Why Is the Night Sky Dark? Olbers' Paradox and the Finite Age of the Universe

0. Key points

• If the universe were “infinitely large, uniformly filled with stars, eternally old, and static”, then a line of sight drawn in any direction would eventually strike the surface of some star. The whole night sky would then shine as brightly as the surface of the Sun. This is Olbers’ paradox.
• That brightness is about 180,000 times the solar constant, and it would heat the Earth to roughly 5800 K. Rock and iron alike would vaporize. The utterly ordinary fact that the night sky is dark is therefore an observation: it tells us that one of the four assumptions above is false.
• The escape route “interstellar dust absorbs the light of distant stars” is not available. Given eternity, the dust warms up and eventually glows just as brightly as the stars themselves (the second law of thermodynamics).
• What actually does the work is that light travels at a finite speed and that the universe had a beginning. Blocking a line of sight with stars requires a depth of $5 \times 10^{24}$ light-years, whereas light has been able to travel only 13.8 billion light-years since the universe began. The fraction of the sky covered by stars is of order $10^{-15}$.
• More fundamentally, there is not enough energy. Making the night sky as bright as a stellar surface calls for about $0.84\ \mathrm{J/m^3}$ of radiation energy, but converting every last bit of matter in the universe — dark matter included — into light would supply only one part in three billion of that.
• Yet Olbers was right in a sense. Every line of sight really does end on a hot wall: the fireball of the universe 380,000 years after its birth. What remains after cosmic expansion has diluted its brightness by a factor of about 1.4 trillion is the cosmic microwave background.

1. Motivation: why “the night is dark” is a problem

The night sky is dark. Few facts are more obvious, and few seem to stand in less need of explanation. “The Sun sets, so it gets dark. End of story” — that appears to settle it.

And yet for more than four centuries astronomers have been troubled by this “obvious” fact. In 1610, discussing the countless stars Galileo had found with his telescope, Kepler wrote that the darkness of the night sky is evidence that the universe is not infinite. Edmond Halley (the comet man) discussed the problem before the Royal Society in 1721; the Swiss astronomer de Chéseaux treated it quantitatively in 1744, and the German physician and astronomer Heinrich Olbers did the same in 1823. That the problem carries Olbers’ name today merely reflects which of them the tradition happened to remember.

Their difficulty was the following argument.

Suppose the universe is infinitely large and that stars are scattered throughout it with the same average density everywhere. Then a line of sight extended in any direction whatever must eventually land on the surface of some star. It is the same as standing in a forest: even if the trees are sparse, if you can see far enough some trunk will always block your view. And — this is the crucial point — the surface of a star has the same brightness per unit area no matter how far away it is. So if every line of sight must strike a star, every point of the sky must shine as brightly as the surface of the Sun.

This is not a matter of the sky being “a bit too bright”. We will compute it below, but the energy arriving at the ground would be 180,000 times that of sunlight on a clear day, and the Earth’s temperature would reach about 5800 K, the same as the surface of the Sun. In such a universe there would be no oceans, no continents, and of course no us. The word “night” would never have been coined.

In other words, “the night sky is dark” is the most powerful cosmological observation one can make without a telescope. In this article we put the paradox into equations, close the escape routes one by one, and finally reach the real answer.

flowchart TD
A["Premise 1: the universe is infinitely large"] --> E
B["Premise 2: stars are distributed with the same density everywhere"] --> E
C["Premise 3: the universe has existed forever"] --> E
D["Premise 4: the universe is static (it does not expand)"] --> E
E["Every line of sight strikes the surface of a star"] --> F["The night sky is as bright as the surface of the Sun"]
F --> G["Observed fact: the night sky is dark"]
G --> H["One of premises 1-4 is false"]

The logical structure of Olbers' paradox. The premises entail a conclusion, but the conclusion contradicts observation. Therefore one of the premises is false.

2. Preliminaries: brightness, solid angle, and the inverse-square law

To turn the paradox into equations we must distinguish two kinds of “brightness”. Confusing them hides the crux of the paradox, so we proceed carefully.

Definition 2.1（Flux and surface brightness）

The energy reaching an observer from a source per unit time and unit area is called the flux $F$, measured in $\mathrm{W/m^2}$. If a source of luminosity (total emitted power) $L$ lies at distance $r$ and radiates equally in all directions, the energy spreads over the whole sphere of radius $r$, so

$$F = \frac{L}{4\pi r^2}$$

This is the inverse-square law (the inverse-square law（Proposition 2.2）［How We Know the Distance to a Star］).

For an extended object, on the other hand, the flux per unit solid angle

$$B = \frac{F}{\Omega} \qquad [\mathrm{W/m^2/sr}]$$

is called the surface brightness (also called the specific intensity). Here $\Omega$ is the solid angle the object subtends on the sky.

Note

The steradian (sr), the unit of solid angle, is the area cut out on a sphere divided by the square of the radius. The whole sky is $4\pi \simeq 12.57\ \mathrm{sr}$, and since the Sun’s apparent diameter is about $0.53^\circ$, the Sun subtends about $6.8 \times 10^{-5}\ \mathrm{sr}$. An object covering a mere $5 \times 10^{-6}$ of the sky is what makes daytime as bright as it is.

Proposition 2.2（Surface brightness is independent of distance）

Consider a small patch of the surface of an extended source. Let its area be $A$ and its distance from the observer $r$, with $A$ small compared with $r$ and the patch facing the observer. If there is no absorbing material along the way, the surface brightness $B$ of this patch is independent of $r$.

