View Full Version : Center of the universe

t.khoury

2009-Jan-05, 05:57 PM

Hi I was just wondering - people say the Earth is not in a special place in the universe (like when Copernicus showed the Earth actually goes around the Sun rather than the other way around) but is there actually a centre of the universe, a place where the Big Bang happened, and if so, where?

NEOWatcher

2009-Jan-05, 06:02 PM

Welcome to the board.

While I think it may be a perplexing concept to you, it has been mentioned time and time again on this board.

Google this board for 'Center of the Universe" for further discussion.

In short, no... there is no center.

antoniseb

2009-Jan-05, 06:14 PM

You may as well ask: where on the surface of a playground ball is the center of the ball?

As Neowatcher pointed out this is a much answered question around here. Welcome! You'll find other answers and friends here too.

NEOWatcher

2009-Jan-05, 06:16 PM

You may as well ask: where on the surface of a playground ball is the center of the ball?

I'd probably pick something like a cue ball since a playground ball does have an inflation valve. ;):lol:

Jeff Root

2009-Jan-05, 06:33 PM

There is no indication of any "edge" of the Universe, so we don't know

where or even if such an edge exists. It is theoretically possible, though

I think very unlikely, that the matter of the Universe is infinite in extent.

The description that you will see most often is that the Universe is finite

but unbounded. That is, even though it is finite in size, it has no edge

of any kind. There is nowhere you could go and no direction you could

look that you would see an absence of galaxies beyond.

In order for this to be, spacetime must be curved on the scale of the

entire Universe. Curved in a way similar to the curvature of the surface

of a sphere. You can move around forever on the 2-D surface of a 3-D

sphere without ever coming to an edge. Similarly, you could move

around forever through a 3-D finite but unbounded universe without

ever coming to an edge.

However, no such overall curvature of the Universe has been found.

That doesn't mean that the Universe isn't curved in that way. It just

means that IF it does curve in that way, the curvature is very gradual

and hence the Universe is very big-- even bigger than the part of it

that we can see.

In either an infinite universe or a finite but unbounded universe, there

is no edge, and therefore no center.

I personally think that the Universe most likely is finite and bounded,

but in that case, too, it must be far larger than the already absurdly

enormous part of the Universe we are able see, and there are no

observations to support my view over that of an unbounded universe.

-- Jeff, in Minneapolis

Sweeney

2009-Jan-05, 10:44 PM

I once read that if you could ever travel to the physical end of our universe, and shot an arrow at the boundary, the arrow would enter at another part of our universe's boundary.

I wish I could remember the book, but I'm sure it was just the author's best guess about an open or closed universe.

cosmocrazy

2009-Jan-05, 11:52 PM

Hi I was just wondering - people say the Earth is not in a special place in the universe (like when Copernicus showed the Earth actually goes around the Sun rather than the other way around) but is there actually a centre of the universe, a place where the Big Bang happened, and if so, where?

Based on the Big Bang theory one could say you are the centre, along with every other point in the universe. Basically the universe is the centre continually expanding to a very large size and maybe beyond.

Head banging stuff!

Welcome to BAUT!

alainprice

2009-Jan-06, 12:13 AM

I once read that if you could ever travel to the physical end of our universe, and shot an arrow at the boundary, the arrow would enter at another part of our universe's boundary.

I wish I could remember the book, but I'm sure it was just the author's best guess about an open or closed universe.

If we are more careful with our use of the word universe, then yes, it is expected to be true.

'Our' universe is really 'our observable' universe.

If you move 1 million light years in one direction, the entire 'bubble' that contains your 'observable universe' shifts over as well.

astromark

2009-Jan-06, 12:58 AM

Try not to think of a universe with a central point. Instead think of that central point being the whole universe which is expanding still ever faster and, forever.

Jens

2009-Jan-06, 03:05 AM

You may as well ask: where on the surface of a playground ball is the center of the ball?

As Neowatcher pointed out this is a much answered question around here. Welcome! You'll find other answers and friends here too.

There's something that's always puzzled me about that analogy. Though the surface of a ball lacks a center, there is a center in three dimensions, of the sphere. Does that mean that the universe has a center, but in some other dimension? Or is it simply that there is some important difference with the analogy?

WayneFrancis

2009-Jan-06, 06:22 AM

I once read that if you could ever travel to the physical end of our universe, and shot an arrow at the boundary, the arrow would enter at another part of our universe's boundary.

I wish I could remember the book, but I'm sure it was just the author's best guess about an open or closed universe.

Even scientists can explain things badly or if you try to read into an analogy to far you are bound to hit mistakes.

There is no indication that there is "boundary" in our universe. There is no edge that you say "hey look nothing past this point"

What the book probably was talking about was that the universe is finite and unbounded.

All indications we have shows the universe to be unbounded. If it is infinite the it just goes on forever. If it is finite then it some out loops back on itself like sphere. There is no "edge" to a sphere, not in the dimensions of the surface of the sphere at least. If there is an "edge" to our universe it is currently in a dimension we don't know how to observe yet and might never be able to see.

NEOWatcher

2009-Jan-06, 01:36 PM

There's something that's always puzzled me about that analogy.

Yep; Analogies break down fast when you take them literally.

Does that mean that the universe has a center, but in some other dimension? Or is it simply that there is some important difference with the analogy?

We don't really know, just like the 2D person can't see the shape of the ball (he doesn't see the curvature with enough resolution because his observable ball-iverse is only a small arc of the ball)

I think I mentioned it somewhere else, but I think of us as being in the center, as someone lost in the fog at sea (or other indistinguishable terrain).

That person does not know where they are in the sea, but does see things locally. (like fish swimming around)

Jeff Root

2009-Jan-06, 02:58 PM

Jens,

When I asked an astronomer who seemed to know this subject, whether

the Universe was curved in a fourth spatial dimension, he said no, three

dimensions of space and one dimension of time are required. That is what

general relativity describes. Spacetime. You don't find a fourth spatial

dimension in relativity. But somehow the time dimension is required in

order for the curvature to exist.

One person suggested that the third dimension in the balloon analogy

could be used to represent time: The Universe expands like a balloon

being blown up, so the interior of the balloon represents the past, when

the Universe was smaller, and the exterior represents the future when

it will be larger. But I think this is a particularly poor analogy, and do

not recommend using it for anything.

-- Jeff, in Minneapolis

Jeff Root

2009-Jan-06, 03:17 PM

Sweeney,

In the situation you describe, of a finite but unbounded Universe, which

seems to be considered by a majority of cosmologists as the most likely

of the currently-known possibilities, there is no place that one could go

to that would be an edge or end or boundary. Nomatter where you are

when you shoot your arrow, the arrow, travelling in a straight line, will

eventually pass through a part of the Universe that it was in previously.

