# De Sitter space, basic questions

1. Feb 4, 2015

### binbagsss

So for $\lambda > 0$, these 3 solutions for $k=-1,0,1$ differ only by a coordinate transformation, and describe the same space-time.

$\lambda <0$: $k=-1$ :
$a=\sqrt{\frac{-3}{\Lambda}}sin(\sqrt {\frac{-\Lambda}{3}}t)$

$\lambda>0$ :

$k=-1$: $a=\sqrt{\frac{3}{\Lambda}}sinh(\sqrt{\frac{\Lambda}{3}}t)$ \\

$k=0$: $a\propto e^{(\pm\sqrt{\frac{\Lambda}{3}}t)}$ \\

$k=1$: $a=\sqrt{\frac{3}{\Lambda}}cosh(\sqrt{\frac{\Lambda}{3}}t)$

Questions

I've read on a few sources that the scale factor grows exponentially, but surely this is just in a particular coordinate chart, as it is only apparent , above, in the $k=0$ solution.

Probably a stupid question, but different coordinate charts suggest different dynamics of $a(t)$, so which is the 'true' relation?

I.e- Is the de-sitter space geometrically open, closed or flat?
Anti de-sitter space geometrically? I'm guessing it's called anti for a reason

The exponential solution for $k=0$ has a $\pm$, how is this possible , don't they describe completely different things - contraction and expansion?

Thanks .

Last edited: Feb 4, 2015
2. Feb 4, 2015

### Staff: Mentor

Yes. Its dynamics are different in other charts. There is even a chart which is static. The Wikipedia page gives a decent summary of the key charts:

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

There's no such thing. The scale factor and its dynamics depend on your choice of chart; that's all there is to it.

If you're using de Sitter spacetime to model something in particular, such as a universe with dark energy in it, then a particular chart might match up with some physical fact about the universe. For example, in a dark energy-dominated universe, the chart in which the scale factor grows exponentially is the "comoving" chart, i.e., it is the chart in which observers who see the universe as homogeneous and isotropic are at rest. So saying the scale factor grows exponentially in that chart is the same as saying it grows exponentially for those observers.

Open, closed, or flat are properties of spacelike slices, not of the spacetime as a whole. Spacelike slices of all three types can be "cut" out of de Sitter spacetime (this is what you are doing by choosing the appropriate type of chart). The geometry of de Sitter spacetime as a whole can't be described by a single term of that sort.

Again, the geometry of AdS spacetime can't be described by a single simple term.

Yes, because the cosmological constant is negative instead of positive.

Do you have a source for this? I have only seen the positive sign in the sources I'm familiar with. From looking at Friedmann equations for this case, I'm not sure I see how the minus sign gives a valid solution, because it would make the signs of $\dot{a}$ and $\ddot{a}$ opposite, and they have to be the same.

3. Feb 5, 2015

### binbagsss

Source: Lecture Notes on GR 1997, Sean M Carroll

K=1,0,-1 are associated with closed, open and open space-time, regardless of the matter dominating the universe, e.g the $k=0$ solution in the OP is described as flat, so the three solutions for k correspond to a cutting of spacelike slices that are closed, flat, open ?

Why is it that the flat geometry corresponds to the co moving observers being at rest?

4. Feb 5, 2015

### Staff: Mentor

Hm, yes, I see equation (8.58) has a $\pm$ in it. I'll have to take a closer look at that section, since I'm still not seeing how the minus sign is consistent with the Friedmann equations.

Yes.

At rest in the comoving chart, not at rest relative to each other. Comoving observers are obviously moving away from each other in this universe. The term "comoving" as it is used here just means "sees the universe as homogeneous and isotropic"; it doesn't mean "always at rest relative to each other".

5. Feb 5, 2015

### Staff: Mentor

I've figured out what I was missing on this one--the $\dot{a}$ term is squared, so the sign of $\dot{a}$ doesn't matter. I plead lack of caffeine.

This does make it appear that there are two possible flat slicings of de Sitter spacetime, one expanding and one contracting. I think they are really the same slicing, though, just with the sign of the time coordinate reversed--in other words, in the "expanding" slicing, the direction in which $t$ increases is opposite from the direction in which it increases in the "contracting" slicing. But both slicings have the same set of surfaces of constant time and the same set of "comoving" worldlines (worldlines in which the spatial coordinates do not change).

