Why do other galaxies have such a low surface brightness?

Question

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.

Answer 1

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.

Answer 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.