There are no seams in a finite but unbounded Universe. Every place is

pretty much like every other place, as on the surface of a billiard ball.

The surface of the ball might be marked with tiny dots representing

galaxies, and those dots might be clustered together the way gravity

clusters galaxies together, but there are no boundaries on the surface

of the billiard ball to shoot your arrow across. Everywhere on the ball

is pretty much the same as everywhere else.

-- Jeff, in Minneapolis

NEOWatcher

2009-Jan-06, 03:25 PM

One person suggested that the third dimension in the balloon analogy could be used to represent time:[...] But I think this is a particularly poor analogy, and do not recommend using it for anything.

That actually makes sense to me, so obviously I'm missing something that makes it a poor analogy.

Ken G

2009-Jan-06, 03:33 PM

The description that you will see most often is that the Universe is finite

but unbounded.Where do you get this from? Citation? I am unaware of any way to determine whether or not the universe is finite but unbounded, based on any existing observation or supported theory. It's pure 100% assumption in any theory that you might be basing that statement on. Do we make assumptions in our theories, and then point to them as conclusions of our theories?

I personally think that the Universe most likely is finite and bounded,

but in that case, too, it must be far larger than the already absurdly

enormous part of the Universe we are able see, and there are no

observations to support my view over that of an unbounded universe.

Yes, that is the fact here. There was a time (not all that long ago) when scientists could not fathom an origin to the universe either, and I'm sure many of them made statements like "I personally think the universe has been here forever, though no observations support that". What are these types of statements beyond simple assertions of the limitations of our own imagination?

Jeff Root

2009-Jan-06, 03:59 PM

Ken,

When I saw that you posted right after NEOWatcher's post #15, I hoped

it was a reply to him. I'm sure the interior-of-the-balloon-as-past analogy

is bad, but I'm not sure I can say why. I bet you can.

I'll see if I can find some support for my assertion that the "finite but

unbounded" description of the Universe is treated as more likely than

any other description by a majority of people expressing an opinion.

-- Jeff, in Minneapolis

grant hutchison

2009-Jan-06, 04:08 PM

I'm sure the interior-of-the-balloon-as-past analogy

is bad, but I'm not sure I can say why.It's good enough to be used, in diagrammatic form, quite frequently in Edward Harrison's highly respected book Cosmology: The Science of the Universe. In Harrison's hands, one can gain quite a few useful insights from it.

Grant Hutchison

Ken G

2009-Jan-06, 04:14 PM

That actually makes sense to me, so obviously I'm missing something that makes it a poor analogy.I think the analogy does have some pretty good advantages, and it also has some disadvantages that should be recognized. To see these advantages and disadvantages, compare a stretching balloon to a stretching chessboard. The stretching balloon is finite and unbounded, the stretching chessboard is either bounded or infinite. So if you insist on imagining a universe that is finite and bounded, you favor the balloon over the chessboard. The problem there is, the balloon has a feature that its radius of curvature is directly related to how far it has been stretched, and this means that it is a spatially curved 2D object embedded in a flat 3D space. If you choose time as the radial coordinate to flesh out your 3D model of spacetime in the analogy, the problem is you now have a flat spacetime with a curved space embedded in it, but what you really want is a curved spacetime. The chessboard accomplishes the latter by relaxing the constraint on the curvature of the spatial component, but at the expense of being able to imagine that the universe is finite and unbounded.

So although I am no expert on GR, it seems to me we must choose our poison here if we are to use either analogy. If we take the chessboard, we can have a curved spacetime, but not a finite and unbounded universe. If we take the balloon with radial time, we can have a finite and unbounded universe, but spacetime is flat, and GR wouldn't give you that. For my own part, I simply reject the assumption that the universe is finite and unbounded, so the chessboard works fine. For those who insist on a finite and unbounded universe, they can use neither the balloon nor the chessboard. To be perfectly honest, I'm not clear on how embedding curved spacetime into our flat-space physical analogies actually works, so I can't say what the limits of such analogies really are-- that takes a real GR expert.

Ken G

2009-Jan-06, 04:23 PM

I'll see if I can find some support for my assertion that the "finite but

unbounded" description of the Universe is treated as more likely than

any other description by a majority of people expressing an opinion.

Yes, I'd like to hear those arguments, keeping in mind of course that science is not done by majority opinion (again I mention the majority view of cosmology in 1900, for example).

NEOWatcher

2009-Jan-06, 05:12 PM

I think the analogy does have some pretty good advantages, and it also has some disadvantages that should be recognized.

Thanks for explaining. It obvious that it has to do with what aspect of the universe you are trying to get at.

For the purposes of "center of the universe", I think we can forgive the curvature and boundedness with the assumption that the recipient will learn that when they get to that point.

pzkpfw

2009-Jan-06, 07:25 PM

To see these advantages and disadvantages, compare a stretching balloon to a stretching chessboard. The stretching balloon is finite and unbounded, the stretching chessboard is either bounded or infinite.

If the chessboard isn't infinite, wouldn't it have a centre?

(It is of course "only" an analogy - but I guess where it takes me is the follow-up question "Could there be a centre of the Universe - but it's beyond our visible Universe?", as in, our visible Universe is one square plus the 8 around it, and we are in C3 so see neither edges nor centre...)

publius

2009-Jan-07, 02:19 AM

So although I am no expert on GR, it seems to me we must choose our poison here if we are to use either analogy. If we take the chessboard, we can have a curved spacetime, but not a finite and unbounded universe. If we take the balloon with radial time, we can have a finite and unbounded universe, but spacetime is flat, and GR wouldn't give you that. For my own part, I simply reject the assumption that the universe is finite and unbounded, so the chessboard works fine. For those who insist on a finite and unbounded universe, they can use neither the balloon nor the chessboard. To be perfectly honest, I'm not clear on how embedding curved spacetime into our flat-space physical analogies actually works, so I can't say what the limits of such analogies really are-- that takes a real GR expert.

Yes, it's a real problem. The trouble is the mixing of space and time in that non positive definite way. You can draw an x-t diagram for a simple 2D space-time, which is done all the time for SR purposes. However, what you're looking at is a 2D Euclidean plane, which just doesn't capture the essence of a 1-1 Minkowski space. Indeed, there is no way to draw a picture of some crazy non-positive definite space, even a flat one. If you realize that, you're okay, but if you don't, it's misleading. Now, start curving things, and it can get even more misleading.

So if you have a basic understanding of just how "really different" the geometry is, such constructs can be useful. But if you don't, they can be very misleading. Lorentzian manifolds play weird tricks with the notion of spherical vs hyperbolic, because of the the difference in signs between the space-like and time-like parts. So a manifold of constant curvature, which looks spherical in terms of the equations, also looks hyperbolic.

