<p>When we look at the Milky Way, we see a limited distance (several thousand light years) into the plane of our Galaxy's disk because of dust extinction.</p>
<p>The Andromeda galaxy is not edge on, it is inclined at around 77 degrees. For the most part, we are looking <em>through</em> the disk, which is only a few hundred light years thick, at this angle. Thus, given a similar stellar volume density (and actually, the stellar density in the Andromeda disk is lower), I would not expect the surface brightness of the Andromeda disk to be as big as for a typical sightline in the Milky Way. Or to put it another way, it should be like looking at the Milky Way surface brightness at <span class="math-container">$\sim 10$</span> degrees out of the Galactic plane.</p>
<p>The situation is different for the Andromeda bulge, where we do have a sightline going through thousands of light years of stars. And indeed, the bulge of Andromeda <em>is</em> a naked eye object despite being several hundred times further away than the stars that make up the naked eye Milky Way.</p>
<p>For more distant galaxies there are two factors which lead to them being fainter in surface brightness terms. The first, which only applies at cosmological distances, is that surface brightness goes down as <span class="math-container">$(1+z)^{-4}$</span> (<a href="https://en.wikipedia.org/wiki/Tolman_surface_brightness_test" rel="nofollow noreferrer">for a flat universe and uniform expansion</a>).</p>
<p>The second is that galaxies are not uniformly bright, so one has to calculate how many stars are actually visible within the &quot;point spread function&quot; of the eye - which is roughly an arcminute in diameter (for good eyesight). Once the &quot;scale-length&quot; (a distance scale on which the surface brightness falls) becomes smaller than 1 arcminute, then the <em>perceived</em> surface brightness is no longer constant, because the number of stars within a 1 arcminute diameter falls faster than the reciprocal of the distance squared. For a &quot;typical&quot; big disk galaxy, then a scale length of a few kpc is appropriate and this will become smaller than 1 arcminute at distances beyond 10 Mpc (i.e. beyond the local group). For smaller galaxies in the local group (i.e. not M31), this may also be a contributing factor.</p>
<p>Finally, another factor to consider is the mixture of stars that make up a galaxy. Galaxies like the LMC have a much higher star formation rate than the Milky Way and therefore have a higher proportion of young, massive and extremely luminous stars. If all other things are equal (i.e. the same number of stars per cubic parsec and a similar path length through the galaxy) then this would increase their surface brightness.</p>

