Has there ever been a statistical analysis of the distribution of stars in the sky (on the surface of the celestial dome), by brightness? I want to know if they are uniformly distributed.

For example, all stars with Apparent Magnitude from 1.0 to 2.0. Are these uniformly distributed? Then all stars from 2.0 to 3.0, etc.

EDIT: To avoid bias for the Milky Way, there should be some cutoff magnitude. I don't know exactly what's the best cutoff, but I would guess somewhere around 4.0 to 5.0 apparent magnitude. If the cutoff is 4.0, then that gives us 3 tiers to look at (1 to 2, 2 to 3, 3 to 4). Are each of those tiers uniformly distributed?

Btw, the "obvious" distribution of theta from 0 to 2pi and phi from 0 to pi will NOT produce a uniform distribution. That will produce clustering near the poles. Wolfram Alpha explains it.)

I did google around for this. The results are all about star clusters or binaries, not the actual starfield of the night sky, or sometimes about how to program a random set of stars on the sphere.

That's not what I want. I'm looking for a statistical analysis on our own starfield. Has it ever been done before? What were the results?

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    $\begingroup$ Before or after correcting for the Milky Way? $\endgroup$
    – Phiteros
    Commented Dec 16, 2019 at 8:44
  • $\begingroup$ @Phiteros Before. Milky Way stars get pretty dim. I shuda put that in the OP, will edit. $\endgroup$
    – DrZ214
    Commented Dec 16, 2019 at 8:46
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    $\begingroup$ Why would you expect it to be spatially uniform? $\endgroup$
    – ProfRob
    Commented Dec 16, 2019 at 10:26
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    $\begingroup$ @RobJeffries I don't actually expect it to be spatially uniform. I don't know what to expect. If you assume the distribution of bright stars to be uniform random thruout space, then i suppose you would expect the distribution in the night sky to appear uniform too. But I dont think the first assumption is safe. The shape of the galaxy will obviously influence things, i just dont know how much for the "bright" stars, and if there will be influence in the vertical directions too. In other words, im not expecting one way or the other, just asking which way it is. $\endgroup$
    – DrZ214
    Commented Dec 16, 2019 at 20:20
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    $\begingroup$ Does this answer your question? Aren't there more naked-eye-visible stars in the Milky Way plane? $\endgroup$
    – James K
    Commented Jul 10, 2021 at 20:46

2 Answers 2


Naked eye stars are not distributed uniformly in the sky. That is because the median naked eye star is at a distance of 440 light years, and this is far enough away that some of the details of Galactic structure start to become apparent. Most importantly, the density of stars increase towards the Galactic midplane and has a scale height of a few hundred light years and thus there is an over-density of stars towards the Galactic plane.

I did an analysis of this in https://astronomy.stackexchange.com/a/10260/2531 , but the important plot is repeated below. It shows the normalised density of stars (per square degree) as a function of Galactic latitude. (Note that I have correctly handled the changing area of the latitude strips). You can see the peak at low Galactic latitude and even work out that our viewing point is probably slightly above the Galactic plane, given that the peak is at about $-10^{\circ}$. The peak is still there in 3rd magnitude stars. You could repeat for a sample limited to even brighter stars, but you run into number statistics trouble.

The distribution of bright stars with Galactic latitude. The y-axis is the normalised number of stars per square degree. i.e. The integral sums to 1.

Galactic latitude distribution of bright stars

Below I show the Hipparcos data, divided into 4 apparent magnitude bins. Points are plotted on a RA vs Dec Aitoff projection. Superimposed in red is a Galactic coordinate grid. There are too few stars with $V<2$ to form much of an opinion, but for fainter stars, a ring-like structure, coincident with the Galactic coordinates equator (i.e. Galactic latitude of zero, or just below, as I showed in the 1D projection above) is clearly seen.

Spatial distribution of stars on an Aitoff RA, Dec projection, with Galactic coordinate grid in red

