<p>Firstly, the galaxy is only about <a href="http://imagine.gsfc.nasa.gov/docs/ask_astro/answers/980317b.html" rel="noreferrer">1000ly thick</a>. We are fairly close to the galactic plane, <a href="http://www.nature.com/nature/journal/v316/n6030/abs/316706a0.html" rel="noreferrer">maybe around 65 ly 'above' it</a> if we call the direction we are moving away from 'down'. So on your assumption that visible objects are all within about 1000ly, we can suppose that we should see more stars in the plane than above and below the plane, as there is only about 500ly of stars in the galaxy above and below. </p>

<p>We are perhaps <a href="http://iopscience.iop.org/1538-4357/597/2/L121/" rel="noreferrer">25kly</a> from the galactic centre. Therefore there are about 25kly of stars outwards, and 75kly of stars inwards. If we could see all of them with the naked eye, we would see more stars to one side than the other. While many of the stars that we see are within 1000ly, that doesn't mean that the stars within this region of visibility are evenly distributed. <a href="http://www.astronomy.ohio-state.edu/~ryden/ast162_7/notes31.html" rel="noreferrer">Stars are packed closer together nearer the centre</a>, which implies that we will see more towards one side than the other.</p>

<p>I don't know where you get the idea you seek confirmation for that the density of stars is not higher in the galactic plane, because it isn't true. Density is greater toward the centre of the galaxy; and orthogonal to the galactic plane, density decreases as you move away from it.</p>

<p>Plus, we see far more than merely dots of light from other stars. We see other objects that are aggregations of stars - clusters, and galaxies like Andromeda, the LMC and the SMC.  We see nebulae, clouds of gas and dust. In fact, dust even obscures what we might otherwise see toward the galactic centre. </p>

<p>So when you ask 'what is it we see?' the answer is: lots of things. Because there is more of that 'stuff' in the plane of the galaxy, we see the majority of that stuff as a ring around us. Because there is more of that stuff towards the centre of the galaxy, we see that the ring is brighter, fuller, more complex, and dominant on one side.</p>

CC BY-SA 3.0
2014-08-06T03:42:10.863Z
I guess, you meant $25 kly = 8 kpc$.
Py-ser
CC BY-SA 3.0
2014-08-06T11:04:15.780Z
Eeeek yup - I'll update, and perhaps to can remove comment.
Jeremy
CC BY-SA 3.0
2014-08-06T12:47:27.840Z
Almost all stars visible to the naked eyes are within 1000 light years. Most of them in turn within a fraction of that distance. Is that disputed? Any density differences on the sky is a function of the internal shape of the Orion arm and our location inside it. I doubt that the Orion arm is denser towards the galactic center. The structure of the Milky Way hardly affects the density of visible stars which are all located between 24 and 26 kly from the center. "Local" effects certainly dominate. And some other claims online say that the Orion arm has a diameter of 3500 ly, not 1000.
LocalFluff
CC BY-SA 3.0
2014-08-07T01:28:23.933Z
It is 3500ly across, not in diameter. Furthermore, I don't get you. Are you seriously disputing the reality of easily made observations, are you seriously claiming that stars should all appear evenly distributed across the sky, because in your mind you've decided that it ought to be this way, based on a couple of things you have read and misunderstood?
Jeremy
CC BY-SA 3.0
2014-08-05T22:52:07.240Z
2014-08-06T11:05:20.680Z
2014-08-06T11:05:20.680Z
Jeremy
<p>This is known as <a href="http://en.m.wikipedia.org/wiki/Olbers%27_paradox" rel="nofollow">Olbers' paradox</a>. An (in my opinion) better explanation can be found <a href="http://curious.astro.cornell.edu/question.php?number=375" rel="nofollow">here</a>. </p>

<p>Note that these links answer the question why the universe (not the milky way alone) isn't blazing our night sky. Now if the universe (including the milky way) can't do it, the milky way can't do it on its own... </p>

<p>There are just not enough stars giving enough light. Brightness is reduced quadratically by distance. That counts up very quickly with astronomical distances. Even for stars that are our neighbors. </p>

