<p>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.</p>
<p>I did an analysis of this in <a href="https://astronomy.stackexchange.com/a/10260/2531">https://astronomy.stackexchange.com/a/10260/2531</a> , 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 <span class="math-container">$-10^{\circ}$</span>. 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.</p>
<p><em>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.</em></p>
<p><a href="https://i.sstatic.net/lOpa1.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/lOpa1.png" alt="Galactic latitude distribution of bright stars" /></a></p>
<p>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 <span class="math-container">$V&lt;2$</span> 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.</p>
<p><a href="https://i.sstatic.net/RoZcx.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/RoZcx.png" alt="Spatial distribution of stars on an Aitoff RA, Dec projection, with Galactic coordinate grid in red" /></a></p>

CC BY-SA 4.0
2019-12-16T20:17:21.320Z
I read the linked answer too, and it's not what I want. You're plotting by galactic latitude. I want to plot by position of the star, so RA and DEC. One cheap way to do this, just by "eyballing" it, is to find a star map that only shows stars of mag 1 to 2, then a map that only shows stars of 2 to 3, etc. Then you could just look at the map and see if it looks uniform, of if there is more density somewhere. I searched but cudnt find those, but i could make one myself. But I'm not sure if I can do a true statistical analysis to say if its uniform or normal or something else. I will try tho.
DrZ214
CC BY-SA 4.0
2019-12-16T21:06:02.173Z
@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.
ProfRob
CC BY-SA 4.0
2019-12-16T21:57:02.553Z
I think there's a fundamental misunderstanding here. "Spacially uniform" with respect to the surface of the sphere (the night sky dome). Not actual volume. Sorry for the confusion, will edit somehow to be more clear. I should not have used the word spatially. My question is fundamentally 2-dimensional, like those statistical questions where you throw darts at a circle and see if the points are uniformly distributed..
DrZ214
CC BY-SA 4.0
2019-12-16T22:06:36.547Z
@DrZ214 And I have shown you that they are not distributed uniformly on a sphere. The surface density varies with Galactic latitude.
ProfRob
CC BY-SA 4.0
2019-12-16T22:35:00.243Z
@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.
Peter Erwin
CC BY-SA 4.0
2019-12-17T00:11:35.113Z
Is Aitoff Projection an equal area projection? Wikipedia is ambiguous about it. Looks kinda weird to me, hard to eyeball it. Maybe if u remove those red lines? BTW i see what you are saying now. Galactic latitude threw me off, thinking that your chart is just saying there are more stars in that disk of the Galaxy, which is ofc true too. Now i see it is also saying there are more in the galactic latitude regions of the sky we see. However, I am still doing my own research for this. I just finished the Mag 1 to 2 stars, in a Lambert Cylindrical Projection, but there are too few as you said.
DrZ214
CC BY-SA 4.0
2019-12-17T00:11:48.860Z
Sadly the dumb wiki list of brightest stars only goes to mag 2.5. Can you show me a catalogue in order of apparent magnitude going to at least 5.0?
DrZ214
CC BY-SA 4.0
2019-12-17T05:43:57.193Z
@DrZ214 [ArcGIS](https://desktop.arcgis.com/en/arcmap/10.3/guide-books/map-projections/aitoff.htm) says "Neither conformal nor equal area."
Keith McClary
CC BY-SA 4.0
2019-12-17T05:48:07.620Z
@DrZ214 But [Caltech/NASA](https://irsa.ipac.caltech.edu/IRASdocs/surveys/coordproj.html) says "The Aitoff equal area projection ...".
Keith McClary
CC BY-SA 4.0
2019-12-17T07:16:32.160Z
@DrZ214 and why would I remove the lines that show you that the stars (as totally expected) are concentrated towards the Galactic plane?
ProfRob
CC BY-SA 4.0
2019-12-18T01:38:08.167Z
@RobJeffries It absolutely does matter. The whole point of the question was about uniform distribution on a sphere, so an equal area projection is vital if we're going to "eyeball" it. I posted my own answer after doing some work, and explained it more there. I found your answer useful, but flagged your comment as rude. I am sorry if there was some misunderstanding earlier, but if it gets to the point of all caps and exclamation marks, my advice is to just forget about it for a while.
DrZ214
CC BY-SA 4.0
2019-12-18T21:19:36.757Z
It's also worth noting that many of the brightest stars in the night sky are part of the Gould Belt, which is very much not uniform: https://en.wikipedia.org/wiki/Gould_Belt
J. O'Brien Antognini
CC BY-SA 4.0
2019-12-16T10:58:16.837Z
2021-07-10T17:40:27.703Z
2021-07-10T17:40:27.703Z
ProfRob
<p>Alright I finally finished this program so I could take a look at each tier individually and see for myself.</p>

