<p>Zero. At the limit, there's the Cosmic Microwave Background which represents the content of the Universe not long after it came into being.</p>
<p>Of course, some would no doubt propose that since the CMB has statistical fluctuations there has to be some infinitesimal point where it is zero. Presumably they'd say that if you looked carefully you'd see God lighting the blue touchpaper and retiring.</p>

CC BY-SA 4.0
2023-02-04T21:15:32.803Z
Is there any physical meaning to the notion of an infinitesimal point where CMB is zero?  I understand CMB to be a wall of ancient fire which surrounds us from all sides.  If I will look in a particular direction, I will receive a faint radio photon coming to me from that direction.
Jirka Hanika
CC BY-SA 4.0
2023-02-05T10:35:31.327Z
God only knows :-)  But it's certainly an interesting question, particularly since so much modern physics relies heavily on the mathematical principles of statistics.
Mark Morgan Lloyd
CC BY-SA 4.0
2023-02-02T21:18:24.297Z
2023-02-03T07:53:18.333Z
2023-02-03T07:53:18.333Z
Mark Morgan Lloyd
<p>Following on @mlk's suggestion to do Olbers' paradox in reverse, I'll try to estimate how much of the sky ends at a visible star (i.e., emitting something in the visible range, so that it could literally be &quot;seen&quot; with a sufficiently large telescope.)</p>
<ul>
<li><a href="https://astronomy.stackexchange.com/questions/11494/">This question &amp; answer</a> says that the total amount of light from the stars is estimated at a total of -6.5 magnitudes integrated over the sky.</li>
<li>Assuming that the above figure is in the visual band, and using the standard reference fluxes on <a href="https://en.wikipedia.org/wiki/Apparent_magnitude" rel="nofollow noreferrer">Wikipedia</a>, this corresponds to about <span class="math-container">$1.5 \times 10^{6} \text{ Jy}$</span>, or (dividing by 4π steradians) an average &quot;surface brightness&quot; of the sky of about <span class="math-container">$1.2 \times 10^{5} \text{ Jy/sr}$</span>.</li>
<li>A blackbody at 3000 K (typical of a red dwarf, which is the most common type of star) has a surface brightness at 550 nm (545000 GHz) of <span class="math-container">$1.3 \times 10^{18} \text{ Jy/sr}$</span>.</li>
</ul>
<p>So we can conclude that the &quot;fraction of the sky&quot; in which there is a star is the ratio of these two numbers, or about <span class="math-container">$10^{-13}$</span>.</p>
<p>This is a very crude estimate, and one could quibble about my choices &amp; simplifications above.  It also doesn't take into account cosmological effects, such as redshift or the focusing/defocusing of light rays by cosmological expansion.  I wouldn't be entirely surprised if it's off by a factor of 100 in either direction.</p>
<p>But even so, the fraction of the sky that &quot;has a star in it&quot; is utterly negligible.  The fraction of the sky that contains something visible will be dominated by the Sun and the Moon (which together take up about <span class="math-container">$10^{-4}$</span> of the sky), with maybe a small correction from the planets in the fourth or fifth decimal place.</p>