CC BY-SA 3.0
2014-08-05T18:08:14.323Z
My impression is that Olbers' paradox is about diffuse light from far far away. Like that which reaches us from the Milky way. Which it indeed does and it glows. But beyond the ability of human unaided eyes to resolve individual stars. My question is about such stars of about 6 magnitude or more visible. Are naked eye visible stars more plentiful towards to Milky Way, or not? I presume, based on their distance and the structure of the MW, that they should not. But I'm open to be wrong about this. That's why I ask.
LocalFluff
CC BY-SA 3.0
2014-08-05T19:08:48.443Z
Olber's Paradox is usually used in regard to the infinite universe model. I suppose, though, that this is a related question. I don't know if it's a paradox, though.
HDE 226868
CC BY-SA 3.0
2014-08-05T15:24:22.787Z
2014-08-05T15:24:22.787Z
agtoever
<p>You can tell a lot about Galactic structure by just looking and indeed the distribution of bright stars is non-spherical and non uniform.</p>
<p>The ~5000 stars that can be seen with the naked eye have a roughly &quot;lognormal&quot; distribution of distance. I show plots below which were generated from the most recent version of the Hipparcos parallax catalogue. Fig.1 shows results for all stars with <span class="math-container">$5.5&lt;V&lt;6.5$</span>  (i.e. very faint naked eye stars). The Hipparcos catalogue is almost complete for these stars, though some stars are so far away that the distance is highly uncertain - nevertheless, the general picture should be ok.</p>
<p>The median distance is about 440 light years. But the premise of your question is I think, that you dispute that this is far enough for the non-spherical distribution of stars in our Galaxy to become apparent. The answer is actually that it is, but only just. The Sun is very close to the plane of the disk of our Galaxy. The scale height of stars above this disk varies depending on stellar age and mass. Very approximately, the exponential scale height is 300-500 light years for most of the stars in our Galaxy.</p>
<p>This is just small enough that if we look at the demographics of stars in two Galactic latitude regions we <em>do</em> see a difference. Fig. 2 below shows the distance distribution of stars with <span class="math-container">$5.5&lt;V&lt;6.5$</span> at low Galactic latitudes (within 15 degrees of the plane in green) and more than 45 degrees from the plane (in blue). There is a clear and significant difference. More stars can be seen at greater distances towards the plane, despite the greater dust extinction in those directions,betraying the non spherically-symmetric nature of stars in our Galaxy.</p>
<p>In other words the structure of the Galaxy <em>is not</em> too small to affect the distribution of naked eye stars.</p>
<p><img src="https://i.sstatic.net/nrkJX.png" alt="Distribution of stellar distances for naked eye stars" /></p>
<p><em>Fig 1: Probability distribution of distance for naked eye stars in the Hipparcos catalogue</em></p>
<p><img src="https://i.sstatic.net/I5pyX.png" alt="Distribution of distances for naked eye stars in two Galactic latitude zones" /></p>
<p><em>Fig 2: Probability distribution of distance for naked eye stars divided into low and high Galactic latitude regions (i.e. towards and away from the Galactic plane)</em>.</p>
<p>I then split the Hipparcos sample into bright (<span class="math-container">$V&lt;6$</span>, 5000 stars) and very bright (<span class="math-container">$V&lt;3$</span>, 173 stars) and looked at the distribution of stars per unit area on the sky as a function of Galactic latitude (Fig. 3). The results are shown below and more obviously make the point. It turns out that the asymmetry between in an out of the plane is clearly seen, even in the brighter sample. The reason is that the very bright stars are not <em>all that much</em> closer than the bright stars. Perhaps a median of 200 light years vs 350 light years. Thus the scale height of the Galactic disc must be small enough that the difference from spherical symmetry is already apparent at these kinds of distances.</p>
<p><img src="https://i.sstatic.net/lOpa1.png" alt="Distribution of stars vs Galactic latitude" /></p>
<p><em>Fig. 3: Normalised plot of number of stars per unit area versus Galactic latitude for a sample of bright and very bright stars. Note the concentration towards the Galactic plane</em>.</p>
<p>You can even see from the plot above that there is also a slight concentration towards negative latitudes; the peak density is around -5 to -10 degrees in both cases. This is likely because the Sun is above the Galactic plane at present (though I also wonder if dust plays a role). The Sun is currently 60 light years above the plane and heading upwards (see <a href="https://astronomy.stackexchange.com/questions/822/how-far-is-the-earth-sun-above-below-the-galactic-plane-and-is-it-heading-towar">How far is the Earth/Sun above/below the galactic plane, and is it heading toward/away from it?</a> ). The combination of the results I have shown <em>may</em> be sufficient to conclude that the Galactic disc has a &quot;characteristic thickness* not more than a few hundred light years, but more than 60 light years!</p>
<p>When we look at the Milky Way we are seeing millions of unresolved stars that are in general more distant still. The effects of the disk-like nature of the Galaxy become more apparent as one goes to greater distances. In particular, at high Galactic latitudes one simply runs out of stars and hence there is no uniformly illuminated sky - leading to Olber's paradox, as correctly stated in another answer. But at low latitudes, there are sufficient stars that in general, within the resolution of our eyes there are many stars whose light sums to provide a visual stimulus. This picture is interrupted by <em>dust</em>. Dust in the Galaxy is even more concentrated towards the plane than the stars. It is for this reason that beyond a few thousand light years, dust plays a major role in shaping the Milky Way structures that we can see, effectively blocking light at very low Galactic latitudes beyond a few <em>thousand</em> light years.</p>

