# Shape of the universe

• B
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## Main Question or Discussion Point

This may be a silly question, but how do we give a shape to the universe if it is The same from every point we look?

Arman777
Gold Member
This may be a silly question, but how do we give a shape to the universe if it is The same from every point we look?
We dont give a shape.

We can make an argument that universe looks the same at larger scales (you can further search cosmological principle). This argument reduces the possible geomtries that universe can have.

1-Flat with zero curvature
2-Spherical geometry with positive curvature
3-Hyperbolic geometry with negative curvature

From observation we can conclude or derive the geometry of universe.

Recent data (Plank 2015 results or etc) shows that universe is flat and has zero curvature.

It may have another shape like a torus maybe but even in that case the size of torus must be huge cause obersvable universe seems flat over a large scale.

PeroK
Homework Helper
Gold Member
This may be a silly question, but how do we give a shape to the universe if it is The same from every point we look?
You could ask the same about the surface of the Earth, assuming it were ideally spherical. Its curvature can be defined by its being a 2D surface embedded in 3D space: i.e. it's the surface of a 3D sphere.

Or, its shape can be defined using differential geometry to define the infinitesimal distance in any direction. Using spherical polar coordinates this is:

##dS^2 = R^2(d\theta^2 + \sin^2 \theta d\phi^2)##, where ##0 \le \phi < 2\pi## and ##0 \le \theta \le \pi##, and ##R## is some parameter, which equates to the radius of the 3D sphere above.

This "line element", in fact, encapsulates the shape of the sphere's surface without directly appealing to an embedding in a higher dimension.

Note that this is the distance along the surface, not taking any shortcuts through the body of the Earth!

Gold Member
I get the analogy of the Earth, but as regards to the universe what does flat mean, surly due to gravity and dark matter the universe must be globular, but how can that be flat?

PeroK
Homework Helper
Gold Member
I get the analogy of the Earth, but as regards to the universe what does flat mean, surly due to gravity and dark matter the universe must be globular, but how can that be flat?
It's flat on a large scale. But, locally, where there are galaxies, black holes or simply where there are stars or planets you have locally curved spacetime.

Arman777
Gold Member
surly due to gravity and dark matter the universe must be globular, but how can that be flat?
Why it would be globular ? And how can it be globular and flat at the same time ?

There's something you should mention, are you talking about the universe or observable universe.

We cant now the geometry of the universe, (since we cant travel or observe the whole universe etc), but we can claim that observable universe is flat. Becasue that is what we observe (by experiment and measurement)

The universe can have a spherical geomtery or it can be just flat as ours, but If its sphere the R (radius of the sphere) must be so huge that, we observe, the observable universe as flat.

but as regards to the universe what does flat mean
Mathematically it means that curvature is zero, In example a plane is a flat becasue, when you set a triangle, and you measure the angles it gives you ##π##. But in spherical geometry it gives you more then ##π##.

Of course in metric terms it would be more different

Try to think that you are a small ant, and you are travelling on a piece of paper. Wherever you walk or go it feels like you are on a flat surface and when you draw a triagle and measure the angles you get ##π##. Observable universe is a piece of paper and you are an ant.

If we were bigger then ant, then we could have notice the curvature, but since we are not, we cannot know the real geometry of the universe.

(thats why our ancestors thought that earth is flat in the first place casue it was hard to observe that earth is spherical, in our case we cannot observe or dont have tools to see the real geometry of the universe, we are more like a bacteria respect to the universe)

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phinds
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2019 Award
The universe can have a spherical geomtery or it can be just flat as ours, but If its sphere the R (radius of the sphere) must be so huge that, we observe, the observable universe as flat.
But if the universe were a sphere, that would clearly imply a preferred direction and there is zero evidence of such a thing.

PeterDonis
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as regards to the universe what does flat mean
It means spatially flat: a spacelike slice of constant time for a comoving observer is Euclidean 3-space.

PeterDonis
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if the universe were a sphere, that would clearly imply a preferred direction
No, it woudn't; there is no preferred direction in a 3-sphere. A 3-sphere is spatially isotropic.