6. Feb 7, 2015

### binbagsss

How do we know this? Would we expect this physically in $k=0$ as a pose to $k=-1,0$?

7. Feb 7, 2015

### binbagsss

I mean in the notes, it says the 3 solutions for the cosmological constant >0 are the same space-time, but theres no mention about the solution <0,
or what is the anti de-sitter space-time in the frame where the comoving coordinates are at rest?

8. Feb 7, 2015

### Staff: Mentor

"Sees the universe as homogeneous and isotropic" is the definition of "comoving". The fact that observers at rest (i.e., with constant spatial coordinates) in a comoving chart see the universe as homogeneous and isotropic will be evident from the metric. These lecture notes have a good discussion:

http://www.tapir.caltech.edu/~chirata/ph217/lec01.pdf

However, I did misspeak about which charts are "comoving" charts in de Sitter spacetime. See below.

Actually, I misspoke earlier; the flat, closed, and open charts on de Sitter spacetime are all "comoving" charts, so observers at rest in all three of those charts see the spacetime as homogeneous and isotropic. However, the "comoving" observers in each of these charts are a different family of observers, so in de Sitter spacetime, there are multiple families of observers who see the universe as homogeneous and isotropic, even though each family is in a different state of motion.

The reason the flat chart is the one picked out as "comoving" for our universe is that our universe has ordinary matter and energy in it, in addition to the cosmological constant. Roughly speaking, if you put ordinary matter and energy into de Sitter spacetime, it breaks the symmetry between the different possible "comoving" charts (flat, closed and open), picking out only one of them as the one that's "comoving" with respect to the ordinary matter and energy. In our universe, that chart is the flat chart, i.e., "comoving" observers in our universe who see the ordinary matter and energy as homogeneous and isotropic are at rest (constant spatial coordinates) in the flat chart. Sorry for not making that clear earlier.

That solution is indeed a different spacetime, not the same one; it is indeed called anti-de Sitter spacetime. See here:

http://en.wikipedia.org/wiki/Anti-de_Sitter_space

9. Mar 4, 2015

### binbagsss

Sorry to re-bump an old thread, seemed to make more sense...

What exactly are these different state of motions? And the definition of a comoving observer is to be at rest with respect to space, so they all obey this?

'Roughly speaking, if you put ordinary matter and energy into de Sitter spacetime' I'm a bit lost here, as the de Sitter solution I saw to be found from solving Friedmann equation for a dark-energy, w=-1 in the eq of state, universe. I'm not sure what this means?

What is meant by 'breaks the symmetry' - so only the flat space observers now sees the universe as isotropic and homogenous once matter and energy are put in? So the observers in the k=1,-1 no longer view the universe iso & homo. Is this the definition of no longer being a comoving observer? Are they also no longer at rest with respect to space?

10. Mar 4, 2015

### Staff: Mentor

I just meant that members of each family of "comoving" observers are moving relative to members of the other families in their immediate vicinity.

No, that's not the definition of a "comoving" observer. "At rest with respect to space" has no meaning. A "comoving" observer is one who sees the universe as homogeneous and isotropic. A universe with only dark energy in it (which is what "de Sitter spacetime" means) has the unusual property of looking homogeneous and isotropic to several different families of observers, who are each moving with respect to members of the other families in their immediate vicinity. See further comments below.

This was a sloppy way of saying "if we change the spacetime from de Sitter spacetime, which only has dark energy in it, to a spacetime that has the same amount of dark energy, but also has ordinary matter and energy in it". Such a spacetime will no longer be de Sitter spacetime, since, as you correctly note, de Sitter spacetime can only have dark energy in it. Sorry for not making that clear.

It means that, in a spacetime that has ordinary matter and energy in it as well as dark energy, there is only one "comoving" observer at each point of space--i.e., at each point of space, there is only one observer who sees the universe as homogeneous and isotropic. Any observer at that point of space who is moving relative to this "comoving" observer will not see the universe as homogeneous and isotropic. This is different from the case I described above, of a universe with only dark energy in it; in that case, as I said above, there can be multiple observers at the same point of space, all moving relative to each other, who all see the universe as homogeneous and isotropic.