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    $\begingroup$ @DrZ214 you asked whether it is (spatially) uniform. The answer is that it isn't. I fail to see how that does not answer the question you actually asked. I plotted by Galactic latitude because that is what the density depends on. There is no clear dependence on Galactic longitude and no reason at all why there should be a direct dependence on RA and Dec. $\endgroup$
    – ProfRob
    Commented Dec 16, 2019 at 21:06
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    $\begingroup$ @DrZ214 And I have shown you that they are not distributed uniformly on a sphere. The surface density varies with Galactic latitude. $\endgroup$
    – ProfRob
    Commented Dec 16, 2019 at 22:06
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    $\begingroup$ @DrZ214 -- Galactic latitude is a coordinate on the surface of the celestial sphere (paired with Galactic longitude as the orthogonal coordinate); it has nothing to do with volume. They are many different ways you can draw coordinates on the celestial sphere: RA and Dec, ecliptic latitude and longitude, Galactic latitude and longitude, etc. $\endgroup$ Commented Dec 16, 2019 at 22:35
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    $\begingroup$ @DrZ214 ArcGIS says "Neither conformal nor equal area." $\endgroup$ Commented Dec 17, 2019 at 5:43
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    $\begingroup$ @DrZ214 and why would I remove the lines that show you that the stars (as totally expected) are concentrated towards the Galactic plane? $\endgroup$
    – ProfRob
    Commented Dec 17, 2019 at 7:16

Alright I finally finished this program so I could take a look at each tier individually and see for myself.

First of all, the projection type does indeed matter, so I will explain it here. It needs to be an equal-area projection. The whole point of the question was about a uniform distribution of stars over the surface of a sphere. In other words, each area of the sphere would tend to have the same amount of stars as each other area. Therefore, a projection that preserves equal areas is necessary.

Every map projection type causes some kinds of distortion. It is inevitable. It's in the nature of geometry when taking a sphere and flattening it to a rectangle, or other flat shape. You can see this in the equi-rectangular map of Earth, where Greenland appears the same size as South America.

enter image description here

But that's of course wrong, an artifact of distortion. Take a look at the globe, or an area-preserving projection such as this cylindrical equal-area one, and you'll see that Greenland is actually much smaller than South America.

enter image description here

I illustrated this because, when it comes to points of light, distortions are not so obvious. You don't have the luxury of familiar terrain shapes like Greenland or continents, to tell distortion. So now I'll show a randomly generated starfield, in two projections.

This is a random starfield in Cylindrical Equal-Area Projection.

enter image description here

This is the same starfield in Equi-Rectangular Projection.

enter image description here

Both of them look pretty uniform, but the last one has a lack of stars near the poles. Without understanding what projection is used, and their pitfalls, you could think that the second one is not uniform.

From here on out I will use the Cylindrical Equal-Area Projection unless otherwise noted. These put the 0 RA 0 DEC point in the center, so the poles are at the top and bottom.

Now I'll show the actual starfield of our sky, based on the Yale Bright Star Catalog. Thanks to user:RobJeffries for pointing it out. I took stars from Sirius to 4.99 apparent magnitude. This was 1,602 stars.

enter image description here

It's a little difficult to discern, but there is a path where stars tend to cluster around the Milky Way. It's hard to see on the left side of the map. On the right side it's easier.

By the way, I did not draw brighter/dimmer stars differently, such as a smaller or grayer point of light. I wanted each point to show up just as well as each other point, because for one thing, I planned to look at each tier individually and in progressing composites.

Here are the fields only looking at one tier at a time. There are 5. The first includes stars from Sirius to 0.99 apparent magnitude. The second goes from 1.0 to 1.99, etc, ending with 4.0 to 4.99.

enter image description here

enter image description here

enter image description here

enter image description here

enter image description here

In my opinion, the only one that could possibly be close to uniform distribution is tier 4, the stars with apparent mags from 3.0 to 3.99.

So now here are the composites. There are 3 of them. The first has tiers 1 and 2 composited together, The second has tiers 1 to 3, and the last has tiers 1 to 4. (All tiers, 1 to 5, was done in the original screen shot of our starfield.)

enter image description here

enter image description here

enter image description here

I suppose the closest one to uniform is the last one, but it's not as close as tier 4 by itself. I'm not really sure why that is or if it's just a coincidence.

Finally, what I want to do is an actual numerical/statistical analysis on the coordinates, at least for tier 4. Sadly I cannot remember how to do that. It's been too long since my stochastic models class in undergrad, and I'm not even sure if we ever learned how to do that. I will search around and try. If I get it done, I will come back here and edit this answer.

I am satisfied, of course, that the stars in the sky are not uniformly distributed. I wanted to post this answer because I did the actual work and wanted to explain some things. It took maybe 5 hours of work, spread over 2 days. 2 hours were a total waste tho, copying the 1.0 to 1.99 stars from Wikipedias list. Another hour just searching for star catalog in a format or interface that I could understand. And now that I look at the time, it's taken almost an hour just to type up this answer, which seems impossible.