Proof（Proposition 2.2）

Let $P$ be the energy per unit time that the patch of area $A$ emits in the direction of the observer. This is fixed at the source and does not depend on where the observer is. The flux from this patch alone follows from the inverse-square law:

$$F = \frac{P}{4\pi r^2} \propto \frac{1}{r^2}$$

The solid angle the patch subtends at the observer is, by definition,

$$\Omega = \frac{A}{r^2} \propto \frac{1}{r^2}$$

Hence the surface brightness is

$$B = \frac{F}{\Omega} = \frac{P}{4\pi r^2} \cdot \frac{r^2}{A} = \frac{P}{4\pi A}$$

and $r$ has dropped out. Numerator and denominator both fall as $1/r^2$, so their ratio does not depend on distance.

∎

The proposition looks counterintuitive, but it can be checked close to home. Compare a photograph of the lunar surface taken through a telescope with one taken while standing on the Moon: the “brightness per pixel” of the ground is the same in both. The distant Moon does not look dimmer; it looks smaller, and so the light is collected over a smaller area. For the same reason, the surface of a distant star has the same surface brightness as the surface of the Sun. A star looks faint because it looks small, not because its surface is dim.

This is the engine of Olbers’ paradox. Once the sky is completely paved with stellar surfaces, the escape route “it looks smaller” is gone, and the entire sky takes on the brightness of the solar surface.

Definition 2.3（The Olbers universe）

In this article we call an idealized universe satisfying the following four conditions an Olbers universe.

1. Space is infinite and Euclidean.
2. Stars are distributed uniformly and isotropically with number density $n$ (number per unit volume). All stars are identical, with radius $R_*$, luminosity $L$ and surface temperature $T_*$.
3. This state has persisted from the infinite past.
4. Space is static (neither expanding nor contracting). Relative motions of stars and observer are also neglected.

In addition, the space between the stars is perfectly transparent (no absorption, no scattering).

3. Putting the paradox into equations

3.1. The naive form: adding up without ever running down

We first ignore the fact that stars hide one another and simply add up the light.

Proposition 3.1（A spherical shell contributes independently of its distance）

In an Olbers universe（Definition 2.3）, the total flux reaching the observer from the stars contained in a shell of radius $r$ and thickness $\Delta r$ centered on the observer is

$$\Delta F = n L \,\Delta r$$

which does not depend on $r$. Hence the flux from all stars within radius $R$ is $F(R) = nLR$, which diverges as $R \to \infty$.

Proof（Proposition 3.1）

The volume of the shell is $4\pi r^2 \Delta r$ (the area $4\pi r^2$ of the sphere times the thickness). By condition 2 of Definition 2.3, the number density of stars is uniformly $n$, so the number of stars in the shell is

$$\Delta N = 4\pi r^2 n\, \Delta r$$

The flux from a single star is $L/(4\pi r^2)$ by the inverse-square law of Definition 2.1. The total flux from the shell is therefore

$$\Delta F = \Delta N \times \frac{L}{4\pi r^2} = 4\pi r^2 n \Delta r \times \frac{L}{4\pi r^2} = n L\, \Delta r .$$

The factors of $r^2$ cancel neatly and $r$ disappears. It remains only to sum from $r = 0$ to $R$:

$$F(R) = \int_0^R n L \,dr = nLR \xrightarrow[R \to \infty]{} \infty .$$∎

Divide the cone swept out by the line of sight into shells of equal thickness. The number of stars in a shell grows as the square of the distance, while the brightness of each star falls as the square of the distance. The two effects cancel exactly, so every shell delivers the same amount of light.

The conclusion of Proposition 3.1 is that the night sky is infinitely bright, which is going too far. In reality nearer stars hide more distant ones, so the answer is not infinite. Once we include the mutual blocking of stars, the answer settles at the finite value “as bright as the surface of a star”. That is quite catastrophic enough.

3.2. The exact form: every line of sight ends on a star

Definition 3.2（Mean free path of starlight (the sight distance of the universe)）

In an Olbers universe（Definition 2.3）, the cross-section with which a single star blocks a line of sight (its geometric cross-section) is $\sigma = \pi R_*^2$. The quantity

$$\lambda = \frac{1}{n\sigma} = \frac{1}{n \pi R_*^2}$$

is called the mean free path of starlight, or the sight distance of the universe. It is the length a line of sight must travel, on average, before it hits the surface of a star.

Example 3.3（How far can you see in a forest?）

Let us check the meaning of the mean free path in a forest. Treating the ground plan as a two-dimensional problem, suppose trees of diameter $d = 0.3\ \mathrm{m}$ grow with density $n_2 = 0.1$ per square meter. Each tree blocks a “cross-section” of width $d$, so the two-dimensional sight distance is

$$\lambda_2 = \frac{1}{n_2 d} = \frac{1}{0.1 \times 0.3} = 33\ \mathrm{m}$$

Indeed, in a wood of that density your view fills up with trunks some 30 meters ahead. And here is the essential point: even if the trees are far sparser, a sufficiently large forest will always block your view eventually. The Olbers universe is an infinitely large forest.

Theorem 3.4（Olbers' paradox）

In an Olbers universe（Definition 2.3）, suppose the centers of the stars follow a uniform Poisson distribution of number density $n$. If the surface brightness of a stellar surface is $B_*$, then the mean surface brightness of the sky computed from stars within distance $R$ of the observer is

$$B_{\text{sky}}(R) = B_*\left(1 - e^{-R/\lambda}\right), \qquad \lambda = \frac{1}{n\pi R_*^2}$$

Consequently, in the limit $R \to \infty$,

$$B_{\text{sky}} = B_*$$

so that the night sky is exactly as bright as the surface of a star.

Proof（Theorem 3.4）

Draw a single line of sight from the observer in an arbitrary direction. A star of radius $R_*$ blocks this line precisely when its center lies within distance $R_*$ of the line, that is, when its center lies inside the cylinder of radius $R_*$ around the line.

