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Olber's paradox asks a surprisingly easy yet profound question that has confused scientists and scholars for years. Why is the night sky dark? We found the answer in fundamental cosmology, a neat example of science giving answers to apparently mind-blowing questions.

In this article, we'll shed light on the darkness of the sky; keep reading to learn:

  • What is Olber's paradox: shouldn't the night sky be bright?
  • Resolution of Olber's paradox: from dust to cosmology.
  • How to describe Olber's paradox's proposed solutions with math.

And much more!

History of a paradox: describing Olber's paradox form the past to the solution

Humanity has always looked up toward the night sky, but only in the last few centuries have we gained a discrete understanding of the fundamental mechanisms behind our vast and mysterious Universe.

Once Earth found its place in an irrelevant corner of the Universe, and the Sun became one of many stars (around the time Copernicus turned upside-down our conception of the Cosmos), scholars started wondering why the night sky is dark, if it's covered by thousand, million, other Suns.

The amateur astronomer Heinrich Olbers formulated the paradox in 1823, but Kepler, two centuries before, had already asked himself the same question.

Multiple attempts were made trying to break the paradox, with great minds such as Lord Kelvin and Edgar Allan Poe slowly starting to unravel the mystery. However, scientists found a definitive and satisfying answer only by understanding the Nature of our Universe.

Let's find out what the problem was in the first place!

Why is the night sky dark? The formulation of Olber's paradox

Olber's paradox lives in an infinite, static universe, populated by an infinite amount of stars. Earth is placed somewhere in this vastness. We consider Earth the center of a series of spherical shells of a given thickness.

The increasing size of a shell with the distance that would cause stars to get fainter and fainter is effectively balanced by a corresponding increase in the number of stars in each shell. The result is that any shell of identical thickness contributes the same amount of light. Summing each shell's contributions suggests that in this universe, the night sky should be bright (very bright). Where is the catch?

Resolving Olber's paradox: different approaches

We know that in an infinite universe with infinite stars, Earth's night sky should be bright. How can we solve this clear error?

We may try to add some dust, or any material that would absorb some light, between us and the stars. Nebulas would probably do the job.

However, this idea is short-lived. Even if dust absorbed a considerable fraction of the radiation, it would eventually warm and radiate it back, only removing a bit of the brightness.

The next step involves introducing the notion of a finite Universe. In this model, there are simply insufficient stars to fill the sky. This model clashes with physics, and an alternative solution was proposed. Even if the universe is infinite (it's not), it has finite age, and due to this, only a fraction of it is visible from Earth. The speed of light controls this fraction. This is the first theory that gave a solution to Olber's paradox. However, we can add even more layers to it: cosmology tells us, for example, that distant stars shine more weakly when seen from Earth.

Now that we introduced the solutions, let's discover the math used to resolve Olber's paradox. We'll start from the fallacy to end with the correct answer!

Describing Olber's paradox with math: solutions for each step

Let's start with our infinite Universe. Populate it with average stars with a luminosity LL (if you want to learn what luminosity is, visit our luminosity calculator). The luminosity equals the total energy output from the star, while we are interested in the flux at the distance between Earth and the star, ff. The formula for ff involves dividing the luminosity by the area where the energy is "projected". At a distance rr, the energy would be diluted over a shell with area A=4πr2A = 4\pi r^2. The formula for the flux from a single star is:

f=LA=L4πr2f = \frac{L}{A} = \frac{L}{4\pi r^2}

To compute the total flux received by Earth, we need to multiply this quantity by the number of stars in the desired shell. This number NN depends on the shell thickness (dr\text{d}r) and the number of stars in a unit of volume, n0n_0.

N=4πr2drn0N = 4\pi \cdot r^2\cdot \text{d}r \cdot n_0

To find the total flux from that specific shell, compute NfN\cdot f.

fshell=4πr2drn0L4πr2=Ln0dr\begin{split} f_\mathrm{shell} &= 4\pi r^2\cdot \text{d}r \cdot n_0 \cdot \frac{L}{4\pi r^2}\\ &= L\cdot n_0\cdot dr \end{split}

As this is the contribution of a single shell, we must sum the same result for all the other shells. We can do this by integration:

ftot=0n0L dr=f_\mathrm{tot} = \int_0^\infty n_0\cdot L\ \mathrm{d}r = \infty

The night sky in this universe is infinitely bright!

Infinite universe
In an infinite, static, and homogenous Universe, the night sky would be infinitely bright!