CC BY-SA 4.0
2023-02-03T22:19:11.703Z
Calculations like the above are not really in my wheelhouse, so please take the above numbers with a grain of salt, and if someone would be so kind as to check my work I'd be grateful.
Michael Seifert
CC BY-SA 4.0
2023-02-03T23:01:53.920Z
I disagree on your math--you are assuming a star as bright as our own and in reality most stars are red dwarfs and you are neglecting the luminosity drop off as you approach the limbs of the star.  However, I fully agree with the basic concept, you're at most a few zeroes off and that doesn't meaningfully change the basic concept.  The sky must be empty or we cook.
Loren Pechtel
CC BY-SA 4.0
2023-02-03T23:26:56.993Z
Also seems to ignore redshift.
ProfRob
CC BY-SA 4.0
2023-02-04T08:40:20.213Z
By that, I mean that every sight line does end up on something that *had* the surface temperature of a star - it's the cosmic microwave background. The argument presented at the moment is only that the sky cannot be (geometrically) full of the discs of unredshifted stars.
ProfRob
CC BY-SA 4.0
2023-02-04T14:13:28.470Z
@LorenPechtel:  Fair critique.  I've edited the numbers for a 3000-K blackbody instead.  It only changes the result by a factor of 2, though.  (Even a 2000-K blackbody only changes my original result by a factor of 3.)
Michael Seifert
CC BY-SA 4.0
2023-02-04T14:15:56.060Z
@ProfRob:  That's a fair point.  My calculation is focused on the things you could literally "see", since that's what the OP seemed to primarily be talking about.  I've put a caveat at the start to reflect this.
Michael Seifert
CC BY-SA 4.0
2023-02-04T14:27:03.733Z
Well of you have to see it, then a consideration of angular resolution is needed. But your calculation is fine for answering whether an infinitesimally narrow sightline intercepts the surface of a star.
ProfRob
CC BY-SA 4.0
2023-02-04T14:28:22.983Z
@ProfRob:  I tweaked my preamble just now to mention "a sufficiently large telescope."  :-)
Michael Seifert
CC BY-SA 4.0
2023-02-05T01:37:12.280Z
Of course, stars aren't the only things that make light. The interstellar medium is very faint in the optical, but it's present on every line of sight. It's brighter at x-ray wavelengths.
John Doty
CC BY-SA 4.0
2023-02-05T04:22:36.113Z
@MichaelSeifert I think something must be wrong with your energy calculation because radiated power goes at the 4th power of the temperature.  It should have a bigger effect than you're getting.
Loren Pechtel
CC BY-SA 4.0
2023-02-05T13:32:33.287Z
@LorenPechtel:  I'm calculating the spectral power density at 550 nm ≈ 545 THz, which is the nominal center of the visible magnitude band (using the figures [here](https://en.wikipedia.org/wiki/Apparent_magnitude#Standard_reference_values)).  The power density radiated at a fixed frequency/wavelength doesn't go as the fourth power;  it's just proportional to the $1/(e^{h \nu/kT} - 1)$ factor in the Planck law.  (It's only when you integrate over all frequencies that you get the Stefan-Boltzmann law.)
Michael Seifert
CC BY-SA 4.0
2023-02-06T01:13:48.853Z
@MichaelSeifert That -6.5 is looking at the whole visual spectrum rather than one frequency.  I don't really think it matters that much overall, though--it's still going to be less than 1 part in a billion.
Loren Pechtel
CC BY-SA 4.0
2023-02-03T22:18:22.160Z
2023-02-04T14:27:50.387Z
2023-02-04T14:27:50.387Z
Michael Seifert
<p>This is supposed to be a very rough ballpark estimate on what percentage of the sky is covered by stars in the milky way galaxy.</p>
<p>There are <a href="https://en.wikipedia.org/wiki/Milky_Way" rel="nofollow noreferrer">approximately </a> <span class="math-container">$10^{11}$</span> stars in the Milky way. First simplification, I will use the size of our sun as the size for all stars. Radius of the sun is <span class="math-container">$7 \cdot 10^{5} km$</span> which gives a disk size of <span class="math-container">$A= 1.5 \cdot 10^{12}km^2$</span>.</p>
<p>Next simplification, I will pretend all stars in the Milky way are at the same distance from us and use the distance to the center of the Milky way. The center of the Milky Way is around 27.000 light years or <span class="math-container">$2.5\cdot 10^{17}km$</span> from Earth.</p>
<p>Then we can compute that <span class="math-container">$10^{11}$</span> stars of disk size <span class="math-container">$1.5 \cdot 10^{12}km^2$</span> at distance <span class="math-container">$2.5\cdot 10^{17}km$</span> would only occupy around <span class="math-container">$10^{11} \cdot 1.5 \cdot 10^{12} km^2 = 1.5\cdot 10^{23} km^2$</span> out of <span class="math-container">$4\pi (2.5\cdot 10^{17})^2 km^2=7.8\cdot 10^{35} km^2$</span> or around <span class="math-container">$2\cdot 10^{-11} \%$</span>.</p>
<p>One could make this approximation a lot more accurate by using a better distribution of the stars in the Milky way but the key message wouldn't change much, the percentage is very tiny.</p>

CC BY-SA 4.0
2023-02-03T10:40:03.313Z
2023-02-03T10:40:03.313Z
quarague
<p>Approximately 100% if you assume the cosmic microwave background to be actually a background and not an object on its own rights.</p>
<p>This is why the sky is black in the first place. It consists of stars and a background.</p>
<p>With the very few exceptions of nearby objects (Moon, Sun, ISS and some nebulae that don't add up for a whole tenth of a percent and if you don't consider Earth as &quot;sky&quot;) anything else is point-like.</p>
<p>If you insist on making longer expositions, you will &quot;spill&quot; some initially invisible objects into whole pixels, but it is not fair to count pixels - these objects are geometrically much smaller than the pixel in question.</p>
<p>Galaxies (including, but not limited to, the Milky way) represent themselves as an object with size and shape - again - only because of our limited resolution. Otherwise, they are quite sparse and transparent. Stars themselves have quite a lot of surface brightness, but a whole galaxy is rather faint because it averages &quot;a little&quot; of stars and many orders of magnitude more dark empty space.</p>