The volume of that cylinder out to distance $r$ from the observer is $\pi R_*^2 \, r = \sigma r$. Since stellar centers follow a uniform Poisson distribution of density $n$, the expected number of centers in this volume is $n \sigma r = r/\lambda$ (using the definition in Definition 3.2). For a Poisson distribution with mean $\mu$ the probability of finding zero objects is $e^{-\mu}$, so the probability that the line of sight meets no star at all out to distance $r$ is

$$P(r) = e^{-r/\lambda}$$

The probability that the line of sight strikes some star within distance $R$ is therefore $1 - e^{-R/\lambda}$.

Now for the brightness. If the line of sight strikes a star, the light arriving from that direction left the surface of the star, and by Proposition 2.2 its surface brightness is $B_*$ regardless of distance (here we used the condition in Definition 2.3 that the space between stars is transparent; with absorption along the way the conclusion changes). If the line of sight strikes no star, that direction stays dark.

Averaging over many directions covering the whole sky, a fraction $1 - e^{-R/\lambda}$ of directions have brightness $B_*$ and the rest have $0$. The mean surface brightness is therefore

$$B_{\text{sky}}(R) = B_* \left(1 - e^{-R/\lambda}\right) + 0 \times e^{-R/\lambda} = B_*\left(1 - e^{-R/\lambda}\right).$$

Letting $R \to \infty$ gives $e^{-R/\lambda} \to 0$, hence $B_{\text{sky}} \to B_*$.

∎

Remark 3.5

The “infinity” of Proposition 3.1 and the saturation at $B_*$ in Theorem 3.4 are not in conflict. Proposition 3.1 assumed that stars do not hide one another, so it diverges; Theorem 3.4 includes the hiding, so it saturates. Indeed, for $R \ll \lambda$ we have $1 - e^{-R/\lambda} \approx R/\lambda$, so $B_{\text{sky}} \approx B_* R/\lambda$, growing in proportion to $R$ exactly as in Proposition 3.1. Saturation sets in only when $R$ approaches $\lambda$. The regime $R \ll \lambda$ is precisely where we actually live, as we verify in §5.

3.3. How bad would it be?

Let us put in numbers, taking the Sun as a representative star. Its luminosity is $L_\odot = 3.83 \times 10^{26}\ \mathrm{W}$ and its radius $R_\odot = 6.96 \times 10^8\ \mathrm{m}$. The flux leaving the surface is

$$F_\odot^{\text{surf}} = \frac{L_\odot}{4\pi R_\odot^2} = \frac{3.83\times10^{26}}{4\pi \times (6.96\times10^8)^2} = 6.3 \times 10^7\ \mathrm{W/m^2}$$

Treating the solar surface as very nearly a black body, this flux corresponds through the Stefan–Boltzmann law $F = \sigma_{\mathrm{SB}} T^4$ (with $\sigma_{\mathrm{SB}} = 5.67\times10^{-8}\ \mathrm{W/m^2/K^4}$) to a surface temperature $T_\odot = 5772\ \mathrm{K}$. The surface brightness follows from the relation $B = F^{\text{surf}}/\pi$ derived in the Appendix:

$$B_\odot = \frac{6.3\times10^7}{\pi} = 2.0\times10^7\ \mathrm{W/m^2/sr}$$

If the whole sky ($4\pi\ \mathrm{sr}$) had this brightness, the flux received at the ground would be

$$F_{\text{sky}} = 4\pi B_\odot = 4\pi \times 2.0\times10^7 = 2.5\times10^{8}\ \mathrm{W/m^2}$$

The present solar constant (the flux the Earth receives from the Sun) is $1361\ \mathrm{W/m^2}$, so this is

$$\frac{2.5\times10^8}{1361} \approx 1.8\times10^5$$

times larger — about 180,000 times. Moreover the radiation arrives from every direction, so the Earth is heated with nowhere to hide and finally reaches the temperature at which it is in radiative equilibrium, namely $5772\ \mathrm{K}$. Iron boils at about 3130 K, and tungsten, the metal with the highest melting point, melts at 3695 K. At 5772 K iron evaporates completely and even tungsten cannot remain liquid. The Earth would turn almost entirely into gas.

Theorem 3.4 therefore does not merely say “it would be too bright to sleep”. It says that our very existence refutes one of the four assumptions of the Olbers universe.

4. The escape routes are closed

Several escape routes have been proposed over the years, and each has been shut in turn. Let us look at the main ones.

4.1. Dust absorbs the light (the answer of de Chéseaux and Olbers)

Both de Chéseaux and Olbers supposed that a thin medium between the stars absorbs the light of distant ones. It sounds plausible, and interstellar dust does exist. But this explanation is incompatible with the premise of an eternal universe.

Proposition 4.1（An absorber offers no escape）

Assume conditions 1–4 of the Olbers universe（Definition 2.3）, and suppose in addition that light-absorbing material (dust) is uniformly distributed between the stars. If this material has no way to dispose of heat other than by exchange with the stars and the radiation field, then after infinite time the system reaches thermal equilibrium at temperature $T_*$ and the surface brightness of the sky is again $B_*$. In other words, inserting an absorber cannot darken the night sky.

Proof（Proposition 4.1）

Consider a single dust grain. It absorbs starlight and gains energy, and it loses energy by thermal radiation at its own temperature $T_d$. By condition 3 of Definition 2.3 (the universe has existed forever), the grain has already been heated for an infinite time. A grain of finite heat capacity, given infinite time, ends in the steady state where absorption balances emission.

Kirchhoff’s law states that absorptivity equals emissivity at every wavelength. Hence the grain balances against the surrounding radiation exactly when its temperature equals the temperature of that radiation. The only radiation around comes from stellar surfaces at temperature $T_*$, so in the steady state $T_d = T_*$.