<p>First of all, the projection type does indeed matter, so I will explain it here. It needs to be an <strong>equal-area projection</strong>. 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.</p>

<p>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 <strong>Greenland appears the same size as South America.</strong></p>

<p><a href="https://i.sstatic.net/GJej2.jpg" rel="noreferrer"><img src="https://i.sstatic.net/GJej2.jpg" alt="enter image description here"></a></p>

<p>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 <strong>Greenland is actually much smaller than South America.</strong></p>

<p><a href="https://i.sstatic.net/zA4ga.jpg" rel="noreferrer"><img src="https://i.sstatic.net/zA4ga.jpg" alt="enter image description here"></a></p>

<p>I illustrated this because, <strong>when it comes to points of light, distortions are not so obvious</strong>. 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.</p>

<p>This is a <strong>random starfield</strong> in <a href="https://en.wikipedia.org/wiki/Cylindrical_equal-area_projection" rel="noreferrer">Cylindrical Equal-Area Projection</a>.</p>

<p><a href="https://i.sstatic.net/6kTJZ.png" rel="noreferrer"><img src="https://i.sstatic.net/6kTJZ.png" alt="enter image description here"></a></p>

<p>This is the <strong>same starfield</strong> in <a href="https://en.wikipedia.org/wiki/Equirectangular_projection" rel="noreferrer">Equi-Rectangular Projection</a>.</p>

<p><a href="https://i.sstatic.net/YOd2b.png" rel="noreferrer"><img src="https://i.sstatic.net/YOd2b.png" alt="enter image description here"></a></p>

<p>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.</p>

<hr>

<p>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.</p>

<p>Now I'll show the actual starfield of our sky, based on the <a href="https://ian.macky.net/pat/yalebsc/index.html" rel="noreferrer">Yale Bright Star Catalog</a>. Thanks to user:RobJeffries for pointing it out. I took stars from Sirius to 4.99 apparent magnitude. This was 1,602 stars.</p>

<p><a href="https://i.sstatic.net/FI4IW.png" rel="noreferrer"><img src="https://i.sstatic.net/FI4IW.png" alt="enter image description here"></a></p>

<p>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.</p>

<p>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.</p>

<p>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.</p>

<p><a href="https://i.sstatic.net/UZuUB.png" rel="noreferrer"><img src="https://i.sstatic.net/UZuUB.png" alt="enter image description here"></a></p>

<p><a href="https://i.sstatic.net/oLunH.png" rel="noreferrer"><img src="https://i.sstatic.net/oLunH.png" alt="enter image description here"></a></p>

<p><a href="https://i.sstatic.net/NBTqW.png" rel="noreferrer"><img src="https://i.sstatic.net/NBTqW.png" alt="enter image description here"></a></p>

<p><a href="https://i.sstatic.net/VJEgd.png" rel="noreferrer"><img src="https://i.sstatic.net/VJEgd.png" alt="enter image description here"></a></p>

<p><a href="https://i.sstatic.net/fKvSF.png" rel="noreferrer"><img src="https://i.sstatic.net/fKvSF.png" alt="enter image description here"></a></p>

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

<p>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.)</p>

<p><a href="https://i.sstatic.net/6OGk3.png" rel="noreferrer"><img src="https://i.sstatic.net/6OGk3.png" alt="enter image description here"></a></p>

<p><a href="https://i.sstatic.net/yxDKq.png" rel="noreferrer"><img src="https://i.sstatic.net/yxDKq.png" alt="enter image description here"></a></p>

<p><a href="https://i.sstatic.net/paoot.png" rel="noreferrer"><img src="https://i.sstatic.net/paoot.png" alt="enter image description here"></a></p>

<p>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.</p>

<hr>

<p>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.</p>

<p>I am satisfied, of course, that the stars in the sky are <strong>not</strong> 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.</p>

<hr>

<p>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.</p>

<pre><code>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, "");
}
</code></pre>

<p>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 <strong>not</strong> form a uniform distribution. <a href="http://mathworld.wolfram.com/SpherePointPicking.html" rel="noreferrer">Wolfram Alpha</a> has more information and I'm not sure I understand it fully myself. For some reason you have to use that arccos.</p>

<p>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.</p>

<pre><code>// 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;
</code></pre>

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

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

<p>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.</p>

<pre><code>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, "");
</code></pre>

<p>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). <strong>Warning,</strong> the minutes are <strong>NOT</strong> 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.</p>