CC BY-SA 4.0
2021-08-22T14:13:42.943Z
"the stellar density in the Andromeda disk is lower" -- What's your basis for that statement? (And in any case, since both disks have densities that decay exponentially with radius, the inner disk of M31 will have a higher density than the outer disk of our galaxy.)
Peter Erwin
CC BY-SA 4.0
2021-08-22T14:18:46.317Z
I'll admit I'm unsure, in the context of this argument, as to why the LMC and SMC are significantly brighter than M31's bulge, given that in both cases we're looking through low-density and (at least in the case of LMC) low-inclination disks.
Peter Erwin
CC BY-SA 4.0
2021-08-22T19:24:22.770Z
M31 has about the same number of stars as the Milky Way, but is considerably bigger. The M31 inner disk is still a naked eye object (for some). The LMC and SMC are not much less dense than our local disk. The LMC and I think the SMC have lots of vigorous star formation and so a larger brightness for a given stellar density - it is the high mass stars and red supergiants that give most of the light.
ProfRob
CC BY-SA 4.0
2021-08-22T20:18:29.167Z
I think the main problem I have with your answer is that it never mentions *distance*, which is crucial for answering the question "why do other galaxies appear so dim?"
Peter Erwin
CC BY-SA 4.0
2021-08-22T20:23:58.377Z
M31 has about twice the stellar mass of the Milky Way, though it is more extended, yes. But: the central surface brightness of M31 is about 15 mag arcsec$^{-2}$ in $g$ (Tempel+2011); the $g$-band central surface brightness of the SMC is about 22.5 (Massana+2020) -- about a thousand times fainter.
Peter Erwin
CC BY-SA 4.0
2021-08-22T22:02:33.960Z
I probably should have clarified that when I said "dim" I meant low brightness relative to solid angle, as in square degrees.
zucculent
CC BY-SA 4.0
2021-08-22T09:38:32.487Z
2021-08-23T12:19:00.040Z
2021-08-23T12:19:00.040Z
ProfRob
<p>The key to understanding what's going on is that the apparent surface brightness (observed flux per unit solid angle) of galaxies at different distances depends on two main things: how many stars are packed into a given <em>angular</em> area on the sky (which depends on the intrinsic density of stars in the galaxy <em>and</em> on the distance), and how faint the individual stars become as a function of distance.</p>
<p>Any image will have a limiting resolution: the size of the seeing or the telescope's point spread function, the size of pixels in the camera, etc. For example, the human eye has a limiting resolution of about 1 arc minute. We can crudely talk about the surface brightness of galaxies in terms of how much stellar flux comes from an angular region (solid angle) with the size of the limiting resolution.</p>
<p>Let's work with a simple model: a circularly symmetric galaxy, without dust, containing identically bright stars with a radial exponential distribution. That is, the galaxy has a certain central surface density (call it <span class="math-container">$n_{0}$</span> stars per square kiloparsec), and that density falls off as you go away from the center with an exponential function that has a scale length of 1 kiloparsec (kpc):</p>
<p><span class="math-container">$n(r) = n_{0} e^{-r}$</span></p>
<p>For a central, circular resolution element corresponding to a <em>physical</em> radius <span class="math-container">$R$</span>, the total number of stars <span class="math-container">$N$</span> within that radius will be</p>
<p><span class="math-container">$N = \int_{0}^{R} 2 \pi r n(r) dr = 2 \pi n_{0} (1 - (1 + R) e^{-R})$</span></p>
<p>Consider three cases: the first where the galaxy is only 1 kpc away from us, the second where the galaxy is 1 Mpc away from us (1000 times further away), and the third where it is 10 Mpc away.</p>
<p>So we're going to look at the question of how many stars are packed into the central circle with radius = 1 arc minute, and how much light we get from just those stars. When the galaxy is only 1 kpc away, the 1-arcmin-radius circle has a physical radius of <span class="math-container">$R = 0.29$</span> pc; for 1 and 10 Mpc away, <span class="math-container">$R = 290$</span> pc (0.29 kpc) and 2.9 kpc, respectively.</p>
<p>If we assume that stars at 1 kpc distance have an observed brightness of 1 (in whatever the appropriate units are), then the same stars at a distance of 1 Mpc will be <em>one million times fainter</em>, and will have individual brightnesses of <span class="math-container">$10^{-6}$</span>; at 10 Mpc, they will have a brightness of <span class="math-container">$10^{-8}$</span>. (This is just the inverse-square law: the same star <span class="math-container">$x$</span> further away will be <span class="math-container">$x^{2}$</span> times fainter.)</p>
<p>For the three cases, we have:</p>
<ol>
<li><p>(<span class="math-container">$D = 1$</span> kpc) There are <span class="math-container">$2.7 \times 10^{-7} n_{0}$</span> stars in the central <span class="math-container">$r = 1$</span> arcmin region, each with flux = 1; total flux in the central region is <span class="math-container">$2.7 \times 10^{-7} n_{0}$</span>.</p>
</li>
<li><p>(<span class="math-container">$D = 1$</span> Mpc) There are <span class="math-container">$0.22 n_{0}$</span> stars in the central <span class="math-container">$r = 1$</span> arcmin region, each with flux <span class="math-container">$= 10^{-6}$</span>; total flux in the central region is <span class="math-container">$2.2 \times 10^{-7} n_{0}$</span>.</p>
</li>
<li><p>(<span class="math-container">$D = 10$</span> Mpc) There are <span class="math-container">$4.9 n_{0}$</span> stars in the central <span class="math-container">$r = 1$</span> arcmin region, each with flux <span class="math-container">$= 10^{-8}$</span>; total flux in the central region is <span class="math-container">$4.9 \times 10^{-8} n_{0}$</span>.</p>
</li>
</ol>
<p>The precise details will depend on the actual structure of the galaxy and the intrinsic brightness of its stars, but the general rule is that more distant galaxies are fainter because the density of stars falls off as you go away from their centers, and so the number of stars within a given angular patch doesn't go up fast enough as you increase the distance to compensate for the decrease in individual stars' brightnesses due to their being at larger distances from us.</p>