What is an n-sphere in Euclid? It's just x^2 + y^2 +.... = constant^2, the locus of all points a fixed distance from an origin. Now, do that for space-time:

t^2 - (x^2 +y^2 +..) = constant.

That's the same thing, a "sphere", yet because of the difference of squares, it has an hyperbolic looking form.

And that latter hyperbolic looking form does indeed give you a deSitter space, which can be visualized as an expanding sphere. Yet add the time part in the diagram, and you get an hyperbolic structure. An expanding hyperballoon works fine for deSitter, in terms of thinking of "space" in certain coordinates -- a balloon expands with time, and therefore the interior is earlier times. However, add time to the thing, and you're dealing with an hyperbolic surface.

The trouble is the balloon part is thinking about space alone (and in certain coordinates) and is not invariant. The hyperbola captures the true invariant parts of the structure, and you can see how some coordinates along that surface can see the space part as flat, while others will see it as curved.

As we discussed before at length, the notion of what space is doing just has no invariant meaning. You've got to think of space-time together. And until one realizes that, until that clicks, one is lost.

And another technical problem with general space-times is that *more than one higher time dimension* may be required to embed. And that just blows one's mind, of course. That's true for Schwarzschild. If you supress the angular coordinates, and just look at the r-t space-time, you can successfully embed that in a 1-2 higher space-time, and draw a 2D curved surface that represents a cuved 1-1 space-time. However, the full Schwarszchild required a 2-4 space-time to embed.

-Richard

WayneFrancis

2009-Jan-07, 02:43 AM

Ken,

When I saw that you posted right after NEOWatcher's post #15, I hoped

it was a reply to him. I'm sure the interior-of-the-balloon-as-past analogy

is bad, but I'm not sure I can say why. I bet you can.

I'll see if I can find some support for my assertion that the "finite but

unbounded" description of the Universe is treated as more likely than

any other description by a majority of people expressing an opinion.

-- Jeff, in Minneapolis

if it is finite then let it be known that the curvature is SOOO small that in our hubble volume we can detect no curvature which means the actual universe to the size of our hubble volume could easily comparable to the size of our hubble volume when compared to an atom. So say that the universe is finite doesn't really mean much. To the lay person everything outside our hubble volume is inaccessible. There is no chance of looking out into the stars and "seeing the back of your head"

Jeff Root

2009-Jan-07, 07:27 AM

if it is finite then let it be known that the curvature is SOOO small that

in our hubble volume we can detect no curvature which means the actual

universe to the size of our hubble volume could easily comparable to the

size of our hubble volume when compared to an atom.

That is way, way, way, way, way overstating it.

No overall curvature has been detected. That means the curvature

in our visible part of the Universe is no more than a couple of percent.

Interpreting that very literally and simplistically, that implies that the

Universe must be at least 50 times the size of what we can see.

Compare that to the ratio you suggest, which is roughly (using the

wrong figure for the size of the Universe, but the right figure would

make the result even more dramatic)

13.7 billion light-years : 100 picometres, or

1,296,093,432,000,000,000,000,000,000,000,000,000.

There is no chance of looking out into the stars and "seeing the

back of your head"

I agree. I have read that if the Universe is closed in such a way

that light could theoretically circle the Universe, in which case the

expansion would eventually come to a halt and reverse, finally

ending in a Big Crunch, then light emitted at the instant of the

Big Bang would only have enough time to get halfway around the

Universe before the Big Crunch.

-- Jeff, in Minneapolis

Justy

2009-Jan-07, 02:08 PM

Based on the Big Bang theory one could say you are the centre, along with every other point in the universe. Basically the universe is the centre continually expanding to a very large size and maybe beyond.

Head banging stuff!

Welcome to BAUT!

This quote is the most sense that anybody has ever made on this subject. :dance:

The balloon analogy doesn't work because we are not ants on a balloon. On a balloon you can draw a dot then continue walking forward in a "straight" line. You'll eventually return to your dot. I don't think that the same could happen to us in our reality.

In addition with a balloon or ball you can define a point inside the ball to designate the center. :whistle:

max8166

2009-Jan-07, 04:49 PM

In addition with a balloon or ball you can define a point inside the ball to designate the center. :whistle:

Using the balloon analogy, and stating that the 2D surface of the balloon represents the 3D space within which we live then, the interior of the balloon then represents the past and so the center of this analogy's Universe would be the so called "big bang".

A time and place was the center. Rather than just a place.

NEOWatcher

2009-Jan-07, 05:28 PM

The balloon analogy doesn't work because we are not ants on a balloon.

That's why it's called an analogy and not a synonymology.

I don't think that the same could happen to us in our reality.

Don't know...but it has been discussed in this thread.

In addition with a balloon or ball you can define a point inside the ball to designate the center. :whistle:

No... in the world of the analogy, the inside doesn't even exist (if you don't use the time analogy that has been discussed)

Ken G

2009-Jan-07, 06:47 PM

And that latter hyperbolic looking form does indeed give you a deSitter space, which can be visualized as an expanding sphere. Yet add the time part in the diagram, and you get an hyperbolic structure. An expanding hyperballoon works fine for deSitter, in terms of thinking of "space" in certain coordinates -- a balloon expands with time, and therefore the interior is earlier times. However, add time to the thing, and you're dealing with an hyperbolic surface.Thanks, I think that clarifies things a lot. There's such a pervasive problem of taking analogies too literally, that they invariably end up creating more confusion than they resolve. It seems that an analogy is a good way to "put off" a question from someone who doesn't really want to know the "truth", but for those who truly do, there's just no substitute for thinking in terms of the proper mathematical structures. Maybe we should be looking for the clearest and easiest ways to describe these actual mathematical structures, rather than bogus analogies.

And another technical problem with general space-times is that *more than one higher time dimension* may be required to embed. And that just blows one's mind, of course. That's true for Schwarzschild. If you supress the angular coordinates, and just look at the r-t space-time, you can successfully embed that in a 1-2 higher space-time, and draw a 2D curved surface that represents a cuved 1-1 space-time. However, the full Schwarszchild required a 2-4 space-time to embed.I had a feeling I didn't understand how to embed curved spacetimes, and now I'm sure I don't.

matthewota

2009-Jan-07, 07:51 PM

Hi I was just wondering - people say the Earth is not in a special place in the universe (like when Copernicus showed the Earth actually goes around the Sun rather than the other way around) but is there actually a centre of the universe, a place where the Big Bang happened, and if so, where?

The easiest way for me to explain this is that since the entire universe began from a singularity (which is an infinitely small point) and expanded from that point, that the center of the universe is everywhere. You cannot centralize or locate a point in a point that takes up no volume.

WayneFrancis

2009-Jan-08, 02:35 AM

That is way, way, way, way, way overstating it.