Now I will show source code for a few things. First, the code to generate a uniformly random starfield, because it's not as simple as you wold think.

repeat (1000)
  hdeg = 360 * random(1);
  //vdeg = 180 * random(1) - 90;
  vdeg = arccos(2*random(1) - 1)*180/pi - 90;

  scr_ini_star2(hdeg, vdeg, 0, "");

I left that middle line commented out, to show that it's wrong. The "obvious" solution is just a longitude from 0 to 360, and a latitude from -90 to 90. But that does not form a uniform distribution. Wolfram Alpha has more information and I'm not sure I understand it fully myself. For some reason you have to use that arccos.

Next is the code to translate the spherical coords to xy coords, or in other words, map a projection. Notice that I made maps in an 800 x 400 pxl format. I chose this to make it easier to switch between equi-rectangular and cylindrical equal-area.

// cylindrical equal-area

x = (400 + RA/360*800) mod 800;
y = 200 - sin(degtorad(DEC))*200;

// equirectangular projection

x = (400 + RA*800/360) mod 800;
y = 200 - DEC*200/90;

The RA/DEC translation code, from hms or dms to degrees:

  RA = (argument0*3600 + argument1*60 + argument2)*360/86400;
  DEC = argument3 + argument4/60;

I wanted to post the entire body of star list initialize code, but discovered a character limit of 30,000 on stack exchange. It was only 1,602 stars... Learn something new every day, but here is a small sample.

scr_ini_star(18, 36, 56, +38, +47, 0.04, "");
scr_ini_star(05, 16, 41, +46, +00, 0.06, "");
scr_ini_star(05, 14, 32, -08, -12, 0.15, "");
scr_ini_star(14, 39, 36, -60, -50, 0.33, "");
scr_ini_star(07, 39, 18, +05, +14, 0.36, "");
scr_ini_star(01, 37, 42, -57, -15, 0.49, "");
scr_ini_star(14, 03, 50, -60, -22, 0.61, "");
scr_ini_star(19, 50, 47, +08, +52, 0.75, "");
scr_ini_star(12, 26, 36, -63, -06, 0.80, "");
scr_ini_star(05, 55, 10, +07, +24, 0.80, "");
scr_ini_star(04, 35, 55, +16, +30, 0.86, "");
scr_ini_star(13, 25, 11, -11, -09, 0.97, "");

The first 3 fields are the hours, minutes, and seconds of the Right Ascension (RA). The next 2 fields are the degrees and minutes for the Declination (DEC). Warning, the minutes are NOT the same thing there. The minutes in DEC are minutes of arc, where 60 minutes is in 1 degree. The minutes in RA are actual minutes of an hour. 1 hour is 15 degrees, and 60 minutes in an hour.

Another warning. With negative declinations, you have to put the minus sign on the minutes too...and seconds if they're there. The database file I found did not have seconds in DEC, but I don't think that level of accuracy was needed for an 800x400 map.

I made this little project in Game Maker 8.1, an old IDE from the late 2000s and early 2010s, but still the fastest way to do small graphical calcs for me. It treats every variable as a double, so I never had to cast anything as a float or double before doing division.

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    $\begingroup$ I don't understand. The question is well-defined in the celestial sphere. Why do you need to think of any kinds of projections at all? $\endgroup$
    – JiK
    Commented Dec 18, 2019 at 13:19
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    $\begingroup$ Oh, apparently the point of this answer is just to eyeball the projection and see if it looks uniformly distributed? In that case, the answer is fine (although I'd remove the discussion about projections and coordinate transforms and just link to stats.SE or something). But the question explicitly asks about statistical analysis, so eyeballing doesn't seem to be the best answer. $\endgroup$
    – JiK
    Commented Dec 18, 2019 at 13:36
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    $\begingroup$ astronomy.stackexchange.com/questions/22388/… $\endgroup$
    – ProfRob
    Commented Dec 18, 2019 at 13:59
  • $\begingroup$ @JiK It's not the best answer possible. It definitely needs a numerical analysis. The point of "eyeballing" it was, if it's just binary question (yes/no), then a non-uniform distribution can usually be seen right away just by looking at the cluster areas. It was how I wanted to start the answer, but I want to go deeper of course. I need to learn more about statistics to do that, tho. $\endgroup$
    – DrZ214
    Commented Dec 18, 2019 at 21:50
  • $\begingroup$ This is a great answer and illustrates the issue nicely. More fun maps in answers to Is there a map that displays every country at its correct relative size? in Earth Science SE. $\endgroup$
    – uhoh
    Commented Jul 11, 2021 at 0:17

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