CC BY-SA 4.0
2023-02-05T04:25:10.640Z
Indeed, far away there is a mist of hot hydrogen at 3000K, its not very dense though, 10`18 times LESS than the atmosphere at sea level, 10^13 particles per cm3 of hydrogen.
bandybabboon
CC BY-SA 4.0
2023-02-03T16:37:39.337Z
2023-02-03T16:49:32.767Z
2023-02-03T16:49:32.767Z
fraxinus
<p>It is really quite hard to answer the question as posed because as you observe deeper and deeper (e.g. using a larger telescope or observing for longer) then more and more (fainter) objects become apparent.</p>
<p>Every telescope that you use (and indeed your eye) has a finite angular resolution - the smallest angle between two objects that can be resolved. i.e. The closest two objects can be where there might still be some perceived &quot;gap&quot; between them.</p>
<p>Since every telescope has a finite resolution, but you could in principle just observe deeper and deeper, then eventually you reach the stage where the whole sky is almost full of very faint objects with few discernable gaps between them (at least for the telescopes in use today).</p>
<p>As a rough idea, there are often said to be <em>at least</em> <span class="math-container">$10^{11}$</span> galaxies in the observable universe and at leat <span class="math-container">$10^{11}$</span> stars in our own galaxy. If we were to spread these evenly over the sky (probably ok for galaxies), then the separation between adjacent galaxies or stars is about 2 arcseconds. For the stars, most big telescopes at good observing sites would be able to resolve these (though the stars are unevenly distributed and telescopes are incapable of resolving all the stars towards the plane of the Milky Way for example because the density is much higher). Galaxies though have a finite size - e.g. a galaxy of diameter 10 kpc seen at a distance of 1 Gpc has an angular size of 2 arcseconds.</p>
<p>Thus with the best telescopes and the deepest exposures, if you look very closely there is a galaxy (or at least the blurred image of a galaxy) intercepting almost every line of sight.</p>
<p>However, if you were to define some brightness limit to your pictures and ignore the likelihood that there <em>were</em> fainter objects in the &quot;gaps&quot; then you <em>could</em> attempt to put some percentage figure on it. e.g. Here is an image from the Hubble Ultra Deep Field.</p>
<p><a href="https://i.sstatic.net/5HJqI.jpg" rel="noreferrer"><img src="https://i.sstatic.net/5HJqI.jpg" alt="Hubble Ultra Deep Field" /></a></p>
<p>You might estimate (by eye) that about 20% of the pixels are filled with a galaxy of some sort. There are about 10,000 identified galaxies in this 3.1x3.1 arcmin<span class="math-container">$^2$</span> image, so each galaxy is actually only separated by about 2 arcsec (see the calculation above) and you would be hard pressed to count those 10,000 galaxies by eye since most of them are extremely faint blurs that occupy what you might perceive initially as gaps.</p>
<p>Finally, the answer you get will depend not only on the depth of your image but the resolution of the instrument taking it. To quote your own comment:</p>
<blockquote>
<p>For a given resolution, driving the depth to infinity sends the proportion of &quot;nothing&quot; to 0, but for a given depth (i.e. exposure time), increasing the resolution increases the proportion of sky that has nothing?</p>
</blockquote>
<p>That is an accurate summary, at least with current instrumentation.</p>
<p><strong>Edit (for the dedicated reader)</strong></p>
<p>To further explain a few things. The answer above considers an &quot;object&quot; to be a resolved thing in the sky. Clearly galaxies, consisting of unresolved stars, are mostly empty space and so the vast majority of sightlines will not intercept the surface of a star or anything else. However, that does not mean that sightline is &quot;dark&quot; because <em>all</em> instrumentation we have has a finite resolution that blurs the light from these stars into an image of a galaxy.</p>
<p>Some have commented on the finite age of the universe, Olber's paradox and possibly misunderstood what is meant by &quot;depth&quot; in the quote above. Galaxies and stars have a <em>vast</em> range of luminosities and the least luminous things are much more common than the more luminous. Even if you can only observe to a set distance (e.g. set by the finite time since stars and galaxies were first formed), then increasing the exposure time, or &quot;depth&quot; of your image will still reveal more and more of the less luminous objects.</p>
<p>If there were a lower limit to the luminosity of a galaxy, then in principle yes, there might come a time when instrumentation was so good that increasing exposure time would not reveal more objects, but we aren't there yet. Even if that were the case, there is no guarantee at all, even with excellent angular resolution, that sightlines will not intercept any galaxies, because they have a finite angular size - <em>and</em> angular size actually <em>increases</em> with large redshifts in the currently accepted cosmological model.</p>
<p>Finally we should talk about wavelength. It is far easier to find &quot;empty sky&quot; in the optical (e.g. the Hubble Deep Field), because the light from distant galaxies gets redshifted out of the visible range. It will be interesting to see how crowded JWST deep fields will be in the infrared at an equivalent depth. They will certainly be more crowded, but whether they present an &quot;infrared wall&quot; will depend on the uncertain details of the formation timescale, size scale and star formation history of early galaxies and the shape of the bottom end of the galaxy luminosity function with redshift.</p>