CC BY-SA 4.0
2021-08-23T10:06:51.953Z
I don't understand the maths. When something is one thousand times further away (1 kpc vs 1 Mpc), it is a thousand times smaller (angularly speaking) and a million times fainter. Also, 1 arcminute at a distance of 1 kpc is 0.29 pc.
ProfRob
CC BY-SA 4.0
2021-08-23T10:55:44.263Z
Ack -- yes, I seem to have done most of the math assuming a factor of 10 difference in distance rather than a factor of 1000; thanks for catching that. Let me see about fixing it...
Peter Erwin
CC BY-SA 4.0
2021-08-23T12:03:49.337Z
There is still something wrong here. Unless you get out to more than a scale-length (which your examples don't), then $N$ should scale as $\sim R^2$ and the surface brightness is then of course constant. Your expression for $N$ scales linearly for small $R$ which is why you seem to find the surface brightness goes down by a factor of 1000. I believe your integral may be incorrect. This explanation is only a factor *beyond* distances where the scale length becomes smaller than $\sim 1$ arcminute.
ProfRob
CC BY-SA 4.0
2021-08-23T17:21:12.747Z
@ProfRob Sigh, yes... That'll teach me to do math before having enough coffee. Let me try fixing it again...
Peter Erwin
CC BY-SA 4.0
2021-08-28T00:10:39.630Z
I think I understand the idea, can I confirm? Galaxies are nonuniform; they have "bright spots"in the middle. When they are close enough we can resolve, see, and perceive the surface brightness of their bright spots, but when they are far, an averaging takes place, so there's sort of a "step" at some distance where they (on average) just start looking dimmer. Is that *roughly* the idea?
uhoh
CC BY-SA 4.0
2021-08-29T14:43:07.417Z
@uhoh It's a transition rather than a "step", but I think that's a reasonable way to think about it.
Peter Erwin
CC BY-SA 4.0
2021-08-22T22:06:24.467Z
2021-08-23T18:56:08.893Z
2021-08-23T18:56:08.893Z
Peter Erwin
<p>Why do we need to take exposure photos of other galaxies, considering their stars should have roughly the same luminosity per square degree as the ones in the milky way? I'm not saying it should be like a mini-sun, but it doesn't make sense unless we're mostly seeing through them.</p>
<p>Edit: It looks like I didn't do a good job of making my question clear, so I'll try rephrasing it.
Let's take the Andromeda Galaxy. Its stars should have roughly the same <a href="https://en.wikipedia.org/wiki/Surface_brightness" rel="nofollow noreferrer">surface brightness</a> (flux per unit <a href="https://en.wikipedia.org/wiki/Solid_angle" rel="nofollow noreferrer">solid angle</a>) as the Milky Way's. With this in mind, you would think the Andromeda Galaxy would have a high enough apparent surface brightness that it would be able to be seen with the naked eye.</p>
<p>Also, as I was editing, I realized that I hadn't considered how far apart stars are. I'm pretty sure now that the reason galaxies don't have an apparent surface brightness equal to that of a star is that the vast majority of lines of sight aimed at a galaxy would go right through it without hitting a star. In other words, much more of a galaxie's &quot;area&quot; in the night sky is space than stars. I hope this makes things clearer.</p>

