The Shape of Spacetime.

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  • #126
bapowell
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Are you sure that the torus does not cumulate positive and negative curvatures?
It has positive and negative principle curvatures, but the Gaussian curvature is zero.
n-spheres are locally flat as well
They are not -- they are positively curved everywhere.

it is simply making the effort to not introduce hypotheses that are not strictly riquired by new observations.
Right, but none of your criteria are strictly required by observations. We have no data supporting the size (finite vs. infinite) of the universe, whether it is globally compact or bounded, or whether it is globally homogeneous. What we know from observations of the observable universe is that it is approximately homogeneous and close to flat locally. Empirically, all three manifolds of constant curvature -- Euclidean (including toroidal and other flat geometries), spherical, and hyperbolic spaces -- are equally in the running.
 
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  • #127
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It has positive and negative principle curvatures, but the Gaussian curvature is zero.
My feeling is that this curvature is more mathematical than concrete, but it is just a feeling

They are not -- they are positively curved everywhere.
to faint to be mesurable. Sir Eddington calculated that for a small circle with a radius of 5 m with a mass of 5 tonnes at its center, one would deviate from pi only at the level of its 24th decimal..

Right, but none of your criteria are strictly required by observations. We have no data supporting the size (finite vs. infinite) of the universe, whether it is globally compact or bounded, or whether it is globally homogeneous. What we know from observations of the observable universe is that it is approximately homogeneous and close to flat locally. Empirically, all three manifolds of constant curvature -- Euclidean (including toroidal and other flat geometries), spherical, and hyperbolic spaces -- are equally in the running.
I believed that night would not have been black in an infinite universe? (necessarily forever..)
 
  • #128
phinds
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I believed that night would not have been black in an infinite universe? (necessarily forever..)
No, that was dealt with conclusively quite some time ago as not being the case. I don't have a reference offhand but I assure you it is true. It has to do with the expansion of the universe. For one thing, it is irrelevant whether the universe is infinite since the only light that reaches us is from the OBSERVABLE universe.
 
  • #129
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No, that was dealt with conclusively quite some time ago as not being the case. I don't have a reference offhand but I assure you it is true. It has to do with the expansion of the universe. For one thing, it is irrelevant whether the universe is infinite since the only light that reaches us is from the OBSERVABLE universe.
OK thanks. The observable universe is the part for which the redshifted wavelengths remain in the visible range?
 
  • #130
bapowell
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I believed that night would not have been black in an infinite universe? (necessarily forever..)
In addition to phinds' response, I'll add that the finite age of the universe also resolves this paradox.
 
  • #131
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In addition to phinds' response, I'll add that the finite age of the universe also resolves this paradox.
something infinite has necessarily always been infinite. The finite age proves that the euclidean 3D space cannot be retained
 
  • #132
phinds
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something infinite has necessarily always been infinite. The finite age proves that the euclidean 3D space cannot be retained
You are talking about infinite in space, which may or may not be true. The post you were commenting on was refering to the finite AGE of the universe.
 
  • #133
phinds
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OK thanks. The observable universe is the part for which the redshifted wavelengths remain in the visible range?
Technically, that is exactly correct. In practise, "visible" requires extraordinarily strong telescopes for a long period (see the Hubble Deep Field).
 
  • #134
bapowell
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something infinite has necessarily always been infinite. The finite age proves that the euclidean 3D space cannot be retained
By finite age, I mean finite time since the big bang. Obviously, if the big bang occurred 13 billion years ago, then there are still CMB photons, as well as starlight, that have simply not had time to reach us yet. This is why the finite age resolves the paradox. It has nothing to do with the size of the spatial geometry.
 
  • #135
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By finite age, I mean finite time since the big bang. Obviously, if the big bang occurred 13 billion years ago, then there are still CMB photons, as well as starlight, that have simply not had time to reach us yet. This is why the finite age resolves the paradox. It has nothing to do with the size of the spatial geometry.
OK I understand
 
  • #136
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Technically, that is exactly correct. In practise, "visible" requires extraordinarily strong telescopes for a long period (see the Hubble Deep Field).
I asked this question because I read an other interpretation (with which I have some concerns) that the unobservable universe located behind the horizon, corresponds to galaxies for which the recession velocities exceed light speed. Is it erroneous according to you?
 
  • #137
phinds
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I asked this question because I read an other interpretation (with which I have some concerns) that the unobservable universe located behind the horizon, corresponds to galaxies for which the recession velocities exceed light speed. Is it erroneous according to you?
Not only is it NOT erroneous, it is an understatement in that galaxies IN the observable universe are "now" ("now" gets a bit tricky) are already receding from us FTL. In fact, those at the edge of the OU are receding at about 3c
 
  • #138
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Not only is it NOT erroneous, it is an understatement in that galaxies IN the observable universe are "now" ("now" gets a bit tricky) are already receding from us FTL. In fact, those at the edge of the OU are receding at about 3c
I definitely cannot understand this point. My intuitive expectation was that the expansion rate cannot exceed c because it it was the case, everything would disconnect at once, even between the sun and earth and between your eyes and your screen..

Did you read the problem of the ants on a rubber rope <http://en.wikipedia.org/wiki/Ant_on_a_rubber_rope> [Broken]
 
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  • #139
phinds
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I definitely cannot understand this point. My intuitive expectation was that the expansion rate cannot exceed c because it it was the case, everything would disconnect at once, even between the sun and earth and between your eyes and your screen..

Did you read the problem of the ants on a rubber rope <http://en.wikipedia.org/wiki/Ant_on_a_rubber_rope>[/QUOTE] [Broken]

The first thing you need to do when studying either cosmology or quantum mechanics it TOTALLY get rid of the concept that your intuition is worth squat. It is not. It's something we all have to get used to.

What you are not understanding is that nothing is traveling FTL in the same reference frame. The universe is expanding. The recession rate has nothing to do with c.
 
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  • #140
Drakkith
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I definitely cannot understand this point. My intuitive expectation was that the expansion rate cannot exceed c because it it was the case, everything would disconnect at once, even between the sun and earth and between your eyes and your screen..

Did you read the problem of the ants on a rubber rope <http://en.wikipedia.org/wiki/Ant_on_a_rubber_rope>[/QUOTE] [Broken]

The expansion is a rate, meaning that the recession velocity increases over distance. The further away something is the faster it recedes. So a galaxy can receded from us at 0.01c that is relatively nearby while another that is very very far away can recede at 2c or 3c or whatever. Note that this also includes light emitted from those galaxies. A galaxy receding from us at 2c that emits light away from us would NOT catch up to the light. In the galaxies frame of reference it is stationary and the light moves at 1c away from it.
 
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  • #141
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I know that v=H*D. If the Hubble constant was constant, this would lead to an exponential recession.
I understood how H is determined (through the variations of redshifts as fonction of distance). I also see how distances can be determined.
But I dont know how recession velocities are measured? do you knwo?
 
  • #142
Chronos
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There are no galaxies residing beyond the surface of last scattering whose photons have not yet reached us. Nothing but hot plasma exists beyond z~1100 [source of the cmb].
 
  • #143
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Recession velocities are measured via redshift. Research the Doppler Effect to understand why Redshift comes about. We can determine which spectral emission lines we can expect to see in stars of certain materials (I can't remember how,) and see how redshifted the stars are. Plug into our easily derivable formula for the relativistic doppler effect, and we're done.
 

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