No overall curvature has been detected. That means the curvature

in our visible part of the Universe is no more than a couple of percent.

Interpreting that very literally and simplistically, that implies that the

Universe must be at least 50 times the size of what we can see.

Compare that to the ratio you suggest, which is roughly (using the

wrong figure for the size of the Universe, but the right figure would

make the result even more dramatic)

13.7 billion light-years : 100 picometres, or

1,296,093,432,000,000,000,000,000,000,000,000,000.

I agree. I have read that if the Universe is closed in such a way

that light could theoretically circle the Universe, in which case the

expansion would eventually come to a halt and reverse, finally

ending in a Big Crunch, then light emitted at the instant of the

Big Bang would only have enough time to get halfway around the

Universe before the Big Crunch.

-- Jeff, in Minneapolis

Can you explain why would wouldn't see any curvature if the universe was only 50x bigger then what we can see? From my understanding of the current interpretation of data from the WMAP is that we detect absolutely curvature to the universe. When I am on the beach on a clear day I can detect the curvature of the earth by looking out at ships in the distance and I'm pretty sure the Earth is much bigger then 50x what I can see around me and this is with the naked eye.

Everything I've heard about the size of the universe points to a HUGE universe. Wish I could remember Alex Filippenko quote on describing the size of the universe. It was along the lines of what I said. I'll see if I can find it.

publius

2009-Jan-08, 03:25 AM

I had a feeling I didn't understand how to embed curved spacetimes, and now I'm sure I don't.

It's not all that conceptually difficult, but the math of the details will get complex and tedious.

The trouble is that space-time is non-positive definite with a difference between time and space-like parts, and that makes things more complicated.

For example, consider a flat Euclidean N-space. N dimensions. Write some expression in the coordinates, f(x1, x2, .....xN) = 0, and you've expressed a subspace, an N-1 dimensional structure.

You can do that with a space-time. f(t, x.....) = 0. However the resulting sub-space needs to be a proper space-time, with a time-like and space-like part.

For example, consider a 1, 3 space-time, and some trivial f(t, x, y, z) = 0on that. t = constant does not give you a proper space-time. It gives you a 3D Euclidean sub-space. But z = constant gives you a 1, 2 space-time.

So not all general f(t, x, y, z)'s will give you a proper sub space-time.

And then the other thing: consider a curve. A curve is a 1D structure, but there are certainly more types of curves than will embed in a 2D plane. A helical coil of a curve requires 3D. And you can imagine going to 4D and constructing a curve there that requires all 4 dimensions.

So, an N-D structure can curve in ways that require more than N+1 higher flat dimensions to embed in. Add the space and time-like distinction in, and you can see that you might need more than one higher time dimension into which to embed as well as higher space-like dimensions, depending on just how the lower space-time curves.

GR is about (1, 3) space-times that obey the EFE. The EFE is a restriction on all the (1, 3) structures you can imagine, and how that constrains all the possible ways things can curve, I don't know. Are there any limits to how many higher time and space-like dimensions are required imposed by the EFE condition, I don't know. I would guess there are, but just don't know.

But I do know that Schwarzschild requires a (2, 4) flat space-time to embed, 6 total dimensions, with one extra time as well as space. :) So Schwarzschild is like the coil, it requires 2 higher dimensions to emded, and the nature of the space vs time dynamic requires one more time and one more space dimension.

When many people think of "curved space" they're thinking only in terms of spatial curvature, and embedding curved positive definite spaces in higher positive definite spaces. Time, and the curvature of it, thrown in the mix throws a monkey wrench in all that.

And indeed, the spatial part of any coordinate system in a space-time is arbitrary anyway. What matters is space-time, not space and time. That is something that just has to click in your mind. Once it does, you can go on. But until it does, one is sort of stuck in the mud with misconceptions dancing around.

-Richard

publius

2009-Jan-08, 03:48 AM

The question of the "curvature of the universe" is one of these misconceptions. When people say the universe appears to be flat, they are talking about *spatial curvature* (although they might not appreciate that as well as they should!). When you're talking spatial curvature, you have dropped down to some coordinatization, and are therefore not talking invariants. There is however, space-time curvature. And that is all that matters, and that's very different from any spatial curvature.

There is no better way to see that than deSitter. You can coordinatize in what amouts to a co-moving form and you have a closed hypersphere that is expanding with time. You have a closed expanding space. However, you can go to other coordinates and you have a flat, unbounded space that is expanding.

That is the nature of that hyperboloid structure. Consider a 3D hyperboloid. Well Wiki has pictures that are worth of thousands of my rambling words:

http://en.wikipedia.org/wiki/Hyperboloid

What we're interested in is a hyperboloid of one sheet. Nuclear reactor cooling towers are familiar example. Those things looks curved and they certainly have curvature, but they can be constructed entirely from straight lines! It is a "ruled surface", a curved surface that can be constructed by piecing together straight lines. At first blush you don't think it can be done, but it's very enlightening to have the realization of how it's done "click" in your mind.

Anyway, that ruled surface property is what allows the flat spatial coordinatization. You can construct the hyperboloid with "slices" of circles of increasing radius, or you can construct it with "slices" of straight lines skewed around each other.

In this universe, when you're talking about what space is doing, you're not talking about anything that means anything. Alas, many peole think it does, but it doesn't.

-Richard

Ken G

2009-Jan-08, 04:26 AM

What we're interested in is a hyperboloid of one sheet. Nuclear reactor cooling towers are familiar example. Those things looks curved and they certainly have curvature, but they can be constructed entirely from straight lines! It is a "ruled surface", a curved surface that can be constructed by piecing together straight lines. At first blush you don't think it can be done, but it's very enlightening to have the realization of how it's done "click" in your mind.

That's pretty cool, I did not know that.

In this universe, when you're talking about what space is doing, you're not talking about anything that means anything. Alas, many peole think it does, but it doesn't. Yes, many people think that a bent piece of paper has curvature, but it is a "ruled surface" too-- if the paper doesn't need to be rubber, then it is still flat, no matter what you do to it, because no distances change for ants crawling around on the paper. Even if you bend the paper all the way around to make a closed sheet, that's a topological change, but the paper is still flat. So finite and unbounded could still be flat, whether one is talking about spacetime or just space in some coordinatization that respects the "rulers" of the ruled surface.

