4
$\begingroup$

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 (flux per unit 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.

$\endgroup$
18
  • 1
    $\begingroup$ 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 $(1+z)^4$. $\endgroup$
    – pela
    Aug 19 at 11:20
  • 1
    $\begingroup$ I probably didn't ask it very clearly. I'll make an edit to rephrase it. $\endgroup$
    – zucculent
    Aug 21 at 19:57
  • 1
    $\begingroup$ @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 $\endgroup$
    – uhoh
    Aug 21 at 22:30
  • 1
    $\begingroup$ @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. $\endgroup$
    – uhoh
    Aug 21 at 22:30
  • 1
    $\begingroup$ @uhoh "if we defocused the Milky Way such that individual stars were no longer noticeable as such" yes this is what I was imgaining $\endgroup$
    – zucculent
    Aug 21 at 22:51
2
+100
$\begingroup$

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.

$\endgroup$
6
  • 1
    $\begingroup$ 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. $\endgroup$
    – ProfRob
    Aug 23 at 10:06
  • $\begingroup$ 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... $\endgroup$ Aug 23 at 10:55
  • 1
    $\begingroup$ 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. $\endgroup$
    – ProfRob
    Aug 23 at 12:03
  • $\begingroup$ @ProfRob Sigh, yes... That'll teach me to do math before having enough coffee. Let me try fixing it again... $\endgroup$ Aug 23 at 17:21
  • 1
    $\begingroup$ @uhoh It's a transition rather than a "step", but I think that's a reasonable way to think about it. $\endgroup$ Aug 29 at 14:43
1
$\begingroup$

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).

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.

$\endgroup$
6
  • $\begingroup$ "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.) $\endgroup$ Aug 22 at 14:13
  • $\begingroup$ 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. $\endgroup$ Aug 22 at 14:18
  • $\begingroup$ 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. $\endgroup$
    – ProfRob
    Aug 22 at 19:24
  • $\begingroup$ 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?" $\endgroup$ Aug 22 at 20:18
  • $\begingroup$ 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. $\endgroup$ Aug 22 at 20:23

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.