CC BY-SA 4.0
2021-08-19T06:48:05.543Z
An answer might begin: "Yes, surface brightness should be relatively constant for a uniform distribution; a wall doesn't get brighter when we walk towards it. But the Milky Way we see is not uniform or smeared out like a more distant galaxy is. Instead much of what we see by eye is the result of individual stars and whatnot..." but it would finish in a more conclusive way than I can muster.
uhoh
CC BY-SA 4.0
2021-08-19T11:20:43.537Z
Note that the surface brightness of galaxies is only ~constant for ~nearby galaxies. Because the relation between luminosity distance and angular diameter distance, the SB decreases quite fast for high-z sources, namely as $(1+z)^4$.
pela
CC BY-SA 4.0
2021-08-19T11:46:37.607Z
If you're asking about why the night sky is dark, this is a duplicate of [What exactly is the “paradox” in Olber's Paradox](https://astronomy.stackexchange.com/questions/21843). If you're asking about something else, it's not clear what you are asking.
David Hammen
CC BY-SA 4.0
2021-08-20T16:18:07.683Z
@DavidHammen so are the stars so dispersed that if we zoomed in enough, we'd see that most lines of sight would go right through the galaxy without hitting any stars, but we can't see individual stars, so it just looks like a dim smear?
zucculent
CC BY-SA 4.0
2021-08-20T16:35:19.977Z
@zucculent The most distant stars visible to the naked eye are but a few thousand light years away. The Andromeda Galaxy is 2.5 million light years away. The Andromeda Galaxy is visible to the naked eye because it contains roughly as many stars as does the Milky Way.  But seeing a specific star in the Andromeda galaxy? You'll need a very good telescope to do that. A very, very good telescope. Light intensity from a star drops with the square of distance.
David Hammen
CC BY-SA 4.0
2021-08-21T03:49:53.077Z
Sorry I @-ed the wrong person.
zucculent
CC BY-SA 4.0
2021-08-21T07:39:11.450Z
Isn't that every light source gets dimmer as its distance grow? See comment by David Hammen. Perhaps I didn't understand your question...
Alchimista
CC BY-SA 4.0
2021-08-21T19:57:12.930Z
I probably didn't ask it very clearly. I'll make an edit to rephrase it.
zucculent
CC BY-SA 4.0
2021-08-21T22:30:19.417Z
@zucculent your question was clear to me from the beginning (see my first comment in which I tried to support it) and is still clear now. Sometimes people see questions written by low reputation users and without taking the time to read and think just assume that it means that they don't know what they are talking about . Your edit may or may not help those that don't take the time to read and think, but it's a good edit nonetheless. Spurious close votes happen from time to time; **voting to leave open**
uhoh
CC BY-SA 4.0
2021-08-21T22:30:24.753Z
@pela it's a good point; the op has clarified further that they're probably mostly interested in the "~nearby galaxies" part but an answer should point that out (as well as that long exposures provide resolution and detail). I think that the answer to why the Milky Way seems or feels to have a larger surface brightness than Andromeda has to do with the statistics of nearby stars; if we *defocused the Milky Way* such that individual stars were no longer noticeable as such, perhaps it and Andromeda and other nearby galaxies of similar types could then be compared on an equal footing.
uhoh
CC BY-SA 4.0
2021-08-21T22:51:14.130Z
@uhoh "if we defocused the Milky Way such that individual stars were no longer noticeable as such" yes this is what I was imgaining
zucculent
CC BY-SA 4.0
2021-08-22T09:42:51.403Z
@uhoh "if we defocused the Milky Way such that individual stars were no longer noticeable as such" -- but this is arguably exactly what's going on with naked-eye observations of "the Milky Way" -- that is, the bright, fuzzy, elongated structure visible from suitably dark sites. Your eyes defocus things so that you can't notice the individual stars (you need a telescope to seem them individually, as Galileo demonstrated).
Peter Erwin
CC BY-SA 4.0
2021-08-22T12:38:19.520Z
@PeterErwin I was on a trip to Indonesia, we were at a high elevation on a crystal clear night and there were no cities anywhere near us. This was years ago, but I remember the *impression* that I was seeing many, many individual stars within the milky way. It did not account for all the light, but it certainly looked "peppered" (okay "salted") with points of light. However, I wonder if this is my mind playing tricks on me? Is it nowhere in the world seeable that the milky way at least contains lots and lots of stars?
uhoh
CC BY-SA 4.0
2021-08-22T14:06:51.340Z
@uhoh Well, there are always going to be brighter, individually distinguishable stars anywhere you look in the sky, and there will be more of them in the direction of the Milky Way. It’s the indistinguishable background “haze” due to all the fainter stars that I’m talking about.
Peter Erwin
CC BY-SA 4.0
2021-08-23T08:36:35.623Z
Most of the Milky Way is actually invisible to us because we have to look through the plane of its disk. A large percentage of our own galaxy's Milky Way is on the other side of the our own galaxy. There is so much dust, and so many bright stars that are actually closer to us that we can't even see the stars behind all of this matter. Just a small part of a thought to your question.
Buckn4u2
CC BY-SA 4.0
2021-08-24T08:06:21.533Z
It seems directed to me. I didn't intend anything if not that I didn't understand the question. And the title doesn't help, even. The straight answer to the title questions is certainly "because they are far away / no bright enough". I have spent more time on reading it than what you can suppose, not to say that I have commented twice. Now it is clearer tough the title could be about a kind of averaged "surface brightness". Quite a curious way to remark that you understood it at a first glance. Bravo :) And OP clearing the Q seems to have got the answer, at least qualitatively.
Alchimista
CC BY-SA 4.0
2021-08-24T12:23:48.730Z
@Alchimista thx will do
zucculent
CC BY-SA 4.0
2021-08-24T13:13:59.103Z
Looking at your edited question, it seems like part of the problem is that you assume the surface brightness of *individual stars* is relevant. But unless you can *resolve* the star (not possible with the eye or most telescopes for any star other than the Sun), it isn't. There are hundred of millions of stars in the Milky Way that aren't significantly obscured by dust -- but the vast majority of them are too faint to be seen without telescopes (and lengthy exposures), because they are too far away.
Peter Erwin
<galaxy>
Why do other galaxies have such a low surface brightness?
zucculent
# Why do other galaxies have such a low surface brightness?

