
#1
Jun1207, 01:11 PM

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This is a very basic question but what is the difference between isotropic and homogeneous? I mean, I can imagine a universe that would be isotropic but not homogeneous although this seems to select a prefered frame (am I worng on this?) But I don't understand how it would be possible to have a universe which would be homogeneous without being isotropic. It seems to me that homogeneous implies isotropic. And yet, I must be wrong because people would not insist on having both conditions satisfied.
Thanks 



#2
Jun1207, 01:35 PM

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As an analogy, consider an electric field given by
[tex]\vec{E} = E \hat{x}[/tex] at every point in flat space. 



#3
Jun1207, 04:19 PM

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nrqed, this is straight out of the glossary section of the book,
"Universe", 6th ed. by Freedman and Kaufmann... isotropic Having the same property in all directions. homogeneous Having the same property in one region as in every other region. 



#4
Jun1207, 04:49 PM

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Isotropic vs homogeneousSee http://xxx.lanl.gov/abs/grqc/9812046 for an excellent review of cosmological models, including the question of symmetry. See the website in my sig (happy b'day to John B, happy b'day to he!) for many more citations to good review papers and textbooks. 



#5
Jun1207, 06:04 PM

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We normally 'insist' on both conditions since that is what is observed. Reasonable models that do not satisfy these conditions need to show how we can be spoofed into thinking this is the case, on example being that the Universe may be inhomogeneous on scales large that the observable Universe. In principle it's possible, and may solve some current conundrums, but obviously testing these types of theories it difficult!




#6
Jun1207, 06:26 PM

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http://relativity.livingreviews.org/...11/index.html The dipole anisotropy is removable by switching to a comoving frame field, as used in standard presentations of simple cosmological models. I discussed this in detail in at least one expository post to sci.physics.* many years ago. But the observed inhomogeneities in the CMB are not removeable. To be sure, this is a very small effect, so that gross properties of our universe are reasonably well modeled on a very large scale by a simpleminded FRW model (with nonzero Lambda, as it turns out). 



#7
Jun1207, 06:58 PM

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SO I initially thought that homogeneous was more stringent than isotropic. And now you give me the impression that isotropic is more stringent than homogeneous. But you say that there can be models which are homogeenous and yet not isotropic. Does that require vector fields? George made me think about vector fields (I was only thinking about scalar fields when I initially posted). I can see how a vector field may be homogeneous but not isotropic (actually, it seems to me that a homogeneous vector field is necessarily anisotropic). So my questions are now (I ask the questions in such a way that the answers all seem to be Yes to me): for a scalar field, does isotropy imply homogeneity? For a nonzero vector field, does homogeneity imply anisotropy? For a vector field, does isotropy implies that the vector field be zero? As for a tensor field, this is not as clear to me. I know that the usual [itex] T_{\mu \nu} [/itex] of the basic cosmological models is isotropic and homogeneous but at first sight, it would have seem to me that a tensor (like a vector field) would have to be zero to be both isotropic and homogeneous. Maybe that's not the case because it it decomposed into its trace plus other stuff, the trace is a scalar and may be nonzero. or maybe I am going at it completely the wrong way. Thansk again 



#9
Jun1307, 01:19 AM

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#10
Jun1307, 01:45 AM

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PF Gold
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You can explain cosmic acceleration by the hypothesis that the Earth is in the very centre of a vast relative void, but this would require us to occupy a unique position in the universe. Garth 



#11
Jun1307, 01:55 AM

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No. We observe homogeneity. This is seen in large galaxy surveys such as 2DFGRS and SDSS.
As I mentioned in post #5 we may, as you suggest, be able to explain acceleration by inhomogeneities on scales larger than the observable Universe (known as 'super horizon sized perturbations'). The problem would be explaining why the universe is inhomogeneous below ~10 Mpc, becomes homogeneous between scales of ~101000's of Mpc and then suddenly becomes inhomogeneous on even larger scales. There is also the issue of testing of such a theory. Of course with more work on these ideas (which is underway) it may turn out that there is an observable consequence that could discriminate between this mechanism for acceleration and others, such as negative pressure energy components. To my knowledge such a discriminant is yet to be established, and a plausible physical origin of these 'super horizon' perturbations has not been proposed. 



#12
Jun1307, 02:05 AM

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PF Gold
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That is inference from observations made locally. To observe homogeneity would need to obtain readings of density, pressure, temperature, etc. at all points in the universe. Garth 



#13
Jun1307, 02:12 AM

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Correct, we don't see 'the Universe is Homogeneous, signed God.' written in the sky. We infer that the Universe is homogeneous, since when we look at it it appears to be homogeneous...
Can you suggest an observation that tells us some information without the need for 'inference'? 



#14
Jun1307, 05:36 AM

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PF Gold
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You might be able to argue that if the universe were not homogeneous to a high degree then everything would have collapsed into a few hypermassive black holes and biological evolution would be impossible anywhere. This might be countered by the anthropic reasoning that life is only possible at a 'unique' homogeneous spot in an otherwise inhomogeneous universe, or in a 'unique' universe in a multiverse ensemble. If that is not a valid deduction then I would argue that to observe homogeneity otherwise you would have to travel around the universe. The precise pattern of atomic spectra observed from objects near and far could be taken as evidence that the Copernican Principle holds for laws of physics, therefore there is no reason to suppose it cannot be applied generally to this universe. Garth 



#15
Jun1307, 11:29 AM

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Cosmic coordinate time [itex]t[/itex] is used to split spacetime into space and time. Space at time [itex]t[/itex] is the set of all events that all have this value of [itex]t[/itex], i.e., an instant in time. Galaxies (with zero peculiar velocities) follow worldlines that 4velocities [itex]u[/itex] orthogonal to space at any instant in time. From the point of view of these galaxies, the cosmological stressenergy tensor field is that of a perfect fluid, so, at any event [itex]p[/itex],
[tex]T\left(p\right) = \left( \rho\left(p\right) + P\left(p\right) \right) u \otimes u + P\left(p\right) g.[/tex] Let [itex]p[/itex] and [itex]q[/itex] be distinct events in the same space part of spacetime, and so are labelled by the same value of [itex]t[/itex] (two galaxies considered at the same time). If a galactic observer at [itex]p[/itex] chooses an othonormal frame (tetrad) with respect to which measurements are made, then [itex]e_0[/itex] must be [itex]u[/itex], but has complete spatial rotational freedom for the remaining members of the tetrad. As long as, at event [itex]p[/itex], [itex]\rho[/itex] and [itex]P[/itex] don't depend on spatial directions (spatially isotropic at event [itex]p[/itex]), this rotational freedom doesn't affect the tetrad components of [itex]T[/itex]. Roughy, spacetial rotations don't effect: [itex]u[/itex], since it's time; [itex]g[/itex], because, restricted to space, [itex]g[/itex] is a constant multiple of the identity (with respect to tetrads). Now consider the tetrad of a galactic observer at event [itex]q[/itex]. As long as [itex]\rho[/itex] and [itex]P[/itex] depend only on [itex]t[/itex], the observer at [itex]q[/itex] will measure the same stressenergy components as the observer at [itex]p[/itex]. This stress energy tensor is homogeneous in space. 



#16
Jun1307, 11:44 AM

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Regards 


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