CC BY-SA 4.0
2023-02-02T20:49:54.040Z
If I fpcus on one of the black bits in the Hubble Ultra Deep Field for long enough with my Super-Double-Hubble Ultra Ultra Deep Field experiment, would I find that all those black bits are actually filled with stars and galaxies?
Valorum
CC BY-SA 4.0
2023-02-02T21:30:39.000Z
After increasing contrast to 100% (to reduce artifacts), ImageMagick says that there are 921,600 pixels in that image, of which 9722 are entirely black, so only 1% is *pure* black. If I instead look for pixels that are *close* to black (by checking if the color RGB value starts with #000), there's 58144 *close to black* pixels or #00080D or 6.3%.
JonathanReez
CC BY-SA 4.0
2023-02-03T06:58:03.823Z
Stars and galaxies in the photo look much bigger and “solid” than they are. A galaxy in the photo looks like a solid disc hundreds of thousands of light years wide.
Michael
CC BY-SA 4.0
2023-02-03T22:55:59.790Z
But those objects are limited by the pixel size of the sensor (and implicitly, also, by the resolution of the lens.)  In practice you can't have every line terminating in a star because that would make the night sky glow with the brightness of the surface of a star.  In practice the vast majority of the sky must truly have nothing.
Loren Pechtel
CC BY-SA 4.0
2023-02-03T23:27:19.620Z
@user253751 I know, this was a typo, I meant to write #000***
JonathanReez
CC BY-SA 4.0
2023-02-04T11:02:00.347Z
"That is an accurate summary." It's not possible to actually set the depth of view to infinity, though, because there's an edge to the observable universe?
nick012000
CC BY-SA 4.0
2023-02-04T11:07:30.140Z
@nick012000 you are labouring under the misapprehension that brightness is only determined by distance. Astronomical objects have a many order of magnitude range of intrinsic luminosities, and there are more of the less luminous things.
ProfRob
CC BY-SA 4.0
2023-02-04T11:09:06.983Z
@ProfRob No? I'm just saying that if you draw a line out from the Earth to the edge of the observable universe, there's going to be some spots where there isn't anything between the two, even if there would be something if you drew the line out to infinity.
nick012000
CC BY-SA 4.0
2023-02-04T11:35:35.287Z
@nick01200 my response is now included as an edit.
ProfRob
CC BY-SA 4.0
2023-02-04T14:42:24.220Z
@JonathanReez #010000 is closer to black than #00080D, and wouldn’t have shown up in your search.
Todd Wilcox
CC BY-SA 4.0
2023-02-04T15:58:50.580Z
@ToddWilcox what is "black" in the picture does not mean zero flux, it is just where someone has set the zero point of the colour scale/contrast. You would need to analyse the science images (and none of the pixels will have zero signal).
ProfRob
CC BY-SA 4.0
2023-02-04T16:28:25.673Z
@ProfRob I don’t think I understand your comment. The subject in question is a computer image analyzed by Johnathan Reez using RGB hex values as a substitute for actual telescope data. In this image, some pixels **do** have the hex value of #000000, or 0, 0, 0 (R, G, B). The total brightness ("distance from black") of hex color #010000 is something like 20 times **less** than the brightness of hex color #00080D. We’re attempting to explain that the criterion for "close to black" that Jonathan Reez used doesn’t seem useful.
Todd Wilcox
CC BY-SA 4.0
2023-02-02T12:24:18.007Z
2023-02-04T11:34:28.367Z
2023-02-04T11:34:28.367Z
ProfRob
<p>From my persepective here on Earth, the sky seems to look like a few large-ish things and a bunch of tiny things.  <a href="https://esahubble.org/science/deep_fields/" rel="noreferrer">Hubble teaches us that</a> even the apparent void between the tiny things has many <em>very</em> tiny things.  Stuff that is very near takes up a large portion of what I see.  For example, the Earth takes up about 50% of what I can see, because I'm right next to it.  The Moon takes up <a href="https://en.wikipedia.org/wiki/Angular_diameter#Use_in_astronomy" rel="noreferrer">a circle that is about half a degree in diameter</a> or so, depending on what day it is.  A small-angle approximation gives that this takes up about <span class="math-container">$\frac{\pi(0.25)^2}{41253} \approx 0.00048\%$</span> of the sky. (<a href="https://en.m.wikipedia.org/wiki/Square_degree" rel="noreferrer">there are about 41253 square degrees in the sky</a>)</p>
<p>The percent of the sky that &quot;has stuff in it&quot; changes dramatically hour by hour as the Earth spins round, since it occupies so much of our view.  This change in time also occurs due to the motions of all of the heavenly bodies, but minus the Earth I would imagine the value would change relatively little over large periods of time, except during eclipses.</p>
<p>Minus the Earth, and barring an eclipse, when we add up all of the Sun, the Moon, the planets and the stars, what percent of the night sky is still just empty or nearly-empty space?</p>