You might be confusing this with the case of a spherically symmetric, but not homogeneous, spacetime such as Schwarzschild spacetime. In this case, yes, the radial direction is different from the tangential directions. But that's due to the fact that there is a mass at the center, but not anywhere else; i.e., the spacetime is not homogeneous. In a closed FRW universe, which has the spatial geometry of a 3-sphere, that's not the case: the density is the same everywhere, and all directions are the same

Gold Member
Thank you for your replies, I can see now that when we talk about the universe having shape we only mean the observable universe

phinds
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2019 Award
Thank you for your replies, I can see now that when we talk about the universe having shape we only mean the observable universe
That's not very exciting since the OU is just a sphere centered on you.

Using the balloon analogy, what's inside the balloon skin? Is it the Universe from a moment ago?

phinds
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2019 Award
Using the balloon analogy, what's inside the balloon skin? Is it the Universe from a moment ago?
You are misunderstanding the analogy. I recommend the link in my signature.

Using the balloon analogy, what's inside the balloon skin? Is it the Universe from a moment ago?
That's one of the biggest flaws of such analogies.
Ther balloon does not represent the universe nor anything about the shape, topology or geometry of the universe. The SURFACE of the balloon 'skin', where drawings of galazy clusters might be made, represents the Relative distances between those clusters, that's about it.

That's not very exciting since the OU is just a sphere centered on you.
I find a universe centred on !e particularly exciting :)

- a joke only.

I find a universe centred on !e particularly exciting :)

- a joke only.
But my research shows its centered on me. :)

A Play. Title: Unseen Shapes (the "balloon analogy" revisited)

Scene 1: You are in what appears locally to be an isotropic Flatland (E²). You look around in every direction. Far away objects
are rather dim, and you suspect that beyond your range of vision, there is more. Indeed, the gods of mathematics have
revealed that if you could see far enough (which you cannot), you would see the same small point-like object in every direction. If these gods
are correct, then you must live in S² (a 2-sphere, "the balloon").

Scene 2: You are now in what appears locally to be an isotropic Solid-land (E³). You look around in every direction. Far away objects
are again rather dim, and you suspect that beyond your range of vision, there is more. Indeed, the gods of mathematics have
revealed that if you could see far enough (which you cannot), you would see the same small point-like object in every direction. If these gods
are correct, then you must live in S³ (a 3-sphere).

End of Play. Do you see any problem with its logic?

Tom McFarland

PeterDonis
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2019 Award
Do you see any problem with its logic?
Not as far as it goes, but in our real universe there are no "gods of mathematics" to tell us whether in fact we could see the same object in all directions if we could see far enough. And we have no evidence that we could. So we have no evidence that your play is describing our actual universe.

PeterDonis:

Sorry to post this question twice. I feared that I had posted incorrectly the first time.

On the issue......

Is it not true that the big bang thesis posits a point-like start to our universe which we are
currently researching, limited by our ability to "see" before the origin of CMB. Indeed,
Even the universe at "last scattering" (when the CMB was produced) would qualify as
sufficiently point-like and Omni-directional to force the current shape to be close to S³, via the above logic.

Thus, in scene 1 above, if an ideal point P is seen in all directions, you get S² for sure, but if P is merely
very small compared to all of space (a big point, like the "last scattering" surface), this still forces the topology to be S²,
like a balloon with your finger plugging its hole (P being your finger). The same applies to S³.

Peter Donis:

Here is another neat topological way to convince yourself that the universe must have the topology of S³.

First, consider the universe starting with us now, extending out (and backward in time) to the surface S of last scattering (at which CMB was produced). We see this surface isotropically in all directions in microwave light. This is a 3-ball (the interior of a big 2-sphere plus boundary S). Hold this in your mind.

Secondly, consider the postulated universe from the big bang out to the same surface of last scattering S. Currently we cannot see this region, but it would also be a 3-ball with the same boundary S as above, if the gods of mathematics are correct. Hold this in your mind.

Now, topologically merge these two 3-balls along their common boundary S. This union is a topological 3-sphere, the universe from big bang to now.