Let's introduce some dust out there and see how this changes. The math is straightforward. Take the equation for the flux of a single star, and multiply it by a function that describes a constant extinction due to the presence of a uniform, absorbing background:

f=LA=L4πr2ec0rf = \frac{L}{A} = \frac{L}{4\pi\cdot r^2}\cdot e^{-c_0\cdot r}

Proceed by substituting this quantity in the previous formulas:

fshell=Ln0drec0r\begin{split} f_\mathrm{shell} = L\cdot n_0\cdot dr \cdot e^{-c_0\cdot r} \end{split}


ftot=0n0Lec0r dr=n0Lc0(1ec0r)0=n0Lc0\begin{split} f_\mathrm{tot}& = \int_0^\infty n_0\cdot L\cdot e^{-c_0\cdot r}\ \mathrm{d}r \\[1em] &=\frac{n_0\cdot L}{c_0}\cdot(1-e^{-c_0\cdot r})\Bigl\rvert_0^\infty\\[1em] &=\frac{n_0\cdot L}{c_0} \end{split}

This result may suggest a finite brightness of the sky, even darkness, but as we know, all that absorbed energy must go somewhere.

Olbers with dust
Adding dust clouds (like nebulae) in an infinite Universe doesn't solve the problem, but can attenuate it.

We better find another solution. Let's start by refining our model regarding our infinite universe: we don't expect to have a star in our closest neighborhood. What is the average distance between us and a star, assuming a constant density of these bodies?

We first calculate the percentage of each shell occupied by a star. Define an average radius RR for these bodies. To find this quantity, we multiply the number of stars per shell, N=4πr2drn0N = 4\pi\cdot r^2\cdot \text{d}r \cdot n_0 by the cross-section of a star πR2\pi \cdot R^2, and we normalize over the area of the shell 4πr24\pi \cdot r^2:

P=(4πr2drn0)πR24πr2=πn0R2dr\begin{split} P &= \frac{ (4\pi\cdot r^2\cdot \text{d}r \cdot n_0 )\cdot \pi \cdot R^2}{4\pi \cdot r^2}\\[1em] & = \pi\cdot n_0\cdot R^2 \cdot \mathrm{d}r \end{split}

To have a completely occupied sky, we need to integrate this quantity until it reaches the value 11 for a given distance:

0dP=1\int_0^d P = 1


1=0dπn0R2dr=πn0R20ddr=πn0R2d\begin{split} 1& = \int_0^d\pi\cdot n_0\cdot R^2 \cdot \mathrm{d}r\\ &=\pi\cdot n_0\cdot R^2 \int_0^d \mathrm{d}r\\[.9em] &= \pi\cdot n_0\cdot R^2\cdot d \end{split}

We can find the value of dd assuming that n0=1010 1/ly3n_0 = 10^{-10}\ \mathrm{1/ly^3}, where the unit reads as "per cubic light year", and R=6.957×108 mR =6.957\times 10^{8}\ \mathrm{m}:

d=1πn0R2=1023 lyd = \frac{1}{\pi\cdot n_0\cdot R^2} = 10^{23}\ \mathrm{ly}

Use this distance as a boundary in the integration to find the total flux received by Earth:

ftot=0dn0L dr=n0L1πn0R2=LπR2\begin{split} f_\mathrm{tot} &= \int_0^d n_0\cdot L\ \mathrm{d}r \\ &=n_0\cdot L\cdot \frac{1}{\pi\cdot n_0\cdot R^2} \\[.5em] &=\frac{L}{\pi\cdot R^2} \end{split}

Assigning an average value of LL equal to L=3.828×1026 WL = 3.828\times10^{26}\ \mathrm{W} (the Solar luminosity), we can find the total flux received by Earth:

ftot ⁣= ⁣LπR2 ⁣= ⁣3.828×1026 Wπ7×108 m=2.49×108 W/m2\begin{split} f_\mathrm{tot}\! &=\!\frac{L}{\pi\cdot R^2}\! =\! \frac{3.828\times10^{26}\ \mathrm{W}}{\pi \cdot 7\times 10^{8}\ \mathrm{m}}\\[1em] &=2.49\times10^{8}\ \mathrm{W/m^2} \end{split}

This result is staggeringly high. But it comes with a relieving catch: the distance we used in the computation is several orders of magnitude larger than the event horizon of our observable Universe, which lays at a mere 101310^{13} light years away. Most stars that contribute to the incredible brightness of this night sky would not even be visible from Earth.

Is this enough to resolve Olber's paradox? Partly. Let's dig deeper and consider two final contributions:

  • In an expanding universe, stars move away from us: the redshift we observe reduces the flux.
  • Time dilation contributes to the perceived shift in the frequency of the light.

These contributions are identical and modify the formula for the flux from a star to:

f=LA=L4πr21(1+z)2f = \frac{L}{A} = \frac{L}{4\pi r^2}\cdot\frac{1}{(1+z)^2}

where zz is the coefficient used in astronomy to describe the shift in wavelengths, z=(λobλem)/λemz = (\lambda_\mathrm{ob}-\lambda_\mathrm{em})/\lambda_\mathrm{em}, with λob\lambda_\mathrm{ob} the observed wavelength, and λem\lambda_\mathrm{em} the emitted one.