CC BY-SA 4.0
2015-03-25T12:31:38.183Z
2021-07-10T17:34:22.497Z
2021-07-10T17:34:22.497Z
ProfRob
<p>Most stars which are visible to the naked eye are within 1,000 light years. The Sun is inside the Orion arm which has a diameter of about 3,500 light years. Thus, all stars (with very few exceptions) we can see unaided are inside the Orion arm and should be equally distributed in all directions. The structure of the galaxy is too large to affect the distribution of visible stars.</p>

<p>Still, we do see the Milky Way disk. What is it we see? The dim light of stars we can't make out individually? Or star light reflected by gas clouds? And is the density of stars to the unaided eye really not higher in the plane of the Milky Way? (Since I'm living in a light polluted city I can't easily check it out myself)</p>

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2014-08-01T20:51:57.537Z
Please provide references for your claims. I think first we need to address some misunderstandings
Jeremy
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2014-08-02T06:54:31.060Z
Here are some claims about the Sun's place in the Milky Way http://www.atlasoftheuniverse.com/5000lys.html
LocalFluff
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2014-08-02T20:29:51.307Z
I couldn't see anything there claiming there is 1000ly of visible stars in all directions of our position. Actually, while the Orion Arm is 3500ly across, the galaxy is only about 1000ly thick, and probably less where we are, so your initial assumption is incorrect.
Jeremy
CC BY-SA 3.0
2014-08-03T03:56:33.987Z
@Jeremy, I assumed the Orion arm was cylindrical with 3,500 ly diameter. If we are near the outer edge of the galaxy (top or bottom), then there are much fewer near and hence visible stars in one direction, the one outwards from the Milky Way, right? Is it the northern or southern side of the galactic disk we are at?
LocalFluff
CC BY-SA 3.0
2014-08-05T18:32:48.767Z
@LocalFluff, what do you mean by "northern" and southern"? I don't think there is a preferred orientation of the galaxy that would warrant such labels.
HDE 226868
CC BY-SA 3.0
2014-08-05T18:56:39.980Z
@HDE 226868, if the Milky Way is disk shaped, then I mean near one of the two large circular surfaces of that disk. Within a disk the disk shape is irrelevant, you'll see the same number of visible stars in all directions. As if inside a sphere. But if on the surface of a dish, most visible stars would be concentrated in one direction, towards the galactic disk equator. North south here is my reference to the poles of the galactic center.
LocalFluff
CC BY-SA 3.0
2014-08-05T19:06:05.897Z
Thanks, @LocalFluff. As the solar system orbits the center of the galaxy, it oscillates up and down in the galactic plane (Wikipedia (https://en.wikipedia.org/wiki/Milky_Way) says this happens 2.7 times during each galactic year, but I don't know the amplitude of the oscillations, or our current position within the oscillations.
HDE 226868
CC BY-SA 3.0
2015-03-25T23:10:55.260Z
@HDE226868 See http://astronomy.stackexchange.com/questions/822/how-far-is-the-earth-sun-above-below-the-galactic-plane-and-is-it-heading-towar
ProfRob
CC BY-SA 3.0
2015-03-25T23:12:26.740Z
@RobJeffries It's not every day that I get a response to a 7-month-old comment. Thanks; I saw that when you answered in December.
HDE 226868
<star-gazing><naked-eye>
Aren't there more naked-eye-visible stars in the Milky Way plane?
LocalFluff
# Aren't there more naked-eye-visible stars in the Milky Way plane?

