Vocabulary

Colours. These images are usually made through a broad-band filter, typically^1.1 V, Bor R, and so should be just black and white. The blackness being a measure of how much, or how little of the light in that wavelength band fell on that part of the photograph/CCD. One often combines images in different filters to produce a colour image, which tries to mimic the colour your eye would see. Most of the galaxy images you'll see are of this type. However, often one of the filters is a narrow-band filter, e.g. in H$\alpha$ or O[III]. Combing such an narrow-band image with images made with others filters gives a false colour image, i.e. the colour of the object is not how your eye would see it, but the colour coding has been chosen to bring out a particular feature. Those really pretty colour pictures of Planetary Nebulae are false colour images.

Surface brightness. The brightness of the images depends, apart from the size and quality of the optics of the camera, the integration time, and the transparency of the atmosphere (all of which we should be able to calibrate), on the apparent brightness of the galaxy. So this is what we want to characterise. Since galaxies are extended, the brightness varies over the image, and so what we want to measure is the surface brightness (SB). The SB measures how much light we receive from the galaxy, per unit solid angle (usually per arcsec$^2$) so it should have units of flux (e.g. in W), per unit wavelength range, per unit solid angle on the sky. So for example in units $L_{{\hbox{\sc V}}\odot}{\hbox{\rm arcsec$^{-2}$}}$, in units of the solar V luminosity. Usually, SB is expressed in magnitudes, by taking $-2.5\log_{10}$ of the previous number - we say (incorrectly, or at least confusingly) that we measure SB in magnitudes per square arc seconds. (It's confusing, because this suggests we divide the magnitude by arcsec$^2$, but no: we first divide by the solid angle, and then take the logarithm). As is usual with magnitudes, we add a constant to it.

The number that characterises the intrinsic brightness distribution of the galaxy (as opposed to apparent) is how much light the galaxy emits per unit surface area, so in for example units of $L_{\odot{\hbox{\sc V}}}{\hbox{\rm kpc}}^{-2}$. Astronomers do not use separate names for these two types of SB, so check the units!

Star counts To quantify this a bit more, assume you observe a distribution of stars, all of the same absolute luminosity $L$, and with uniform number density $n$ (in stars per $^3$, say). The flux you receive from a single star, when at distance $r$, is $l=L/4\pi r^2$ (in Wm$^{-2}$). The number of stars at distance between $r$ and $r+dr$, and within a solid angle $d\Omega$ on the sky, is $dN(r,d\Omega)=nr^2drd\Omega$. Combining^1.2 these two, the luminosity of all stars within $r$ and $r+dr$, and within $d\Omega$, is

$$dl = l dN(r,d\Omega)={L n \over 4\pi}d\Omega dr .$$ (1.1)

The surface brightness is $\propto dl/d\Omega=L ndr/4\pi$. For an external galaxy, you could integrate $\int ndr$ along the line of sight through the galaxy, to get the surface density of stars, $\int ndr=\Sigma$ of stars. The SB is then $L\Sigma$. The important thing to notice is that it is independent of the distance $r$. Going back through the calculation, you can see what has happened: the luminosity from a single star at distance $r$ decreases $\propto 1/r^2$, but the number of stars within $d\Omega$ increases $\propto r^2$. And since the SB is proportional to the product of these two, it is independent^1.3 of $r$. (You might object saying we've assumed all stars to have the same luminosity. So just redo the calculation with the average luminosity then!)

When looking-out into a stellar distribution - for example when counting stars in the Milky Way - we should be able to infer $n(r)$ by counting how many stars we see as function of their luminosity. Here is how to do it: the number of stars within a solid angle $d\Omega$, and with distance between $r$ and $r+dr$, $dN(r,d\Omega)=nr^2drd\Omega$, as we had before. Now again assume them to have all the same intrinsic luminosity $L$. The trick is to change variables from distance $r$ to apparent magnitude $m$, using

$$m=-2.5\log(L/r^2)=M+5\log(r) ,$$ (1.2)

up to a constant. Recall where this came from: since the observed luminosity $\propto L/r^2$, the apparent magnitude $m=-2.5\log(L/r^2)=-2.5\log(L)-5\log(r)=M-5\log(r)$, up to a constant.^1.4 $m-M$ is called the distance modulus.) Now we're almost done, since rearranging

$$r\propto 10^{0.2(m-M)} ,$$

and by differentiating

$$dr/dm\propto 10^{0.2(m-M)}\propto r .$$

Substituting this in the expression for $dN$ gives

$$dN(m,d\Omega)\propto n(m) 10^{0.6(m-M)} dm d\Omega .$$ (1.3)

For example, if the number density is constant (and hence also $n(m)$ is constant), then, $dN(m+1,d\Omega)/dN(m,d\Omega)=10^{0.6}\approx 4$. In other words: by going one magnitude fainter, we get around 4 times as many stars. If we get fewer, then either the more distant stars are intrinsically fainter, or light has been lost on the way, or the density of stars drops (or some combination).

As a last exercise: we found that the luminosity of all stars in a shell, within a given solid angle, was independent of the radius of the shell. So, integrating over all radii, we get an infinite luminosity. Which clearly conflicts with the fact that the night sky is dark. This is called Olber's paradox. What is wrong with this reasoning?