❗ We talked in detail about the redshift at our redshift calculator!

We rewrite the contributions as follows:

1+z=1+λobλemλem=λobλem=νemνob\begin{split} 1+z &=1+\frac{\lambda_\mathrm{ob}-\lambda_\mathrm{em}}{\lambda_\mathrm{em}}\\[1em] & = \frac{\lambda_\mathrm{ob}}{\lambda_\mathrm{em}}= \frac{\nu_\mathrm{em}}{\nu_\mathrm{ob}} \end{split}

We apply the rules of time dilation:

1+z=νemνob=1+v/c1v/c\begin{split} 1+z &= \frac{\nu_\mathrm{em}}{\nu_\mathrm{ob}} = \sqrt{\frac{1+v/c}{1-v/c}} \end{split}

We define the speed thanks to Hubble's law (meet it at our Hubble's law calculator): v=H0rv = H_0\cdot r, where H0H_0 is Hubble's constant, and rr the distance.

1+z=1+H0r/c1H0r/c1+z =\sqrt{\frac{1+H_0\cdot r/c}{1-H_0\cdot r/c}}

Substitute this quantity in the formula for the flux:

f=LA=L4πr21H0r/c1+H0r/cf = \frac{L}{A} = \frac{L}{4\pi r^2}\cdot\frac{1- H_0\cdot r/c}{1+H_0\cdot r/c}

Insert this expression in the flux for the shell:

fshell=4πr2drn0L4πr21H0r/c1+H0r/c=n0Ldr1H0r/c1+H0r/c\begin{split} f_\mathrm{shell} &= 4\pi r^2\cdot \text{d}r \cdot n_0 \cdot \frac{L}{4\pi r^2}\\ &\quad \cdot \frac{1- H_0\cdot r/c}{1+H_0\cdot r/c}\\[1em] &=n_0\cdot L\cdot \text{d}r\cdot \frac{1- H_0\cdot r/c}{1+H_0\cdot r/c} \end{split}

We now integrate over all the shells up to a limit dictated by the rate of expansion of the Universe:

ftot=n0L0cH01H0r/c1+H0r/cdrf_\mathrm{tot} = n_0\cdot L\int_0^\frac{c}{H_0}\frac{1- H_0\cdot r/c}{1+H_0\cdot r/c}\text{d}r

This integral is rather complex. It results in:

ftot=n0LcH0(2ln(2)1)f_\mathrm{tot} = n_0\cdot L\cdot\frac{c}{H_0}\left(2\ln(2)-1\right)

We can estimate this value to about ftot=2×106 W/m2f_\mathrm{tot} = 2\times10^{-6}\ \mathrm{W/m^2}, a reasonably low quantity!

Finite, expanding universe
Only assuming a finite observable Universe (marked by the black boundary), that is expanding, allows us to solve Olber's paradox.

So, why is the sky dark? Olber's paradox resolved

The sky is dark because our Universe is:

  • Finite (as opposed to the infinite Universe postulated in the paradox); and
  • Dynamic (while the first assumption was a static Universe).

The only assumption that still stands is the one of homogeneity: there is no preferred position in the Universe, so eventually, at large scales, every point of the sky looks the same.


What is Olber's paradox?

Olber's paradox is an apparent contradiction that questions the darkness of the night sky. In a vast and possibly infinite Universe, why isn't the night sky "accumulating" the brightness of infinite stars and shining as bright as day?

Scientists tried many solutions, but only modern cosmology provided the tools (finiteness of the Universe and expansion) necessary to resolve Olber's paradox.

Why is Olber's paradox not true?

Olber's paradox is not true because the observable Universe is finite in size and expanding. Both these factors contribute to reducing the brightness of the night sky.

  • The finiteness of the Universe cuts the number of stars that would saturate the night sky with light.
  • The expansion of the Universe causes light from distant sources to progressively dim.

The result is the darkness we are familiar with.

Why is the night sky dark?

The night sky is dark because the light flux coming in our direction from stars and distant celestial objects is finite. The finite size of the observable universe, joined by the fact that distant objects fade toward lower energies, makes sure that most of the night sky is dark, let alone for the bright lights of stars close to us.

Is the Universe infinite?

We don't know: many theories tend to agree on the possibility that the Universe is infinite in size, but this is probably a question without an answer. On the other hand, the observable Universe is finite in size, as we are bound by the distance traveled by light from an "initial" moment (the Big Bang) in our direction. As the universe expands, and it does so accelerating, some of the Universe is already beyond our "visual range".

Davide Borchia
Which theory?
Infinite, static, homogeneous Universe
Luminosity (L)
Star density (n0)
Total flux (f)
This is Olber's paradox: in an infinite Universe, the light flux received by Earth should be infinite, making the night sky infinitely bright. This clearly doesn't happen! Keep exploring!
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