
#1
Jun1110, 08:56 AM

P: 2,889

I understand that accordingt to GR mass curves the spacetime (I'm not referring to spatial curvature k), so that the universe globally considered is a manifold with constant curvature, is this right?
If so, is this curvature positive or negative in the current cosmological model? 



#2
Jun1110, 09:19 AM

P: 48

Good question, I'd actually just logged on to ask a similar one myself. As such I shall not directly answer you but I've just been reading this wiki page that seems pretty descriptive: http://en.wikipedia.org/wiki/Shape_of_the_Universe




#3
Jun1110, 09:50 AM

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The latest data from WMAP7 is that the universe is flat to within a few percent.




#4
Jun1110, 10:03 AM

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doubt about our spacetime manifold 



#5
Jun1110, 10:09 AM

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Could you provide a source for this? In reading through WMAP's most recent findings (arXiv:1001.4538), they report: WMAP+BAO+SN (95% CL): [tex]0.0178 < \Omega_k < 0.0063[/tex] WMAP+BAO+H (95% CL): [tex]0.0133 < \Omega_k < 0.0084[/tex] 



#6
Jun1110, 10:18 AM

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#7
Jun1110, 12:57 PM

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http://books.google.com/books?id=IyJ...page&q&f=false. 



#8
Jun1110, 01:00 PM

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#9
Jun1110, 01:46 PM

P: 2,889

Thanks for the answers.
I was thinking in terms of curvature R, as in this models from cosmoogy books: a de Sitter spacetime, and an Einstein spacetime have R>0, Anti de Sitter spacetime has R<0 , Minkowski spacetime has R=0. But of course all of these models are of static universes, I didn't realize that in our dynamical (expanding) universe the curvature is not so straightforward as is it is dynamical and I guess it can vary (noncostant and nonzero) as GeorgeJones pointed out. Am I on the right track? 



#10
Jun1110, 01:59 PM

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Yes, you are on the right track. The scalar curvature, R, cannot be measured directly, but can be related to measurable dynamical quantities in a FriedmannRobertsonWalker universe:
[tex]R \propto \dot{H} + 2H^2 + \frac{k}{a^2}[/tex] where H is the Hubble parameter, k the curvature of spatial slices (the thing that WMAP constrains to be close to zero), and 'a' the scale factor. Using the Friedmann equations, this can be recast in terms of the energy content of the universe: [tex]R \propto \frac{1}{3}\rho  p = \rho\left(\frac{1}{3}  w\right)[/tex] where [tex]\rho[/tex] is the energy density of the universe and [tex]p[/tex] the pressure. The final equality is written in terms of observable parameters that are actively being constrained by current observations. 



#11
Jun1110, 02:05 PM

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bapowell, what's w stand for in the last equation?
Thanks 



#12
Jun1110, 02:10 PM

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[tex]p = w\rho[/tex] so nothing new...just a retooling of the previous equation. I write it this way because you frequently see [tex]w[/tex] constrained in experiments  it is referred to as the equation of state parameter, or simply the equation of state. For reference, [tex]w = 1[/tex] is de Sitter expansion, [tex]w = 0[/tex] is pressureless dust, and [tex]w = 1/3[/tex] is radiation. You'll notice that a universe that is filled with radiation (uniformly) has R = 0. 



#13
Jun1210, 01:11 AM

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#14
Jun1310, 05:27 AM

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Thanks 



#15
Jun1310, 05:33 AM

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#16
Jun1310, 03:12 PM

P: 2,889

[tex]R \propto  p [/tex] I know this is simple math but I'm not sure what I did wrong. Thanks in advance. 



#17
Jun1310, 05:00 PM

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