"Scientists have measured space to be flat to high accuracy"

[sevensages]

TL;DR

Space is three dimensional. So how can space be flat?

In Max Tegmark's book Our Mathematical Universe, Tegmark wrote the following: "As we saw in the last chapter, we've measured our space to be flat to high accuracy" (99). Since space is three dimensional, I am totally baffled by Tegmark's statement that scientists have measured space to be flat to high accuracy. Since space is three dimensional, how can space be flat?

 

[Ibix]

Flat in this context means that it obeys the rules of Euclidean geometry - for example, if you draw a triangle its internal angles will add to 180°. If we lived in a "closed" or "open" universe (the other two possibilities aside from flat) the angles in sufficiently large triangles would add to more than or less than 180°.

 

[sevensages]

Flat in this context means that it obeys the rules of Euclidean geometry - for example, if you draw a triangle its internal angles will add to 180°. If we lived in a "closed" or "open" universe (the other two possibilities aside from flat) the angles in sufficiently large triangles would add to more than or less than 180°.

That definitely makes sense. Thank you for your answer. I would have never figured it out on my own.

 

[sevensages]

Flat in this context means that it obeys the rules of Euclidean geometry - for example, if you draw a triangle its internal angles will add to 180°. If we lived in a "closed" or "open" universe (the other two possibilities aside from flat) the angles in sufficiently large triangles would add to more than or less than 180°.

If the angles of a large triangle added up to more than 180 degrees or less than 180 degrees, meaning that we live in a close or an open universe, would that mean that the universe does not have three dimensions?

 

[Ibix]

If the angles of a large triangle added up to more than 180 degrees or less than 180 degrees, meaning that we live in a close or an open universe, would that mean that the universe does not have three dimensions?

No. You can measure curvature without reference to any higher-dimensional space that the curved surface might be embedded in - so the fact you can measure curvature by measuring angles in triangles doesn't necessarily mean that the surface is embedded in a higher dimensional space.

Spacetime is a four dimensional structure (a "manifold"). Space at one instant of time is a three dimensional "slice" through that four dimensional manifold. There is a lot of flexibility in how you choose to do the slicing, so "space" is a lot more subjective than spacetime. However, specifically in the kinds of spacetimes used in cosmology, there's only one sensible way to do the slicing and the geometry of the slices you get (flat, closed, or open) turns out to depend on the densities of matter and other stuff like radiation and dark energy.

Space, therefore, is a 3d manifold embedded in a 4d manifold. Spacetime is a 4d manifold that isn't embedded in anything as far as we know. Both can be curved, but it turns out that spacetime is curved and space (at least in cosmological models and cosmological conventions) isn't.

So to answer your question, depending on whether you meant space or spacetime when you wrote "universe", the universe is either three or four dimensional. If you meant space, yes it's part of a 4d structure, but if you meant spacetime then no it's not embedded in any higher dimensional space. Either way, the curvature or lack thereof doesn't imply anything about higher dimensions.

Sorry that's such a delightful mess of a yes-and-no.

(Note: there are a lot of theories that do require more than four dimensions, but we don't know which of them, if any, is correct. My answers are all based on standard GR cosmology - anybody telling you about 5+ dimensions is talking about much more speculative theories, however confident they sound.)

 

[sevensages]

No. You can measure curvature without reference to any higher-dimensional space that the curved surface might be embedded in - so the fact you can measure curvature by measuring angles in triangles doesn't necessarily mean that the surface is embedded in a higher dimensional space.

Spacetime is a four dimensional structure (a "manifold"). Space at one instant of time is a three dimensional "slice" through that four dimensional manifold. There is a lot of flexibility in how you choose to do the slicing, so "space" is a lot more subjective than spacetime. However, specifically in the kinds of spacetimes used in cosmology, there's only one sensible way to do the slicing and the geometry of the slices you get (flat, closed, or open) turns out to depend on the densities of matter and other stuff like radiation and dark energy.

Space, therefore, is a 3d manifold embedded in a 4d manifold. Spacetime is a 4d manifold that isn't embedded in anything as far as we know. Both can be curved, but it turns out that spacetime is curved and space (at least in cosmological models and cosmological conventions) isn't.