(Asked by: username_0 on 2014-08-01)

Most stars which are visible to the naked eye are within 1,000 light years. The Sun is inside the Orion arm which has a diameter of about 3,500 light years. Thus, all stars (with very few exceptions) we can see unaided are inside the Orion arm and should be equally distributed in all directions. The structure of the galaxy is too large to affect the distribution of visible stars.

Still, we do see the Milky Way disk. What is it we see? The dim light of stars we can't make out individually? Or star light reflected by gas clouds? And is the density of stars to the unaided eye really not higher in the plane of the Milky Way? (Since I'm living in a light polluted city I can't easily check it out myself)

**Tags:** star-gazing, naked-eye

> **Comments:**
> **username_1**: Please provide references for your claims. I think first we need to address some misunderstandings  
> **username_0**: Here are some claims about the Sun's place in the Milky Way http://www.atlasoftheuniverse.com/5000lys.html  
> **username_1**: I couldn't see anything there claiming there is 1000ly of visible stars in all directions of our position. Actually, while the Orion Arm is 3500ly across, the galaxy is only about 1000ly thick, and probably less where we are, so your initial assumption is incorrect.  
> **username_0**: @username_1, I assumed the Orion arm was cylindrical with 3,500 ly diameter. If we are near the outer edge of the galaxy (top or bottom), then there are much fewer near and hence visible stars in one direction, the one outwards from the Milky Way, right? Is it the northern or southern side of the galactic disk we are at?  
> **username_2**: @username_0, what do you mean by "northern" and southern"? I don't think there is a preferred orientation of the galaxy that would warrant such labels.  
> **username_0**: @username_2, if the Milky Way is disk shaped, then I mean near one of the two large circular surfaces of that disk. Within a disk the disk shape is irrelevant, you'll see the same number of visible stars in all directions. As if inside a sphere. But if on the surface of a dish, most visible stars would be concentrated in one direction, towards the galactic disk equator. North south here is my reference to the poles of the galactic center.  
> **username_2**: Thanks, @username_0. As the solar system orbits the center of the galaxy, it oscillates up and down in the galactic plane (Wikipedia (https://en.wikipedia.org/wiki/Milky_Way) says this happens 2.7 times during each galactic year, but I don't know the amplitude of the oscillations, or our current position within the oscillations.  
> **username_3**: @username_2 See http://astronomy.stackexchange.com/questions/822/how-far-is-the-earth-sun-above-below-the-galactic-plane-and-is-it-heading-towar  
> **username_2**: @RobJeffries It's not every day that I get a response to a 7-month-old comment. Thanks; I saw that when you answered in December.  

---

## 3 answers

### Answer by username_3 on 2015-03-25. Score: 12 | Accepted answer

You can tell a lot about Galactic structure by just looking and indeed the distribution of bright stars is non-spherical and non uniform.