<p>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.</p>

<p>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.</p>

CC BY-SA 4.0
2019-12-18T13:19:37.630Z
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?
JiK
CC BY-SA 4.0
2019-12-18T13:36:08.470Z
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.
JiK
CC BY-SA 4.0
2019-12-18T13:59:03.027Z
https://astronomy.stackexchange.com/questions/22388/generate-an-uniform-distribution-on-the-sky
ProfRob
CC BY-SA 4.0
2019-12-18T21:50:52.473Z
@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.
DrZ214
CC BY-SA 4.0
2021-07-11T00:17:30.393Z
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?](https://earthscience.stackexchange.com/q/8242/6031) in Earth Science SE.
uhoh
CC BY-SA 4.0
2021-12-01T09:11:20.267Z
Very cool analysis - I have derived from a similar argument the distribution of the brightness of the stars in https://laurentperrinet.github.io/sciblog/posts/2021-03-27-density-of-stars-on-the-surface-of-the-sky.html
meduz
CC BY-SA 4.0
2019-12-18T01:30:51.400Z
2019-12-18T01:30:51.400Z
DrZ214
<p>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.</p>
<p>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.</p>
<p><strong>EDIT:</strong> 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?</p>
<p>Btw, the &quot;obvious&quot; distribution of theta from 0 to 2pi and phi from 0 to pi will <strong>NOT</strong> produce a uniform distribution. That will produce clustering near the poles. <a href="http://mathworld.wolfram.com/SpherePointPicking.html" rel="nofollow noreferrer">Wolfram Alpha</a> explains it.)</p>
<p>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.</p>
<p>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?</p>

CC BY-SA 4.0
2019-12-16T08:44:28.093Z
Before or after correcting for the Milky Way?
Phiteros
CC BY-SA 4.0
2019-12-16T08:46:03.177Z
@Phiteros Before. Milky Way stars get pretty dim. I shuda put that in the OP, will edit.
DrZ214
CC BY-SA 4.0
2019-12-16T08:46:56.847Z
True. But just the fact that there are more of them would throw off the results.
Phiteros
CC BY-SA 4.0
2019-12-16T08:51:31.817Z
@Phiteros Yeah. But you can't just cut off a certain slice of area when talking about uniform distribution on a sphere. You have to cut off by brightness. Made the edit. And honestly there is gonna be a cutoff point by brightness somewhere, otherwise you are talking about a near-infinite number of stars. Magnitude 1 to 4, or 1 to 5, is what I want, but i don't know the best cutoff magnitude to avoid Milky Way bias.
DrZ214
CC BY-SA 4.0
2019-12-16T10:26:34.093Z
Why would you expect it to be spatially uniform?
ProfRob
CC BY-SA 4.0
2019-12-16T19:41:08.973Z
Seeing as we defined magnitude for some time to be based off an arbitrary star (Vega) and even though we no longer directly define it there, it is scaled to be similar to that, it seems unlikely that breaking it up into bands of 1.0 and 2.0 etc would give any particular scientific insight.
corsiKa
CC BY-SA 4.0
2019-12-16T20:20:36.090Z
@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.
DrZ214
CC BY-SA 4.0
2021-07-10T20:46:14.640Z
Does this answer your question? [Aren't there more naked-eye-visible stars in the Milky Way plane?](https://astronomy.stackexchange.com/questions/6005/arent-there-more-naked-eye-visible-stars-in-the-milky-way-plane)
James K
<night-sky><mathematics><sky-survey>
Are the stars distributed in uniform distribution, on the celestial dome, with respect to brightness?
DrZ214
# Are the stars distributed in uniform distribution, on the celestial dome, with respect to brightness?

(Asked by: username_0 on 2019-12-16)

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](http://mathworld.wolfram.com/SpherePointPicking.html) 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?

**Tags:** night-sky, mathematics, sky-survey
> **Comments:**
> **username_1**: Before or after correcting for the Milky Way?  
> **username_0**: @Phiteros Before. Milky Way stars get pretty dim. I shuda put that in the OP, will edit.  
> **username_1**: True. But just the fact that there are more of them would throw off the results.  
> **username_0**: @Phiteros Yeah. But you can't just cut off a certain slice of area when talking about uniform distribution on a sphere. You have to cut off by brightness. Made the edit. And honestly there is gonna be a cutoff point by brightness somewhere, otherwise you are talking about a near-infinite number of stars. Magnitude 1 to 4, or 1 to 5, is what I want, but i don't know the best cutoff magnitude to avoid Milky Way bias.  
> **username_2**: Why would you expect it to be spatially uniform?  
> **username_3**: Seeing as we defined magnitude for some time to be based off an arbitrary star (Vega) and even though we no longer directly define it there, it is scaled to be similar to that, it seems unlikely that breaking it up into bands of 1.0 and 2.0 etc would give any particular scientific insight.  
> **username_0**: @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.  
> **username_4**: Does this answer your question? [Aren't there more naked-eye-visible stars in the Milky Way plane?](https://astronomy.stackexchange.com/questions/6005/arent-there-more-naked-eye-visible-stars-in-the-milky-way-plane)  

