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Why does the apparent density of galaxies drop off at larger distances?

Well, in general, I'm trying to reconcile two concepts. So, basically, my question is what does the edge of the universe look like compared to relatively closer "areas" of space? Is it more dense, or less dense? The maps of the universe I have seen show that the density of galaxies drops off at very large distances, but we also know something else--as we look further away we are looking "longer ago" and closer to the big bang--and those images look more dense. So is this to say that more distant galaxies, or galaxy-clusters are "more dense" than closer galaxies and galaxy clusters...but there are just fewer galaxies and galaxy-clusters that we can see very far away? And not all slivers of space that we map show the most distant galaxies? Or does this simply have to do with the interdepedency of light and time? Meaning that we see something different than we mathematically extrapolate to map? I'm just curious as to how these facts are reconciled. If you don't underst! and the question, just notify me that I'm not being clear and I'll rephrase it. But to be honest, I'm simply trying to reconcile the pictures on pages 22,37,40,and 72 of Hawking's Universe in a Nutshell.

The short answer is that it's harder to see things that are farther away. So while we can see almost all the galaxies nearby, we can only see the very brightest ones far away. This effect overwhelms everything else, and is responsible for the density of galaxies in those maps dropping off at large distances. So if you look at one of those maps, you can imagine that there are actually many more galaxies on the outskirts, but we just can't see them.

What if you weren't limited by this effect? What if you could see *everything* out to the edge of the observable universe? If you looked out to the edge of the universe, you'd see the universe at a time when it was very young. You would see the pieces of what would eventually become galaxies. These would appear more densely spaced together than galaxies are today because there are more of them (they haven't had the chance to merge together to reduce their number). There's also the fact that the universe was physically smaller at this early time, because it hadn't had as much time to expand. You might expect that this would translate into an even higher apparent density, but actually it doesn't. The "mini-galaxies" are projected on the sky in such a way that this doesn't happen. (An apparent distance projected on the sky turns out to be different from what you would expect from Euclidean geometry.)

The answer on this page is somewhat confusing. Probably by use of the word 'apparent'. I'm not sure what the difference between 'apparent density' and just regular density is.

If we define the "density of galaxies" as the number of galaxies per unit volume, then the density does in fact decrease as time goes on (it was greater in the past than it is now). But the question specifically related to what we observe in galaxy catalogs. Can you tell that the density of the universe was greater in the past than it is now by looking at the distribution of objects on the sky? No. Why not?

Imagine that you're looking at a very distant galaxy in one part of the sky, and then compare it to another very distant galaxy in another part of the sky. The angular separation of those two galaxies can be very large. So you could say that it "looks" like they're billions of light-years apart. But yet in the very distant past, when the universe was much much smaller than it is now, they were physically very close together. So you can't really measure the density of the universe at that early time by counting up galaxies and dividing by the volume they appear to occupy just as you would in a universe that wasn't expanding. The expansion of the universe means that objects that were very close together at the time they emitted the light that we're now seeing are spread out over the sky in a way that wouldn't happen in a universe that wasn't expanding.

June 2003, Christopher Springob (more by Christopher Springob) (Like this Answer)

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