How bright is the night sky?

Question

Olbers' paradox says that the night sky should be bright if the universe is infinite and full of stars. If you look in any direction, eventually you will come to a star. All directions should be emitting light. Minute Physics has a description of the paradox and the resolution that is hard to improve on. Why is it Dark at Night?.

Don Lincoln made a $2019$ video describing why we expect the universe is at least $250$ times bigger than the visible universe. How far is the edge of the universe? Measurements since then improved that number to $500$. The universe might be infinite. Distant parts might be much the same as the part we can see. Or not.

You might imagine waiting until distant light reaches us to see a bright sky. There are a number of problems with that. The expansion of the universe. The finite lifetime of all the stars in the universe. For that matter, my finite lifetime. Veritasium has a video that describes how the expansion of the universe would affect this. Misconceptions About the Universe

So my question is what fraction of the night sky is bright? (The CMB doesn't count.)

I have always heard that the Cosmic Microwave Background comes from all directions. No light could propagate because it would hit a free charge before long. Also the universe has expanded by a factor of $1100$ since the CMB was emitted. At first glance, it sounds like at least $1$ part in $1100$ of the night sky must be obscured. But since then matter has concentrated into galaxies and stars. Is it correct that most (normal) matter is in a galaxy, and most matter in a galaxy is in dust or gas? How much of the view behind a galaxy does dust or gas obscure? What fraction of the view of a galaxy is stars?

What fraction of the sky is covered by galaxies. JWST deep sky images make it look significant. But they look back to a time when the universe was smaller. That decreases the distance between galaxies.

Another way to ask the question: How big would the visible universe as it is now have to be for the night sky to be half bright? In other words, ignore pesky issues like the speed of light and the expansion of the universe and whose reference frame defines "now". Assume space is flat and homogeneous and like what we see now. Pick a direction. What is the distance until you have a $50$% chance of hitting something opaque? What fraction of that something is bright and what fraction dark? (Dark means darker than a star.)

Answer 1

There are about $10^{22}$ stars in the observable universe right now, within a radius of about 40 billion light years. Most of these stars are smaller than the Sun; a small fraction are much larger. Let's assume that a solar radius is the average.

The mean free path of a photon, ignoring further expansion, is $(n\sigma)^{-1}$, where $n$ is the number density of stars and $\sigma = \pi r^2$ is the cross-sectional area of one star.

Using the estimates above, the mean free path is $1.6\times 10^{24}$ light years. Travelling for 40 billion light years there is roughly a chance of $10^{-14}$ of light (or any line of sight) intercepting a stellar surface.

Thinking about JWST images gives a false impression because the stars are not resolved. Their light is blurred and sight lines passing through a galaxy of such stars are of course nothing like as bright as the surface of a star.

Absorption by dust is far more likely. Typical extinction for visible light over large distances in the universe is perhaps one percent over a billion light years (Menard et al. 2009), suggesting that a significant fraction of light would encounter something in 40 billion light years.

Of course the evolution of the universe complicates the details of these calculations - the universe is expanding, but the number of stars is increasing but at an ever-slowing rate; the dust is being more dilute with the expansion and is being produced by massive stars etc.

These calculations would be for a random spot in intergalactic space. From our position in a galaxy and near the Sun, the probabilities of interaction are higher. Notably, the Sun occupies a fraction $5\times 10^{-6}$ of the sky (as does the Moon). The local stellar density is around 1 per cubic parsec, leafing to a local mean free path of $10^{15}$ light years, but only over the $\sim 1000$ light years it takes to get out of the Galactic plane, so this amounts to a probability of $\sim 10^{-12}$ for a sight line intercepting a star in our Galaxy. There are more stars towards the Galactic plane, but your sight line there will much more likely encounter cold dust with a mean free path of a few thousand light years.

Answer 2

If all emitting sources were identical stars with the same radiance in visible range as the Sun (measure of brightness of the observed emitting area), and no radiation was absorbed by dark absorbers, we could find the fraction of the sky occupied by stars in the following way:

$$ f = R_{galaxies} / R_{Sun}, $$

where $R_{galaxies}$ is visible radiance due to galaxies, and $R_{Sun}$ is visible radiance of the Sun.

Sun's solid angle on the sky is 68 $\mu \text{sr}$, and 43% of the average radiation power on Earth 1370 W is in visible range, so visible radiance of the Sun is on average

$$ R_{Sun} = 8 700 000 ~~\text{W.m}^{-2}.\text{sr}^{-1}. $$

Radiance of the galaxies $R_{galaxies}$ is a lot harder to come by, but there are some determinations based on measurements of radiation far from the Sun. However, numbers in literature vary a lot.

According to the paper

Postman et al. Synoptic Observations of the Cosmic Optical Background with New Horizons ApJ 972 95 (2024) , https://iopscience.iop.org/article/10.3847/1538-4357/ad5ffc

$$ R_{galaxies} = 8.17 ± 1.18 ~~\text{nW. m}^{-2}.\text{sr}^{−1}. $$

Adopting that value, for the fraction of the sky occupied by stars we get approximately $$ f = R_{galaxies} / R_{Sun} = 9.4\cdot 10^{-16} \approx 10^{-15}. $$

Of course, this whole calculation is an oversimplified fiction. Stars have varying radiance and absorbing matter exists. Also, it is hard to determine $R_{galaxies}$ that is purely due to star surfaces (from measurement of total radiance, radiance due to reflections or other kinds of emissions has to be excluded).