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Metric Expansion of Space 
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#1
Mar112, 11:54 PM

P: 381

I'm coming with a good background in Big Bang expansion as the following sciam 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. 


#2
Mar212, 12:25 AM

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#3
Mar212, 12:36 AM

P: 381

"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 submanifolds. 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? 


#4
Mar212, 01:32 AM

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P: 4,782

Metric Expansion of Space
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 spacetime around a nonrotating 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 spacetime: a black hole is about as far as you can get from flat spacetime. 


#5
Mar212, 12:02 PM

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P: 2,193

[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. 


#6
Mar212, 12:34 PM

Astronomy
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P: 23,108

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 


#7
Mar512, 08:16 AM

P: 939

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


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