CC BY-SA 4.0
2023-02-02T11:23:25.320Z
It's very hard to answer your question unless you provide a definitive list of stuff and how far away/how bright it needs to be. Obviously as you go deeper and deeper them more and more things become apparent at greater distances.
ProfRob
CC BY-SA 4.0
2023-02-02T11:31:34.543Z
@ProfRob do you mean to imply that the proportion of sky that appears to have nothing in it approaches 0 as the telescope we use to measure it gets better and better resolution?
Him
CC BY-SA 4.0
2023-02-02T11:59:48.313Z
Yes - that is indeed what I mean. Although here, "resolution" is the wrong word. Better resolution means that you can more easily resolve objects from each other so that there is a gap between them. I really mean "depth", in the sense that as your telescope collects more and more light it can see more and more (fainter) objects and eventually your finite resolution means that there are no gaps between them.
ProfRob
CC BY-SA 4.0
2023-02-02T12:01:49.857Z
So the question really can only be posed as what fraction of the sky is effectively empty of resolved objects down to some brightness limit. Even then it is quite hard to define what is exactly meant by that.
ProfRob
CC BY-SA 4.0
2023-02-02T12:04:06.690Z
@ProfRob so depth and resolution have opposite effects, then?  For a given resolution, driving the depth to infinity sends the proportion of "nothing" to 0, but for a given depth (i.e. exposure time), increasing the resolution increases the proportion of sky that has nothing?
Him
CC BY-SA 4.0
2023-02-02T12:09:40.320Z
@ProfRob what if we phrase it in the opposite direction: we shine a photon in a random direction.  What is the probability that it hits something?  This is similar, but seems to skirt these problems with resolution and exposure time.
Him
CC BY-SA 4.0
2023-02-02T12:25:08.483Z
"For a given resolution, driving the depth to infinity sends the proportion of "nothing" to 0, but for a given depth (i.e. exposure time), increasing the resolution increases the proportion of sky that has nothing? " this is an accurate summary.
ProfRob
CC BY-SA 4.0
2023-02-02T12:29:03.023Z
See also https://astronomy.stackexchange.com/questions/17990/what-happens-to-the-99-9-of-the-sun-rays-that-do-not-fall-on-any-planets-or-an but note that galaxies are mostly empty space and that most photons would just go straight through them.
ProfRob
CC BY-SA 4.0
2023-02-03T13:52:13.653Z
I guess, you could try running Olber's paradox in reverse and compare the average brightness of the nights-sky with that of the surface of an average star. This would give you a reasonable estimate on the proportion of directions that directly face a star. I haven't been able to find any reliable numbers that aren't about terrestrial light pollution though. But it is certain that the resulting number will be tiny.
mlk
CC BY-SA 4.0
2023-02-03T22:21:27.310Z
@mlk:  I took a crack at running the numbers on your idea in my answer below, but I have no idea if I did it right.
Michael Seifert
CC BY-SA 4.0
2023-02-04T02:46:38.843Z
This question is reminiscent of Mandelbrot's "How long is the coast of Britain?" (https://users.math.yale.edu/mandelbrot/web_pdfs/howLongIsTheCoastOfBritain.pdf)  It depends on how closely you look.
Syntax Junkie
CC BY-SA 4.0
2023-02-04T03:05:41.453Z
@SyntaxJunkie the "looking out" analogue to the "looking in" coast of Britain paradox is apparently Olber's paradox pointed out by @ mlk above.  It seems to be resolved by our current understanding that we can only look so far out.  Possibly the coast of Britain paradox is resolved by our current understanding that we can only look so far in.
Him
CC BY-SA 4.0
2023-02-05T20:24:58.550Z
Noting the comments about Mandelbrot etc., I'd like to highlight one possible wrong answer. When I was studying elementary science in the late 1960s and early 1970s, the syllabus textbooks stated that stars were too small to be resolved as other than points in any telescope, and completely omitted any mention of nebulae and of galaxies other than our own. There is a risk that somebody whose education or reading stopped at that level would argue that since the stars appear as points, the visible part of the sky (i.e. the part not obscured by the Earth) is empty of everything other than planets.
Mark Morgan Lloyd
<observable-universe><angular-diameter>
Approximately what percent of the sky has nothing in it?
Him
# Approximately what percent of the sky has nothing in it?

(Asked by: username_0 on 2023-02-02)

From my persepective here on Earth, the sky seems to look like a few large-ish things and a bunch of tiny things. [Hubble teaches us that](https://esahubble.org/science/deep_fields/) even the apparent void between the tiny things has many *very* tiny things. Stuff that is very near takes up a large portion of what I see. For example, the Earth takes up about 50% of what I can see, because I'm right next to it. The Moon takes up [a circle that is about half a degree in diameter](https://en.wikipedia.org/wiki/Angular_diameter#Use_in_astronomy) or so, depending on what day it is. A small-angle approximation gives that this takes up about $\frac{\pi(0.25)^2}{41253} \approx 0.00048\%$ of the sky. ([there are about 41253 square degrees in the sky](https://en.m.wikipedia.org/wiki/Square_degree))

The percent of the sky that "has stuff in it" changes dramatically hour by hour as the Earth spins round, since it occupies so much of our view. This change in time also occurs due to the motions of all of the heavenly bodies, but minus the Earth I would imagine the value would change relatively little over large periods of time, except during eclipses.

Minus the Earth, and barring an eclipse, when we add up all of the Sun, the Moon, the planets and the stars, what percent of the night sky is still just empty or nearly-empty space?