---

## 2 answers

### Answer by username_2 on 2019-12-16. Score: 44

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](https://i.sstatic.net/lOpa1.png)](https://i.sstatic.net/lOpa1.png)

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](https://i.sstatic.net/RoZcx.png)](https://i.sstatic.net/RoZcx.png)

> **Comments:**
> **username_0**: I read the linked answer too, and it's not what I want. You're plotting by galactic latitude. I want to plot by position of the star, so RA and DEC. One cheap way to do this, just by "eyballing" it, is to find a star map that only shows stars of mag 1 to 2, then a map that only shows stars of 2 to 3, etc. Then you could just look at the map and see if it looks uniform, of if there is more density somewhere. I searched but cudnt find those, but i could make one myself. But I'm not sure if I can do a true statistical analysis to say if its uniform or normal or something else. I will try tho.  
> **username_2**: @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.  
> **username_0**: I think there's a fundamental misunderstanding here. "Spacially uniform" with respect to the surface of the sphere (the night sky dome). Not actual volume. Sorry for the confusion, will edit somehow to be more clear. I should not have used the word spatially. My question is fundamentally 2-dimensional, like those statistical questions where you throw darts at a circle and see if the points are uniformly distributed..  
> **username_2**: @DrZ214 And I have shown you that they are not distributed uniformly on a sphere. The surface density varies with Galactic latitude.  
> **username_5**: @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.  
> **username_0**: Is Aitoff Projection an equal area projection? Wikipedia is ambiguous about it. Looks kinda weird to me, hard to eyeball it. Maybe if u remove those red lines? BTW i see what you are saying now. Galactic latitude threw me off, thinking that your chart is just saying there are more stars in that disk of the Galaxy, which is ofc true too. Now i see it is also saying there are more in the galactic latitude regions of the sky we see. However, I am still doing my own research for this. I just finished the Mag 1 to 2 stars, in a Lambert Cylindrical Projection, but there are too few as you said.  
> **username_0**: Sadly the dumb wiki list of brightest stars only goes to mag 2.5. Can you show me a catalogue in order of apparent magnitude going to at least 5.0?  
> **username_6**: @DrZ214 [ArcGIS](https://desktop.arcgis.com/en/arcmap/10.3/guide-books/map-projections/aitoff.htm) says "Neither conformal nor equal area."  
> **username_6**: @DrZ214 But [Caltech/NASA](https://irsa.ipac.caltech.edu/IRASdocs/surveys/coordproj.html) says "The Aitoff equal area projection ...".  
> **username_2**: @DrZ214 and why would I remove the lines that show you that the stars (as totally expected) are concentrated towards the Galactic plane?  
> **username_0**: @RobJeffries It absolutely does matter. The whole point of the question was about uniform distribution on a sphere, so an equal area projection is vital if we're going to "eyeball" it. I posted my own answer after doing some work, and explained it more there. I found your answer useful, but flagged your comment as rude. I am sorry if there was some misunderstanding earlier, but if it gets to the point of all caps and exclamation marks, my advice is to just forget about it for a while.  
> **username_7**: It's also worth noting that many of the brightest stars in the night sky are part of the Gould Belt, which is very much not uniform: https://en.wikipedia.org/wiki/Gould_Belt  

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### Answer by username_0 on 2019-12-18. Score: 11

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.

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.

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.

This is the same starfield in Equi-Rectangular Projection.

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.

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.





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.)



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.

> **Comments:**
> **username_8**: 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?  
> **username_8**: 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.  
> **username_2**: https://astronomy.stackexchange.com/questions/22388/generate-an-uniform-distribution-on-the-sky  
> **username_0**: @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.  
> **username_9**: 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?](https://earthscience.stackexchange.com/q/8242/6031) in Earth Science SE.  
> **username_10**: Very cool analysis - I have derived from a similar argument the distribution of the brightness of the stars in https://laurentperrinet.github.io/sciblog/posts/2021-03-27-density-of-stars-on-the-surface-of-the-sky.html  

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