So to answer your question, depending on whether you meant space or spacetime when you wrote "universe", the universe is either three or four dimensional. If you meant space, yes it's part of a 4d structure, but if you meant spacetime then no it's not embedded in any higher dimensional space. Either way, the curvature or lack thereof doesn't imply anything about higher dimensions.

Sorry that's such a delightful mess of a yes-and-no.

(Note: there are a lot of theories that do require more than four dimensions, but we don't know which of them, if any, is correct. My answers are all based on standard GR cosmology - anybody telling you about 5+ dimensions is talking about much more speculative theories, however confident they sound.)

That is mighty interesting and informative.

What do the letters GR mean in "standard GR cosmology"?

 

[Jaime Rudas]

What do the letters GR mean in "standard GR cosmology"?

The most widely accepted cosmological model today is based on General Relativity (GR) and the cosmological principle that assumes that, at very large scales, the universe is homogeneous and isotropic.

 

[Hornbein]

That definitely makes sense. Thank you for your answer. I would have never figured it out on my own.

It might have been better to call it "straight space", because if not influenced by anything light goes in a straight line. Or maybe "parallel space", because two such beams of light that start out parallel remain parallel and the same distance apart. Or Euclidean space, because this is what was postulated by Euclid way back when.

 

[bob012345]

A mental model one can use is this, imagine you are standing on a smooth desert plain. It looks flat, very flat. But you are actually standing on a very large sphere, the Earth. The ground deviates from flat by about 8 cm per Km. If you walked a quarter way around the Earth, turned 90 degrees, walked the same distance, turned 90 degrees and walked again, you would close a triangle but the sum of the angles would be 270 degrees, not 180 degrees. A shorter walk of say 1 Km would make a triangle deviate from 180 degrees by a tiny fraction of a degree.

 

[Johannes k]

I am exploring a conceptual model where dark energy behaves like a fluid in a closed system. How does Fluid Gravity math overlap with this?

 

[Johannes k]

I do have to ask if you are in space and you use a telescope and look in every direction is there stars in each and every direction

 

[Ibix]

I am exploring a conceptual model where dark energy behaves like a fluid in a closed system. How does Fluid Gravity math overlap with this?

Please do re-read the site rules, specifically the bits about personal and non-mainstream theories.

I do have to ask if you are in space and you use a telescope and look in every direction is there stars in each and every direction

No. Things far away are seen as they were in the past, and redshifted. Far enough away and you see the universe as it was before star formation and ultimately you see the last scattering surface, so you do not see stars in every direction. This is the resolution to Olber's paradox.

You can extend spacelike lines from a point in all directions and I suspect they will all eventually intersect with a star if the universe is infinite (although I haven't checked this and one should be a little cautious about statements about infinities without checking the maths) and on the back of your head if it's closed. But that is not what you see through a telescope.

 

[Jaime Rudas]

No. Things far away are seen as they were in the past, and redshifted. Far enough away and you see the universe as it was before star formation and ultimately you see the last scattering surface, so you do not see stars in every direction. This is the resolution to Olber's paradox.

In the steady-state model, is Olers' paradox resolved by redshift?

 

[Ibix]

In the steady-state model, is Olers' paradox resolved by redshift?

It can't be, because it doesn't expand.

A point about the Einstein static universe - AFAIK it's unstable to small perturbations, so it can't have stars anyway without becoming an expanding or contracting FLRW universe. In fact, it can only be filled with an ideal FLRW medium - i.e. an ideal fluid everywhere in eternal thermal equilibrium with itself. So I think that the resolution of Olber's paradox in that spacetime is that the sky is equally bright everywhere.

 

[Jaime Rudas]

It can't be, because it doesn't expand.

A point about the Einstein static universe

Well, I don't know if Einstein's original model can be considered as steady-state model; I would say it's static. The one I'm referring to is this steady-state model that does expand.

 

[PeterDonis]

The one I'm referring to is this steady-state model that does expand.

That model obviously violates General Relativity because of its "continuous creation of matter", which is impossible by the Einstein Field Equation. AFAIK this criticism has never been fully addressed by proponents of the steady-state model (which has morphed several times as new observations have been made).