The ~5000 stars that can be seen with the naked eye have a roughly "lognormal" distribution of distance. I show plots below which were generated from the most recent version of the Hipparcos parallax catalogue. Fig.1 shows results for all stars with $5.5<V<6.5$ (i.e. very faint naked eye stars). The Hipparcos catalogue is almost complete for these stars, though some stars are so far away that the distance is highly uncertain - nevertheless, the general picture should be ok.

The median distance is about 440 light years. But the premise of your question is I think, that you dispute that this is far enough for the non-spherical distribution of stars in our Galaxy to become apparent. The answer is actually that it is, but only just. The Sun is very close to the plane of the disk of our Galaxy. The scale height of stars above this disk varies depending on stellar age and mass. Very approximately, the exponential scale height is 300-500 light years for most of the stars in our Galaxy.

This is just small enough that if we look at the demographics of stars in two Galactic latitude regions we *do* see a difference. Fig. 2 below shows the distance distribution of stars with $5.5<V<6.5$ at low Galactic latitudes (within 15 degrees of the plane in green) and more than 45 degrees from the plane (in blue). There is a clear and significant difference. More stars can be seen at greater distances towards the plane, despite the greater dust extinction in those directions,betraying the non spherically-symmetric nature of stars in our Galaxy.

In other words the structure of the Galaxy *is not* too small to affect the distribution of naked eye stars.

![Distribution of stellar distances for naked eye stars](https://i.sstatic.net/nrkJX.png)

*Fig 1: Probability distribution of distance for naked eye stars in the Hipparcos catalogue*

![Distribution of distances for naked eye stars in two Galactic latitude zones](https://i.sstatic.net/I5pyX.png)

*Fig 2: Probability distribution of distance for naked eye stars divided into low and high Galactic latitude regions (i.e. towards and away from the Galactic plane)*.

I then split the Hipparcos sample into bright ($V<6$, 5000 stars) and very bright ($V<3$, 173 stars) and looked at the distribution of stars per unit area on the sky as a function of Galactic latitude (Fig. 3). The results are shown below and more obviously make the point. It turns out that the asymmetry between in an out of the plane is clearly seen, even in the brighter sample. The reason is that the very bright stars are not *all that much* closer than the bright stars. Perhaps a median of 200 light years vs 350 light years. Thus the scale height of the Galactic disc must be small enough that the difference from spherical symmetry is already apparent at these kinds of distances.

![Distribution of stars vs Galactic latitude](https://i.sstatic.net/lOpa1.png)

*Fig. 3: Normalised plot of number of stars per unit area versus Galactic latitude for a sample of bright and very bright stars. Note the concentration towards the Galactic plane*.

You can even see from the plot above that there is also a slight concentration towards negative latitudes; the peak density is around -5 to -10 degrees in both cases. This is likely because the Sun is above the Galactic plane at present (though I also wonder if dust plays a role). The Sun is currently 60 light years above the plane and heading upwards (see [How far is the Earth/Sun above/below the galactic plane, and is it heading toward/away from it?](https://astronomy.stackexchange.com/questions/822/how-far-is-the-earth-sun-above-below-the-galactic-plane-and-is-it-heading-towar) ). The combination of the results I have shown *may* be sufficient to conclude that the Galactic disc has a "characteristic thickness\* not more than a few hundred light years, but more than 60 light years!

When we look at the Milky Way we are seeing millions of unresolved stars that are in general more distant still. The effects of the disk-like nature of the Galaxy become more apparent as one goes to greater distances. In particular, at high Galactic latitudes one simply runs out of stars and hence there is no uniformly illuminated sky - leading to Olber's paradox, as correctly stated in another answer. But at low latitudes, there are sufficient stars that in general, within the resolution of our eyes there are many stars whose light sums to provide a visual stimulus. This picture is interrupted by *dust*. Dust in the Galaxy is even more concentrated towards the plane than the stars. It is for this reason that beyond a few thousand light years, dust plays a major role in shaping the Milky Way structures that we can see, effectively blocking light at very low Galactic latitudes beyond a few *thousand* light years.