**Tags:** observable-universe, angular-diameter
> **Comments:**
> **username_1**: It's very hard to answer your question unless you provide a definitive list of stuff and how far away/how bright it needs to be. Obviously as you go deeper and deeper them more and more things become apparent at greater distances.  
> **username_0**: @ProfRob do you mean to imply that the proportion of sky that appears to have nothing in it approaches 0 as the telescope we use to measure it gets better and better resolution?  
> **username_1**: Yes - that is indeed what I mean. Although here, "resolution" is the wrong word. Better resolution means that you can more easily resolve objects from each other so that there is a gap between them. I really mean "depth", in the sense that as your telescope collects more and more light it can see more and more (fainter) objects and eventually your finite resolution means that there are no gaps between them.  
> **username_1**: So the question really can only be posed as what fraction of the sky is effectively empty of resolved objects down to some brightness limit. Even then it is quite hard to define what is exactly meant by that.  
> **username_0**: @ProfRob so depth and resolution have opposite effects, then?  For a given resolution, driving the depth to infinity sends the proportion of "nothing" to 0, but for a given depth (i.e. exposure time), increasing the resolution increases the proportion of sky that has nothing?  
> **username_0**: @ProfRob what if we phrase it in the opposite direction: we shine a photon in a random direction.  What is the probability that it hits something?  This is similar, but seems to skirt these problems with resolution and exposure time.  
> **username_1**: "For a given resolution, driving the depth to infinity sends the proportion of "nothing" to 0, but for a given depth (i.e. exposure time), increasing the resolution increases the proportion of sky that has nothing? " this is an accurate summary.  
> **username_1**: See also https://astronomy.stackexchange.com/questions/17990/what-happens-to-the-99-9-of-the-sun-rays-that-do-not-fall-on-any-planets-or-an but note that galaxies are mostly empty space and that most photons would just go straight through them.  
> **username_2**: I guess, you could try running Olber's paradox in reverse and compare the average brightness of the nights-sky with that of the surface of an average star. This would give you a reasonable estimate on the proportion of directions that directly face a star. I haven't been able to find any reliable numbers that aren't about terrestrial light pollution though. But it is certain that the resulting number will be tiny.  
> **username_3**: @mlk:  I took a crack at running the numbers on your idea in my answer below, but I have no idea if I did it right.  
> **username_4**: This question is reminiscent of Mandelbrot's "How long is the coast of Britain?" (https://users.math.yale.edu/mandelbrot/web_pdfs/howLongIsTheCoastOfBritain.pdf)  It depends on how closely you look.  
> **username_0**: @SyntaxJunkie the "looking out" analogue to the "looking in" coast of Britain paradox is apparently Olber's paradox pointed out by @ mlk above.  It seems to be resolved by our current understanding that we can only look so far out.  Possibly the coast of Britain paradox is resolved by our current understanding that we can only look so far in.  
> **username_5**: Noting the comments about Mandelbrot etc., I'd like to highlight one possible wrong answer. When I was studying elementary science in the late 1960s and early 1970s, the syllabus textbooks stated that stars were too small to be resolved as other than points in any telescope, and completely omitted any mention of nebulae and of galaxies other than our own. There is a risk that somebody whose education or reading stopped at that level would argue that since the stars appear as points, the visible part of the sky (i.e. the part not obscured by the Earth) is empty of everything other than planets.  

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## 5 answers

### Answer by username_1 on 2023-02-02. Score: 25 | Accepted answer

It is really quite hard to answer the question as posed because as you observe deeper and deeper (e.g. using a larger telescope or observing for longer) then more and more (fainter) objects become apparent.

Every telescope that you use (and indeed your eye) has a finite angular resolution - the smallest angle between two objects that can be resolved. i.e. The closest two objects can be where there might still be some perceived "gap" between them.

Since every telescope has a finite resolution, but you could in principle just observe deeper and deeper, then eventually you reach the stage where the whole sky is almost full of very faint objects with few discernable gaps between them (at least for the telescopes in use today).

As a rough idea, there are often said to be *at least* $10^{11}$ galaxies in the observable universe and at leat $10^{11}$ stars in our own galaxy. If we were to spread these evenly over the sky (probably ok for galaxies), then the separation between adjacent galaxies or stars is about 2 arcseconds. For the stars, most big telescopes at good observing sites would be able to resolve these (though the stars are unevenly distributed and telescopes are incapable of resolving all the stars towards the plane of the Milky Way for example because the density is much higher). Galaxies though have a finite size - e.g. a galaxy of diameter 10 kpc seen at a distance of 1 Gpc has an angular size of 2 arcseconds.

Thus with the best telescopes and the deepest exposures, if you look very closely there is a galaxy (or at least the blurred image of a galaxy) intercepting almost every line of sight.

However, if you were to define some brightness limit to your pictures and ignore the likelihood that there *were* fainter objects in the "gaps" then you *could* attempt to put some percentage figure on it. e.g. Here is an image from the Hubble Ultra Deep Field.

[![Hubble Ultra Deep Field](https://i.sstatic.net/5HJqI.jpg)](https://i.sstatic.net/5HJqI.jpg)

You might estimate (by eye) that about 20% of the pixels are filled with a galaxy of some sort. There are about 10,000 identified galaxies in this 3.1x3.1 arcmin$^2$ image, so each galaxy is actually only separated by about 2 arcsec (see the calculation above) and you would be hard pressed to count those 10,000 galaxies by eye since most of them are extremely faint blurs that occupy what you might perceive initially as gaps.

Finally, the answer you get will depend not only on the depth of your image but the resolution of the instrument taking it. To quote your own comment:

> For a given resolution, driving the depth to infinity sends the proportion of "nothing" to 0, but for a given depth (i.e. exposure time), increasing the resolution increases the proportion of sky that has nothing?

That is an accurate summary, at least with current instrumentation.

**Edit (for the dedicated reader)**

To further explain a few things. The answer above considers an "object" to be a resolved thing in the sky. Clearly galaxies, consisting of unresolved stars, are mostly empty space and so the vast majority of sightlines will not intercept the surface of a star or anything else. However, that does not mean that sightline is "dark" because *all* instrumentation we have has a finite resolution that blurs the light from these stars into an image of a galaxy.

Some have commented on the finite age of the universe, Olber's paradox and possibly misunderstood what is meant by "depth" in the quote above. Galaxies and stars have a *vast* range of luminosities and the least luminous things are much more common than the more luminous. Even if you can only observe to a set distance (e.g. set by the finite time since stars and galaxies were first formed), then increasing the exposure time, or "depth" of your image will still reveal more and more of the less luminous objects.

If there were a lower limit to the luminosity of a galaxy, then in principle yes, there might come a time when instrumentation was so good that increasing exposure time would not reveal more objects, but we aren't there yet. Even if that were the case, there is no guarantee at all, even with excellent angular resolution, that sightlines will not intercept any galaxies, because they have a finite angular size - *and* angular size actually *increases* with large redshifts in the currently accepted cosmological model.