(Asked by: username_0 on 2021-08-19)

Why do we need to take exposure photos of other galaxies, considering their stars should have roughly the same luminosity per square degree as the ones in the milky way? I'm not saying it should be like a mini-sun, but it doesn't make sense unless we're mostly seeing through them.

Edit: It looks like I didn't do a good job of making my question clear, so I'll try rephrasing it.
Let's take the Andromeda Galaxy. Its stars should have roughly the same [surface brightness](https://en.wikipedia.org/wiki/Surface_brightness) (flux per unit [solid angle](https://en.wikipedia.org/wiki/Solid_angle)) as the Milky Way's. With this in mind, you would think the Andromeda Galaxy would have a high enough apparent surface brightness that it would be able to be seen with the naked eye.

Also, as I was editing, I realized that I hadn't considered how far apart stars are. I'm pretty sure now that the reason galaxies don't have an apparent surface brightness equal to that of a star is that the vast majority of lines of sight aimed at a galaxy would go right through it without hitting a star. In other words, much more of a galaxie's "area" in the night sky is space than stars. I hope this makes things clearer.

**Tags:** galaxy
> **Comments:**
> **username_1**: An answer might begin: "Yes, surface brightness should be relatively constant for a uniform distribution; a wall doesn't get brighter when we walk towards it. But the Milky Way we see is not uniform or smeared out like a more distant galaxy is. Instead much of what we see by eye is the result of individual stars and whatnot..." but it would finish in a more conclusive way than I can muster.  
> **username_2**: Note that the surface brightness of galaxies is only ~constant for ~nearby galaxies. Because the relation between luminosity distance and angular diameter distance, the SB decreases quite fast for high-z sources, namely as $(1+z)^4$.  
> **username_3**: If you're asking about why the night sky is dark, this is a duplicate of [What exactly is the “paradox” in Olber's Paradox](https://astronomy.stackexchange.com/questions/21843). If you're asking about something else, it's not clear what you are asking.  
> **username_0**: @DavidHammen so are the stars so dispersed that if we zoomed in enough, we'd see that most lines of sight would go right through the galaxy without hitting any stars, but we can't see individual stars, so it just looks like a dim smear?  
> **username_3**: @zucculent The most distant stars visible to the naked eye are but a few thousand light years away. The Andromeda Galaxy is 2.5 million light years away. The Andromeda Galaxy is visible to the naked eye because it contains roughly as many stars as does the Milky Way.  But seeing a specific star in the Andromeda galaxy? You'll need a very good telescope to do that. A very, very good telescope. Light intensity from a star drops with the square of distance.  
> **username_0**: Sorry I @-ed the wrong person.  
> **username_4**: Isn't that every light source gets dimmer as its distance grow? See comment by David Hammen. Perhaps I didn't understand your question...  
> **username_0**: I probably didn't ask it very clearly. I'll make an edit to rephrase it.  
> **username_1**: @zucculent your question was clear to me from the beginning (see my first comment in which I tried to support it) and is still clear now. Sometimes people see questions written by low reputation users and without taking the time to read and think just assume that it means that they don't know what they are talking about . Your edit may or may not help those that don't take the time to read and think, but it's a good edit nonetheless. Spurious close votes happen from time to time; **voting to leave open**  