This merger is analogous to merging two 2-dimensional discs (like the upper and lower crusts of a pie) along their circular boundaries, to obtain a 2-sphere.

PeterDonis
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2019 Award
Is it not true that the big bang thesis posits a point-like start to our universe
No, it is not true. The "point-like start" is an artifact of a particular class of idealized models. It is not a feature of our actual current best-fit model of the universe.

However, let's put that aside and consider the idealized models I just referred to. Even in those idealized models, it is not true that the spatial topology of the universe must be ##S^3##. It is perfectly possible to have such an idealized model, with a "point-like start", with spatial topology ##R^3## (i.e., infinite). The reason is that the "point-like start" is not a point in space; it is a moment of time. Your arguments for the spatial topology having to be ##S^3## are only valid if the "point" that is visible in all directions is a point in space. (The technical way of putting it is that, for your argument to be valid, the "point" must be a timelike curve--it's a curve when you include the time dimension. But the "point-like start" in the idealized models I referred to is spacelike, not timelike.)

Here is another neat topological way to convince yourself that the universe must have the topology of S³.
Nope. This argument also fails for the reason I gave above. Even if the "point" is actually a small region, it would have to be a timelike "world tube" (a small bundle of timelike worldlines) for your argument to be valid; but the "tube" occupied by, for example, the region of last scattering is spacelike, not timelike.

PeterDonis
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2019 Award
First, consider the universe starting with us now, extending out (and backward in time) to the surface S of last scattering (at which CMB was produced). We see this surface isotropically in all directions in microwave light. This is a 3-ball (the interior of a big 2-sphere plus boundary S).
No, it isn't. You're leaving out the time dimension. Our observable universe back to the surface of last scattering is a 4-ball, not a 3-ball. More precisely, it's the past light cone of the Earth at its present point in spacetime, which is a 4-volume bounded by a spacelike surface in the past (the surface of last scattering) and a null cone whose apex is the Earth's present event.

However, this only applies to the observable universe. It does not in any way rule out the possibility that the full spacelike surface of last scattering is spatially infinite. We can't rule that out because we can only see a portion of that surface, and that surface, as I have said, is spacelike, not timelike, so we can't think of it as a "point-like region visible in all directions" in the sense your arguments require.

Thank you for your patience.....I am not a trained cosmologist. I have replied to you point-for-point, but the time dimension is a struggle for me.

You write:
>>The "point-like start" ..... is not a feature of our actual current best-fit model

Accepted. The logic in favor of S³ does require a particular starting moment, or a point-like start, which was used to simplify the argument.

You write:
>>argument (in favor of S³ topology) also fails ....... Even if the "point" is actually a small region

Again, the "small point-like region" device was a rhetorical convenience, but not a logical requirement in support of S³ topology

You write:
>>....Your arguments for the spatial topology having to be S³ are only valid if the "point" that is visible in all directions is a point in space

This is not true. The argument is valid as long as the same object is visible in all directions. The object needn't be a point. This object
then becomes the "point at infinity" and ties off the balloon, making it a sphere. However, the second argument (merging balls) avoids this issue

You write:
>>You're leaving out the time dimension. Our observable universe back to the surface of last scattering is a 4-ball, not a 3-ball.

Accepted, but when we view the CMB, we are looking at a very narrow interval on the time axis, so I have treated this surface of last scattering (S) as a purely space-like surface, albeit at a particular time. You have also agreed to this in (*) below.

Given the above paragraph, I think we are agreed that whatever the initial conditions of the universe (point or smear), the surface S is a space-like 2-sphere which contains a 4-ball universe from time t=0 up to the time of last scattering , right?

Now you write:
>>we can only see a portion of that surface (S)

Has this claim actually been verified, or just assumed. If you are correct here, my argument is not falsified, but it is no longer complete, and I give up.
However, I have 2 reasons to think you are incorrect here. Firstly , since your claim would require mapping a portion of S continuously onto the surface
which we see today as a spherical map of the CMB, then such a mapping would necessarily display discontinuities which we do not see. Thus,
if you take Massachusetts alone, and try to stretch and paste it to cover the entire globe....you will see discontinuities. Secondly, even if some part
of S were not visible to us today, how would we know this? The needed data is inaccessible ! Indeed, you seem to recognize this as you write.....