A grain at $T_d = T_*$ shines as a black body at temperature $T_*$ (more precisely, as a body radiating with the same efficiency with which it absorbs). Since it glows at the same temperature as the stars, it may block a line of sight but it does not darken it: the solid angle the grain covers shines with surface brightness $B_*$ in place of the star behind it.

Restated in the language of the second law of thermodynamics: if the dust stayed forever colder than the stars, heat would flow perpetually in one direction from stars to dust without ever raising the dust’s temperature. That violates the second law for the entropy of an isolated system. Given infinite time, a closed system must settle at a uniform temperature, and inside it black-body radiation fills everything.

∎

Caution

Real interstellar dust manages to stay cold (a few tens of kelvin) because the universe is not an eternal static box: it is still young, and it is expanding. So Proposition 4.1 does not say “the dust hypothesis is wrong”; it says that the very failure of the dust hypothesis is evidence that the universe is neither eternal nor static. This point was made by John Herschel in the middle of the nineteenth century.

4.2. What if the number of stars is finite?

The escape route “the universe may be infinite but the stars are finite in number” amounts to discarding condition 2 of Definition 2.3. Logically it works, but it does not match observation. Large galaxy surveys have confirmed that the distribution of galaxies is uniform and isotropic on scales larger than roughly 300 million light-years. If instead the universe contained only finitely many galaxies — an “island universe” — we would have to sit near its center, which conflicts with the modern cosmological refusal to grant ourselves a special place (the cosmological principle).

4.3. What if stars were distributed hierarchically (fractally)?

From the end of the nineteenth century into the early twentieth, Charlier and others proposed a picture in which stars form a hierarchy whose density thins out on larger scales. This does in fact evade the paradox. As we compute in Exercise 8.4, if the number of stars within radius $r$ behaves as $N(r) \propto r^D$, the total light stays finite precisely when $D$ is less than $2$.

The observed distribution of galaxies does show fractal behavior with $D \approx 2$ on scales below a few tens of millions of light-years, but on larger scales it crosses over to uniformity ($D = 3$). So in the real universe this escape route is unavailable as well.

5. The real answer (1): light travels at a finite speed and the universe had a beginning

The remaining assumptions are 3 and 4, “eternal” and “static”. What actually does the work is mainly assumption 3.

If the universe had a beginning and the speed of light $c$ is finite, then there is a limit to how far we can see. Writing $t_0$ for the age of the universe, light reaches us from roughly a distance $c t_0$. Stars farther away contribute nothing to the night sky, because their light has not yet arrived, even if they exist.

Proposition 5.1（The night sky of a static universe of finite age）

Drop only condition 3 of the Olbers universe（Definition 2.3）, so that the stars have been shining only for a past interval $t_0$ (all other conditions unchanged). If $c t_0 \ll \lambda$, the mean surface brightness of the sky is

$$B_{\text{sky}} \approx B_* \cdot \frac{c t_0}{\lambda}$$

That is, the fraction of the night sky covered by stars is $f = c t_0/\lambda$.

Proof（Proposition 5.1）

If the stars have been shining for only $t_0$, then light from stars farther than $R = c t_0$ has not yet arrived. Only stars within $R = ct_0$ contribute to the night sky, so the intermediate formula of Theorem 3.4 applies directly:

$$B_{\text{sky}} = B_*\left(1 - e^{-ct_0/\lambda}\right)$$

For $x \ll 1$ we have $e^{-x} = 1 - x + x^2/2 - \cdots$, hence $1 - e^{-x} \approx x$, and under the assumption $x = ct_0/\lambda \ll 1$ we obtain

$$B_{\text{sky}} \approx B_* \cdot \frac{ct_0}{\lambda}$$∎

It remains to insert the values of $\lambda$ and $ct_0$.

Example 5.2（Computing the sight distance of the universe）

Let us estimate the mean free path $\lambda = 1/(n\pi R_*^2)$ with the values of the actual universe.

Step 1: the number density $n$ of stars. The reference point for the mean density of the universe is the critical density（Definition 5.5）［Dark Matter and Dark Energy］. Taking the Hubble constant to be $H_0 = 67.4\ \mathrm{km/s/Mpc}$ (the Planck value), and using $1\ \mathrm{Mpc} = 3.086\times10^{22}\ \mathrm{m}$ so that $H_0 = 2.18\times10^{-18}\ \mathrm{s^{-1}}$, we get

$$\rho_c = \frac{3H_0^2}{8\pi G} = \frac{3\times(2.18\times10^{-18})^2}{8\pi\times 6.674\times10^{-11}} = 8.5\times10^{-27}\ \mathrm{kg/m^3}$$

Ordinary matter (baryons) makes up about 4.9 % of this, so $\rho_b = 4.2\times10^{-28}\ \mathrm{kg/m^3}$, and of that a little under a tenth (we take 7 % here) is locked up in stars, so

$$\rho_* \approx 3\times10^{-29}\ \mathrm{kg/m^3}$$

Dividing by the solar mass $M_\odot = 1.99\times10^{30}\ \mathrm{kg}$,

$$n = \frac{3\times10^{-29}}{1.99\times10^{30}} = 1.5\times10^{-59}\ \mathrm{m^{-3}} .$$

That is $10^{-59}$ stars per cubic meter, which shows just how empty the universe is.

Step 2: the cross-section.

$$\sigma = \pi R_\odot^2 = \pi\times(6.96\times10^8)^2 = 1.52\times10^{18}\ \mathrm{m^2}.$$

Step 3: the mean free path.

$$\lambda = \frac{1}{n\sigma} = \frac{1}{1.5\times10^{-59}\times1.52\times10^{18}} = 4.4\times10^{40}\ \mathrm{m}.$$

Dividing by 1 light-year $= 9.46\times10^{15}\ \mathrm{m}$,

$$\lambda = 4.6\times10^{24}\ \text{light-years} \approx 5\times10^{24}\ \text{light-years}.$$

The estimate of the stellar number density is uncertain by a factor of a few, but the order of magnitude stays at $10^{24}$–$10^{25}$ light-years.