 

[Ibix]

Well, I don't know if Einstein's original model can be considered as steady-state model; I would say it's static. The one I'm referring to is this steady-state model that does expand.

Oh, I see.

I don't know. I suspect you'd need a complete theory of such a universe to answer that authoritatively, and I'm not sure there is one. I do believe that extinction is an important part of how steady state proponents propose to get redshift, so they probably propose that same mechanism to dim distant stars. Faster-than-inverse-square dimming would resolve Olber's paradox in favour of a dark sky.

 

[Jaime Rudas]

That model obviously violates General Relativity because of its "continuous creation of matter", which is impossible by the Einstein Field Equation. AFAIK this criticism has never been fully addressed by proponents of the steady-state model (which has morphed several times as new observations have been made).

The steady-state cosmological model was widely accepted by a large part of the scientific community in the 1950s, especially by those who considered any model implying a beginning for the universe unacceptable, and even more so if that model was formulated by a Catholic priest.

Observations made from the mid-1960s to the present have consistently supported the Big Bang model and contradicted the steady-state model.

But my question wasn't about whether that model fit reality, but rather whether or not it could explain Olbers' paradox through redshift.

 

[PeterDonis]

The steady-state cosmological model was widely accepted by a large part of the scientific community in the 1950s

I wouldn't say "a large part", but it was a signficant number, yes.

whether or not it could explain Olbers' paradox through redshift.

I don't know if the mathematical development of the model was sufficiently detailed to know whether it would make a prediction for this, and if so, what it would be.

 

[Jaime Rudas]

I don't know if the mathematical development of the model was sufficiently detailed to know whether it would make a prediction for this, and if so, what it would be.

The steady-state model implies that the density of the universe and the rate of expansion are constant in space and time. The question, then, is whether, under these conditions, Olbers' paradox is resolved by the effect of redshift. In other words, my question is whether redshift alone could explain Olbers' paradox.

 

[PeterDonis]

The steady-state model implies that the density of the universe and the rate of expansion are constant in space and time.

But that alone is not sufficient to evaluate the effects of redshift. To do that you also need a spacetime geometry, and, as I've said, the steady-state model is inconsistent with the Einstein Field Equation (because of the continuous creation of matter), so it's not clear what spacetime geometry we would use. Again, I'm not aware of any actual math for this that's been proposed by any steady state model proponents. And without that, the question is simply not answerable.

 

[Jaime Rudas]

But that alone is not sufficient to evaluate the effects of redshift. To do that you also need a spacetime geometry, and, as I've said, the steady-state model is inconsistent with the Einstein Field Equation (because of the continuous creation of matter), so it's not clear what spacetime geometry we would use. Again, I'm not aware of any actual math for this that's been proposed by any steady state model proponents. And without that, the question is simply not answerable.

But if spacetime is homogeneous and isotropic, isn't the metric necessarily FLRW?
And if the density is constant, isn't the Hubble parameter H necessarily also constant?
And if H is constant, doesn't that imply that $a(t)=a_0 e^{Ht}$?

 

[renormalize]

Again, I'm not aware of any actual math for this that's been proposed by any steady state model proponents. And without that, the question is simply not answerable.

Actually, a few different mathematical models have been proposed for the steady-state universe. For example, take a look at this 1960 paper by Bonnor and the references therein: The relativistic model of the steady-state universe.

 

[Jaime Rudas]

Actually, a few different mathematical models have been proposed for the steady-state universe. For example, take a look at this 1960 paper by Bonnor and the references therein: The relativistic model of the steady-state universe.

Equation 1.1 of the cited paper indicates that it is the FLRW metric with constant H.

 

[Ibix]

Equation 1.1 of the cited paper indicates that it is the FLRW metric with constant H.

Unfortunately, the paper then shows that the equations of motion of the cosmological fluid in such a system are completely undetermined. At the end of section 3 (about half way down p 477) it points out that formulae for redshift and other such phenomena in this universe therefore include an unknown function $dr/ds$ that would have to be supplied by some other physical theory. (And much of the rest of the paper is dedicated to pointing out how implausible any such theory would have to be.)