> **username_1**: @pela it's a good point; the op has clarified further that they're probably mostly interested in the "~nearby galaxies" part but an answer should point that out (as well as that long exposures provide resolution and detail). I think that the answer to why the Milky Way seems or feels to have a larger surface brightness than Andromeda has to do with the statistics of nearby stars; if we *defocused the Milky Way* such that individual stars were no longer noticeable as such, perhaps it and Andromeda and other nearby galaxies of similar types could then be compared on an equal footing.  
> **username_0**: @uhoh "if we defocused the Milky Way such that individual stars were no longer noticeable as such" yes this is what I was imgaining  
> **username_5**: @uhoh "if we defocused the Milky Way such that individual stars were no longer noticeable as such" -- but this is arguably exactly what's going on with naked-eye observations of "the Milky Way" -- that is, the bright, fuzzy, elongated structure visible from suitably dark sites. Your eyes defocus things so that you can't notice the individual stars (you need a telescope to seem them individually, as Galileo demonstrated).  
> **username_1**: @PeterErwin I was on a trip to Indonesia, we were at a high elevation on a crystal clear night and there were no cities anywhere near us. This was years ago, but I remember the *impression* that I was seeing many, many individual stars within the milky way. It did not account for all the light, but it certainly looked "peppered" (okay "salted") with points of light. However, I wonder if this is my mind playing tricks on me? Is it nowhere in the world seeable that the milky way at least contains lots and lots of stars?  
> **username_5**: @uhoh Well, there are always going to be brighter, individually distinguishable stars anywhere you look in the sky, and there will be more of them in the direction of the Milky Way. It’s the indistinguishable background “haze” due to all the fainter stars that I’m talking about.  
> **username_6**: Most of the Milky Way is actually invisible to us because we have to look through the plane of its disk. A large percentage of our own galaxy's Milky Way is on the other side of the our own galaxy. There is so much dust, and so many bright stars that are actually closer to us that we can't even see the stars behind all of this matter. Just a small part of a thought to your question.  
> **username_4**: It seems directed to me. I didn't intend anything if not that I didn't understand the question. And the title doesn't help, even. The straight answer to the title questions is certainly "because they are far away / no bright enough". I have spent more time on reading it than what you can suppose, not to say that I have commented twice. Now it is clearer tough the title could be about a kind of averaged "surface brightness". Quite a curious way to remark that you understood it at a first glance. Bravo :) And OP clearing the Q seems to have got the answer, at least qualitatively.  
> **username_0**: @Alchimista thx will do  
> **username_5**: Looking at your edited question, it seems like part of the problem is that you assume the surface brightness of *individual stars* is relevant. But unless you can *resolve* the star (not possible with the eye or most telescopes for any star other than the Sun), it isn't. There are hundred of millions of stars in the Milky Way that aren't significantly obscured by dust -- but the vast majority of them are too faint to be seen without telescopes (and lengthy exposures), because they are too far away.  

---

## 2 answers

### Answer by username_7 on 2021-08-22. Score: 2 | Accepted answer

When we look at the Milky Way, we see a limited distance (several thousand light years) into the plane of our Galaxy's disk because of dust extinction.