Your point about the de Sitter space looking like a spatially closed universe that expands in time, and a bunch of infinitely long spatial slices along the "rulers" of the hyperboloid, is a good one indeed. I really should have learned some differential geometry! I presume the time coordinate in the "infinite and straight" spatial coordinatization is one that wraps up and around the hyperboloid like the stripes on a barber pole, so the time cordinate appears curved in the 3D Euclidean space that the 1+1 hyperboloid is embedded in. So it seems that in the embedding space, one can trade off curvature between space and time coordinate axes.

publius

2009-Jan-08, 07:04 AM

Yes, many people think that a bent piece of paper has curvature, but it is a "ruled surface" too-- if the paper doesn't need to be rubber, then it is still flat, no matter what you do to it, because no distances change for ants crawling around on the paper. Even if you bend the paper all the way around to make a closed sheet, that's a topological change, but the paper is still flat. So finite and unbounded could still be flat, whether one is talking about spacetime or just space in some coordinatization that respects the "rulers" of the ruled surface.

Yes, and there's a lot more to these concepts, and I forget the exact terminology -- we'd need a real expert to properly explain all this. But there's more to it that ruled vs. non ruled surfaces. In your example, we take a piece of flat plane and roll it around into a cylinder. But we can't take a sphere and unroll into a flat plane without distortion, as map makers lament. And that corresponds to two basic "ways to curve". We can loosely define that as those surfaces you can get by rolling and bending a flat sheet without stretching it, and those you can't.

In lower dimensions, there are simple ways to quantify all that, a set of numbers that tell the information. But at some point, you have to go to tensors, and realize the simpler ways are just parts of those tensors.

Ricci curvature, which is represented by a rank 2 tensor is about the distorting kinds of curvature. If the Ricci curvature is non-zero, then you can't "unroll" into into a flat sheet. The full curvature tensor is the Riemann tensor, which is a *rank-4* tensor. That tells you all the information. So our cylinder would have non-zero Riemann curvature, but zero Ricci curvature.

The way to think about is in terms of the generalized "hypervolume". Say we have a N-D space. We have a concept of N-D hypervolume. If we have Ricci curvature, then our hypervolume of structures in that N-D space will be distorted, and the Ricci curvature is a measure of that distortion.

For example, if we draw a "square" on a flat sheet, the area is x^2, where x is the length of a side. If we roll the flat sheet into the cylinder, our square gets curved relative to the embedding space, of course, but the area remains x^2. An observer on the sheet would see no distortion in area. But now, do that on the surface of a sphere, and the area of the "square" is no longer x^2.

In terms of tangent plane to the cylinder, the square we've drawn "curves out of" the tangent plane, but it's area is not distorted. On the surface of sphere, it curves out of the tangent plane and is distorted. The Riemann curvature is all about what things are doing relative to that tangent plane, while the Ricci curvature is a contraction of that tells you if the area is going to be distorted. So "Ricci flat" means no distortion, even though it could be Riemann curved.

In highly symmetric cases (or low enough dimensions), you can pull a single number out of the Ricci tensor, a scalar curvature number. In the common convention, positive scalar curvature means the area will be greater than Ricci flat, while negative means it will be less than Ricci flat. A sphere is a surface of positive Ricci curvature, while a hyperboloid is constant negative curvature.

Note you can have a ruled surface with Ricci curvature! A hyperboloid is ruled, but it is constant negative curvature. And I think you can see that in terms of straight lines vs. planes. You can't bend a plane into an hyperboloid without distorting areas, but you can still build the surface with straight lines. I'm sure you could see the Ricci curvature as something you'd have to do with a bundle of straight lines. Imagine a plane constructed with a bundle of parallel straight lines. You have to "skew" those straight lines in a way that you can't do without distorting the plane somehow. You don't have to "bend" any individual line, but you do have to "bend" or "cock" them relative to each other in some way.

And just imagine how the possibilities expand as you go into higher and higher dimensions. You can imagine some hypervolume being "ruled" with planes, yet still having Ricci curvature.

Now, we've seen that we can roll a flat sheet into a cylinder, and all sorts of other structures. So, somehow all of those should be in one class, some generalized "plane class". I think that's "isomorphism", all the surfaces that are isomorphic to a plane. And you can do the same thing with a sphere, imagining all the surfaces that are isomorphic to a sphere. Just as we can imagine bending the flat sheet, we can imagine bending a sphere without "undoing" the basic Ricci curvature. And that's as far as my knowledge goes, to dimly see the genesis of the joke that a topologist doesn't know the difference between a doughnut and a coffee mug. :)

-Richard

Ken G

2009-Jan-08, 11:02 AM

Ricci curvature, which is represented by a rank 2 tensor is about the distorting kinds of curvature. If the Ricci curvature is non-zero, then you can't "unroll" into into a flat sheet. The full curvature tensor is the Riemann tensor, which is a *rank-4* tensor. That tells you all the information. So our cylinder would have non-zero Riemann curvature, but zero Ricci curvature.

That clarifies some things, indeed.

A sphere is a surface of positive Ricci curvature, while a hyperboloid is constant negative curvature.

I see, so although a hyperboloid is a ruled surface, the rulers aren't parallel, so it's not Ricci flat like a cylinder is. So the difference between a sphere and a hyperboloid is more than just the sphere having positive Ricci curvature and the hyperboloid having negative-- the fact that the latter is a ruled surface means that you can find spatial foliations that are straight if you are willing to let time curl around.

And you can do the same thing with a sphere, imagining all the surfaces that are isomorphic to a sphere. Just as we can imagine bending the flat sheet, we can imagine bending a sphere without "undoing" the basic Ricci curvature. And that's as far as my knowledge goes, to dimly see the genesis of the joke that a topologist doesn't know the difference between a doughnut and a coffee mug.I actually think that topology is a bit different (maybe simpler?) than differential geometry, such that the topology of the mug and the doughnut are the same, but they would spawn different metrics and tangent spaces and so forth. I believe this is the difference between a homeomorphism and a homomorphism-- the former depicts a topological equivalence like the mug and doughnut, but the latter depicts a deeper structural sameness, like preserving distances between points or dot products or whatever is the structure of interest that is implying the need for homomorphisms. I think we might say that two spacetimes that are homeomorphic back and forth are "topologically isomorphic" so are topologically equivalent, thus are sort of grossly similar, whereas two spacetimes that are homomorphic back and forth under the algebraic operations used in physics are just plain "isomorphic", meaning they are "physically identical." Given our penchant for equating physics with reality, we might then also say they are really the same (whereas physics had better be able to distinguish why a doughnut is so much tastier than a mug).

publius

2009-Jan-08, 11:23 PM

Ken,

Yes, I may have gotten my "-morphisms" mixed up. An expert would probably be wincing and groaning at what I was trying to say, but I hope and think I'm getting the general conceptual picture right. Anyway, a flat plane and a cylinder are some sort of 'x'-morphic to each other, meaning than can be bent and curved back and forth without distortion. :lol: And that forms natural groups/classes.

I was meaning to continue with this last night, but it got late. Anyway, a confusing point I wanted to make was about positive vs negative curvatures, hyperbolic vs spherical for space-times as compared to regular positive-definite spaces.