The age of the universe is $t_0 = 13.8$ billion years $= 1.38\times10^{10}$ years, so $ct_0 = 1.38\times10^{10}$ light-years. The fraction in Proposition 5.1 is

$$f = \frac{ct_0}{\lambda} = \frac{1.38\times10^{10}}{4.6\times10^{24}} = 3.0\times10^{-15}$$

so stars cover about three parts in a thousand trillion of the night sky. To reach the state demanded by the Olbers universe, in which every line of sight strikes a star, the universe would have to wait another factor of $10^{15}$ — some $5\times10^{24}$ years. Since even the longest-lived stars last only about $10^{13}$ years, that waiting time is entirely unrealistic.

Example 5.3（Comparing the estimate with observation）

Putting numbers into the formula of Proposition 5.1, the predicted surface brightness of the night sky is

$$B_{\text{sky}} \approx 3.0\times10^{-15}\times 2.0\times10^7 = 6.0\times10^{-8}\ \mathrm{W/m^2/sr} = 60\ \mathrm{nW/m^2/sr}$$

The observed extragalactic background light — the sum of all light emitted by stars from the birth of the universe to today, combining the optical through the far infrared — is measured to be roughly $50$–$100\ \mathrm{nW/m^2/sr}$.

An estimate built on a stack of crude assumptions lands in the same order of magnitude as the observation. This is no accident: it is direct confirmation that the physics of Proposition 5.1 — the night sky is dark because the universe has a finite age — is correct.

Note that the estimate ignores dimming by redshift, the history of stars being born and dying, and the contribution of stars brighter than the Sun. Including them moves the answer by a factor of a few, which is negligible against a ratio of $10^{15}$.

For how distances to stars are measured, see How do we know the distances to the stars? (the starting point is the idea of a standard candle（Definition 2.3）［How We Know the Distance to a Star］); for where the figure of 13.8 billion years comes from, see The edge and the age of the universe (Theorem 3.6［The Edge and the Age of the Universe］).

6. The real answer (2): there is simply not enough energy

The argument of the previous section was geometric: the light has not arrived yet. But there is a cruder and more decisive argument, based on conservation of energy. The cosmologist Edward Harrison pressed this point hard.

Theorem 6.1（The energy needed to light up the night sky does not exist in the universe）

If the whole sky shone with the same surface brightness as a black-body surface at temperature $T_*$, the energy density of that radiation would be

$$u_{\text{req}} = \frac{4\pi B_*}{c} = \frac{4\sigma_{\mathrm{SB}} T_*^4}{c} = a T_*^4, \qquad a = 7.566\times10^{-16}\ \mathrm{J/m^3/K^4}$$

For $T_* = 5772\ \mathrm{K}$ (the solar surface) this gives $u_{\text{req}} = 0.84\ \mathrm{J/m^3}$. The rest energy density of all matter in the universe (dark matter included), on the other hand, is only $u_{\text{mat}} = \Omega_m \rho_c c^2 = 2.4\times10^{-10}\ \mathrm{J/m^3}$. Hence

$$\frac{u_{\text{req}}}{u_{\text{mat}}} = 3.5\times10^{9}$$

so that converting every last particle of matter in the universe into light would supply only one part in three billion of what is required.

Proof（Theorem 6.1）

First we compute $u_{\text{req}}$. The relation $u = 4\pi B/c$ between the energy density of isotropic radiation and its surface brightness, and the surface brightness $B = \sigma_{\mathrm{SB}}T^4/\pi$ of a black-body surface, are both derived in the Appendix. Combining them,

$$u_{\text{req}} = \frac{4\pi}{c}\cdot\frac{\sigma_{\mathrm{SB}}T_*^4}{\pi} = \frac{4\sigma_{\mathrm{SB}}}{c}T_*^4 = aT_*^4 .$$

Inserting numbers, $T_*^4 = (5772)^4 = 1.11\times10^{15}\ \mathrm{K^4}$, so

$$u_{\text{req}} = 7.566\times10^{-16}\times1.11\times10^{15} = 0.84\ \mathrm{J/m^3}.$$

Next $u_{\text{mat}}$. Multiplying the critical density $\rho_c = 8.5\times10^{-27}\ \mathrm{kg/m^3}$ obtained in Example 5.2 by the matter fraction (baryons plus dark matter) $\Omega_m = 0.315$ gives $\rho_m = 2.7\times10^{-27}\ \mathrm{kg/m^3}$, and multiplying by $c^2 = 8.99\times10^{16}\ \mathrm{m^2/s^2}$,

$$u_{\text{mat}} = 2.7\times10^{-27}\times8.99\times10^{16} = 2.4\times10^{-10}\ \mathrm{J/m^3}.$$

The ratio is $0.84 / (2.4\times10^{-10}) = 3.5\times10^{9}$.

∎

Example 6.2（How far short does nuclear fusion fall?）

Theorem 6.1 assumes, absurdly generously, that all matter is converted into light with 100 % efficiency. The means actually available to stars is hydrogen fusion, and turning four hydrogen nuclei into one helium nucleus releases only about 0.7 % of the original rest energy. Dark matter, moreover, does not fuse at all.

With only baryons usable and an efficiency of 0.7 %, the maximum radiation energy density the universe can produce over its whole lifetime is

$$u_{\text{avail}} = 0.007 \times \rho_b c^2 = 0.007 \times 4.2\times10^{-28}\times8.99\times10^{16} = 2.6\times10^{-13}\ \mathrm{J/m^3}$$

The ratio to what is required is

$$\frac{u_{\text{req}}}{u_{\text{avail}}} = \frac{0.84}{2.6\times10^{-13}} = 3.2\times10^{12}$$

that is, one part in three trillion.