The Andromeda galaxy is not edge on, it is inclined at around 77 degrees. For the most part, we are looking *through* the disk, which is only a few hundred light years thick, at this angle. Thus, given a similar stellar volume density (and actually, the stellar density in the Andromeda disk is lower), I would not expect the surface brightness of the Andromeda disk to be as big as for a typical sightline in the Milky Way. Or to put it another way, it should be like looking at the Milky Way surface brightness at $\sim 10$ degrees out of the Galactic plane.

The situation is different for the Andromeda bulge, where we do have a sightline going through thousands of light years of stars. And indeed, the bulge of Andromeda *is* a naked eye object despite being several hundred times further away than the stars that make up the naked eye Milky Way.

For more distant galaxies there are two factors which lead to them being fainter in surface brightness terms. The first, which only applies at cosmological distances, is that surface brightness goes down as $(1+z)^{-4}$ ([for a flat universe and uniform expansion](https://en.wikipedia.org/wiki/Tolman_surface_brightness_test)).

The second is that galaxies are not uniformly bright, so one has to calculate how many stars are actually visible within the "point spread function" of the eye - which is roughly an arcminute in diameter (for good eyesight). Once the "scale-length" (a distance scale on which the surface brightness falls) becomes smaller than 1 arcminute, then the *perceived* surface brightness is no longer constant, because the number of stars within a 1 arcminute diameter falls faster than the reciprocal of the distance squared. For a "typical" big disk galaxy, then a scale length of a few kpc is appropriate and this will become smaller than 1 arcminute at distances beyond 10 Mpc (i.e. beyond the local group). For smaller galaxies in the local group (i.e. not M31), this may also be a contributing factor.

Finally, another factor to consider is the mixture of stars that make up a galaxy. Galaxies like the LMC have a much higher star formation rate than the Milky Way and therefore have a higher proportion of young, massive and extremely luminous stars. If all other things are equal (i.e. the same number of stars per cubic parsec and a similar path length through the galaxy) then this would increase their surface brightness.

> **Comments:**
> **username_5**: "the stellar density in the Andromeda disk is lower" -- What's your basis for that statement? (And in any case, since both disks have densities that decay exponentially with radius, the inner disk of M31 will have a higher density than the outer disk of our galaxy.)  
> **username_5**: I'll admit I'm unsure, in the context of this argument, as to why the LMC and SMC are significantly brighter than M31's bulge, given that in both cases we're looking through low-density and (at least in the case of LMC) low-inclination disks.  
> **username_7**: M31 has about the same number of stars as the Milky Way, but is considerably bigger. The M31 inner disk is still a naked eye object (for some). The LMC and SMC are not much less dense than our local disk. The LMC and I think the SMC have lots of vigorous star formation and so a larger brightness for a given stellar density - it is the high mass stars and red supergiants that give most of the light.  
> **username_5**: I think the main problem I have with your answer is that it never mentions *distance*, which is crucial for answering the question "why do other galaxies appear so dim?"  
> **username_5**: M31 has about twice the stellar mass of the Milky Way, though it is more extended, yes. But: the central surface brightness of M31 is about 15 mag arcsec$^{-2}$ in $g$ (Tempel+2011); the $g$-band central surface brightness of the SMC is about 22.5 (Massana+2020) -- about a thousand times fainter.  
> **username_0**: I probably should have clarified that when I said "dim" I meant low brightness relative to solid angle, as in square degrees.  

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### Answer by username_5 on 2021-08-22. Score: 2

The key to understanding what's going on is that the apparent surface brightness (observed flux per unit solid angle) of galaxies at different distances depends on two main things: how many stars are packed into a given *angular* area on the sky (which depends on the intrinsic density of stars in the galaxy *and* on the distance), and how faint the individual stars become as a function of distance.

Any image will have a limiting resolution: the size of the seeing or the telescope's point spread function, the size of pixels in the camera, etc. For example, the human eye has a limiting resolution of about 1 arc minute. We can crudely talk about the surface brightness of galaxies in terms of how much stellar flux comes from an angular region (solid angle) with the size of the limiting resolution.

