \frac{3.0 \times 10^{4}\ \mathrm{km/s}}{70\ \mathrm{km/s/Mpc}} = 4.3 \times 10^{2}\ \mathrm{Mpc}$$

Converting to light years, $4.3 \times 10^{2} \times 3.26 \times 10^{6} = 1.4 \times 10^{9}$ light years, about $1.4$ billion.

(Caution: at $z = 0.1$ the effect of the expansion history begins to appear at the level of a few percent, so this is only an estimate.)

Exercise 7.2Standard

The present temperature of the CMB is $T_0 = 2.7255\ \mathrm{K}$.

(1) Find the redshift $z$ of the era when the temperature of the universe equalled the boiling point of water, $373\ \mathrm{K}$.

(2) By what factor was the volume of the universe smaller then than it is now?

(3) Find the peak wavelength of the CMB at that era. Take the constant in Wien’s displacement law to be $2.898 \times 10^{-3}\ \mathrm{m \cdot K}$.

Solution

(1) By Proposition 4.2, $T = T_0(1+z)$, so

$$1 + z = \frac{373}{2.7255} = 136.9,\qquad z = 135.9 \approx 136$$

(2) Lengths were smaller by a factor $1/(1+z) = 1/136.9$, so the volume was

$$\left(\frac{1}{136.9}\right)^{3} = \frac{1}{2.57 \times 10^{6}}$$

about one $2.6$-millionth. The same matter was contained in a volume $2.6$ million times smaller, so the density was higher by that factor.

(3) By Wien’s displacement law,

$$\lambda_{\max} = \frac{2.898 \times 10^{-3}\ \mathrm{m \cdot K}}{373\ \mathrm{K}} = 7.8 \times 10^{-6}\ \mathrm{m}$$

about $7.8\ \mu\mathrm{m}$, in the infrared. Note that this is exactly $1/136.9$ of the present value $1.06\ \mathrm{mm}$ found in Example 4.1, consistent with the claim of Remark 3.6 that wavelengths stretch with the scale factor.

Exercise 7.3Hard

Suppose the mean lifetime of the neutron were far longer than it actually is, so that not a single neutron decayed during the deuterium bottleneck. That is, suppose the freeze-out ratio $n_n/n_p = 1/5$ were preserved unchanged until synthesis began.

(1) Find the helium mass fraction $Y$ in this case.

(2) Conversely, what must $n_n/n_p$ be at the onset of synthesis in order to reproduce the observed $Y_p = 0.245$?

(3) From the results of (1) and (2), state what the observed helium mass fraction tells us about the lifetime of the neutron.

Solution

(1) Substitute $n_n/n_p = 1/5 = 0.2$ into the general formula of Proposition 5.4.

$$Y = \frac{2 \times 0.2}{1 + 0.2} = \frac{0.4}{1.2} = 0.333$$

about $33\ \%$. That is $9$ percentage points above the observed $24.5\ \%$, a difference vastly larger than the observational error $\pm 0.003$. Such a universe is decisively excluded.

(2) Put $r = n_n/n_p$ and solve $Y = 2r/(1+r) = 0.245$.

$$2r = 0.245(1 + r) \quad\Longrightarrow\quad 2r - 0.245 r = 0.245 \quad\Longrightarrow\quad 1.755\, r = 0.245$$$$r = \frac{0.245}{1.755} = 0.1396 \approx \frac{1}{7.2}$$

in good agreement with the $1/7$ obtained in Example 5.3 by including the decay.

(3) The reduction from $1/5$ at freeze-out to $1/7$ at the onset of synthesis was caused by the beta decay of the neutron (Example 5.3). The observed helium mass fraction therefore demands that the lifetime of the neutron be of the same order of magnitude as the waiting time between freeze-out and the onset of synthesis, about $250$ seconds. The mean lifetime measured in the laboratory is $880$ seconds — precisely that order.

Had the lifetime been $100$ seconds, the neutrons would have decayed away almost entirely and hardly any helium would have formed; had it been $10^{6}$ seconds, we would have $Y \approx 0.33$. That our universe holds hydrogen and helium in the ratio $3:1$ is owed to a property of an elementary particle that can be measured in a terrestrial laboratory.

References

• Satō Katsuhiko, Uchūron Nyūmon: Tanjō kara Mirai e (Introduction to Cosmology: From the Beginning to the Future), Iwanami Shinsho, 2008 (in Japanese) — the chapters on the early universe and nucleosynthesis.
• Steven Weinberg, The First Three Minutes, Basic Books, 1977 (Japanese translation by Obi Shinya, Chikuma Gakugei Bunko, 2008) — the classic popular account of the material of §5 of this article.
• Barbara Ryden, Introduction to Cosmology, 2nd ed., Cambridge University Press, 2017 — the chapters on cosmic expansion, the cosmic microwave background and Big Bang nucleosynthesis. A standard textbook accessible from the first year of university.
• A. A. Penzias and R. W. Wilson, “A Measurement of Excess Antenna Temperature at 4080 Mc/s”, Astrophysical Journal 142 (1965), 419 — the discovery paper for the cosmic microwave background.
• Planck Collaboration, “Planck 2018 results. VI. Cosmological parameters”, Astronomy & Astrophysics 641 (2020), A6. arXiv:1807.06209 — the current standard measurements of $H_0$, the baryon density and the CMB fluctuations.
• Particle Data Group, “Big-Bang Nucleosynthesis” (Review of Particle Physics), https://pdg.lbl.gov/ — predicted and observed abundances of the light elements, and the status of the lithium problem.

Appendix: The numbers used in this article

Constants and conversions. For readers who wish to redo the calculations, here are the values used.

Quantity	Value
Hubble constant $H_0$	$70\ \mathrm{km/s/Mpc}$ (used as a representative value in the text)
$1\ \mathrm{Mpc}$	$3.086 \times 10^{19}\ \mathrm{km} = 3.26 \times 10^{6}$ light years
$1$ year	$3.156 \times 10^{7}\ \mathrm{s}$
CMB temperature $T_0$	$2.7255\ \mathrm{K}$
CMB photon number density	$411\ \mathrm{cm^{-3}}$
constant in Wien’s displacement law	$2.898 \times 10^{-3}\ \mathrm{m \cdot K}$
temperature-energy conversion	$kT = 1\ \mathrm{MeV} \leftrightarrow T = 1.16 \times 10^{10}\ \mathrm{K}$
neutron-proton mass difference	$1.29\ \mathrm{MeV}$
mean lifetime $\tau$ of the neutron	$880\ \mathrm{s}$
binding energy of deuterium	$2.22\ \mathrm{MeV}$
baryon-to-photon ratio $\eta$	$6 \times 10^{-10}$

Differences in sensitivity. The freeze-out temperature $0.8\ \mathrm{MeV}$ of Example 5.3 varies between $0.7$ and $0.9$ in careful calculations, but the effect on $Y$ stays at the level of a few percent. The reason the influence is small even though the quantity sits in an exponent is that the subsequent neutron decay automatically trims away any excess. The baryon-to-photon ratio $\eta$, by contrast, acts sharply on the amount of deuterium left over: doubling $\eta$ reduces $\mathrm{D}/\mathrm{H}$ by roughly a factor of three. It is this high sensitivity that makes deuterium usable as a baryometer.