Consider the defintion of a (n-)sphere, the locus of all points some constant distance from a reference point, the center. In some positive definite flat embedding space, that is:

x1^2 + x2^2 +....xN^2 = constant.

But in a space-time, the "constant distance" gives this:

t^2 - (x1^2 + x2^2 + ....xN^2) = constant (and this constant must be negative to define a proper time-like congruence of world lines).

So a space-time sphere is really hyperbolic looking! That can be really confusing. The sign of the Ricci curvature scalar is just a convention, and that's related to time-like sign convention. In positive-definite land, I think the sign of curvature sort of comes naturally-- that is, it makes sense to for a sphere to be positive and a hyperboloid negative -- but in non-positive definite spaces, the sign convention is not so apparent. I'm not sure of all the details of that, but the common definition defines that hyperbolic space-time definition as constant *positive* curvature, to agree with a sphere.

That can be confusing. Anyway, a positive cosmological constant (in otherwise empty space-time) gives you a deSitter space, with constant positive curvature. That "sphere" looks hyperbolic :) -- deSitter space is the Lorentzian equivalent of a sphere in positive definite land.

Negative space-time curvature is something else. Constant negative curvature gives you an anti-deSitter space, which is very weird. You can construct anti-deSitter space in much the same way as deSitter, but your embedding space has to have one more time dimension.

IOW, a "sphere" in a (1, 4) embedding space-time gives use a (1, 3) deSitter space-time, and that "sphere" looks hyperbolic. But a similiar condition in a (2, 3) space-time gives us a

(1, 3) anti-deSitter space-time with constant negative curvature. What that looks like, I don't know.

-Richard

publius

2009-Jan-08, 11:50 PM

I hope I'm not making that more confusing than it is, but this sphere vs. hyperbola thing in space-times vs regular positive-definite spaces is important.

Note we can only "draw" and visualize positive-definite spaces. When draw a Minkowski x-t diagram, we use a positive-definite plane. We have no way to draw that non-positive behavior. That effects what "perpendicular" is. For example, take a time-like world line with some slip v/c < 1. The spatial axis for that path, u/c is determined by the relation uv = c^2, or

(u/c)*(v/c) = 1. Thus, the line "perpendicular" to a slope of v/c = 30 degrees, is u/c = 90 - 30, no 90 + 30 degrees! They look acute on the x-t plane, but they are perpendicular according to the non-positive definite definitions.

And that brings up your comment about what the world lines on the deSitter hyperboloid with the flat spatial slices look like. You'd have to look at a proper diagram, but they may not look exactly like the barber pole stripes because the angle between them and the "straight ruler" lines may look acute on a diagram because of this behavior.

Any good text exploring deSitter space should show what all the various world lines you build coordinate systems around look like on the hyperboloid. A good one will show the "static coordinates" and the "cosmological horizon" lines relative to that as well as the other common coordinate systems.

Anyway, when you draw these things, the non-positive definite nature just can't be captured graphically. You have to appreciate that to really know what you're looking at. :) That is, you need to know that what you're looking at is really different than what you think you're looking at. :lol:

-Richard

Ken G

2009-Jan-09, 12:46 AM

Anyway, when you draw these things, the non-positive definite nature just can't be captured graphically. You have to appreciate that to really know what you're looking at. :) That is, you need to know that what you're looking at is really different than what you think you're looking at.That actually made sense.

publius

2009-Jan-09, 03:43 AM

And one more interesting thing while we're on this.

The EFE is a statement about the Ricci curvature of space-time. It associates the Ricci curvature of a point with the stress-energy at that point. What this means is that vacuum regions (with zero Lambda, of course) are always Ricci flat. Space-time only has the "distorting curvature" in the presence of mass-energy-momentum.

So when you plop mass down in empty space-time, the surrounding vacuum regions will curve in a way similiar to the ways you can bend the flat sheet. The flat space-time "in between" the mass curves itself in a Ricci flat way to match up with the Ricci curvature of the mass regions.

And note that is a statement about the invariant space-time curvature, not the curvature of any spatial notions you have in a given coordinate system.

For example, in Schwarzschild coordinates, the spatial slices are Ricci curved. Notions of space, at constant coordinate time, can be considered as separate geometric structures. But those structures are just arbitrary spatial slices.

But the space-time itself, in the vacuum regions, is Ricci flat. It has Riemann curvature, but not Ricci.

I don't know if there's any interesting physical insight in any of that -- it may just be merely abstract mathematical properties -- but I always thought it was fascinating.

I forget the details, but when Einstein was working out GR, he originally set the Ricci curvature directly equal to the stress-energy tensor. But that didn't work -- something to do with the proper action -- and he had to add the term with the Ricci scalar to get it right, which became the Einstein tensor. David Hilbert had something to do with that, and the "Einstein-Hilbert Action" bears his name as well because of it.

All this stuff leads to the concept of an "Einstein Manifold". An Einstein manifold is one where the Einstein Tensor is zero, which means the Ricci curvature tensor is equal to a constant times the metric, ie:

Ric = k*g, where k is constant

All Ricci flat regions have k = 0. Any vacuum space-time is trivially an Einstein manifold with k = 0, since T, the stress-energy, is zero. However, the properties of the cosmological constant allow Einstein manifolds with non-zero k.

As I'm sure you've realized, deSitter space is an example with a positive k (using the proper sign convention), which is proportional to Lambda. And that constant k is therefore a simple scalar that tells you all you need to know about the Ricci curvature.

Back in normal positive definite land, a sphere is an Einstein manifold with positive k, while an hyperboloid is one with negative k.

-Richard

Jeff Root

2009-Jan-09, 05:21 AM

In the balloon analogy, as usually presented, the surface of the

balloon is what is being considered; everything inside or outside

the balloon is simply ignored. Those things are ignored because

they aren't relevant. They are not part of the analogy, just as

the color of the balloon is not part of the analogy.

The surface of the balloon is two-dimensional. It is curved in

three-dimensional space. That curvature is analogous to the

possible overall curvature of spacetime of the entire Universe.

The two-dimensional surface of the balloon represents a slice of

spacetime through the Universe having two spatial dimensions and

one time dimension. There is no way to illustrate in full detail three

spatial dimensions using a two-dimensional surface, and mapping

three spatial dimensions onto a two-dimensional surface gives a

messy, confusing map. So the surface of the balloon represents

only a sample slice through the Universe, not the entire Universe.

-- Jeff, in Minneapolis

Jeff Root

2009-Jan-09, 06:07 AM

Can you explain why would wouldn't see any curvature if the universe

was only 50x bigger then what we can see? From my understanding

of the current interpretation of data from the WMAP is that we detect

absolutely curvature to the universe.

It is because of what is required to "see" the curvature.

