# Can someone explain what it is meant by a 'Flat Universe'?

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1. Mar 21, 2015

### Kiyal

I have been reading and looking at videos regarding cosmic inflation and it said that the universe is flat. I am no expert and have no mathematical background - I love reading about space and learning new things about it. It just really irritates me when things go right over my head and this idea of a 'Flat Universe' is out my grasp, so please try and enlighten me.

I would appreciate it if someone could explain it to me in a simple way or direct me to material that will give me an answer to a young laymen, like myself.

What is it meant by a 'Flat Universe'?

Thank you.

2. Mar 21, 2015

### phinds

3. Mar 21, 2015

### wabbit

It just means that space is "Euclidian". Parallel lines don't meet, the sum of angles in a triangle is 180° and so on. While this might seem obvious at short scales, there is a priori no reason for it to hold at cosmological scales. General Relativity predicts that "in general" space is curved, so the fact that it is observed to be flat is remarkable.

Actually, the more careful statement and the only thing derived from observations is that space is "nearly flat", i.e. its curvature is too small to be detected with current observations. It may very well turn out to have some small curvature after all once we get higher precision measurements.

Last edited: Mar 21, 2015
4. Mar 21, 2015

### Kiyal

I think I have some understanding here, so the universe is so massive that we can't detect any curvature, so geometry doesn't pick up any shape? Excuse my ignorance.

5. Mar 21, 2015

### phinds

What does "geometry doesn't pick up any shape" mean ?

Did you read the article I pointed you to?

6. Mar 21, 2015

### Kiyal

"When physicists describe the Universe as being flat or nearly flat, they're talking geometry: how space and time are warped according to general relativity." - Can you explain that to me please?

7. Mar 21, 2015

### wabbit

It is not a question of being massive. And I don't think we know (I sure don't) why the universe is nearly flat. It is just that, metaphorically, we've been measuring bigger and bigger triangles, and the sum of their angles has so far turned out to be close to 180° - not differing by more than the measurement precision.

8. Mar 21, 2015

### wabbit

General Relativity is just the standard physical theory of the geometry of time and space, and what causes this geometry to be flat or not and in which way it isn't.
Beyond that "just", the link you point to is probably as good a place to start as any. GR is not in any way easy however, be warned.

9. Mar 21, 2015

### Kiyal

10. Mar 21, 2015

### wabbit

I should still add - don't take that "is just" as dismissive in any way - arriving at a consistent and general theory of spacetime was an immense achievement, and part of that was to establish that gravity is an aspect of geometry - yes, everything we know about gravity is included in that theory.

11. Mar 21, 2015

### Kiyal

I get that mass affects space-time, in that it bends/distorts it. So when they say the universe is flat, are they referring to it on a wider scale?

12. Mar 21, 2015

### Ibix

I see what you are getting at. Yes, every triangle we've been able to draw (and some of them have sides 13 billion light years long) looks really like a regular triangle. That means that the universe is flat, or else so big that the curvature is too slight to detect on that scale.

I'd avoid words like "massive" in this context. That means "has lots of mass", where as I suspect you meant "really, really big".

The FRW (or FLRW) metric describes the universe on the kind of scale where galaxies are too small to notice - like atoms in every day life. Space-time can be very curved on the scale of a galaxy (it's what gravity is) yet flat on the large scale. Think of hammering a dent into the side of an aircraft carrier. The dent could be quite big up close, but looked at as part of the whole carrier, the side of the ship is still pretty much flat.

13. Mar 21, 2015

### wabbit

At any scale at which we can measure the curvature currently, big or small. One thing to note is that as you say mass curves spacetime - which isn't flat at all, even here on earth. It is only space, considered separetely from time, that is nearly flat.

14. Mar 21, 2015

### wabbit

@Ibix - I'm afraid I must take issue with what you said here : as I understand it, the key distinction isn't one of scale, but of space vs spacetime - the curvature of spacetime is detectable easily on earth from daily experience.

15. Mar 21, 2015

### Staff: Mentor

Yes.

No. As wabbit said, they are referring to the curvature (or lack thereof) of space, not spacetime. More precisely, they are referring to the curvature of space as seen by observers who see the universe as being, on average, homogeneous and isotropic, i.e., the same everywhere and in all directions. The "on average" is necessary because obviously the universe is not exactly the same everywhere and in all directions. But averaged over large enough distance scales, it is.

16. Mar 21, 2015

### Kiyal

Thanks, this all really makes sense to me now! Yes! I meant really, really big!!

I'm pretty taken aback with this information, just amazing - really does make me feel so lucky to be able to find out the answers to these questions, thanks all.

17. Mar 21, 2015

### Staff: Mentor

No. As wabbit said, the curvature of space is not the same as the curvature of spacetime. When cosmologists talk about the universe being flat, they are talking about spatial curvature (on a large enough distance scale). The spacetime curvature of the universe is actually easier to observe on large scales than on small scales.