In other words, even if the universe were infinitely old and stars lived arbitrarily long, the fuel would run out and the night sky would never grow bright. The Olbers universe fails on its energy budget before geometry even enters.

Remark 6.3

One often reads that the night sky is dark because the universe expands and the light is redshifted. This is not the main reason. The quantitative analysis of Wesson and collaborators shows that a universe with the expansion switched off but the same age would have a background light brighter by only about a factor of two. The overwhelming part of the ratio $10^{15}$ is carried by the youth of the universe and the finiteness of stellar fuel. Redshift is only the finishing touch.

That said, as the next section shows, there is a stage on which redshift does play the leading role.

7. Olbers was right after all: the light that really does fill the sky

There is a mischievous punchline to all of this.

The geometric conclusion of Theorem 3.4 — that every line of sight ends on a hot surface — actually holds in the real universe. What the line of sight hits, however, is not a star.

For about the first 380,000 years the universe was filled with a plasma of free electrons and protons, and light could not travel straight because it was scattered by the electrons. Just as in fog, the universe of that era was opaque. When the temperature fell to about 3000 K, electrons and protons combined into hydrogen atoms, the fog lifted, and light could fly freely (an estimate of this epoch appears in Example 4.3［Was There Really a Big Bang? Expansion, Background Radiation, and the Ratio of the Elements］). The surface marking the moment the fog cleared is called the surface of last scattering.

Whatever direction we look, a sufficiently long line of sight always ends on this surface. And that surface glowed at 3000 K, exactly like the surface of a star. So, just as Olbers said, the sky is literally paved, without a single gap, with “stellar surface”.

Example 7.1（The cosmic microwave background: a night sky diluted by a factor of 1.4 trillion）

Why, then, are we not being roasted by that 3000 K glow? Here at last cosmic expansion takes the leading role.

Expansion stretches the wavelength of light by a factor $(1+z)$, where $z$ is the redshift（Definition 2.1）［Was There Really a Big Bang? Expansion, Background Radiation, and the Ratio of the Elements］. Black-body radiation remains black-body radiation under expansion, and its temperature falls as $T \propto (1+z)$ (Proposition 4.2［Was There Really a Big Bang? Expansion, Background Radiation, and the Ratio of the Elements］). The redshift of the surface of last scattering is $z \simeq 1090$, so radiation that was at 3000 K then has cooled today to

$$T_0 = \frac{2973\ \mathrm{K}}{1 + 1090} = 2.73\ \mathrm{K}$$

The measured value is $T_0 = 2.7255 \pm 0.0006\ \mathrm{K}$. Using Wien’s displacement law $\lambda_{\max} = 2.898\times10^{-3}\ \mathrm{m\cdot K} / T$, the wavelength at which this radiation is strongest is

$$\lambda_{\max} = \frac{2.898\times10^{-3}}{2.7255} = 1.06\ \mathrm{mm}$$

which lies in the microwave band. This is the cosmic microwave background (CMB).

Now let us see by how much the brightness has been diluted. The energy density of radiation is $u = aT^4$, so if the temperature drops by $1/(1+z)$,

$$\frac{u_{\text{today}}}{u_{\text{then}}} = \frac{1}{(1+z)^4} = \frac{1}{1091^4} = \frac{1}{1.4\times10^{12}}$$

that is, one part in 1.4 trillion. Indeed, the present energy density of the CMB is

$$u_{\text{CMB}} = a T_0^4 = 7.566\times10^{-16}\times(2.7255)^4 = 4.2\times10^{-14}\ \mathrm{J/m^3}$$

which is $1.4\times10^{12}$ times smaller than the value $a\times(2973)^4 = 5.9\times10^{-2}\ \mathrm{J/m^3}$ at the surface of last scattering.

Incidentally, $5.9\times10^{-2}\ \mathrm{J/m^3}$ is one fourteenth of the $u_{\text{req}} = 0.84\ \mathrm{J/m^3}$ of Theorem 6.1. The sky paved with solar surfaces that Olbers imagined and the actual surface of last scattering differ in brightness by only a little more than one order of magnitude. What saved us was the subsequent dilution by a factor of 1.4 trillion.

Tip

In summary, the darkness of the night sky has a two-part explanation.

Wavelength band	Does it fill the sky?	Why it is dark
Visible light (starlight)	No (covering fraction $3\times10^{-15}$)	The universe is young, so light reaches us only from nearby. On top of that, there is not enough fusion fuel
Microwaves (light of the Big Bang)	Completely (covering fraction 1)	Redshift from cosmic expansion has dimmed it by a factor of $1.4\times10^{12}$

To say “the night sky is dark” is, strictly speaking, to say “the night sky is dark in visible light”. To a creature with microwave eyes, the night sky glows uniformly without a single gap.

How the CMB is used as evidence for the Big Bang is treated in Did the Big Bang really happen?. A question of the same shape — “why is it so quiet?” — asked about extraterrestrials is Definition 5.4［How Many Aliens Are There? The Drake Equation and the Fermi Paradox］ in How many aliens are there? (Fermi’s paradox). Narrowing down the structure of the universe from an apparently trivial observation is one of the most powerful tools cosmology has.

8. Exercises

Exercise 8.1Easy

In a static infinite universe with stars of luminosity $L$ distributed with uniform density $n$, neglect the effect of stars hiding one another. Find the total flux reaching the observer from the stars lying between distances $r_1$ and $r_2$. Then compare the case $r_1 = 10$ light-years, $r_2 = 20$ light-years with the case $r_1 = 1000$ light-years, $r_2 = 1010$ light-years.

