Metric Expansion of Space

waterfall

I'm coming with a good background in Big Bang expansion as the following sci-am article shows (which I've mastered):

http://space.mit.edu/~kcooksey/teach...icAmerican.pdf

What I'd like to understand is this. Expansion can only be felt in unbound system. Meaning brooklyn doesn't expand because matter are bounded with one another. But in depth of space in between super galactic clusters where there are no matters. Space expand. Can we say the space there is Minkowski flat (since there is no matter to cause spacetime curvature)? If so.. then this minkowski flat space is expanding? But they said space expansion automatically means curve spacetime. Is this true? There seems to be some contradictions. Can anyone help clear up this confusion? Thanks.

Nabeshin

There's a difference between flat space and flat spacetime. The universe, as a whole, has approximately flat space, but spacetime is not flat -- of course, we know it is expanding. The FRW metric, which describes the geometry of the universe as a whole, is only applicable on scales large enough that the universe appears homogeneous and isotropic. So, it doesn't make sense to apply it to a small region where there is either an overdensity (Earth, sun, galaxy) or an underdensity (cosmic void). Strictly speaking, it only makes sense when you have a uniform density.

waterfall

But according to Dalespam in thread http://www.physicsforums.com/showthread.php?t=582440 at msg # 4:

"When we say that space is expanding we are talking about a foliation of the spacetime manifold along the time coordinate. We are then comparing different distances in different foliated sub-manifolds.

Since there is only one spacetime manifold I don't know what meaning could be ascribed to the phrase "expanding spacetime". What comparison is possible?"

Hence he meant "spacetime expanding" is an invalid term. The right term is space expanding. So can you please reformulate what you said above using his context? (and do you agree with him)

Also why can't the space in "flat spacetime" expand? Why is there always curvature when there is expansion?

Chalnoth

The thing to be careful of here is that you don't need to have matter at every point in space to have space-time curvature. To take a very simple example, go much above the Earth's atmosphere, and there's basically no matter up there. But the space-time around the Earth is still rather curved, which is why satellites orbit the Earth, and why the Moon does as well.

So if you just take out a local region of space, a region with zero or very little matter, you are missing a good part of the picture, because you don't know how the surrounding matter (if any) is impacting the curvature of the local region.

Just to drive the point home: the Schwarzschild metric which describes the space-time around a non-rotating black hole is a vacuum solution to Einstein's equations: there is no matter in the metric (note: there is a singularity, but it would be dividing by zero to include that in the metric, which would screw up the math). But it would be silly to conclude that just because there is no matter in the Schwarzschild metric that it is the same as flat space-time: a black hole is about as far as you can get from flat space-time.

Nabeshin

OK, I think I see the point here. Let's look for a minute at the metric for a flat space universe:
[tex]ds^2=-dt^2+a^2(t)[dr^2+r^2d\Omega^2][/tex]

What Dalespam was saying is that we can foliate this metric into individual slices, each labeled by a unique time. (This is generally true, but especially easy here). So at time [itex]t=t_1[/itex], the spatial slice of the metric looks like:
[tex]ds^2=a^2(t_1)[dr^2+r^2d\Omega^2][/tex]
While at some later time [itex]t=t_2[/itex], it looks like:
[tex]ds^2=a^2(t_2)[dr^2+r^2d\Omega^2][/tex]
In the case of an expanding universe, [itex]a(t_2)>a(t_1)[/itex], and someone measuring distances will observe things to be getting farther apart. The spatial portion of the metric is expanding. I hope that clears it up, my original choice of word was confusing and I think Dalespam's is superior.

Now, to see why you can't have expansion in flat spacetime, the metric would be simply:
[tex]ds^2=-dt^2+[dr^2+r^2d\Omega^2][/tex]
Performing the same foliation procedure as before will yield equal distances at nonequal times, i.e. the universe is static. You need the a(t) factor, which breaks from flat minkowski spacetime, in order to have an expanding universe.

marcus

You are short-changing yourself--cheating yourself out of reading the full SciAm article of which your link just gives an abridged versions without several illustrated special text boxes and stuff.
I have a link in my signature to the full SciAm article. I would encourage you to "master" that as well
http://www.mso.anu.edu.au/~charley/p...DavisSciAm.pdf

clamtrox

You need to be careful here again - Minkowski spacetime is of course flat, and in the usual coordinates also the 3-dimensional spatial slices are flat. However, if you use FRW-metric and take the limit of rho->0, you end up with different coordinates where the spatial slices are actually expanding as a~t. The spacetime curvature is same as before (as it is a geometric quantity and therefore coordinate independent) but spatial curvature depends on your coordinate choice and therefore is not coordinate-independent.

You can even find weirder coordinate transformations, like the Rindler coordinates.