18. Mar 21, 2015

### Ibix

I think I see. Barring a careful choice of a(t) (which is presumably either not a valid solution of the Einstein field equations, or not consistent with observation) the Ricci scalar for the FRW metric is non-zero even when k=0. However, when k=0 the spatial part of the metric is simply the Euclidean metric with a time-dependant scale factor.

Crudely, then, in the k=0 case am I talking about replacing a Minkowski diagram with one whose lines of constant x aren't straight vertical lines, but rather curve away from the t axis as in this sketch?

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19. Mar 21, 2015

### Staff: Mentor

I'm not sure what this means.

Yes, the Ricci scalar for spacetime is. That's not the same as the Ricci scalar for a spacelike slice. See below.

I'm not sure what you're proposing, but it looks like you're talking about changing coordinates. The statement that our universe is spatially flat (i.e., that it is described by an FRW spacetime with k = 0) is not a statement about coordinates; it's a statement about the spacelike slices that are orthogonal to the worldlines of "comoving" observers (observers who see the universe as homogeneous and isotropic). Those spacelike slices can be defined independently of coordinates, and their curvature is also independent of coordinates.

Of course, you could pick a different set of spacelike slices, and their curvature might be nonzero even in a k = 0 FRW spacetime. But the spacelike slices that are flat in our universe are the "comoving" ones, and those are the ones cosmologists talk about.

(Also, your use of the term "Minkowski diagram" makes me wonder if you are assuming the diagram will follow the usual rules for spacetime diagrams in SR. It won't. Even if the spacelike slices are flat, spacetime is not, and you need flat spacetime, not just flat spacelike slices, to have a Minkowski diagram that works the way spacetime diagrams work in SR.)

20. Mar 22, 2015

### Ibix

I was just noting that I can pick a(t)=constant, to give flat space-time. But that isn't the universe we see around us.

The bit about being a solution to the EFE is just plain wrong, I think. My understanding is that I can pick any a(t) I like and the EFE will tell me the stress-energy tensor I need to get that metric. That may or may not be a plausible tensor.

I was attempting to illustrate a 2d t-r slice through an FLRW space-time, including lines of constant t and r (which I labelled x, for some reason). As you note, though, that's a co-ordinate dependant picture, so at the least I need to specify which co-ordinates I'm working in. I don't think that's illegitimate, just incomplete (?).

Nope - you picked me up on that a few months ago, when I tried to Lorentz transform Schwarzschild co-ordinates. But I guess that knowledge hasn't completely bedded in since I ought to have realised the points you made above. There is a globally significant frame (the comoving one) that isn't present in SR, so you can't casually take a 2d slice through spacetime without specifying its relationship to that frame. And I need to stop thinking in co-ordinate terms where possible.

21. Mar 22, 2015

### wabbit

Just a side note : a(t)=constant with k=0 gives (locally) Minkowski spacetime, a valid GR solution. I suspect that conversely any flat spacetime solution will have a(t)=constant, but I haven't checked the calculation to prove this.

22. Mar 22, 2015

### George Jones

Staff Emeritus
No. The Milne universe is a curved space ($k=-1$) flat spacetime that has $a \left( t \right) = t$. It is a portion of Minkowski spacetime.

23. Mar 22, 2015

### wabbit

Interesting, thanks. Curved space flat spacetime... So no gravity, just intrinsic spatial curvature... and expanding too. Nice, hadn't thought of that.

You make me doubt now: might that be possible for k=1 as well? An expanding empty 3-sphere?
Edit: doesn't seem possible. Ricci flat if empty yes, but it looks like geodesics of comoving observers on the expanding sphere must diverge, so not flat overall for k=1 (?)

Last edited: Mar 22, 2015
24. Mar 22, 2015

### Staff: Mentor

Only if $k = 0$. If you set $a(t)$ constant, but have $k = 1$ or $k = -1$, you don't get flat spacetime. (You do get a stress-energy tensor that many people will consider to be physically unreasonable; see below.)

Yes. More generally, you can pick any metric you like, compute its Einstein tensor, divide it by $8 \pi$ (or $8 \pi G / c^4$ if you are using conventional units), and call that the stress-energy tensor for that metric. But, as you say, this procedure may or may not give you a stress-energy tensor that is physically reasonable. That's why, for most applications, physicists work the other way: they first obtain an expression for the stress-energy tensor of some physically reasonable substance (such as a perfect fluid), and then try to figure out what metric will lead to that SET through the EFE. This is how the FRW metrics are obtained.

Yes, but as you say, you need to specify which coordinates you are working in.

25. Mar 22, 2015

### bahamagreen

Latex... later...

Last edited: Mar 22, 2015