>>(*) Our observable universe back to the surface of last scattering is a 4-ball, not a 3-ball. ..... bounded by a spacelike surface in the past (the surface of last scattering S) and a null cone whose apex is the Earth's present event

At this point, we can now return to my 2nd argument (merging balls) modified to accommodate your protests.

We have a 4-ball bounded by a space-like S containing the (invisible) early universe. and another 4-ball containing the visible universe, also bounded by S. As before, we merge these two objects by identifying their common boundaries S. We now have a universe with one time dimension (t = 0 to now), and whose spatial context is S³ (positive curvature, not flat).

PeterDonis
Mentor
2019 Award
when we view the CMB, we are looking at a very narrow interval on the time axis, so I have treated this surface of last scattering (S) as a purely space-like surface, albeit at a particular time.
You're not addressing my argument. My argument is not that you need to look at a larger "time slice" around the surface of last scattering. My argument is that your argument requires that the "point" or "region" (you are correct that it doesn't have to be a single spatial point) being looked at is timelike, not spacelike. Since the surface of last scattering is spacelike (as you agree), your argument does not apply to our observations of it.

Has this claim actually been verified
The surface of last scattering was a finite time ago, and light travels at a finite speed. That means we are only seeing a finite intersection of our past light cone with the surface of last scattering.

If you're asking how we know that finite intersection does not comprise the entirety of the surface of last scattering, we know that if the spatial topology of our universe were in fact ##S^3##, its radius of curvature would have to be much, much larger than the size of our observable universe. That ensures that the portion we see of the surface of last scattering is only a portion; it can't be the whole surface. If it were, we would see strong evidence of positive spatial curvature, and we don't.

We have a 4-ball bounded by a space-like S containing the (invisible) early universe. and another 4-ball containing the visible universe, also bounded by S.
In other words, S is the surface of last scattering. If that is the case, then we have two 4-balls, separated by a spacelike 3-surface, yes. But you have not shown that that spacelike 3-surface is an ##S^3##. See below.

As before, we merge these two objects by identifying their common boundaries S. We now have a universe with one time dimension (t = 0 to now)
Yes. But what topology does this 4-manifold have? You are claiming that it has topology ##S^3 \times R##; but you haven't shown that. See below.

and whose spatial context is S³ (positive curvature, not flat).
Nope. The construction of an ##S^3## by merging two 3-balls works like this: you have two 3-balls, each with a 2-sphere as a boundary. You identify the two boundaries and "glue" the two 3-balls together at their common boundary; the resulting manifold is then an ##S^3##.

But in the case you are describing, you have two 4-balls, each bounded by the same 3-surface. But you can't assume that that 3-surface (the surface of last scattering) is an ##S^3##; that's what you are trying to prove. So you can't show that the 4-manifold you get when you "glue" the two 4-balls together at their common boundary has topology ##S^3 \times R##, which is what you are claiming.

Peter:

I am grateful for your feedback, but I feel I should take a couple of cosmology courses here at UW-Madison before continuing this thread. I am currently taking a related physics course. However, here are some parting comments.

[1] I had assumed that the surface of last scattering (S) was a 2-sphere, not a 3-sphere, like the colorful maps of the CMB we have seen.

[2] I had intended to glue the two 3-balls exactly as you did. Indeed, I describe this in my original post, to which you agreed.

[3] I don't think I understand your use of the terms "time-like" and "space-like", in spite of my attempting to also use them.

[4] I still feel that the spatial topology of our universe will be S³, that is, we live within the surface of a 4-ball with a very large diameter. I have other reasons for this "feeling", but currently these reasons are fanciful, without evidence, so I won't waste your time with them. However, if the topology were indeed S³, I can envision several big currently unsolved problems acquiring elegant solutions, such as "what is dark matter?" and "Why is space expanding?" When the James Webb wakes up, I expect a big change in the standard model.

Cheers, Tom