Solution

By Proposition 3.1, the contribution of a shell of thickness $dr$ is $nL\,dr$, independent of $r$. Hence

$$F(r_1, r_2) = \int_{r_1}^{r_2} nL\,dr = nL\,(r_2 - r_1)$$

so the answer depends only on the thickness of the shell.

The shell from $r_1 = 10$ to $r_2 = 20$ light-years is 10 light-years thick, and so is the shell from $r_1 = 1000$ to $r_2 = 1010$ light-years, so the two deliver equal amounts of light. The second shell is 100 times farther away, so each star appears $100^2 = 10^4$ times fainter; but its volume is $(1000/10)^2 = 10^4$ times larger, so it contains $10^4$ times as many stars, and the two effects cancel exactly.

This is where one sees why the intuition “distant stars are faint, so we can ignore them” fails in this problem.

Exercise 8.2Standard

Suppose stars of the size of the Sun (radius $R_\odot = 6.96\times10^8\ \mathrm{m}$) are distributed uniformly with number density $n = 1.5\times10^{-59}\ \mathrm{m^{-3}}$.

1. Find the mean free path $\lambda$ in light-years.
2. When light has traveled a distance $ct_0 = 1.38\times10^{10}$ light-years, find the probability that a line of sight has struck a star (that is, the covering fraction of the night sky).
3. How long must the universe wait for the covering fraction to reach $50\ \%$?

Solution

1. The cross-section is $\sigma = \pi R_\odot^2 = \pi\times(6.96\times10^8)^2 = 1.52\times10^{18}\ \mathrm{m^2}$. Hence

$$\lambda = \frac{1}{n\sigma} = \frac{1}{1.5\times10^{-59}\times1.52\times10^{18}} = 4.4\times10^{40}\ \mathrm{m}.$$

Dividing by $1$ light-year $= 9.46\times10^{15}\ \mathrm{m}$ gives $\lambda = 4.6\times10^{24}$ light-years.

2. From the formula in the proof of Theorem 3.4, the probability of a hit is $1 - e^{-R/\lambda}$. Here

$$\frac{R}{\lambda} = \frac{1.38\times10^{10}}{4.6\times10^{24}} = 3.0\times10^{-15}$$

is extremely small compared with $1$, so $1 - e^{-x} \approx x$ applies and the covering fraction is $3.0\times10^{-15}$, about three parts in a thousand trillion.

3. Solving $1 - e^{-R/\lambda} = 0.5$ gives $e^{-R/\lambda} = 0.5$, that is $R/\lambda = \ln 2 = 0.693$. Hence

$$R = 0.693 \times 4.6\times10^{24} = 3.2\times10^{24}\ \text{light-years}$$

and light needs $3.2\times10^{24}$ years to travel that far — $2.3\times10^{14}$ times the age of the universe. Since stars live at most about $10^{13}$ years, not a single shining star would remain by then (and, as Example 6.2 shows, the fuel would not suffice in any case).

Exercise 8.3Standard

The present temperature of the cosmic microwave background is $T_0 = 2.7255\ \mathrm{K}$. In an expanding universe the temperature of black-body radiation obeys $T = T_0(1+z)$.

1. Find the temperature of the radiation at the surface of last scattering ($z = 1090$) and its peak wavelength from Wien’s displacement law, taking the constant to be $2.898\times10^{-3}\ \mathrm{m\cdot K}$.
2. The energy density of radiation is $u = aT^4$ with $a = 7.566\times10^{-16}\ \mathrm{J/m^3/K^4}$. Find $u$ at the surface of last scattering and today, and check their ratio.
3. Compare the present energy density of the CMB with that of starlight (the extragalactic background light, surface brightness $60\ \mathrm{nW/m^2/sr}$). The energy density of isotropic radiation is $u = 4\pi B/c$.

Solution

1. $T = 2.7255\times(1+1090) = 2.7255\times1091 = 2973\ \mathrm{K}$, about 3000 K. The peak wavelength is

$$\lambda_{\max} = \frac{2.898\times10^{-3}}{2973} = 9.75\times10^{-7}\ \mathrm{m} = 975\ \mathrm{nm}$$

which is near infrared (just outside the range of human vision, at wavelengths slightly longer than red). The universe of that era was a fog glowing at this wavelength.

2. At the surface of last scattering $T^4 = (2973)^4 = 7.81\times10^{13}\ \mathrm{K^4}$, so

$$u_{\text{then}} = 7.566\times10^{-16}\times7.81\times10^{13} = 5.9\times10^{-2}\ \mathrm{J/m^3}.$$

Today $T_0^4 = (2.7255)^4 = 55.2\ \mathrm{K^4}$, so

$$u_{\text{now}} = 7.566\times10^{-16}\times55.2 = 4.2\times10^{-14}\ \mathrm{J/m^3}.$$

The ratio is $5.9\times10^{-2} / (4.2\times10^{-14}) = 1.4\times10^{12}$. Since $(1+z)^4 = 1091^4 = 1.42\times10^{12}$, the relation $u \propto T^4 \propto (1+z)^{-4}$ indeed holds.

3. The energy density of starlight is

$$u_{\text{star}} = \frac{4\pi B}{c} = \frac{4\pi\times 6.0\times10^{-8}}{3.00\times10^8} = 2.5\times10^{-15}\ \mathrm{J/m^3}.$$

Hence

$$\frac{u_{\text{CMB}}}{u_{\text{star}}} = \frac{4.2\times10^{-14}}{2.5\times10^{-15}} \approx 17 .$$

Of the light energy filling the universe, the embers of the Big Bang contribute more than a dozen times as much as everything the stars have emitted in 13.8 billion years. If we are to speak of the “brightness” of the night sky, the leading role belongs to microwaves, not to visible light.

Exercise 8.4Hard

Suppose stars are distributed hierarchically, so that the number of stars within radius $r$ of the observer can be written, for $r \ge r_0$, as $N(r) = A r^D$ (with $A$ a positive constant and $0 < D \le 3$). The case $D = 3$ corresponds to a uniform distribution. All stars have luminosity $L$, and the effect of stars hiding one another is neglected.

