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What % of the sky is occupied?

Intuitively, when you look up at night most of the sky is black "space". Of course, our eyes are not very sensitive, so that's a dubious measurement.

Imagine if our eyes worked as some ancients believed, with rays shooting out of them. When you observe the sky (the entire $4\pi r$ of it) what fraction of those rays would eventually hit some celestial object, excluding those in our solar system?

I know that the observable universe is not necessarily the whole universe, and there may be objects whose light never reached us yet. For this question, feel free to make the (over) simplifying assumption that we have observed the entire universe, and there is nothing that has not been observed thus far.

Yes, the title says "star", it's for poetic reasons :) I'm wondering about any object, luminous or not. Although feel free to answer for just stars if that's easier.

Justification for assumptions:

  • Objects in our solar system are trivial to calculate, and not that interesting. They also needlessly complicate the result with syzygys and such.
  • If we knew the number for our own universe, it would be straightforward to generalize it to as yet unseen objects with some estimate of their distribution.

Responses to comments:

gomennathan
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    Are you counting the Cosmic Microwave Background? because that energy was generated way back when the entire universe was dense enough to fuse hydrogen everywhere, and radiated back when the universe expanded enough that it stopped being opaque to light, and it's coming from every direction., behind all the objects we can see. – notovny Oct 12 '23 at 00:28
  • Related: https://en.wikipedia.org/wiki/Hubble_Deep_Field – PM 2Ring Oct 12 '23 at 02:41
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    Of note is Olber’s paradox https://en.m.wikipedia.org/wiki/Olbers%27s_paradox – Topcode Oct 12 '23 at 04:14
  • There's a cool, previous question here somewhere - it's something like slicing the universe like a loaf of raisin bread - how many raisins (stars) would you cut in half... These are also related Sky density of milky-way stars vs external galaxies and Density of stars on celestial sphere See also all posts referencing Olber per Topcode's comment. – uhoh Oct 13 '23 at 00:21
  • @notovny I'm not sure, I'm not an expert on cosmology. It sounds like if we did count CMB, the answer would be "100%". So probably we shouldn't count it, so that we can get a more interesting answer. – gomennathan Oct 13 '23 at 02:13
  • @uhoh you mean https://physics.stackexchange.com/questions/388802/if-i-sliced-the-universe-in-half-would-the-slice-go-through-a-star The answer is lots of stars would be sliced. But a plane is not a line. – ProfRob Oct 13 '23 at 16:03
  • @ProfRob yes that's it, thanks! And yes it's not :-) – uhoh Oct 13 '23 at 20:14

1 Answers1

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Only about 1 in $10^{12}$ sightlines end on the surface of a star (or 1 in 50,000 if you include the Sun). The rest end where/when the cosmic microwave background formed.

Details

There are thought to be of order $10^{23}$ stars in the observable universe (e.g.,here). I'll make the assumption that this has always been the case for the last 13 billion years (doubtful - the number of stars has increased with time). I'll then take 13 billion light years as a radius to calculate a Euclidean volume these stars are in. This is dubious, but using the proper distance of around 25 billion light years (to the first stars) would not account for the fact that the universe was denser in the past.

This gives an average stellar density of $10^{-8}$ per cubic light year. If we multiply this by the projected area of a star (use the Sun as an average) in square light years and take the reciprocal, we have the mean free path of a photon of light before it encounters a star.

The result is about $10^{22}$ light years. Since this is 12 orders of magnitude larger than the size of the universe then, once the Sun is avoided, sightlines do not, in general, end on the surface of a star. Instead they end at the last scattering surface of the cosmic microwave background.

To work out a fraction of the sky, we can work out approximately the fraction of sightlines that are intercepted. This will be $\sim 1-\exp(-x/l)$, where $x$ is the size assumed for the observable universe in this calculation (I assumed 13 billion light years) and $l$ is the calculated mean free path. When $x/l$ is small then $1-\exp(-x/l) \simeq x/l = 10^{-12}$ or $10^{-10}$%.

NB whether a ray of outgoing light would hit a star or whether all sightlines end on a stellar surface are different questions. The former requires specification of a timescale and knowledge of the future evolution of the universe.

Also note I'm ignoring scattering by dust or free electrons, which is far more important and would-be at the level of a few per cent prior to reaching 13 billion light years.

Finally, planets, white dwarfs, black holes, rocks etc. are quite negligible compared with stars because they have many orders of magnitude smaller cross sections but are not many orders of magnitude more abundant.

ProfRob
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  • What about the angular size of the starts though? Doesn't it matter that distant stars look smaller, so fewer rays would hit them? – gomennathan Oct 13 '23 at 13:06
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    @gomenathan No, because there are more of them. The two effects cancel (if the density is roughly uniform). – ProfRob Oct 13 '23 at 13:12
  • I see, that makes sense. I'm still not able to connect this question to a percentage like what I was wondering though. Maybe I'm just too dumb... – gomennathan Oct 13 '23 at 15:07
  • The answer implies that, as a percentage, 0.0000000001% of sight-lines end on a star (this ignores the sun, which is so close it would skew the calculation) - I hope I can count the zeros correctly – James K Oct 13 '23 at 21:51
  • @gomennathan Also, this calculation presumes the rays are perfect mathematical lines. But the image of any real light source is always diffused, see https://en.wikipedia.org/wiki/Airy_disk – PM 2Ring Oct 14 '23 at 02:05
  • @PM2Ring The image of a source depends on the imaging system. – ProfRob Oct 14 '23 at 06:12
  • Of course. :) I'm not implying that there's anything wrong with your answer. The OP asks about the "eye ray" area, and that's what your answer calculates. I'm just pointing out that the image of real light rays captured by some optical system will necessarily cover a larger area. – PM 2Ring Oct 14 '23 at 07:29