Let's work with a simple model: a circularly symmetric galaxy, without dust, containing identically bright stars with a radial exponential distribution. That is, the galaxy has a certain central surface density (call it $n\_{0}$ stars per square kiloparsec), and that density falls off as you go away from the center with an exponential function that has a scale length of 1 kiloparsec (kpc):

$n(r) = n\_{0} e^{-r}$

For a central, circular resolution element corresponding to a *physical* radius $R$, the total number of stars $N$ within that radius will be

$N = \int\_{0}^{R} 2 \pi r n(r) dr = 2 \pi n\_{0} (1 - (1 + R) e^{-R})$

Consider three cases: the first where the galaxy is only 1 kpc away from us, the second where the galaxy is 1 Mpc away from us (1000 times further away), and the third where it is 10 Mpc away.

So we're going to look at the question of how many stars are packed into the central circle with radius = 1 arc minute, and how much light we get from just those stars. When the galaxy is only 1 kpc away, the 1-arcmin-radius circle has a physical radius of $R = 0.29$ pc; for 1 and 10 Mpc away, $R = 290$ pc (0.29 kpc) and 2.9 kpc, respectively.

If we assume that stars at 1 kpc distance have an observed brightness of 1 (in whatever the appropriate units are), then the same stars at a distance of 1 Mpc will be *one million times fainter*, and will have individual brightnesses of $10^{-6}$; at 10 Mpc, they will have a brightness of $10^{-8}$. (This is just the inverse-square law: the same star $x$ further away will be $x^{2}$ times fainter.)

For the three cases, we have:

1. ($D = 1$ kpc) There are $2.7 \times 10^{-7} n\_{0}$ stars in the central $r = 1$ arcmin region, each with flux = 1; total flux in the central region is $2.7 \times 10^{-7} n\_{0}$.
2. ($D = 1$ Mpc) There are $0.22 n\_{0}$ stars in the central $r = 1$ arcmin region, each with flux $= 10^{-6}$; total flux in the central region is $2.2 \times 10^{-7} n\_{0}$.
3. ($D = 10$ Mpc) There are $4.9 n\_{0}$ stars in the central $r = 1$ arcmin region, each with flux $= 10^{-8}$; total flux in the central region is $4.9 \times 10^{-8} n\_{0}$.

The precise details will depend on the actual structure of the galaxy and the intrinsic brightness of its stars, but the general rule is that more distant galaxies are fainter because the density of stars falls off as you go away from their centers, and so the number of stars within a given angular patch doesn't go up fast enough as you increase the distance to compensate for the decrease in individual stars' brightnesses due to their being at larger distances from us.

> **Comments:**
> **username_7**: I don't understand the maths. When something is one thousand times further away (1 kpc vs 1 Mpc), it is a thousand times smaller (angularly speaking) and a million times fainter. Also, 1 arcminute at a distance of 1 kpc is 0.29 pc.  
> **username_5**: Ack -- yes, I seem to have done most of the math assuming a factor of 10 difference in distance rather than a factor of 1000; thanks for catching that. Let me see about fixing it...  
> **username_7**: There is still something wrong here. Unless you get out to more than a scale-length (which your examples don't), then $N$ should scale as $\sim R^2$ and the surface brightness is then of course constant. Your expression for $N$ scales linearly for small $R$ which is why you seem to find the surface brightness goes down by a factor of 1000. I believe your integral may be incorrect. This explanation is only a factor *beyond* distances where the scale length becomes smaller than $\sim 1$ arcminute.  
> **username_5**: @ProfRob Sigh, yes... That'll teach me to do math before having enough coffee. Let me try fixing it again...  
> **username_1**: I think I understand the idea, can I confirm? Galaxies are nonuniform; they have "bright spots"in the middle. When they are close enough we can resolve, see, and perceive the surface brightness of their bright spots, but when they are far, an averaging takes place, so there's sort of a "step" at some distance where they (on average) just start looking dimmer. Is that *roughly* the idea?  
> **username_5**: @uhoh It's a transition rather than a "step", but I think that's a reasonable way to think about it.  

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