1. Express the total flux from all stars beyond distance $r_0$ as an integral, and find the condition on $D$ for it to be finite.
2. Is that condition satisfied in the real universe?

Solution

1. From $N(r) = Ar^D$, the number of stars between radii $r$ and $r+dr$ is

$$dN = N'(r)\,dr = A D r^{D-1}\,dr .$$

The flux from a single star is $L/(4\pi r^2)$, so the flux from all stars beyond $r_0$ is

$$F = \int_{r_0}^{\infty} ADr^{D-1}\cdot\frac{L}{4\pi r^2}\,dr = \frac{ADL}{4\pi}\int_{r_0}^{\infty} r^{D-3}\,dr .$$

The integral $\int_{r_0}^{\infty} r^{s}\,dr$ converges exactly when $s < -1$, so the condition is $D - 3 < -1$, that is,

$$D < 2 .$$

In that case the value is explicitly

$$F = \frac{ADL}{4\pi}\cdot\frac{r_0^{D-2}}{2-D}$$

For $D = 3$ (uniform distribution) it diverges, in agreement with Proposition 3.1. The case $D = 2$ is exactly the boundary, where $\int r^{-1}dr = \ln r$ diverges logarithmically.

2. The distribution of galaxies is known to show fractal behavior with $D \approx 2$ up to scales of a few tens of millions of light-years, but large galaxy surveys have confirmed that on larger scales (beyond roughly 300 million light-years) it becomes uniform, that is $D = 3$. So the condition $D < 2$ fails in the real universe, and a hierarchical cosmos is not the answer to Olbers’ paradox.

(Note also that in a universe with $D < 2$ the view changes drastically with the observer’s position, which sits badly with the cosmological principle.)

References

• Harrison, E. R., Darkness at Night: A Riddle of the Universe, Harvard University Press, 1987. — The standard book-length treatment of the history and physics of Olbers’ paradox. A Japanese translation exists (Chijin Shokan).
• Harrison, E. R., Cosmology: The Science of the Universe, 2nd ed., Cambridge University Press, 2000. — The chapter “Darkness at Night” contains the arguments corresponding to Theorem 3.4 and Theorem 6.1 of this article.
• Wesson, P. S., Valle, K., Stabell, R., “The extragalactic background light and a definitive resolution of Olbers’s paradox”, The Astrophysical Journal 317 (1987), 601. — Separates the contributions of finite age and redshift and compares them quantitatively, showing that the redshift effect amounts to only about a factor of two. This is the basis for Remark 6.3.
• Hauser, M. G., Dwek, E., “The Cosmic Infrared Background: Measurements and Implications”, Annual Review of Astronomy and Astrophysics 39 (2001), 249. — On the measured values of the cosmic background light, used for the comparison in Example 5.3.
• Fixsen, D. J., “The Temperature of the Cosmic Microwave Background”, The Astrophysical Journal 707 (2009), 916. — The source of the CMB temperature $T_0 = 2.7255 \pm 0.0006\ \mathrm{K}$.
• Planck Collaboration, “Planck 2018 results. VI. Cosmological parameters”, Astronomy & Astrophysics 641 (2020), A6. — The source of the values of $H_0$, $\Omega_m$, $\Omega_b$, the age of the universe, and the redshift of the surface of last scattering.

Appendix: Surface brightness and radiation energy density

Let us derive the two relations used in Theorem 6.1. Both go slightly beyond high-school physics, but all that is needed is to keep track of units.

(A-1) The energy density of isotropic radiation, $u = 4\pi B / c$

Consider a small plane of area $A$, and let radiation of surface brightness $B$ arrive from a solid angle $d\Omega$ around the direction perpendicular to it. By the definition of surface brightness (Definition 2.1), the energy crossing this plane in time $dt$ is

$$dE = B \, A \, d\Omega \, dt$$

Just before $dt$, this energy was contained in the column in front of the plane with base area $A$ and length $c\,dt$. The volume of the column is $A c\,dt$, so the energy density carried by radiation from this direction is

$$du = \frac{dE}{A c \, dt} = \frac{B\, d\Omega}{c}$$

Summing over all directions, and using isotropy ($B$ the same in every direction),

$$u = \frac{1}{c}\int B \, d\Omega = \frac{B}{c}\times 4\pi = \frac{4\pi B}{c} .$$

(A-2) The surface brightness of a black-body surface, $B = \sigma_{\mathrm{SB}} T^4/\pi$

The energy leaving a black-body surface at temperature $T$ per unit area and unit time is $F = \sigma_{\mathrm{SB}}T^4$ by the Stefan–Boltzmann law. A black-body surface is Lambertian, meaning its surface brightness $B$ is the same from every viewing angle. For radiation emitted at angle $\theta$ from the surface normal, the “visible” effective area is reduced by $\cos\theta$, so summing over the outward hemisphere gives the flux

$$F = \int_{\text{hemisphere}} B\cos\theta \, d\Omega = B\int_0^{2\pi}\!\!\int_0^{\pi/2} \cos\theta \sin\theta \, d\theta \, d\varphi = B \cdot 2\pi \cdot \left[\frac{\sin^2\theta}{2}\right]_0^{\pi/2} = \pi B$$

Therefore

$$B = \frac{F}{\pi} = \frac{\sigma_{\mathrm{SB}}T^4}{\pi}.$$

This $\pi$ (rather than $4\pi$) is the result of integrating over the hemisphere with the $\cos\theta$ weight. It is this relation that we used in the main text when computing the surface brightness of the Sun as $6.3\times10^7/\pi = 2.0\times10^7\ \mathrm{W/m^2/sr}$.