<p>Only about 1 in <span class="math-container">$10^{12}$</span> 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.</p>
<p><strong>Details</strong></p>
<p>There are thought to be of order <span class="math-container">$10^{23}$</span> stars in the observable universe (e.g.,<a href="https://skyandtelescope.org/astronomy-resources/how-many-stars-are-there/" rel="nofollow noreferrer">here</a>). 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.</p>
<p>This gives an average stellar density of <span class="math-container">$10^{-8}$</span> 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 <a href="https://en.wikipedia.org/wiki/Mean_free_path" rel="nofollow noreferrer">mean free path</a> of a photon of light before it encounters a star.</p>
<p>The result is about <span class="math-container">$10^{22}$</span> 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.</p>
<p>To work out a fraction of the sky, we can work out approximately the fraction of sightlines that are intercepted. This will be <span class="math-container">$\sim 1-\exp(-x/l)$</span>, where <span class="math-container">$x$</span> is the size assumed for the observable universe in this calculation (I assumed 13 billion light years) and <span class="math-container">$l$</span> is the calculated mean free path. When <span class="math-container">$x/l$</span> is small then <span class="math-container">$1-\exp(-x/l) \simeq x/l = 10^{-12}$</span> or <span class="math-container">$10^{-10}$</span>%.</p>
<p><strong>NB</strong>
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.</p>
<p>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.</p>
<p>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.</p>

CC BY-SA 4.0
2023-10-13T13:06:23.540Z
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
CC BY-SA 4.0
2023-10-13T13:12:35.507Z
@gomenathan No, because there are more of them. The two effects cancel (if the density is roughly uniform).
ProfRob
CC BY-SA 4.0
2023-10-13T15:07:23.333Z
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
CC BY-SA 4.0
2023-10-13T21:51:57.487Z
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
CC BY-SA 4.0
2023-10-14T02:05:01.323Z
@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
CC BY-SA 4.0
2023-10-14T06:12:29.160Z
@PM2Ring The image of a source depends on the imaging system.
ProfRob
CC BY-SA 4.0
2023-10-14T07:29:52.120Z
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
CC BY-SA 4.0
2023-10-13T05:33:02.913Z
2023-10-13T22:25:11.693Z
2023-10-13T22:25:11.693Z
ProfRob
<p>What % of the sky is occupied?</p>
<p>Intuitively, when you look up at night most of the sky is black &quot;space&quot;. Of course, our eyes are not very sensitive, so that's a dubious measurement.</p>
<p>Imagine if our eyes worked as some ancients believed, with rays shooting out of them. When you observe the sky (the entire <span class="math-container">$4\pi r$</span> of it) what fraction of those rays would <em>eventually</em> hit some celestial object, excluding those in our solar system?</p>
<p>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 <em>entire</em> universe, and there is nothing that has not been observed thus far.</p>
<p>Yes, the title says &quot;star&quot;, it's for poetic reasons :) I'm wondering about any <em>object</em>, luminous or not. Although feel free to answer for just stars if that's easier.</p>
<hr />
<p>Justification for assumptions:</p>
<ul>
<li>Objects in our solar system are trivial to calculate, and not that interesting. They also needlessly complicate the result with syzygys and such.</li>
<li>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.</li>
</ul>
<hr />
<p>Responses to comments:</p>
<ul>
<li><a href="https://en.m.wikipedia.org/wiki/Olbers%27s_paradox" rel="noreferrer">Olber's paradox</a> seems related, but I can't really connect it to my question. The paradox is about an infinite universe, while I am asking about currently observable only.</li>
<li><a href="https://astronomy.stackexchange.com/questions/51364">Sky density of milky-way stars vs external galaxies</a> is close, but I am asking about what area they occupy in the sky, not number. That question does not take angular size into account.</li>
<li><a href="https://astronomy.stackexchange.com/questions/43147">Density of stars on celestial sphere</a> does not specify, but the answers appear to have treated stars as points rather than consider their angular size.</li>
</ul>

CC BY-SA 4.0
2023-10-12T00:28:56.760Z
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
CC BY-SA 4.0
2023-10-12T02:41:26.847Z
Related: https://en.wikipedia.org/wiki/Hubble_Deep_Field
PM 2Ring
CC BY-SA 4.0
2023-10-12T04:14:07.963Z
Of note is Olber’s paradox https://en.m.wikipedia.org/wiki/Olbers%27s_paradox
Topcode
CC BY-SA 4.0
2023-10-13T00:21:28.810Z
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](https://astronomy.stackexchange.com/q/51364/7982) and [Density of stars on celestial sphere](https://astronomy.stackexchange.com/q/43147/7982) See also [all posts referencing Olber](https://astronomy.stackexchange.com/search?q=Olber+) per Topcode's comment.
uhoh
CC BY-SA 4.0
2023-10-13T02:13:40.457Z
@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
CC BY-SA 4.0
2023-10-13T16:03:21.397Z
@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
CC BY-SA 4.0
2023-10-13T20:14:37.233Z
@ProfRob yes that's it, thanks! And yes it's not :-)
uhoh
<night-sky>
Is there a star everywhere you look?
gomennathan
# Is there a star everywhere you look?

(Asked by: username_0 on 2023-10-11)

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:

* [Olber's paradox](https://en.m.wikipedia.org/wiki/Olbers%27s_paradox) seems related, but I can't really connect it to my question. The paradox is about an infinite universe, while I am asking about currently observable only.
* [Sky density of milky-way stars vs external galaxies](https://astronomy.stackexchange.com/questions/51364) is close, but I am asking about what area they occupy in the sky, not number. That question does not take angular size into account.
* [Density of stars on celestial sphere](https://astronomy.stackexchange.com/questions/43147) does not specify, but the answers appear to have treated stars as points rather than consider their angular size.

**Tags:** night-sky
> **Comments:**
> **username_1**: 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.  
> **username_2**: Related: https://en.wikipedia.org/wiki/Hubble_Deep_Field  
> **username_3**: Of note is Olber’s paradox https://en.m.wikipedia.org/wiki/Olbers%27s_paradox  
> **username_4**: 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](https://astronomy.stackexchange.com/q/51364/7982) and [Density of stars on celestial sphere](https://astronomy.stackexchange.com/q/43147/7982) See also [all posts referencing Olber](https://astronomy.stackexchange.com/search?q=Olber+) per Topcode's comment.  
> **username_0**: @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.  
> **username_5**: @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.  
> **username_4**: @ProfRob yes that's it, thanks! And yes it's not :-)  

---

## 1 answers

### Answer by username_5 on 2023-10-13. Score: 9

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](https://skyandtelescope.org/astronomy-resources/how-many-stars-are-there/)). 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](https://en.wikipedia.org/wiki/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.

> **Comments:**
> **username_0**: What about the angular size of the starts though? Doesn't it matter that distant stars look smaller, so fewer rays would hit them?  
> **username_5**: @gomenathan No, because there are more of them. The two effects cancel (if the density is roughly uniform).  
> **username_0**: 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...  
> **username_6**: 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  
> **username_2**: @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  
> **username_5**: @PM2Ring The image of a source depends on the imaging system.  
> **username_2**: 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.  

---