The spatial curvature seen by observers who see the universe as homogeneous and isotropic and are thus "comoving" (in the case where ordinary matter and energy are present) depends on the details of how much ordinary matter and energy there is. There are solutions where the "comoving" observers see k=1, solutions where they see k=-1, and solutions where they see k=0. It just happens that the solution that (as far as we can tell) describes our actual universe is the one in which comoving observers see k=0.

11. Mar 5, 2015

### binbagsss

Okay thanks.
So this is my understanding so far:

- Changing from solely dark energy to dark energy, matter and energy, is changing to a more physically realistic cosmological fluid, and compared to a solely dust-filled universe, or solely radiaiton-filled universe comoving observers are not only at rest - among these families of different states of motion, I assume the at rest case is still there?
- Apologies, I have yet to look at the link you sent me (due to assignment deadlines), but i believe this expains why comoving observers are at rest, and so I'm guessing this reasoning holds only for radiaiton and dust, not dark energy - would it hold for a mixture of dust and radiaiton i.e its not the the cmsological fluid being a mix that causes the families of comoving observers but rather the dark energy ?

- And in terms of what you input into the model, for dust-filled, radiaiton-filled , we can input any value of k, and a comoving observer for each k is one which is at rest with respect to space, and even for a solely dark-energy filled universe there exists a observer who views the universe as isotropic and homogenous and for each k - i.e who observers the universe with this respective geometry of the universe, but, for any mixture, that includes dark energy, this does not hold and (depending upon the exact distribution between matter, radiaiton, dark-energy), there only exists observers with a certain state of motion who view the universe as isotropic and comoving, and these view the universe with a certain geometry - a certain k.

12. Mar 5, 2015

### Staff: Mentor

Yes.

"At rest" is a relative term; it depends on your choice of coordinates. That's why, as you'll note, I was careful to define "comoving" observers in a coordinate-independent way: they see the universe as homogeneous and isotropic.

As for what families of "comoving" observers exist for the different possible universes, here's a quick summary:

* For a universe with either dust or radiation in it, there is only one family of "comoving" observers--that is, at each spatial point, there is only one "comoving" observer, and any other observer at the same spatial point who is moving relative to that observer is not "comoving". (Such a universe can also have dark energy in it; that doesn't change what I just said.)

* For a universe with only dark energy in it, there are multiple families of "comoving" observers--that is, at each spatial point, there are multiple observers, moving relative to each other, who all are "comoving"--they all see the universe as homogeneous and isotropic.

Meaning, we can construct a valid model with any value of $k$? Yes, we can, but we will have to choose a density for the dust or radiation that is consistent with that value of $k$ (meaning, has the right relationship to the critical density).

No. I said in my previous post that "at rest with respect to space" has no meaning. I also carefully defined "comoving" in a coordinate-independent way. See above.

Observes this respective spatial geometry of the universe. (The spacetime geometry is the same in all three cases.) Yes, this is correct.

Yes. Which $k$ the comoving observers will see depends on the densities of matter, radiation, and dark energy that are present. In our actual universe, as best we can tell, the densities add up to just the right "critical" density so that "comoving" observers see $k = 0$.

13. Mar 15, 2015

### binbagsss

Also looking at your post ealier
'The fact that observers at rest (i.e., with constant spatial coordinates*) in a comoving chart see the universe as homogeneous and isotropic ....'

So it's not at rest with respect to space, but once in a comoving chart and with constant spatial coordinates? I.e- not all observers in a comoving chart see the universe as homo genus and isotropic, only those with constant spatial coordinates.

Sorry I'm still confused with what the different states of motion means for dark energy. So I understand that for each k there exists comoving charts and observers who view the universe as isotropic and homogenous within these charts, and that, compared to the typical case, where these comoving observers are not in motion with respect to each other, here they are. But going with *, are these observers within the subset of constant spatial coordinates, but also in motion with respect to each other?

Thanks.

(Lastly, why when I google comoving , does the definition of constant spatial coordinates come up alot? )

14. Mar 15, 2015

### Staff: Mentor

Please stop using the term "at rest with respect to space". It has no meaning. Trying to give it meaning will only cause you further confusion.

The term "at rest" should always be qualified with what something is at rest with respect to. For a given coordinate chart, an observer "at rest" in that chart is one with constant spatial coordinates in the chart. (Note that this assumes that the chart has one timelike and three spacelike coordinates, which not all charts do. But all of the FRW charts used in cosmology do.)

