Olber's Paradox

Estimate the mean free path to a star's surface from stellar density and radius, then compare it to the light-travel-time cosmic horizon to see why Olbers' paradox predicts a blazing night sky that we don't actually observe.

Quick Facts

The paradox
Infinite + uniform + static ⇒ bright sky
If the universe were infinite, uniformly filled with stars, and unchanging, every line of sight would end on a star's surface, and the night sky would blaze as bright as the Sun.
Mean free path
λ = 1 / (nσ), σ = πR²
Average distance a light ray travels before striking a star's disk, given number density n and stellar radius R.
Local stellar density
≈ 0.1–0.14 stars/pc³
Typical number density of stars in the Sun's neighborhood of the Milky Way.
The resolution
Horizon ≈ c × age of universe
A finite, ~13.8-billion-year-old universe limits how far we can see — far short of the mean free path.

Your Results

Calculated
Mean Free Path
-
Average distance to a star's surface, λ = 1/(nσ)
Observable Horizon
-
Light-travel-time distance, ≈ c × age
Path-to-Horizon Ratio
-
Mean free path ÷ observable horizon distance
Sky Coverage at Horizon
-
Fraction of sky that would be starlit, 1 − e^(−horizon/λ)

Ready

Enter a stellar density, radius, and cosmic age, then press Calculate.

How This Olbers' Paradox Calculator Works

Olbers' paradox is the puzzle named after astronomer Heinrich Olbers: if the universe is infinite, populated with stars at a roughly uniform density, and has existed forever, every line of sight from Earth should eventually end on the surface of some star. Adding up the light from every direction should make the entire night sky as bright as the surface of a star like the Sun — yet the sky is dark. This calculator puts numbers on both sides of that argument: the average distance light must travel before hitting a star (the "mean free path"), and the much shorter distance we can actually see given the universe's finite age.

Deriving the mean free path

Treat each star as an opaque disk with cross-sectional area σ = πR², scattered through space with a uniform number density n (stars per unit volume). Along any line of sight, the probability that no star has been intersected within a distance r follows the same exponential law as radioactive decay or light attenuation: P(no hit) = e−nσr. The distance at which this probability falls to 1/e — roughly the average distance to the first star surface — is the mean free path, λ = 1/(nσ). Raising either the stellar density or the stellar radius shrinks λ, because there are more (or bigger) obstacles for a photon's path to run into.

Comparing the mean free path to the observable horizon

Light travels at a finite speed, so we can only see objects whose light has had time to reach us: roughly c × (age of the universe), or about 13.8 billion light-years using the light-travel-time approximation (the true particle horizon is larger, around 46.5 billion light-years, once cosmic expansion is included). This calculator divides the mean free path by that horizon distance to get a ratio, and uses the optical-depth relation 1 − e−horizon/λ to estimate what fraction of the sky would be covered by stellar disks if you could see all the way out to the horizon. Because the mean free path for ordinary stars is many orders of magnitude larger than the horizon, only a tiny fraction of the sky is actually covered — which is why the night sky is dark despite Olbers' argument.

Where this simplified model breaks down

  • It ignores cosmic expansion: the true observable-universe radius (~46.5 billion light-years) is larger than the simple c × age estimate used here, and distant starlight is redshifted, which further dims — rather than brightens — the sky.
  • It treats all stars as identical: real populations span a huge range of radii and luminosities; using one average radius is fine for an order-of-magnitude estimate but not for precision cosmology.
  • It ignores galaxy structure: stars cluster into galaxies separated by enormous voids, so the number density averaged over the whole universe is far lower than the local stellar density near the Sun, which changes the results substantially if you swap one for the other.

Frequently Asked Questions

What is Olbers' paradox?
It is the puzzle that if the universe were infinite, uniformly filled with stars, and unchanging in time, then every line of sight would eventually terminate on a star's surface, making the entire night sky as bright as a star instead of dark. Since the sky is obviously dark, one or more of those assumptions must be wrong.
How is Olbers' paradox actually resolved?
Mainly because the universe has a finite age (about 13.8 billion years) and is expanding. Light from stars farther away than the observable horizon has not had time to reach us yet, and light from the most distant sources is redshifted out of the visible band. The distance light would need to travel to guarantee hitting a star (the mean free path) is vastly larger than that horizon, so only a tiny fraction of the sky is actually covered by starlight.
What does "mean free path" mean in this calculator?
It is the average distance a light ray travels through a field of stars, with number density n and cross-sectional area σ = πR² each, before it strikes a star's surface: λ = 1/(nσ). A larger mean free path means space is emptier and light has to travel farther, on average, before hitting something.
Why does the calculator use "age of the universe" as a distance?
Because light travels at a fixed speed c, the farthest light can have reached us since the universe formed is roughly c × age, so one light-year of distance corresponds to one year of travel time. This calculator uses that light-travel-time distance (about 13.8 billion light-years) as a simplified stand-in for the observable horizon; the true particle horizon is larger, about 46.5 billion light-years, once cosmic expansion is accounted for.