Science > Physics > Both flat and finite, yet oddly homogenous and isotropic in nature?
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Science > Physics |
| User: |
"" |
| Date: |
03 Apr 2006 04:59:22 PM |
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Both flat and finite, yet oddly homogenous and isotropic in nature? |
Is it possible to make a spacetime which is flat and finite, but also
homogenous and isotropic in nature? By homogenous and isotropic, I mean
there is one velocity for every point in spacetime which makes it
appear both homogenous and isotropic.
What I'm asking is, can you take a sphere, which is by definition NOT
flat, but finite, yet make it so that by accelerating (either linearly
or by turning) you affect spacetime in such a way that you could not
measure it as nonflat, but you would measure it as flat? Or is that
itself a contradiction? I'm wondering about this because it seems that
a universe cannot be all three of these: Finite, Flat, and
homogenous/isotropic.
Thank you for reading this!
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| User: "Sam Wormley" |
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| Title: Re: Both flat and finite, yet oddly homogenous and isotropic in nature? |
04 Apr 2006 12:35:11 AM |
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wrote:
Is it possible to make a spacetime which is flat and finite, but also
homogenous and isotropic in nature?
That sounds like a fair description of the universe... isotropic, homogeneous
and flat on the large scale. The extent appears to be at least 24 Gpc in
extent. Infinite? No way to know.
No Center
http://www.astro.ucla.edu/~wright/nocenter.html
Also see Ned Wright's Cosmology Tutorial
http://www.astro.ucla.edu/~wright/cosmolog.htm
http://www.astro.ucla.edu/~wright/cosmology_faq.html
WMAP: Foundations of the Big Bang theory
http://map.gsfc.nasa.gov/m_uni.html
WMAP: Tests of Big Bang Cosmology
http://map.gsfc.nasa.gov/m_uni/uni_101bbtest.html
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| User: "Henning Makholm" |
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| Title: Re: Both flat and finite, yet oddly homogenous and isotropic innature? |
04 Apr 2006 02:09:50 PM |
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Scripsit
Is it possible to make a spacetime which is flat and finite, but also
homogenous and isotropic in nature? By homogenous and isotropic, I mean
there is one velocity for every point in spacetime which makes it
appear both homogenous and isotropic.
I don't think so. My differential geometry skillz are not strong
enough to prove it rigorously, but here is a kind of hand-wavy
argument by contradiction:
1. Select an event A and a distinguished velocity V (i.e., timelike
unit vector) at that event.
2. Consider the space S spanned by all geodesics through A orthogonal to V.
3. Because the spacetime is flat, S must be a geodesic (hyper)surface
with purely Riemannian signature (that is, locally Euclidic, so we
do not have to worry about null vectors in the following steps).
4. While waving hands vigorously, assert that finiteness of the
original spacetime means that S must be finite too, and that if it
isn't it is because we chose V wrongly in step 1, so go back and
choose a better V.
5. Thus we have reduced the problem to finding out whether there is a
3D _Riemannian_ manifold that is flat everywhere, yet finite and
isotropic at at least one point (namely A).
6. Find a geodesic L in S that connects A with A nontrivially:
a) Define distance within S conventionally as the length of the
shortest curve between two points. (Such a curve will be
geodesic).
b) Because S is finite, the distance between any two points is
bounded by d, the diameter of S.
c) Choose a random geodesic L0 from A and go along it for a
distance of 2d, reaching point B. B is connected to S by
another geodesic K of length at most d, so we have a 2-gon
in S with corners A and B and different sides. Because
everything is smooth, we can adjust the direction and
length of L0 such that its far endpoint moves smoothly
along K from B to A. Its length cannot shrink by more than
the length of K, so it ends up as a nontrivial geodesic L
from A to A.
7. A was assumed isotropic, så if one geodesic of length |L| from A
ends at A itself, it must be that _all_ geodesics of length |L|
from A end at A itself.
8. Now consider two neighbouring geodesics H and K radiating out
from A, both parameterised by curve length. We have that
dist(H(t),K(t)) is 0 for t=0, nonzero for small t's, and 0
again for t=|L|. Thus
d²/dt² dist(H(t),K(t))
cannot stay zero forever no matter how close we choose H and K.
This directly says that S has positive Gaussian curvature, a
contradiction.
--
Henning Makholm "Jeg køber intet af Sulla, og selv om uordenen griber
planmæssigt om sig, så er vi endnu ikke nået dertil hvor
ordentlige mennesker kan tillade sig at stjæle slaver fra
hinanden. Så er det ligegyldigt, hvor stærke, politiske modstandere vi er."
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