The simplest, most straightforward way I know of to measure any

curvature would be to count galaxies at different distances. If you

could see galaxies at all distances equally well; if new galaxies didn't

form and galaxies never merged; and if you could accurately measure

the distances to galaxies, then the number of galaxies seen in any

given area of the sky (such as would be measured in square degrees

or steradians) at any given distance would depend on the curvature.

In the case of perfectly "flat" spacetime, the number of galaxies

seen would increase as the square of the distance. In the case

of "spherical" spacetime, curved like a balloon, the number of

galaxies would increase with distance at a decreasing rate until,

at some distance, it would stop increasing and begin decreasing.

The number of galaxies would decrease with distance at an

increasing rate after that.

Those two cases are practically impossibe to distinguish if you can

see only 1/50 of the whole Universe.

When I am on the beach on a clear day I can detect the curvature

of the earth by looking out at ships in the distance and I'm pretty

sure the Earth is much bigger then 50x what I can see around me

and this is with the naked eye.

The Earth has a global surface. You can draw geodesics on the

surface, and compare them to straight lines which are not confined

to that surface.

The Universe has no global surface, so you can't draw geodesics

on it. All you can do is shoot beams of light through it and see

what happens to them.

You can see that the surface of the Earth curves, because you

can see that things disappear below the horizon. The Universe

has no analogous feature.

Everything I've heard about the size of the universe points to a

HUGE universe.

And everyone agrees. The questions is: How huge?

-- Jeff, in Minneapolis

Ken G

2009-Jan-09, 03:14 PM

But the space-time itself, in the vacuum regions, is Ricci flat. It has Riemann curvature, but not Ricci.

I don't know if there's any interesting physical insight in any of that -- it may just be merely abstract mathematical properties -- but I always thought it was fascinating.

I'm sure there is physical insight there. People like Feynman are good ones for finding it! For example, I think Feynman never really liked the geometrical interpretation of general relativity. He preferred to think in equivalent terms that he thought were more physical. That means, he wanted a physical reason why straight lines should curve, and one can then take all those curved lines and interpret the geometric structure they delineate, rather than taking the geometric structure as the fundamental thing that "causes" the lines to be curved. So he tended to focus on the gravitaitonal potential, and say that the potential caused time to slow down where the potential was large and negarive. Then the straight lines are like Snell's law in refraction-- they are the lines of minimum phase change through a gradient in the rate of time.

Personally, I don't like imagining that the rate of time is different, because the concept of a "rate of time" seems self-contradictory. But I can see Feynman's point that geometry is something you infer from the paths, the paths don't conform to the geometry-- the physics acts on the paths themselves. So either view seems OK from the point of view of physical insight.

I forget the details, but when Einstein was working out GR, he originally set the Ricci curvature directly equal to the stress-energy tensor. But that didn't work -- something to do with the proper action -- and he had to add the term with the Ricci scalar to get it right, which became the Einstein tensor. David Hilbert had something to do with that, and the "Einstein-Hilbert Action" bears his name as well because of it.

Action is basically another way to think of phase. So it sounds like you are saying that Ricci curvature alone does not do the right things with the phases, and you need something that ends up acting more like the gravitational potential. I don't know any details.

Back in normal positive definite land, a sphere is an Einstein manifold with positive k, while an hyperboloid is one with negative k.

Yes, and most introductory explanations of this do everything in positive definite land, and pretend that the curvature of space itself is a meaningful thing. They choose the comoving frame, typically, to get the time conventions, and then what's left of spacetime becomes a spatial curvature that now is in "positive definite land". But I think you are right that this fails to embed the physics in its proper setting by not dealing with invariant quantities, just as it fails to embed positive definite land in the correct spacetime. As you said, you have to know that what you are looking at is not what you are looking at.

sabianq

2009-Jan-09, 04:25 PM

i see at least two possible answers to the question about where the center of the universe is.

and the two are based on assumptions...

one, the center of the "observable" universe is the place from where it is being observed. meaning that if i look in any direction, i can only see the same distance in which ever direction i look, the observer is in the center of the universe. and that goes for looking at distant and very large objects to looking at the small and very tiny world of the quantum.

and second, if the universe is indeed infinite in size, then every single point in the universe is actually the exact center.

speedfreek

2009-Jan-09, 05:35 PM

There are indeed more possibilities, one of which is that the universe is finite but unbounded and has a topology that renders the idea of a centre to be arbitrary, the same as the idea of having a centre for the surface of the Earth.

publius

2009-Jan-10, 03:08 AM

Ken and all,

Here's the Wiki article on deSitter space-time, which is pretty good.

http://en.wikipedia.org/wiki/De_Sitter_space

Note the expanding hyperballoon from the point of view of the embedding space. These are the "co-movers", those riding that balloon, and the radius is accelerating in Rindler fashion, with constant proper acceleration. Note this defines a "bouncing universe", if we let the time go from -infinity to +infinity. The surface of the balloon contracts in from -infinity, but slowing down, stops (the neck of the hyperboloid), then expands back out, all at constant proper acceleration.

They first show the comoving form, which gives you a metric of the form:

ds^2 = dt^2 - cosh^2(t)*d(closed3sphere_metric)^2

They then introduce the static coordinates. However, they don't show the "flat space" coordinates. This paper, however, does:

http://www.bourbaphy.fr/moschella.pdf

This is the "deSitter sightseeing tour" paper. THat shows the flat space world lines (note there is some coordinate funny business there at the middle). Note you can also get an open (spatially hyperbolic) space as well! You can coordinatize deSitter space-time according to all three FLRW forms, closed, flat, and open! But note it is the same invariant space-time, just different coordinate games with different notions of "space". And then there's the static coordinates..........

When one finally appreciates this, one will realize just how silly the question of is space open, closed, or flat is, not to mention the question of "Is space expanding?" This is just one of things that has to click in your mind, I'm afraid.

-Richard

DrRocket

2009-Jan-10, 06:32 PM

Yes, it's a real problem. The trouble is the mixing of space and time in that non positive definite way. You can draw an x-t diagram for a simple 2D space-time, which is done all the time for SR purposes. However, what you're looking at is a 2D Euclidean plane, which just doesn't capture the essence of a 1-1 Minkowski space. Indeed, there is no way to draw a picture of some crazy non-positive definite space, even a flat one. If you realize that, you're okay, but if you don't, it's misleading. Now, start curving things, and it can get even more misleading.

So if you have a basic understanding of just how "really different" the geometry is, such constructs can be useful. But if you don't, they can be very misleading. Lorentzian manifolds play weird tricks with the notion of spherical vs hyperbolic, because of the the difference in signs between the space-like and time-like parts. So a manifold of constant curvature, which looks spherical in terms of the equations, also looks hyperbolic.