Finally we should talk about wavelength. It is far easier to find "empty sky" in the optical (e.g. the Hubble Deep Field), because the light from distant galaxies gets redshifted out of the visible range. It will be interesting to see how crowded JWST deep fields will be in the infrared at an equivalent depth. They will certainly be more crowded, but whether they present an "infrared wall" will depend on the uncertain details of the formation timescale, size scale and star formation history of early galaxies and the shape of the bottom end of the galaxy luminosity function with redshift.

> **Comments:**
> **username_12**: If I fpcus on one of the black bits in the Hubble Ultra Deep Field for long enough with my Super-Double-Hubble Ultra Ultra Deep Field experiment, would I find that all those black bits are actually filled with stars and galaxies?  
> **username_13**: After increasing contrast to 100% (to reduce artifacts), ImageMagick says that there are 921,600 pixels in that image, of which 9722 are entirely black, so only 1% is *pure* black. If I instead look for pixels that are *close* to black (by checking if the color RGB value starts with #000), there's 58144 *close to black* pixels or #00080D or 6.3%.  
> **username_14**: Stars and galaxies in the photo look much bigger and “solid” than they are. A galaxy in the photo looks like a solid disc hundreds of thousands of light years wide.  
> **username_7**: But those objects are limited by the pixel size of the sensor (and implicitly, also, by the resolution of the lens.)  In practice you can't have every line terminating in a star because that would make the night sky glow with the brightness of the surface of a star.  In practice the vast majority of the sky must truly have nothing.  
> **username_13**: @user253751 I know, this was a typo, I meant to write #000***  
> **username_15**: "That is an accurate summary." It's not possible to actually set the depth of view to infinity, though, because there's an edge to the observable universe?  
> **username_1**: @nick012000 you are labouring under the misapprehension that brightness is only determined by distance. Astronomical objects have a many order of magnitude range of intrinsic luminosities, and there are more of the less luminous things.  
> **username_15**: @ProfRob No? I'm just saying that if you draw a line out from the Earth to the edge of the observable universe, there's going to be some spots where there isn't anything between the two, even if there would be something if you drew the line out to infinity.  
> **username_1**: @nick01200 my response is now included as an edit.  
> **username_16**: @JonathanReez #010000 is closer to black than #00080D, and wouldn’t have shown up in your search.  
> **username_1**: @ToddWilcox what is "black" in the picture does not mean zero flux, it is just where someone has set the zero point of the colour scale/contrast. You would need to analyse the science images (and none of the pixels will have zero signal).  
> **username_16**: @ProfRob I don’t think I understand your comment. The subject in question is a computer image analyzed by Johnathan Reez using RGB hex values as a substitute for actual telescope data. In this image, some pixels **do** have the hex value of #000000, or 0, 0, 0 (R, G, B). The total brightness ("distance from black") of hex color #010000 is something like 20 times **less** than the brightness of hex color #00080D. We’re attempting to explain that the criterion for "close to black" that Jonathan Reez used doesn’t seem useful.  

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### Answer by username_5 on 2023-02-02. Score: 17

Zero. At the limit, there's the Cosmic Microwave Background which represents the content of the Universe not long after it came into being.

Of course, some would no doubt propose that since the CMB has statistical fluctuations there has to be some infinitesimal point where it is zero. Presumably they'd say that if you looked carefully you'd see God lighting the blue touchpaper and retiring.

> **Comments:**
> **username_6**: Is there any physical meaning to the notion of an infinitesimal point where CMB is zero?  I understand CMB to be a wall of ancient fire which surrounds us from all sides.  If I will look in a particular direction, I will receive a faint radio photon coming to me from that direction.  
> **username_5**: God only knows :-)  But it's certainly an interesting question, particularly since so much modern physics relies heavily on the mathematical principles of statistics.  

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### Answer by username_3 on 2023-02-03. Score: 8

Following on @mlk's suggestion to do Olbers' paradox in reverse, I'll try to estimate how much of the sky ends at a visible star (i.e., emitting something in the visible range, so that it could literally be "seen" with a sufficiently large telescope.)

* [This question & answer](https://astronomy.stackexchange.com/questions/11494/) says that the total amount of light from the stars is estimated at a total of -6.5 magnitudes integrated over the sky.
* Assuming that the above figure is in the visual band, and using the standard reference fluxes on [Wikipedia](https://en.wikipedia.org/wiki/Apparent_magnitude), this corresponds to about $1.5 \times 10^{6} \text{ Jy}$, or (dividing by 4π steradians) an average "surface brightness" of the sky of about $1.2 \times 10^{5} \text{ Jy/sr}$.
* A blackbody at 3000 K (typical of a red dwarf, which is the most common type of star) has a surface brightness at 550 nm (545000 GHz) of $1.3 \times 10^{18} \text{ Jy/sr}$.

So we can conclude that the "fraction of the sky" in which there is a star is the ratio of these two numbers, or about $10^{-13}$.

This is a very crude estimate, and one could quibble about my choices & simplifications above. It also doesn't take into account cosmological effects, such as redshift or the focusing/defocusing of light rays by cosmological expansion. I wouldn't be entirely surprised if it's off by a factor of 100 in either direction.