Just to clarify, observers are not "in" one chart but not others; all observers' worldlines are always there in every chart. The only difference between charts is the specific equations describing different worldlines. In a comoving FRW chart, observers at rest with respect to the chart (i.e., with constant spatial coordinates) see the universe as homogeneous and isotropic. Observers not at rest with respect to the chart (i.e., with spatial coordinates that vary with time), in the general case, do not see the universe as homogeneous or isotropic. However, in the special case of de Sitter spacetime (i.e,. a spacetime with nothing but dark energy in it), there can be observers not at rest in a comoving chart who still see the universe as homogeneous and isotropic; see below.

More precisely, in de Sitter spacetime (i.e., a spacetime with nothing but dark energy in it), there is a comoving chart for each k (+1, 0, and -1), and observers at rest with respect to each of these charts see the universe as homogeneous and isotropic. Since these are three different charts, the observers at rest in them are three different sets of observers, and these observers are not at rest with respect to each other. So for each comoving chart, one of the three sets of observers is at rest in the chart, and the other two are not; but all three sets of observers still see the universe as homogeneous and isotropic. This only happens in de Sitter spacetime.

15. Mar 16, 2015

### binbagsss

How does this compare to the solely dust and solely radation case? Here the comoving chart is the same for each k and so observers are not in motion relative to each other, they are the same family? So can any value of k be observed here by a comoving observer?

And once the symmetry is broken , adding radiation/dust to the dark-energy a comoving chart only exists for one particular value of k. Differeing from de sitter space-time as here there was a DIFFERENT comoving chart for each k, and differing from the solely radiation/dust-filled case as here there was a single (same for each k) comoving chart?

16. Mar 16, 2015

### Staff: Mentor

No, it isn't. The metric is different for the three different values of $k$; that means each value of $k$ has a different comoving chart. When references talk about "the" FRW chart, they are being sloppy; strictly speaking there are three, one for each $k$.

The key point is that, in an FRW spacetime with solely dust or radiation, there is only one set of observers who see the universe as homogeneous and isotropic. What spatial curvature those observers see (i.e., which value of $k$ gives the correct comoving chart for the observers who see the universe as homogeneous and isotropic) depends on the density of the dust or radiation relative to the critical density (larger than critical -> $k = 1$; equal to critical -> $k = 0$; less than critical -> $k = -1$ ).

17. Mar 20, 2015

### binbagsss

Okay, so 3 different values of k implies 3
different coordinate charts - in general, when is a coordinate chart called a comoving chart, is it when there exists equations describing the worldlines of some observer,that views the universe as isotropic and homogenous, as being at rest in that chart?

Why is it that observers at rest in the comoving charts for each value of $k$ are at rest wrt each other in a solely dust/radiation-filled universe but a different state of motion for dark energy -so each coordinate chart differing means that in general the equations describing the the worldlines of observers are different in each chart- does this mean that whilst this is true for the worldlines of observes who are at rest in the dark energy universe, the worldlines describing the observers at rest in the different comoving charts in the solely dust/radiation filled universe are given by the same equation?

18. Mar 20, 2015

### Staff: Mentor

In cosmology, a "comoving" chart is one in which "comoving" observers--observers who see the universe as homogeneous and isotropic--are at rest (have constant spatial coordinates). (In other applications, the term "comoving" can have other meanings, but we're concerned with the cosmology meaning here.)

I think you're still misunderstanding the dust/radiation case. In the dust/radiation case, the charts with the three different values of $k$ are charts of three different spacetimes, i.e., three different universes (or possible universes). In each of these three different spacetimes, there is one family of "comoving" observers. The three different families of "comoving" observers are in three different spacetimes, so it makes no sense to ask whether they are at rest relative to each other or not. There is no comparison between their states of motion.

In the case of a solely dark energy-filled universe, the three different families of "comoving" observers (one for each value of $k$ ) are all in the same spacetime, i.e., in the same universe (or possible universe). So at any given point in that spacetime, you can compare the states of motion of "comoving" observers from the three different families, and find that they are moving relative to each other.

19. Mar 21, 2015

### binbagsss

Okay, thanks very much for your help,
do you know of any useful sources for anything discussed in this thread?

20. Mar 27, 2015

### binbagsss

Apologies, what is meant by correct here?