What is an n-sphere in Euclid? It's just x^2 + y^2 +.... = constant^2, the locus of all points a fixed distance from an origin. Now, do that for space-time:

t^2 - (x^2 +y^2 +..) = constant.

That's the same thing, a "sphere", yet because of the difference of squares, it has an hyperbolic looking form.

And that latter hyperbolic looking form does indeed give you a deSitter space, which can be visualized as an expanding sphere. Yet add the time part in the diagram, and you get an hyperbolic structure. An expanding hyperballoon works fine for deSitter, in terms of thinking of "space" in certain coordinates -- a balloon expands with time, and therefore the interior is earlier times. However, add time to the thing, and you're dealing with an hyperbolic surface.

The trouble is the balloon part is thinking about space alone (and in certain coordinates) and is not invariant. The hyperbola captures the true invariant parts of the structure, and you can see how some coordinates along that surface can see the space part as flat, while others will see it as curved.

As we discussed before at length, the notion of what space is doing just has no invariant meaning. You've got to think of space-time together. And until one realizes that, until that clicks, one is lost.

And another technical problem with general space-times is that *more than one higher time dimension* may be required to embed. And that just blows one's mind, of course. That's true for Schwarzschild. If you supress the angular coordinates, and just look at the r-t space-time, you can successfully embed that in a 1-2 higher space-time, and draw a 2D curved surface that represents a cuved 1-1 space-time. However, the full Schwarszchild required a 2-4 space-time to embed.

-Richard

In this context, I believe that we can, however, take care of the "unbounded" part. Whether space-time is compact or not, all models treat it as a 4-manifold without boundary, which is simply one more way of saying that all points admit a local chart that is 4-dimensional.

Ken G

2009-Jan-13, 12:09 AM

In this context, I believe that we can, however, take care of the "unbounded" part. Whether space-time is compact or not, all models treat it as a 4-manifold without boundary, which is simply one more way of saying that all points admit a local chart that is 4-dimensional.Right, but I think there is a subtle but important distinction between saying that all observers we would care to treat can use a 4D local chart, and saying something "real" about the universe. We see no evidence of any boundary, so we put none in our model of the universe. That is all that may be said, and it's quite a bit different from saying "most cosmologists don't think the universe has a boundary". The latter statement doesn't mean a heck of a lot, frankly-- as history demonstrates all too clearly.

Ken G

2009-Jan-13, 12:16 AM

When one finally appreciates this, one will realize just how silly the question of is space open, closed, or flat is, not to mention the question of "Is space expanding?" This is just one of things that has to click in your mind, I'm afraid. Even before that "click" occurs, I'd say that one can see the difference between a statement about a picture that one can use to get the answers right, and a statement about how the universe "is". One example of that phenomenon is that if you use polar coordinates in a plane, and you go around 360 degrees in the theta coordinate, you get back to the same place. Does this mean the universe is suddenly "closed" when one uses polar coordinates, whereas in Cartesian coordinates it isn't? Apparently, even in something as simple to picture as a plane, you can still be looking at something that is not what you think you are looking at. For a hyperboloid, yikes.

publius

2009-Jan-13, 03:31 AM

One example of that phenomenon is that if you use polar coordinates in a plane, and you go around 360 degrees in the theta coordinate, you get back to the same place. Does this mean the universe is suddenly "closed" when one uses polar coordinates, whereas in Cartesian coordinates it isn't?

Yes, that's a good example. However, let's look at slightly trickier example. Consider a infinitely long cylinder. We take a piece of plane, infinite in one direction, but finite in the other, and roll it around and make a cylinder.

Is that open or closed? It has infinite surface area. That could represent a "non-expanding" space-time, like the hyperbolic surface, but not increasing diameter with the time coordinate (Whether such a cylindrical type surface could be a solution of the EFE, I don't know, but for our purposes we can imagine such a space-time). The time coordinate is along the infinite direction, while the space coordinate is the closed one. Is there any coordinatization where both coordiantes could be of infinite extent? I don't think so.

Note the space-like vs time-like distinction would come into play there into which direction is which.

I think there will be some sort of invariant notion of spatially closed, which would depend on if any time-like geodesics could get back to where they started. That is, of all possible spatial foliations, none were open. (I'm sure a rigorous defintion would be tedious, and certainly beyond my talents, but there's got to be some condition like that).

And there is an interesting rub on this. In the spatially closed FLRW style coordinatization of the deSitter hyperboloid, it's not possible to get back where you started. If you shoot a light beam off in any direction, even though at any instant of time the spatial hypersurfaces are finite and closed, they increase faster than light can make up the difference.

So I would say that only the "real" spatial closure is one where time-like geodesics can get back to where they started in finite time. And that definition gets tricky -- consider an orbit, a trivial time-like geodesic that gets back to where it started spatially. :) You've got define it such that there's no way not to get back where you started.

And then, on the time-like side, you can imagine temporal closure -- the time axis wraps back on itself. Now, that would lead to closed time-look curves. But then there's the other kind, where world lines just "start and stop". Imagine a space-time where all possible world lines have a beginning and an end.

Our Big Bang space-time has a beginning. Now, imagine it has an end. Now, make it spatially closed as well. You have a finite, closed surface of space-time. Big-Bang --> Big Crunch is an example. And considering that Schwarzschild requires a 2,4 embedding space, I'm not making any bets that the BB-BC universe would embed in any 1,4 space where we could compare with the deSitter hyperboloid! But one can imagine such a thing -- the hyperballoon starts out as point, fans out a bit, then contracts back to a point. Time-like world lines would be of finite length. Again, whether such a thing in 1,4 embedding space would be a solution, I don't know. I would guess it isn't and something more complicated is required. :)

And heck, you can imagine one that is spatially infinite, but the time part has a beginning and end. (And again, if all these possibilities could be solutions of the EFE, I don't know).

-Richard

Ken G

2009-Jan-13, 11:56 PM

Yeah, it's pretty tricky. First you have general concepts of space and time and what their geometry might be, then you have the subsets that solve the EFE. What geometrical restrictions does that imply? I tend to believe that the curvature restrictions are separate from the topological ones, so the EFE is not by itself complete. Perhaps you also have to make a topological assumption to generate a global model. Can we ever test the latter? Probably not, insofar as we would probably have already gotten the result if there was one to get. Unsatisfying though it may be, the answer to "what is the real topology of the universe" might always be "we have no idea".

publius

2009-Jan-14, 01:33 AM

I have heard it said that GR is topologically incomplete -- or was it topologically indeterminate. I'm not exactly sure what it means rigorously, but it indeed means we can never be sure of the exact topology of the universe.

We'd need a real expert to properly explain the details of all that. As you say, are the topological restrictions something different than the curvature restrictions.

-Richard

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