---

### Answer by username_1 on 2014-08-05. Score: 5

Firstly, the galaxy is only about [1000ly thick](http://imagine.gsfc.nasa.gov/docs/ask_astro/answers/980317b.html). We are fairly close to the galactic plane, [maybe around 65 ly 'above' it](http://www.nature.com/nature/journal/v316/n6030/abs/316706a0.html) if we call the direction we are moving away from 'down'. So on your assumption that visible objects are all within about 1000ly, we can suppose that we should see more stars in the plane than above and below the plane, as there is only about 500ly of stars in the galaxy above and below.

We are perhaps [25kly](http://iopscience.iop.org/1538-4357/597/2/L121/) from the galactic centre. Therefore there are about 25kly of stars outwards, and 75kly of stars inwards. If we could see all of them with the naked eye, we would see more stars to one side than the other. While many of the stars that we see are within 1000ly, that doesn't mean that the stars within this region of visibility are evenly distributed. [Stars are packed closer together nearer the centre](http://www.astronomy.ohio-state.edu/~ryden/ast162_7/notes31.html), which implies that we will see more towards one side than the other.

I don't know where you get the idea you seek confirmation for that the density of stars is not higher in the galactic plane, because it isn't true. Density is greater toward the centre of the galaxy; and orthogonal to the galactic plane, density decreases as you move away from it.

Plus, we see far more than merely dots of light from other stars. We see other objects that are aggregations of stars - clusters, and galaxies like Andromeda, the LMC and the SMC. We see nebulae, clouds of gas and dust. In fact, dust even obscures what we might otherwise see toward the galactic centre.

So when you ask 'what is it we see?' the answer is: lots of things. Because there is more of that 'stuff' in the plane of the galaxy, we see the majority of that stuff as a ring around us. Because there is more of that stuff towards the centre of the galaxy, we see that the ring is brighter, fuller, more complex, and dominant on one side.

> **Comments:**
> **username_4**: I guess, you meant $25 kly = 8 kpc$.  
> **username_1**: Eeeek yup - I'll update, and perhaps to can remove comment.  
> **username_0**: Almost all stars visible to the naked eyes are within 1000 light years. Most of them in turn within a fraction of that distance. Is that disputed? Any density differences on the sky is a function of the internal shape of the Orion arm and our location inside it. I doubt that the Orion arm is denser towards the galactic center. The structure of the Milky Way hardly affects the density of visible stars which are all located between 24 and 26 kly from the center. "Local" effects certainly dominate. And some other claims online say that the Orion arm has a diameter of 3500 ly, not 1000.  
> **username_1**: It is 3500ly across, not in diameter. Furthermore, I don't get you. Are you seriously disputing the reality of easily made observations, are you seriously claiming that stars should all appear evenly distributed across the sky, because in your mind you've decided that it ought to be this way, based on a couple of things you have read and misunderstood?  

---

### Answer by username_5 on 2014-08-05. Score: 1

This is known as [Olbers' paradox](http://en.m.wikipedia.org/wiki/Olbers%27_paradox). An (in my opinion) better explanation can be found [here](http://curious.astro.cornell.edu/question.php?number=375).

Note that these links answer the question why the universe (not the milky way alone) isn't blazing our night sky. Now if the universe (including the milky way) can't do it, the milky way can't do it on its own...

There are just not enough stars giving enough light. Brightness is reduced quadratically by distance. That counts up very quickly with astronomical distances. Even for stars that are our neighbors.

> **Comments:**
> **username_0**: My impression is that Olbers' paradox is about diffuse light from far far away. Like that which reaches us from the Milky way. Which it indeed does and it glows. But beyond the ability of human unaided eyes to resolve individual stars. My question is about such stars of about 6 magnitude or more visible. Are naked eye visible stars more plentiful towards to Milky Way, or not? I presume, based on their distance and the structure of the MW, that they should not. But I'm open to be wrong about this. That's why I ask.  
> **username_2**: Olber's Paradox is usually used in regard to the infinite universe model. I suppose, though, that this is a related question. I don't know if it's a paradox, though.  

---