But even so, the fraction of the sky that "has a star in it" is utterly negligible. The fraction of the sky that contains something visible will be dominated by the Sun and the Moon (which together take up about $10^{-4}$ of the sky), with maybe a small correction from the planets in the fourth or fifth decimal place.

> **Comments:**
> **username_3**: Calculations like the above are not really in my wheelhouse, so please take the above numbers with a grain of salt, and if someone would be so kind as to check my work I'd be grateful.  
> **username_7**: I disagree on your math--you are assuming a star as bright as our own and in reality most stars are red dwarfs and you are neglecting the luminosity drop off as you approach the limbs of the star.  However, I fully agree with the basic concept, you're at most a few zeroes off and that doesn't meaningfully change the basic concept.  The sky must be empty or we cook.  
> **username_1**: Also seems to ignore redshift.  
> **username_1**: By that, I mean that every sight line does end up on something that *had* the surface temperature of a star - it's the cosmic microwave background. The argument presented at the moment is only that the sky cannot be (geometrically) full of the discs of unredshifted stars.  
> **username_3**: @LorenPechtel:  Fair critique.  I've edited the numbers for a 3000-K blackbody instead.  It only changes the result by a factor of 2, though.  (Even a 2000-K blackbody only changes my original result by a factor of 3.)  
> **username_3**: @ProfRob:  That's a fair point.  My calculation is focused on the things you could literally "see", since that's what the OP seemed to primarily be talking about.  I've put a caveat at the start to reflect this.  
> **username_1**: Well of you have to see it, then a consideration of angular resolution is needed. But your calculation is fine for answering whether an infinitesimally narrow sightline intercepts the surface of a star.  
> **username_3**: @ProfRob:  I tweaked my preamble just now to mention "a sufficiently large telescope."  :-)  
> **username_8**: Of course, stars aren't the only things that make light. The interstellar medium is very faint in the optical, but it's present on every line of sight. It's brighter at x-ray wavelengths.  
> **username_7**: @MichaelSeifert I think something must be wrong with your energy calculation because radiated power goes at the 4th power of the temperature.  It should have a bigger effect than you're getting.  
> **username_3**: @LorenPechtel:  I'm calculating the spectral power density at 550 nm ≈ 545 THz, which is the nominal center of the visible magnitude band (using the figures [here](https://en.wikipedia.org/wiki/Apparent_magnitude#Standard_reference_values)).  The power density radiated at a fixed frequency/wavelength doesn't go as the fourth power;  it's just proportional to the $1/(e^{h \nu/kT} - 1)$ factor in the Planck law.  (It's only when you integrate over all frequencies that you get the Stefan-Boltzmann law.)  
> **username_7**: @MichaelSeifert That -6.5 is looking at the whole visual spectrum rather than one frequency.  I don't really think it matters that much overall, though--it's still going to be less than 1 part in a billion.  

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### Answer by username_9 on 2023-02-03. Score: 3

This is supposed to be a very rough ballpark estimate on what percentage of the sky is covered by stars in the milky way galaxy.

There are [approximately](https://en.wikipedia.org/wiki/Milky_Way)  $10^{11}$ stars in the Milky way. First simplification, I will use the size of our sun as the size for all stars. Radius of the sun is $7 \cdot 10^{5} km$ which gives a disk size of $A= 1.5 \cdot 10^{12}km^2$.

Next simplification, I will pretend all stars in the Milky way are at the same distance from us and use the distance to the center of the Milky way. The center of the Milky Way is around 27.000 light years or $2.5\cdot 10^{17}km$ from Earth.

Then we can compute that $10^{11}$ stars of disk size $1.5 \cdot 10^{12}km^2$ at distance $2.5\cdot 10^{17}km$ would only occupy around $10^{11} \cdot 1.5 \cdot 10^{12} km^2 = 1.5\cdot 10^{23} km^2$ out of $4\pi (2.5\cdot 10^{17})^2 km^2=7.8\cdot 10^{35} km^2$ or around $2\cdot 10^{-11} \%$.

One could make this approximation a lot more accurate by using a better distribution of the stars in the Milky way but the key message wouldn't change much, the percentage is very tiny.


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### Answer by username_10 on 2023-02-03. Score: 1

Approximately 100% if you assume the cosmic microwave background to be actually a background and not an object on its own rights.

This is why the sky is black in the first place. It consists of stars and a background.

With the very few exceptions of nearby objects (Moon, Sun, ISS and some nebulae that don't add up for a whole tenth of a percent and if you don't consider Earth as "sky") anything else is point-like.

If you insist on making longer expositions, you will "spill" some initially invisible objects into whole pixels, but it is not fair to count pixels - these objects are geometrically much smaller than the pixel in question.

Galaxies (including, but not limited to, the Milky way) represent themselves as an object with size and shape - again - only because of our limited resolution. Otherwise, they are quite sparse and transparent. Stars themselves have quite a lot of surface brightness, but a whole galaxy is rather faint because it averages "a little" of stars and many orders of magnitude more dark empty space.

> **Comments:**
> **username_11**: Indeed, far away there is a mist of hot hydrogen at 3000K, its not very dense though, 10`18 times LESS than the atmosphere at sea level, 10^13 particles per cm3 of hydrogen.  

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