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Isotropic vs homogeneous

by nrqed
Tags: homogeneous, isotropic
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nrqed
#1
Jun12-07, 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
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George Jones
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Jun12-07, 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.
Nasher
#3
Jun12-07, 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.

Chris Hillman
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Jun12-07, 04:49 PM
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Isotropic vs homogeneous

Quote Quote by nrqed View Post
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.
Wrong way around. Think about a scalar field in [itex]E^3[/itex]. If it's isotropic, its gradient (which you can think of as a vector) must not pick out any direction at any point. This can happen only if the gradient vanishes everywhere. Thus, the scalar field must be constant. Adding a time variable, the same argument shows that the scalar field must depend only on time, i.e. it must be homogeneous in surfaces of "constant time". In a cosmological model, these surfaces will be the hyperslices orthogonal to the world lines of the matter.

Quote Quote by nrqed View Post
And yet, I must be wrong because people would not insist on having both conditions satisfied.
They don't. There are plenty of cosmological models (dust solutions) which are homogeneous but not isotropic, and there are models which are not even homogeneous. In another thread I recently gave a simple explicit example of a homogeneous but anisotropic model, the Bianchi II dust, an exact dust solution which exhibits Kasner epochs similar to those exhibited by the Mixmaster model, aka the Bianchi IX dust. I've also given the line element defining the Stephani dust, an simple exact dust solution which is an inhomogeneous model (a perturbation of an FRW model). The Kantowski-Sachs dusts provide further popular examples of anisotropic but homogeneous cosmological models.

See http://xxx.lanl.gov/abs/gr-qc/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.
Wallace
#5
Jun12-07, 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!
Chris Hillman
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Jun12-07, 06:26 PM
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Quote Quote by Wallace View Post
We normally 'insist' on both conditions since that is what is observed.
Er, COBE? Dipole anisotropy? (That's anisotropy.) Fluctuations in the CMB? (That's inhomogeneity.)
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 simple-minded FRW model (with nonzero Lambda, as it turns out).
nrqed
#7
Jun12-07, 06:58 PM
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Quote Quote by Chris Hillman View Post
Wrong way around. Think about a scalar field in [itex]E^3[/itex]. If it's isotropic, its gradient (which you can think of as a vector) must not pick out any direction at any point.
yes, I understand. I was thinking of examples which would be inhomogeneous but isotropic relative to a single point only (which is why I mentioned that a special frame would be picked, in my post). I understand that imposing isotropy at all points implies a constant scalar field.

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 non-zero 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.



This can happen only if the gradient vanishes everywhere. Thus, the scalar field must be constant. Adding a time variable, the same argument shows that the scalar field must depend only on time, i.e. it must be homogeneous in surfaces of "constant time". In a cosmological model, these surfaces will be the hyperslices orthogonal to the world lines of the matter.



They don't. There are plenty of cosmological models (dust solutions) which are homogeneous but not isotropic, and there are models which are not even homogeneous. In another thread I recently gave a simple explicit example of a homogeneous but anisotropic model, the Bianchi II dust, an exact dust solution which exhibits Kasner epochs similar to those exhibited by the Mixmaster model, aka the Bianchi IX dust. I've also given the line element defining the Stephani dust, an simple exact dust solution which is an inhomogeneous model (a perturbation of an FRW model). The Kantowski-Sachs dusts provide further popular examples of anisotropic but homogeneous cosmological models.

See http://xxx.lanl.gov/abs/gr-qc/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.
Thank you.... I was talking about the very basic standard textbooks cosmological models. Those more sophisticated models must have inhomogeneities that are pretty constrained...


Thansk again
Chronos
#8
Jun13-07, 01:08 AM
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Quote Quote by nrqed
As for a tensor field, this is not as clear to me. I know that the usual 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.
Correct. Rethink your position.
Wallace
#9
Jun13-07, 01:19 AM
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Quote Quote by Chris Hillman View Post
Er, COBE? Dipole anisotropy? (That's anisotropy.) Fluctuations in the CMB? (That's inhomogeneity.)
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 simple-minded FRW model (with nonzero Lambda, as it turns out).
Thanks Chris. Just out of curiosity, did anyone else read my post to mean that we observe no CMB power spectrum and have no peculiar velocity? I guess saying 'the Universe is observed to be statistically isotropic and homogeneous below a certain size scale defined by [tex]\sigma_8[/tex] which is the RMS density fluctuations in spheres of radius 8Mpc' get to be a bit of a mouthful after a while
Garth
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Jun13-07, 01:45 AM
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Quote Quote by Wallace View Post
We normally 'insist' on both conditions since that is what is observed.
Actually, up to the limit of the CMB anisotropies, we observe isotropy (in the CMB - having subtracted the dipole due to our own motion wrt the SLS) and infer homogeneity using the Copernican Principle.

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
Wallace
#11
Jun13-07, 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 ~10-1000'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.
Garth
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Jun13-07, 02:05 AM
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Quote Quote by Wallace View Post
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 ~10-1000'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.
Wallace,

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
Wallace
#13
Jun13-07, 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'?
Garth
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Jun13-07, 05:36 AM
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Quote Quote by Wallace View Post
Can you suggest an observation that tells us some information without the need for 'inference'?
The observation that we exist?

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 hyper-massive 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
George Jones
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Jun13-07, 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 4-velocities [itex]u[/itex] orthogonal to space at any instant in time. From the point of view of these galaxies, the cosmological stress-energy 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 stress-energy components as the observer at [itex]p[/itex]. This stress energy tensor is homogeneous in space.
nrqed
#16
Jun13-07, 11:44 AM
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Quote Quote by George Jones View Post
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 4-velocities [itex]u[/itex] orthogonal to space at any instant in time. From the point of view of these galaxies, the cosmological stress-energy 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 member 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, spacetime 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 stress-energy components as the observer at [itex]p[/itex]. This stress energy tensor is homogeneous in space.
Very informative, George. Thank you for taking the time to make this clear!

Regards


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